{ "Tag": [ "vector", "calculus" ], "Problem": "what is the acceleration of a car that moves from a speed of 5.0 m/s to a speed of 15 m/s during a time of 6.0 s? Thanks for any help.", "Solution_1": "Acceleration is defined as change in velocity divided by time. So $ a=(15-5)/6=10/6=5/3\\mbox{ m/s^{2}}$.", "Solution_2": "a is vector is suppose! :) $ a ^$", "Solution_3": "Do you know what acceleration is? It is basically a change in velocity. It is measured in $ ms^{\\minus{}2}$ (m being metre, the standard). The $ s^{\\minus{}2}$ might be confusing, and $ s^{\\minus{}1}$ may also be confusing, if you can imagine velocity as a rate, or in other words a fraction, a fraction of speed on time, such like this $ \\frac{speed}{time}$, we could ask our self what the units maybe, lets say they are the second and metre; then the rate is $ m/s$ or, in fraction form, $ \\frac{m}{s}$.\r\n\r\nNow, the second is in the bottom, and so if this were two multiples, then it would be $ m \\times \\frac{1}{s}$, and with the $ \\frac{1}{s}$, as we know from the basic indice law $ a^{\\minus{}1} \\equal{} \\frac{1}{a}$, we know that now $ \\frac{1}{s}$ can be written as $ s^{\\minus{}1}$. So thats basically it.\r\n\r\nWith your question, you must know a simple formula, that is $ a \\equal{} \\frac{v\\minus{}u}{t}$. Where\r\n\r\n$ a$ is acceleration\r\n$ v$ is the final velocity\r\n$ u$ is the initial velocity\r\n$ t$ is time\r\n\r\nNow let me explain this formula. This formula is for average acceleration, the difference between average acceleration and just acceleration is that average acceleration is just simply the average, because, as we know, if the acceleration is not constant, then it is always changing, and this gives us a curve, which we use calculus to deal with. Now we should understand that acceleration is the rate of change of velocity, the chance in velocity is the top part in this equation, that is the change, and the bottom is the other change, as acceleration is in metres per second... per second.\r\n\r\n\r\n[quote=\"slowlearner\"][size=117]what is the acceleration of a car that moves from a speed of 5.0 m/s to a speed of 15 m/s during a time of 6.0 s? Thanks for any help.[/size][/quote]\r\n\r\nIn this question, it is obvious that the initial velocity is $ 5.0 ms^{\\minus{}1}$, the final velocity is $ 15.0 ms^{\\minus{}1}$ and the time is 6.0 seconds. Lets sum this up\r\n\r\n$ v \\equal{} 15.0 ms^{\\minus{}1}$ \r\n$ u \\equal{} 5.0 ms^{\\minus{}1}$\r\n$ t \\equal{} 6.0 s$\r\n\r\nIf we plug this into our equation to find the velocity we get\r\n\r\n$ a \\equal{} \\frac{v\\minus{}u}{t}$\r\n\r\n$ a \\equal{} \\frac{15.0 ms^{\\minus{}1}\\minus{}5.0 ms^{\\minus{}1}}{6.0 s}$\r\n\r\n$ a \\equal{} \\frac{10.0ms^{\\minus{}1}}{6.0 s}$\r\n\r\n$ a \\equal{} 1\\frac{2}{3} ms^{\\minus{}2}$ \r\n\r\nThis is roughly $ 1.7 ms^{\\minus{}2}$", "Solution_4": "wat r u trying to explain or especially to whom?\r\n :mad:" } { "Tag": [], "Problem": "A circular clock with centre C has twelve hour marks. The one o'clock and two o'clock hour marks are O and T respectively. The circle centre O thorugh T cuts CT again at D. Where does OD cut the circumference of the clockface? Explain!", "Solution_1": "[hide][color=indigo]This was hard! After I did the stuff in the problem, I named the point where OD intersected the circle again F. Then I drew OC, CF, OT, and TF.\n1. COT is an isosceles triangle (all radii of a circle are equal). :ang: OCT=1/12*360 = 30 :^o: so :ang: COT and :ang: OTC equal 75 :^o:\n2. OTD is an isosceles triangle (same reason as #1). :ang: OTD = :ang: OTC = 75 :^o:\n3. CT is a straight line so 180 :^o: - 75 :^o: = :ang: CDO. :ang: CDO = 105 :^o:\n4. DOT = 30 :^o: , and 75 - 30 = 45, so :ang: COF = 45 :^o:\n5. COF is another isosceles triangle (still the same reason). CFO = COF = 45 :^o: , so OFC = 90.\n6. 90 = 1/4*360 so it is three hours from O to F. That means F is at the 4 o'clock mark. :D [/color][/hide]", "Solution_2": "[hide]Point $ F $[/hide]", "Solution_3": "[hide]point f :lol: [/hide]", "Solution_4": "[hide=\"da answer\"]\npoint F[/hide]", "Solution_5": "correct :P", "Solution_6": "This wasnt as easy as I thought but I got the ans too\r\npoint F :first:" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "[b] Let [/b] $ A \\in M_{n} \\mathbb{(R)}$ , I is identity nxn matrix ,$ m \\in N^{*}$ ,\r\n$ m \\geq 2$ \r\nProve that : \r\n $ Lim_{n \\to \\infty} \\sum_{k = 0}^{2008n} {[(A + \\frac {I}{n + k})^{m} - A^{m}]} = m.ln(2009).A^{m - 1}$", "Solution_1": "$ \\sum_{k \\equal{} 0}^{2008n} {[(A \\plus{} \\frac {I}{n \\plus{} k})^{m} \\minus{} A^{m}]}\\equal{}mA^{m\\minus{}1}\\sum_{k \\equal{} 0}^{2008n}\\dfrac{1}{n\\plus{}k}\\plus{}\\dfrac{m(m\\minus{}1)}{2}A^{m\\minus{}2}\\sum_{k \\equal{} 0}^{2008n}\\dfrac{1}{(n\\plus{}k)^2}\\plus{}\\cdots$.\r\n$ \\sum_{k \\equal{} 0}^{2008n}\\dfrac{1}{n\\plus{}k}\\equal{}ln(2009n)\\plus{}\\gamma\\minus{}ln(n)\\minus{}\\gamma\\plus{}O(1/n)\\equal{}ln(2009)\\plus{}O(1/n)$.\r\n$ \\sum_{k \\equal{} 0}^{2008n}\\dfrac{1}{(n\\plus{}k)^2}<\\sum_{k \\equal{} n}^{\\infty}\\dfrac{1}{(k)^2}$ converges to $ 0$ when $ n$ tends to infinity.\r\nThe factors of $ A^{m\\minus{}3},\\cdots$ converge also to $ 0$. QED" } { "Tag": [ "trigonometry" ], "Problem": "I am stuck... need help on how to start this, or something... anything... \r\n\r\nquestion:\r\n\r\nIt is a common experience to hear the sound of a low flying airplane, and look at the wrong place in the sky to see the plane. Suppose that a plane is traveling directly at you at a speed of 200 mph and an altitude of 3000 feet, and you hear the sound at what seems to be an angle of inclination of 20 degrees. At what angle should you actually look in order to see the plane? Let the speed of sound be 1100 ft/sec.", "Solution_1": "Is this homework? If so, AoPS is not the place to post it.\r\n\r\n[hide=\"Hint\"]Calculate the distance the plane flies as the sound comes to you, and it's position relative to where it was before.[/hide]", "Solution_2": "no its not homework....\r\n\r\n its a challenge problem for a class I am taking... its not graded... just a challenge... I dont want the answer, just some help getting it started...", "Solution_3": "[hide=\"To Start\"]Draw a line $ m$ on your paper. Place a point, $ P$, on the line that represents yourself. Draw a line $ d$ that goes $ 20^{\\circ}$ (relative to $ m$) either to the left or right. Construct a line $ n$ parallel to $ m$ and at a distance of 3000 ft (scaled down, of course) from it. The intersection of $ n$ and $ d$ (call this $ Q$) is the place at which you looked. Now calculate how long it took for the sound to get to $ P$ from $ Q$, and then find how much the plane would have flown in this time. Finally, mark off a point on $ n$ where the plane is when you hear it's sound coming from $ Q$, and calculate the angle.[/hide]" } { "Tag": [], "Problem": "how to solve this equation by hand\r\n2x^3-3x^2-12x+12=0", "Solution_1": "You can use Horner's Scheme.", "Solution_2": "hello, the solutions are\r\n$ \\{\\{x\\to \\frac{1}{2}+\\frac{9}{2 \\sqrt[3]{-11+4 i \\sqrt{38}}}+\\frac{1}{2} \\sqrt[3]{-11+4 i \\sqrt{38}}\\}$,$ \\{x\\to \\frac{1}{2}-\\frac{9 (1+i\r\n \\sqrt{3})}{4 \\sqrt[3]{-11+4 i \\sqrt{38}}}-\\frac{1}{4} (1-i \\sqrt{3}) \\sqrt[3]{-11+4 i \\sqrt{38}}\\},\\{x\\to \\frac{1}{2}-\\frac{9\r\n (1-i \\sqrt{3})}{4 \\sqrt[3]{-11+4 i \\sqrt{38}}}-\\frac{1}{4} (1+i \\sqrt{3}) \\sqrt[3]{-11+4 i \\sqrt{38}}\\}\\}$\r\nSonnhard.", "Solution_3": "[quote=\"GunUltimateID\"]how to solve this equation by hand\n2x^3-3x^2-12x+12=0[/quote]\r\n\r\nAs Dr. Graubner pointed out, the solutions are very ugly. To attempt to solve this by hand, you can try memorizing (Although I would [b]not[/b] recommend it) the [url=http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots]Cubic Formula[/url].", "Solution_4": "Instead of using the \"cubic formula\" we can try to use the rational zero theorem, Descarte's rule of signs. If that doesn't work, you can always try to take a look at the how the sum of the roots, product of the roots, and the sum of the product of the roots taken 2 at a time (in this case), relate to your roots. \r\n(intuition)", "Solution_5": "[quote=\"Neil30z\"]Instead of using the \"cubic formula\" we can try to use the rational zero theorem, Descarte's rule of signs. If that doesn't work, you can always try to take a look at the how the sum of the roots, product of the roots, and the sum of the product of the roots taken 2 at a time (in this case), relate to your roots. \n(intuition)[/quote]\r\n\r\nYes, but this is clearly not a case for intuition (Look at the solutions)." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "geometry", "perpendicular bisector" ], "Problem": "I am looking for someone to solve the question below using steps as you go through the lesson. I would like to learn how to solve similar questions on my own. Ready? \r\n\r\nHow many points are equidistant from point R(3,1) \r\nand S(3, 6) and also 4 units from the origin?", "Solution_1": "As people explained in another topic, the locus of points equidistant from two points is the perpendicular bisector of those two points.\r\n\r\nThe perpendicular bisector of $RS$ is the line $y=3\\frac{1}{2}$ (see why?).\r\n\r\nThe locus of points distance 4 from the origin is a circle. \r\n\r\nWe need to find the number of points that this line and this circle meet at. The maximum number that any circle and line can meet at is 2- and if we draw the circle (compass point on the origin, circle passing through $(0,4)$) and the line we see that this circle and this line do meet in exactly two places.\r\n\r\nThere is no need to find the points (!) but if we did want to find them, we could draw a right-angled triangle with one corner at the origin and the other end of the hypotenuse on one of the two points, and do pythagoras.", "Solution_2": "How did you get y = 3.5? \r\n\r\nWhere did 3.5 come from?\r\n\r\nYou stated:\r\n\r\n\"There is no need to find the points (!) but if we did want to find them, we could draw a right-angled triangle with one corner at the origin and the other end of the hypotenuse on one of the two points, and do pythagoras.\"\r\n\r\nCan you give me an example using the steps in the above statement?", "Solution_3": "$y=3.5$ is the line that contains the points equidistant from $(3,1)$ and $(3,6).$ To find it, you first find the midpoint of the two points, find their slope, and then use point-slope form to find the perpendicular bisector. The midpoint is at $\\frac72$ and the slope is $\\text{undefined},$ so the perpendicular bisector is $y=\\frac72,$ or $y=3.5.$\r\n\r\nTo find the points you are looking for, you construct a circle with radius $4$ centered at the origin, and find it's intersections with the line $y=3.5.$\r\n\r\nTim was saying you could construct a triangle with one side being a radius of the circle (from the origin to one of the intersections with the line), the second being a segment on the $x$-axis, and the third being a vertical segment from the point of intersection to the $x$-axis.\r\n\r\n$\\left(\\frac72\\right)^{2}+x^{2}=4^{2}\\Rightarrow x=\\frac{\\sqrt{15}}2$\r\n\r\nThe $x$ coordinates are $\\pm\\frac{\\sqrt{15}}2,$ so the points of intersection are $\\left(\\frac{\\sqrt{15}}2,\\frac72\\right)$ and $\\left(-\\frac{\\sqrt{15}}2,\\frac72\\right).$", "Solution_4": "Strange-looking points but I follow your reasoning." } { "Tag": [ "geometry", "perpendicular bisector", "angle bisector", "geometry proposed" ], "Problem": "Let a triangle $ ABC$, a point $ P$ inside triangle so that $ \\angle{ABP}\\equal{}\\angle{ACP}$. Denote $ D,E$ are midpoint of $ AP,BC$. Prove that: $ DE$ pass through perpendicular feet of $ P$ on bisector of angle $ \\angle{BAC}$.", "Solution_1": "Circle (E) with center E and diameter AP cuts AB, AC, ED at $ U \\in AB,\\ V \\in AC,\\ Q \\in ED.$ Since AP is a diameter of (E), the points U, V are perpendicular projections of P on AB, AC and $ PQ \\perp AQ.$ By solution of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=174227]Two opposite angles of a quadrilateral equal[/url] also discussed in [url]http://www.mathlinks.ro/viewtopic.php?t=172417[/url], posts 6-7, ED is the perpendicular bisector of UV. This makes the $ \\triangle QUV$ is isosceles with QU = QV. From the circle (E), $ \\angle QAC \\equiv \\angle QAV \\equal{} \\angle QUV \\equal{} \\angle QVU \\equal{} \\angle QAU \\equiv QAB,$ AQ bisects the $ \\angle CAB.$", "Solution_2": "Let $ M$, $ N$ and $ K$ the feet of perpendicular lines from $ P$ to $ AB$, $ AC$ and the angle bisector of angle $ BAC$, respectively. We can easily to prove that $ DK$ is the perpendicular bisector of the segment $ MN$ $ (1)$.\r\nNow, let $ R$, $ S$ be the mid-points of the segments $ PB$, $ PC$, respectively. We have $ \\triangle MRE\\equal{}\\triangle ESL$ (S.A.S). It implies $ EM\\equal{}EN$, i.e. $ E$ lies on the perpendicular bisector of the segment $ MN$ $ (2)$.\r\nFrom $ (1)$ and $ (2)$, we conclude that $ D$, $ E$ and $ K$ are collinear as desired." } { "Tag": [], "Problem": "Noah was walking down Elm St. He then made 2 left turns and 3 right turns and is now facing east. Every turn is at a $ 90^\\circ$ angle. What direction was Noah originally going on Elm St.?", "Solution_1": "2 left turns+3 right turns=1 90 degree turn to the right. If you turn 90 degrees right once and face east, you originally faced north.\r\n\r\nAnswer: Noah was originally going $ \\boxed{\\text{north}}$ on Elm St." } { "Tag": [ "geometry", "3D geometry", "pyramid", "sphere" ], "Problem": "Given the triangle $ ABC$ such that $ (AB)^2\\plus{}(AC)^2\\equal{}(BC)^2$ and $ (AB)\\equal{}(AC)\\equal{}10$, consider the line $ t$ , passing through the midpoint of the larger side and perpendicular to the plane $ \\alpha$ of the given triangle. Given the point $ P$, such that $ P\\in t$ , calculate the volume of the pyramid $ PABC$ $ (P\\cancel{\\in}\\alpha)$, knowing that the radius of the inscribed sphere is $ \\sqrt{2}$ .", "Solution_1": "Let $ M$ be the midpoint of $ BC$, and $ h: \\equal{} MP$.\r\n\r\nSince $ MA\\equal{}MB\\equal{}MC\\equal{}5\\sqrt{2}$, we get $ PA\\equal{}PB\\equal{}PC\\equal{}\\sqrt{h^2\\plus{}50}$\r\n\r\nIf $ V$ is the volume and $ S$ the surface area of the pyramid, then\r\n\r\n$ V\\equal{}{Sr\\over 3}$\r\n\r\nwhere $ r$ is the inradius.\r\n\r\nThe triangles $ PAB$ and $ PAC$ are isosceles, with the sides $ 10,\\sqrt{h^2\\plus{}50},\\sqrt{h^2\\plus{}50}$, so we easily find that their combined area is $ S'\\equal{}10\\sqrt{h^2\\plus{}25}$\r\n\r\nThe area of $ \\triangle PBC$ is $ S''\\equal{}{5h\\sqrt{2}}$\r\n\r\nThe area of $ \\triangle ABC$ is $ S'''\\equal{}50$\r\n\r\nSo we get $ S\\equal{}S'\\plus{}S''\\plus{}S'''\\equal{}50\\plus{}5h\\sqrt{2}\\plus{}10\\sqrt{h^2\\plus{}25}$\r\n\r\nOn the other hand, $ V\\equal{}{S'''h\\over 3}\\equal{}{50h\\over 3}$\r\n\r\nTherefore we get the equation\r\n\r\n$ 50h\\equal{}(50\\plus{}5h\\sqrt{2}\\plus{}10\\sqrt{h^2\\plus{}25})\\sqrt{2}$\r\n\r\nAfter simplyfing and rearranging, we get\r\n\r\n$ 4h\\minus{}5\\sqrt{2}\\equal{}\\sqrt{2h^2\\plus{}50}\\implies h\\equal{}{20\\sqrt{2}\\over 7}$\r\n\r\nNow $ V\\equal{}{50h\\over 3}\\equal{}{1000\\sqrt{2}\\over 21}$" } { "Tag": [ "quadratics", "geometry", "perpendicular bisector", "algebra" ], "Problem": "Let $S_1$ be a semicircle with centre $O$ and diameter $AB$.A circle $C_1$ with centre $P$ is drawn, tangent to $S_1$, and tangent to $AB$ at $O$. A semicircle $S_2$ is drawn, with centre $Q$ on $AB$, tangent to $S_1$ and to $C_1$. A circle $C_2$ with centre $R$ is drawn, internally tangent to $S_1$ and externally tangent to $S_2$ and $C_1$. Prove that $OPRQ$ is a rectangle.", "Solution_1": "Consider the semicircles $S_1, S_2$ as full circles. Let $R_1 = \\dfrac{AB}{2}$ be the radius of the initial circle $S_1$ and $r_1, R_2, r_2$ radii of the remaining circles $C_1, S_2, C_2$. Obviously, the circle $C_1$ is centered on the perpendicular bisector of the segment AB and its radius is $r_1 = PO = \\dfrac{R_1}{2}$. Let U be the tangency point of the circles $S_1, C_1$ and let D be the other intersection of the circle $S_2$ with the line AB different from the point B. Invert the circles $C_1, S_2$ in the circle $S_1$. The circle $C_1$ tangent to the inversion circle and passing through the inversion center O is carried into a line tangent to the inversion circle $S_1$ at the point U, which is parallel to the line AB. The tangency point B of the semicircles $S_1, S_2$ is on the inversion circle and it stays in place. The circle $S_2$ is carried into a circle $S_2'$ tangent to the inversion circle $S_1$ at the point B and centered on the line AB. The other intersection D of the circle $S_2$ with the line AB is carried into the other intersection D' of the inverted circle $S_2'$ with the line AB. Since the circle $S_2'$ is centered on the line AB and tangent to the line parallel to AB through the point U, i.e., at a distance $OU = R_1$, its radius is equal to $R_2' = R_1$. Thus the distance of the inverted point D' from the inversion center is equal to ${OD' = OB + BD' = R_1 + 2R_2' = 3R_1}$. By the basic property of inversion, $OD \\cdot OD' = R_1^2$ and consequently,\r\n\r\n$OD = \\dfrac{R_1^2}{OD'} = \\dfrac{R_1^2}{3R_1} = \\dfrac{R_1}{3}$\r\n\r\nThis allows to calculate the radius $R_2$ of the circle $S_2$:\r\n\r\n$R_2 = \\dfrac{DB}{2} = \\dfrac{OB - OD}{2} = \\dfrac{R_1 - \\dfrac{R_1}{3}}{2} = \\dfrac{R_1}{3}$\r\n\r\nThe pairwise tangent circles $S_1, C_1, S_2, C_2$ form a quadruple of Soddy's circles and their radii $R_1, r_1, R_2, r_2$ are bound by the relation\r\n\r\n$\\dfrac{1}{(-R_1)^2} + \\dfrac{1}{r_1^2} + \\dfrac{1}{R_2^2} + \\dfrac{1}{r_2^2} = \\frac 1 2 \\left(\\dfrac{1}{-R_1} + \\dfrac{1}{r_1} + \\dfrac{1}{R_2} + \\dfrac{1}{r_2}\\right)^2$\r\n\r\nThe radius of the circle $S_1$ has to be taken with the negative sign, because its tangencies with the remaining 3 circles are internal. (If the 4 circles had all tangencies external, all radii would have positive signs.) This yields an equation for the radius of the circle $C_2$:\r\n\r\n$\\dfrac{1}{R_1^2} + \\dfrac{4}{R_1^2} + \\dfrac{9}{R_1^2} + \\dfrac{1}{r_2^2} = \\frac 1 2 \\left(-\\dfrac{1}{R_1} + \\dfrac{2}{R_1} + \\dfrac{3}{R_1} + \\dfrac{1}{r_2}\\right)^2$\r\n\r\n$\\dfrac{14}{R_1^2} + \\dfrac{1}{r_2^2} = \\dfrac{1}{2} \\left(\\dfrac{4}{R_1} + \\dfrac{1}{r_2}\\right)^2$\r\n\r\n$14 \\dfrac{r_2^2}{R_1^2} + 1 = \\frac 1 2 \\left(4 \\dfrac{r_2}{R_1} + 1\\right)^2$\r\n\r\nDenoting $x = \\dfrac{r_2}{R_1}$, we get the following quadratic equation for x:\r\n\r\n$2(14x^2 + 1) = (4x + 1)^2$\r\n\r\n$28x^2 + 2 = 16x^2 + 8x + 1$\r\n\r\n$12x^2 - 8x + 1 = 0$\r\n\r\n$x = \\dfrac{8 \\pm \\sqrt{64 - 48}}{24} = \\dfrac{8 \\pm \\sqr{16}}{24} = \\dfrac{8 \\pm 4}{24} = \\dfrac{2 \\pm 1}{6} = \\frac 1 2\\ \\text{or}\\ \\frac 1 6$\r\n\r\nThe first root obviously corresponds to a circle congruent to the circle $C_1$ symmetrical to it with respect to the line AB. By the problem conditions (the semicircles), this root is not acceptable. Thus $r_2 = \\dfrac{R_1}{6}$. Now we calculate the sides of the quadrilateral OPRQ:\r\n\r\n$OP = r_1 = \\dfrac{R_1}{2}$\r\n\r\n$QR = R_2 + r_2 = \\dfrac{R_1}{3} + \\dfrac{R_1}{6} = \\dfrac{R_1}{2}$\r\n\r\n$PR = r_1 + r_2 = \\dfrac{R_1}{2} + \\dfrac{R_1}{6} = \\frac 2 3\\ R_1$\r\n\r\n$OQ = OD + R_2 = \\dfrac{R_1}{3} + \\dfrac{R_1}{3} = \\frac 2 3\\ R_1$\r\n\r\nSince the opposite sides of the quadrilateral OPRQ are equal, it is a parallelogram. Since one of its internal angles $\\angle POQ = 90^\\circ$ is right, it is a rectangle.\r\n\r\nHi shobber, nice to meet you again! :clap:", "Solution_2": "I think that this sangaku-like problem can be solved in a simple way:\r\n\r\nLet $OA=OB$ be $r$. We have $OP = \\dfrac{r}{2}$. If $OQ=t$ we have\r\n\\[ \\sqrt{t^2 + \\dfrac{{r^2 }} {4}} = PQ = \\dfrac{r} {2} + (r - t) = \\dfrac{{3r}} {2} - t. \\]\r\nHence $OQ = t = \\dfrac{2r}{3}$ and $QB=\\dfrac{r}{3}$. We draw\r\nparallel lines $PR$ and $QR$ to $OQ$ and $OP$, respectively, and\r\nwe get the intersection point $R$ such that $RP=QO=\\dfrac{2r}{3}$\r\nand $RQ=PO=\\dfrac{r}{2}$. Being $\\dfrac{{2r}} {3} - \\dfrac{r} {2} = \\dfrac{r} {6} = \\dfrac{r} {2} - \\dfrac{r} {3}$, the circle\r\n $C_2$ with center $R$ and radius $\\dfrac{r}{6}$ is\r\ntangent to $C_1$ y $S_2$. On the other way, the relations \\[ r - OR = r - \\sqrt{\\dfrac{{4r^2 }} {9} + \\dfrac{{r^2 }} {4}} = r - \\dfrac{{5r}} {6} = \\dfrac{r} {6}, \\] tell us that the circle $C_2$ is tangent to $S_1$.\r\n\r\nSo, by construction, $OPRQ$ is a rectangle. :o" } { "Tag": [], "Problem": "A sequence of integers is obtained by starting with three digits $ d_1$,$ d_2$, and $ d_3$. The fourth term of the sequence is the units digit of $ d_1 \\plus{} d_2 \\plus{} d_3$. Each succeeding term is the units digit of the sum of the three previous terms. What numbers belong in the first, second and third positions respectively to complete the sequence. \n\n\\[ \\text{ \\_\\_ },\\text{ \\_\\_ },\\text{ \\_\\_ },1,1,1\\]", "Solution_1": "[quote=\"GameBot\"]A sequence of integers is obtained by starting with three digits $ d_1$,$ d_2$, and $ d_3$. The fourth term of the sequence is the units digit of $ d_1 \\plus{} d_2 \\plus{} d_3$. Each succeeding term is the units digit of the sum of the three previous terms. What numbers belong in the first, second and third positions respectively to complete the sequence.\n\\[ \\text{ \\_\\_ },\\text{\\_\\_ },\\text{\\_\\_ },1,1,1\n\\]\n[/quote]\r\n\r\nLet them be a,b,c\r\n\r\nit is clear that a+b+c>10\r\n\r\nif a+b+c=11:\r\n\r\nThis means that b+c+1=11 so b+c=10.\r\n\r\nIf b=4 c=6 which fails\r\nb=3 c=7 fails\r\nb=2 c=8 fails\r\nb=1 c=9 SUCEEDS!\r\n\r\nThus we have 1,1,9." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "How can I prove that\r\nin any triangle the following is true\r\n\r\n$(cosA+cosB+cosC)^2\\leq sin ^2 A+sin^ 2 B+sin ^2 C$ ??\r\n\r\n\r\nThanks for your time and assistance!\r\n\r\n :) :)", "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 (some of the latest posts).\r\n\r\n darij", "Solution_2": "i have a ineq (like that) :\r\nprove that:\r\n (cosA/2)^2+(cosB/2)^2+(cosC/2)^2 = >(sinA/2 +sinB/2 + sinC/2)^2", "Solution_3": "[quote=\"tmbtw\"]i have a ineq (like that) :\nprove that:\n (cosA/2)^2+(cosB/2)^2+(cosC/2)^2 = >(sinA/2 +sinB/2 + sinC/2)^2[/quote]\r\nthey are equal more and less :P", "Solution_4": "[quote=\"zhaobin\"][quote=\"tmbtw\"]i have a ineq (like that) :\nprove that:\n (cosA/2)^2+(cosB/2)^2+(cosC/2)^2 = >(sinA/2 +sinB/2 + sinC/2)^2[/quote]\nthey are equal more and less :P[/quote]\r\n\r\nWhat do you mean?? :? \r\n\r\nProbably that the ineq is not valid in any triangle??", "Solution_5": "try let\r\n$x=\\frac{\\pi}2 -\\frac A 2$\r\nAnd y,z similar\r\nyou will find they are equal :)" } { "Tag": [ "function", "induction", "algebra proposed", "algebra" ], "Problem": "A sequence of real numbers $\\{a_n\\}_n$ is called a [i]bs[/i] sequence if $a_n = |a_{n+1} - a_{n+2}|$, for all $n\\geq 0$. Prove that a bs sequence is bounded if and only if the function $f$ given by $f(n,k)=a_na_k(a_n-a_k)$, for all $n,k\\geq 0$ is the null function.\r\n\r\n[i]Mihai Baluna - ISL 2004[/i]", "Solution_1": "The condition simply says that the set $\\{a_n|n\\in\\mathbb N^*\\}$ has at most one non-null value. It's clear that if this condition hiolds, then the sequence is bounded, so the other implication is the interesting one.\r\n\r\nWe assume the sequence is bounded. There are two cases:\r\n\r\n[b]Case 1[/b]: the maximum $M$ of $\\{a_n\\}$ is reached. \r\n\r\nWe can see that if $a_n=M$, then $a_{n+1}$ is either $M$ or $0$, and the same holds for $a_{n-1}$. Moreover, $a_{n+1}=M\\iff a_{n-1}=0$, and we can construct the sequence in two possible ways, in both directions, according to whether $a_{n+1}=M$ or $0$. In both these situations, the sequence contains only two values, in both directions, $0$ and $M$, so the conclusion is proven.\r\n\r\n[b]Case 2[/b]: the maximum $M$ is never reached.\r\n\r\nLet $\\varepsilon>0$ be small enough (small enough that $2(M-\\varepsilon)>M$, for instance) so that we can find $a_n=M-\\varepsilon$. It's easy to show that we can also find $n$ s.t. $a_n=M-\\varepsilon$ and $a_{n+1}=t,\\ 0M$ , which is a contradiction. This shows that, in fact, $a_{n+3}=M-\\varepsilon+2t, a_{n+4}=t$, and we can repeat the procedure with $\\varepsilon$ replaced with $\\varepsilon-2t$ (which must necessarily be $>0$). We find that $\\varepsilon-2nt>0,\\ \\forall n\\ge 0$, which is absurd.\r\n\r\nHope it's Ok.", "Solution_2": "I hope my memory still works (I try to stay beyond Alzheimer's club of Pierre, but it becomes harder and harder), but I think this problem was already solved on the forum. I think Darij posted it, after a kind of TST of Germany. Probably I'm wrong, but it sounds too familiar to me and I do not have the ISL 2004. ;)", "Solution_3": "Yes, I believe that problem asked whether such a sequence could be bounded :?.", "Solution_4": "yea it was with the conditions given that $x_0$ and $x_1$ and positive and not equal to each other. And the answer is no. Its in the ISL 2004, and came up in the British FST.", "Solution_5": "Only the main idea: write all $x_i-s$ in terms of the first 2 distinct $x_i-s$. By induction one can easily obtain that all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\\in \\mathbb{N}$ . There are finitely many $(m,n)-s$ with $m<\\frac M{x_1}, n<\\frac M{x_2}$, with $M$ being the bound so $M$ is attained . Let $x_{i+1}$ be the first term that achieves the maximum, so $0x_{i+1}=M$, false.\r\n\r\nThere are still some cases to be considered for the first few terms of the sequence and for the case $0$ occurs in the sequence.", "Solution_6": "[quote=\"ikap\"]all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\\in \\mathbb{N}$[/quote]\r\n\r\nWhy is that? ($m,n \\in \\mathbb{N}$) I can only prove it with $m,n \\in \\mathbb{Z}$ and that doesn't help very much..." } { "Tag": [ "logarithms", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Prove that the sequence [tex](x_n)_{n\\geq 2}[/tex], defined by [tex]x_n=\\log_n n![/tex] does not contain arithmetic progressions with infinitely many terms.", "Solution_1": "First of all, prove that $x_{n+1}-x_n<1,\\ \\forall n\\ge 2$ and $\\displaystyle \\lim_{n\\to\\infty}(x_{n+1}-x_n)=1$. Then try to show that a sequence which satisfies these conditions can\u2019t contain infinite arithmetic progressions. It\u2019s not hard at all." } { "Tag": [ "topology", "limit", "advanced fields", "advanced fields unsolved" ], "Problem": "$f: \\mathbb{C}\\to \\mathbb{C}$ continuous and ${\\lim_{z \\to \\infty}f(z) = 0}$. \r\nShow that $f$ has a fixed point.\r\n\r\n[mod edit]: use \\infty for $\\infty$", "Solution_1": "This is a version of the Brouwer theorem; $f$ is bounded, so it maps some sufficiently large disk into itself.\r\n\r\nTo prove this version, look at the winding number of $f(z)-z$ around zero along the circle of radius $r$ centered at zero. For small $r$, either zero is a fixed point or the winding number is zero. For large $r$, the winding number is 1. The winding number is continuous in $r$ as long as $f(z)-z$ is never zero on the circle; since it can't be globally continuous, $f(z)-z=0$ somewhere." } { "Tag": [ "induction", "linear algebra", "matrix", "geometry", "rectangle", "strong induction" ], "Problem": "Hello\r\n\r\nProve the following Identity\r\n\r\n$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$", "Solution_1": "[quote=\"mathgeniuse^ln(x)\"]Hello\n\nProve the following Identity\n\n$(F_{n+1})^{2}+(F_{n-1})^{2}=F_{2n}$[/quote]\r\n[hide=\"idea\"]Induction[/hide]\n[hide=\"uglier idea\"]Binet's formula[/hide]", "Solution_2": "Oh, I forgot to mention.\r\n\r\nIs it possible to do it without the ugly, but useful formula of Binet?\r\n\r\nThanks", "Solution_3": "[quote=\"mathgeniuse^ln(x)\"]Oh, I forgot to mention.\n\nIs it possible to do it without the ugly, but useful formula of Binet?\n\nThanks[/quote]\r\nThis seems definitely like an induction problem.", "Solution_4": "I have not thought about this problem (excluding induction and Binet's formula) yet, but the idea that first came to my mind was to simplify the problem by subtracting $2F_{n-1}F_{n+1}$ and using $\\left( \\begin{matrix}1&1\\\\1&0\\end{matrix}\\right)^{n}=\\left(\\begin{matrix}F_{n+1}&F_{n}\\\\F_{n}&F_{n-1}\\end{matrix}\\right)$ along with $\\det \\left(\\prod A_{i}\\right)=\\prod\\det (A_{i})$.", "Solution_5": "hey good idea!\r\n\r\n :P", "Solution_6": "[quote=\"iamagenius\"]hey good idea!\n\n :P[/quote]\r\n\r\nThat was totally retarded, annoying, unnecessary, and spammy.", "Solution_7": "... Am I being totally stupid or wut\r\n\r\n$n=3$\r\n$(F_{4})^{2}+(F_{2})^{2}=F_{6}$\r\n$3^{2}+1^{2}=8$\r\n$9+1=8$\r\n$10=8$\r\n\r\n??????????", "Solution_8": "[quote=\"tjhance\"]... Am I being totally stupid or wut\n\n$n=3$\n$(F_{4})^{2}+(F_{2})^{2}=F_{6}$\n$3^{2}+1^{2}=8$\n$9+1=8$\n$10=8$\n\n??????????[/quote]\r\nIt seems that the identity should be \r\n\r\n$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$.", "Solution_9": "A well known Fibonacci formula is:\r\n\r\n$F_{2n+1}=F_{n}^{2}+F_{n+1}^{2}$\r\n\r\nI don't know if that's what you were going for. I don't think your formula works.\r\n\r\nEdit: I lied, I think rnwang is right.", "Solution_10": "For induction (this is for the identity I had in my previous post), it looks like you'll have to prove \r\n\r\n$F_{n+1}\\cdot (F_{n+2}+F_{n})=F_{2n+2}$.", "Solution_11": "Use the definition and you should get a more general formula:\r\n\r\n\\[F_{2n}=F_{j+1}F_{2n-j}+F_{j}F_{2n-(j+1)}\\]\r\n\r\nSo for $j=n$, we have\r\n\r\n\\[\\begin{aligned}F_{2n}&=F_{n+1}F_{n}+F_{n}F_{n-1}\\\\ &=(F_{n+1})^{2}-(F_{n-1})^{2}\\end{aligned}\\]", "Solution_12": "[quote=\"rnwang2\"][quote=\"tjhance\"]... Am I being totally stupid or wut\n\n$n=3$\n$(F_{4})^{2}+(F_{2})^{2}=F_{6}$\n$3^{2}+1^{2}=8$\n$9+1=8$\n$10=8$\n\n??????????[/quote]\nIt seems that the identity should be \n\n$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$.[/quote]\r\nNo wonder, I tried proving it by induction and I went in circles...", "Solution_13": "[quote=\"rnwang2\"][quote=\"tjhance\"]... Am I being totally stupid or wut\n\n$n=3$\n$(F_{4})^{2}+(F_{2})^{2}=F_{6}$\n$3^{2}+1^{2}=8$\n$9+1=8$\n$10=8$\n\n??????????[/quote]\nIt seems that the identity should be \n\n$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$.[/quote]\r\n\r\nYeah, my mistake.\r\n\r\nBut I have a question, is there any way that might be possible to prove without any formulae.", "Solution_14": "[quote=\"mathgeniuse^ln(x)\"]But I have a question, is there any way that might be possible to prove without any formulae.[/quote]\r\n\r\nYou might be able to use the fact that $F_{n}$ represents the number of ways to tile a board of length $n$ with $1\\times 1$ squares and $2\\times 1$ rectangles", "Solution_15": "[hide]First, rewrite the equality as\n\n$F_{2n}= F_{n+1}^{2}-(F_{n+1}-F_{n})^{2}= F_{n}(2F_{n+1}-F_{n})$.\n\nThe base case is easy, since we have $F_{4}= 3 = 2^{2}-1^{2}= F_{3}^{2}-F_{1}^{2}$. Now we will show that it is true for $n = k+1$ assuming it is true for $n \\le k$. We have\n\n$F_{2k+2}= F_{2k+1}+F_{2k}= 3F_{2k}-F_{2k-2}$.\n\nNow we apply the inductive hypothesis to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-F_{k-1}(2F_{k}-F_{k-1})$,\n\nwhich we can simplify with the substitution $F_{k-1}= F_{k+1}-F_{k}$ to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-(F_{k+1}-F_{k})(3F_{k}-F_{k+1}) = 2F_{k}F_{k+1}+F_{k+1}^{2}$\n\nand finally from $F_{k}= F_{k+2}-F_{k+1}$ we get\n\n$F_{2k+2}= F_{k+1}(2F_{k+2}-F_{k})$\n\nas desired.[/hide]\r\n\r\nThe interpretation of the Fibonnaci sequence that renatzu mentioned leads to a simple proof that $F_{2n}= F_{n}^{2}+F_{n-1}^{2}$, but I'm not quite sure how to apply it to this problem.", "Solution_16": "The so-called \"formula\" in your post is incorrect (try n=2).", "Solution_17": "[quote=\"paladin8\"]$F_{2n}= F_{n}^{2}+F_{n-1}^{2}$[/quote]$F_{2n-1}= F_{n}^{2}+F_{n-1}^{2}$", "Solution_18": "whats wrong with binet's formula...it is not messy computation at all", "Solution_19": "[quote=\"boxedexe\"]The so-called \"formula\" in your post is incorrect (try n=2).[/quote]\r\n\r\nSorry, that formula has the indices shifted (i.e. $F_{0}= 1$, $F_{1}= 1$, etc.).", "Solution_20": "[quote=\"paladin8\"][quote=\"boxedexe\"]The so-called \"formula\" in your post is incorrect (try n=2).[/quote]\n\nSorry, that formula has the indices shifted (i.e. $F_{0}= 1$, $F_{1}= 1$, etc.).[/quote]But, if you shift the index, you will make this statement false: $F_{5n}$ is divisible by 5 for all $n\\in\\mathbb{N}$ :D", "Solution_21": "[quote=\"paladin8\"][hide]First, rewrite the equality as\n\n$F_{2n}= F_{n+1}^{2}-(F_{n+1}-F_{n})^{2}= F_{n}(2F_{n+1}-F_{n})$.\n\nThe base case is easy, since we have $F_{4}= 3 = 2^{2}-1^{2}= F_{3}^{2}-F_{1}^{2}$. Now we will show that it is true for $n = k+1$ assuming it is true for $n \\le k$. We have\n\n$F_{2k+2}= F_{2k+1}+F_{2k}= 3F_{2k}-F_{2k-2}$.\n\nNow we apply the inductive hypothesis to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-F_{k-1}(2F_{k}-F_{k-1})$,\n\nwhich we can simplify with the substitution $F_{k-1}= F_{k+1}-F_{k}$ to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-(F_{k+1}-F_{k})(3F_{k}-F_{k+1}) = 2F_{k}F_{k+1}+F_{k+1}^{2}$\n\nand finally from $F_{k}= F_{k+2}-F_{k+1}$ we get\n\n$F_{2k+2}= F_{k+1}(2F_{k+2}-F_{k})$\n\nas desired.[/hide]\n\nThe interpretation of the Fibonnaci sequence that renatzu mentioned leads to a simple proof that $F_{2n}= F_{n}^{2}+F_{n-1}^{2}$, but I'm not quite sure how to apply it to this problem.[/quote]\r\n\r\nOkay that is a good proof.\r\n\r\nAre we allowed to assume the base cases are true and then prove that for $n=k+1$, we can use the cases for $n=k-1$ and $n=k$ and prove our statement?\r\n\r\nIs that okay?\r\n\r\nI didn't know that, that is why I didn't approach it like that.", "Solution_22": "[quote=\"mathgeniuse^ln(x)\"][quote=\"paladin8\"][hide]First, rewrite the equality as\n\n$F_{2n}= F_{n+1}^{2}-(F_{n+1}-F_{n})^{2}= F_{n}(2F_{n+1}-F_{n})$.\n\nThe base case is easy, since we have $F_{4}= 3 = 2^{2}-1^{2}= F_{3}^{2}-F_{1}^{2}$. Now we will show that it is true for $n = k+1$ assuming it is true for $n \\le k$. We have\n\n$F_{2k+2}= F_{2k+1}+F_{2k}= 3F_{2k}-F_{2k-2}$.\n\nNow we apply the inductive hypothesis to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-F_{k-1}(2F_{k}-F_{k-1})$,\n\nwhich we can simplify with the substitution $F_{k-1}= F_{k+1}-F_{k}$ to get\n\n$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-(F_{k+1}-F_{k})(3F_{k}-F_{k+1}) = 2F_{k}F_{k+1}+F_{k+1}^{2}$\n\nand finally from $F_{k}= F_{k+2}-F_{k+1}$ we get\n\n$F_{2k+2}= F_{k+1}(2F_{k+2}-F_{k})$\n\nas desired.[/hide]\n\nThe interpretation of the Fibonnaci sequence that renatzu mentioned leads to a simple proof that $F_{2n}= F_{n}^{2}+F_{n-1}^{2}$, but I'm not quite sure how to apply it to this problem.[/quote]\n\nOkay that is a good proof.\n\nAre we allowed to assume the base cases are true and then prove that for $n=k+1$, we can use the cases for $n=k-1$ and $n=k$ and prove our statement?\n\nIs that okay?\n\nI didn't know that, that is why I didn't approach it like that.[/quote]\r\n\r\nYes, though my proof is incomplete because I need two base cases :oops:. That's ok, though, you can fill that part in. Formally, it's called strong induction (assuming it is true for $n \\le k$ and then proving it for $n = k+1$). One flawed \"proof\" using strong induction is this.\r\n\r\nProve: $a^{n}= 1$ for all positive reals $a$ and integers $n \\ge 0$.\r\n\r\nBase Case: $a^{0}= 1$ trivially.\r\n\r\nInduction Step: $a^{k+1}= \\frac{a^{k}\\cdot a^{k}}{a^{k-1}}= \\frac{1 \\cdot 1}{1}= 1$ by the induction hypothesis that $a^{n}= 1$ for all $n \\le k$.", "Solution_23": "Were teh two base cases n=2 and n=k?\r\n\r\nThanks\r\n\r\nAnd paladin, I think since we know that my formula is true, we can go to the next case and subtract the two to get it, I don't konw.", "Solution_24": "[quote=\"mathgeniuse^ln(x)\"]Were teh two base cases n=2 and n=k?[/quote]\r\n\r\nThe base case is n=1, since he assumes it holds true for k and k+1.." } { "Tag": [ "geometry", "perimeter" ], "Problem": "Find the number of distinct integer-sided triangles with the perimeter of 2007.", "Solution_1": "[hide=\"hint\"]the length of each of the sides is less than 1004[/hide]" } { "Tag": [], "Problem": "\u0388\u03c7\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03c4\u03b5 \u03ad\u03bd\u03b1 \u03ba\u03b1\u03bb\u03cc \u03b8\u03ad\u03bc\u03b1 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 , \u03bd\u03b1 \u03c0\u03b5\u03c1\u03b9\u03ad\u03c7\u03b5\u03b9 (\u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03b3\u03b9\u03bd\u03cc\u03bc\u03b5\u03bd\u03bf \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03c4\u03b5\u03c4\u03b1\u03b3\u03bc\u03ad\u03bd\u03b5\u03c2) :lol:", "Solution_1": "Parathetw me tin eukairia mia wraia askisi pou eixa dei pro kairou.\r\nDinontai dio statheres eutheies sto epipedo oi opoies temnontai sto simeio O. Apo to O ksekinoun stis tesseris kateuthinseis twn aksonwn ta kinita A,B,C,D me diaforetikes taxitites. Na apodeiksete oti i posotita\r\n\r\n$\\frac{AB^{2}-BC^{2}+CD^{2}-DA^{2}}{AC\\cdot BD}$, paramenei statheri.\r\n\r\nAlexandros", "Solution_2": "Check \u03bc\u03b5\u03c1\u03b9\u03ba\u03ad\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b5\u03b4\u03ce:\r\n\r\nhttp://www.telemath.gr/mathematical_lycee/mathematical_lycee_second/pos_tech/index.htm", "Solution_3": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03b5\u03bc\u03bd\u03bf\u03c5\u03c3\u03ce\u03bd.", "Solution_4": "\u0398\u03cd\u03bc\u03b9\u03c3\u03ad \u03bc\u03bf\u03c5 \u03c0\u03bf\u03b9\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03b5\u03bc\u03bd\u03bf\u03c5\u03c3\u03ce\u03bd ?? \u03c4\u03af \u03bb\u03ad\u03b5\u03b9 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2???", "Solution_5": "\u0391\u03c0\u03bf \u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03cc \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u039c \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c4\u03ad\u03bc\u03bd\u03bf\u03c5\u03c3\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03b5\u03bc\u03bd\u03bf\u03c5\u03bd \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u0391, \u0392 , \u0393 ,\u0394 \r\n\u03c4\u03cc\u03c4\u03b5 \u039c\u0391 \u039c\u0392 =\u039c\u0393 \u039c\u0394", "Solution_6": "\u03c3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b1\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c4\u03b5\u03bc\u03bd\u03bf\u03c5\u03c3\u03ce\u03bd \u03bd\u03b1 \u03be\u03ad\u03c1\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03ce\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03ad\u03ba\u03c6\u03c1\u03b1\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u0398\u0395\u03a9\u03a1\u0397\u039c\u0391 \u03a4\u0395\u039c\u039d\u039f\u03a5\u03a3\u03a9\u039d \u03a4\u039f\u03a5 \u039a\u03a5\u039a\u039b\u039f\u03a5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03be\u03ad\u03c7\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03ac\u03bb\u03bb\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03bc\u03b1\u03c4\u03b1 \u03c4\u03b5\u03bc\u03bd\u03bf\u03c5\u03c3\u03ce\u03bd \u03bf\u03ba??\r\n\u03c0\u03c7 \u039c\u03b5\u03bd\u03ad\u03bb\u03b1\u03bf\u03c2", "Solution_7": "[img]http://img433.imageshack.us/img433/6535/askisiib4.png[/img]", "Solution_8": "[b][size=109]giannis,[/size][/b] \u03ba\u03b1\u03bb\u03c9\u03c3\u03cc\u03c1\u03b9\u03c3\u03b5\u03c2 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ad\u03b1 \u03bc\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ae \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1.\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2." } { "Tag": [ "logarithms", "function" ], "Problem": "$f(x)=a\\log_{b}(x)$, where $a>0$, $b>0$, and $b\\not=1$.\r\nIf $f \\left(\\frac{16}{81}\\right)=-2$ and $a=\\log_{k}b$, determine the positve value of $k$.", "Solution_1": "This is what i got:\r\n[hide] $f\\left(\\frac{16}{81}\\right)=2a\\log_{b}\\left(\\frac{4}{9}\\right)=-2$\n$a=1, b=\\frac{9}{4}$\n$1=\\log_{k}\\left(\\frac{9}{4}\\right)$\n$k=\\frac{9}{4}$[/hide]", "Solution_2": "[quote=\"tiredepartment\"]$ f(x) \\equal{} a\\log_{b}(x)$, where $ a > 0$, $ b > 0$, and $ b\\not \\equal{} 1$.\nIf $ f \\left(\\frac {16}{81}\\right) \\equal{} \\minus{} 2$ and $ a \\equal{} \\log_{k}b$, determine the positve value of $ k$.[/quote]\r\n[hide]\n$ \\minus{} 2 \\equal{} a\\frac {\\log(16) \\minus{} log(81)}{\\log b}$\n$ a \\equal{} \\frac {\\log b }{\\log k}$\n$ \\minus{} 1 \\equal{} \\log_k (\\frac {4}{9})$\n$ k \\equal{} \\left( \\frac {4}{9} \\right)^{ \\minus{} 1}$\n$ k \\equal{} \\frac {9}{4}$\n[/hide]" } { "Tag": [ "USAMTS", "AMC", "AIME" ], "Problem": "We just got the graded papers back from the NSA today. It'll take us a couple of days to get the scores and comments scanned in, so we hope to have the Round 3 scores and feedback released on Thursday or Friday of this week.", "Solution_1": "sweet thanks for your hard work", "Solution_2": "Round 3 scores and feedback are now available on the USAMTS website. \r\n\r\nWe will make an announcement about AIME qualification during the week of Feb 23-27. We will develop a procedure so that students who did not resubmit Round 2 by mail will still be eligible to qualify for the AIME. [b]Please do not inquire about this prior to our announcement[/b]." } { "Tag": [], "Problem": "Can someone tell me why the answer is E?!?\r\nPeterson's online test told me it was E, however I dont see how that is possible. \r\n\r\nIf \u20132 < a < 3 and 5 < b < 10, which of the following represents the range of possible values for ab ?\r\n\r\n\r\na. 2 < ab < 13 \r\n\r\nb. \u20132 < ab < 11 \r\n\r\nc. \u201310 < ab < 30 \r\n\r\nd. \u201320 < ab < 15 \r\n\r\ne. \u201320 < ab < 30", "Solution_1": "[quote=\"woohooxd\"]Can someone tell me why the answer is E?!?\nPeterson's online test told me it was E, however I dont see how that is possible. \n\nIf \u20132 < a < 3 and 5 < b < 10, which of the following represents the range of possible values for ab ?\n\n\na. 2 < ab < 13 \n\nb. \u20132 < ab < 11 \n\nc. \u201310 < ab < 30 \n\nd. \u201320 < ab < 15 \n\ne. \u201320 < ab < 30[/quote]\r\n\r\n[hide]To find the range of possible values, try to find the smallest possible value for $ab$, and then the largest possible value. \n\nPretending that the $>$ and $<$ signs were $\\ge$ and $\\le$ instead, the smallest possible value of $ab$ would be when $a=-2$ and $b=10$, when $ab=-20$. \n\nThe largest possible value of $ab$ would be when $a=3$ and $b=10$, when $ab=30$\n\nSo the range RIGHT NOW would be $-20\\le{ab}\\le30$\n\nChanging back the signs wouldn't do us any different, so the answer must be E. [/hide]", "Solution_2": "[hide]We want the smaller number ab to be negative. The more neagative it is, the better, so pick -2 for a (to make it negative) and 10 from b to make it -20. \nIf you picked -2 and 5, you would've gotten -10, not smaller than -20. \n\nThe biggest number ab should be positive and the biggest numbers, 3 and 10, which make 30. \n\nso -200$ and changed sign", "Solution_2": "Since $ \\left(\\frac {2n}{n \\plus{} 1}\\right)^{\\pi}$ tends to a number other than $ 0$ or $ \\infty,$ then we should just start looking at it as if it were that constant. We need to \"de-clutter\" and just remove it from our sight.\r\n\r\nThe workhorse test for \"de-cluttering\" is the limit comparison test: if $ a_n$ and $ b_n$ are both eventually positive and if $ \\lim_{n\\to\\infty}\\frac {a_n}{b_n} \\equal{} L,$ then $ \\sum a_n$ and $ \\sum b_n$ either both converge or both diverge.\r\n\r\nHence in this case, $ \\sum_{n \\equal{} 2}^{\\infty}|a_n|$ is limit comparable to $ \\sum \\frac1{\\sqrt {n}},$ which is known to diverge. \r\n\r\nSo the original series does not converge absolutely. Since the alternating series test does apply to it, then the series converges conditionally.\r\n\r\nAs part of checking that the alternating series test holds, we do have to show that $ |a_n|$ is an eventually decreasing function of $ n.$ In this case, that's a little annoying. For completeness, let me accept the annoyance and show that:\r\n\r\nLet $ f(x) \\equal{} \\frac1{\\sqrt {x}}\\left(\\frac {2x}{x \\plus{} 1}\\right)^{\\pi} \\equal{} x^{ \\minus{} 1/2}\\left(2 \\minus{} \\frac2{x \\plus{} 1}\\right)^{\\pi}.$\r\n\r\nThen $ f'(x) \\equal{} \\minus{} \\frac12x^{ \\minus{} 3/2}\\left(2 \\minus{} \\frac2{x \\plus{} 1}\\right)^{\\pi} \\plus{} x^{ \\minus{} 1/2}\\pi\\left(2 \\minus{} \\frac2{x \\plus{} 1}\\right)^{\\pi \\minus{} 1}\\left(\\frac2{(x \\plus{} 1)^2}\\right)$\r\n\r\n$ \\equal{} x^{ \\minus{} 1/2}\\left(2 \\minus{} \\frac2{x \\plus{} 1}\\right)^{\\pi \\minus{} 1} \\left[ \\minus{} \\frac1{2x}\\left(2 \\minus{} \\frac2{x \\plus{} 1}\\right) \\plus{} \\frac {2\\pi}{(x \\plus{} 1)^2}\\right].$\r\n\r\nExamination of the quantity in the brackets to find the dominant terms (namely $ \\minus{} \\frac1x$ ) shows that this derivative is negative for all sufficiently large $ x.$ Hence $ f(x)$ is an eventually decreasing function and thus $ |a_n| \\equal{} f(n)$ is an eventually decreasing sequence.\r\n\r\nIf this question were to appear on one of my tests, I would not expect my students to go to that much trouble to show the monotonicity.", "Solution_3": "I think there's an old theorem (Hardy?) that says that any rational function of finitely many terms like $ x^p$, $ \\log x$, and $ a^x$ is monotone for sufficiently large $ x$. Not particularly interesting or surprising, but would be helpful here.\r\nOf course, the monotonicity part was needed for AST, not for LCT.\r\n\r\n[color=green][Oops. Thanks for pointing that out. I've edited. - K.M.][/color]", "Solution_4": "Thanks a lot.I haven't known about de-cluttering method." } { "Tag": [ "percent" ], "Problem": "The hour hand of a broken clock points to $ 5$ and the minute hand to $ 12$. To the nearest percent what part of the clock face is shaded?\n\n[asy]draw(circle((0,0),1));\nfilldraw(arc((0,0),1,-75,90)--(0,1)--(0,0)--(0.25,-0.96)--cycle,gray);\ndraw((0,0)--(0,1),Arrow);\ndraw((0,0)--(0.25,-0.96),Arrow);[/asy]", "Solution_1": "The shaded portion represents $ \\frac{5}{12}$ of the whole area. $ \\frac{1}{12} \\equal{} 8.\\overline{3}\\%$. Multiplying this by $ 5$ gives $ 41.\\overline{6}\\% \\approx \\boxed{42\\%}$" } { "Tag": [ "modular arithmetic", "abstract algebra", "number theory", "relatively prime", "group theory" ], "Problem": "Find the remainder when the integer 1x3x5x7x....x2003x2005 is divided by 1000", "Solution_1": "obviously , the product include the number 125 so we only need to find the remainder of $ 1x3x...x2005$ when we divided it by 8\r\nnotice that every numbers in the product can be rewritten as $ 2k\\plus{}1$ where k from 0 to 1002\r\nif k is divided by 4 we found $ 2k\\plus{}1 \\equiv 1 (mod 8)$\r\nif k \\equiv 1 (mod 4) we found $ 2k\\plus{}1 \\equiv 3(mod 8)$\r\nif k \\equiv 2(mod 4) we found $ 2k\\plus{}1 \\equiv 5 (mod 8)$\r\nif k \\equiv 3 (mod 4) we found $ 2k\\plus{}1 \\equiv 7 (mod 8)$\r\ntake a look at the product $ 1x3x5x7....2005 \\equal{} (1x3x5x7)x..x(1993x1995x1997x1999)$$ 2001x2003x2005 \\equiv (1x3x5x7)^{250}x2001x2003x2005 \\equiv 7 (mod 8)$\r\nand now you got the answer ...", "Solution_2": "The factor or $ 125$ is a very good observation! However, your answer is incorrect. We can fix that by cleaning up the rest of the proof a bit.\r\n\r\n[hide]Now, note that every odd number is $ \\pm 1, \\pm 3$ in mod 8.\n\nFor ease of counting, let us consider all odds less than 2008 first. There are 2008/8 = 251 sets of $ \\pm1, \\pm 3$ factors in mod 8 within this product.\n\nHowever:\n- one factor of $ \\minus{} 1 \\equiv 2007$ was artificially added for the above calculation; we will remove this now\n- one factor of $ \\minus{} 3 \\equiv 125$ will be removed for now; we will multiply in the 125 later\nThus, these two will only have 250 factors. So, we have, in $ \\pmod{8}$:\n\n$ \\frac {\\Pi}{125}$ $ \\equiv ( \\minus{} 1)^{250} (1)^{251} ( \\minus{} 3)^{250} (3)^{251} \\equiv 3^{250} 3^{251} \\equiv 9^{250}(3) \\equiv 1^{250} (3) \\equiv 3$ $ \\pmod{8}$\n\nHence, our product is of the form $ \\prod \\equal{} 125 (8k \\plus{} 3) \\equiv 125 \\cdot 3 \\equiv 375 \\pmod {1000}$\n\n\n[b]@HTA[/b] You cannot count the 125 towards the mod 8 calculation, since it is already being set aside to facilitate the final multiplication.\n[/hide]", "Solution_3": "Am I missing something or is $ 375\\equiv7\\pmod8$, hence HTA's solution is fine", "Solution_4": "[b]Follow-up:[/b] Find the remainder when the integer $ 1 \\times 3 \\times 7 \\times ... \\times 2003 \\times 2007$ is divided by $ 1000$, where the product is taken over all integers from $ 1$ to $ 2007$ not divisible by $ 2$ or by $ 5$.", "Solution_5": "[quote=\"t0rajir0u\"][b]Follow-up:[/b] Find the remainder when the integer $ 1 \\times 3 \\times 7 \\times ... \\times 2003 \\times 2007$ is divided by $ 1000$, where the product is taken over all integers from $ 1$ to $ 2007$ not divisible by $ 2$ or by $ 5$.[/quote]\r\n\r\n[hide] Hmm... if we take this as 2001 x 2003 x 2007 x ( 1 x 3 x...999) x (1001 x 1003 x 1999)\n\nnote that the last two terms are the same mod 1000, and each congruent to the product of all of the number relatively prime to 1000. And note that since the numbers relatively prime to 1000 form an abelian group under multiplication (mod 1000) we can just pair each number up with its inverse. Either a number has an inverse other than itself, in which case we can match it up with the number in its term (ie match 3 with 667 and 1003 with 1667) or if it is its own inverse we can match it up with itself from the other term (ie 999 and 1999). From there we have 2001 x 2003 x 2007 x (a bunch of 1's mod 1000) which comes out to be 21 mod 1000\n[/hide]", "Solution_6": "[quote=\"Hamster1800\"]Am I missing something or is $ 375\\equiv7\\pmod8$, hence HTA's solution is fine[/quote]\r\n\r\nHe found the remainders modulo 125, which is zero, and modulo 8, which is 7 - but what's asked for is COMBINED remainder modulo 1000, so we must find a number which is divisible by 125, but gives remainder 7 when divided by 8, and that's 375.", "Solution_7": "[quote=\"Farenhajt\"][quote=\"Hamster1800\"]Am I missing something or is $ 375\\equiv7\\pmod8$, hence HTA's solution is fine[/quote]\n\nHe found the remainders modulo 125, which is zero, and modulo 8, which is 7 - but what's asked for is COMBINED remainder modulo 1000, so we must find a number which is divisible by 125, but gives remainder 7 when divided by 8, and that's 375.[/quote]\r\n\r\nAs far as I've seen on these forums, finishing a solution with the Chinese Remainder Theorem is often left out of these types of problems. Therefore, I felt like there was nothing lacking in his solution.", "Solution_8": "If that's the case, that habit should be fought against, since abandoning a problem without giving a final answer is worth almost nothing.", "Solution_9": "It's not abandoning so much as just knowing that you just have to check $ 0, 125, 250, 375, 500, 625, 750, 875$ and check which one gives you the correct remainder upon division by $ 8$. Alternatively, you could algorithmically arrive at the answer, but that is less common.\r\n\r\nAlso, why do you say it has almost no value? The key insights were made, all that was left out was some computation. Clearly, on a competition you would finish the problem, but when it's practice, I see nothing wrong with leaving out computation.", "Solution_10": "Sketch of a solution is one thing, but going all the way through, and then abandoning it at the very end out of sheer laziness (no other reason possible) really looks bad.", "Solution_11": "I'm sorry, I misunderstood HTA's intention. Personally, I think that even if a proof is not fully completed, the full intent of the solution sketch should not have to be implied. A better choice of words and neater typesetting on his part (or more time spent on my part) would have made the intent more obvious.", "Solution_12": "i'm sorry , i didn't give the final answer because i was running out of time , i had to go to school , so here is the rest of the solution \r\nas Hamster1800 said , we have to check 0;125;250;375;500;625;750;875 to see which one gives the correct remainder , but notice that it gives the remainder 7 upon division by 8 so it must be an odd number,\r\n$ 125 \\equiv 5 (mod 8)$\r\n$ 375 \\equiv 7 (mod 8)$\r\nso the answer is 375 \r\n\r\np/s it's no need to check all of them , stop when you find the desire answer" } { "Tag": [ "search" ], "Problem": "1)A piece of copper of density 8.9g/cm^3 floats on mercury of density 13.6g/cm^3.What fraction of its volume will be above the surface?\r\n\r\n2)A helium ballom is tied to the back seat by a massless string of a car and at rest,the bllon floats upwards.Now,the car is closed.If the car is suddenly accelerated,where will the helium ballon go?(Front,back,sideways?)", "Solution_1": "[quote=\"shadysaysurspammed\"]\n2)A helium ballom is tied to the back seat by a massless string of a car and at rest,the bllon floats upwards.Now,the car is closed.If the car is suddenly accelerated,where will the helium ballon go?(Front,back,sideways?)[/quote]\r\n[color=blue]The answer is [b]FRONT[/b][/color]", "Solution_2": "1) I would do this one by assuming a mass of copper, calculating it's volume, than the volume of an equal weight of mercury and the percentage of copper's volume that is the mercury's volume is the percentage that is submerged in mercury when the copper if floating in mercury.\r\n\r\n2) The balloon will move back. An object at rest remains at rest until a force acts on it. When the car starts accelerating, no force has had an opporutnity to act on the balloon, so it will intially seem to move back, as the car moves forward. (I've actually done this one, so they theoritical answer is confirmed by experiement.)", "Solution_3": "I'm sorry i have less time to thiunk on the 2nd one bcos i have xams right now,but answer is front(i think) bcos of archimedes.There is some air in the car that will go back and thus push the ballon forward.", "Solution_4": "Let me explain the second one:-\r\n\r\nThere is some air in the car as the question says that the balloon is floating in the car.Now we close the doors.\r\n\r\nWhen we acclelrate,the air moves to the back of the car.Since the helium balloon is lighter than air and has a massless string(so no inertia,same for rubber skin),the helium balloon will move forward by buoyant force (Consider it like a cork on water).", "Solution_5": "1) We must have $F_A = G$.\r\n\r\n$\\Longleftrightarrow \\rho_{\\textrm{liquid}} V_{\\textrm{submerged}} g = \\rho_{\\textrm{body}} V_{\\textrm{total}}g$;\r\n\r\n$\\Longleftrightarrow \\, \\frac{V_{\\textrm{submerged}}}{V_{\\textrm{total}}} = \\frac{\\rho_{\\textrm{body}}}{\\rho_{\\textrm{liquid}}}$. Right, guys?\r\n\r\n2) I don't really understand what force pushes the balloon to the front :blush: I didn't learn physics when I should have.", "Solution_6": "[quote=\"perfect_radio\"]1) We must have $F_A = G$.\n\n$\\Longleftrightarrow \\rho_{\\textrm{liquid}} V_{\\textrm{submerged}} g = \\rho_{\\textrm{body}} V_{\\textrm{total}}g$;\n\n$\\Longleftrightarrow \\, \\frac{V_{\\textrm{submerged}}}{V_{\\textrm{total}}} = \\frac{\\rho_{\\textrm{body}}}{\\rho_{\\textrm{liquid}}}$. Right, guys?\n\n2) I don't really understand what force pushes the balloon to the front :blush: I didn't learn physics when I should have.[/quote]\r\n\r\nYou are right about the first one.\r\n\r\nIn the second one,[b]buoyant force[/b] pushes the balloon \r\nforward\r\n\r\nLike a cork floats up on water due to buoyant force,you know about this right?So in the same way,the air pushes the balloon forward by buoyant force.\r\n\r\nWhy does a balloon float in air in the first place?Due o buyant force exerted by the air around it.If you take a balloon filled with air to the moon(where there's no atmosphere),it won't float up since there's no atmosphere.", "Solution_7": "Concerning the second problem about a helium balloon in an accelerating vehicle, you can find a useful discussion at this site:\r\n\r\n\r\nhttp://www.amherst.edu/physicsqanda/Helans.htm", "Solution_8": "The link doesn't work ... :?", "Solution_9": "The link is NOW CORRECT .I am sorry about this mistake.\r\n\r\nAlso a Google search under, inertial frames helium balloon,will turn up a lot of other similar sites.", "Solution_10": "Another dumb question, related to the link provided above: when an elevator is going upwards, you feel heavier because of what? The ??? force, or what?\r\n\r\nSorry for the stupid question \r\n :D", "Solution_11": "If you stand on a weigjhing machine,your wieght will be more than usaual if the elevator is accelerating upwards because now the downward force would be $m(g+a)$ instead of $mg$ where $a$ is the $net$ acceleration with which the elevator is going upwards.The elevator is actually going up with acceleration $g+a$,so by newton's third law,you should get weight=$m(g+a)$ :)", "Solution_12": "I searched on the net about this subject and I found that weight still stays equal to $mg$. You feel heavier because of the normal force. Kind of logical, if you think about it.", "Solution_13": "Yes,that is what i said,the weiht on the scale will be more but you won't get fatter or anything.Didn't i say newton's third law?That is basically normal force." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $T$ be a point inside a square $ABCD$. The lines $TA, TB, TC, TD$ meet the circumcircle of $ABCD$ again at $A',B',C',D'$, respectively\r\n. Prove that\r\n$A'B'.C'D'=A'D'.B'C'$.", "Solution_1": "We have:\r\n\r\n$\\triangle TA'B'\\sim \\triangle TBA \\Longrightarrow \\frac{A'B'}{a}=\\frac{TB'}{TA}$,\r\n$\\triangle TC'D'\\sim \\triangle TDC \\Longrightarrow \\frac{C'D'}{a}=\\frac{TC'}{TD}$,\r\n\r\n$\\triangle TA'D' \\sim \\triangle TDA \\Longrightarrow \\frac{A'D'}{a}=\\frac{TA'}{TD}$,\r\n$\\triangle TB'C' \\sim \\triangle TCB \\Longrightarrow \\frac{B'C'}{a}=\\frac{TB'}{TC}$,\r\n\r\nwhere $AB=a$.\r\n\r\nIt follows that $A'B' \\cdot C'D'=a^{2}\\frac{TB' \\cdot TC'}{TA \\cdot TD}$, and $B'C' \\cdot A'D'=a^{2}\\frac{TA' \\cdot TB'}{TC \\cdot TD}$. Now it's enogh to show $\\frac{TC'}{TA}=\\frac{TA'}{TC}$, which is trivial w.r.t. power of the point theorem." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Consider all the ways of writing exactly ten times each of the numbers $0, 1, 2, \\ldots , 9$ in the squares of a $10 \\times 10$ board.\r\nFind the greatest integer $n$ with the property that there is always a row or a column with $n$ different numbers.", "Solution_1": "$4$ acording to Iran's problems, and beautiful Fiachra solution:\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16445[/url]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "M = sqrt (4k^2 + (a1)^2) + sqrt (4k^2 + (a2)^2) +.....+ sqrt (4k^2 + (a9)^2) + sqrt (4k^2 + (a10)^2)\r\n\r\nwhere a1 + a2 + ... a9 + 10 = 50... Find M^2.\r\n\r\nFind the minimum value of \r\n(loga1 (2004)+ loga2 (2004) + .... loga10 (2004))/(log a1a2a3a4...a10 (2004)) as a1,a2,a3,a4.. a10 range over the set of real number greater than 1...", "Solution_1": "I think I have solved the second one (it is late, so perhaps my thinking is bad):\r\n\r\nLet b[n] = log[2004](a[n]) (b[n] will always be positive, because a[n]>1)\r\n\r\n(log[a[1]](2004)+log[a[2]](2004)++log[a[10]](2004))/log[a[1]a[2]a[10]](2004) = \r\n(1/log[2004](a[1])+1/log[2004](a[2])+)/(1/(log[2004](a[1])+log[2004](a[2])+)) =\r\n(1/b[1]+1/b[2]+1/b[3]++1/b[10])/(1/(b[1]+b[2]+b[3]++b[10])) =\r\n(b[1]+b[2]++b[10])*(1/b[1]+1/b[2]++1/b[10]).\r\nApplying Cauchy, we get\r\n(b[1]+b[2]++b[10])*(1/b[1]+1/b[2]++1/b[10]) >= (b[1]/b[1]+b[2]/b[2]++b[10]/b[10])^2 = (10)^2 = 100. So, the minimum value is [b]100[/b]. \r\n\r\nSayoonara.", "Solution_2": "Anyone know how to solve the first question..." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "hey, \r\nHere are a few problems I'v tried to solve but with no success. \r\n\r\nCould anyone help? \r\n\r\nintegrate (x^5)cos(x^3) dx (First make a substitution then use integration by parts) \r\n\r\nintegrate within the limits of 9 and 5 the expession (sin(5x))^2 (cos(5x))^2 dx (Hint: you need to use some tig identities such as the double angle formulas for sine and cos.)\r\n\r\nIntegrate within the limits of pi/2 and 0 the expression (sinx)^5 (cosx)^20 dx\r\n\r\n\r\nIntegrate within the limits of pi/15 to 0 the expression ((sec(5x))^12)/(cot(5x)) dx \r\n\r\nThanks a lot", "Solution_1": "[quote=\"someoneELSE\"] \n\nintegrate (x^5)cos(x^3) dx [/quote]\r\n$\\int x^{5}\\cos{x^{3}}dx=\\frac{1}{3}\\int x^{3}\\cos{x^{3}}d(x^{3})$. Now, $\\int t\\cos{t}dt=\\int td\\sin{t}=t\\sin{t}-\\int\\sin{t}dt=...$" } { "Tag": [], "Problem": "Rusty ate an average of two hamburgers a day in January. What is the largest possible number of hamburgers that he could have eaten on January 15?", "Solution_1": "[quote=\"chantown\"]Rusty ate an average of two hamburgers a day in January. What is the largest possible number of hamburgers that he could have eaten on January 15?[/quote]\r\n[hide] There are 31 days in January. So, the total number of hamburgers he at must me (average)(number of days)=2(31)=62. So, he ate 62 hamburgers. Now, if he doesn't have to eat any hamburgers in a day, then he could have simply ate all 62 on January 15th. If he must eat at least once a day, then say that he ate 1 each day for the other 30 days. So, there are 62-30(1)=32 hamburgers he ate on January 15th.\n[/hide]" } { "Tag": [], "Problem": "CGMO\u597d\u50cf\u662f\u57288\u670810\u65e5\u81f315\u65e5\u8209\u884c... \u8acb\u554f\u6709\u6c92\u6709\u4eba\u77e5\u9053\u984c\u76ee\u5462\uff1f", "Solution_1": "http://gifted.hkedcity.net/Gifted/ActReview/2005CGMO/index.html\r\n\r\n\u4e0a\u9762\u6709\u8bd5\u9898\u3001\u89e3\u7b54\u53ca\u4f5c\u8005", "Solution_2": "\u8bf7\u95ee\u6709\u6ca1\u67092004\u5e74\u5973\u5b50\u6570\u5b66\u5965\u6797\u5339\u514b(\u5357\u660c)\u7684\u524d\u4e09\u540d\u7684\u83b7\u5956\u540d\u5355?", "Solution_3": "\u9806\u4fbf\u554f\u4e00\u4e0b\u6709\u6c92\u670905\u5e74CGMO\u7684\u7372\u734e\u540d\u55ae\uff1f", "Solution_4": "\u8fd9\u4e9b\u5370\u9898\u7684\u957f\u6625\u4eba\uff0c\u7adf\u7136\u5728\u5377\u7eb8\u4e0a\u5199\u534e\u7f57\u5e9a\u7684\u540d\u8a00\uff1f\u8111\u5b50\u957f\u4e86\u866b\u5b50\u3002\u591a\u5f71\u54cd\u9009\u624b\u5fc3\u60c5\u3002[\u81f3\u5c11\u662f\u6211\u4e0d\u723d\uff0c\u8fd9\u4e0d\u7740\u56db\u516d\u7684\u2026\u2026]\r\n\u4e0d\u8fc7\u611f\u89c9\u4eca\u5e74\u6bd4\u53bb\u5e74\u96be\uff1f", "Solution_5": "\u5475\u5475. \u534e\u8001\u7684\u8bdd\u5230\u6ca1\u90a3\u4e48\u5012\u80c3\u53e3.\r\n\r\nBTW: \u53bb\u5e74\u7684\u9898\u80fd\u4e0d\u80fd\u53d1\u4e0a\u6765\u770b\u770b?", "Solution_6": "\u53bb\u5e74\u7684\u9898\u89c1\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=18483", "Solution_7": "\u6211\u5011\u6fb3\u9580\u4e5f\u53d6\u5f97\u4e00\u9285\u734e :winner_third: , \u96d6\u4e0d\u7b97\u592a\u597d, \u4f46\u4ea6\u4e0d\u932f\u5427! :D", "Solution_8": "[quote=\"ifai\"]\u6211\u5011\u6fb3\u9580\u4e5f\u53d6\u5f97\u4e00\u9285\u734e :winner_third: , \u96d6\u4e0d\u7b97\u592a\u597d, \u4f46\u4ea6\u4e0d\u932f\u5427! :D[/quote]\r\n\u61c9\u8a72\u662f\u53bbimo \u90a3\u500b\u5973\u5b69\u5427", "Solution_9": "[quote=\"Soarer\"][quote=\"ifai\"]\u6211\u5011\u6fb3\u9580\u4e5f\u53d6\u5f97\u4e00\u9285\u734e :winner_third: , \u96d6\u4e0d\u7b97\u592a\u597d, \u4f46\u4ea6\u4e0d\u932f\u5427! :D[/quote]\n\u61c9\u8a72\u662f\u53bbimo \u90a3\u500b\u5973\u5b69\u5427[/quote]\r\n\u662f\u7684. \u5979\u5728imo\u4ea6\u5f97\u9285\u734e." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Who is Kevin Chen? Everybody seems to be talking about him. What is up???? :?:", "Solution_1": "Kevin Chen was the 2007 Mathcounts national champion." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "quadratics" ], "Problem": "How would I solve the following:\r\n\r\n$\\vec{x'}= \\begin{bmatrix}\\; \\; \\; 0 & \\; \\; \\; 1 & \\; \\; \\; 2 \\\\-5 &-3 &-7 \\\\ \\; \\; \\; 1 & \\; \\; \\; 0 & \\; \\; \\; 0 \\end{bmatrix}\\vec{x}$?\r\n\r\nI found 3 eigenvalues using Maple.", "Solution_1": "[url=http://www.efunda.com/math/ode/syslinearode1.cfm]Engineering fundamentals[/url]", "Solution_2": "I got it down to $\\lambda^{3}+3\\lambda^{2}+7\\lambda-1 = 0$\r\n\r\nHow would I factor this?", "Solution_3": "I don't see where you got the polynomial from. The matrix has eigenvalues $-1,-1,-1$. (Which means that my link above is not so helpful; it assumes distinct eigenvalues.) Anyway, the general solution is $\\vec x=C\\exp(tA)$, and the [url=http://en.wikipedia.org/wiki/Ordinary_differential_equation]Wikipedia page[/url] explains how to use this formula.", "Solution_4": "How did you get those eigenvalues? Did you use $|A-\\lambda I| = 0$ and find the determinant by expansion by minors? When I did expansion by minors, I got:\r\n\r\n$-\\lambda \\left|\\begin{matrix}-3-\\lambda &-7 \\\\ 0 &-\\lambda \\end{matrix}\\right|-\\left|\\begin{matrix}-5 &-7 \\\\ 1 &-\\lambda \\end{matrix}\\right|+2 \\left| \\begin{matrix}-5 &-3-\\lambda \\\\ 1 & 0 \\end{matrix}\\right|$", "Solution_5": "I ended up getting:\r\n\r\n$\\vec{x}(t) = e^{-t}\\begin{bmatrix}1 \\\\ 1 \\\\-1 \\end{bmatrix}+\\left(t\\begin{bmatrix}1 \\\\ 1 \\\\-1 \\end{bmatrix}+\\begin{bmatrix}-3 \\\\ 0 \\\\ 2 \\end{bmatrix}\\right)e^{-t}+\\left(\\frac{1}{2}t^{2}\\begin{bmatrix}1 \\\\ 1 \\\\-1 \\end{bmatrix}+t\\begin{bmatrix}-3 \\\\ 0 \\\\ 2 \\end{bmatrix}+\\begin{bmatrix}7 \\\\ 0 \\\\-5 \\end{bmatrix}\\right)e^{-t}$\r\n\r\nDoes this look right?", "Solution_6": "Looks about right: the solution should have components $p_{i}(t)e^{-t}$, where $p_{i}$, $i=1,2,3$, is a quadratic polynomial." } { "Tag": [], "Problem": "Three pencils and two erasers cost $ \\$0.60$. Two pencils and three erasers cost $ \\$0.55$. How much will it cost to buy seven pencils and seven erasers?", "Solution_1": "We can set up two equations:\r\n\r\n1) 3p + 2e = 0.60\r\n2) 2p + 3e = 0.55\r\n\r\nIf we add them together, we get 5p + 5e = 1.15. Dividing both sides by 5, we get p + e = 0.21. Since we want 7p + 7e, we have 0.21 times 7 = 1.47.", "Solution_2": "If we add the equations together we get $ 5p \\plus{} 5e \\equal{} \\$1.15$ Now we get $ p \\plus{} e \\equal{} \\$.23$. So $ 7p \\plus{} 7e \\equal{} \\$1.61$.\r\n\r\nEdit: beaten.. :( And I think your solution is incorrect izzy." } { "Tag": [ "group theory", "abstract algebra", "number theory", "relatively prime", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ diving $n$. Prove that $G$ is not simple.", "Solution_1": "It's been shown here before that if $p|n$ is the minimal prime factor of $n$, then a subgroup of index $p$ in $G$ must be normal. Hence the result.", "Solution_2": "A proof of the result grobber quoted:\r\n\r\nSuppose $H$ is a subgroup of $G$ of index $p$ and order $n$, with every prime factor of $n$ at least $p$. Let $C$ be the set of all conjugates of $H$ in $G$; $G$ acts transitively on $C$ by conjugation. $C$ has at most $p$ elements since $aHa^{-1}=(ah)H(ah)^{-1}$ whenever $h\\in H$.\r\nThis action of $G$ gives a homomorphism $\\phi$ from $G$ to $S_{|C|}$. If $|C|= 2^n - 1 and every ai > 0\r\nprove that 1/a1 + 1/a2 +... + 1/an < 2", "Solution_1": "are you sure it's true, J\r\ni can prove it with the condition\r\nx_i {i=1->n} > 0\r\nx_1+x_2+..+x_i \\geq 2^i - 1", "Solution_2": "yes i apologize, i actually forgot to add that every ai > 0", "Solution_3": "what about the condition x_1+x_2+...+x_i \\geq 2^i - 1" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x, y>0$ such that $x\\neq e$ and $x^{y}=y^{x}$. Prove that $x^{y} >e^{e}.$ ;)", "Solution_1": "For $x=y=\\frac{1}{2},$ results $x^{y}n. find all polinomyal P with that condition such that the list 1,P(1),PP(1),PPP(1)........contains a multiple of m for every m.", "Solution_1": "It is Iran TST 2004\r\nI had sent it 2 months ago", "Solution_2": "but do you have a solution? i think its to easy for being iran tst 2004, what do you think?", "Solution_3": "I think I solved it when Omid posted it. Let's denote those nasty compositions of $P$ by $a_n=P^{(n)}(1)$. Let $k=P(1)-1>0$. There is some $n$ s.t. $k|a_n$. However, from the fact that, in general, $m-n|P(m)-P(n)$, we get $k|a_2-a_1|a_3-a_2|\\ldots|a_n-a_{n-1}$, so $k|a_{n-1}$. We repeat the process until we get $k|a_1=P(1)\\Rightarrow P(1)-1|P(1)\\iff P(1)=2$. As you can see, the method works perfectly fine if we take $k=a_{t+1}-a_t$ for any $t\\ge 1$. In the end we get $P(n)=n+1$.", "Solution_4": "Grobber: Indeed your solution is nicer than mine. Thanks.", "Solution_5": "Of course this problem had the least mark form the other questions.", "Solution_6": "I know this problem for ages, and it was proposed for 9th grade here in Russia." } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Let the series (a_n)n from N be difined like so: a_(n+2)=(a_(n+1))^(1/a_n) for any n natural with a_0 , a_1 from (0, oo). Prove that the series (a_n) converges and compute lim n-->oo (a_n).\r\n\r\nbye", "Solution_1": "I don't think it's too hard: just consider the cases a_1<1, a_1=1, a_1>1. If a_1<1 then a_n<1 for any n>=1. Because of this the sequence will become strictly decreasing for n>=2, so the sequence converges to L. Because 0<=L<1 and L=L^(1/L) we find L=0. If a_1=1 then a_n=1 for any n>=1. The sequence converges to L=1. If a_1>1 a_n>1 for any n>=1. The sequence becomes strictly decreasing for n>=2 so the sequence converges to L>=1. Because L=L^(1/L) we get L=1. Think it's correct?" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "Harvard", "college", "MIT" ], "Problem": "(Cackle)", "Solution_1": "Is it really worth it? Sure, you'll qualify for USAMO. But that's about it. You won't have any other mathematicians your age within a few hundred miles. I think most of the time people do better when they have competition around them. I certainly benefited tremendously by living in one of the acknowledged mathematical centers of the country. The alternative would be to work a bit harder over the next year and qualify for USAMO the way most people do -- by getting a high USAMO index. Then you get a whole lot more -- you'll be better at math, you'll have mathematical friends, and you'll still make the USAMO. Isn't that better?", "Solution_2": "I live in the same region that you do, ComplexZeta, and I've had enough...", "Solution_3": "[quote=\"tAsDoOm\"]I live in the same region that you do, ComplexZeta, and I've had enough...[/quote]\r\n\r\nCome on over to Minnesota. Heck, I could probably drum up a job for you coaching the young mathletes of my acquaintance (unless the local guru, Sly Si, locks it up first).", "Solution_4": "After I returned from my first year of MOP (this was the year with 10 qualifiers from SFBA), Upstate NY seemed like the middle of nowhere math-people-wise. I didn't have any local math clubs or teams for high school students, and the nearest USAMO qualifier I knew of lived 3 hours away. Things have improved since, but at the time I would have killed to live in the Bay Area.", "Solution_5": "Exactly. After all, when you have the AoPS books and online community, what else do you need? :D The Calif. Bay Area is, in my personal opinion, not at all what it's often hyped up to be. The area is plagued with rich, intelligent snobs, a lot of competition, and a load of other problems that I don't even want to get into...", "Solution_6": "[quote=\"ComplexZeta\"]Is it really worth it? Sure, you'll qualify for USAMO. But that's about it. You won't have any other mathematicians your age within a few hundred miles. I think most of the time people do better when they have competition around them. I certainly benefited tremendously by living in one of the acknowledged mathematical centers of the country. The alternative would be to work a bit harder over the next year and qualify for USAMO the way most people do -- by getting a high USAMO index. Then you get a whole lot more -- you'll be better at math, you'll have mathematical friends, and you'll still make the USAMO. Isn't that better?[/quote]\r\n\r\nI'll agree. Nevada is no fun if you're interested in maths - no one else likes math, you get made fun of (sometimes), you lose motivation, etc.", "Solution_7": "I'll admit, I was quite silly at the time. (And I had ulterior motives.) I was highly deprived as a freshman -- there was no AoPS community then :-P But having friendly smart math people to work with in person has made a big difference for me, when I've had the chance to do so.\r\n\r\n[quote=\"tAsDoOm\"]Exactly. After all, when you have the AoPS books and online community, what else do you need[/quote]", "Solution_8": "Just curious, if you are an US citizen living overseas who happen to qualify for USAMO, which region do you compare yourself with?", "Solution_9": "Ditto. What do you mean by region Oasis? Well if you wanted to know, it's APO AP or FPO AP. Something of that sort. I'm not sure though.", "Solution_10": "What would you do if some other good mathematicians moved to Wyoming? Now you're stuck in a hellhole.", "Solution_11": "Ha! A hellhole full of cheese! At least the Great Lakes are nearby so you could douse yourself if you were on fire.", "Solution_12": "I think you're confusing Wyoming with Wisconsin ... Wisconsin isn't a bad place. Wyoming has more cows than people.", "Solution_13": "hey! don't make fun of the USACO! the cows will have their revenge!", "Solution_14": "Which state is generally better at math?\r\nCalifornia or Massachussetts\r\nThey both seem pretty good although I feel California is better.", "Solution_15": "California does have the advantage of having more than 5 times the population of Massachusetts.", "Solution_16": "haha i'm from MA so i'll just chime in and say that massachusetts clearly is better at math.\r\n\r\nbtw the 2nd level massachusetts olympiad was today.. here was one of the problems....\r\n\r\nProve that among any 14 consecutive positive integers there is at least 1 not divisible by any of {2,3,5,7,11}.\r\n\r\nI dont have any formal education in number theory. I was able to write a proof without any BS, but it was over a page long, does anyone know a quick way to do it?", "Solution_17": "Can you just do it by inclusion-exclusion? There are 7 divisible by 2, 4 by 3, and so on, then do number divisible by 2 of them, then by 3 of them, and so on. You'd have to be very careful about what the best-case scenarios are, but they're definitely provably best-case. Was this your method?", "Solution_18": "[quote=\"beta\"]Which state is generally better at math?\nCalifornia or Massachussetts\nThey both seem pretty good although I feel California is better.[/quote]\r\n\r\nCalifornia is by far the best math state.", "Solution_19": "namstap wrote:Prove that among any 14 consecutive positive integers there is at least 1 not divisible by any of {2,3,5,7,11}. \n\n\n\nHint: [hide]Prove that among 7 consecutive odd integers, there are at least three not divisible by 3 or 5.[/hide]", "Solution_20": "[quote]Which state is generally better at math? \nCalifornia or Massachussetts \nThey both seem pretty good although I feel California is better.[/quote]\r\n\r\nAre we talking about high school math competitions or math in general.\r\n\r\nFor math in general, I would definitely say MA is preferable, with all of the math activities present at HArvard and MIT in Cambridge.\r\n\r\nFor high school math competitions, it is hard to say. Going by ARML stats, SF Bay beat MA by a few places in 2003, MA beat SF Bay by a few places in 2002, and the two were right next to each other in the standings in 2001.", "Solution_21": "[quote=\"tAsDoOm\"]Exactly. After all, when you have the AoPS books and online community, what else do you need? :D The Calif. Bay Area is, in my personal opinion, not at all what it's often hyped up to be. The area is plagued with rich, intelligent snobs, a lot of competition, and a load of other problems that I don't even want to get into...[/quote]\r\n\r\nis not. i live here too. what are you talking about? i realize some people are incredibly arrogant, but yanno, then again, people elsewhere are too. what are you talking about? who is this?", "Solution_22": "[quote=\"beta\"]Which state is generally better at math?\nCalifornia or Massachussetts\nThey both seem pretty good although I feel California is better.[/quote]\r\n\r\ndefinitely Cali. especially SFBA. we hardly train for arml, and we almost beat TJ last year. what can I say? and the weather is nicer. San Diego's pretty good, too... especially now that Richard is training the soucal people...", "Solution_23": "yea, i would have to say cali too... of course, i don't have much MA experience, but the Bay Area is pretty nice and relaxed.\r\n\r\nOf course, losing 6 or 7 seniors is going to hurt our ARML team, so we may or may not place higher than MA this year...", "Solution_24": "Wow! I was completely tricked... When someone said a hellhole full of cheese, I automatically thought of Wisconsin instead of Wyoming! lol", "Solution_25": "Hey hey hey, y'all better watch out, SoCal shall bloom into one of the foremost mathematical centres in the coming years. Just give us a few years.", "Solution_26": "It's scaring me. SFBA keeps getting weaker, and other regions keep getting stronger. That means that if you are from SFBA and you are reading this, do more math, and convince everyone else within a 50-mile radius to do more also, especially if I usually tell you to do more math anyway.", "Solution_27": "Ah...here in the great state of Texas there's loads of good maths people (sorta), lots of space (if u live outside of houston and dallas/ft worth), and great food (if u live inside of houston and dallas/ft worth and austin and san antonio).\r\n\r\nHowever, I think it would benefit us greatly (as a mathematical society) if Mr. Ruzczyk and Mr. Crawford moved down here :-)", "Solution_28": "[quote=\"ComplexZeta\"]It's scaring me. SFBA keeps getting weaker, and other regions keep getting stronger. That means that if you are from SFBA and you are reading this, do more math, and convince everyone else within a 50-mile radius to do more also, especially if I usually tell you to do more math anyway.[/quote]\r\n\r\nscares me too. i'm going to hold faux arml practices at my school. what years should i use?", "Solution_29": "[quote]For high school math competitions, it is hard to say. Going by ARML stats, SF Bay beat MA by a few places in 2003, MA beat SF Bay by a few places in 2002, and the two were right next to each other in the standings in 2001.[/quote]\r\n\r\nUm, I'd like to note that the SFBay area is a pretty small portion of California as a whole. I would guess the population of that area is smaller than MA, as well. So, I'd argue that if MA = Bay Area, then MA << CA in terms of math.\r\n\r\nAs for MA having MIT and Cambridge, do keep in mind that CA has Caltech, Stanford, and Berkeley. The Bay Area also has many of the top Olympiad educators (Zeitz, Davis, Vakil, Poonen, women at San Jose State and Berkeley, etc).\r\n\r\nRegardless, I don't think you can go wrong as a student in either the Bay Area or Boston.", "Solution_30": "What? More California / Massachusetts rivalry? I thought the question was settled nearly three years ago with the creation of the grand state of Calimass.\r\n\r\nMOPperwise, Massachusetts has really gone down over the past few years. (Then again, CA isn't what it was in 2000 either.)" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Can you prove\r\n\r\n$ \\sum_{n\\equal{}0}^\\infty \\left(\\sqrt{n\\plus{}1}\\minus{}\\sqrt{n}\\right)^3\\equal{}\\frac{3}{2\\pi} \\, \\zeta\\left(\\frac32\\right)$\r\n\r\n$ \\sum_{n\\equal{}0}^\\infty \\left(\\sqrt{n\\plus{}1}\\minus{}\\sqrt{n}\\right)^5\\equal{}\\frac{15}{2\\pi^2} \\, \\zeta\\left(\\frac52\\right)$", "Solution_1": "Now I know how to show it.\r\n\r\n$ (\\sqrt{k}\\minus{}\\sqrt{k\\minus{}1})^3\\equal{}k^\\frac32\\minus{}3k(k\\minus{}1)^\\frac12\\plus{}3 k^\\frac12 (k\\minus{}1)\\minus{}(k\\minus{}1)^\\frac32$\r\n\r\n$ \\Rightarrow \\sum_{k\\equal{}1}^n \\left(\\sqrt{k}\\minus{}\\sqrt{k\\minus{}1}\\right)^3\\equal{}\\sum_{k\\equal{}1}^n k^\\frac32\\minus{}\\sum_{k\\equal{}1}^n (k\\minus{}1)^\\frac32\\plus{}3\\sum_{k\\equal{}1}^n k^\\frac12 (k\\minus{}1)\\minus{}3\\sum_{k\\equal{}2}^n k(k\\minus{}1)^\\frac12$ \r\n\r\n$ \\equal{}n^\\frac32\\plus{}3(n\\plus{}1)n^\\frac12\\minus{}6\\sum_{k\\equal{}1}^n k^\\frac12\\equal{}\\minus{}6\\left(\\sum_{k\\equal{}1}^n k^\\frac12\\minus{}\\frac23 n^\\frac32\\minus{}\\frac12 n^\\frac12\\right)\\to \\minus{}6\\,\\zeta\\left(\\minus{}\\frac12 \\right)$\r\n\r\nand this is $ \\frac{3}{2\\pi}\\zeta\\left(\\frac32\\right)$ after Riemann's functional equation.\r\n\r\nThe next series goes similar.\r\n\r\n\r\nMore general one can show\r\n\r\n$ \\sum_{k\\equal{}0}^\\infty \\left(\\sqrt{k\\plus{}1}\\minus{}\\sqrt{k}\\right)^{2m\\plus{}1}$\r\n\r\n$ \\equal{}\\sum_{k\\equal{}0}^m \\sum_{j\\equal{}k}^m \\Big((\\minus{}1)^j\\minus{}1\\Big)\\, {j\\choose k} {2m\\plus{}1\\choose 2j} \\zeta\\left(\\minus{}\\left(m\\minus{}k\\plus{}\\frac12\\right)\\right)$." } { "Tag": [ "inequalities" ], "Problem": "Suppose $ a,b,c,d$ are positive real numbers.\r\n\r\nProve that\r\n\r\n$ \\frac{ab}{a\\plus{}b}\\plus{}\\frac{cd}{c\\plus{}d}\\le\\frac{(a\\plus{}c)(b\\plus{}d)}{a\\plus{}b\\plus{}c\\plus{}d}$\r\n\r\n\r\n :P", "Solution_1": "hello, we have $ \\frac{(a\\plus{}c)(b\\plus{}d)}{a\\plus{}b\\plus{}c\\plus{}d}\\minus{}\\frac{ab}{a\\plus{}b}\\minus{}\\frac{cd}{c\\plus{}d}\\equal{}\\frac{(ad\\minus{}bc)^2}{(a\\plus{}b)(c\\plus{}d)(a\\plus{}b\\plus{}c\\plus{}d)}\\geq0$.\r\nSonnhard." } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra solved" ], "Problem": "How to prove directly from the definitions of maxiaml and prime ideals that every maximal ideal of a commutative ring R with unity is a prime ideal?", "Solution_1": "suppose M is maximal but not prime\r\nthen there a and b not in M such that $a b \\in M$\r\n\r\na is not in M and M is maximal : we can write :$a r_{1}+m_{1}=1$ and $b r_{2}+m_{2}=1$ for $r_1,r_2 \\in R$ and $m_1,m_2 \\in M$\r\n\r\nmultiplying both equations gives you that $ab r_{1} r_{2}-1$ is in M\r\nbut we assumed $a b \\in M$ , so $1\\in M$\r\n\r\ncontradiction : a maximal ideal cannot contain unity", "Solution_2": "[quote=\"fredbel6\"]suppose M is maximal but not prime\nthen there a and b not in M such that $a b \\in M$\n\na is not in M and M is maximal : we can write :$a r_{1}+m_{1}=1$ and $b r_{2}+m_{2}=1$ for $r_1,r_2 \\in R$ and $m_1,m_2 \\in M$\n\nmultiplying both equations gives you that $ab r_{1} r_{2}-1$ is in M\nbut we assumed $a b \\in M$ , so $1\\in M$\n\ncontradiction : a maximal ideal cannot contain unity[/quote]\r\n\r\nThanks a lot! :)", "Solution_3": "Here is another proof.\r\nLet $A$ be a commutative rings and $M$ a maximal ideal of $A$.\r\nLet $x, y \\in A$ be elements of $A$ such that $xy \\in M$.\r\nWe will prove that $x \\not \\in M \\Longrightarrow y \\in M$ (which is equivalent to saying that $M$ is a prime ideal).\r\n\r\nSuppose that $x \\not \\in M.$\r\n[b]Note that $M$ is proprely contained in $M + Ax$[/b]\r\nSince $M$ is maximal, any proper ``overset`` of $M$ is equal to $A$.\r\nTherefore $M + Ax = A$.\r\nHence there is some $u \\in M$ and $a \\in A$ such that\r\n\\[ 1 = u + ax \\]\r\nMultiply both sides by $y$ to obtain\r\n\\[ y = yu + yax \\]\r\n\\[ y = yu + axy \\]\r\n(this is were commutativity comes into play).\r\nBut $yu \\in M$ since $u \\in M$ and $axy \\in M$ since $xy \\in M$ (just recall the definition of an ideal).\r\nMoreoever $y = yu + axy$ and both summands $yu$, and $axy$belongs to $M$. Hence $y \\in M$." } { "Tag": [ "symmetry", "analytic geometry", "function", "calculus" ], "Problem": "Can someone please explain how to derive \"-b/a\" as the x value for the (x,y) that is the origin of symmetry in a cubic equation?", "Solution_1": "What is $a$ and what is $b$?\r\nSo you want to find k where $-f(x+k) = f(k-x)$, or am I confused?", "Solution_2": "Sorry, I meant \t$-b/3a$ as the $x$ value of the $(x,y)$ that is the coordinate of the origin of symmetry for a cubic function of the form $ax^3 + bx^2 + cx + d$", "Solution_3": "[hide]If $f(x)=ax^3+kx$, then $f(-x)=-f(x)$, so symmetric about origin. Then $a(x-p)^3+k(x-p)+q$ is symmetric about $(p,q)$. Set equal to $ax^3+bx^2+cx+d$ and set $x^2$ coefficients equal, get $-3ap=b$, so $p=-\\frac{b}{3a}$.[/hide]", "Solution_4": "Why does $a(x-p)+kx+q$ have its point of symmetry at $(p, q)$?\r\n\r\nIs point of symmetry when ever $f(x) = f(-x)$ in a function where $-f(x) = f(-x)$?\r\n\r\nIf so, then I understand.", "Solution_5": "A point of symmetry is a point $(p,q)$ such that $f(p-x)-q=q-f(p+x)$ for all $x$. When $p=q=0$ this just becomes $f(-x)=-f(x)$, and if $f(x)$ is symmetric about the origin, $g(x)=f(x-p)+q$ will be symmetric about $(p,q)$ since it's translating the graph.\r\n\r\no wait that should read $a(x-p)^3+k(x-p)+q$. But this only changes the constant, which is not used anyway.", "Solution_6": "you could also prove this with some basic calculus. but i like the proof that was provided more than the calculus one.", "Solution_7": "Nice explanation scorpius, I understand this now. \r\n\r\nHowever, can you please elaborate on why $f(p-x)-q=q-f(x-p)$ is true about any point of symmetry $(p,q)$?", "Solution_8": "[quote=\"breez\"]Nice explanation scorpius, I understand this now. \n\nHowever, can you please elaborate on why $f(p-x)-q=q-f(x-p)$ is true about any point of symmetry $(p,q)$?[/quote]\r\n\r\nOuch, I typed the wrong thing again! It now reads $f(p+x)$ instead of $f(x-p)$. Now this is accurate. If you start at $(p,q)$ and move over $x$ to the left, you will move $f(p-x)-q$ up, while if you move $x$ to the right, you will move $f(p+x)-q$ up, or $q-f(p+x)$ down. The distance moved up when traveling $x$ to the left will be the same as the distance moved down when traveling $x$ to the right (by symmetry of $(p,q)$).", "Solution_9": "Yeah, I was wondering about that, cause like $f(x)$ and $f(-x)$ wouldn't be related as an ordered pair unless an odd function.\r\n\r\nThanks for the explanations anyhow, they were very helpful." } { "Tag": [], "Problem": "Find all positive integers $ a$ such that\r\n\r\n$ \\frac{a^2\\plus{}2a\\minus{}2}{a^3\\minus{}5a\\plus{}3}$\r\n\r\nis a fraction in reduced form. Prove your answer.", "Solution_1": "[hide=\"Hint\"] Euclidean Algorithm. :) [/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c,d > 0$ and $ a\\plus{}b\\plus{}c\\plus{}d \\equal{} 1$. Prove that $ abc\\plus{}bcd\\plus{}cda\\plus{}dab\\leq \\frac{1}{27}\\plus{}\\frac{176}{27}abcd$", "Solution_1": "ISL 1993 :wink:", "Solution_2": "[quote=\"Erken\"]ISL 1993 :wink:[/quote]\r\nYes , you are true http://www.mathlinks.ro/Forum/viewtopic.php?p=463192#463192 :wink:" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let P be a p-Sylow subgroup of $G$. Show that $N_{G}[N_{G}(P)]=N_{G}(P)$\r\n\r\nI also need help with this one\r\nSuppose H and K are subgroups of G and that $K\\triangleleft H$. Show that $H\\leq N_{G}(K)$\r\n\r\n$N_{G}(K)$ is the normalizer.", "Solution_1": "$P$ is a normal $p$-sylow subgroup of $N(P)$. Hence, it's the unique one. Now suppose there exists $x \\in N(N(P))$ s.t. $x \\notin N(P)$. Then $x P x^{-1}\\neq P$, and $x P x^{-1}\\subseteq x N(P) x^{-1}= N(P)$. So, $x P x^{-1}$ is another $p$-sylow subgroup of $N(P)$, a contradiction.\r\n \r\nThe second exercise is a trivial calculation ... this should not make any problems.", "Solution_2": "the second one is straight from the definition of the normalizer\r\nthe first :\r\nsuppose that $x \\notin N_{G}(P)$ but $x \\in N_{G}[N_{G}(P)]$ thus\r\n$xN_{G}(P)x^{-1}=N_{G}(P)$ so $\\phi_{x}(g)=xgx^{-1}$ is an automorphism\r\non $N_{G}(P)$ so $\\phi_{x}(P)=P$ since $P$ is the only Sylow $p$-subgroup in $N_{G}(P)$ (since it's normal in $N_{G}(H)$), thus $xPx^{-1}=P$ - contradiction with our assumption.", "Solution_3": "Thanks, got the first one!\r\n\r\nHmmm, regarding the second one, I had $<$ floating around in my mind when I was thinking of how to solve it and not $\\leq$. Well, it makes sense now!" } { "Tag": [ "induction" ], "Problem": "Unweighted Grade point average is calculated as follows $\\frac{4x+3y+2z+1t+0f}{x+y+z+t+f}$\r\n\r\nWhere...\r\n$x=$number of A's\r\n$y=$number of B's\r\n$z=$number of C's\r\netc.\r\n\r\nWhat are the minimum number of total grades required so that every tenths of a GPA can be represented [no rounding] (any combination of grades totaling the amount is allowed; from the range of $0 \\le g \\le 4$ where $g=$grade point average)?", "Solution_1": "One needs ten classes because if the denominator is 10, one can guarantee every tenths place will be picked.\r\nAny smaller number would skip at least one number in the denominator simply because there's not enough decimal representations of fractions to cover every tenth.", "Solution_2": "clearly, the answer is $10$.", "Solution_3": "It is not finished unless you show that each possible point can in fact be achieved using only ten classes.", "Solution_4": "[quote=\"JBL\"]It is not finished unless you show that each possible point can in fact be achieved using only ten classes.[/quote]\r\nWell, $4x+3y+2z+1t+0f$ must be all the numbers between 0 and 40. Starting at 40, $x=10$ and $y=z=t=f=0$. Then going to 39, we can move one of the grades down to a B with $x=9$, $y=1$, $z=t=f=0$. Therefore, we can keep moving grades down one letter grade to get all the numbers between 0 and 40. \r\n\r\nAs you can see, I'm horrible at writing proofs. I think this is call induction or something." } { "Tag": [ "geometry", "rectangle" ], "Problem": "Consider the six by six grid of unit squares below. How many\nrectangles of area 3 square units can be formed using only the\nline segments of the grid as the sides of the rectangles?\n\n[asy]unitsize(0.1inch);\ndraw((0,0)--(6,0)--(6,6)--(0,6)--cycle);\ndraw((1,0)--(1,6));\ndraw((2,0)--(2,6));\ndraw((3,0)--(3,6));\ndraw((4,0)--(4,6));\ndraw((5,0)--(5,6));\ndraw((0,1)--(6,1));\ndraw((0,2)--(6,2));\ndraw((0,3)--(6,3));\ndraw((0,4)--(6,4));\ndraw((0,5)--(6,5));[/asy]", "Solution_1": "$ (2)(5\\minus{}3\\plus{}1)(5\\minus{}1\\plus{}1)\\equal{}(2)(3)(5)\\equal{}30$.", "Solution_2": "Where in the world does this come from?", "Solution_3": "[quote=\"Maxima\"]Where in the world does this come from?[/quote]\ni know, right? How did you get this", "Solution_4": "they are alcumas problems", "Solution_5": "[hide=Solution]4 rectangles on each row\n4*6=24 on the horizontal rows\nBut we also need to count the Vertical rows\nSo: $(4\\times 6)\\times 2 = 48$[/hide]", "Solution_6": "[quote=venusalien][quote=\"Maxima\"]Where in the world does this come from?[/quote]\ni know, right? How did you get this[/quote]\n\nSource: MathCounts 2003 State Team\n", "Solution_7": "[quote=arnolde1234][quote=venusalien][quote=\"Maxima\"]Where in the world does this come from?[/quote]\ni know, right? How did you get this[/quote]\n[/quote]\nSource: MathCounts 2003 State Team\n\nEDIT: Sorry for revive. just thought u still wanted to know :blush: \n\n", "Solution_8": "[quote=Ridvik]they are alcumas problems[/quote]\n\nYeah, it probably is, but it's not the problem stated above." } { "Tag": [], "Problem": "For which primes x,y,z is (x + y + z)xyz a square ?", "Solution_1": "x=y=3 , z=2 ..", "Solution_2": "What a lousy answer, no proof. :D :D :D", "Solution_3": "if x|(x+y+z),y|(x+y+z),z|(x+y+z),so x|(y+z),y|(x+z),z|(x+y)\r\nassume that x+z=sy x+y=tz\r\nso y=(1+t)x/(ts - 1 ), z = (s+1)x/(ts-1)\r\n(y+z)/x = (s+t+2)/(ts-1)\r\ns+t+2 >= st - 1\r\nso we can know that (t-1)(s-1) <= 4\r\nif t=1 because (s+t+2)/(ts-1)=(s+3)/(s-1) is a integer,(s-1)|4\r\n if s = 2,y=2x,not a prime\r\n if s = 3,y=x,z=2x,not a prime\r\n if s = 5,y=x/2,y=3x/2,so x = 2,y=1,not a pime\r\nelse if t=2,because (s+t+2)/(ts-1)=(s+4)/(2s-1) is a integer,s+4>=2s-1\r\n if s = 2,y=x,z=x,so (x+y+z)xyz=3*x^4,not a square\r\n if s = 3,4,(s+4)/(2s-1) not a integer\r\n if s = 5,y=x/3,z=2x/3,so x = 3,y=1,not a prime\r\nelse if t = 3,because (s+t+2)/(ts-1)=(s+5)/(3s-1) is a integer,s+5>=3s-1\r\n or s <= 3,that s = 3,y=x/2,not a prime\r\nelse if t = 4,5,the same we can know there don't exit the (x,y,z)\r\n\r\nso there is x,y,x=y\r\n(x+y+z)xyz=(2x+z)z*x^2\r\nthen (2x+z)*z must be a square\r\nwe know that z|(2x+z),\r\nso z = 2,and (x+y+z)xyz = 4*(x+1)*x^2\r\nthen (x+1) must be a square,assume x+ 1 = n^2\r\nx = (n-1)*(n+1)\r\nso n - 1 = 1,x=3\r\n\r\nAbove prove that (3,3,2),(3,2,3)and so on is the all(x,y,z)", "Solution_4": "Wow that's a long proof :)\r\n\r\nI got this question from arne long time ago, I think i solved in 4 lines? I'll see if i can find it back, and post it.", "Solution_5": "If we have x=y then z(z+2x) is a square and z|z+2x, so z=2.\r\nThen 4x+4 is a square and x=3.\r\nIf x,y,z are distinct, WLOG let xx 2 z>3z>x+y+z.\r\n\r\nQ. E. D. (Exactly 4 lines :D )" } { "Tag": [], "Problem": "On three tests, Molly scored 84, 83, and 91. On the same three tests, Milly scored 86, 87 and 88. What is the positive difference between their average scores?", "Solution_1": "Molly's average is $ \\frac{84\\plus{}83\\plus{}91}{3}\\equal{}\\frac{258}{3}\\equal{}86$. Milly's average, by inspection, is $ 87$. The difference is $ 87\\minus{}86\\equal{}\\boxed{1}$.", "Solution_2": "Faster way.\nCompute the difference of each element then divide by 3\n$\\frac{(84-86)+(83-87)+(91-88)}{3}=-1$\nPositive is 1", "Solution_3": "how is that faster the first one is faster ", "Solution_4": "Note that now you only have to compute small numbers (-2, -4, 3) rather than adding three large ones together." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find sum $ S \\equal{} \\sqrt {1 \\plus{} \\frac{1}{{2^2 }} \\plus{} \\frac{1}{{3^2 }}} \\plus{} \\sqrt {1 \\plus{} \\frac{1}{{3^2 }} \\plus{} \\frac{1}{{4^2 }}} \\plus{} \\sqrt {1 \\plus{} \\frac{1}{{4^2 }} \\plus{} \\frac{1}{{5^2 }}} \\plus{} ... \\plus{} \\sqrt {1 \\plus{} \\frac{1}{{2009^2 }} \\plus{} \\frac{1}{{2010^2 }}}$", "Solution_1": "[quote=\"thanhnam2902\"]Find sum $ S = \\sqrt {1 + \\frac {1}{{2^2 }} + \\frac {1}{{3^2 }}} + \\sqrt {1 + \\frac {1}{{3^2 }} + \\frac {1}{{4^2 }}} + \\sqrt {1 + \\frac {1}{{4^2 }} + \\frac {1}{{5^2 }}} + ... + \\sqrt {1 + \\frac {1}{{2009^2 }} + \\frac {1}{{2010^2 }}}$[/quote]\r\n\r\n$ \\sqrt {1 + \\frac {1}{{n^2 }} + \\frac {1}{{(n+1)^2 }}}$\r\n$ =\\sqrt{\\frac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}}$\r\n$ =\\sqrt{\\frac{[n(n+1)+1]^2}{[n(n+1)]^2}}$\r\n$ =1+\\frac{1}{n}-\\frac{1}{n+1}$\r\n\r\nSo the sum $ =2008+(\\frac{1}{2}-\\frac{1}{3})+(\\frac{1}{3}-\\frac{1}{4})+...+(\\frac{1}{2009}-\\frac{1}{2010})$\r\n$ =2008 \\frac{502}{1005}$", "Solution_2": "We can use\r\n$ \\sqrt{1\\plus{}\\frac{1}{n^2}\\plus{}\\frac{1}{(n\\plus{}1)^2}}\\equal{}1\\plus{}\\frac{1}{n}\\minus{}\\frac{1}{n\\plus{}1}$" } { "Tag": [ "inequalities" ], "Problem": "1. Prove that $ \\frac{1*3*5*7*...*9997*9999}{2*4*6*8*...*9998*10000}<1/100.$\r\n\r\n2. Prove that: $ 1/1^2\\plus{}1/2^2\\plus{}1/3^2\\plus{}1/4^2\\plus{}....\\plus{}1/9999^2\\plus{}1/10000^2<2\\minus{}1/10001.$", "Solution_1": "[quote=\"mathwizarddude\"]1. Prove that $ \\frac {1*3*5*7*...*9997*9999}{2*4*6*8*...*9998*10000} < 1/100.$\n\n2. Prove that: $ 1/1^2 \\plus{} 1/2^2 \\plus{} 1/3^2 \\plus{} 1/4^2 \\plus{} .... \\plus{} 1/9999^2 \\plus{} 1/10000^2 < 2 \\minus{} 1/10001.$[/quote]\r\n\r\nmathwizarddude, \r\n\r\nI amended your subject description; I left\r\nit blank. You can just let other readers decide\r\nhow THEY feel about the problems' difficulty \r\nlevel, and you don't need to set out a lure \r\nfor the problems. Let the problems speak \r\nfor themselves.", "Solution_2": "1)$ (\\frac{1*3*5*7*...*9997*9999}{2*4*6*8*...*9998*10000})^2<\\frac{1*2*3*...*9999*10000}{2*3*...*10000*10001}$\r\n$ \\equal{}\\frac{1}{10001}<\\frac{1}{100^2}$, which gives the desired result.\r\n\r\n2)$ \\frac{1}{2^2}\\plus{}\\frac{1}{3^2}\\plus{}\\frac{1}{4^2}\\plus{}....\\plus{}\\frac{1}{9999^2}\\plus{}\\frac{1}{10000^2}$\r\n$ <\\frac{1}{1*3}\\plus{}\\frac{1}{2*4}\\plus{}...\\plus{}\\frac{1}{9998*10000}\\plus{}\\frac{1}{9999*10001}$\r\n$ \\equal{}\\frac{1}{2}*[(\\frac{1}{1}\\minus{}\\frac{1}{3})\\plus{}(\\frac{1}{2}\\minus{}\\frac{1}{4})\\plus{}...\\plus{}(\\frac{1}{9998}\\minus{}\\frac{1}{10000})\\plus{}(\\frac{1}{9999}\\minus{}\\frac{1}{10001})]$\r\n$ \\equal{}\\frac{1}{2}*(\\frac{1}{1}\\plus{}\\frac{1}{2}\\minus{}\\frac{1}{10000}\\minus{}\\frac{1}{10001})$\r\n$ <\\frac{3}{4}\\minus{}\\frac{1}{10000}$\r\n$ <1\\minus{}\\frac{1}{10001}$, which gives the desired result." } { "Tag": [ "trigonometry", "LaTeX", "analytic geometry", "algebra", "polynomial", "quadratics", "quadratic formula" ], "Problem": "This problem is in volume 2, page29, #27.\r\n\r\n If 0 0 with complex coefficients. Show that there exists a complex number [tex]z_0[/tex] such that [tex]f(z_0)=0[/tex].", "Solution_1": "Yep, I'd sure like to see the steps of an elementary proof of the Fundamental Theorem of Algebra (posted in spoiler, of course).", "Solution_2": "I think the same proof appears in Michael Spivak's CALCULUS text book\r\n(chapter 25).", "Solution_3": "[quote=\"josejacobi\"]I think the same proof appears in Michael Spivak's CALCULUS text book\n(chapter 25).[/quote]\r\n\r\nI haven't seen Spivak's proof. Could you post a sketch in spoiler?", "Solution_4": "Bump.", "Solution_5": "Hint:\n\nStep 1. [hide]based on the fact in analysis I stated above, show that |f(z)| attains minimum value on the entire complex plane C.[/hide]\n\n\n\nStep 2. [hide]Assume that the Fundamental Theorem of Algebra is false, let m = |f(z_0)| = min |f(z)| on C, then m>0. Define g(z) based on f(z) such that g(0)=0 and |g(z)-1| :ge: 1 for any z in C.[/hide]", "Solution_6": "Here's a proof using the function g. \r\nI still don't know TeX. I hope no one has trouble\r\nopening the attached file. \r\n\r\nIf there's interest in the proof I might\r\nwrite a new version using TeX. \r\n(I guess that would be good exercise in TeX)" } { "Tag": [ "MATHCOUNTS", "limit", "geometry", "rectangle", "perimeter", "special factorizations" ], "Problem": "How many pairs $(m,n)$ of integers satisfy the equation $m+n=m \\cdot n$ ?", "Solution_1": "This can't be a Mathcounts problem, but this is how I solved it.\r\n\r\nLet $m$ be $x$ and $n$ by $y$.\r\n$x+y=xy$\r\n$x=xy-y$\r\n$x=y(x-1)$\r\n$y=\\frac{x}{x-1}$\r\nNow you can see that $\\lim\\limits_{x\\rightarrow\\infty}\\left(\\frac{x}{x-1}\\right)=1$ and the same with $x\\rightarrow -\\infty$, which means that $y$ cannot be less than 1 while $x$ increases and $y$ cannot be greater than 1 while $x$ decreases. So using this knowlage and a graphing calculator, the pairs are (0,0), (2,2), and eventually, $(\\infty,1)$ and $(-\\infty,1)$, so that's 4 pairs.", "Solution_2": "I think it can be a Mathcounts problem as 0,0 and 2,2 are common sense solutions, though I'm not sure if it can be proved with Mathcounts level material. To ln(dx/dy): you can't really treat infinity as a number, it's not an actual number, much less an integer, just a concept.", "Solution_3": "There is the \"complete the rectangle\" factoring trick.\r\n\r\n$mn=m+n$\r\n\r\n$mn-m-n=0$\r\n\r\n$mn-m-n+1=1$\r\n\r\n$(m-1)(n-1)=1$.\r\n\r\nNow you're looking for pairs of integers whose product is 1. I'll let others finish it from here.\r\n\r\nHere's a related question: Find all rectangles with integer sides such that the perimeter has the same numerical value as the area.\r\n\r\nThe same factoring idea works for this one.", "Solution_4": "Certainly looks like MATHCOUNTS level. By Kent Merryfield's advise, you need only know what makes up an integer and the possible ways to obtain that product, i.e. that the product of two integers is equal to 1.", "Solution_5": "Sorry to say this but infinity is not a number, and this problem can be solved [b]without[/b] using limits.\r\n\r\n[size=75][Edited by rcv to remove sarcasm and excessive quoting][/size]", "Solution_6": "[quote=\"3X.lich\"]\nSorry to say this but infinity is not a number, and this problem can be solved [b]without[/b] using limits.[/quote]\r\nEven if we're working in the extended real numbers, it's certainly a stretch to say that its an integer.", "Solution_7": "A standard trick (or tool in your tool box as my son used to call these tricks).\r\n\r\n[hide]we have $mn -m -n = 0 $\nrewrite it as:\n\n$ (m-1) (n-1) - 1 = 0 $\nor $(m-1)(n-1) = 1$ \n\nThis reduces into finding all the factors of 1 , which , obviously, are (+1, +1) and (-1, -1)\n\nand $(m-1)=(n-1) = +1 $ gives $ m=n=2 $ \nwhiel the other $m-1 = n-1 = -1 $ gives $m = n = 0 $\n\nThe t method can be applied to the type of equation: $mn + $ (somthing)*$m$ (anothersomething)*$n$ $etc$ \n\nHope this helps. (There are many such problems which reduces to this type - so knowing this method is helpful)[/hide]", "Solution_8": "[quote]The method can be applied to the type of equation:\n$ mn + \\mathrm{(something)}*m +\\mathrm{(anothersomething)}*n$ etc [/quote]\r\nBy analogy to the algebraic technique of completing the square, I like to refer to this as \"completing the rectangle.\"", "Solution_9": "You will also see the same factoring trick refered to around this site as \"Simon's Favorite Factoring Trick.\"" } { "Tag": [ "geometry" ], "Problem": "Just curious how hard this is. Any comments or anything are appreciated.", "Solution_1": "So, my honest opinion: I would describe it as Intermediate, but it feels very contrived. That is, it seems like you introduced a lot of stuff just for the sake of it, and to some extent it just serves to make the problem \"not pretty.\" I don't know if you did your diagram to scale, but it was quite messy, and ideally you would spread things out more so that they were more visible. I wouldn't say it was a bad problem -- just that it didn't feel natural. Maybe I would say it seems like you constructed the problem in the wrong way -- instead of starting with a diagram and trying to find things out about it, it seems you started with the answer to the problem and tried to make a question for it, a messy prospect.\r\n\r\n\r\nI attached an alternate diagram (without labels), which I thought looked clearer.\r\n\r\nNote: diagram edited due to eariler error.", "Solution_2": "JBL in your picture it looks like you made :ang: D automatically a right angle but the original problem isn't marked that way.", "Solution_3": "Sorry, you're right. That was one thing I was going to say about the diagram, actually -- that angle looked right -- but instead I convinced myself it really was right :) My mistake." } { "Tag": [ "geometry", "perimeter", "rectangle" ], "Problem": "[color=red]Given the convex quadrilateral ABCD, inscribe another quadrilateral with minimal perimiter.[/color]\r\n\r\nThe following is what I tried.\r\n\r\n(1) The problem must have a solution: ABCD is compact and the \"perimiter mapping\" is continuous.\r\n\r\n(2) Let XYZT an arbitrary quadrilateral inscribed in ABCD. $X\\in [DA], Y\\in [CD], Z\\in [BC]$\r\nLet X' be the symmetric of X with respect of CD. Suppose ZX' meets SEGMENT [CD] in U.\r\nThen perimiter(XYZU)<=perimiter(XYZT).\r\n\r\n(3) Consider the former case (2) and see what happens if $ZX'\\notin [CD]$. Then perimiter(XYZU))<=perimiter(XYZT)\r\nwhere U=D.\r\n\r\n(4) We now apply the same trick used in (2) and (3) to the new inscribed quadrilateral XYZU.\r\nWe now consider two cases: (a) the trick fails because of the reason explained in (2) and (3). (b) the trick does not fail.\r\n\r\n(a) Apply (3) again.\r\n(b) Perimiter is strictly decreasing in general unless we could find an inscribed quadrilateral XYZT such that $\\widehat{AXY}=\\widehat{DXT}$, $\\widehat{DTX}=\\widehat{CTZ}$ etc.\r\n\r\nIf that happens I proved that ABCD must be cyclic. Then I think there are infinitely many solutions.\r\n\r\nBut what happens if ABCD is NOT cyclic? Recall we know there is solution so I think this implies that if ABCD is not cyclic then \"the trick\" fails, ie, we must consider the case explained in (3).\r\nI think that my argument is \"weird\" if it is correct...", "Solution_1": "wait, i'm trying the problem, but i don't understand the part \"Let U be the symmetric of Z with respect to the side [D,A].\" please explain.", "Solution_2": "drunner2007\r\n\r\nI edited my first message :oops: but now I explained my reasoning :)\r\n\r\nMaybe this problem is not for gettin started, sorry", "Solution_3": "Connecting midpoints intuitively seems right...", "Solution_4": "[quote=\"Iron Fist\"]Connecting midpoints intuitively seems right...[/quote]\r\n\r\nNo... Thats not true in general. Try it. It works for a rectangle as I proved but it isnt the only solution (in fact, there are infintely many others).\r\nEven if ABCD is cyclic the Varignon's quadrilateral is not a solution." } { "Tag": [], "Problem": "\ubb38\uc11c\ub97c \uc5b4\ub5bb\uac8c \ud558\uba74 \uc9c0\uc6b8\uc218 \uc788\uc8e0? -_-\r\n\r\nHello.\r\n\r\nHow can i delete my postings?\r\n\r\n-_-\r\n\r\nonly edit??", "Solution_1": "In most cases, you can't delete your posts, after next reply is posted or many hours have passed. Contact a moderator or an administrator.\r\n\r\n\ub300\ubd80\ubd84\uc758 \uacbd\uc6b0\uc5d0\ub294 \uc790\uc2e0\uc758 \uae00\uc740 \uc9c0\uc6b8 \uc218 \uc5c6\uc73c\uba70 \uac8c\uc2dc\ud310 \uad00\ub9ac\uc790 \ud639\uc740 \uc0ac\uc774\ud2b8 \uc6b4\uc601\uc790\uc5d0\uac8c \uc5f0\ub77d\ud558\uc5ec\uc57c \ud569\ub2c8\ub2e4.", "Solution_2": "\uac10\uc0ac\ud569\ub2c8\ub2e4. \ub808\uc774\ud14d \uc2dc\ud5d8\uc744 \ud574\ubcfc\ub824\uace0 \ud588\ub294\ub370, \uc9c0\uc6cc\uc9c0\uc9c0 \uc54a\uc544\uc11c \uc218\uc815\ubc16\uc5d0 \ud560\uc218\uac00 \uc5c6\uc5c8\ub124\uc694.\r\n\r\n\ub808\uc774\ud14d\ubcf4\ub2e4\ub294 \ud55c\uae00\uc758 \uc218\uc2dd\ud3b8\uc9d1\uae30\uac00 \uc775\uc219\ud574\uc11c \ubcc0\ud658\ubc29\ubc95\uc744 \ucc3e\ub2e4\uac00, \uacb0\uad6d\uc740 \ub808\uc774\ud14d\uc744 \uc774\uc6a9\ud560\uc218\ubc16\uc5d0 \r\n\r\n\uc5c6\ub2e4\ub294 \uacb0\ub860\uc744 \ub0b4\ub838\uc2b5\ub2c8\ub2e4.\r\n\r\n\ub808\uc774\ud14d... \uc815\ub9d0 \ub300\ud559\uad50 \uc2dc\uc808\uc5d0 \uc368\ubcf4\uace0 \uc624\ub79c\ub9cc\uc5d0 \ub2e4\uc2dc \uc368\ubcf4\ub124\uc694. (\uc804 93\ud559\ubc88. -_-)" } { "Tag": [ "limit", "logarithms", "function", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "x>0 \r\n\r\ncompute \r\n\r\n$\\lim _{n\\rightarrow + \\infty}n\\ln(1+\\ln(1+(...+\\ln(1+x/n))...)$\r\n\r\n$\\ln $appear n-times", "Solution_1": "Choose $x\\in\\mathbb{R}^+$ and let $f(t)=\\log(1+t)$; now, fix $\\epsilon>0$ and take $\\bar{n}$ such that $x/\\bar{n}<\\epsilon$. Obviously, $f(t)\\bar{n}$.\r\nWe have that\r\n\\[f^{(n+1)}(x/(n+1)) 0$. Rearranging that inequality gives\r\n\\[\r\n f(t) > \\frac{1}{1/t + 1/2}\r\n\\]\r\nfor $t > 0$. Iterating, we get\r\n\\[\r\n f^{(n)}(t) \\ge \\frac{1}{1/t + n/2}\r\n\\]\r\nfor $t > 0$. That gives us a lower bound matching jmerry's answer.\r\n\r\nNext, we will aim for an upper bound. Given $\\epsilon > 0$, we have $g(t) \\ge g(\\epsilon)$ for $0 < t \\le \\epsilon$. Rearranging that inequality gives\r\n\\[\r\n f(t) \\le \\frac{1}{1/t + g(\\epsilon)} \r\n\\] \r\nfor $0 < t \\le \\epsilon$. Iterating, we get\r\n\\[\r\n f^{(n)}(t) \\le \\frac{1}{1/t + g(\\epsilon) n}\r\n\\]\r\nfor $0 < t \\le \\epsilon$. That gives us an upper bound on the limit (superior)\r\nof\r\n\\[\r\n \\frac{x}{1 + g(\\epsilon) x} \\, .\r\n\\]\r\nAs $\\epsilon$ approaches $0$, that bound approaches jmerry's answer." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "How many cubic numbers are also perfect squares under 1000?", "Solution_1": "[hide]Perfect squares that are also perfect cubes are powers of six because lcm(2,3) = 6.\n\nClearly, $1000 = 10^{3}$.\n\n$1000 = 10^{3}= (\\sqrt{10}){}^{2}{}^{3}= (\\sqrt{10})^{6}$\n\n$\\sqrt{10}\\approx 3.16$\n\nThus, where $n$ is an integer, $\\forall n \\text{ s.t. }000. This means that the limit we're looking for is L=lim {n->00} (a_n^(1/2^n)). \r\n\r\nThere exists an a s.t. e^a+e^(-a)=3=a_0. Then a_1=a_0^2-2=e^(2a)+e^(-2a), .., a_n=e^(2^n*a)+e^(-2^n*a), which means that L=e^a=e^(arccosh(3/2)). Is it OK orl?", "Solution_3": "Hi grobber,\r\n\r\nI don't know whether your solution is correct as I wonder how arccosh is defined. I know cosh(x) is defined as 1/2*(e^x+e^(-x)). So what is the inverse function ? Anyway the limit is the square of the golden mean: ((1+5^0.5)/2)^2.", "Solution_4": "Well, I don't know what the inverse function is either, but could you check my solution and see whether there's something wrong with it or not? And could you please post you solution or send it to me by PM?", "Solution_5": "arccosh(x) = ln(x \\pm \\sqrt(x^2 - 1)) for real x\r\n\r\nmathworld.wolfram.com is a good mathematical resource for that sort of stuff.", "Solution_6": "Well, we now know what arccosh(x) is, but there's still a problem: I got e^arccosh(3/2), and orl says it's ((1+5^0.5)/2)^2, so are these 2 equal? Or maybe I made some mistake, but in this case I'd like to know what it is. My soln looks Ok to me... Could you people check this out?", "Solution_7": "let us denote arccosh 3/2 = x. Then cosh x = 3/2 :) which means that \r\n\r\nex+1/ex = 3 => y2 - 3y + 1 = 0 => y = ( 3 +\\sqrt 5) / 2 = ( 1+\\sqrt 5 )2 / 4, which is exactly the number orl indicated. :) \r\n\r\nI have denoted by y grobber's result.", "Solution_8": "Oh... So it WAS correct! :D Thanks Valentin!", "Solution_9": "(3)\r\n\r\nS = \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+2)) = \\sum^{infinity}_{ n=1} (-1)^n (f_(n+2)f(n))/(f_n*f_(n+1)*f_(n+2))\r\n\r\n = \\sum^{infinity}_{ n=1} (-1)^n [1/(f_n*f_(n+1))- 1/(f_(n+1)*f_(n+2))]\r\n\r\n= \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+1)) + = \\sum^{infinity}_{ n=1} (-1)^(n+1)/(f_(n+1)*f_(n+2))\r\n\r\n= \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+1)) + = \\sum^{infinity}_{ n=2} (-1)^n/(f_n*f_(n+1))\r\n\r\n= \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+1)) + (= \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+1)) (-1)/(f_1*f_2) )\r\n\r\n= 2 * \\sum^{infinity}_{ n=1} (-1)^n/(f_n*f_(n+1)) + 1\r\n\r\n= 2 * \\sum^{infinity}_{ n=1} ((f_(n+1))^2-f_n*f_(n+2))/(f_n*f_(n+1)) + 1\r\n\r\n= 2 * \\sum^{infinity}_{ n=1} (f_(n+1)/f_n-f_(n+2)/f_(n+1)) + 1\r\n\r\n= 1 lim {n->00} f_(n+2) / f_(n+1) = 1 [(1+5^0.5)/2] = 2 5^0.5\r\n\r\nHmm... so you can find the golden mean again. :)", "Solution_10": "Here's another way to do it:\r\n\r\ne^(arccosh(3/2)) = e^(ln(3/2 \\pm \\sqrt((3/2)^2 - 1))) = 3/2 \\pm \\sqrt (9/4 - 1) = 3/2 \\pm \\sqrt(5/4) = (3 \\pm \\sqrt(5))/2\r\n\r\nTake the addition sign, and you get your answer." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Let's say i qualified for USAMO. (although im not so good)\r\nit says in the qualification that i have to be either\r\n\r\n1. a citizen of U.S.\r\n\r\nor\r\n\r\n2. possess a green card.\r\n\r\nWhat happens if I am in the process of green card and INS\r\nhas issued an alien number so that i am legally in the United states?", "Solution_1": "The person to ask about things like that is someone from the AMC. Go e-mail them. Steve Dunbar's e-mail address is available here: http://www.unl.edu/amc/whoswho.html" } { "Tag": [ "quadratics", "arithmetic sequence", "algebra", "geometric sequence", "geometric series" ], "Problem": "Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?", "Solution_1": "[quote=\"IntrepidMath\"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]\r\n\r\n102 times ... if the Albatross could pick 102 apples at once though, why didn't he do that first? [i]Sheesh[/i] ... talk about a bird brain.", "Solution_2": "[hide=\"Solution\"]\n\n Albatross drops apples in the following arithmetic progression $ 1, 2 ,3 ,4 ,5\\cdots$ so when they ask you how many times he has repeated this process there are simply asking you to find $n$ in the equation:\n\n$ 1 + 2 + 3 + 4 + 5 + 6 \\cdots + n = 5253 $\n\n Applying Gauss's formula we see that $\\frac{n(n+1)}{2}=5253$\n\n Thus,$ n(n+1)=10506 $\n\n Now we can solve this equation two ways.\n\n The first solving as a quadratic equation: $ n^2 + n -10560=0$, but this becomes very messy, and is unesessary for such a simple problem.\n\n The other solution involves noticing that $ n$ and $ n+1$ are so close to together, we could just as easily take the square root of 10506.\n\n$\\sqrt{10506}=102.4987805\\cdots$\n\n So lets check:\n\n$ \\frac{102(103)}{2}= \\frac{10506}{2} = 5253$.\n\n And it works! Thus, the number of times the process was repeated is $\\boxed{102}$ times.[/hide]", "Solution_3": "[quote=\"Dr. No\"][quote=\"IntrepidMath\"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]\n\n102 times ... if the Albatross could pick 102 apples at once though, why didn't he do that first? [i]Sheesh[/i] ... talk about a bird brain.[/quote]\r\n\r\nIts better to actually post a solution than pointlessly spamming like this.", "Solution_4": "good work G-UNIT. Your posts have been good.", "Solution_5": "I'm afraid I must disagree with the previous answers and solutions.", "Solution_6": "[quote=\"rcv\"]I'm afraid I must disagree with the previous answers and solutions.[/quote]\r\nI agree with rcv.\r\n[hide]He repeated 5253 times!!![/hide]", "Solution_7": "Hi everyone:\r\n\r\n[hide]\n I used the formula for an arithmetic progression for this problem.\n\n [tex]s = \\frac {n}{2}[2a + (n - 1)d][/tex]\n\n The sum [tex]s[/tex] is 5253\n The value of [tex]a[/tex] is 1\n The difference [tex]d[/tex] is 1\n \n We are trying to find the value of [tex]n[/tex]\n\n Substituting the known values into the formula we have:\n\n [tex]5253 = \\frac {n}{2}[2(1) + (n - 1)(1)] = \\frac {n}{2}(n + 1)[/tex]\n\n We end up with the quadratic equation:\n\n [tex]n^2 + n - 10506 = 0[/tex]\n\n solving for [tex] n [/tex] we have [tex] n = 102 \\ or \\ -103[/tex]\n\n since [tex] n [/tex] can't be negative the answer must be [tex] 102 [/tex]\n\n So I agreed with G-unit up to the quadratic equation, but obtained it a different way.\n\n Also, I didn't see solving the quadratic as \"messy\".[/hide]", "Solution_8": "[quote=\"10000th User\"][quote=\"rcv\"]I'm afraid I must disagree with the previous answers and solutions.[/quote]\nI agree with rcv.\n[hide]He repeated 5253 times!!![/hide][/quote]\r\n\r\n I guess the solution to this problem depends on how we define \"drops 5253 apples\". Whether we interpret this in \"one go\" meaning that was the last drop he made was the 5253rd, then yes the answer is 5253. However, if you define \"drops 5253 apples\" as the sum of all the apples he has already dropped on the floor, then I would disagree and say the answer is 102. Also note that the word \"total\" in \"dropped a total of 5253 apples\" also points in the direction that we are looking for the sum of all the apples rather than the last one made. Well, thats just how I see it, mabye theres another angle im not seeing..............", "Solution_9": "I see the word total, which all the previous too, I guess. \r\nI still think the answer is 102 times.", "Solution_10": "[quote=\"A+MATH\"][quote=\"Dr. No\"][quote=\"IntrepidMath\"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]\n\n102 times ... if the Albatross could pick 102 apples at once though, why didn't he do that first? [i]Sheesh[/i] ... talk about a bird brain.[/quote]\n\nIts better to actually post a solution than pointlessly spamming like this.[/quote]\r\n\r\nI resent the accusation that anything I post is [i]spam[/i].\r\n\r\nThis problem is easily solved but given that I was the first to post a reply I didn't wish to undermine anyone else's learning by posting a full solution. In other words, the joy of doing this stuff is found in doing it for one's self.", "Solution_11": "I still disagree with all previous answers and solutions.", "Solution_12": "I have a feeling it's 5253/3 or 1751 times. he repeated it 1750 times. If the process mentioned is drop one, then drop two.", "Solution_13": "With regards to rcv's posts, perhaps the pattern is not clear? We immediately assumed Albatross would then drop 3, then 4 etc. Many sequences begin with 1, 2, ...", "Solution_14": "[quote=\"rcv\"]I still disagree with all previous answers and solutions.[/quote]\r\nhmm maybe [hide=\"this?\"]5253 divided by 3apples per repeat of dropping 1,2,1,2,1,2...?[/hide]", "Solution_15": "[quote=\"Palytoxin\"][quote=\"rcv\"]I still disagree with all previous answers and solutions.[/quote]\nhmm maybe [hide=\"this?\"]5253 divided by 3apples per repeat of dropping 1,2,1,2,1,2...?[/hide][/quote]\r\n\r\nI agree...", "Solution_16": "[quote=\"xenon_nightmare152\"][quote=\"Palytoxin\"][quote=\"rcv\"]I still disagree with all previous answers and solutions.[/quote]\nhmm maybe [hide=\"this?\"]5253 divided by 3apples per repeat of dropping 1,2,1,2,1,2...?[/hide][/quote]\n\nI agree...[/quote]\r\n\r\n\r\n So if you assume the amount of apples dropped ends at 2, then I suppose 1751 is right. Unless you are counting the individual steps in which case it would be twice the amount mentioned above, 3502.", "Solution_17": "[quote=\"IntrepidMath\"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]\r\nI think 236factorial has found the chief fault in the previous answers. The question states that he \"repeats this process\".\r\n\r\nI believe the question defines \"the process\" as \"pick an apple; drop it; pick 2 apples; drop them\". (I see nothing in the problem statement to suggest that 1 and 2 are the first two terms of a sequence. Also, why do we want to assume this is an arithmetic sequence, rather than a geometric sequence? If we change the problem from \"5253 apples\" to \"4095 apples\", do we now assume that a geometric sequence is involved? Or should common sense lead us to ask if he can hold 2048 apples at once. Or 102 apples at once? If Albatross is a human, he can't hold that many apples. If Albatross is a mythical beast, perhaps he can hold that many apples.)\r\n\r\nWe're not quite done. I'm not sure which answer 236factorial is claiming -- 1751 or 1750? What is the problem asking for?\r\n\r\nIntrepidMath: If you have the source for this problem, this would be a good time to share the souce and the solution with this thread. If you created this problem on your own, did you intentionally choose the wording so this would be a tricky question? Or was the wording unintentionally ambiguous?", "Solution_18": "If he repeated the process 5253 times, then the albatross would have had to pick up only one apple each time. But the problem states that after he picked up one apple and dropped it, he then picked up [tex] 2 [/tex] apples and dropped those.\r\nIt was only an assumption on my part that each time he picked up more apples, it was always one more from the time before. But if that wasn't the case, then what did he do? But just due to that one statement I can't say that it was 5253 times.", "Solution_19": "[quote=\"rcv\"]We're not quite done. I'm not sure which answer 236factorial is claiming -- 1751 or 1750? What is the problem asking for?[/quote]\r\n\r\nI see that he [i]repeated[/i] it 1750 times.", "Solution_20": "The way it works is an arithmetic sequence, 1, 2,3, 4,5 ,6, 7,8,9,10..., sorry for any confusion. [hide=\"the answer\"] is 102[/hide]", "Solution_21": "As I see it, because the problem was rather vague and did not specify on what exactly the process was; to solve this problem one had to use his common sense and make assumption concerning the \"proccess\" given. We are given that the first two terms are 1 and 2. From this I believe we can make 3 common conclusions. The first, it repeats in a geometric progression, 1, 2 ,4 , 8......., and so on. The second, it repeats in a arithmetic progression, 1, 2, 3, 4, 5, 6......, and so on. Lastly, the only two terms in the sequence are 1 and 2, thus' repeating as 1, 2, 1, 2, 1, 2.....,and so on. In my case, I felt the arithemetic progression was most suitable. Why? I will explain. I immediately ruled out the geometric series in my mind. Why? Well, assumptions, and conclusions can only come from you experience or rather background, with what you are dealing with. In this case it problem solving; and in the bulk of geometric series, concerning 2, I have seen, almost 99% of them give at least 3 or 4 terms in the series before ending with continued line of dots. For example, in a problem concerning geometric serieses you might see:\r\n\r\n1, 2, 4,........ OR 1, 2, 4, 8............. But in my personal experience I have never seen a problem give only two terms, 1,2....... ,and expect that you can tell they are reffering to an geometric progression. The other two I found equaly likely to occur. However, I chose the arithmetic progression over the re-occuring 1,2 progression. Why? Simply because I have seen more problem, where the key principle is arithmetic progression, and the number of term that are given are generally limited. For example:\r\n\r\n1, 2, 3............. OR 1, 2, 3 ,4...................... I do understand that only the first two terms are given, and if INTREPIDMATH had only written a few more lines concerning how many apples Albatross was dropping, it would have become obvious what the problem was asking for and we could have avoided this delimma. Nontheless, in problems we have to work with whats given and in this case I made an assumption based on my personal experience and backround with problems, since my backround is likely to differ from yours, it is most likely that different answers will occur (which they have). However, let this be a learning lesson to users posting problems, to [b] specify[/b] things that are crucial in the problem,(in this case it was the repreating pattern) and not force the problem solver to make controversial assumptions, which can turn out wrong.", "Solution_22": "I don't think we received enough information to know exactly what the \"process\" is.", "Solution_23": "[hide]\n\nI started out by writing the equation: $1+2+...n = 5253$ Then, I converted the equation into $\\frac{n(n+1)}{2} = 5253$ \nMultiply by 2 on each side: $n(n+1) = 10506$\nSo:\n$n = \\framebox{102}$[/hide]", "Solution_24": "[quote=\"pascal_1623\"][hide]\n\nI started out by writing the equation: $1+2+...n = 5253$ [/hide][/quote]\r\nwhy does it have to equal 5253?", "Solution_25": "[quote=\"math92\"][quote=\"pascal_1623\"][hide]\n\nI started out by writing the equation: $1+2+...n = 5253$ [/hide][/quote]\nwhy does it have to equal 5253?[/quote]\r\n\r\nThat's how many apples he dropped total." } { "Tag": [ "geometry", "3D geometry", "sphere", "tetrahedron", "calculus", "integration" ], "Problem": "You can find all of the 2007 PUMaC test documents [url=http://www.princeton.edu/~nsavir/PUMaC/]here[/url].\r\n\r\nTests labeled \"Subject A\" or \"Subject B\" are the documents that we gave out at the contest. If I remember correctly, we issued the following errata before the respective tests began: \r\nGeometry A1/B6: In the diagram, A and B are switched. \r\nGeometry A6: Change 80-whatever to 37. \r\nTeam 7: CP = PQ should be CQ = PQ \r\nTeam 9: r should be positive. \r\n\r\nTests labeled \"Subject 3\" contain all 15 questions for the subject and have solutions. \r\n\r\nFeel free to discuss the problems and solutions here!", "Solution_1": "Here is what I tried to do for combo #8, though I ended up getting the wrong answer somehow. This time, I got the right answer.\r\n[hide=\"Combo 8\"]Let f(n) be the number of pairs of n-digit numbers with even dot-product of digits (this is what I am calling the operation in the problem). We may verify that f(1)=56 and there are 75 ways for the product of the ith digits to be even, and 25 ways for the product to be odd, i>1. Then $ f(n \\plus{} 1) \\equal{} 75f(n) \\plus{} 25((9*10^{n \\minus{} 1})^2 \\minus{} f(n)) \\equal{} 50f(n) \\plus{} 2025*10^{2n \\minus{} 2}$. We may see that this recurrence is satisfied by $ f(n) \\equal{} 775*50^{n \\minus{} 2} \\plus{} 4050*100^{n \\minus{} 2}$.[/hide]\nAnd here is our awesome solution to #7 on the team round:\n[hide=\"Team 7\"]Take Menelaus on triangle ABQ and colinear points R, P, C. We find AQ=BP. Now choose a point R' on ray CR so that BR'=BR. Note BR'=AR, BP=AC, R'PB=RCA. We have a SSA congruence, which is satisfied by triangles RBP and R'BP. One of these is congruent to ARC. Obviously, RBP is not, so R'BP is, and in particular R'P=RC, so R'P=PC=RB. Hence R'RB is equilateral, so R'BR=60 and CRB=120.[/hide]", "Solution_2": "Are you going to put up official solutions to the Power Test?", "Solution_3": "Yeah WindingFunction will post them some time soon I think.", "Solution_4": "In number four on the team round, it says that x can not be larger than 3, but how does that tell us that there is not a nonintegral value of x?", "Solution_5": "[quote=\"diophantient\"]In number four on the team round, it says that x can not be larger than 3, but how does that tell us that there is not a nonintegral value of x?[/quote]\r\nWe never showed that $ x$ must be an integer. When I was solving the problem, I first evaluated at $ x\\equal{}0, 1, 2, 3$ to check what it looked like. I got $ x\\equal{}1, 2$ as solutions. Then I showed that solutions are at most 3. Then I realized that in fact there can only be 2, so I was already done. \r\n\r\nAn alternative sequence of thoughts might be to first check that $ x$ can't be too small or big and show that $ 0 < x < 3.$ Then check how many possible solutions there are (2). Then you might want to get a better idea of where exactly they are, so you can check $ x\\equal{}1, 2,$ and you get lucky. \r\n\r\nTo be honest I don't know how I would approach the problem if there was a non-integer solution.", "Solution_6": "[quote=\"yjneb\"]Are you going to put up official solutions to the Power Test?[/quote]\r\n\r\nSorry it's been slow...I have to fix a couple of things in the document, but I also have to catch up on the work I neglected last week due to PUMaC preparations, and that's unfortunately taking priority. Please be patient. :(", "Solution_7": "What was the breakdown of points for the problems?\r\n\r\nWas it 3,3,3,4,4,4,4,5,5,5", "Solution_8": "It was 3334444555 as expected.", "Solution_9": "Can someone post the Power Test problems for those of us that did not participate?", "Solution_10": "[quote=\"CatalystOfNostalgia\"]Can someone post the Power Test problems for those of us that did not participate?[/quote]\r\n\r\nThat I can do.", "Solution_11": "[hide=\"win solution for geometry 9\"]Put stuff in the 4-D plane. $ A\\equal{}(\\sqrt{2},0,0,0),B\\equal{}(0,\\sqrt{2},0,0),C\\equal{}(0,0,\\sqrt{2},0),D\\equal{}(0,0,0,\\sqrt{2})$.The center of the sphere in the middle is the center of this tetrahedron, $ O\\equal{}(\\sqrt{2}/4,\\sqrt{2}/4,\\sqrt{2}/4,\\sqrt{2}/4)$. The answer is just $ OA\\minus{}1$.[/hide]", "Solution_12": "For the 2009 pumac, are you still going to have 15 questions for each subject, and give #1-10 to Division B, and # 6-15 to division A?", "Solution_13": "[hide=\"Win Solution for NT #7 better than functionology\"]$ 8(x^3 \\plus{} x^2y \\plus{} xy^2 \\plus{} y^3) \\equal{} 15(x^2 \\plus{} y^2 \\plus{} xy \\plus{} 1) \\implies gcd(15,8) \\equal{} 1 \\implies (*)x^3 \\plus{} x^2y \\plus{} xy^2 \\plus{} y^3 \\equal{} 15 , (**)x^2 \\plus{} y^2 \\plus{} xy \\plus{} 1 \\equal{} 8$Multiplying (**) by y and subtracting from (*) we have $ x^3 \\equal{} 15 \\minus{} 7y, x > 0 \\implies y \\equal{} 1,2$ both of which yield integrals solutions for x and give our only solutions $ (1,2) (2,1)$[/hide]", "Solution_14": "[quote=\"mathemonster\"][hide=\"Win Solution for NT #7 better than functionology\"]$ 8(x^3 \\plus{} x^2y \\plus{} xy^2 \\plus{} y^3) \\equal{} 15(x^2 \\plus{} y^2 \\plus{} xy \\plus{} 1) \\implies gcd(15,8) \\equal{} 1 \\implies (*)x^3 \\plus{} x^2y \\plus{} xy^2 \\plus{} y^3 \\equal{} 15 , (**)x^2 \\plus{} y^2 \\plus{} xy \\plus{} 1 \\equal{} 8$Multiplying (**) by y and subtracting from (*) we have $ x^3 \\equal{} 15 \\minus{} 7y, x > 0 \\implies y \\equal{} 1,2$ both of which yield integrals solutions for x and give our only solutions $ (1,2) (2,1)$[/hide][/quote]\r\n\r\nYour solution is false; from $ 8(x^3 \\plus{} x^2y \\plus{} xy^2 \\plus{} y^3) \\equal{} 15(x^2 \\plus{} y^2 \\plus{} xy \\plus{} 1)$ we can only conclude that there is some integer k such that $ x^3\\plus{}x^2y\\plus{}xy^2\\plus{}y^3 \\equal{} 15k$ and $ x^2\\plus{}y^2\\plus{}xy\\plus{}1 \\equal{} 8k.$", "Solution_15": "yes i realize. You just have to test a few cases though :D", "Solution_16": "Can we get the problem from the 2009 competition(it was yesterday)?", "Solution_17": "Be patient. The results aren't even up yet, and I'm sure the organizers are pretty tired from all the difficulties they had yesterday" } { "Tag": [ "AMC" ], "Problem": "So I read the FAQ on the AMC site and it said that you could take the AMC 12A and 12B if your school registered for both. However, I know my school doesn't do both (we only do B) - so could I take the \"A\" at another school? How would I could about finding a local school which offers it?", "Solution_1": "Usually, if schools want to offer only one of the dates, they do the A. For that reason, there is a list of places [url=http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml]here[/url] that offer the B test. I am not sure about where you can take the A test though.", "Solution_2": "You are at an advantage, as PowerOf$ \\pi$ said, most schools do A. So do the A at one of those schools, and take B at your regular school" } { "Tag": [ "inequalities", "linear algebra", "matrix" ], "Problem": "Let $ A,B,C,D \\in S_n{(\\mathbb{R})},n,m \\in \\mathbb{N^{*}}$\r\nProve inequality determinant of matrix :\r\n $ \\sqrt [n]{\\det(A^{2m} \\plus{} B^{2m} \\plus{} C^{2m} \\plus{} D^{2m})}\\geq\\sqrt [n]{\\det(A^{2m} \\plus{} B^{2m})} \\plus{} \\sqrt [n]{\\det(C^{2m} \\plus{} D^{2m})}$\r\n$ S_n(\\mathbb{R})$ are set $ nxn$ real symmetric matrices.\r\n[hide=Answer][/hide]", "Solution_1": "It is a kind of Minkowski's determinantal inequality." } { "Tag": [ "geometry", "geometric transformation", "homothety", "parallelogram", "IMO Shortlist", "geometry proposed" ], "Problem": "Let $ ABC$ be a triangle with incircle $ (I)$.$ (I)$ touch $ BC;CA;AB$ at $ D;E;F$,respectively.$ DF$ cuts $ BE$ at $ M$;$ DE$ cuts $ CF$ at $ N$\r\n$ K;L$ are midpoints of $ ME;NF$,respectively.\r\nProve that $ AI;FK;EL$ are concurrent.", "Solution_1": "It is suffice to show $ \\angle FEL\\equal{}\\angle KFE$ which is imo shortlist problem.\r\n\r\n[color=blue][b]Lemma[/b]: Let $ ABC$ be a triangle and $ D$ be a point on $ CA$. $ X,Y$ are lie on $ BD,BC$ such $ AX,AY$ are isogonal line wrt $ \\angle A$ and $ E,F$ be on $ BD,BC$ such $ XF\\parallel YE$. Then $ AF,AE$ are isogonal line wrt $ \\angle A$.[/color]\r\nproof\r\nLet $ P$ be a point of $ AB$ such triangle $ PXF$ and $ AEY$ are homothety with homothety center $ B$. It is obvious that $ P$ is well defined. From $ \\angle PAX\\equal{}\\angle BAX\\equal{}\\angle YAC\\equal{}\\angle AYE\\equal{}\\angle PFX$, $ A,P,X,F$ are cyclic. So $ \\angle BAF\\equal{}\\angle PAF\\equal{}\\angle PXF\\equal{}\\angle AEY\\equal{}\\angle EAC$. So $ AF,AE$ are isogonal line wrt $ \\angle A$.\r\nI remember that this lemma is in crux.\r\n\r\nLet return to above problem.\r\nLet $ G$ be intersection of $ BE,CF$(or equivalently, $ G$ is symedian point of triangle $ DEF$),$ P$ be a midpoint of $ FD$ and $ X,Y,$ be a point on $ DF,DE$ such $ GX\\parallel DE,GY\\parallel DF$. By well known theorem,$ GE$ and $ PE$ are isogonal line wrt $ \\angle DEF$. Hence by lemma, we have $ \\angle LEF\\equal{}\\angle DEX$ and similarly, $ \\angle KFE\\equal{}\\angle YFD$. So it is suffice to show $ E,F,X,Y$ are cyclic. Let $ Z$ be a point on $ DF$ such $ E,F,Z,Y$ are cyclic. I will show $ Z\\equal{}X$. Since $ GXDY$ is parallelogram, $ GD$ pass midpoint of $ XY$ and $ GD$ pass midpoint of $ YZ$ because triangle $ DYZ$ and $ DFE$ are inverserly similarl and $ GD$ is isogonal line of $ D\\minus{}$median. Hence midpoint of $ YZ,YX$ and $ D$ are collinear which imply $ Z\\equal{}X$. So we are done.", "Solution_2": "The Lemma is correct for $ XF//EY//$[color=darkred]$ AC$[/color] and not only for $ XF//YE$. Besides you take that to your proof.", "Solution_3": "[quote=\"kritikos\"]The Lemma is correct for $ XF//EY//$[color=darkred]$ AC$[/color] and not only for $ XF//YE$. Besides you take that to your proof.[/quote]\r\n You are right. Sorry for my mistake. I forgot writing $ \\parallel AC$ in lemma." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "let $n\\geq 2$ an integer and $x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}$ real numbers such that : $\\underset{i=1}{\\overset{n}{\\sum }}x_{i}^{2}=\\underset{i=1}{\\overset{n}{\\sum }}y_{i}^{2}=1$\r\nProve that : $\\left( x_{1}y_{2}-x_{2}y_{1}\\right) ^{2}\\leq 2\\left| 1-\\underset{i=1}{\\overset{n}{\\sum }}x_{i}y_{i}\\right| $", "Solution_1": "It is Korea 2001.", "Solution_2": "really ? do you have the solution ? :lol:", "Solution_3": "Easy,I think\r\n\r\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum{x_i-y_j)^2 \\ge (x_1y_2-x_2y_1)^2}$\r\n\r\nThen $(1+\\sum{x_iy_i})(1-\\sum{x_iy_i}) \\ge (x_1y_2-x_2y_1)^2$\r\n\r\nBecause $\\sum{x_iy_i} \\le 1$ then we have the result!", "Solution_4": "[quote=\"hungkhtn\"]Easy,I think\n\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum{x_i-y_j)^2 \\ge (x_1y_2-x_2y_1)^2}$\n\nThen $(1+\\sum{x_iy_i})(1-\\sum{x_iy_i}) \\ge (x_1y_2-x_2y_1)^2$\n\nBecause $\\sum{x_iy_i} \\le 1$ then we have the result![/quote]\r\n\r\nhungkhtn, do you mean\r\n\r\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum(x_iy_j-x_jy_i)^2 \\ge (x_1y_2-x_2y_1)^2$", "Solution_5": "[quote=\"siuhochung\"][quote=\"hungkhtn\"]Easy,I think\n\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum{x_i-y_j)^2 \\ge (x_1y_2-x_2y_1)^2}$\n\nThen $(1+\\sum{x_iy_i})(1-\\sum{x_iy_i}) \\ge (x_1y_2-x_2y_1)^2$\n\nBecause $\\sum{x_iy_i} \\le 1$ then we have the result![/quote]\n\nhungkhtn, do you mean\n\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum(x_iy_j-x_jy_i)^2 \\ge (x_1y_2-x_2y_1)^2$[/quote]\r\n\r\nA bit mismath is typing:\r\n\r\n$\\sum{x_i^2}\\sum{y_i^2}-(\\sum{x_iy_i})^2=\\sum_{i<>j}{(x_iy_j-x_jy_i)}^2 \\ge (x_1y_2-x_2y_1)^2$\r\n\r\nIt's very simple!", "Solution_6": "Indeed, this one is the solution that I have posted some time ago on the forum and that appears in Old and New Inequalities. It is really simple problem." } { "Tag": [ "trigonometry", "inequalities unsolved", "inequalities" ], "Problem": "In triangle $ABC$, prove that\r\n$\\frac{(b+c)cosA}{a^2} + \\frac{(c+a)cosB}{b^2} + \\frac{(a+b)cosC}{c^2} \\geq \\frac{9}{a+b+c}$", "Solution_1": "write $cosA = \\frac{b^2+c^2-a^2}{2bc}$ and so on,\r\n\\[\r\n\\begin{array}{l}\r\n (a + b + c)\\sum {\\frac{{\\left( {b + c} \\right)\\cos A}}{{a^2 }}} \\\\ \r\n = (a + b + c)\\left( {\\frac{1}{{2abc}}\\sum {\\frac{{b + c}}{a}} \\left( {b^2 + c^2 - a^2 } \\right)} \\right) \\\\ \r\n = (a + b + c)\\left( {\\frac{1}{{2abc}}\\sum {\\frac{{b + c}}{a}} \\left( {b^2 + c^2 - a^2 } \\right)} \\right) \\\\ \r\n = (a + b + c)\\left( {\\frac{1}{{2abc}}\\left( {\\left( {a + b + c} \\right)\\left( {\\sum {\\frac{{b^2 + c^2 - a^2 }}{a}} } \\right) - \\sum {a^2 } } \\right)} \\right) \\\\ \r\n \\ge \\frac{{a + b + c}}{{2abc}}\\left( {\\left( {a + b + c} \\right)\\left( {a + b + c} \\right) - \\sum {a^2 } } \\right) \\\\ \r\n = \\frac{{a + b + c}}{{abc}}\\left( {ab + bc + ca} \\right) \\\\ \r\n \\ge 9 \\\\ \r\n \\end{array}\r\n\\]" } { "Tag": [ "AMC 10", "AMC 10 A", "AMC 12 A", "AMC" ], "Problem": "[quote=\"yif man 12 (in the College forum)\"]The SAT problems are probably even easier than that. I'd say they're mainly AMC 8 level, with a couple that might be on the AMC 10[b]B[/b].[/quote]\r\n\r\nAre the problems in AMC 10B generally easier than those in AMC 10A?", "Solution_1": "The AMC 10 A and B are supposed to be the same difficulty. Of course, the exams are different so you would probably find one to be easier than the other depending on what questions appear on each one. The same goes for the AMC 12 A and B.", "Solution_2": "Yes, I know, but I was wondering why he mentioned 10[b]B[/b]..." } { "Tag": [], "Problem": "Alright, so it seems that the Open response (Part II) of the USNCO is worth 100 points.\r\nHow many points is the MC worth? I am guessing 60, but it could also be 120. And I have no idea about how much the lab worth but some of you say it is worth very little. \r\nDoes anyone have a clearer idea on how it all works? There's like a few days left before the exam, so I want to get as much information as possible. Thanks! :)", "Solution_1": "I maintain my stated belief that the lab doesn't count for beans, based on the fact that I completely blew it last year and it ended up not mattering.", "Solution_2": "Yes. I screwed it up too. Actually, I've heard that you can get a good chunk of the points for the lab just by having organized work (i.e. a chart) even if your method/results suck. Sorry though, I have no idea how its all weighted. I would just say, try your best on all of it:). Yes, I have a feeling that won't help at all." } { "Tag": [ "calculus", "integration", "probability and stats" ], "Problem": "Hello, \r\n\r\nI've come across this integral when trying to find an expression for $ \\displaystyle\\int_t^Tr(u) \\, du$ with $ dr$ following the Vasicek model and $ W_s$ being a standard Brownian motion: \r\n\r\n$ \\sigma\\displaystyle\\int_t^Te^{-au} \\displaystyle\\int_t^ue^{as} dW_s \\, du\r\n= \\dfrac{\\sigma}{a}\\displaystyle\\int_t^T(1 - e^{-a(T-s)}) \\, dW_s$\r\n\r\nThe expression given above is supposed to be correct. I tried to get this answer using standard integration by parts as follows: \r\n\\begin{align*}\r\n\\sigma\\displaystyle\\int_t^Te^{-au} \\displaystyle\\int_t^ue^{as} dW_s \\, du\r\n&=& \r\n\\left[-\\dfrac{\\sigma}{a}e^{-au}\\displaystyle\\int_t^ue^{as}\\, dW_s\\right]_t^T + \r\n\\dfrac{\\sigma}{a}\\displaystyle\\int_t^Te^{-au} \\dfrac{d}{du}\\displaystyle\\int_t^u e^{as} \\, dW_s\\, du\r\n\\\\&=& \r\n\\dfrac{\\sigma}{a}\\displaystyle\\int_t^Te^{-au} \\dfrac{d}{du}\\displaystyle\\int_t^u e^{as} \\, dW_s\\, du\r\n-\\dfrac{\\sigma}{a}\\displaystyle\\int_t^Te^{-a(T-s)}\\, dW_s\r\n\\end{align*}\r\n\r\nNow clearly I would like $ e^{-au}\\dfrac{d}{du}\\displaystyle\\int_t^u e^{as} \\, dW_s\\, du = 1dW_s$ but I have no idea how to justify this. Any help?", "Solution_1": "$ \\equal{} \\sigma \\int_t^T e^{ \\minus{} au} \\displaystyle \\int_t^T 1_{s < u} e^{as} dW_s \\, du$ and then integrate first along $ du$ then $ dW_s$", "Solution_2": "So the interchange of $ dW_s$ and $ du$ can be justified?\r\n\r\nDoing this, I get: \r\n\r\n$ \\equal{}\\sigma\\int_{t}^{T}e^{\\minus{}au}\\int_{t}^{T}1_{s < u}e^{as}dW_{s}\\, du \r\n\\\\\\equal{} \\sigma\\int_{t}^{T}e^{as}\\int_{s}^{T}1_{s < u}e^{\\minus{}au}\\, du\\, dW_{s}\r\n\\\\\\equal{} \\dfrac{\\sigma}{a}\\int_{t}^{T}e^{as}(e^{\\minus{}as} \\minus{} e^{\\minus{}aT})\\, dW_{s}\r\n\\\\\\equal{} \\dfrac{\\sigma}{a}\\int_{t}^{T}(1 \\minus{} e^{\\minus{}a(T\\minus{}s)})\\, dW_{s}$\r\n\r\nThanks.", "Solution_3": "[quote=\"jshen\"]So the interchange of $ dW_s$ and $ du$ can be justified?\n[/quote]\r\n\r\nThis is the \"Stochastic Fubini\" theorem. I don't remember the proof though, and I think that it could be interesting to try to prove it again. Let's do it :)", "Solution_4": "I think that it can be done via Monotone Class Theorem" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "$ a$ and $ b$ are positive real numbers with $ a\\le b\\le1$. Show that\r\n\\[ 1\\minus{}a^6b^6\\ \\ge\\ a^2b^3(1\\minus{}a^2)\\plus{}a^3b(1\\minus{}b^4)\r\n\\]\r\n:?:", "Solution_1": "Let\r\n\r\n $ a \\equal{} \\frac {1}{1 \\plus{} x \\plus{} xy}, b \\equal{} \\frac {1 \\plus{} x}{ 1 \\plus{} x \\plus{} xy} (x,y\\geq0)$\r\n\r\nthen\r\n\r\n$ 1 \\minus{} a^6b^6 \\minus{} a^2b^3(1 \\minus{} a^2) \\minus{} a^3b(1 \\minus{} b^4)$\r\n\r\n$ \\equal{} \\frac {x(4 \\plus{} 6y \\plus{} 34x \\plus{} 72xy \\plus{} \\ldots \\plus{} 16611x^9y^5)}{(1 \\plus{} x \\plus{} xy)^{12}}\\geq0$.", "Solution_2": "$ 1 \\minus{} a^6 b^6 \\; \\ge \\;a^2 b^3 (1 \\minus{} a^2 ) \\plus{} a^3 b(1 \\minus{} b^4 ) \\Leftrightarrow \\frac{1}{a} \\plus{} \\frac{1}{{b^2 }} \\plus{} a^3 b^3 \\le a \\plus{} b^2 \\plus{} \\frac{1}{{a^3 b^3 }}$.\r\nThat seems easier.", "Solution_3": "[quote=\"Mary Barton\"]$ a$ and $ b$ are positive real numbers with $ a\\le b\\le1$. Show that\n\\[ 1 \\minus{} a^6b^6\\ \\ge\\ a^2b^3(1 \\minus{} a^2) \\plus{} a^3b(1 \\minus{} b^4)\n\\]\n:?:[/quote]\r\nNice inequality. Here is my solution :)", "Solution_4": "I think I used to learn Vietnamese at past......but forgot due to long gap!!! :P \r\nSo, can anyone translate them for me, PLEASE!\r\n\r\n[P/s: Ami ekjon veto bangali, vietnami vasha jani na, doya kore ektu keu onubaad kore deben!!] :P", "Solution_5": "[quote=\"ZHANGWENZHONGKK\"]$ 1 \\minus{} a^6 b^6 \\; \\ge \\;a^2 b^3 (1 \\minus{} a^2 ) \\plus{} a^3 b(1 \\minus{} b^4 ) \\Leftrightarrow \\frac {1}{a} \\plus{} \\frac {1}{{b^2 }} \\plus{} a^3 b^3 \\le a \\plus{} b^2 \\plus{} \\frac {1}{{a^3 b^3 }}$.\nThat seems easier.[/quote]\r\nBut that was where I took the problem from!\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=219432\r\n\r\nI narrowed it down to the case $ a\\le b\\le1\\le c$, which is equivalent to this problem.\r\n\r\nSo if this problem is solved, so is the other inequality! [img]http://i41.photobucket.com/albums/e271/Carduelis_carduelis/Tongue.gif[/img]\r\n\r\nThanks fjwxcsl and can_hang2007 for your solutions! :omighty:", "Solution_6": "[quote=\"Mary Barton\"][quote=\"ZHANGWENZHONGKK\"]$ 1 \\minus{} a^6 b^6 \\; \\ge \\;a^2 b^3 (1 \\minus{} a^2 ) \\plus{} a^3 b(1 \\minus{} b^4 ) \\Leftrightarrow \\frac {1}{a} \\plus{} \\frac {1}{{b^2 }} \\plus{} a^3 b^3 \\le a \\plus{} b^2 \\plus{} \\frac {1}{{a^3 b^3 }}$.\nThat seems easier.[/quote]\nBut that was where I took the problem from!\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=219432\n\nI narrowed it down to the case $ a\\le b\\le1\\le c$, which is equivalent to this problem.\n\nSo if this problem is solved, so is the other inequality! [img]http://i41.photobucket.com/albums/e271/Carduelis_carduelis/Tongue.gif[/img]\n\nThanks fjwxcsl and can_hang2007 for your solutions! :omighty:[/quote]\r\nhaha, it's equivalent to my inequality but I dont see it. :blush: \r\nBy the way, the original inequality on that link, there still has another simpler solution for it. :)" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ 0 < a_1 < a_2 < ... < a_n$ be integer.A positive integer $ m$ such that: we have find the integer $ 0 < b_1 < b_2 < ... < b_m$ such that:\r\n $ \\sum_{k \\equal{} 1}^{n}2^{a_k} \\equal{} \\sum_{k \\equal{} 1}^{m}b_k$ and $ \\prod_{k \\equal{} 2}^{n}(2^{a_k})! \\equal{} \\prod_{k \\equal{} 1}^{m}b_k!$\r\nProve that: $ n \\equal{} m$ and $ b_i \\equal{} 2^{a_i}$ $ \\forall i \\equal{} 1,2,...,n$", "Solution_1": "It's quite easy.\r\nFrom the second indenty, we have:\r\n$ \\begin{array}{l}\r\n \\sum\\limits_{k \\equal{} 1}^n {v_2 (2^{a_k } !) \\equal{} \\sum\\limits_{k \\equal{} 1}^m {v_2 (b_k !)} } \\\\ \r\n \\Leftrightarrow n \\equal{} \\sum\\limits_{k \\equal{} 1}^m {s_2 (b_k )} \\\\ \r\n \\end{array}$. \r\nThus $ n\\equal{}m.$ Because $ 00 \\rightarrow f(x)>0}$ and $ x_{0}=3$ we take $ x_{\\infty}=x_{\\infty,2}=\\sqrt {\\frac {3}{2}}$", "Solution_4": "milin, that's a clever technique! I'll keep it in mind. :)", "Solution_5": "Tnx.. :) The theory around it is something I didn't have time to do (and probably wouldn't know how to do), but for nice, continues functions it work's OK!", "Solution_6": "[quote=\"milin\"]Tnx.. :) The theory around it is something I didn't have time to do (and probably wouldn't know how to do), but for nice, continues functions it work's OK![/quote]\r\nIt's called the Newton-Raphson method, and will work for almost any differentiable function in a suitable region around the zero you are approximating to.\r\n\r\nAnother way of solving this question would be to use the Contraction Mapping theorem on a suitable interval.", "Solution_7": "Can anyone give some additional information (or post a link) on how the Newton Raphson method is rigorously justified in application to problems with sequences, such as this one. What I'm seeing doesn't have to do with sequences so I'm a little confused about motivation and the proof that it works.\r\n\r\nBut it is beautiful. Such a handy thing! Where did you learn about it, Milin?", "Solution_8": "It's a standard method for finding solutions for equations, and it's very hard to use for anything in proofs. You always get convergence on some interval when trying to solve $ f(x)\\equal{}0$ for a nice enough function $ f$- but the size of that interval depends on $ f'$ and $ f''$. The Banach contraction principle gives convergence on the largest symmetrical interval around a root where $ \\left|\\frac{f(x)f''(x)}{(f'(x))^2}\\right|\\le 1$.\r\n\r\nThe reason it's a useful method is that the convergence is [i]fast[/i] once you're close enough.\r\n\r\nThis particular application of the method (finding square roots) is also very old; it goes back to the Babylonians. In this case, the convergence is much more reliable and works from any starting point, since you're above the root after one step, and it's a contraction out to $ \\infty$ on that side.", "Solution_9": "milin can you give more information about your method pls ?", "Solution_10": "Ok, I don't want to sound stupid, but I didn't found out about it.. I made it up by my self.. So, sorry.. I can't give any clues in where to find some informations about is.", "Solution_11": "The general Babylonian square root iteration for $ \\sqrt{a}$: Start with any positive $ x_0$ (preferably close to $ \\sqrt{a}$), and let $ x_{n\\plus{}1}\\equal{}\\frac{x_n\\plus{}\\frac{a}{x_n}}{2}$. This sequence always converges to $ \\sqrt{a}$, and the convergence is rapid once we get close; $ |x_{n\\plus{}1}\\minus{}\\sqrt{a}|\\le C|x_n\\minus{}\\sqrt{a}|^2$ for some constant $ C$.\r\n\r\nThis is a specialization of Newtons method; this problem behaves better than the general case, since it doesn't blow up for large $ x$.\r\n\r\nFor $ a\\equal{}\\sqrt{\\frac32}$, this iteration would be $ x_{n\\plus{}1}\\equal{}\\frac{x_n\\plus{}\\frac{3}{2x_n}}{2}\\equal{}\\frac{2x_n^2\\plus{}3}{4x_n}$. The problem's iteration is an extrapolation; $ \\frac43$ of this and $ \\minus{}\\frac13$ of staying put. That drastically slows convergence; the new error is about $ \\minus{}\\frac13$ of the old, rather than its square.\r\nIt's easy to prove local convergence; that's all about the derivative of the itaration function at the fixed point, which is less than $ 1$ in absolute value. To make it work globally, we need more about the derivative so we can apply the contraction principle on a larger interval.\r\n\r\nLet $ f(x)\\equal{}\\frac{x^2\\plus{}3}{3x}\\equal{}\\frac{x}{3}\\plus{}\\frac1x$; $ f(x)\\ge\\frac{2}{\\sqrt{3}}$ for all $ x>0$, with equality iff $ x\\equal{}\\sqrt{3}$. We have $ f'(x)\\equal{}\\frac13\\minus{}\\frac1{x^2}$, so $ \\minus{}1\\le f'(x)\\le \\frac13$ for all $ x$ in the range of $ f$. That's almost enough; if that $ \\minus{}1$ were $ c$ for any $ c<1$, we could apply the Mean Value Theorem and the contraction principle directly.\r\n\r\nIn this case, the simplest fix is probably to look at $ f(f(x))\\equal{}\\frac{x}{9}\\plus{}\\frac1{3x}\\plus{}\\frac{3x}{x^2\\plus{}3}$ instead; its derivative is less than $ \\frac19\\plus{}\\frac34$ for all $ x>1$, which includes all $ x$ in the range of $ x$." } { "Tag": [], "Problem": "A rapidly growing tree grows 24 inches in one week. On average, how many inches does the tree grow in one hour? Express your answer as a common fraction.", "Solution_1": "There are 7 days in a week and 24 hours in a day, so the answer is $ 24/(24*7)\\equal{}\\boxed{1/7}$" } { "Tag": [ "geometry", "integration", "calculus", "calculus computations" ], "Problem": "Use Green's Theorem to find the area of the region between the graphs of $ y \\equal{} x^2$ and $ y \\equal{} x^3$.\r\n\r\nSo $ A \\equal{} \\oint_{C} x \\ dy \\equal{} \\minus{} \\oint_{C} y \\ dx \\equal{} \\frac{1}{2} \\oint_{C} x \\ dy \\minus{} y \\ dx$\r\n\r\nI used $ A \\equal{} \\minus{} \\oint_{C} y \\ dx$\r\n\r\nSo $ A \\equal{} \\minus{} \\int_{1}^{0} x^2 \\minus{} x^3 \\ dx \\equal{} \\frac{1}{12}$\r\n\r\nIs this the correct way of doing the problem?", "Solution_1": "How about this way:\r\n$ A\\equal{}\\frac{1}{2}\\oint_{\\partial D}\\minus{}y\\mbox{ d}x\\plus{}x\\mbox{ d}y\\equal{}\\frac{1}{2}\\iint_{D}2\\mbox{ d}x\\mbox{ d}y$\r\nSo we need to find the intersection points, which are $ (0,0)$ and $ (1,1)$. So then we have\r\n$ \\frac{1}{2}\\int_{0}^{1}\\int_{x^3}^{x^{2}}2\\mbox{ d}x\\mbox{ d}y\\equal{}\\int_{0}^{1}y^{2}\\minus{}y^{3}\\mbox{ d}y\\equal{}\\frac{1}{12}$\r\nTwo different methods, same result.", "Solution_2": "[b]ragingbullrocky[/b]: your solution is correct, although on a test I'd like to see a little explanation of the limits and signs. \r\n\r\n[b]JRav[/b]: you did not really use Green's theorem - you just calculated the double integral that represents the area.", "Solution_3": "My solution is an application of Green's Theorem (at least according to my textbook, in which this formula is listed in the section devoted to Green's Theorem).", "Solution_4": "This is actually an interesting point. If an exam problem asks to compute a double integral (such as $ \\iint_D 1\\,dA$) using Green's theorem, should a solution with $ \\iint_D 1\\,dA\\equal{}\\minus{}\\oint_C y\\,dx\\equal{}\\iint_D 1\\,dA$ count? After all, Green's theorem was used - even twice!", "Solution_5": "Probably not, because its probably means using only Green's Theorem once and using a line integral.", "Solution_6": "I guess its at the discretion of the professor then." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $a,b,c$ be distinct real numbers such that\r\n$a^4+b^4+c^4+ab^3+bc^3+ca^3= 2(a^3b+b^3c+c^3a)$ .\r\nProve that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=0$.", "Solution_1": "what if a=b=-2 and $c=\\sqrt{5}+1$?", "Solution_2": "[quote=\"Vasc\"]Let $a,b,c$ be distinct real numbers such that\n$a^4+b^4+c^4+ab^3+bc^3+ca^3= 2(a^3b+b^3c+c^3a)$ .\nProve that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=0$.[/quote]\n\nThe equation $a^4+b^4+c^4+ab^3+bc^3+ca^3=2(a^3b+b^3c+c^3a)$ rewrites as $(a^4+b^4+c^4)+(b^3a+c^3b+a^3c)-2(a^3b+b^3c+c^3a)=0$. Because of the identity\n\n$(a^4+b^4+c^4)+(b^3a+c^3b+a^3c)-2(a^3b+b^3c+c^3a)$\n$=\\frac12((a^2-b^2+bc-ab)^2+(b^2-c^2+ca-bc)^2+(c^2-a^2+ab-ca)^2)$,\n\nthis equation is equivalent to $(a^2-b^2+bc-ab)^2+(b^2-c^2+ca-bc)^2+(c^2-a^2+ab-ca)^2)=0$. But since the numbers a, b, c are real and squares of reals are always $\\geq 0$, this yields $a^2-b^2+bc-ab=0$, $b^2-c^2+ca-bc=0$ and $c^2-a^2+ab-ca=0$. Now, $a^2-b^2+bc-ab=0$ rewrites as $a^2-ab=b^2-bc$, or, equivalently, a (a - b) = b (b - c), and similarly we get b (b - c) = c (c - a). Thus, a (a - b) = b (b - c) = c (c - a). If we denote a (a - b) = b (b - c) = c (c - a) = k, we thus have $a-b=\\frac{k}{a}$, $b-c=\\frac{k}{b}$, $c-a=\\frac{k}{c}$, so the trivial equality (a - b) + (b - c) + (c - a) = 0 becomes $\\frac{k}{a}+\\frac{k}{b}+\\frac{k}{c}=0$, so that $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=0$, and we are done.\n\nActually, the above argument is not 100% complete, since difficulties might occur in the case when k = 0. However, this case is easy to filter out: Since a (a - b) = b (b - c) = c (c - a) = k, when k = 0, we must have a (a - b) = b (b - c) = c (c - a) = 0. Thus, either the numbers a - b, b - c and c - a are all 0, what yields a = b = c and thus contradicts the condition that the numbers a, b, c are distinct, or (at least) one of the numbers a, b, c is 0; in let's WLOG assume that a = 0. In this case, the equation a (a - b) = b (b - c) = c (c - a) = 0 becomes 0 (0 - b) = b (b - c) = c (c - 0) = 0; hence, $c^2=c(c-0)=0$, so that c = 0, and thus b (b - c) = 0 becomes $b^2=0$, so that b = 0, what leads to a = b = c = 0, what is impossible again since a, b, c must be distinct. Thus, the proof is completed.\n\n[quote=\"siuhochung\"]what if a=b=-2 and $c=\\sqrt{5}+1$?[/quote]\r\n\r\nThis triple doesn't satisfy $a^4+b^4+c^4+ab^3+bc^3+ca^3=2(a^3b+b^3c+c^3a)$, does it?\r\n\r\n Darij", "Solution_3": "My solution is just same.\r\nCongratulations Darij! :)", "Solution_4": "[quote=\"Vasc\"]My solution is just same.\nCongratulations Darij! :)[/quote]\r\nBut if i cannot find mr. stefan's identity, what can i do? :?" } { "Tag": [ "geometry" ], "Problem": "Hello everyone. \r\n\r\nFor years, geometry has been most challenging me of all the subjects in olympiad-style questions. What good books are out there that go over Olympiad-level geometry concepts and have good explanations and problems? \r\n\r\nThank you so much for your help. \r\n\r\nJB", "Solution_1": "Theres a good book on how to...\r\n\r\nnevermind.\r\n\r\nI can't even try to make this look like a real post. See you tomorrow.\r\n\r\nSeahawks v. Steelers!\r\nSUPER BOWL EXTRA LARGE! (XL)", "Solution_2": "For online stuff, try this:\r\n\r\nhttp://math.mit.edu/~kedlaya/geometryunbound/\r\n\r\nFor TONS AND TONS of problems, you can dig through the files here:\r\n\r\nhttp://www.math.su.se/~mleites/Prasolov/\r\n\r\nA compiled version of the plane geo:\r\n\r\nhttp://students.imsa.edu/~tliu/Math/planegeo.pdf\r\n\r\nYou've probably heard of books like Geometry Revisited, etc -- those are pretty good too. The AoPS class is also pretty amazing. Hmmm.. yeah hope that helps. :)" } { "Tag": [ "ratio", "geometry", "similar triangles", "power of a point" ], "Problem": "circle C has a radius of 4 and circle A has a radius of 6. if the length of segment AC is 15, and segment BE is tangent to both circles, what is the length of segment BE?\r\n\r\ncan i check my answer?\r\n\r\n[hide=\"answer?\"]$5\\sqrt{5}$[/hide]", "Solution_1": "could someone post a correct diagram? im not sure im drawing this out correctly...\r\n\r\nAC is a line segment connecting the centers of circles A and C correct? there is more than one way to have a line tangent to both circles it seems, are we looking for the shortest?", "Solution_2": "There are both internal and external tangents :( , but we can find both :) \r\n\r\nWhat iamagenius gives as an answer is for the internal tangent:\r\n\r\n[hide=\"internal\"]Sketch the diagram and create two right triangles that are similar by Vertical Angle Theorem. Two triangles have side ratio of $4/6=2/3$. Hypotenuse has length $15 \\cdot \\frac{2}{5}=6$ and $15 \\cdot \\frac35=9$ each. So the other legs have length $\\sqrt{36-16}$ and $\\sqrt{81-36}$. Add these together to get $2\\sqrt{5}+3 \\sqrt{5}=\\boxed{5 \\sqrt{5}}$.[/hide]\nbut there is also external one\n[hide=\"external\"]Again, sketch the diagram and draw radii segments perpendicular to the external tangent. Draw altitude from $C$ to the radius drawn from $A$ perpendicular to the tangent. If we let that point $H$, we can find the length of $CH$ from $\\triangle ACH$, which is equivalent to $\\sqrt{15^{2}-2^{2}}=\\sqrt{221}$. $221=13 \\cdot 17$ has no square factors , so the answer is just $\\boxed{\\sqrt{221}}$. [/hide]", "Solution_3": "okay I did this out and got the same answer as the original poster. I used the fact that there were similar triangles and then I used power of a point to find the segment tangent length..\r\n\r\nkohjhsd your solution seems much simpler though, and Id like to know your reasoning behind how you found the hypotenuse length so easily, where did $\\frac{3}{5}$ and $\\frac{2}{5}$ come from? and then you find the other lengths so simple.. is it a case of not showing your work or is there another theorem you used that I am unaware of?\r\n\r\nthanks :D", "Solution_4": "[quote=\"Glen Mann\"]okay I did this out and got the same answer as the original poster. I used the fact that there were similar triangles and then I used power of a point to find the segment tangent length..\n\nkohjhsd your solution seems much simpler though, and Id like to know your reasoning behind how you found the hypotenuse length so easily, where did $\\frac{3}{5}$ and $\\frac{2}{5}$ come from? and then you find the other lengths so simple.. is it a case of not showing your work or is there another theorem you used that I am unaware of?\n\nthanks :D[/quote]\r\n\r\nWell, $\\frac35$ and $\\frac25$ is from the similarity. Because the triangles have ratio of sides 2 to 3, their hypotenuses will also be in ratio 2/3. Because the sum is 15, one is 2/5 and the other is 3/5. There's no need to apply these fancy power of a point methods, I think.\r\n\r\nThe hypotenuse lengths are just from Pythagorean Theorems. Nothing unusual.\r\n\r\nHowever, this is the faster way for the internal one:\r\n\r\n[hide]There are 2 similar triangles. Extend the radius 4 to have radius 10, and move the 15 segment down and you have a right triangle whereas you can find the hypotenuse with simple $5 \\sqrt{3^{2}-2^{2}}$.\n\nSee picture below.\n[/hide]", "Solution_5": "oh right, right angle triangles are formed.. DOH!\r\n\r\n:blush:\r\n\r\nyour second method is much quicker", "Solution_6": "great! sorry, i was referring to the external tangent. the problem has a diagram with it, but i couldn't provide it :(" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all the function $ f: R\\minus{}\\minus{}\\minus{}>R$ such that:\r\n$ f(f(x)\\plus{}y)\\equal{}f(x^2\\minus{}y)\\plus{}4f(x)y$", "Solution_1": "Setting $ y\\equal{}\\frac{x^2\\minus{}f(x)}{2}$ yields (*) $ 0\\equal{}f(x)[x^2\\minus{}f(x)]$. Assume for a moment the existence of some $ z,w\\not\\equal{}0$ with $ f(z)\\equal{}0,f(w)\\not\\equal{}0$. (*) and the original equation imply $ 0\\not\\equal{}w^2\\equal{}f(w)\\equal{}f(f(z)\\plus{}w)\\equal{}f(z^2\\minus{}w)\\Longrightarrow (z^2\\minus{}w)^2\\equal{}w^2\\Longrightarrow 2w\\equal{}z^2$. \r\n\r\nHence, either the set $ \\Omega\\equal{}\\{x\\not\\equal{}0|f(x)\\equal{}0\\}$ or the set $ \\Omega'\\equal{}\\{x\\not\\equal{}0|f(x)\\not\\equal{}0\\}\\equal{}\\{x\\not\\equal{}0|f(x)\\equal{}x^2\\}$ is empty. This corresponds to the solutions $ f(x)\\equal{}x^2$ and $ f(x)\\equal{}0$.", "Solution_2": "[quote=\"hjbrasch\"]Setting $ y \\equal{} \\frac {x^2 \\minus{} f(x)}{2}$ yields (*) $ 0 \\equal{} f(x)[x^2 \\minus{} f(x)]$. Assume for a moment the existence of some $ z,w\\not \\equal{} 0$ with $ f(z) \\equal{} 0,f(w)\\not \\equal{} 0$. (*) and the original equation imply $ 0\\not \\equal{} w^2 \\equal{} f(w) \\equal{} f(f(z) \\plus{} w) \\equal{} f(z^2 \\minus{} w)\\Longrightarrow (z^2 \\minus{} w)^2 \\equal{} w^2\\Longrightarrow 2w \\equal{} z^2$. \n\nHence, either the set $ \\Omega \\equal{} \\{x\\not \\equal{} 0|f(x) \\equal{} 0\\}$ or the set $ \\Omega' \\equal{} \\{x\\not \\equal{} 0|f(x)\\not \\equal{} 0\\} \\equal{} \\{x\\not \\equal{} 0|f(x) \\equal{} x^2\\}$ is empty. This corresponds to the solutions $ f(x) \\equal{} x^2$ and $ f(x) \\equal{} 0$.[/quote]\r\nA nice solution gjbrasch.But i want to ask you a question:\r\n$ 2w\\equal{}z^2$ it's used for ???", "Solution_3": "It is used to show that exactly one of the sets $ \\Omega$ or $ \\Omega'$ is empty, since $ \\forall z\\in\\Omega\\,,\\,w\\in\\Omega': 2w\\equal{}z^2$ holds (and note, that $ \\Omega\\cup\\Omega'\\equal{}\\mathbb{R}^*$)" } { "Tag": [ "function" ], "Problem": "Consider all functions $f$ from the positive integers to the positive integers such that\r\n\r\ni) For each positive integer $m$, there is a unique positive integer $n$ such that $f(n)=m$;\r\nii) For each positive integer $n$, we have\r\n$f(n+1)$ is [b]either[/b] $4f(n)-1$ [b]or[/b] $f(n)-1$.\r\n\r\nFind the set of positive integers $p$ such that $f(1999)=p$ for some function $f$ with properties i) and ii).", "Solution_1": "Should be pretty easy to calculate... there are only two possible functions.", "Solution_2": "I dont think so. It can be different for each $n$ surely?", "Solution_3": "I don't have any experience about functional equations, but I'll try making some considerations...\r\n\r\n [hide] From i) we can deduce that $f$ is injective.\nFrom ii) we can say that either the function is increasing or it's decreasing. In fact, since the function is defined on the positive integers, we see that $f(n+1)=4f(n)-1$ is monotonically increasing, while $f(n+1)=f(n)-1$ is monotonically decreasing. Obviously, $f$ can't be increasing and decreasing at the same time, so we have to distinguish between two cases...\nI) $f$ is increasing\nII) $f$ is decreasing\n[/hide]\r\n\r\nAfter this, I don't know how to proceed... :blush:\r\n\r\nEDIT: Ok, after what tim wrote next I must cancel my (confused!) ideas... :wink:", "Solution_4": "To clarify, for each different $n$ we can choose one of the two possibilities independently, as seamusoboyle suggested.\r\n\r\nThus for any particular function, we might have \r\n\r\n$f(3)=4f(2)-1$ but $f(7)=f(6)-1$" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "I believe that this is more a real analysis problem than complex and that's why i post it here..\r\nI need some help on proving that the complex function f(z)=|z|/(i+Imz) is continuous.\r\nI multiplied the numerator and denom. with the conjugate, but i should know that Imz is continuous..Can i be sure about that or is there any other way?\r\nThank you very much.", "Solution_1": "$ |\\text{Im}(z_1)\\minus{}\\text{Im}(z_2)|\\equal{}|\\text{Im}(z_1\\minus{}z_2)|\\leq |z_1\\minus{}z_2|$." } { "Tag": [ "integration", "conics", "function", "complex analysis", "complex analysis unsolved" ], "Problem": "let $ \\gamma$ be a piecewise smooth curve in disk $ |z| < 1$ then $ \\int_{\\gamma} \\frac {|dz|}{1 \\minus{} |z|^2}$ is called [i]conic lenght(?)[/i] of $ \\gamma$ prove that an analytic function from a disk to it self doesn't increase conic length of curves.", "Solution_1": "I have not heard the term conic length, I had always heard hyperbolic length. (Hopefully they are the same :) ).\r\n\r\nAhlfors' book \"Complex Analysis\" (3rd edition) takes you through the proof on pages 135 and 136. He then concludes by giving a first exercise which is precisely what your question is asking for. It results from using formula (36) on page 136 to compute expression." } { "Tag": [ "topology", "geometry", "3D geometry", "sphere", "function", "calculus", "integration" ], "Problem": "I am trying to write up the solutions to the problems below, but I often find that I have problems explicating clearly to the grader, either analytically or verbally my reasoning, so I was hoping that any comments would serve to refine my approach at reasoning.\r\n\r\n1. Consider a sequence $ (x_{n})$ in the metric space $ \\mathbb{R}$. If $ (x_{n})$ converges in $ \\mathbb{R}$ prove that the sequence of absolute values $ (\\mid x_{n}\\mid)$ converges in $ \\mathbb{R}$. \r\n\r\n[hide]\nmy attempt:\nI can argue this pretty well for positive sequences that converge to a positive limit or a sequence of negative numbers that converges to a negative limit. But for the case where the limit $ x < 0$, how would I show analytically that $ d(x_{n}, x) < d(\\mid x_{n} \\mid , \\mid x \\mid) < \\epsilon$? Should I just write that and then give an example of how this would be true?\n[/hide]\n\n2. A map $ f: M \\rightarrow N$ is open if for each open set $ U \\subset M$, the image set $ f(U)$ is open in $ N$. If $ f$ is open, is it continuous? If $ f$ is an opn , continuous bijection, is it a homeomorphism?\n\n[hide]\nmy attempt:\nWhat would be solid counterexamples for both these questions?\n[/hide]\n\n3.Prove that the 2-sphere is not homeomorphic to the plane.\n\n[hide]\nmy attempt:\nI can see that for some radius r, I claim that the 2-sphere is not homeomorphic because then our 2-sphere is bounded and the plane is not. However, couldn't someone say if my r is infinity then my 2-sphere is not bounded anymore? How can I counter this last case?\n[/hide]\n\n4. Consider the identity map $ id: C_{max} \\rightarrow C_{int}$ where $ C_{max}$ is the metric space $ C([a,b], \\mathbb{R})$ of continuous real valued functions defined on [a,b], equipped with the max metric $ d_{max}(f,g) \\equal{} max \\mid f(x) \\minus{} g(x) \\mid$, and $ C_{int}$ is $ C([a,b],\\mathbb{R})$ equipped with the integral metric, \n\n$ d_{int}(f,g) \\equal{} \\int_{a}^{b} \\mid f(x) \\minus{} g(x) \\mid dx$.\n\nShow that id is a continuous linear bijection.\n\n[hide]\nmy attempt:\nThe linear bijection is easy to prove, but I am having trouble explaining clearly the continuity part. My rough idea was to show that $ \\forall \\epsilon > 0$ and $ \\forall f,g \\in C_{max} \\exists \\delta > 0$ such that $ d_{max}(f,g) \\equal{} max \\mid f(x) \\minus{} g(x) \\mid < \\delta$. I started with showing:\n\n$ d_{int}(id(f),id(g)) \\equal{} \\int_{a}^{b} id(f) \\minus{} id(g) dx \\equal{} \\int_{a}^{b} f \\minus{} g dx \\equal{} [b \\minus{} a] (f_{1} \\minus{} g_{1}) \\cdots (f_{i} \\minus{} g_{i})$ somehow $ < d_{max}(f,g) < \\delta \\equal{} \\epsilon \\cdot [b \\minus{} a]$\n\nI 'm not sure if we can use the Riemannian definition of integrals yet, but this was the only way I could think of showing it.\n[/hide]", "Solution_1": "1. Just use the reverse triangle inequality $ d(|x_n|, |x|) \\leq d(x_n, x)$. You should be able to fill in the details from there.\r\n\r\n2. Make up a simple map $ f$ that maps at least one closed set to an open set to show that openness of a map does not imply continuity.\r\n\r\n3. The sphere is compact while the plane is not. However, I do not know if you have studied compactness in your course.\r\n\r\n4. Use the fact that for a continuous function $ h$ on a compact interval $ [a,b]$, $ \\int_a^b h(x)\\mbox{ }dx \\leq (b \\minus{} a) \\mbox{max }_{t \\in [a,b]} h(t)$ and the result will follow almost immediately.\r\n\r\n5. Use the hide tags:\r\n[code][hide][/hide][/code]\r\n\r\nPost your solutions after considering my hints and I can tell you how they are.", "Solution_2": "1. I was able to figure out, I used triangle inequality as well.\r\n\r\n2. Can I just make up some [a,b] - in $ M$ and call that a closed set within the open set $ M$? Or can I use $ \\mathbb{Q}$ and say that it's closed in $ \\mathbb{R}$- which is open and if $ f$ maps from $ \\mathbb{R} \\rightarrow \\mathbb{R}$, then a mapping from the rationals to the reals would show that it's not continuous? Can I use this same example to disprove that $ f$ being an open, continuous bijection is not a homeomorphism?\r\n\r\n3. Yes, we studied compactness. But how do I show that it's closed and bounded which I think would really help in showing that it's compact. From above, I have a hard time trying to argue the 2-sphere is closed especially if we have a 2 sphere that was formed from a sphere of radius infinity, because then I feel that I can run into the problem, with a 2-sphere of that \"size\" , by being able to map every point on my 2-sphere onto $ \\mathbb{R}^{2}$", "Solution_3": "3. For simplicity, consider the unit sphere. It is closed because it is the inverse image of the closed set $ \\{1\\}$ under the norm induced by the metric, which is a continuous function. It is bounded because it can be contained within the ball of radius $ 2$.", "Solution_4": "The sphere is the inverse image of a one-point set under the continuous function $ \\parallel{}x\\parallel{}$. Therefore the sphere is closed. $ S_{n}$ is also bounded. Therefore it is compact.\r\n\r\nEdit: sorry Zhylliolom, I was writing a reply at the same time you submited your post and it was exactly the same.... :P", "Solution_5": "I see what you're saying with the unit ball, but in general would I have to argue ad infinitum that there will always be a 2-sphere formed from radius r that will always be bounded by a 2-sphere formed from r-1?\r\n\r\nEdit: ok, I will take some time to look into the inverse image.", "Solution_6": "[quote]I can see that for some radius r, I claim that the 2-sphere is not homeomorphic because then our 2-sphere is bounded and the plane is not.[/quote]\r\n\r\nThis is wrong, I mean, if $ A$ is bounded and $ B$ is not, this is not enough to claim that are not homeomorphic, in fact, consider for example $ \\mathbb{I} \\equal{} (0,1)$ and $ \\mathbb{R}$, you can see that $ \\mathbb{I}$ is bounded and is! homeomorphic to $ \\mathbb{R}$, can you find the homeomorphism?", "Solution_7": "But, technically isn't $ I$ an open interval in $ \\mathbb{R}$ ? I thought that means it is open. However if $ I$ were in the $ \\mathbb{Z}$ then it is closed. Perhaps I'm wrong, any thoughts or correction on this would be helpful.\r\n\r\n4. Do I take $ \\delta \\equal{} \\frac {\\epsilon}{[b \\minus{} a]}$? And what do I use to justify this inequality (name of the rule/thm) $ \\int_{a}^{b}h(x)\\mbox{ }dx\\leq (b\\minus{}a)\\mbox{max }_{t\\in [a,b]}h(t)$?", "Solution_8": "yes, $ \\mathbb{I}$ is open in $ \\mathbb{R}$, why this question? \r\nWhat I wanted to mean is that this is not the way of proving that they're not homeomorphic, I think the properly way is by compactness.", "Solution_9": "I guess I associated openness with boundedness, which I now see is not necessarily true.", "Solution_10": "In fact, in a metric space with norm $ \\parallel{}x\\parallel{}$ to be bounded means that you can put it into a sphere (as a posible definition). but for example in $ \\mathbb{R}$, the interval $ [0,1]$ is a subset of the sphere of center $ 0$ and radius $ 10$, so it's bounded, do you get the point? ...", "Solution_11": "Yes, I think I just need some more time to mull over that. I think it's starting to make sense as I pore over the definition of boundedness.\r\n\r\nAre there any thoughts on what I posted above for 2. and 4.?" } { "Tag": [ "inequalities" ], "Problem": "[a^2+ab+b^2][b^2+bc+c^2][c^2+ca+a^2] >= {ab+bc+ca}^3", "Solution_1": "[hide=\"One line solution\"]\nBy Holder's inequality,\n$ (ab\\plus{}a^2\\plus{}b^2)(b^2\\plus{}c^2\\plus{}bc)(a^2\\plus{}ac\\plus{}c^2) \\ge (ab\\plus{}bc\\plus{}ca)^3$[/hide]", "Solution_2": "@akashram if you wanted the proof for this inequality then you can find it at this link: http://www.artofproblemsolving.com/Wiki/index.php/Holder%27s_Inequality", "Solution_3": "Ive never used Holders. Please provide details on the transformation a what you set the variables to be :blush:", "Solution_4": "To avoid using confusing notation,\r\nHolder's says that if you have n parentheses of sums you can take the nth root of the product of all the first terms in each parentheses, plus the nth root of the product of the second terms, etc, add it up and take it to the nth power and it will be less than the original.", "Solution_5": "I see... Now I just have to put it into good use." } { "Tag": [], "Problem": "Hey here are some problems that i am enjoying from a book our teacher gave us to do out of class like in spare time. see if you find them difficult:\r\n\r\n[b]1. [/b]Jack, John and Jill go up the hill with five empty pails to a place where three springs are. Each of the five pails has a $8$ litre capacity. One of the springs gives $2$ litres of water in one minute while each of the other two give $1$ litre of water in one minute. It is possible to use one spring to fill two pails at the same time, and it takes less than $2$ minutes but more than $1$ minute to take a pail from one spring to another one. What is the shortest time Jack, John, and Jill need to fill all five pails to their capacity? \r\n\r\n[b]2. [/b]When Adam saw the diagram below, he wrote a number in each of the empty circles so that the sum of all the numbers in the circles equals 25. Then he noticed that whenever three circles are such that every two of them are connected with a single line, the numbers in these three circles add up to 12. What number did he put in the central circle? Find all possible answers. [b](refer to the picture i have attatched).[/b]\r\n\r\n[b]3. [/b]When a positive integer $N$ is written in base $9$, it is a two-digit number. When $6N$ is written in base $7$, it is a three-digit number obtained from the two-digit number by writing digit $4$ to its right. Find the decimal representations of all such numbers $N$.\r\n\r\n[b]4.[/b] Find all pairs of positive integers $a$ and $b$ such that\r\n $a^2 + b^2 - 7 = ab$\r\n\r\n[b]5.[/b] A bus route from town $A$ to town $D$ goes along a straight line and passes first through town $B$ and then through town $C$. Ann has a positioning device that can tell her the distance from her current location to any of the four towns on the route. When she is somewhere between $A$ and $B$, from the information supplied by the device Ann learns that the sum of the distances to $A$, to $B$ and to $D$ is greater than the distance to $C$ by $5.9$ km. When Ann is between $B$ and $C$, she finds out that the sum of the distances to all four towns in $16.7$ km. When she is between $C$ and $D$, Ann discovers that the sum of the distances to $B$ and to $D$ is $8.1$ km. What is the distance between $C$ and $D$?\r\n\r\nThese are meant to be cryptic and challenging, lets see how you all go :) \r\n\r\nEnjoy,\r\nKesh[img]My%20Documents\\question%202%20pic[/img]", "Solution_1": "Yor diagram hasn't come. Can you please post it!!!", "Solution_2": "[quote=\"keshin\"]\n[b]4.[/b] Find all pairs of positive integers $a$ and $b$ such that\n $a^2 + b^2 - 7 = ab$\n[/quote]\r\n\r\nI 'll try this problem... :) \r\nrewrite the given equation as $(a-b)^2+a^2+b^2=14$\r\nso $a^2=1$ or $4$ or $9$ and then you find the pairs easily.. :)", "Solution_3": "And how did you conclude the last step????", "Solution_4": "because we must have $a^2\\leq14$ and $a \\in N$..... :) \r\nok?", "Solution_5": "[quote=\"socrates\"]because we must have $a^2\\leq14$ and $a \\in N$..... :) \nok?[/quote]\r\n\r\n$\\displaystyle (a-b)^2+a^2+b^2=14$\r\nadd that all out :P" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Determine all pairs of positive integers $ a$ and $ n$ for which each prime divisor of $ a^{\\phi (n)} \\minus{} 1$ divides $ n$", "Solution_1": "[quote=\"jbmorgan\"]Determine all pairs of positive integers $ a$ and $ n$ for which each prime divisor of $ a^{\\phi (n)} - 1$ divides $ n$[/quote]\r\nHmm may be I have some mistakes ... \r\nOnly need to consider $ a > 1$\r\nThe first we have an old result : Let $ p$ is an odd prime divisor of $ a^{2^k} + 1$ then $ 2^{k + 1}|p - 1$\r\nIt guesses us to consider prime divisors of number ${ a^{\\frac {\\varphi(n)}}{2}} + 1$ . \r\nStep 1 : $ n$ has at most one odd prime divisor . If not we consider a odd prime divisor q of $ a^{\\frac {\\varphi(n)}{2}} + 1$ (Notice that if $ a^2 + 1$ can't be a power of 2 for $ a > 1$ .From the lemma we have : ${ v_2(q - 1)\\geq v_2(\\varphi(n))\\geq \\sum_{p|n} v_2(p - 1}$ it gives contradiction . \r\nStep 2 : Similarity to step 1 ,it is easy to show that $ n$ must be one of following forms $ {p,2p}$\r\nNow consider two case \r\ni) $ n = p$ then we must have :\r\n\\[ a^{p - 1} - 1 = p^k\r\n\\]\r\nIt is not hard to shows that $ p$ must be equal 2 or 3 . \r\nFor the case $ p = 2$ we have $ a = 2^{t} + 1$\r\nFor the case $ p = 3$ then we have $ a = 2$\r\nii ) $ n = 2p$ then we have :\r\n\\[ a^{p - 1} - 1 = 2^k.p^t\r\n\\]\r\n\\[ \\equiv alent (a^{\\frac {p - 1}{2} + 1)(a^{\\frac {p - 1}{2}} - 1) = 2^k.p^t\r\n}\\]\r\nThis equation can be solved by some calculations ..." } { "Tag": [ "topology", "function", "advanced fields", "advanced fields unsolved" ], "Problem": "Let g:X->Y be a continuous mapping of a separable metric space X onto a compact Hausdorff space Y. Show that Y is metrizable. Also what is an example of a continuous mapping of a separable metric space onto a non-metrizable normal space?\r\n\r\nAl", "Solution_1": "First the counterexample:\r\n\r\nAny countable topological space is a continuous image of a separable metric space, namely a discrete countable space. This means that in order to find a counterexample, it suffices to find a countable $T_{4}$ (normal and Hausdorff) space $X$ which is not metrizable. \r\n\r\nLet $X$ be the set of non-negative integers. All points are open except for $0$, and the neighborhoods of $0$ are $\\{0\\}\\cup A$, as $A$ ranges through the elements of a [url=http://en.wikipedia.org/wiki/Ultrafilter#Types_and_existence_of_ultrafilters]free ultrafilter[/url] on $X\\setminus\\{0\\}$. $X$ is clearly Hausdorff, since all points different from $0$ are open. It's also normal, because among any two disjoint closed sets, one will also be open, being contained in $X\\setminus\\{0\\}$. Finally, $X$ is not metrizable because it's not first-countable at $0$, as a simple argument shows.\r\n\r\n\r\nNow the first part (the proof of the fact that compact Hausdorff continuous images of separable metric spaces are metrizable):\r\n\r\nThe conclusion follows from these two results:\r\n\r\n\r\n[i]Proposition 1[/i]\r\n\r\nA compact Hausdorff space $X$ is metrizable if the diagonal $\\Delta=\\{(x,x)\\ |\\ x\\in X\\}$ is a $G_\\delta$ subset of $X\\times X$. \r\n\r\n(the converse is also true, but that's obvious)\r\n\r\nEven though this is not difficult, I'll just outline the proof (just don't feel like typing that much :)). As is well-known, a compact Hausdorff space is metrizable iff it has a countable basis of open sets. If $\\Delta=\\bigcap_{n\\ge 1}G_{n}$ where $G_{n}$ are open in $X\\times X$, take, for each $n$ and each $x\\in X$, and open neighborhood $U_{n,x}$ of $x$ such that $x\\times U_{n,x}\\subset G_{n}$. For each $n$ the sets $U_{n,x},\\ x\\in X$ cover $x$, so we can extract a finite subcovering. The union of all these finite coverings as $n$ ranges over the positive integers is a countable family of open sets, and the idea is to prove that it's actually a basis. \r\n\r\n\r\n[i]Proposition 2[/i]\r\n\r\nIf $f: T\\to Y$ is a continuous function of the separable metric space $T$ onto the T_4 topological space $Y$, then every closed subset of $Y$ is a $G_\\delta$. \r\n\r\n[i]Proof[/i]\r\n\r\nLet $F$ be closed in $Y$, and let $(F_\\alpha)$ be a family of closed neighborhoods of $F$ in $Y$ with intersection $Y$ (such a family does exist, because $Y$ is T_4; less than the T_4 condition may be enough, but I'm not interested in that :)). Now $f^{-1}(F)$ is a closed subset of $T$, and $(f^{-1}(F_\\alpha))$ is a family of closed neighborhoods of $f^{-1}(F)$ with intersection $f^{-1}(F)$. The open complements $D_\\alpha=T\\setminus f^{-1}(F_\\alpha)$ cover the separable and hence Lindelof metric space $T\\setminus f^{-1}(F)$, so countably many cover it. If these are, say, $D_{n},\\ n\\in\\mathbb N$, then $F$ is the intersection of the closed neighborhoods $f(T\\setminus D_{n})$ (these are some of the $F_\\alpha$ we considered initially) and hence of the interiors of $f(T\\setminus D_{n})$. This makes $F$ a $G_\\delta$, as desired.\r\n\r\n\r\nIn order to complete the proof, consider the continuous map $g: S\\to X$ of the separable metric space $S$ onto the compact Hausdorff space $X$. It induces, in an obvious way, a continuous onto map $f: S\\times S\\to X\\times X$. Apply Proposition 2 to $T=S\\times S$ (which is clearly separable metrizable) and $Y=X\\times X$ to conclude that every closed subset of $X\\times X$ is a $G_\\delta$. Since $X$ is Hausdorff, $\\Delta$ is closed in $X\\times X$ and hence a $G_\\delta$. Now an application of Proposition 1 finishes the whole thing." } { "Tag": [], "Problem": "Crystal has 3 nickels, 2 dimes, and 1 quarter. How many different sums of money can Crystal obtain by using exactly four of these coins?\r\n\r\nBob has 3 pennies 2 nickels and 1 quarter. How many different sums of money can he make using these coins?\r\n\r\nWhat are the odds of you flipping two heads and a tails with a coin?", "Solution_1": "[quote=\"Farazy\"]What are the odds of you flipping two heads and a tails with a coin?[/quote]\r\n\r\n[hide=\"the answer is\"] 1/8?[/hide]", "Solution_2": "[quote=\"Farazy\"]Crystal has 3 nickels, 2 dimes, and 1 quarter. How many different sums of money can Crystal obtain by using exactly four of these coins?\n\nBob has 3 pennies 2 nickels and 1 quarter. How many different sums of money can he make using these coins?\n\nWhat are the odds of you flipping two heads and a tails with a coin?[/quote]\r\n\r\nIn exactly 3 flips, right?", "Solution_3": "Yeah, three filps. I know the answer...\r\n\r\n[hide=\"Here it is\"]3/8[/hide]\r\n\r\nBUT\r\n\r\nThere should be a really quick method of solving this." } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "Let $A,B$ be two matrices with eigenvalues being in ${\\mathbb R}_+$ and $det(AB)=1$. Show that $Tr(A) + Tr(B) \\geq 2n$.\r\n :ninja:", "Solution_1": "Yes it is nice \r\n\r\n$a_k,b_k$ eigenvalues of $A,B$ \r\n\r\n$a_1...a_nb_1...b_n=1$ \r\n\r\n$\\frac{tr(A)+tr(B)}{n}=\\frac{a_1+...+a_n}{n}+\\frac{b_1+...+b_n}{n}\\geq \r\n(a_1...a_n)^{1/n}+(b_1...b_n)^{1/n}\\geq 2\\sqrt{(a_1...a_n)^{1/n}(b_1...b_n)^{1/n}}=2$ \r\n\r\nremind $x+y\\geq 2\\sqrt{xy}$ if $x,y\\geq 0$", "Solution_2": "no is to easy :D :P :lol: :blush:", "Solution_3": "i edited", "Solution_4": "[quote=\"Moubinool\"]$a_1...a_nb_1...b_n=1$ \n[/quote]\r\n\r\nI can't understand that ?", "Solution_5": "He just used that $det(AB)=detA detB$, alekk." } { "Tag": [ "trigonometry", "circumcircle", "geometry", "trig identities", "Law of Sines" ], "Problem": "If one side of a triangle is $ 12$ inches and the opposite angle is $ 30$ degrees, then the diameter of the circumscribed circle is:\r\n$ \\textbf{(A)}\\ 18\\text{ inches} \\qquad\\textbf{(B)}\\ 30\\text{ inches} \\qquad\\textbf{(C)}\\ 24\\text{ inches} \\qquad\\textbf{(D)}\\ 20\\text{ inches}\\\\\r\n\\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide]By the extended Law of Sines, we have $ \\frac{a}{\\sin A}\\equal{}2R$, so, $ \\frac{12}{\\frac{1}{2}}\\equal{}24\\equal{}2R$. So, the diameter is $ 24$, or $ \\textbf{(C)}$[/hide]", "Solution_2": "[hide=\"Without trig\"]\nDraw a diagram and label the triangle $ ABC$, where $ B$ is the vertex of the $ 30 ^\\circ$ angle and $ AC\\equal{}12$. We know that arc $ AC\\equal{}60^\\circ$, so $ AC\\equal{}r \\implies 2r\\equal{}d\\equal{}24$, or $ \\boxed{\\textbf{(C)}}$.\n[/hide]" } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "Calculate the sum $\\frac{1}{1.2.3.4}+\\frac{1}{2.3.4.5}+\\frac{1}{3.4.5.6}+\\frac{1}{4.5.6.7}+...+\\frac{1}{2004.2005.2006.2007}$", "Solution_1": "$\\frac{1}{n(n+1)(n+2)(n+3)}={1\\over 6n}-{1\\over 2(n+1)}+{1\\over 2(n+2)}-{1\\over 6(n+3)}$ using [url=http://en.wikipedia.org/wiki/Partial_fraction]partial fraction decomposition[/url].\r\nThe rest is easy.\r\nThe answer is ${1\\over 18}-{1\\over 3\\cdot 2005\\cdot 2006\\cdot 2007}$.", "Solution_2": "Or by $\\frac{1}{n(n+1)(n+2)(n+3)}=\\frac{1}{3}(\\frac{1}{n(n+1)(n+2)}-\\frac{1}{(n+1)(n+2)(n+3)})$ and telescopic sum.", "Solution_3": "olorin's is textbook method, while n.t.tuan's is simpler solution using \"insight method\"... nice solution n.t.tuan (it generalizes, right?)", "Solution_4": "[quote=\"me@home\"]olorin's is textbook method, while n.t.tuan's is simpler solution using \"insight method\"... nice solution n.t.tuan (it generalizes, right?)[/quote]\r\n$\\frac{1}{n(n+1)(n+2)...(n+k)}=\\frac{1}{k}(\\frac{1}{n(n+1)...(n+k-1)}-\\frac{1}{(n+1)(n+2)...(n+k)})$ :D", "Solution_5": "How about $\\frac{1}{[n(n+1)(n+2)...(n+k)}]^{2})$\r\n\r\n(note. Can you always find the sum of squares of a sequence given its sum of linears? Third powers? etc? (I think you can't))", "Solution_6": "No, I Can't :D", "Solution_7": "I have question :\n\n$ \\sum_{i=0}^{n}\\frac{1}{(4i+1)(4i+2)(4i+3)(4i+4)} $", "Solution_8": "[quote=\"smart of math\"]I have question :\n\n$ \\sum_{i=0}^{n}\\frac{1}{(4i+1)(4i+2)(4i+3)(4i+4)} $[/quote]\n\nFor $n=\\infty$ my CAS yields $\\frac{1}{24}\\left(6\\ln 2-\\pi\\right)$, so I don't expect that there would be a closed form for $n$ finite (unless you can find a closed form whose $n\\to\\infty$ limit is $\\frac{1}{24}\\left(6\\ln 2-\\pi\\right)$).", "Solution_9": "here is his original post : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=427525&p=2422336#p2422336\nhe asks for the limit too! so, i think you have the solution he is asking for." } { "Tag": [ "search", "inequalities unsolved", "inequalities" ], "Problem": "$(a,b\\geq0)$\r\n\r\nprove $\\frac{a+\\sqrt{ab}+b}{3}\\leq\\sqrt{\\frac{(a^{\\frac{2}{3}}+b^{\\frac{2}{3}})^3}{8}}$\r\n\r\n :roll: \r\nplz help", "Solution_1": "It's here\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/topic-55164.html[/url]\r\n\r\nPlease search before posting" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "let A and B be two n x n complex matrices. let C=AB-BA. if CA=AC , then show that C is nilpotent", "Solution_1": "It follows that each C^k is a commutator, where k>0, and hence has trace zero. Thus C is nilpotent.", "Solution_2": "[quote=\"prime\"]It follows that each C^k is a commutator, where k>0, and hence has trace zero. Thus C is nilpotent.[/quote]\r\n\r\nCould you explain more please ??? :?", "Solution_3": "sorry, i dint quite follow your proof because i dont no what commutator means. when u say that $C^k$ is a commutator, what does that mean?", "Solution_4": "For every k>0, one easily checks that C^k=AC^(k-1)B - C^(k-1)BA. Thus TrC^k=0 for all k>0. It is known (and a nice exercise) that this last condition is equivalent to A being nilpotent.", "Solution_5": "[quote=\"prime\"]It follows that each C^k is a commutator, where k>0, and hence has trace zero. Thus C is nilpotent.[/quote]\r\n\r\nI'll try to explain what he probably meant: for every to matrix $A,B$ you can define the matrix $[A,B]=AB-BA$ which is called a commutator of $A,B$. Notice that every matrix which is commutator of some $A,B$ has zero trace since $tr(AB)=tr(BA)$. Now, according to our problem, since $C=[A,B]$ multiply both sides by $C$ from left and use $[A,C]=0$ you'll get $C^2=CAB-CBA=ACB-CBA=[A,CB]$. Now multiply this by $C$ from right and you'll get $C^3=[A,CBC]$ and so on... So you see that $tr(C^k)=0$ for $k=1,2,...,n,...$. Now try to prove that this follows that all \r\nthe eigenvalues of $C$ are zero and thus $C$ is nilpotent.\r\n\r\nEdit. I didn't see that prime replied,", "Solution_6": "Thanks both of you :)", "Solution_7": "ya i got it too." } { "Tag": [ "inequalities" ], "Problem": "[img]http://webwork2.math.utah.edu/math1050fall2007-4/tmp/l2h/set2/42-19754108/19754108img1.gif[/img]", "Solution_1": "This is too easy for Intermediate and should probably go in Classroom Math.\r\n\r\n[hide=\"Solution\"]$ \\frac{2x\\minus{}7}{x\\minus{}5}\\ge3$\n\n$ 2x\\minus{}7\\ge3x\\minus{}15$\n\n$ \\boxed{x\\le8\\text{ and }x\\ne5}$[/hide]", "Solution_2": "$ x\\equal{}0$ doesn't work...your solution set is wrong\r\n\r\nit should be\r\n\r\n$ 50$ \r\n2 case\r\n$ 8\\minus{}x\\leq0$\r\n$ x\\minus{}5<0$\r\nFrom the first case we take $ 5 < x\\leq8$ and the second has no solution\r\nSo the only solution to the problem is\r\n$ 5 < x\\leq8$", "Solution_5": "[quote=\"i_like_pie\"]\n$ \\frac{2x\\minus{}7}{x\\minus{}5}\\ge3$\n\n$ 2x\\minus{}7\\ge3x\\minus{}15$\n\n$ \\boxed{x\\le8\\text{ and }x\\ne5}$[/quote]\r\nActually, you can't multiply by $ x\\minus{}5$ 'cause we don't know if it less or great than zero.", "Solution_6": "[quote=\"Hetidemek\"][quote=\"i_like_pie\"]\n$ \\frac{2x\\minus{}7}{x\\minus{}5}\\ge3$\n\n$ 2x\\minus{}7\\ge3x\\minus{}15$\n\n$ \\boxed{x\\le8\\text{ and }x\\ne5}$[/quote]\nActually, you can't multiply by $ x\\minus{}5$ 'cause we don't know if it less or great than zero.[/quote]\r\nThat's correct :wink:", "Solution_7": "[quote=\"galois01\"]1st case \n$ 8\\minus{}x\\geq0$ \n$ x\\minus{}5\\geq0$ \n2 case\n$ 8\\minus{}x\\leq0$\n$ x\\minus{}5\\leq0$[/quote]\r\n\r\nYou mean $ x\\minus{}5>0$ and $ x\\minus{}5<0$ :wink: :P", "Solution_8": "[quote=\"Anaxerzia\"][quote=\"galois01\"]1st case \n$ 8\\minus{}x\\geq0$ \n$ x\\minus{}5\\geq0$ \n2 case\n$ 8\\minus{}x\\leq0$\n$ x\\minus{}5\\leq0$[/quote]\n\nYou mean $ x\\minus{}5 > 0$ and $ x\\minus{}5 < 0$ :wink: :P[/quote]\r\nThank's for correcting me you are right I forget it I edit immediately :wink: :P", "Solution_9": "[quote=\"i_like_pie\"]This is too easy for Intermediate and should probably go in Classroom Math.\n\n[hide=\"Solution\"]$ \\frac{2x\\minus{}7}{x\\minus{}5}\\ge3$\n\n$ 2x\\minus{}7\\ge3x\\minus{}15$\n\n$ \\boxed{x\\le8\\text{ and }x\\ne5}$[/hide][/quote]\r\n\r\nmaybe it is not so easy :P ?\r\n\r\n\r\n$ \\frac{2x\\minus{}7}{x\\minus{}5}\\geq 3$\r\n\r\n$ 2\\plus{}\\frac{3}{x\\minus{}5}\\geq3$\r\n\r\n$ \\frac{3}{x\\minus{}5}\\geq 1$\r\n\r\nClearly $ 02 has no solution in Z+.\r\n2.With p=3,(1) has a solution x=1,y=1,z=2\r\n3.With p=5,(1) has a solution x=1,y=2,z=9", "Solution_1": "Do you have a solution?", "Solution_2": "yes , and I think this problem is not difficult." } { "Tag": [ "geometry", "circumcircle" ], "Problem": "1. IN TRIANGLE ABC TWO STRAIGHT LINES ARE DRAWN FROM AN ARBITARY POINT D ON SIDE AB WHICH ARE PARRALLEL TO AC AND BC .THESE LINES MEET BC AND AC AT F AND G RESPECTIVELY .PROVE THAT SUM OF CIRCUMFERENCES OF CIRCUMSCRIBED CIRCLES AROUND TRIANGLE ADG AND TRIANGLE BDF IS EUQAL TO THE CIRCUMFERENCES OF THE CIRCUMCIRCLE AROUND THE TRIANGLE ABC.", "Solution_1": "$ \\triangle ACB \\sim \\triangle AGD \\sim \\triangle DFB$\r\n\r\nDenote by $ l(C)$ the length of the circumference of the circle $ C$\r\n\r\n$ \\frac {l(C_{ADG})} {l(C_{ABC})}\\equal{}\\frac {AD} {AB}$\r\n\r\n$ \\frac {l(C_{DFB})} {l(C_{ACB})}\\equal{}\\frac {DB} {AB}$\r\n\r\n$ \\Rightarrow l(C_{ADG})\\plus{}l(C_{DFB})\\equal{}l(C_{ACB})$" } { "Tag": [ "number theory", "greatest common divisor", "modular arithmetic", "quadratics", "number theory solved" ], "Problem": "Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.", "Solution_1": "Posted, more exactly: yesterday!", "Solution_2": "Let a_1, b_1 be positive integers satisfying the required condition.\r\nThen a_1=4a_2 and b_1=4b_2 for some a_2 and b_2 as a_1 and b_1 divisible by 4.\r\nSo smaller a_2 and b_2 also satisfies the condition. \r\nBut there are not infintely many positive integers less than a_1 and b_1, contradiction.", "Solution_3": "We see that d=gcd (a,b) is a power of 2. Let a=dA, b=dB. Cancelling both sides by d and observing, we get A and B are both even, contradiction.", "Solution_4": "Another approach - the system $ 36a \\plus{} b \\equal{} 2^s, 36b \\plus{} a \\equal{} 2^t$ has integer solutions iff $ 36^2 \\minus{} 1 | 36 \\cdot 2^t \\minus{} 2^s, 362^s \\minus{} 2^t$ (just solve the system explicitly).\r\n$ 36^2 \\minus{} 1 \\equal{} 35 \\cdot 37$. $ 35, 37$ are coprime.\r\n$ 35 | 36 \\cdot 2^t \\minus{} 2^s, 36 \\cdot 2^s \\minus{} 2^t \\leftrightarrow 2^{s \\minus{} t} \\equiv 1 \\pmod {35}$\r\n$ 37 | 36 \\cdot 2^t \\minus{} 2^s, 36 \\cdot 2^s \\minus{} 2^t \\leftrightarrow 2^{s \\minus{} t} \\equiv \\minus{} 1 \\pmod {37}$\r\n\r\n$ 2^{s \\minus{} t} \\equiv 1 \\pmod {35} \\implies s \\equiv t \\pmod {12}$ (because the order of $ 2$ is $ 12$)\r\n$ 2^{s \\minus{} t} \\equiv \\minus{} 1 \\pmod {37} \\implies s \\equiv t \\pmod {18}$ but $ s \\neq t \\pmod {36}$ (because the order of $ 2$ mod $ 37$ is $ 36$, and $ 2$ is not a quadratic residue)\r\nYet, $ 12,18 |a \\implies lcm(12,18) \\equal{} 36 |a$, a contradiction.\r\n\r\n[EDIT: Now when I think of it, it reminds me of some IMO problem. See this - [url]http://projectpen.files.wordpress.com/2008/12/pen-16sol.pdf[/url] ]", "Solution_5": "[quote=\"shobber\"]Show that for any positive integers $ a$ and $ b$, $ (36a \\plus{} b)(a \\plus{} 36b)$ cannot be a power of $ 2$.[/quote]\r\nAssume that there exists a pair. Then there must a exist a pair such that $ (a,b) \\equal{} 1$. Then $ 36a \\plus{} b \\equal{} 2^n, 36b \\plus{} a \\equal{} 2^m$. So $ n,m > 5$. Thus $ 2^5 \\mid (36a\\plus{}b)\\plus{}(36b\\plus{}a) \\equal{} 37(a\\plus{}b) \\iff 2^5 \\mid a\\plus{}b$. Thus $ a \\equiv b \\equiv 1 \\bmod 2$. And thus $ 36a \\plus{} b \\equiv 1 \\bmod 2$. Contradiction..", "Solution_6": "let $ (a\\plus{}36b)(36a\\plus{}b)\\equal{}2^z$. so $ 36a\\plus{}b\\equal{}2^x$ and $ a\\plus{}36b\\equal{}2^{x\\plus{}y}$. (assume that a+36b is not less thsn 36a+b). So $ 36a.2^y\\plus{}b.2^y\\equal{}2^{x\\plus{}y}$ . Hence $ y<6$. but $ 37|2^x\\plus{}2^{x\\plus{}y}$.absurd.", "Solution_7": "Assume there is a solution. Take $a < b$ and the smallest possible $a$. Now $(36a + b)$ and $(a + 36b)$ must each be powers of $2$. Hence $4|b$ and $4|a$. So $a/2$ and $b/2$ is a smaller solution with $a/2 < b/2$. Contradiction.", "Solution_8": "First $a$ and $b$ must be distint\nso WLOG $a>b$ so $36b+a$ divides $36a+b$\nhence $36b+a$ divides $a-b$ (impossible)", "Solution_9": "WLOG assume that $v_2(b)\\ge v_2(a)$. Then we can divide by by $2^{2v_2(a)}$ to get $(36m+2^km)(2^k\\cdot 36n+m)$ where $m$ and $n$ are odd positive integers. Since $2^k\\cdot 36n+m$ is odd and not equal to $1$, then $(36a+b)(a+36b)$ cannot be a power of $2$.", "Solution_10": "[quote=subham1729]First $a$ and $b$ must be distint\nso WLOG $a>b$ so $36b+a$ divides $36a+b$\nhence $36b+a$ divides $a-b$ (impossible)[/quote]\n\nHow do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?", "Solution_11": "Would this solution be acceptable?\n\n[hide=Solution]Suppose for the sake of contradiction that there is a solution. Consider the solution such that $a+b$ takes on the smallest value, we will use infinite descent to show that there is always a smaller ordered pair that works. If $(36a+b)(36b+a)=2^k$ where $k>10$, then we must have that both $36a+b$ and $36b+a$ are even. This means both $a$ and $b$ must be even. Let $a=2x$ and $b=2y$. Our equation is now $(72x+2y)(72y+2x)=2^k$ or when divided by 4, $(36x+y)(36y+x)=2^{k-2}$. Clearly this is another solution in which $x+y=\\frac{a+b}{2}$ contradicting the minimality of $a+b$. Thus, we are done $\\blacksquare$. [/hide]", "Solution_12": "It looks fine, just make sure the grader knows that you know that $2^k\\geq 2^2$ :)", "Solution_13": "[quote=mathlete3141592][quote=subham1729]First $a$ and $b$ must be distint\nso WLOG $a>b$ so $36b+a$ divides $36a+b$\nhence $36b+a$ divides $a-b$ (impossible)[/quote]\n\nHow do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?[/quote]\n\nDoes anybody see why this is true?", "Solution_14": "i think it should be $35a-35b$ by Euclid's Algorithm, I can't get any farther :(", "Solution_15": "[quote=rafayaashary1]i think it should be $35a-35b$ by Euclid's Algorithm, I can't get any farther :([/quote]\n\nThen why can't 36b+a divide 35a-35b? I can't understand subham's solution.", "Solution_16": "[quote=mathlete3141592][quote=subham1729]First $a$ and $b$ must be distint\nso WLOG $a>b$ so $36b+a$ divides $36a+b$\nhence $36b+a$ divides $a-b$ (impossible)[/quote]\n\nHow do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?[/quote]\n\nRemember we are assuming that $36b+a$ is a power of two. Then $(36b+a,35)=1$ so $36b+a|35a-35b$ implies $36b+a|a-b$... ", "Solution_17": "$(36a+b)(36b+a)=2^{m}$. Let\n$36a+b=2^{t}$\n$36b+a=2^{g}$\nTherefore $37a+37b=2^{t}+2^{g}=2^{\\min{(g,t)}}(2^{|g-t|}+1)$. Therefore $2^{\\min{(g,t)}} | a+b$. Contradiction", "Solution_18": "[u][b][i]Solution[/i][/b][/u]\n[b]WLOG[/b] suppose $ b \\le a $\nThen clearly $ (36b +a ) | (36a +b) $\nTherefore $ gcd({ 36a + b , 36b + a }) = gcd({ 36a +b , 35(a-b) }) = 36b + a$\n$ \\implies $ $ (36b + a) | 35(a-b)$\nThus either $ (36b +a) | 35$ ........$ (1) $\n OR\n$ (36b + a ) | (a-b) $..........$(2) $\n$1$ is absurd as $ a,b \\in Z+ $\nand \n$2$ is clearly impossible.", "Solution_19": "Just write $a=2^{\\alpha}a',b=2^{\\beta}b'$ to get $2^{\\alpha +2} \\cdot 9a' +2^{\\beta}b'=2^m$, and $2^{\\beta+2} \\cdot 9b' +2^{\\alpha}a'=2^n$- thus, as $a,b$ are positive integers, $\\alpha+2=\\beta, \\beta+2=\\alpha$ which is absurd.", "Solution_20": "I'm not quite sure about this, can someone :help: or Point any flaws?\n[quote=APMO 1998 P2]Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.[/quote]\n[color=#000][b]Solution:[/b][/color] Since, $(36a+b)(a+36b)$ is symmetric. Hence, Suppose, $a \\geq b$. Now suppose, there exists a $k$\n$$(36a+ ~b)(a+36b)=2^k=2^{x+y}$$\nSuppose, $$\\begin{cases} 36a+ ~b=2^x \\\\ 36b+ ~a=2^y \\end{cases}$$\n\nHence, $a,b$ are even $\\implies$ $a=2a_1$ and $b=2b_1$ for $a_1,b_1 \\in \\mathbb{Z}^+$ [size=50](#27) Thanks[/size]\n$$\\begin{cases} 36a+b=2(36a_1+b_1)=2^x \\implies 36a_1+b_1=\\frac{2^x}{2} =2^{x-1}=2^{x_1} \\in \\mathbb{Z}^+ \\\\ 36b+a=2(36b_1+a_1)=2^y \\implies 36b_1+a_1=\\frac{2^y}{2} =2^{y-1}=2^{y_1} \\in \\mathbb{Z}^+ \\end{cases}$$\nHence, now,\n$$(36a_1+b_1)(36b_1+a_1)=2^{x_1+y_1}$$\nRepeating the Same Procedure, we find, $a_1=2a_2$ and $b_1=2b_2$ for $a_2,b_2 \\in \\mathbb{Z}^+$. We can continue This till $x_n+y_n$ doesn't become negative and Since, $a,b$ are positive integers, we get a Contradiction! $\\qquad \\blacksquare$\n\n---------------------------------------------\n@Jupiter and agastya, Thanks", "Solution_21": "Redacted", "Solution_22": "[hide=possible fault][quote=AlastorMoody]I'm not quite sure about this, can someone :help: or Point any flaws?\n[quote=APMO 1998 P2]Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.[/quote]\n[color=#000][b]Solution:[/b][/color] Since, $(36a+b)(a+36b)$ is symmetric. Hence, Suppose, $a \\geq b$. Now suppose, there exists a $k$\n$$(36a+ ~b)(a+36b)=2^k=2^{x+y}$$\nSuppose, $$\\begin{cases} 36a+ ~b=2^x \\\\ 36b+ ~a=2^y \\end{cases}$$\nNow, $a \\geq b \\implies x \\geq y$\n$$35(a ~ - ~ b)=36a ~ + ~ b -(36b ~ + ~a)=2^x ~ - ~2^y \\implies \\qquad a ~ - ~b=\\frac{2^x-2^y}{35} \\qquad - - - \\text{[1]}$$\n$$37(a ~ + ~b)=36a ~ + ~b + ~36b ~+~ a=2^x~ +~ 2^y \\implies \\qquad a ~ + ~b=\\frac{2^x+2^y}{37} \\qquad - - - \\text{[2]}$$\nHence,\n$$\\text{[1]} + \\text{[2]} \\implies a = \\frac{36 \\cdot 2^x ~ - ~ 2^y}{35 \\cdot 37} \\in \\mathbb{Z}^+$$\nHence, $a,b$ are even $\\implies$ $a=2a_1$ and $b=2b_1$ for $a_1,b_1 \\in \\mathbb{Z}^+$\n$$\\begin{cases} 36a+b=2(36a_1+b_1)=2^x \\implies 36a_1+b_1=\\frac{2^x}{2} =2^{x-1}=2^{x_1} \\in \\mathbb{Z}^+ \\\\ 36b+a=2(36b_1+a_1)=2^y \\implies 36b_1+a_1=\\frac{2^y}{2} =2^{y-1}=2^{y_1} \\in \\mathbb{Z}^+ \\end{cases}$$\nHence, now,\n$$(36a_1+b_1)(36b_1+a_1)=2^{x_1+y_1}$$\nRepeating the Same Procedure, we find, $a_1=2a_2$ and $b_1=2b_2$ for $a_2,b_2 \\in \\mathbb{Z}^+$. We can continue This Infinitely, But Since, $a,b$ are positive integers, Hence by FMID we get a Contradiction! $\\qquad \\blacksquare$\n\n\n---------------------------------------------\n\n\n\n@below Am I making any mistake :maybe: ? Also, where shall I fix my solution? :maybe: Pls Helpp :help: I thought FMID works in this way? (Pls correct, me a total noob)[/quote]\n[/hide]\nNote that $x_1+y_10$. However, the sequence $\\{x_i+y_i\\}_{i=1}^{\\infty}$ is strictly decreasing by a gap of $2$. Thus, once $x_i+y_i$ becomes negative (for some $i$), you lose the power to prove that $a_i\\equiv b_i\\equiv 0\\pmod{2}$", "Solution_26": "^^ try solving this with geo\n\n@below thanks for reminding", "Solution_27": "Apmo 1998", "Solution_28": "https://artofproblemsolving.com/community/c2113h1042655", "Solution_29": "$\\spadesuit \\ \\textbf{\\textrm{A different solution without infinte decent:}}$\nLet $a=2^c \\cdot p, b=2^d \\cdot q$ where $p,q$ are odd numbers and $c,d$ are non-negative integers, WLOG, assume that $c\\ge d$, \\[(36a+b)(a+36b)=2^d(36\\cdot 2^{c-d} \\cdot p+q)(36b+a)\\] since $36\\cdot 2^{c-d} \\cdot p+q$ is a non-trivial odd positive integer, $(36a+b)(a+36b)$ cannot be a power of $2$. $\\quad \\blacksquare$", "Solution_30": "36a+b and 36b+a are both powers of 2 we can assume WLOG that : a+36b|b+36a \nThat gives a+36b|35(a-b) but a+36b is a power of 2 so a+36b|a-b that gives a=b and that also is a contradiction" } { "Tag": [ "trigonometry", "Support" ], "Problem": "Help required :- :? \r\n\r\n1. A cork floating in a calm water executes SHM of frequency $\\nu$ when a wave generated by a boat passes by it. The frequency of the wave is \r\n(a) $\\nu$ (b) $\\frac{\\nu}{2}$ (c) $2\\nu$ (d) $\\sqrt{2} \\nu$\r\n\r\n2. A sonometer wire of length $l$ vibrates in fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length\r\nis doubled keeping other things same, the string will vibrate with a frequency of\r\n(a) 416 Hz (b) 208 Hz (c) 832 Hz (d) stop vibrating\r\n\r\n3. A wave moving in a gas \r\n(a) must be longitudinal (b) may be longitudinal\r\n(c) must be transverse (d) may be transverse \r\n\r\n4. Two particles $A$ and $B$ have same phase difference of $\\pi$ when a sine wave passes through a region \r\n\r\n(a) $A$ oscillates at half the frequency of $B$.\r\n(b) $A$ and $B$ move in opposite directions.\r\n(c) $A$ and $B$ must be separated by half of a wavelength\r\n(d) The displacements at $A$ and $B$ have equal magnitudes.\r\n\r\n5. In a stationary wave,\r\n(a) all the particles of the medium vibrate in phase \r\n(b) all the antinodes vibrate in phase\r\n(c) the alternate antinodes vibrate in phase \r\n(d) all the particles between consecutive nodes vibrate in phase.\r\n\r\nTry to support your answers with reasons.", "Solution_1": "I'd say that number 4 is obviously (c), a little less obviosly (d) and could be (b) if we assume that they [i]are[/i] moving at all.", "Solution_2": "[quote=\"bus\"]I'd say that number 4 is obviously (c), a little less obviosly (d) and could be (b) if we assume that they [i]are[/i] moving at all.[/quote]\r\n\r\nTry to provide reasons. The anwer given is (b) & (d).\r\nShocked at not having (c), I too was.\r\nTry for other problems also.\r\nThanks for replying.", "Solution_3": "OK, I get it... they don't have to be separated by 1/2 wavelength, it can be k + 1/2 wavelength for any integer k. (b) and (d) are kinda obvious, the displacements are given by equations:\r\n\r\n$x_1=A\\sin\\phi_1$\r\n$x_2=A\\sin\\phi_2=A\\sin(\\phi_1+\\pi)=-A\\sin\\phi_1=-x_1$\r\n\r\nand the velocities are:\r\n\r\n$v_1=v_\\text{max}\\cos\\phi_1$\r\n$v_2=v_\\text{max}\\cos\\phi_2=v_\\text{max}\\cos(\\phi_1+\\pi)=-v_\\text{max}\\cos\\phi_1=-v_1$.", "Solution_4": "[quote=\"bus\"]OK, I get it... they don't have to be separated by 1/2 wavelength, it can be k + 1/2 wavelength for any integer k. (b) and (d) are kinda obvious, the displacements are given by equations:\n\n$x_1=A\\sin\\phi_1$\n$x_2=A\\sin\\phi_2=A\\sin(\\phi_1+\\pi)=-A\\sin\\phi_1=-x_1$\n\nand the velocities are:\n\n$v_1=v_\\text{max}\\cos\\phi_1$\n$v_2=v_\\text{max}\\cos\\phi_2=v_\\text{max}\\cos(\\phi_1+\\pi)=-v_\\text{max}\\cos\\phi_1=-v_1$.[/quote]\r\n\r\nThe equation of a travelling wave is $y = A\\sin (kx - wt)$.\r\nSo, we cannot stick only to path difference $\\frac{\\lambda}{2}$\r\nwe can also have some condition for $t$.\r\nI have got the solutions of all the problems which were submitted by me. If you stick to some of them then you can consult me. \r\nBye... :lol:", "Solution_5": "I'd say: 1a, actually the cork is always following the wave and both of them have the same frequency\r\n2b, actually if the lenght of the wire is doubled the lenghtwave will be doubled too (in foundamental mode) so the frequency'll be the half.\r\n3a(i think)\r\n5: i don't know what an antinode is! I think a,d are wrong, if i understand right what an antinode is the answer should be b (if an antinode is a point that moves with the maximum allowed amplitude)", "Solution_6": "[quote=\"Bacco\"]I'd say: 1a, actually the cork is always following the wave and both of them have the same frequency\n2b, actually if the lenght of the wire is doubled the lenghtwave will be doubled too (in foundamental mode) so the frequency'll be the half.\n3a(i think)\n5: i don't know what an antinode is! I think a,d are wrong, if i understand right what an antinode is the answer should be b (if an antinode is a point that moves with the maximum allowed amplitude)[/quote]\r\n\r\nYou are true about Q1, but for Q2 the frequency will be the same as was before because the wire is in resonance with a tuning fork.\r\nWhat you are thinking about an antinode is correct, but the answer is (c) & (d).\r\nThanks for your reply.\r\nBye... :lol:", "Solution_7": "Sorry...\r\nthe fact is that i disn't exactly understand what's a tuning fork...\r\nSorry for 5 too, actually d is true because all particles have the same phase but different amplitudes" } { "Tag": [], "Problem": "Walker Middle School sells graphing calculators to raise funds. The school pays $ \\$90$ for each calculator and sells them for $ \\$100$ apiece. They hope to earn enough money to purchase an additional set of 30 calculators. How many calculators must they sell to reach their goal?", "Solution_1": "If they want to buy an additional 30 calculators, they need another $ \\$2700$. Since they make $ \\$100$ on each calculator sold, they need to sell 27.", "Solution_2": "[quote=\"JRav\"]If they want to buy an additional 30 calculators, they need another $ \\$2700$. Since they make $ \\$100$ on each calculator sold, they need to sell 27.[/quote]\nActually, they make $\\$10$ on each calculator sold since $\\$100-\\$90=\\$10$ so they need to sell $\\frac{2700}{10}=\\boxed{270}$ calculators." } { "Tag": [ "logarithms", "algebra unsolved", "algebra" ], "Problem": "$e \\leq x \\leq y$ . Prove that: $x^{y}\\geq y^{x}$", "Solution_1": "You are to examine the behaviour of the graph of $y=\\frac{\\ln x}{x}.$", "Solution_2": "[quote=\"kunny\"]You are to examine the behaviour of the graph of $y=\\frac{\\ln x}{x}.$[/quote]\r\n[u][b]Similar[/b][/u] \r\nFind all $x,y \\in$[b]Z[/b] : $m^{n}=n^{m}$" } { "Tag": [], "Problem": "Calculate $\\left\\{\\left[i^{100}-\\left(\\frac{1-i}{1+i}\\right)^5\\right]^2+\\left(\\frac{1+i}{\\sqrt 2}\\right)^{20}\\right\\}(1+2i)$.", "Solution_1": "[quote=\"ifai\"]Calculate $\\left\\{\\left[i^{100}-\\left(\\frac{1-i}{1+i}\\right)^5\\right]^2+\\left(\\frac{1+i}{\\sqrt 2}\\right)^{20}\\right\\}(1+2i)$.[/quote]\r\n[hide]First, $i^{100}=1$. Now $\\frac{i-1}{i+1}$ must have a length of one as $i+1$ and $i-1$ have the same length. And it must have an angle of $\\frac{- \\pi}{2}$ since $i-1$ has an angle of $\\frac{-\\pi}{4}$ and you're subtracting another $\\frac{\\pi}{4}$, so it equals $-i$. $(-i)^5 = -i$, so we have ${\\left[i^{100}-\\left(\\frac{1-i}{1+i}\\right)^5\\right]^2=(1+i)^2 = 2i}$. Now $\\displaystyle (\\frac{i+1}{\\sqrt 2}) ^2 = \\frac{2i}{2} = i$. $i^{10} = -1$. We end up with $(2i-1)(2i+1) = -4-1 = -5$. So $-5$ is our answer.[/hide]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hey folks!!!\r\nI had trouble understanding the explanation at the back of the mathcounts handbook for this problem.\r\nIf you have the handbook, can you plz explain to me, in layman's terms, how to solve the problem?\r\nI would VERY much appreciate it.", "Solution_1": "Can some1 plz help me???\r\nOnly 8 views, come on!", "Solution_2": "I think most people here don't have the handbook with them right now.\r\n\r\nIf you can post the problem, that would be great. :)", "Solution_3": "I have the handbook. The problem is\r\n\r\nA bird collection has exactly four types of birds (eagles, doves, crows and sparrows). The eagles and doves make up 60% of the collection, and the doves and crows make up 20% of the collection. If the 18 crows in the collection represent 5% of the total number of birds, how many of the birds are sparrows?\r\n\r\n*checks watch* *counts words* YAY typing record! :lol: \r\n\r\nIf 18 crows is 5% of the total number of birds, then there are 360 birds total. Since doves and crows are 20% of the total birds together, the doves make up 15% of the total (20-5). And eagles and doves make up 60% together, so the eagles make up 45% of the total birds. So crows (5%), doves (15%), and eagles (45%) make up 65% of the total birds. What's left for the sparrows is 35%.\r\n\r\n35% of 360 is 126, so there are 126 sparrows.\r\n\r\nHope this helps." } { "Tag": [ "ratio", "LaTeX", "logarithms", "calculus" ], "Problem": "Compare the value of a,b,c:\r\n$a=1-\\frac12+\\frac13-\\frac14+......+\\frac1{2003}-\\frac1{2004}+\\frac1{2005}$\r\n$b=\\frac1{1003}+\\frac1{1004}+......+\\frac1{2005}$\r\n$c=1+\\frac12-\\frac13-\\frac14+\\frac15+\\frac16-\\frac17-\\frac18+......+\\frac1{2001}+\\frac1{2002}-\\frac1{2003}-\\frac1{2004}+\\frac1{2005}$", "Solution_1": "That's an excellent question. What contest is this from?", "Solution_2": "By compare, do you mean find the ratio between?\r\nYou know that c= 1+1/6-1/20+1/42-1/72...\r\na= 1-1/6-1/20-1/42...\r\nBut that is an interesting problem if you're comparing the ratios...", "Solution_3": "I believe compare means put in order of magnitude i.e. largest to smallest or vice versa.", "Solution_4": "well we know c>a...", "Solution_5": "see if this helps:\r\n$a=\\sum_{n=1}^{2005}\\frac{(-1)^{n+1}}{n}$\r\n$b=\\sum_{n=1003}^{2005}\\frac{1}{n}$\r\n$c=\\sum_{n=1}^{2005}\\frac{(-1)^\\frac{n(n+1)+2}{2}}{n}$", "Solution_6": "we also know that b1,2^nk<2005}\\frac1{2^n}\\right)$ The inner sum is $\\frac1{2^mk}$, where $2^mk$ is the largest number of that form less than or equal to 2005. Rearranging the sum, it becomes the sum of the reciprocals of the integers between $\\frac{2005}2$ and 2005, which is $b$.[/hide]", "Solution_12": "yea it is great \r\nbut during a contest i think that I would've just done\r\n[hide] subtract b from a and find out if its < or > 0\nso then you get all the odd terms minus the evens till 1002 minus 2*evens till 2004\nthen I noticed that the averages even out so it =0[/hide]", "Solution_13": "Thanks for your excellent solution! Nice job! :)", "Solution_14": "This suggests a nice problem. \r\n\r\nLet $H_k = 1/1 + 1/2 + \\dots + 1/k$ be the k'th harmonic number.\r\n\r\nLet $a_n, b_n, c_n$ be the above sums with $2005$ replaced by any $n = 4m+1$. \r\n\r\nShow that $a_n$ and $b_n$ can be expressed simply in terms of the $H_k$, and explain why this is not true for $c_n$." } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "y is it called that? i mean its not really partial is it?", "Solution_1": "Because you're taking the derivative with respect to one variable of many.", "Solution_2": "That method is used in Multivariable Calculus, when you have multiple independent variables and one dependent variable, in which when you do derivatives, you can only do it with one at a time, so that's why it's called partial differentiation.", "Solution_3": "well, well i didn't know you could do gauss1181 could do calculus!" } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Is it true that if we have a series of independent positive random variables $\\sum \\xi_{k}$ then the sufficient condition for convergence a.s. is convergence of $\\sum E\\xi_{k}$? I think it is because we can switch $E$ and $\\sum$, but i also know that Kolmogorov theorem says that there must be also $\\sum D\\xi_{k}<\\infty$ :?:", "Solution_1": "Kolmogorov's necessary and sufficient condition (also known as \"the three series theorem\") says that in general the following three series must converge: $\\sum_{k}P(|\\xi_{k}|>1), \\quad \\sum_{k}E\\xi'_{k}, \\quad \\sum_{k}D\\xi'_{k}$ where $\\xi'_{k}=\\begin{cases}\\xi_{k}, |\\xi_{k}|\\le 1\\\\0, |\\xi_{k}|>1\\end{cases}$\r\nI leave it to you to check that, for positive $\\xi_{k}$, the condition $\\sum_{k}E\\xi_{k}<+\\infty$ implies these three. The condition $\\sum_{k}D\\xi_{k}<+\\infty$ is not necessary for convergence a.s." } { "Tag": [ "function", "limit", "integration", "inequalities", "trigonometry", "calculus", "geometry" ], "Problem": "A recent attempt at proving the Riemann Hypothesis can be found here.\r\n\r\n[url=http://arxiv.org/PS_cache/arxiv/pdf/0806/0806.0892v1.pdf]Zhang's Proof[/url]\r\n\r\nThe proof relies on showing that the function:\r\n\\[ \\Phi(t)\\equal{}4\\pi\\sum_{n\\equal{}1}^{\\infty}\\left\\{2n\\pi^4 e^{9t/2}\\minus{}3n^2e^{5t/2}\\right\\}\\exp(\\minus{}n^2 \\pi e^{2t})\\]\r\nsatisfies:\r\n\\[ \\lim_{t\\to\\infty}\\frac{\\ln|\\Phi^{(n)}(t)|}{t}\\equal{}\\minus{}\\infty;\\quad n\\in\\mathbb{Z}^{\\plus{}}\\]\r\nand the entire function:\r\n\\[ \\Xi(z)\\equal{}\\int_0^{\\infty}\\Phi(t)\\cos(zt)dt\\]\r\nhas genus at most of 1. He doesn't explain this. Can someone help explain how these are satisfied?\r\n\r\n[edit] corrected statement\r\n\r\nThanks,\r\nShaw", "Solution_1": "$ \\Xi$ has order 1, and hence has genus $ \\leq$ 1 by Hadamard's factorization theorem", "Solution_2": "Ok, thanks Kalle. The proof is at a level I believe I can follow (with some additional work). May try to take it apart to see if I can find a flaw.", "Solution_3": "I find the zeta function interesting to study. Some background is needed first to understand what Zhang is attempting to show above. We start with Jensen's Theorem that each of the following inequalities is necessary and sufficient for the entire function $ f(z)\\neq 0$ with genus $ 0$ or $ 1$ to have only real zeros:\r\n\\[ y\\frac {\\partial}{\\partial y}\\big|f(x + iy)\\big|^2\\geq 0\\quad\\quad \\frac {\\partial^2}{\\partial y^2}\\big|f(x + iy)\\big|^2\\geq 0;\\quad - \\infty < x,y < \\infty\r\n\\]\r\nTake for example $ f(z) = sin(x + iy)$:\r\n\\begin{align*} y\\frac {\\partial}{\\partial y}\\big|\\sin(x + iy)\\big|^2 & = y\\frac {\\partial}{\\partial y}\\big(\\sin^2 x + \\sinh^2 y\\big) = y\\sinh 2y\\geq 0 \\\\\r\n\\frac {\\partial^2}{\\partial y^2}\\big|\\sin(x + iy)\\big|^2 & = \\frac {\\partial}{\\partial y}\\sinh 2y = 2\\cosh2y > 0 \\end{align*}\r\nand thus by Jensen's Theorem, $ f(z) = \\sin(z)$ has only real zeros.\r\nNow consider the function defined as $ \\displaystyle\\xi(z) = \\frac {z(z - 1)}{2\\pi^{z/2}}\\Gamma(z/2)\\zeta(z)$ and a variation of this function given by $ \\Xi(z) = \\xi(1/2 + iz)$. The Riemann Hypothesis can be formulated in terms of the $ \\Xi$ function: The Zeta function has non-trivial zeros only along the critical line if the $ \\Xi$ function has only real zeros. Now the $ \\Xi$ function can be written as:\r\n\\[ \\Xi(z) = \\int_{ - \\infty}^{\\infty}\\Phi(t)\\exp(izt)dt\r\n\\]\r\nwith\r\n\\[ \\Phi(t) = 2\\pi\\sum_{n = 1}^{\\infty}\\left\\{2n\\pi^4 e^{9t/2} - 3n^2e^{5t/2}\\right\\}\\exp( - n^2 \\pi e^{2t})\r\n\\]\r\nNow, assuming $ \\Xi(z)$ satisfies all the criteria of Jensen's Theorem, then $ \\Xi(z)$ will only have real zeros if the following are true:\r\n\\begin{align*} y\\frac {\\partial}{\\partial y}\\left|\\int_{ - \\infty}^{\\infty} \\Phi(t)\\exp(izt)dt\\right|^2 & \\geq 0 \\\\\r\n\\frac {\\partial^2}{\\partial y^2}\\left|\\int_{ - \\infty}^{\\infty}\\Phi(t)\\exp(izt)dt\\right|^2 & \\geq 0 \\end{align*}\r\nI haven't yet obtained Jensen's reference on expressing these partials as double integrals but I think I can if $ \\displaystyle \\overline{\\Xi(x + iy)} = \\int_{ - \\infty}^{\\infty}\\Phi(t)e^{i(x - iy)t}dt$ is correct? I'm not sure but if so then since $ |u|^2 = u\\overline{u}$, by Leibniz and the product rule:\r\n\\begin{align*}\\frac {\\partial}{\\partial y}\\left\\{\\int_{ - \\infty}^{\\infty}\\Phi(t)e^{i(x + iy)t}dt\\int_{ - \\infty}^{\\infty}\\Phi(s)e^{i(x - iy)s}ds\\right\\} & = \\\\\r\n& \\hspace{ - 30pt}\\int_{ - \\infty}^{\\infty}\\Phi(t)e^{i(x + iy)t}dt\\left(s\\int_{ - \\infty}^{\\infty}\\Phi(s)e^{i(x - iy)s}ds\\right) \\\\\r\n& \\hspace{ - 30pt} + \\int_{ - \\infty}^{\\infty}\\Phi(s)e^{i(x - iy)s}ds\\left( - t\\int_{ - \\infty}^{\\infty}\\Phi(t)e^{i(x + iy)t}dt\\right) \\\\\r\n& \\hspace{ - 40pt} = \\int_{ - \\infty}^{\\infty}\\int_{ - \\infty}^{\\infty}\\Phi(t)\\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)dsdt \\end{align*}\r\nI'd need to justify combining two single integrals into a double one though. And if I take another partial:\r\n\\[ \\frac {\\partial^2}{\\partial y^2}\\big|\\Xi(x + iy)\\big|^2 = \\int_{ - \\infty}^{\\infty}\\int_{ - \\infty}^{\\infty}\\Phi(t)\\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)^2dsdt\r\n\\]\r\nwe can then state one of Jensen's necessary and sufficient condition for the Riemann Hypothesis to hold:\r\n\\[ \\int_{ - \\infty}^{\\infty}\\int_{ - \\infty}^{\\infty}\\Phi(t)\\Phi(s)e^{i(s + t)x}e^{(s - t)y}(s - t)^2dsdt\\geq 0;\\quad - \\infty < x,y < \\infty\r\n\\]\r\nJensen give another necessary and sufficient condition based on a variation of this double integral which I do not currently know how to derive:\r\n\\[ \\int_{ - \\infty}^{\\infty}\\int_{ - \\infty}^{\\infty}\\Phi(t)\\Phi(s)e^{i(s + t)x} (s - t)^{2n}dsdt\\geq 0;\\quad - \\infty < x < \\infty,\\quad n=0,1,2,\\cdots\r\n\\]\r\nIt is this last expression that Zhang is attempting to prove.\r\n\r\n[edit] corrected formulas", "Solution_4": "Wait a minute. Guess that's basic Complex Analysis:\r\n\r\nConsider $ \\displaystyle\\Xi(x)\\equal{}\\int_{\\minus{}\\infty}^{\\infty} \\Phi(t)e^{ixt}dt;\\quad x\\in\\mathbb{R}$. Then $ \\displaystyle\\Xi(x)\\equal{}\\int_{\\minus{}\\infty}^{\\infty}\\Phi(t)\\cos(xt)\\plus{}i\\int_{\\minus{}\\infty}^{\\infty} \\Phi(t)\\sin(xt)dt$.\r\n\r\nNow, it is know that $ \\Phi(t)$ is an even function and therefore $ \\displaystyle \\int_{\\minus{}\\infty}^{\\infty}\\Phi(t)\\sin(xt)dt\\equal{}0$ and thus by the Reflection Principle, $ \\displaystyle\\overline{\\Xi(z)}\\equal{}\\Xi(\\overline{z})\\equal{}\\int_{\\minus{}\\infty}^{\\infty}\\Phi(t)e^{i(x\\minus{}iy)t}dt$", "Solution_5": "Seems like a nice approach. I have never seen Jensen's theorem, looks very interesting.", "Solution_6": "I'm having some difficulty finding an accessible reference for Jensen's work. He died before publishing the results stated here; G. Polya reviewed his work and latter published it but in German. So for now, I'll have to leave some holes in the preliminary lead in to Zhang's work. How about this: Suppose independent of any Riemann Hypothesis, we had the following assignment. Given that the function $ \\Psi(x)$ is a rapidly decreasing, positive, and even function, is:\r\n\\begin{align*} (47) \\quad f_n(x) & = \\int_{ - \\infty}^{\\infty}\\int_{ - \\infty}^{\\infty}\\Psi(s)\\Psi(u)e^{i(s + t)x} (s - t)^{2n}dsdt \\\\\r\n (49) \\hspace{3.5em} & \\overset{?}{=} 2\\int_0^{\\infty}\\cos(ux)\\left\\{\\int_0^{\\infty}\\Psi\\left(\\frac {u + v}{2}\\right)\\Psi\\left(\\frac {u - v}{2}\\right)v^{2n}dv\\right\\}du;\\quad x\\in\\mathbb{R}, n = 0,1,2,\\cdots \\end{align*} where $ u=s+t$ and $ v=s-t$ and I'm using the same equation numbers as Zhang. This part looks ok to me. Note I can write $ (47)$ as:\\begin{align*}\r\nf_n(x)&=\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty}\\Psi(s)\\Psi(u)\\cos\\big[x(s+t)\\big](s-t)^{2n}dsdt \\\\ &+i\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty}\\Psi(s)\\Psi(u)\\sin\\big[x(s+t)\\big](s-t)^{2n}dsdt\r\n\\end{align*} and since the sine integral is odd across the line $ s=-t$, I can rotate the axes by $ \\pi/4$ and make a change of variable in the second integral:\r\n$ \\displaystyle u=\\frac{s}{\\sqrt{2}}+\\frac{t}{\\sqrt{2}}$ and\r\n$ \\displaystyle v=-\\frac{s}{\\sqrt{2}}+\\frac{t}{\\sqrt{2}}$\r\nthen:\r\n$ \\displaystyle s=\\frac{u}{\\sqrt{2}}-\\frac{v}{\\sqrt{2}}$ and\r\n\r\n$ \\displaystyle t=\\frac{u}{\\sqrt{2}}+\\frac{v}{\\sqrt{2}}$\r\n\r\nWhich is then: \\[ \\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty}(1)\\Psi(s)\\Psi(t)\r\n\\sin\\big[xu\\sqrt{2}\\big](-\\sqrt{2}v)^{2n}dudv\\] which looks odd to me across the $ u$-axis and thus the integral would be zero although I'm not sure. The second change of variable is then used by Zhang to arrive at $ (49)$. Anybody see anything wrong with just this much?", "Solution_7": "I've reviewed equations $ (49)$ through $ (57)$ and although the author is a bit sloppy with notation, and equation cross-referencing, I believe the math is OK and illustrates some interesting problems in basic Calculus. Anyone disagree? However, I detected a flaw in Zhang's work at equation $ (58)$:\r\n\r\nConsider the integral:\r\n\\[ I_1(x) = \\int_0^{\\frac {1}{x}} f(ux)\\int_0^{\\infty}\\Psi\\left(\\frac {u + v}{2}\\right)\\Psi\\left(\\frac {u - v}{2}\\right)dvdu;\\quad x > 0\r\n\\]\r\nwhere $ \\Psi$ is even, positive, and rapidly decreasing and $ f$ well-behaved to make everything converge. Letting:\r\n\\[ u = \\frac {\\cos(\\theta)}{x},\\quad\\quad v = v;\\quad 0 < \\theta < \\pi/2\r\n\\]\r\nthe Jacobian is:\r\n\\[ \\frac {\\partial(u,v)}{\\partial(\\theta,v)} = - \\frac {\\sin(\\theta)}{x}\r\n\\]\r\nmaking the other substitutions, I obtain:\r\n\\begin{align*} \\int_0^{\\frac {1}{x}} f(ux)\\int_0^{\\infty}\\Psi\\left(\\frac {u + v}{2} \\right)\\Psi\\left(\\frac {u - v}{2}\\right)dvdu & \\\\\r\n& \\hspace{ - 155pt} = \\int_{0}^{\\pi/2}\\frac {\\sin(\\theta)}{x}f\\big(\\cos(\\theta)\\big)\\int_0^{\\infty}\\Psi\\left(\\frac {x^{ - 1}\\cos(\\theta) + v)}{2}\\right)\\Psi\\left(\\frac {x^{ - 1}\\cos(\\theta) - v}{2}\\right)v^{2n}dvd\\theta \\end{align*}\r\nI'm pretty sure that's right. I checked it numerically with $ f(x) = \\cos(x)$ and $ \\Psi(x) = e^{ - x^2}$. I realize numerical estimates are not enough but it does give some credence I think. Zhang does not have the $ x$ in the denominator and this I think makes his estimates in equation $ (59)$ no longer valid. I could be wrong of course: if this was water, I'd be standing on tip-toes to breathe. \r\nAnyone here interested in this is willing to look at it? I'll contact the author and give him a chance to respond." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "geometry", "similar triangles" ], "Problem": "USAMO Winners: The 12 scores in the range 38-27, inclusive\r\nWinners names, schools and states will be posted on the AMC website later today.\r\n\r\nHonorable Mention: The 13 scores in the range 26-23, inclusive.\r\n\r\nBlue MOSP: The 16 non-12th grade students with scores in the range 26 - 18, inclusive.\r\n\r\nRed MOSP: The 21 9th grade students with scores in the range 8-15, inclusive.\r\n\r\nTraining for CGMO: 8 girls with scores in the range 8-16, inclusive.\r\n\r\nOfficial invitations will be going out today (Mon, May 11) and tomorrow (Tues, May 12)\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_1": "Hmmmm... are CGMO and Red MOSP people mutually exclusive?", "Solution_2": "Don't think so... In fact, I think the CGMO group usually trains as part of Red? (maybe? I have no idea now that I think about it, but it makes sense since no girl got above 16...)\r\n\r\nHmm, I think my grandma who speaks only chinese might have picked up the phone while everyone else was away. That or what I described will happen tomorrow. Any insight on how this will affect my MOP invitation?", "Solution_3": "Just tell you grandma not to pick any phone call, I guess? It probably won't affect anything though, as they should e-mail you information later.", "Solution_4": "HOLY GUANO I GOT A 15!!!\r\nsorry for the trolling, i solved one problem and i though each problem was worth 7 points. ...", "Solution_5": "Um each problem is worth 7 points?", "Solution_6": "yeah, i solved one problem o.0 it was 4\r\ni misread 2 and 6, couldn't think of any observations to put on 3, observed unhelpful similar triangles on #1 and #5\r\nEDIT: shoot firefox automatically took me to 2005. gah\r\nwhen do you find this year's scores?\r\nat least now i have a goal :)", "Solution_7": "So how did you get a 15?", "Solution_8": "he clicked on the wrong link, so he saw his \"score\" in 2005 of that number", "Solution_9": "I got an 8, but I'm a sophomore. I have to study so hard to make blue mop next year... Anyone mind giving me mop notes? :)", "Solution_10": "I got an 20.. but i am canadian. SO does it mean i am done with this year's math fun?.. I want to go to MOP!", "Solution_11": "University applications are not the reason to go to MOP. It's about the fun and the contest training- which in your case would apply toward potentially making the Canadian IMO team." } { "Tag": [ "LaTeX", "calculus", "advanced fields", "advanced fields unsolved" ], "Problem": "Fo each $ a>0$, let $ x_a$ denote the sequence $ \\left(1/n^a \\right)_{n\\equal{}1}^{\\infty}$.\r\nFor which $ a$ do we have $ x_a \\in l^\\infty$? \r\nFor which $ a$ do we have $ x_a \\in l^1$ ? \r\nFor which $ a$ do we have $ x_a \\in l^2$?", "Solution_1": "$ a\\geq 0 \\iff x_a\\in l^\\infty$\r\n$ a>1/2 \\iff x_a\\in l^2$\r\n$ a>1 \\iff x_a\\in l^1$", "Solution_2": "$ \\text{\\LaTeX}$ tip: I prefer \\ell to l. $ \\ell^1,\\ell^2,\\ell^{\\infty}$ rather than $ l^1,l^2,l^{\\infty}.$\r\n\r\nOf course, once you strip off the abstract language, this is pretty much a standard calculus problem." } { "Tag": [ "search" ], "Problem": "I have Linux and WIN XP dual system. i have internet connection through modem in WIN XP. but recently i wanted to make available it in linux as well. but on pack of modem it is said that: supported drivers for windows 98/ME/00/XP. that means it is not possible to connect internet at Linux?", "Solution_1": "It means it's a windmodem. You need to download specific drivers for you modem(if they are available). Try using linuxant drivers and slmodem drivers. Also, try to google for drivers for modem chipset, not the modem itself.", "Solution_2": "I installed Fedora Core the other day and failed to use the Internet. The weird thing is that I can access the modem through an IP (192.168.100.1), which is used for initialization. On the screen I get, it is said to me that all tasks are completed, so it should work (at least, it works when I get such a message under Windows). When I tell it to search proper bandwidth again, the modem is reset (so the connection works, til there at least), but do I need any drivers for the Internet to work?\r\n\r\nI'd be grateful if anyone can answer my question. I'm a total newbie in this Linux stuff.", "Solution_3": "Could you provide some information about your hardware(especially modem)." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all triples $(x,y,z)$ of positive integers such that $(x+1)^{y+1}+1=(x+2)^{z+1}$.", "Solution_1": "Trivial by Mihailescu's Theorem :rotfl:", "Solution_2": "[quote=\"t0rajir0u\"] Mihailescu's Theorem[/quote]\r\n\r\nWhat is it?", "Solution_3": "Mihailescu is the one who proved Catalan's conjecture.", "Solution_4": "[quote=\"ZetaX\"]Mihailescu is the one who proved Catalan's conjecture.[/quote]\r\n\r\nBut what is Mihailescu's Theorem ? :huh:", "Solution_5": "Well, proving a conjecture leads often to them be named after you... ;)\r\n\r\nOh, and there is also google ;)", "Solution_6": "Ok! It here http://mathworld.wolfram.com/CatalansConjecture.html :D \r\n\r\nBut above problem has level is Olympiad?", "Solution_7": "i smel here short listed problem 2000. find all natural numbers such:$a^{m}+1=(a+1)^{n}$. and also there is generalisation for this $a^{m}+1|(a^{k}+1)^{n}$. i will post my solution using elementry number theory. :wink:", "Solution_8": "Both are trivial after using Zsygmoni :P (and it is elementary)", "Solution_9": "[quote=\"ZetaX\"] Zsygmoni [/quote]\r\n\r\nWhat is it? again :D", "Solution_10": "Sorry, it's spelled \"Zsigmondy\", see http://mathworld.wolfram.com/ZsigmondyTheorem.html .", "Solution_11": "[quote=N.T.TUAN]Find all triples $(x,y,z)$ of positive integers such that $(x+1)^{y+1}+1=(x+2)^{z+1}$.[/quote]\nBy Mihailescu's Theorem:\n$\\Rightarrow (x,y,z)=(1,2,1)$\nBy Zsigmondy's Theorem:\n$\\exists$ prime $p/ p|(x+1)^{y+1}+1, p\\nmid x+2$\n$\\Rightarrow 0\\equiv (x+2)^{z+1} \\pmod{p} (\\Rightarrow \\Leftarrow)_\\blacksquare$\nBut there are exceptions:\n$\\Rightarrow (x,y,z)=(1,2,1)$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a.b.c$ be ral numbers such that $ 4(a \\plus{} b \\plus{} c) \\minus{} 9 \\equal{} 0$. Find max\r\n$ S \\equal{} (a \\plus{} \\sqrt {a^2 \\plus{} 1})^b(b \\plus{} \\sqrt {b^2 \\plus{} 1})^c(c \\plus{} \\sqrt {c^2 \\plus{} 1})^a$", "Solution_1": "[quote=\"Algadin\"]Let $ a.b.c$ be ral numbers such that $ 4(a \\plus{} b \\plus{} c) \\minus{} 9 \\equal{} 0$. Find max\n$ S \\equal{} (a \\plus{} \\sqrt {a^2 \\plus{} 1})^b(b \\plus{} \\sqrt {b^2 \\plus{} 1})^c(c \\plus{} \\sqrt {c^2 \\plus{} 1})^a$[/quote]\r\nLock it,mod,please,see here http://www.mathlinks.ro/viewtopic.php?t=168836" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "This problem is from Vol II of \"Introduction from Calculus and Analysis\" by R. Courant and F. John. (this book is my calculus textbook)\r\nI think some problems in the book are interesting (this does not imply that they are difficult)\r\n\r\nFind the values of $a,b,c,d$, such that the equation $Af_{xx}+2Bf_{xy}+Cf_{yy} =0$ becomes\r\n(a) $f_{\\xi\\xi}+f_{\\eta\\eta} = 0$\r\n(b) $f_{\\xi\\eta} = 0$\r\nafter the transfrom of $\\xi = ax+by$ and $\\eta = cx+dy$ ($ad-bc\\neq 0$)", "Solution_1": "liyi, is there supposed to be a condition $AC-B^2>0$?", "Solution_2": "Okay, your are required to discuss yourself.\r\nFor (a) we need $AC>B^2$;\r\nFor (b) we need $AC0 and n is even\r\nf(n) = 2f(n-1), if n >0 and n is odd.\r\nThe \"recursion depth\" of f(n) is the number of other integers m such that the value of f(m) is calculated whilst computing f(n). [The recursion depth of f(4) is 3.]\r\nFor any non-negative integer n,\r\ng(n) = 0, if n = 0,\r\ng(n) =1+ g(n/2), if n>0 and n is even,\r\ng(n) =1+ g(n-1), if n>0 and n is odd.\r\n(i) What is g(2^k) where k is a non-negative integer?\r\n(ii) What is g(2^l + 2^k) when l>k>=0 are integers?\r\n(iii) Why is the value of g(n) = the recursion depth of f(n)?", "Solution_1": "(i)-\r\n $ g(2^k) \\equal{} 1 \\plus{} g(2^{k \\minus{} 1}) \\equal{} 2 \\plus{} g(2^{k \\minus{} 2}) \\equal{} ... \\equal{} k \\plus{} g(1) \\equal{} k \\plus{} 1$\r\n(ii)-\r\n$ g(2^k \\plus{} 2^l) \\equal{} g(2^k(2^{l \\minus{} k} \\plus{} 1)) \\equal{} k \\plus{} g(2^{l \\minus{} k} \\plus{} 1) \\equal{} k \\plus{} 1 \\plus{} g(2^{l \\minus{} k}) \\equal{} k \\plus{} 1 \\plus{} l \\minus{} k \\plus{} 1 \\equal{} l \\plus{} 2$", "Solution_2": "Extension: Find a simple way to describe $ f(n)$ for all $ n$.\r\n\r\nEDIT: Oops, I meant $ g(n)$.", "Solution_3": "Yes, obviously 2^n. But what about (iii)? Is there a connection?", "Solution_4": "Oh, I meant to write $ g(n)$ above.\r\n\r\nFor (iii), there is a connection: if it takes $ n$ steps to find $ f(n/2)$ where $ n$ is even, then we can find $ f(n)$ in just one more step, so $ g(n)\\equal{}g(n/2)\\plus{}1$ if $ n$ is even. The logic for $ n$ odd is exactly the same.\r\n\r\nThe fact that this recursion for $ g(n)$ gives the actual \"recursion depth\" of $ n$, i.e. the minimal number of steps to find $ n$ is less obvious - it helps to have an explicit form for $ g(n)$ if you want to prove this.", "Solution_5": "It seems that if n = 2^k(1) + 2^k(2) + ... + 2^k(r), then g(n) = k(1) + r." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $a,b,c\\geq 0,a+b+c=1$\r\nFind min $A=ab+2bc+3ca$", "Solution_1": "[quote=\"chien than\"]Let $a,b,c\\geq 0,a+b+c=1$\nFind min $A=ab+2bc+3ca$[/quote]\r\n$\\min A=0$ and holds when $a=1,$ $b=c=0.$" } { "Tag": [ "geometry", "Euler", "3D geometry", "prism", "geometry solved" ], "Problem": "Can you prove the incircle theorem of Darij's avatar.\r\n\r\nBomb", "Solution_1": "I didn't get your meaning clearly\r\n\r\nI think you meaned:\r\nwe have two circle with centers $O\\ ,\\ I$ if we have $OI^2=R^2-2Rr$ ($R,r$ are radius of bigger and smaller circles in order) prove that for any point on bigger circle we can construct a triangle that another circle be it's incircle.\r\n\r\nif you meaned this this not hard at all.\r\n\r\nto prove it use Euler theorem", "Solution_2": "[quote=\"bomb\"]Can you prove the incircle theorem of Darij's avatar.[/quote]\r\n\r\nIn fact, the theorem is called Poncelet's porism for triangles (I am saying this since I may change my avatar sometime in future...).\r\n\r\nWe had some topics about it:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=42330\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=31633\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=41614\r\n\r\n Darij" } { "Tag": [ "Gauss", "geometry", "geometric transformation", "search", "calculus" ], "Problem": "I'd really like to read some classic math texts written by the masters (as per Abel's advice) e.g. those written by Gauss, Galois, Legendre, etc, but I don't know where I could find them or even what exact titles to look for. Any help anyone can give me would be greatly appreciated.", "Solution_1": "You may have to speak a foreign language to read some of it. It is possible that someone has just typed some of that up and put it online, so you might try searching for that. Giving Amazon a shot is always worthwhile. Otherwise, your only good chance I think is specialized publishers, and I don't have any idea how you should best go about that. I warn you that it is almost certainly going to be difficult going, even if you can find them in English. Hope this was a little helpful.", "Solution_2": "Disquisitiones Arithmeticae by Gauss is for sale on Amazon for about $40.", "Solution_3": "In what language?", "Solution_4": "[quote=\"ComplexZeta\"]Disquisitiones Arithmeticae by Gauss is for sale on Amazon for about $40.[/quote]\r\n\r\nYes, it's an [url=http://www.amazon.com/exec/obidos/ASIN/0300094736/learninfreed]English translation[/url] by Arthur \r\nA. Clarke. \r\n\r\nThere are a few English translations of part of Euler's vast output, but it's still worthwhile to learn Latin to follow Abel's advice to learn from the masters.", "Solution_5": "What about Euclid's Elements?", "Solution_6": "[quote=\"walrus\"]I'd really like to read some classic math texts written by the masters (as per Abel's advice) e.g. those written by Gauss, Galois, Legendre, etc, but I don't know where I could find them or even what exact titles to look for. Any help anyone can give me would be greatly appreciated.[/quote]\r\n\r\nRound Table is a good forum for this question, but I would personally expect this kind of question in the Other Problem-Solving Topics forum. A good first place to look for info on your quest for books by the masters would be the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0321016181/learninfreed]A History of Mathematics: An Introduction[/url] by Victor J. Katz, which would tell you who the masters are, and probably provide citations to the English translations available for their works. Definitely a lot of the good stuff is still in the original Latin (or French, after a certain period). I also like John Stillwell's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387953361/learninfreed]Mathematics and Its History[/url] 2nd edition, which has some interesting citations to works by the masters, and is just plain a good book to read about mathematics. \r\n\r\nHope this helps!", "Solution_7": "There's almost certainly German and Russian stuff available as well. I'm sure things were written and translated among those four languages -- Latin, French, German, and Russian -- but I'm not sure how much of it would ever have been translated to English.\r\n\r\nReading Newton might be fun.", "Solution_8": "[quote]Reading Newton might be fun.[/quote]\r\n\r\nYeah, you have to read [i]Principia[/i].", "Solution_9": "Many of the \"classic\" texts of mathematics have been translated into several languages. In particular, I know that Dover has published several of them in English translation - including Euclid's [u]Elements[/u] and Descartes' [u]Geometry[/u]. They're not, however, necessarily easy to read. The notation is sometimes bad, and the names of things are different than what we call them today. My advice is to read a good textbook on a subject, and supplement that with the classic texts later on. It's also good to just read History of Mathematics books like E.T. Bell's [u]Men of Mathematics[/u] and Katz's [u]A History of Mathematics: An Introduction[/u].\r\n\r\nSome translations can also be found online. An online version of Euclid's [u]Elements[/u] can be found here: http://www.math.princeton.edu/~cojocaru/art-research-rm.html\r\n\r\nAnd a lot of translations can also be found in bounded collections of author's works, or as parts of other books. References are usually found in books on the History of Mathematics, or you can use these sites to search for them:\r\n\r\nhttp://www.amazon.com/\r\nhttp://www.ericweisstein.com/encyclopedias/books/MathematicsClassics.html\r\nhttp://www.ericweisstein.com/encyclopedias/books/GeneralMathematicsHistory.html\r\nhttp://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html", "Solution_10": "I don't understand why a person would want to read one of these classic books for educational purposes (although the reading may be enjoyable). Why would reading a book by Newton on calculus be more informative or enlightening than reading a good modern calculus textbook?", "Solution_11": "It becomes more important when you become a graduate student or a researcher. But for students of you guys ages I don't think it is a good way to \"learn\" a subject.", "Solution_12": "[quote=\"gauss202\"]\nSome translations can also be found online. An online version of Euclid's [u]Elements[/u] can be found here: http://www.math.princeton.edu/~cojocaru/art-research-rm.html\n[/quote]\r\nDid you perhaps copy the wrong link? That isn't Euclid's Elements. It's \"the Art of Research\"", "Solution_13": "Sorry, I can't resist.\r\n\r\n[quote=\"gauss202\"]And a lot of translations can also be found in bounded collections of author's works.[/quote]\r\n\r\n'bounded collection' -- would that be a collection which is contained entirely in a ball of some radius R, centered at a point x?", "Solution_14": "[quote=\"confuted\"]\nDid you perhaps copy the wrong link? That isn't Euclid's Elements. It's \"the Art of Research\"[/quote]\r\n\r\nOops, yeah. I meant: http://babbage.clarku.edu/~djoyce/java/elements/Euclid.html" } { "Tag": [ "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $n$ be an odd positive integer and let $x_1, \\ldots, x_n, y_1, \\ldots, y_n$ be nonnegative real numbers satisfying $x_1 + \\cdots + x_n = y_1 + \\cdots + y_n$. Show that there exists a proper, nonempty subset $J \\subseteq \\{1,\\ldots,n\\}$ such that \\[\\frac{n-1}{n+1}\\sum_{j\\in J}x_j \\leq \\sum_{j\\in J} y_j \\leq \\frac{n+1}{n-1}\\sum_{j\\in J} x_j.\\]", "Solution_1": "wlog we assume $n>1$ and that $\\sum_i x_i=\\sum_i y_i=1$.if the result were not true then for each $i$, $y_i<\\frac{n-1}{n+1}x_i$ or $x_i<\\frac{n-1}{n+1}y_i$.now if $y_i<\\frac{n-1}{n+1}x_i$ then it must follow that for the complement set with sum $1-x_i,1-y_i$ the otherway inequality must hold(since the sums are the same),i.e. \r\n $y_i<\\frac{n-1}{n+1}x_i\\Rightarrow 1-y_i>\\frac{n+1}{n-1}(1-x_i)$.since $n$ is odd one of these inequalities occurs atleast $\\frac{n+1}{2}$ times,i.e. if $I_0=\\{i|y_i<\\frac{n-1}{n+1}x_i\\},I_1=\\{i|x_i<\\frac{n-1}{n+1}y_i\\}$ then either $|I_0|\\geq\\frac{n+1}{2}$ or $|I_1|\\geq \\frac{n+1}{2}$. say $|I_0|\\ge \\frac{n+1}{2}$. then since we have \r\n $y_i<\\frac{n-1}{n+1}x_i$ and $1-x_i<\\frac{n-1}{n+1}(1-y_i)$ it follows that(solve for the intersection of the lines)\r\n$x_i>\\frac{2}{n+1}$.but then $\\sum_{i\\in I_0} x_i > 1$ and that is a contradiction." } { "Tag": [ "algebra", "polynomial", "algebra unsolved", "function" ], "Problem": "Suppose that $a,b,c,d\\in\\mathbb{R}$ and $f(x)=ax^3+bx^2+cx+d$ such that $f(2)+f(5)<77>g(4)$) there exists $u\\in(4,5)$ such that $g(u)=7$.", "Solution_2": "Oh ,now I find a new solution:\r\n\r\nI think this solution has a little different to Diarmuid's solution\r\n\r\nWe must prove the equality of $f(x)+f(7-x)=7$ has a root in real number. Suppose that $g(x)=f(x)=f(7-x)-7$ we know that $g(x)$ is a polynomial which its degree at most $2$ , $g(2)<0$ and $g(3)>0$ so that the equation $g(x)=0$, then it has at least one root. :) \r\n\r\n[color=red]Excuse me for my bad question [/color]" } { "Tag": [ "algorithm", "videos" ], "Problem": "is possible to train calculating in human memory to calculate any problem in 1 second??? 5687,25 * 22 and others numbers? sorry for my bad english.", "Solution_1": "[quote=\"jiri.otevrel\"]calculate any problem in 1 second???[/quote]\r\n\r\nEr, no, not really. All you have to do is make the problem arbitrarily long; just keep adding digits and eventually the capacity of the problem will be greater than the capacity of the human brain to store data, and then obviously such a calculation is not possible by the human brain.", "Solution_2": "check out vedic arithmetic\r\n\r\nvedic math has lots of different methods of doing basic arithmetic (and even some higher level stuff) that are conducive to faster, mental calculations\r\n\r\nVedic methods involve different, multi-step techniques, but they allow a person to say.. multiply 4 digit numbers in their head.\r\n\r\nit requires a lot of memorizing though. There's no real point when you have calculators... or even paper.\r\n\r\nthe only one that I know that I actually use is a squaring algorithm\r\n\r\n$a^{2}= (a-b)(a+b)+b^{2}$\r\nso something like 28^2 = (20)(36) + 8^2 = 784", "Solution_3": "Sure it is! These guys are true human calculators. They can get the answers faster than you can type problems into the calculator.\r\n\r\n[youtube]EueFhYZ4HxI[/youtube]", "Solution_4": "tnx for video ... it's nice.", "Solution_5": "But that isn't really a human calculator, it is a abacus which is a way to store numbers.", "Solution_6": "[quote=\"chickendude\"]the only one that I know that I actually use is a squaring algorithm\n\n$ a^{2}\\equal{} (a\\minus{}b)(a\\plus{}b)\\plus{}b^{2}$\nso something like 28^2 = (20)(36) + 8^2 = 784[/quote]\r\nHow do you find $ b$?", "Solution_7": "[quote=\"hunter34\"]But that isn't really a human calculator, it is a abacus which is a way to store numbers.[/quote]\r\n\r\ndid u watch all of the video?", "Solution_8": "Wow @ those kids in the video.", "Solution_9": "Art Benjamin is amazing at mental calculation. Here's one video of one of his performances.\r\n\r\n[url]http://video.google.com/videoplay?docid=6687235905416745311&q=art+benjamin&total=3198&start=0&num=10&so=0&type=search&plindex=0[/url]", "Solution_10": "[quote=\"Kyron\"][quote=\"chickendude\"]the only one that I know that I actually use is a squaring algorithm\n\n$ a^{2}\\equal{} (a\\minus{}b)(a\\plus{}b)\\plus{}b^{2}$\nso something like 28^2 = (20)(36) + 8^2 = 784[/quote]\nHow do you find $ b$?[/quote]\r\n$ b$ is any number that he picks that makes the calculation convenient, because for any number $ b$ the equation will work." } { "Tag": [ "factorial", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi,\r\n\r\nI'm looking for pointers to a problem I'm trying to slove. How do I show that there is an isomorphism between an R-Algebra and an factorial ring? Is there always a standard approach to these kind of problems? \r\n\r\nThanks.\r\n\r\nMike", "Solution_1": "Please be more explicit.\r\nWhat kind of a ring is $R$?\r\nIsomorphism with respect to which structure?" } { "Tag": [ "conics", "hyperbola", "quadratics", "analytic geometry", "graphing lines", "slope", "trigonometry" ], "Problem": "[color=darkblue] [u][b]Please (if it is possibly), an algebraical proof ![/b][/u] [/color]\r\n\r\n[quote=\"Virgil Nicula\"]$ \\{\\begin{array}{c}\na\\ge 0\\ ,\\ b\\ge 0\\\\\\\\\na^2+b^2=1\\end{array}\\ \\ \\implies\\ \\ 2\\le (1+a)(1+b)\\le \\frac {3+2\\sqrt 2}{2}$ .[/quote]", "Solution_1": "[color=darkblue]Solution 1[/color]\r\n\r\nSuppose that $ (1 \\plus{} a)(1 \\plus{} b)$ takes some value $ k$, there are more than one intersection points between the graphs of $ a^2 \\plus{} b^2 \\equal{} 1\\ (a\\geq 0,\\ b\\geq 0)$ and $ (a \\plus{} 1)(b \\plus{} 1) \\equal{} k.$ Let $ T\\left(\\frac {1}{\\sqrt {2}},\\ \\frac {1}{\\sqrt {2}}\\right)$ be the point of tangency. Thus when the hyperbola $ (a \\plus{} 1)(b \\plus{} 1) \\equal{} k$ passes through the point $ (a,\\ b) \\equal{} (1,\\ 0),\\ (0,\\ 1)$, we have $ k \\equal{} 2$, then the hyperbola $ (a \\plus{} 1)(b \\plus{} 1)$ touches to the circle $ a^2 \\plus{} b^2 \\equal{} 1\\ (a\\geq 0,\\ b\\geq 0)$ at $ T$, we have $ k \\equal{} \\frac {3 \\plus{} 2\\sqrt {2}}{2}$. Since $ k$ varies continuously betwee the values (reffer to the attached file ''Graph 1''), yeilding $ 2\\leq k\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}\\Longleftrightarrow 2\\leq (1 \\plus{} a)(1 \\plus{} b)\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}$.", "Solution_2": "[color=darkblue]Solution 2[/color]\r\n\r\nLet $ a \\plus{} b \\equal{} s,\\ ab \\equal{} t$, $ a,\\ b$ are the non negative roots of the quadratic equation:$ x^2 \\minus{} sx \\plus{} t \\equal{} 0$.\r\n$ \\therefore a\\geq 0$ and $ b\\geq 0\\Longleftrightarrow a \\plus{} b\\geq 0$ and $ ab\\geq0$ and the discriminant $ D\\geq 0$.\r\n$ \\Longleftrightarrow s\\geq 0$ and $ t\\geq 0$ and $ s^2 \\minus{} 4t\\geq 0\\ \\cdots [1]$.\r\n$ a^2 \\plus{} b^2 \\equal{} 1\\Longleftrightarrow s^2 \\minus{} 2t \\equal{} 1\\Longleftrightarrow t \\equal{} \\frac {1}{2}s^2 \\minus{} \\frac {1}{2}\\ \\cdots [2]$ and \r\n$ k \\equal{} (1 \\plus{} a)(1 \\plus{} b) \\equal{} 1 \\plus{} s \\plus{} t\\Longleftrightarrow t \\equal{} \\minus{} s \\plus{} k \\minus{} 1\\ \\cdots [3]$. Thus our problem can be restated that : On $ s \\minus{} t$ plane the region of the common part of $ [1]$ and $ [2]$, $ s\\geq 0,\\ t\\geq 0$ have intersects with the family of the line $ [3]$ with the constant slope $ \\minus{} 1$. As attached file ''Graph 2'' shows the red line, when the line passes through the point $ A(1,\\ 0)$, we have $ k \\equal{} 2$ and passes through the point $ B\\left(\\sqrt {2},\\ \\frac {1}{2}\\right)$, we have $ k \\equal{} \\frac {3 \\plus{} 2\\sqrt {2}}{2}$. Since $ k$ varies continuously betwee the values (reffer to the attached file ''Graph 2''), yeilding $ 2\\leq k\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}\\Longleftrightarrow 2\\leq (1 \\plus{} a)(1 \\plus{} b)\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}$.\r\n\r\n\r\n[color=darkblue]Solution 3[/color]\r\n\r\nSimilar clue to the problem as Solution 1, 2, we can set the intersection points of the graphs of $ a^2 \\plus{} b^2 \\equal{} 1\\ (a\\geq 0,\\ b\\geq 0)$ and $ (a \\plus{} 1)(b \\plus{} 1) \\equal{} k$ as $ a \\equal{} \\cos \\theta ,\\ b \\equal{} \\sin \\theta\\ \\left(0\\leq \\theta \\leq \\frac {\\pi}{2}\\right)$. Then $ (1 \\plus{} a)(1 \\plus{} b) \\equal{} (1 \\plus{} \\cos \\theta)(1 \\plus{} \\sin \\theta) \\equal{} 1 \\plus{} \\sin \\theta \\plus{} \\cos \\theta \\plus{} \\sin \\theta \\cos \\theta$\r\n\r\n$ \\equal{} 1 \\plus{} t \\plus{} \\frac {t^2 \\minus{} 1}{2} \\equal{} \\frac {1}{2}(t \\plus{} 1)^2 \\equal{} f(t)$, where $ t \\equal{} \\sin \\theta \\plus{} \\cos \\theta$. Since $ 1\\leq t\\leq \\sqrt {2}$, yielding $ f(1)\\leq f(t)\\leq f(\\sqrt {2})\\Longleftrightarrow 2\\leq (1 \\plus{} a)(1 \\plus{} b)\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}$.\r\n\r\n\r\n[color=darkblue]Solution 4[/color]\r\n\r\n$ (a \\plus{} b)^2 \\plus{} (a \\minus{} b)^2 \\equal{} 2(a^2 \\plus{} b^2) \\equal{} 2\\Longrightarrow |a \\plus{} b|\\leq \\sqrt {2}$. Since $ a\\geq 0,\\ b\\geq 0$ we have $ 0\\leq a \\plus{} b\\leq \\sqrt {2}$.\r\n\r\nThus use A.M.-G.M. we have $ (1 \\plus{} a)(1 \\plus{} b)\\leq \\left\\{\\frac {(1 \\plus{} a) \\plus{} (1 \\plus{} b)}{2}\\right\\}^2 \\equal{} \\left(\\frac {a \\plus{} b \\plus{} 2}{2}\\right)^2\\leq \\left(\\frac {\\sqrt {2} \\plus{} 2}{2}\\right)^2$\r\n\r\n$ \\equal{} \\frac {3 \\plus{} 2\\sqrt {2}}{2}$. The equality holds when $ 1 \\plus{} a \\equal{} 1 \\plus{} b$ and $ a^2 \\plus{} b^2 \\equal{} 1\\ (a\\geq 0,\\ b\\geq 0)\\Longleftrightarrow (a,\\ b) \\equal{} \\left(\\frac {1}{\\sqrt {2}},\\ \\frac {1}{\\sqrt {2}}\\right)$.\r\n\r\n\r\n[color=red]Solution 5[/color]\r\n\r\n$ a^2 \\plus{} b^2 \\equal{} 1\\ (a\\geq 0,\\ b\\geq0)\\Longrightarrow \\minus{} 1\\leq a \\minus{} b\\leq 1$ and $ 1\\leq a \\plus{} b\\leq \\sqrt {2}$, \r\n\r\nLet $ f(a,\\ b): \\equal{} (1 \\plus{} a)(1 \\plus{} b) \\equal{} \\frac {1}{4}\\{(a \\plus{} b \\plus{} 2)^2 \\minus{} (a \\minus{} b)^2\\}$, $ f(a,\\ b)$ is maximized when $ a \\minus{} b \\equal{} 0$ and $ a \\plus{} b \\equal{} \\sqrt {2}\\Longleftrightarrow a \\equal{} b \\equal{} \\frac {1}{\\sqrt {2}}$, yielding maximum value $ \\frac {3 \\plus{} 2\\sqrt {2}}{2}$, then when $ |a \\minus{} b| \\equal{} 1$ and $ a \\plus{} b \\equal{} 1\\Longleftrightarrow (a,\\ b) \\equal{} (1,\\ 0),\\ (0,1)$, yielding minimum value $ 2$. That is to say $ 2\\leq (1 \\plus{} a)(1 \\plus{} b)\\leq \\frac {3 \\plus{} 2\\sqrt {2}}{2}$.", "Solution_3": "[hide=\"Algebraic solution for second half\"]\nBy trivial inequality, $ a^2\\plus{}b^2\\ge 2ab$ or $ \\frac{1}{2}\\ge ab$. Note now that $ (1\\plus{}a\\plus{}b)^2\\equal{}2(a\\plus{}1)(b\\plus{}1)$. By AM-GM, $ a\\plus{}b\\plus{}2\\ge \\sqrt{2(2)(a\\plus{}1)(b\\plus{}1)}\\equal{}\\sqrt{2}(1\\plus{}a\\plus{}b)$. Solving for $ a\\plus{}b$, we get $ a\\plus{}b\\le\\sqrt{2}$. Combining, $ ab\\plus{}a\\plus{}b\\plus{}1\\equal{}(a\\plus{}1)(b\\plus{}1)\\le\\frac{3\\plus{}2\\sqrt{2}}{2}$.[/hide]", "Solution_4": "Left inequalty: $ (1\\plus{}a)(1\\plus{}b)\\equal{}1\\plus{}a\\plus{}b\\plus{}ab\\geq 1\\plus{}a\\plus{}b$. As $ 1\\equal{}a^2\\plus{}b^2\\geq a^2$, $ a\\in [0,1]$, similarly $ b\\in [0,1]$ giving $ a\\plus{}b\\geq a^2\\plus{}b^2\\equal{}1\\Rightarrow (1\\plus{}a)(1\\plus{}b)\\geq 2$.\r\n\r\nRight inequality: $ (1\\plus{}a)(1\\plus{}b)\\leq \\frac{(1\\plus{}a)^2\\plus{}(1\\plus{}b)^2}{2}$. By Cauchy, $ (1\\plus{}a)^2\\leq (\\frac{\\sqrt{2}}{2}\\plus{}a^2)(\\sqrt{2}\\plus{}1)$ and a similar inequality for $ b$, giving\r\n\\[ \\frac{(1\\plus{}a)^2\\plus{}(1\\plus{}b)^2}{2}\\leq \\frac{\\sqrt{2}\\plus{}1}{2}(\\sqrt{2}\\plus{}a^2\\plus{}b^2)\\equal{}\\frac{(\\sqrt{2}\\plus{}1)^2}{2}\\equal{}\\frac{3\\plus{}2\\sqrt{2}}{2}\\]", "Solution_5": "[quote=\"Virgil Nicula\"]$ \\{\\begin{array}{c} a\\ge 0\\ ,\\ b\\ge 0 \\\\\n \\\\\na^2 + b^2 = 1\\end{array}\\ \\ \\implies\\ \\ 2\\le (1 + a)(1 + b)\\le \\frac {3 + 2\\sqrt 2}{2}$ .[/quote]\n[color=darkblue][b][u]An algebraical proof.[/u][/b] $ \\{\\begin{array}{c} 0\\le 2ab\\implies 1\\le (a + b)^2\\implies 1\\le a + b \\\\\n \\\\\n(a + b)^2\\le 2(a^2 + b^2) = 2\\implies a + b\\le\\sqrt 2\\end{array}\\|$ $ \\implies$ $ (a + b)\\in [1,\\sqrt 2]$ .\n\nObserve that $ 2(1 + a)(1 + b) =$ $ 1 + 2(a + b) + (a + b)^2 = [1 + (a + b)]^2$ . In conclusion, \n\n$ 2(1 + a)(1 + b)\\in [\\ (1 + 1)^2\\ ,\\ (1 + \\sqrt 2)^2\\ ]$ , i.e. $ 2\\le(1 + a)(1 + b)\\le\\frac {3 + \\sqrt 2}{2}$ .\n\nSimilarly prove easily the following bilateral conditioned inequality with three variables :[/color]\n\n[quote=\"Virgil Nicula\"]$ \\{\\begin{array}{c} a\\ge 0\\ ,\\ b\\ge 0 \\ ,\\ c\\ge 0\\\\\n \\\\\na^2 + b^2 +c^2= 1\\end{array}\\ \\ \\implies\\ \\ 2\\le (1 + a)(1 + b)(1+c)\\le \\frac {18 + 10\\sqrt 3}{9}$ .[/quote]", "Solution_6": "[quote=\"Virgil Nicula\"]$ \\{\\begin{array}{c} a\\ge 0\\ ,\\ b\\ge 0 \\ ,\\ c\\ge 0 \\\\\n \\\\\na^2 + b^2 + c^2 = 1\\end{array}\\ \\ \\implies\\ \\ 2\\le (1 + a)(1 + b)(1 + c)\\le \\frac {18 + 10\\sqrt 3}{9}$ .[/quote]\r\n\r\nLet $ a + b + c = k$, we have $ ab + bc + ca = \\frac {1}{2}\\{(a + b + c)^2 - (a^2 + b^2 + c^2)\\} = \\frac {k^2 - 1}{2}$.\r\n\r\nThus $ a,\\ b,\\ c$ are the roots of the cubic equation : $ (x - a)(x - b)(x - c) = \\boxed{x^3 - kx^2 + \\frac {k^2 - 1}{2}x - abc = 0}$.\r\n\r\nPlugging $ x = - 1$ into the equation gives $ (1 + a)(1 + b)(1 + c) = \\frac {1}{2}(k + 1)^2 + abc$.\r\n\r\nSince $ a\\geq 0,\\ b\\geq 0,\\ c\\geq 0\\Longrightarrow ab + bc + ca\\geq 0\\Longleftrightarrow\\frac {k^2 - 1}{2}\\geq 0\\ (k\\geq 0)$ and \r\n\r\n$ 3(a^2 + b^2 + c^2) \\geq (a + b + c)^2\\Longleftrightarrow (a - b)^2 + (b - c)^2 + (c - a)^2\\geq 0\\Longleftrightarrow 0\\leq k\\leq \\sqrt {3}$. \r\n\r\nThus the range of $ k$ is $ 1\\leq k\\leq \\sqrt {3}$. Furthermore we have $ 0\\leq abc\\leq (\\frac {a + b + c}{3})^3\\Longleftrightarrow 0\\leq abc\\leq \\frac {1}{27}k^3$.\r\n\r\n$ \\therefore \\frac {1}{2}(1 + 1)^2 + 0\\leq (1 + a)(1 + b)(1 + c)\\leq \\frac {1}{2}(\\sqrt {3} + 1)^2 + \\frac {1}{27}\\cdot (\\sqrt {3})^3$, yielding $ 2\\leq (1 + a)(1 + b)(1 + c)\\leq \\frac {18 + 10\\sqrt {3}}{9}$." } { "Tag": [ "function", "algebra", "domain", "Asymptote", "logarithms" ], "Problem": "Determine the domain, range and vertical asymptote of each log function.\r\n\r\n(1) g(x) = -3logx\r\n\r\n(2) G(x) = log(5x)", "Solution_1": "Note that in $ \\log x$,\r\nx must be greater than zero (there is no log of negative numbers or zero)", "Solution_2": "If this is that hard for you that it must be posted on here, please explain what part of the problem you need help with. Otherwise, stop trying to get other people to do your homework." } { "Tag": [ "factorial", "function", "combinatorics proposed", "combinatorics" ], "Problem": "We have a set 1, 2, 3, ..., n. Consider all the r-element subsets of\r\nthis set. Prove that if we take the smallest element of each of these\r\nsubsets, the average of these smallest elements will be (n+1)/(r+1).", "Solution_1": "Maybe we could make this a combinatorics problem:\r\n\r\n$C^{n-1}_{r-1} \\times 1 + C^{n-2}_{r-1} \\times 2 +...$\r\n\r\nNot sure if this'll help.", "Solution_2": "I've managed to get an expression that seems to be the\r\nsame as (n+1)/(r+1), but I don't know how to prove that it's the same,\r\nsince my expression uses factorials (from the Choice function) and\r\nsigma. This expression that I've obtained is the following sum:\r\n \r\nSum of k*Choose(n-k, r-1) for k = 1 to n\r\n\r\ndivided by Choose(n, r). I don't know how to show that my expression\r\nis equal to (n+1)/(r+1) which, through my many test cases, it seems to\r\nbe.", "Solution_3": "Anyone got any ideas???", "Solution_4": "Hm, this isn't that easy, this was a problem at IMO 1981...\r\n\r\nI'll move it to Olympiad Section.", "Solution_5": "There's a very elegant proof on the Kalva website along the following lines:\r\nLet $F(n,r)$ be the desired average. Then ${n\\choose r}F(n,r)=\\sum_{i=1}^{n-r+1}i{n-i\\choose r-1}$ as mann already had.\r\n\r\nBut now consider the set of all $r+1$ element subsets of $1,2,\\cdots,n+1$. There are $i{n-i\\choose r-1}$ such subsets with $i+1$ as the second smallest element, so $\\sum_{i=1}^{n-r+1}i{n-i\\choose r-1}$ must be the total number of such subsets, which is simply ${n+1\\choose r+1}$.\r\n\r\nSo ${n\\choose r}F(n,r)={n+1\\choose r+1}$, and the result $F(n,r)=\\frac{n+1}{r+1}$ follows.\r\n\r\nThis approach also makes is (relatively) simple to tackle a more general problem:\r\nLet $F(n,r,m)$ be the average value of the $m^{th}$ smallest element, taken over all $r$ element subsets of $1,2,\\cdots,n$. Prove that $F(n,r,m)=m\\frac{n+1}{r+1}$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all functions from integers to integers such that\r\n f(x+f(y))=f(x)- f(y) for all integer values of x,y\r\n\r\n [hide] i think anwer is f(x)=f(0)-x or a null function, but it it's rather hard to prove that i've found that if it's not f(0)-x then there is an integer d such that for all x f(x) is divisible by d (d is not 1), otherwise if there is at least one pair of integes x,y such that gcd(f(x),f(y))=1 then it's solved[/hide]", "Solution_1": "Let R be the range of f(x). Also Let x be in R.\r\nthen f(x) = f(0 + f(y)) = f(0) - f(y)= a -x where a=f(0) \r\nNote that this implies that f(a)=0. \r\nAlso if x belongs to R: f(a+x)= -x. so that if x in R so is -x.\r\nIf y belongs to R and x be arbitrary then f(x+y)=f(x)-y.\r\nwe also know that if x1 and x2 belong to the set, so does x1 -x2 and x2 - x1. \r\nHence if the set R be {...,y2,y1,0,x1,x2,...} where 0 B arbitrarily, where A={1,2,3,...,x1-1} and B={...,-2x1,-x1,0,x1,2x1,...}\r\nThen f(x)=g(x(mod x1)) - (x-x1) for all x.\r\nSo f(x) is not nesssacarily a-x for all x. The above construction gives all the solutions.\r\n-Ali", "Solution_2": "thanks for your solution that was a really good function \r\n there's another one here, i 've made it yesterday \r\nit's simpler than this one(i've made in 10 mins) but i liked it \r\n f(x^2-y^2)=xf(x)-yf(y) f is from reals to reals", "Solution_3": "Well, the solution took much less than 10 minutes :lol: \r\n\r\nNote that f(x+y)=f(x)+f(y) and f(x^2)=xf(x)\r\nso f((x+y)^2) = (x+y)f(x+y) = xf(x)+yf(y)+ yf(x)+xf(y)\r\nso xf(x)+y(fy)+2f(xy) = xf(x)+yf(y)+yf(x)+xf(y)\r\n2f(xy)=yf(x)+xf(y)\r\nnow let y be rational:\r\nso 2yf(x)=yf(x)+xf(y)\r\nso yf(x)=xf(y)=xyf(1)\r\nso f(x)=xf(1) so all x.\r\n-Ali", "Solution_4": "Probably, I'm asking something stupid :oops: \r\nBut how did you got f(x+y)=f(x)+f(y) ?", "Solution_5": "put y=0, to get f(x^2)=xf(x)\r\nalso from here it is easy to see that f(-x)=-f(x)\r\nso f(x^2-y^2)=f(x^2) - f(y^2) \r\nso f(a)+f(b)=f(a+b) for all b>=0. and a>=-b\r\nbut even if b<0 f(a) +f(b) = f(a) - f(-b)=f(a-(-b))=f(a+b) \r\nso no matter what , f(a+b)=f(a)+f(b)...\r\n-Ali", "Solution_6": "Ah ! I just didn't relate the fact that f(x^2)=x*f(x) to f(x^2-y^2)=x*f(x)-y*f(y)=f(x^2)-f(y^2)\r\nThanks !", "Solution_7": "i've did it in the same way :) by hte way 10 minutes was a symbolic time unit :) it's too easy for thinking about it 10 mins :)", "Solution_8": "try this one if u want, but this one is really hard \r\nit;s from the IMO 2004 shortlist \r\n f is from reals to reals \r\nf(x^2+y^2+2f(xy))=f(x+y)^2", "Solution_9": "This was discussed in much detail here:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=33734", "Solution_10": "and this one(quite easy\r\nf is from reals to reals \r\nf(x)^f(y)=f(x^y)\r\n\r\n[hide] show that f is multiplicative, additive and increasing[/hide]" } { "Tag": [ "logarithms", "floor function" ], "Problem": "I would like to compile a list of problems (with solutions) using the number 2009. For example:\r\n\r\n1. How many integer solutions, (x,y), does the equation $ x^2 \\minus{} y^2 \\equal{} 2009$ have?\r\n\r\n2. The product of two numbers is 2009 and one number is 8 less than the other. Compute the sum of these two numbers.\r\n\r\n3. How many digits does the 2009th Fibonacci number have? (Hint: Use log and Binet's formula)", "Solution_1": "[quote=\"Thorn\"]I would like to compile a list of problems (with solutions) using the number 2009. For example:\n\n1. How many integer solutions, (x,y), does the equation $ x^2 \\minus{} y^2 \\equal{} 2009$ have?\n\n2. The product of two numbers is 2009 and one number is 8 less than the other. Compute the sum of these two numbers.\n\n3. How many digits does the 2009th Fibonacci number have? (Hint: Use log and Binet's formula)[/quote]\r\n\r\nNumber 2:\r\n\r\n$ x(x \\minus{} 8) \\equal{} 2009\\equal{}x^2 \\minus{} 8x \\minus{} 2009$\r\n\r\nWe solve it and find out that $ x \\equal{} 49$ or $ x \\equal{} \\minus{} 41$.\r\n\r\nIf $ x \\equal{} 49$, then the other number is $ 41$, and the sum is $ \\boxed{90}$\r\n\r\nIf $ x \\equal{} \\minus{} 41$, then the other number is $ \\minus{} 49$, and then the sum is $ \\boxed{ \\minus{} 90}$\r\n\r\nThe answer is either $ 90$ or $ \\minus{} 90$.", "Solution_2": "[hide=\"1\"]\n$ x^2 \\minus{} y^2 \\equal{} (x \\plus{} y)(x \\minus{} y) \\equal{} 41.(7)^2$\n$ (x,y)$ has three intiger solutions[/hide]\n\n[hide=\"2\"]\n$ xy \\equal{} 2009, x \\minus{} y \\equal{} 8$\n$ x^2 \\minus{} 2xy \\plus{} y^2 \\equal{} 64$\n$ x^2 \\plus{} 2xy \\plus{} y^2 \\equal{} 64 \\plus{} 4(2009)$\n$ x \\plus{} y \\equal{} \\sqrt {64 \\plus{} 4(2009)} \\equal{} \\pm 90$[/hide]\n\n[hide=\"3\"]\nFor such a large number, and approximation will work\n$ F_n \\approx \\dfrac{\\Phi^n}{\\sqrt {5}}$\nNumber of digits: $ 1 \\plus{} \\log_{10} (F_n)$\n\n$ 1 \\plus{} \\lfloor 2009\\log_{10}(\\Phi) \\minus{} \\frac {1}{2}\\log_{10}(5) \\rfloor \\equal{} 420$\n[/hide]", "Solution_3": "That's a good start! \r\nBut problem 1 requires more explanation, and did we say x,y are positive?\r\n\r\nIs there anything special or interesting about 2009? Can we turn that interesting thing into a problem?\r\n\r\nThere must be, since all the positive integers are interesting:\r\n\r\nClearly 1 is interesting for many reasons (neither prime nor composite, the multiplicative inverse of itself,...)\r\n2 is the only even prime, which is very interesting.\r\nNow suppose that some integer n>2, n is uninteresting.\r\nThat would make n the least uninteresting number, which is [b]very interesting![/b]\r\nTherefore no such n exists, and natural numbers are interesting. Especially 2009.", "Solution_4": "[hide=\" 1.\"]Since $ 2009 \\equal{} 41 \\cdot 7^2$, it has 6 divisors and $ x^2 \\minus{} y^2 \\equal{} 2009$ has 3 solutions in positive integers. Each of these solutions has 4 permutations when one considers all integers so there are a total of 24 ordered pairs $ (x, y)$ which satisfy the equation.[/hide]", "Solution_5": "[quote=\"Illos\"][hide=\" 1.\"]Since $ 2009 \\equal{} 41 \\cdot 7^2$, it has 6 divisors and $ x^2 \\minus{} y^2 \\equal{} 2009$ has 3 solutions in positive integers. Each of these solutions has 4 permutations when one considers all integers so there are a total of 24 ordered pairs $ (x, y)$ which satisfy the equation.[/hide][/quote]\r\n\r\nYou mean 12, right? :)", "Solution_6": ".... yeah\r\n\r\n\r\n\r\n :wallbash:", "Solution_7": "[quote=\"Thorn\"]\n\nClearly 1 is interesting for many reasons (neither prime nor composite, the multiplicative inverse of itself,...)\n2 is the only even prime, which is very interesting.\nNow suppose that some integer n>2, n is uninteresting.\nThat would make n the least uninteresting number, which is [b]very interesting![/b]\nTherefore no such n exists, and natural numbers are interesting. Especially 2009.[/quote]\r\n\r\n\r\nLinguistic induction. That's new.. haven't heard of that one :wink: \r\n\r\nI like the joke. I think it's been done, but that's a fresh rendition of it :)" } { "Tag": [ "LaTeX" ], "Problem": "Hi guys,\r\nI'd like to better format the following code for a post here. Any way I can make the \\sum appear in \\displaystyle inside the \\begin{align}\\begin{cases} construct? \\displaystyle does not seem to work.\r\n\r\nAlso, any way I can line up the \"for n odd\" and \"for n even\" in both sets? Too much to ask?\r\n\r\n\\[ \\begin{align*}\r\na_{n,k}&\\equal{}\\begin{cases}\\displaystyle 2\\sum_{h\\equal{}1}^{k} (\\minus{}1)^h (h\\pi)^{n\\minus{}1} & \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n\\phantom{llll}0 & \\text{for } n \\text{ even}\r\n\\end{cases}\r\n\\\\\r\n\\\\\r\nb_{n,k}&\\equal{}\\begin{cases}\\cfrac{(\\minus{}1)^{\\frac{3\\plus{}n}{2}} 2 (2^{n\\minus{}1}\\minus{}1) B_{2n}}{(1\\plus{}n)!}\\plus{}\\displaystyle 2\\sum_{h\\equal{}1}^{k} \\cfrac{(\\minus{}1)^h}{(h\\pi)^{n\\plus{}1}} & \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n\\phantom{\\minus{}\\minus{}\\minus{}\\minus{}}0 & \\text{for } n \\text{ even}\r\n\\end{cases}\r\n\\end{align*}\\]\r\n\r\nThanks,\r\nShaw", "Solution_1": "Ok, mega-use of \\phantom for the alignment of \"for n even\" and \"for n odd\", but that's a trial and error thing and if I reformat the sums correctly then I'll have to go back and alter the \\phantom code. No way to do it automatically?\r\n\r\n\r\n\\begin{align*}\r\n a_{n,k}&=\\begin{cases}\\displaystyle 2\\sum_{h=1}^{k} (-1)^h (h\\pi)^{n-1} &\\phantom{xxxxkkjjkkxxxxxxx} \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n \\phantom{llll}0 & \\phantom{xxxxkkjjkkxxxxxxx} \\text{for } n \\text{ even}\r\n \\end{cases}\r\n \\\\\r\n \\\\\r\n b_{n,k}&=\\begin{cases}\\cfrac{(-1)^{\\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+\\displaystyle 2\\sum_{h=1}^{k} \\cfrac{(-1)^h}{(h\\pi)^{n+1}} & \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n \\phantom{----}0 & \\text{for } n \\text{ even}\r\n \\end{cases}\r\n \\end{align*}", "Solution_2": "Ok, I think I have it -- I'm sure you guys would have figured it out though: I used \\mathop\\sum\\limits for the sums.\r\n\r\n\\begin{align*} a_{n,k} & = \\begin{cases} 2\\mathop\\sum\\limits_{h = 1}^{k} ( - 1)^h (h\\pi)^{n - 1} & \\phantom{xxxxkkjjkkxxxxxxx} \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n\\phantom{llll}0 & \\phantom{xxxxkkjjkkxxxxxxx} \\text{for } n \\text{ even} \\end{cases} \\\\\r\n \\\\\r\nb_{n,k} & = \\begin{cases}\\cfrac{( - 1)^{\\frac {3 + n}{2}} 2 (2^{n - 1} - 1) B_{2n}}{(1 + n)!} +2 \\mathop\\sum\\limits_{h = 1}^{k} \\cfrac{( - 1)^h}{(h\\pi)^{n + 1}} & \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n\\phantom{ - - - - }0 & \\text{for } n \\text{ even} \\end{cases} \\end{align*}", "Solution_3": "Do not use \\mathop and \\phantom. Just use \\limits \r\n\r\n\r\n\\[ a_{n,k} \\& = \\begin{cases} 2\\sum\\limits_{h = 1}^{k} ( - 1)^h (h\\pi)^{n - 1} \\& \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n0 \\& \\text{for } n \\text{ even} \\end{cases}\\]", "Solution_4": "Ok. But I still seem to need \\phantom to line up the first set of \"for n odd\" and \"for n even\" with the second set (just a personal preference, I think it makes it a little easier to read and less cluttered):\r\n\r\n\\begin{align*}\r\na_{n,k}&=\\begin{cases}2\\sum\\limits_{h=1}^{k} (-1)^h (h\\pi)^{n-1} & \\phantom{aaaaaaaaaaaaaaaaaa}\\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n0 & \\phantom{aaaaaaaaaaaaaaaaaa}\\text{for } n \\text{ even}\r\n\\end{cases}\r\n\\\\\r\n\\\\\r\nb_{n,k}&=\\begin{cases}\\cfrac{(-1)^{\\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+2\\sum\\limits_{h=1}^{k} \\cfrac{(-1)^h}{(h\\pi)^{n+1}} & \\text{for } n \\text{ odd}\\vspace{10pt} \\\\\r\n0 & \\text{for } n \\text{ even}\r\n\\end{cases}\r\n\\end{align*}", "Solution_5": "I think it's easier to read.", "Solution_6": "[quote=\"Valentin Vornicu\"]I think think it's easier to read.[/quote]\r\n\r\n\"think think\" as in not easier to read? I don't know for sure. Still learning.", "Solution_7": "\\hspace is better than \\phantom here:\r\n$ \\begin{cases}\r\na,&\\hspace{100pt}b\\\\\r\ncxx,&\\hspace{100pt}d\r\n\\end{cases}$\r\nI suspect that stevem can come up with even better solution though :)", "Solution_8": "Thanks for the vote of confidence fedja :)\r\nThe problem here is that the 2 cases environments are not connected so LaTeX can't tell in the first one how wide the second one is going to be. \r\nIt's possible to get round this by abandoning the align and cases environments and putting everything in a table to ensure it all lines up. But it would be complicated and would need the multirow and bigdelim packages so wouldn't work on the forum.\r\nIt's also possible to set up boxes containing some of the formulae, measure the width of the boxes and use that to find the amount of horizontal space required.\r\nBut neither method would be as simple as using \\hspace with a bit of trial and error, so I suggest sticking with that.\r\n\r\nIn a document, instead of \\text{for } n \\text{ odd}, you can write \\text{for $ \\$$n$ \\$$ odd} but that won't work on the forum. However, you can replace $ \\$$...$ \\$$ by \\(...\\) (plus an extra space that has been swallowed up) to get \\text{for \\(\\ n\\) odd}.\r\n\\begin{align*}\r\na_{n,k}&=\\begin{cases}2\\sum\\limits_{h=1}^{k} (-1)^h (h\\pi)^{n-1} & \\hspace{9.5em}\\text{for \\(\\ n\\) odd}\\\\[10pt]\r\n0 & \\hspace{9.5em}\\text{for \\(\\ n\\) even}\r\n\\end{cases}\r\n\\\\\r\n\\\\\r\nb_{n,k}&=\\begin{cases}\\cfrac{(-1)^{\\frac{3+n}{2}} 2 (2^{n-1}-1) B_{2n}}{(1+n)!}+2\\sum\\limits_{h=1}^{k} \\cfrac{(-1)^h}{(h\\pi)^{n+1}} & \\text{for \\(\\ n\\) odd}\\\\[10pt]\r\n0 & \\text{for \\(\\ n\\) even}\r\n\\end{cases}\r\n\\end{align*}", "Solution_9": "I bet a package could be written, say \"multicases\" to do just this. Here's the pseudo-code:\r\n\r\n[code]\\begin{multicases}\na_1=\\begin{cases} long text here & n odd \\\\\n 0& n even\n \\end{cases}\na_2=\\begin{cases} 2x& n odd \\\\\n more long text & n even\n \\end{cases}\n\\end{multicases}[/code]" } { "Tag": [ "Old" ], "Problem": "A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. If the magnitude of the magnetic field is 0.285 mT at a point 4.0 mm from the axis of the wire and 0.200 mT at a point 10 mm from the axis, what is the radius of the wire?\r\n\r\nand\r\n\r\nA long solenoid (n = 1200 turns/m, radius = 2.0 cm) has a current of a 0.30 A in its winding. A long wire carrying a current of 20 A is parallel to and 1.0 cm from the axis of the solenoid. What is the magnitude of the resulting magnetic field at a point on the axis of the solenoid?\r\n\r\nThanks for the help", "Solution_1": "okay its really embarrassing I forgot how to do this? can anybody help, I think you using the line $ \\oint \\vec{B} \\cdot \\vec{dl} \\equal{} \\mu_0 I_{\\text{enclosed}}$", "Solution_2": "Therefore, your problem is almost done except you have to make calculations. Don\u2019t confuse a(distance from the axis) with radius of the wire r. You are given two values of B dependent on the distance from axis so you will have 2 equations of 2 unknowns. Solve the equation and obtain the result." } { "Tag": [], "Problem": "\u0395\u03c0\u03b1\u03bd\u03b5\u03b9\u03bb\u03b8\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03c9 \u03c0\u03b1\u03b9\u03b4\u03b5\u03b9\u03b1 \u03c5\u03c3\u03c4\u03b5\u03c1\u03b1 \u03b1\u03c0\u03bf \u03b4\u03b9\u03b1\u03ba\u03bf\u03c0\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b5\u03b9\u03c7\u03b1 \u03c0\u03b1\u03b5\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1 \u03bf\u03bc\u03bf\u03c1\u03c6\u03bf \u03c7\u03c9\u03c1\u03b9\u03bf \u03b5\u03be\u03c9 \u03b1\u03c0\u03bf \u03c4\u03bf \u0393\u03c5\u03b8\u03b5\u03b9\u03bf....\u039f\u03bc\u03bf\u03bb\u03bf\u03b3\u03c9 \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03c0\u03c9\u03c2 \u03bc\u03bf\u03c5 \u03b5\u03bb\u03b5\u03b9\u03c8\u03b5 \u03b9\u03b4\u03b9\u03b1\u03b9\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u0398.\u0391 \u03c4\u03bf\u03c5 \u03a0\u03bf\u03c5\u03bb\u03bf\u03c5(\u03b5\u03c4\u03c3\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c3\u03c4\u03b7 \u03bc\u03bf\u03bd\u03bf \u03c3\u03c4\u03bf \u03c7\u03b1\u03c1\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03bd \u03c3\u03c4\u03c5\u03bb\u03bf...)\u0395\u03ba\u03b5\u03b9 \u03b2\u03b5\u03b2\u03b1\u03b9\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b9\u03ba\u03b1 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b5\u03b9\u03c2.\u0391\u03c2 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03c9 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03c4\u03b7\u03bd \u03c0\u03c1\u03c9\u03c4\u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7:N\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bd \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03b1\u03b8\u03bf\u03c5\u03bd \u03bc\u03b5 \u03c4\u03b7\u03bd \u03bc\u03bf\u03c1\u03c6\u03b7 (\u03b1^2+1)/(\u03b2^2-1) \u03bf\u03c0\u03bf\u03c5 \u03bf\u03b9 \u03b1,\u03b2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03b9 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03b9.\u03a0\u03b5\u03c1\u03b9\u03bc\u03b5\u03bd\u03c9.... :lol:", "Solution_1": "\u039a\u03b1\u03bc\u03b9\u03b1 \u03b9\u03b4\u03b5\u03b1?When (a^2+1)/(b^2-1) is an integer?(a,b are possitive integers) :huh:", "Solution_2": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03bc\u03bf\u03c5 \u03b8\u03c5\u03bc\u03af\u03b6\u03b5\u03b9\u03c4\u03b1 \u03b5\u03be\u03ae\u03c2 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1....\r\n\u0399\u039c\u039f 1988 \u0391\u03bd $\\frac{a^{2}+b^{2}}{ab+1}=q$ \u03cc\u03c0\u03bf\u03c5 $q$ \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03bf $q$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03ad\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03bf .\r\n\r\n(\u039a\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c0\u03bb\u03ae\u03c3\u03b9\u03bf)\r\n\u0391\u03bd $\\frac{a^{2}+b^{2}}{ab-1}=m$ \u03cc\u03c0\u03bf\u03c5 $m$ \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c2 \u03c4\u03cc\u03c4\u03b5 $m=5$ \r\n\r\nKai \u03c4\u03b1 \u03b4\u03cd\u03bf \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03af\u03b4\u03b9\u03bf \u03c4\u03c1\u03cc\u03c0\u03bf (\u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c5\u03c0\u03b5\u03c1\u03b2\u03bf\u03bb\u03ad\u03c2) .....\r\n\u0392\u03bb\u03ad\u03c0\u03c9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b4\u03b9\u03ba\u03cc \u03c3\u03bf\u03c5 \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03ba\u03b1\u03c4\u03b7\u03b3\u03bf\u03c1\u03af\u03b1 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03b1\u03c1\u03cc\u03bc\u03bf\u03b9\u03b1......\r\n\u0391\u03bb\u03bb\u03ac \u03b1\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03bb\u03ac \u03c3\u03ba\u03ad\u03c8\u03b7 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03b1\u03c7\u03bf\u03bb\u03b7\u03b8\u03b5\u03af \u03b1\u03ba\u03cc\u03bc\u03b1.... :roll:", "Solution_3": "\u03a0\u03bf\u03bb\u03c5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03bd \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03b1\u03bd \u03ba\u03b1\u03b9 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03c9 \u03bc\u03bf\u03bd\u03bf\u03c2 \u03bc\u03bf\u03c5 \u03c4\u03bf \u03b5\u03c6\u03c4\u03b9\u03b1\u03be\u03b1(\u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03c3\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03bd \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03bf) \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03b7.\u0398\u03b1 \u03b7\u03b8\u03b5\u03bb\u03b1 \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9\u03c2 \u03c0\u03b9\u03bf \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03bf\u03c2. :)", "Solution_4": "\u03cc\u03c7\u03b9 \u03bf\u03c4\u03b9 \u03c4\u03bf \u03ad\u03c8\u03b1\u03be\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c0\u03bf\u03ba\u03bb\u03b5\u03af\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03ba\u03b1\u03b8\u03b1\u03c1\u03ae \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd \u03b4\u03b9\u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03c5\u03c0\u03b5\u03c1\u03b2\u03bf\u03bb\u03ad\u03c2 :D ? \r\n\u03b5\u03c1\u03c9\u03c4\u03c9", "Solution_5": ":) \u039f\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b5\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03b9\u03b4\u03b5\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03b1\u03c2 \u03c4\u03b7\u03bd \u03c0\u03b5\u03b9.", "Solution_6": "\u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03b5\u03c7\u03c9 \u03bc\u03b9\u03b1 \u03b9\u03b4\u03b5\u03b1 . \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c9?", "Solution_7": "\u03a6\u03c5\u03c3\u03b9\u03ba\u03b1 \u039a\u03c9\u03c3\u03c4\u03b1!\u03a0\u03b5\u03c1\u03b9\u03bc\u03b5\u03bd\u03c9.... :)", "Solution_8": "\u03bf \u03b1^2+1 \u03b5\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2 4\u03ba+1.\u03b1\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03bf \u03b2^2-1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03bf\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03c9\u03c4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03b9\u03c1. \u03c3\u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03b7 4\u03ba+1. \u03c3\u03c9\u03c3\u03c4\u03b1?", "Solution_9": "\u039f\u03c7\u03b9! \u03a5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b1\u03bd\u03c4\u03b9\u03c0\u03b1\u03c1\u03b1\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03b5\u03b9\u03c0\u03b5\u03c2.\u03c0.\u03c7 \u03b3\u03b9\u03b1 \u03b2=6 \u03c4\u03bf\u03c4\u03b5 \u03bf \u03b2^2-1=6^-1=0mod7 \u03bf \u03c0\u03c9\u03c2 \u03b2\u03bb\u03b5\u03c0\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c5 0 7 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2 4\u03ba+1. :)", "Solution_10": "\u03c1\u03b5 \u03b4\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7 \u03b1\u03bd d \u03b5\u03bd\u03b1\u03c2 \u03c0\u03c1\u03c9\u03c4\u03bf\u03c2 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03b7\u03c2 \u03c4\u03bf\u03c5 \u03b2^2-1 \u03c4\u03bf\u03c4\u03b5 d \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b1^2+1 \u03b1\u03c1\u03b1 d = 4\u03ba+1. \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b4\u03b7\u03bb \u03b7 \u03b5\u03c0\u03b9\u03c0\u03bb\u03b5\u03bf\u03bd \u03c3\u03c5\u03bd\u03b8\u03b7\u03ba\u03b7 \u03b2^2-1 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03b9 \u03b1^2+1.", "Solution_11": "\u03a3\u03c9\u03c3\u03c4\u03bf\u03c2.\u0393\u03b9\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b9\u03c3\u03b5....", "Solution_12": "\u03b1\u03bd \u03bd=\u03b1^2+1/\u03b2^2-1 \u03b8\u03b5\u03c9\u03c1\u03c9 \u03c4\u03bf \u03bd+1 . \u03b5\u03c7\u03c9 \u03bd+1=\u03b1^2+\u03b2^2/\u03b2^2-1 \u03c4\u03bf\u03c4\u03b5 \u03b1^2+\u03b2^2=\u03b1^2+2=\u03bd+1mod4 .\u03bc\u03b7\u03c0\u03c9\u03c2 \u03ba\u03b1\u03bd\u03c9 \u03bb\u03b1\u03b8\u03bf\u03c2?", "Solution_13": "\u0394\u03b5\u03bd \u03c3\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03b9\u03bd\u03c9...", "Solution_14": ":icecream:", "Solution_15": "\u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03c0\u03b1\u03c9. \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03c3\u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03c3\u03bc\u03b1.\u03b5\u03be\u03b1\u03bb\u03bf\u03c5 \u03b9\u03c3\u03c9\u03c2 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03b9 \u03b9\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c0\u03c9 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b1 :)", "Solution_16": "\u039a\u03c9\u03c3\u03c4\u03b1 \u03b1\u03c3\u03b5 \u03c4\u03b1 \u03c5\u03c0\u03bf\u03bd\u03bf\u03bf\u03c5\u03bc\u03b5\u03bd\u03b1! :diablo:", "Solution_17": ":harhar: \u03c0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03b5\u03b9\u03c4\u03b5\u03c4\u03b7\u03bd \u03b5\u03c7\u03c9 \u03bb\u03c5\u03c3\u03b5\u03b9 \u03ae \u03bf\u03c7\u03b9 \u03b1\u03c5\u03c4\u03b7 \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c5\u03c0\u03b5\u03c1\u03bf\u03c7\u03b7.\u03b8\u03b1 \u03b7\u03c4\u03b1\u03bd \u03c0\u03b5\u03c1\u03b9\u03b6\u03b7\u03c4\u03b7\u03c4\u03b7 \u03c3\u03c4\u03bf \u03b1dvanced.\u03bf\u03c3\u03bf \u03b3\u03b9\u03b1 \u03b7 \u03bb\u03c5\u03c3\u03b7... \u03b1\u03c5\u03c4\u03bf \u03ba\u03b9 \u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03bf(\u03c5\u03c0\u03bf\u03bd\u03bf\u03bf\u03c5\u03bc\u03b5\u03bd\u03bf \u039d\u03bf2)", "Solution_18": "\u03a9\u03c3\u03c4\u03b5 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c0\u03b5\u03b9\u03c2 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bb\u03bb\u03bf \u039a\u03c9\u03c3\u03c4\u03b7? :gleam:", "Solution_19": "\u03b5\u03c7\u03c9 \u03b4\u03b9\u03b1\u03b2\u03b1\u03c3\u03bc\u03b1 \u03b4\u03b5\u03bd \u03c0\u03c1\u03bf\u03bb\u03b1\u03b2\u03b1\u03b2\u03b1\u03b9\u03bd\u03c9 \u03bd\u03b1 \u03c0\u03c9 \u03c0\u03bf\u03bb\u03bb\u03b1.\u03bd\u03b1 \u03b4\u03c9\u03c3\u03c9 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b5\u03b3\u03c9 (\u03b1\u03bd \u03c4\u03b7\u03bd \u03b4\u03c9\u03c3\u03c9) \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03c3\u03b7\u03bc\u03b1\u03c3\u03b9\u03b1.\u03c4\u03bf \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03c4\u03b9 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03b9\u03c0\u03b1.\u03b1\u03bd \u03c4\u03b5\u03bb\u03bf\u03c2\u03c0\u03b1\u03bd\u03c4\u03c9\u03bd \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b1\u03c0\u03bf \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b1\u03bd \u03bf\u03c0\u03c9\u03c2 \u03bf \u03c3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03b8\u03b5\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c9 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ba\u03b7 \u03bc\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03b5\u03b3\u03b3\u03b9\u03c3\u03b7 \u03b8\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c9.\u03bf\u03ba? :blush:", "Solution_20": "\u039f.\u039a!\u03a0\u03b1\u03b9\u03b4\u03b5\u03b9\u03b1 \u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03b7\u03bd \u03b5\u03c7\u03b5\u03b9 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03b5\u03b9 \u03b1\u03c2 \u03c0\u03b5\u03b9 \u03ba\u03b1\u03c4\u03b9.... :wink:", "Solution_21": "\u0395\u03b3\u03ce \u03c4\u03b7\u03bd \u03ad\u03c7\u03c9 \u03c8\u03ac\u03be\u03b5\u03b9 \u03bc\u03b5 \u03ba\u03b1\u03b8\u03b1\u03c1\u03ae \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03ba\u03b9 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03ad\u03c2 \u03ba\u03b9 \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03b5\u03b2\u03b3\u03b1\u03b9\u03bd\u03b1\u03bd \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b8\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03bf \u03be\u03b5\u03c7\u03ac\u03c3\u03c9 \u03ad\u03c4\u03c3\u03b9 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac?", "Solution_22": "\u03bf\u03bc\u03c9\u03c2 \u03b1\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03b5 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 \u03c4\u03b7 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03c4\u03c9\u03bd \u03b1,\u03b2.\u03b9\u03c3\u03c9\u03c2 \u03b1\u03bd \u03b4\u03bf\u03ba\u03b9\u03bc\u03b1\u03b6\u03b5\u03c2 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1.", "Solution_23": "\u03a6\u03af\u03bb\u03b5 \u03ba\u03ce\u03c3\u03c4\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03b5\u03b9\u03c4\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03ce \u03c3\u03c7\u03ad\u03c3\u03b7 \u03b5\u03b9\u03c4\u03b5 \u03bd\u03b1 \u03c6\u03c1\u03ac\u03be\u03c9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03ba\u03bb\u03ac\u03c3\u03bc\u03b1 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03c9\u03bd \u03ba\u03b9 \u03b4\u03b5\u03bd \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03b8\u03b1 \u03c4\u03bf \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03c9 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03ba\u03ac\u03c0\u03c9\u03c2 \u03bd\u03b1 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9", "Solution_24": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03b1\u03bd \u03ba\u03b1\u03bd\u03b5\u03b9\u03c2 \u03b5\u03c7\u03b5\u03b9 \u03b2\u03c1\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03b8\u03b1 \u03b7\u03b8\u03b5\u03bb\u03b1 \u03c0\u03bf\u03bb\u03c5 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b5\u03b9. :lol:", "Solution_25": "[size=150][color=darkblue][b]\u03a0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1:[/b][/color][/size] \u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bd \u03cc\u03bb\u03b1 \u03c4\u03b1 $a,b \\in \\mathbb Z$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $N=\\frac{a^{2}+1}{b^{2}-1}$ \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c2.\r\n\r\n\r\n[color=green][b]\u039b\u03ae\u03bc\u03bc\u03b1:[/b][/color] \u0391\u03bd $d|(a^{2}+1)$ ,\u03cc\u03c0\u03bf\u03c5 $a,d \\in \\mathbb Z$ \u03c4\u03cc\u03c4\u03b5 \u03bf $d$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03ae $8k+2$ \u03bc\u03b5 $k \\in \\mathbb Z$. \r\n\r\n[u][color=blue][b]\u039b\u03cd\u03c3\u03b7:[/b][/color][/u]\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 $(b^{2}-1)|(a^{2}+1)$ \u03c4\u03cc\u03c4\u03b5:\r\n\r\n- \u0391\u03bd \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bf $b$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2, \u03b1\u03c6\u03bf\u03cd \u03b1\u03bd \u03ae\u03c4\u03b1\u03bd \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2 \u03b8\u03b1 \u03af\u03c3\u03c7\u03c5\u03b5 \u03cc\u03c4\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc (\u03ac\u03c4\u03bf\u03c0\u03bf). \u0388\u03c3\u03c4\u03c9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd $a=2r$ \u03ba\u03b1\u03b9 $b=2m$ \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $(4m^{2}-1)|(4r^{2}+1)$ \u03bb\u03cc\u03b3\u03c9 \u03cc\u03bc\u03c9\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03bb\u03ae\u03bc\u03bc\u03b1\u03c4\u03bf\u03c2 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $4r^{2}+1$ \u03ad\u03c7\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03ae $8k+2$, \u03ac\u03c1\u03b1 \u03bf $4m^{2}-1$ \u03ad\u03c7\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ad\u03c2 (\u03ac\u03c4\u03bf\u03c0\u03bf).\r\n\r\n- \u0391\u03bd \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $a$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2 \u03b4\u03b9\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b4\u03cd\u03bf \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2:\r\n \r\n 1)\u0391\u03bd \u03bf $b$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2, \u03ad\u03c3\u03c4\u03c9 $a=2r+1$ \u03ba\u03b1\u03b9 $b=2m+1$ \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $(4m^{2}+4m)|(4r^{2}+4r+2)$ \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $4|(4r^{2}+4r+2)$ (\u03ac\u03c4\u03bf\u03c0\u03bf).\r\n \r\n 2)\u0391\u03bd \u03bf $b$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2, \u03ad\u03c3\u03c4\u03c9 $a=2r+1$ \u03ba\u03b1\u03b9 $b=2m$ \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $(4m^{2}-1)|(4r^{2}+4r+2) \\Leftrightarrow (4m^{2}-1)|(2r^{2}+2r+1)$. \u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03cc\u03bc\u03c9\u03c2 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03bb\u03ae\u03bc\u03bc\u03b1 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $4r^{2}+4r+2=(2r+1)^{2}+1$ \u03ad\u03c7\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03ae $8k+2$ \u03ac\u03c1\u03b1 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $2r^{2}+2r+1$ \u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03bf $4m^{2}-1$ \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae (\u03ac\u03c4\u03bf\u03c0\u03bf).\r\n\r\n\u0386\u03c1\u03b1 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $a,b \\in \\mathbb Z$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $N=\\frac{a^{2}+1}{b^{2}-1}$ \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf\u03c2.\r\n\r\nStelios...", "Solution_26": "\u03c3\u03c4\u03b5\u03bb\u03b9\u03bf \u03ba\u03b1\u03b9 \u03b1\u03bd \u03c0\u03b1\u03c1\u03b5\u03b9\u03c2 \u03bc\u03bf\u03bd\u03bf d=4k+1 \u03c0\u03b1\u03bb\u03b9 \u03b4\u03b5 \u03b2\u03b3\u03b1\u03bd\u03b5\u03b9?", "Solution_27": "\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c0\u03ac\u03c1\u03c9 \u03cc\u03c4\u03b9 $d=4k+1$ \u03b1\u03c6\u03bf\u03cd \u03c0.\u03c7. \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $105^{2}+1=11026$ \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd $74$ \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$. \u0393\u03b9' \u03b1\u03c5\u03c4\u03cc \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03cc\u03c4\u03b9 \u03bf $d$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $8k+2$. :wink:", "Solution_28": "\u03bd\u03b1\u03b9 \u03b1\u03bb\u03bb\u03b1 \u03bf 74 \u03b4\u03b5\u03bd \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2 \u03bd^2-1 ? \u03b5\u03c6\u03bf\u03c3\u03bf\u03bd \u03b1\u03c5\u03c4\u03bf \u03c4\u03bf \u03ba\u03bb\u03b1\u03c3\u03bc\u03b1 \u03b5\u03c7\u03c9", "Solution_29": "\u039d\u03b1\u03b9, \u03cc\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c0\u03b5\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03bd \u03cc\u03c4\u03b9 \u03b1\u03bd \u03bf d \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf \u03b1^2+1 \u03c4\u03cc\u03c4\u03b5 \u03bf d \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 4\u03ba+1. \u03a4\u03bf \u03bb\u03ae\u03bc\u03bc\u03b1 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03cc \u03ad\u03c4\u03c3\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03c4\u03bf \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1.", "Solution_30": "\u03b5\u03c7\u03b5\u03b9\u03c2 \u03b4\u03b9\u03ba\u03b9\u03bf.\u03b5\u03b3\u03c9 \u03b5\u03bd\u03bd\u03bf\u03bf\u03c5\u03c3\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03c9\u03c4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03c1\u03b5\u03c4\u03b5\u03c2.\u03c9\u03c1\u03b1\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 :)", "Solution_31": "[quote=\"\u0399\"]\n[color=green][b]\u039b\u03ae\u03bc\u03bc\u03b1:[/b][/color] \u0391\u03bd $d|(a^{2}+1)$ ,\u03cc\u03c0\u03bf\u03c5 $a,d \\in \\mathbb Z$ \u03c4\u03cc\u03c4\u03b5 \u03bf $d$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+1$ \u03ae $8k+2$ \u03bc\u03b5 $k \\in \\mathbb Z$.\n[/quote]\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af \u03bc\u03ae\u03c0\u03c9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3'\u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03bb\u03ae\u03bc\u03bc\u03b1, \u03b3\u03b9\u03b1\u03c4\u03af \u03b1\u03c5\u03c4\u03ae \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03ba\u03bf\u03bb\u03bb\u03ac\u03b5\u03b9... :maybe: .", "Solution_32": "\u039d\u03b1 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03bc\u03bf\u03c5.\r\n\u039f $a^{2}+1$ \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b5\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4k+3$. \u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03b1\u03bd \u03b5\u03af\u03c7\u03b5 \u03b8\u03b1 \u03b5\u03af\u03c7\u03b5 \u03ba\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03ce\u03c4\u03bf \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b7 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 4t+3 \u03b1\u03c6\u03bf\u03cd \u03ba\u03ac\u03b8\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 4k+3 \u03ad\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1\u03bd \u03c4\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd \u03c0\u03c1\u03ce\u03c4\u03bf \u03b4\u03b9\u03b1\u03b9\u03c1\u03ad\u03c4\u03b7 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4t+3=p$ .\u0386\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n $a^{2}\\equiv-1 \\mod{p}$. \u0386\u03c1\u03b1 \u03b1\u03c0\u03cc \u03b5\u03ba\u03b5\u03af \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $(a^{2})^{\\frac{p-1}{2}}=a^{p-1}\\equiv 1 \\equiv (-1)^{\\frac{p-1}{2}}\\mod{p}$.\r\n\u0386\u03c1\u03b1 \u03bf $\\frac{p-1}{2}$ \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03bf \u03b3\u03b9\u03b1 $p=4t+3$ . \r\n\r\n\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03cc\u03c2....", "Solution_33": "\u03a3\u03c9\u03c3\u03c4\u03cc\u03c2... .\r\nThanks! :wink:", "Solution_34": "\u03b7 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b3\u03b9\u03b1 8\u03ba+2 \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03bc\u03bf\u03bd\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf 2? \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03b9 \u03b1\u03bb\u03bb\u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03b5 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03b5\u03c2 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9\u03c2 \u03c4\u03bf \u03bc\u03b9\u03ba\u03c1\u03bf \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03a6\u03b5\u03c1\u03bc\u03b1.please \u03bc\u03b7\u03bd \u03be\u03b5\u03c7\u03bd\u03b1\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9\u03c2 \u03b3\u03b9\u03b1\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03bf \u03b3\u03bd\u03c9\u03c1\u03b9\u03b6\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 .", "Solution_35": "\u039c\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bd\u03b1 \u03c3\u03c5\u03bd\u03c4\u03bf\u03bc\u03ad\u03c8\u03bf\u03c5\u03bc\u03b5 \u03bb\u03af\u03b3\u03bf \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03b7.\r\n\r\n\u0391\u03bd $b =2k+1$ \u03c4\u03cc\u03c4\u03b5 $b^{2}-1 \\equiv0\\mod 4$ \u03b5\u03bd\u03ce $a^{2}+1 \\not \\equiv 0 \\mod 4$ .\r\n\r\n\u0391\u03bd $b$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $b=4k$ \u03ae $b=4k+2$ \u03ac\u03c1\u03b1 \u03c4\u03bf $b^{2}-1=(b-1)(b+1)$ \r\n\r\n\u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 $2$ \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b1\u03c1\u03ac\u03b3\u03bf\u03bd\u03c4\u03b1 \u03c4\u03b7\u03c2 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 $4m+3$ \r\n\r\n\u03b1\u03b4\u03cd\u03bd\u03b1\u03c4\u03bf \u03c3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03bb\u03ae\u03bc\u03bc\u03b1 .\r\n\r\n-------------------------------------------------------------------\r\n\r\n\u039d\u03b1\u03b9 \u039a\u03ce\u03c3\u03c4\u03b1 \u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae $8k+2$ \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03bf $d$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 .", "Solution_36": "\u03bc\u03b9\u03b1 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03bf\u03c5\u03c3\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5781[/url]", "Solution_37": "[quote=\"\u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6\"]\u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03c5\u03bd \u03bf\u03bb\u03bf\u03b9 \u03bf\u03b9 \u03b1\u03ba\u03b5\u03c1\u03b1\u03bf\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03b9 \u03bf\u03b9 \u03bf\u03c0\u03bf\u03b9\u03bf\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b9\u03c2 \u03bc\u03bf\u03c1\u03c6\u03b7\u03c2: $\\frac{a^{2}+1}{b^{2}-1}$, \u03bc\u03b5 $a,b$ \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03b9[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=784285#p784285]\u03b5\u03b4\u03ce[/url]" } { "Tag": [ "irrational number" ], "Problem": "In my class as a bonus my teacher gave 3to the 1.5 power. I didn't know so I guessed. Now that I think of it the answer might be 4.5 because 3 to the 1.5 power + 3 to the 1.5 power = 3 to the 3rd power which is 9. Am I right? and how do you do harder questions like 9 to the 7.8 power?", "Solution_1": "I don't know either, and I am in geometry. sad. \r\n\r\nI hope someone answers this.\r\n\r\nI just know that it has something to do with origins or something but have no idea.", "Solution_2": "$3^{1.5}$\r\n$3^{3/2}$\r\n$\\sqrt{3^{3}}$\r\n$\\sqrt{27}$\r\n$3\\sqrt{3}$\r\n\r\nWhich is roughly $5.196$", "Solution_3": "Let $m,n\\in \\mathbb{Z},n\\ne 0$ and $a\\geq 0$. Then $a^{m/n}=\\sqrt[n]{a^{m}}$. For irrational number $x$ we have defined \\[a^{x}=\\sup \\,\\{a^{q}| q\\in\\mathbb{Q}\\text{ and }q> x\\}.\\]", "Solution_4": "how can I delete this post?", "Solution_5": "[quote=\"puuhikki\"]Let $m,n\\in \\mathbb{Z},n\\ne 0$ and $a\\geq 0$. Then $a^{m/n}=\\sqrt[n]{a^{m}}$. For irrational number $x$ we have defined \\[a^{x}=\\sup \\,\\{a^{q}| q\\in\\mathbb{Q}\\text{ and }q> x\\}.\\] [/quote]\r\n\r\nIf what you are saying is right then this is way beyond my level", "Solution_6": "[quote=\"Ihatepie\"][quote=\"puuhikki\"]Let $m,n\\in \\mathbb{Z},n\\ne 0$ and $a\\geq 0$. Then $a^{m/n}=\\sqrt[n]{a^{m}}$. For irrational number $x$ we have defined \\[a^{x}=\\sup \\,\\{a^{q}| q\\in\\mathbb{Q}\\text{ and }q> x\\}.\\] [/quote]\n\nIf what you are saying is right then this is way beyond my level[/quote]\r\n\r\nThe first two sentences are definitely within grasp - it's saying in math language that when m and n are real and a is greater than (or equal to) 0, a to the $\\frac{m}{n}$ power is the nth root of a to the mth power.", "Solution_7": "[quote=\"Ihatepie\"]In my class as a bonus my teacher gave 3to the 1.5 power. I didn't know so I guessed. Now that I think of it the answer might be 4.5 because 3 to the 1.5 power + 3 to the 1.5 power = 3 to the 3rd power which is 9. Am I right? and how do you do harder questions like 9 to the 7.8 power?[/quote]\r\nA couple of points I wanted to make about your reasoning:\r\n$a^{b}+a^{c}\\neq a^{b+c}$ You can check this with $2^{2}+2^{3}\\neq 2^{5}$\r\nSo $3^{1.5}+3^{1.5}\\neq 3^{3}$. \r\nAlso, $3^{3}\\neq 9$.", "Solution_8": "Basically, $a^\\frac{p}{q}=\\sqrt[q]{a^{p}}$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Let $A\\subset [0,1]$ be the set of all numbers with a decimal expansion that has only digits $0$, $5$, and $9$. (An infinite tail of $9$s is allowed.) Let $B\\subset [0,1]$ be the set of all numbers with a decimal expansion that has only digits $0$, $8$, and $9$. \r\n\r\n#1. Find a Lipschitz map from $A$ onto $B$.\r\n#2. Is there a Lipschitz map from $B$ onto $A$? \r\n\r\nWarning: #2 is a version of an open problem in the book \"Fractured fractals and broken dreams\" by David and Semmes. A strange title, but it fits the book.\r\n\r\nEDIT: Three Chinese mathematicians (Rao, Ruan, and Xi) recently solved #2: there is even a bi-Lipschitz map between $A$ and $B$. :o", "Solution_1": "I think the most obvious approach is to read [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X1B-4HYV07W-6&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1032703163&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=2eaf0af5170e97dccd5e3df3a99569b2]the original article[/url] written by Rao, Ruan, and Xi." } { "Tag": [ "function", "Support", "real analysis", "topology", "real analysis unsolved" ], "Problem": "Hi,\r\n\r\nI came across an interesting problem in Rudin's PMA, for which I haven't been able to find a solution. Help is needed.\r\n\r\nHere goes: Is it possible for each closed subset E of R1, to find a real valued f defined on R1 such that\r\n \r\n (1) f is infinitely differentiable on R1\r\n (2) the zero set of f is exactly E ?\r\n \r\n\r\n\r\nI think I have been able to find an f which is n times continuously differentiable, as follows:\r\n\r\nsuppose E such that E is closed and an arbitrary integer n.\r\n\r\n=> R1\\E is open.\r\n\r\nR1\\E is an open subset of R1 => it is a union of disjoint open segments (ak,bk) ; k=1,2,3...\r\n\r\nDefine f(x)=\r\n 0 ;when x belongs to E\r\n ((x-ak)*(bk-x))^(n+1) ;when x belongs to (ak,bk).\r\n\r\nNow f(x) is n times continuously differentiable on R1 and the zero set of f is exactly E.\r\n\r\n\r\nBut when it's time to find a function that is infinitely differentiable, I can't find one:\r\nI came to the conclusion that if the Taylor's serie of f converges to f, we have for every x belongs to (ak,bk)\r\n\r\nf(x)= Sum[n goes from zero to infinity] ( f(n) (ak)/n!* (x-ak)^n)=0 ,\r\n since f(n)(ak)=0 for every ak,\r\n (because ak belongs to E and f(n) is continuous).\r\n\r\n=> f does not fulfill the conditions wanted because f is zero also on some point of R1\\E. \r\n\r\nSo then this would mean that f must not be representable by its Taylor's serie. But this makes everything pretty difficult...\r\n\r\nThe only infinitely continuosly differentiable function I know which is not representable by it's taylor serie is the famous exp(-1/x\u00b2),\r\nbut is it the only function with such properties?\r\n\r\nCan anyone come up with a function? Or perhaps a mistake in my reasoning?\r\n\r\nThanks a lot in advance for any help\r\n-Laura", "Solution_1": "Here's how to build it:\r\nStep 1: Construct a $C^\\infty$ function $g$ with compact support (zero outside the interval $[0,1]$).\r\nLet $g(x)=0$ if $x\\not\\in(0,1)$. Let $g(x)=e^{-\\frac1x}\\cdot e^{-\\frac1{1-x}}$ if $x\\in(0,1)$. This is basically the example you already know.\r\n\r\nNow, we have $\\mathbb{R}\\setminus E=\\cup_n (a_n,b_n)$, where $(a_n,b_n)$ are disjoint open intervals. Choose nonzero constants $c_n$, and let $f(x)=\\sum_n c_ng\\left(\\frac{x-a_n}{b_n-a_n}\\right)$. That's almost the answer, but it doesn't deal with infinite intervals properly. If $b_n=\\infty$, we use $e^{-\\frac1{x-a_n}}$ instead and if $a_n=-\\infty$, we use $e^{-\\frac1{b_n-x}}$ instead. If the only interval is $(-\\infty,\\infty)$, use $e^x$.", "Solution_2": "Thank you very much jmerry for such a swift response.\r\n\r\nKisses \r\n-Laura :)", "Solution_3": "uhm...\r\nthere's a simpler solution, that can be easierly generalized to functions from other sets to $\\mathbb{R}$:\r\nsince $E$ is closed, the distance function: $d_E(x) = \\inf_{y \\in E} |x-y|$ is well defined, and is $0$ exactly on the set $E$, and positive everywhere else.\r\nnow, we can define $f_E|_E = 0$ (meaning that the function $f_E$ restricted to $E$ is $0$), and $f_E|_{E^c}(x) = e^{-\\frac1{d_E(x)^2}}$.\r\nit's quite easy to check that $f_E \\in C^{\\infty}$.\r\n\r\nanother solution (which i don't know much about, though), would be considering the convolution product of two functions..\r\nif anyone knows more about it, please tell us something :)", "Solution_4": "Convolution \"smears out\" functions and spreads the support around. It's not good for controlling the exact zero set of a function.\r\n\r\nAlso, your example has a problem. Suppose we're working on $\\mathbb{R}$ and the set $E=\\{-1,1\\}.$ Then your example gives a function which is $e^{-\\frac1{(x+1)^2}}$ for $x\\in(-1,0]$ and $e^{-\\frac1{(1-x)^2}}$ for $x\\in[0,1).$ That function fails to be differentiable at $x=0.$", "Solution_5": "In fact, it ahs been shown on this forum (I believe I showed it, in a thread started by sam-n :)) that given a closed subset $E$ of $\\mathbb R^n$, the function $d_E$ defined by ma_go is differentiable at $x\\not\\in E$ iff there is only one point $y\\in E$ with $d_E(x)=\\|x-y\\|$.", "Solution_6": "yet, there is a way to fix it...\r\ntake a closed set $E'$ such that, chosen any two points $x (where v_k is the k-th vector in the V basis and u_j is the j-th vector in the U basis).\r\nIs can be shown that for all k,j v_k and u_j are not orthogonal:\r\nwhen you take any v_k (a vector of the orthonormal V basis) and subtruct a linear combination the same basis vectors: alfa_1*v_1+alfa_2*v_2+...+alfa_n*v_n, where 0 the inner product of this result with any v_k is larger than zero.\r\nThe Gram-shmidt procedure can be viewed as repeating the above procedure multiple n times." } { "Tag": [ "geometry", "perimeter", "geometry solved" ], "Problem": "Given 2 lines intersecting at $O$, a length $z$, and a point $P$. Construct a line passing through the point $P$ that intersect the two lines at $A$ and $B$ such that $OAB$ has a perimeter of $z$.", "Solution_1": "Here is the crux of the construction; the determination is boring and will be omitted.\r\n\r\nLet X and Y be two points on the two given lines such that $OX=OY=\\frac{z}{2}$. (Actually, there are two different possible positions for the point X and two different possible positions for the point Y - try out all of them: sometimes the one, sometimes another will yield a triangle OAB with perimeter z.)\r\n\r\nSince OX = OY, there exists a circle k which touches the line OX at the point X and the line OY at the point Y. Let t be one of the two tangents from the point P to this circle k, and let A and B be the points of intersection of this tangent t with the lines OX and OY. Then, depending of the arrangement of the points, this circle k is either the incircle or one of the excircles of triangle OAB. If the circle k happens to be the excircle of triangle OAB opposite to the vertex O, then triangle OAB has the perimeter z, and thus, the line t is one of the required lines. [The proof of this is given in the spolier.]\r\n\r\n[hide=\"Proof\"][i]Proof:[/i] If the circle k is the excircle of triangle OAB opposite to the vertex O, the point X is the point of tangency of this excircle with the sideline OA of the triangle OAB, and thus, by a well-known theorem about excircles, the length OX equals the semiperimeter of triangle OAB. In other words, the perimeter of triangle OAB equals $2\\cdot OX$. But since $OX=\\frac{z}{2}$, the perimeter of triangle OAB thus equals $2\\cdot\\frac{z}{2}=z$. Hence, the line t is one of the lines we have to construct.[/hide]\r\n\r\nConversely, we can show by redoing our argumentation backwards that each required line can be found as a tangent from the point P to the circle k (don't forget that there are four different circles k, depending on the different possible positions of the points X and Y, and we have to check all of them).\r\n\r\n Darij" } { "Tag": [ "modular arithmetic" ], "Problem": "What is the remainder when $ 13^{51}$ is divided by 5?", "Solution_1": "$ 13^4 \\equiv 1 \\pmod{10}$.\r\nSo, $ 13^{51}\\equal{}13^{4\\times12\\plus{}3}\\equiv13^3\\equiv7 \\pmod{10}$.\r\nThus, $ 13^{51} \\equiv 7 \\equiv \\boxed{2} \\pmod{5}$.", "Solution_2": "Alternative sol : \n$\\phi 5 = 4$\n\n$13^{51} = (13^4) ^{12} * 13 ^ 3 = 1 * 13 ^3$ (mod 5 ) \n$13 = (3)$ (mod 5) \n$(3)^3 = 27$ \n$27 = 2$ (mod 5)", "Solution_3": "Just find the last digit and mod 5 it", "Solution_4": "[quote=coolmath2017]Alternative sol : \n$\\phi 5 = 4$\n\n$13^{51} = (13^4) ^{12} * 13 ^ 3 = 1 * 13 ^3$ (mod 5 ) \n$13 = (-3 )$ (mod 5) \n$(-3)^3 = 27$ \n$27 = 2$ (mod 5)[/quote]\n\nWait how is $(-3)^{3} = 27 ($mod $5)$? Isn't it $-27=3($mod $5)$?", "Solution_5": "[quote=GameBot]What is the remainder when $ 13^{51}$ is divided by 5?[/quote]\n\n$\\phi(5)=4$\n\nThis reduces to $3^3 \\equiv \\boxed{2} \\pmod 5$.", "Solution_6": "@advanture \nI made a mistake. 13 = 3 mod 5, not -3 mod 5 lol. " } { "Tag": [], "Problem": "a 5.00 g sample of aluminum pellets and a 10.00 g sample of iron pellets are placed together in a dry test tube, and the test tube is heated in a boiling water bath to 100 degree celcius,. The mixture of hot iron and aluminum is then poured into 97.3 g of water at 22.5 C. Towhat final temperature is the water heated by the metals?\r\n\r\nthanks", "Solution_1": "Hello,\r\n You could post it in the Physics section as it has more to do with Thermal Physics.Infact,there is no chemistry required to solve the problem.You have to apply the Principle of mixtures.And ofcourse,the specific heats of the metals need to be given.Just apply,heat lost by the metals equals the heat gained by the water and you are through.\r\nSathej", "Solution_2": "In case you need more detail, all three relevant items (alum, iron and water) start at their initial temperature.\r\n\r\nFor each degree of temperature change for each material a certain amount of heat is needed or generated. (This can be calculated by multiplying the heat capacity by the weight.) One then knows, that since no reaction occurs, that there is no net heat change. The amount heat released by the alum and iron equals the increase in heat by the water at some specific temperature (between 100 and 22.5). This is the final temperature of the mixture. The critical aspect is that the alum and iron start from 100 and move down to the final temperature, while the water starts at 22.5 and moves up to the final temperture. The tricky part is geting both changes covered by the same one temperature variable.\r\n\r\nIf you have trouble, try to have the one variable defined that works for a specified temperature you select between 100 and 22.5. It may help to select random temperatures and construct a data table then constuct the temperature variable from that data table." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Given 5 circumferences, every four of them have a common point.\r\nProve that there exist a point that belongs to all five circumferences.", "Solution_1": "Perform an inversion having as pole the common point of four of the circles. It turns four circles into lines and the fifth circle into a circle (if it's also a line then all five of them pass through the pole). If there are three lines which are not concurrent, then the system formed by them and the circle doesn't satisfy the conditions. This means that all the lines are concurrent, and the circle passes through the concurrence point, and that's it.", "Solution_2": "thats also my sol. thanks!" } { "Tag": [ "inequalities", "logarithms", "inequalities proposed" ], "Problem": "Let $ a$ and $ b$ be positive real numbers such that $ a^{13}\\plus{}b^{13}\\equal{}2.$ Prove that\r\n$ \\frac{5a^2}{b}\\plus{}\\frac{3b^3}{a^2} \\ge 8.$", "Solution_1": "[quote=\"can_hang2007\"]Let $ a$ and $ b$ be positive real numbers such that $ a^{13} \\plus{} b^{13} \\equal{} 2.$ Prove that\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 8.$[/quote]\r\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 8\\Leftrightarrow\\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 8\\sqrt [13]{\\frac {a^{13} \\plus{} b^{13}}{2}},$ which is homogeneous inequality.\r\nLet $ b \\equal{} 1.$ Thus, we need to prove that $ f(a)\\geq0,$ where\r\n$ f(a) \\equal{} \\ln(5a^4 \\plus{} 3) \\minus{} 2\\ln a \\minus{} \\frac {\\ln(a^{13} \\plus{} 1)}{13} \\minus{} \\frac {38\\ln2}{13}.$\r\nBut \r\n$ f'(a) \\equal{} \\frac {(a \\minus{} 1)(5a^{16} \\plus{} 5a^{15} \\plus{} 5a^{14} \\plus{} 5a^{13} \\minus{} 4a^{12} \\minus{} 4a^{11} \\minus{} 4a^{10} \\minus{} 4a^9 \\minus{} 4a^8 \\minus{} 4a^7 \\minus{} 4a^6 \\minus{} 4a^5 \\minus{} 4a^4 \\plus{} 6a^3 \\plus{} 6a^2 \\plus{} 6a \\plus{} 6)}{a(5a^4 \\plus{} 3)(a^{13} \\plus{} 1)},$\r\nwhich gives $ a_{min} \\equal{} 1.$ Done! :lol:\r\n\r\n[hide=\"Remark.\"]\n$ 5a^{16} \\plus{} 5a^{15} \\plus{} 5a^{14} \\plus{} 5a^{13} \\minus{} 4a^{12} \\minus{} 4a^{11} \\minus{} 4a^{10} \\minus{} 4a^9 \\minus{}$\n$ \\minus{} 4a^8 \\minus{} 4a^7 \\minus{} 4a^6 \\minus{} 4a^5 \\minus{} 4a^4 \\plus{} 6a^3 \\plus{} 6a^2 \\plus{} 6a \\plus{} 6 \\equal{}$\n$ \\equal{} 4(a^{16} \\minus{} a^{12} \\minus{} a^8 \\plus{} a^4 \\plus{} a^{15} \\minus{} a^{11} \\minus{} a^7 \\plus{} a^3 \\plus{} a^{14} \\minus{} a^{10} \\minus{} a^6 \\plus{} a^2 \\plus{} a^{13} \\minus{} a^9 \\minus{} a^5 \\plus{} a) \\plus{}$\n$ \\plus{} a^{16} \\plus{} a^{15} \\plus{} a^{14} \\plus{} a^{13} \\plus{} 2a^3 \\plus{} 2a^2 \\plus{} 2a \\plus{} 9\\cdot\\frac {2}{3} \\minus{} 8a^4\\geq$ \n$ \\geq4(a^4 \\minus{} 1)^2(a^4 \\plus{} 1)(a^4 \\plus{} a^3 \\plus{} a^2 \\plus{} a) \\plus{} \\left(16\\sqrt [16]{8\\left(\\frac {2}{3}\\right)^9} \\minus{} 8\\right)a^4 > 0.$[/hide]", "Solution_2": "I actually spent all night on this problem and couldn't find any solution :(\r\nCan,I was wondering if you have anything in your mind different than Arqady's solution?(which is nice but it has too many computations).\r\nCould you give us a hint?Could anyone think a solution of this inequality just using Cauchy-Schwarz etc.Thanks\r\n\r\n~Kostas", "Solution_3": "My proof is not very nice. I used the inequality:\r\n$ \\frac {a \\plus{} b}{2} \\plus{} \\frac {3(a \\minus{} b)^2}{a \\plus{} b} \\ge \\sqrt [13]{\\frac {a^{13} \\plus{} b^{13}}{2}},$ (*)\r\nto reduce the problem into proving that\r\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {24(a \\minus{} b)^2}{a \\plus{} b},$\r\nwhich is true and it is not hard to prove.\r\n\r\nBy the way, I think (*) has a proof by AM-GM. I hope someone will find it out.", "Solution_4": "The following inequality is also true.\r\nLet $ a$ and $ b$ are positive numbers such that $ a^{14}\\plus{}b^{14}\\equal{}2.$ Prove that:\r\n\\[ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 8\\]", "Solution_5": "[quote=\"can_hang2007\"]My proof is not very nice. I used the inequality:\n$ \\frac {a \\plus{} b}{2} \\plus{} \\frac {3(a \\minus{} b)^2}{a \\plus{} b} \\ge \\sqrt [13]{\\frac {a^{13} \\plus{} b^{13}}{2}},$ (*)\nto reduce the problem into proving that\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {24(a \\minus{} b)^2}{a \\plus{} b},$\nwhich is true and it is not hard to prove.\n\nBy the way, I think (*) has a proof by AM-GM. I hope someone will find it out.[/quote]\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {24(a \\minus{} b)^2}{a \\plus{} b}$\n\n[b]Proof?[/b]\n\nThanks.", "Solution_6": "[quote=\"can_hang2007\"]\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {24(a \\minus{} b)^2}{a \\plus{} b},$\n.[/quote]\n$\\Leftrightarrow 5{{x}^{5}}-23{{x}^{4}}+40{{x}^{3}}-28{{x}^{2}}+3x+3\\ge 0\\Leftrightarrow {{\\left( x-1 \\right)}^{2}}\\left( 5{{x}^{3}}-13{{x}^{2}}+9x+3 \\right)\\ge 0$,\nwhere $x=\\frac{a}{b}$", "Solution_7": "Thanks.\nA simple proof ??", "Solution_8": "$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {25(a \\minus{} b)^2}{a \\plus{} b},a,b>0$", "Solution_9": "[quote=\"ionbursuc\"]$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {25(a \\minus{} b)^2}{a \\plus{} b},a,b>0$[/quote]\n$\\Leftrightarrow 5{{x}^{5}}-24{{x}^{4}}+42{{x}^{3}}-29{{x}^{2}}+3x+3\\ge 0$\n$\\Leftrightarrow {{\\left( x-1 \\right)}^{2}}\\left( 5{{x}^{3}}-14{{x}^{2}}+9x+3 \\right)\\ge 0$,\nwhere $x=\\frac{a}{b}$.", "Solution_10": "\\[ \\frac {5a^2}{b} + \\frac {3b^3}{a^2} \\ge 4(a + b) + \\frac {51(a - b)^2}{2(a + b)},\\]\nwhere $a,b>0$.\n\nIf $ \\frac {5a^2}{b} + \\frac {3b^3}{a^2} \\ge 4(a + b) + \\frac {\\lambda(a - b)^2}{a + b},$ then ${\\lambda_{max}=?.}$", "Solution_11": "[quote=\"sqing\"]\\[ \\frac {5a^2}{b} + \\frac {3b^3}{a^2} \\ge 4(a + b) + \\frac {51(a - b)^2}{2(a + b)},\\]\nwhere $a,b>0$.\n\nIf $ \\frac {5a^2}{b} + \\frac {3b^3}{a^2} \\ge 4(a + b) + \\frac {\\lambda(a - b)^2}{a + b},$ then ${\\lambda_{max}=?.}$[/quote]\n${{\\lambda }_{\\max }}=11+5x+\\frac{9}{x}+\\frac{3}{{{x}^{2}}}$,where $5{{x}^{3}}-9{{x}^{2}}-6=0,x>0$", "Solution_12": "[quote=\"ionbursuc\"][quote=\"can_hang2007\"]\n$ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 4(a \\plus{} b) \\plus{} \\frac {24(a \\minus{} b)^2}{a \\plus{} b},$\n.[/quote]\n$\\Leftrightarrow 5{{x}^{5}}-23{{x}^{4}}+40{{x}^{3}}-28{{x}^{2}}+3x+3\\ge 0$ \n$\\Leftrightarrow {{\\left( x-1 \\right)}^{2}}\\left( 5{{x}^{3}}-13{{x}^{2}}+9x+3 \\right)\\ge 0$,\nwhere $x=\\frac{a}{b}$[/quote]\n$5{{x}^{5}}-23{{x}^{4}}+40{{x}^{3}}-28{{x}^{2}}+3x+3= {{\\left( x-1 \\right)}^{2}}\\left( 5{{x}^{3}}-13{{x}^{2}}+9x+3 \\right)$\n$\\ge {{\\left( x-1 \\right)}^{2}}\\left( 2\\sqrt{5{{x}^{3}}\\cdot 9x}-13{{x}^{2}}+3\\right)={{\\left( x-1 \\right)}^{2}}\\left( (6\\sqrt{5}-13)x^2+3\\right) \\ge 0.$", "Solution_13": "[quote=can_hang2007]Let $ a$ and $ b$ be positive real numbers such that $ a^{13}\\plus{}b^{13}\\equal{}2.$ Prove that\n$ \\frac{5a^2}{b}\\plus{}\\frac{3b^3}{a^2} \\ge 8.$[/quote]\n[quote=arqady]The following inequality is also true.\nLet $ a$ and $ b$ are positive numbers such that $ a^{14}\\plus{}b^{14}\\equal{}2.$ Prove that:\n\\[ \\frac {5a^2}{b} \\plus{} \\frac {3b^3}{a^2} \\ge 8\\][/quote]\nSolution of kuing:", "Solution_14": "Let $ a,b $ are positive real numbers such that $a^{12}+b^{12}=2. $ [url=https://artofproblemsolving.com/community/c6h2869319p25486803]Prove that[/url]$$\\frac{3a^2}{b}+\\frac{b^5}{a^4} \\geq 4$$", "Solution_15": "[quote=can_hang2007]Let $ a$ and $ b$ be positive real numbers such that $ a^{13}\\plus{}b^{13}\\equal{}2.$ Prove that\n$ \\frac{5a^2}{b}\\plus{}\\frac{3b^3}{a^2} \\ge 8.$[/quote]\n\nSplendid" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let [tex] a,b,c [/tex] be three positive real numbers. Prove that:\r\n[tex] a^2+b^2+c^2+2abc+1 \\geq 2(ab+ac+bc) [/tex] :)", "Solution_1": "WLOG $ (1-b)(1-c)>=0$\r\n\r\n$ a^2 +b^2 +c^2 +2abc+1-2(ab+bc+ca)=(a^2 +1)+(b^2 +c^2)+2abc-2(ab+bc+ca) >= 2a+2bc+2abc-2ab-2bc-2ca=2a+2abc-2ab-2ca =2a(1-b)(1-c) \r\n>=0$", "Solution_2": "This problem is posted on ML many times . It's mentioned in my recent topic . \r\n By AM-GM : $ 2abc + 1 \\geq 3 {^3}\\sqrt{a^2b^2c^2} $ \r\n After that , use Schur 's ineq ;) \r\n the equality holds for $ a=b=c=1 $" } { "Tag": [ "geometry", "geometric transformation", "Columbia", "articles", "reflection" ], "Problem": "Hello,\r\n\r\nFirst than all, if I say something that is not totally correct, I confess: [b]I know basically nothing about asian languages[/b]. But one of my office mates and my advisor are chineses (and some other friends too), and they answer some of my questions. Now, from the next quote comes from [url=http://www2.ignatius.edu/faculty/turner/languages.htm]This Webpage[/url]:\r\n\r\n[quote]The Summer Institute for Linguistics (SIL) Ethnologue Survey (1999) lists the following as the top languages by population:\n(number of native speakers in parentheses)\n\n 1. Chinese* (937,132,000)\n 2. Spanish (332,000,000)\n 3. English (322,000,000)\n 4. Bengali (189,000,000)\n 5. Hindi/Urdu (182,000,000)\n 6. Arabic* (174,950,000)\n 7. Portuguese (170,000,000)\n 8. Russian (170,000,000)\n 9. Japanese (125,000,000)\n 10. German (98,000,000)\n 11. French* (79,572,000)\n\n* The totals given for Chinese, Arabic, and French include more than one SIL variety[/quote]\nAnd latter on they make the aclaration\n[quote]* \"Chinese\" for this chart is composed of the total of the following SIL designated languages:\n\tMandarin \t\t836,000,000 \t\n\tXiang \t\t36,015,000 \t\n\tHakka \t\t34,000,000 \t\n\tGan \t\t20,580,000 \t\n\tMinbei \t\t10,537,000 \t[/quote]\r\nWell, if you consider that Mandarin is [b]spoken[/b] by 836 millions, then, it would be without any doubt the principal native language of the world. Nevertheless chineses that I know (including my advisor) say that Mandarin Speakers that comes from different places of China, do not understand each other if they are speaking (even when they understand each other perfectly if they are writting), but just locally they really can [b]speak[/b] their Mandarin. Therefore, depending on your definition of speakers, it would not be the most widely speaking mother language, but the most widely writing mother language.\r\n\r\nThe interesting part is that, as a native english speaker, and as a second-language english speaker, I can say that in both of them, even when you can spend some time trying to [i]get the accent[/i] (sometimes I have problem understanding Australians, but my spoken english is very American), nevertheless I can speak fluently in spanish with basically every native spanish speaker, and I claim I could do the same in english.\r\n\r\nHence, would you say that Mandarin (or Chinese) is really [u]the world's most widely spoken native language[/u]?\r\n\r\nI would like to hear the opinion of other people in the forum, and I am craving to read specifically the opinion of chinese people, because as I already said, I know basically nothing about chinese, and everything I am saying about it is second hand information. \r\n\r\n[b]Remark[/b]: I honestly spect that chinese MathLinkers don't feel offended by anything I said here, if so, I appologize, I tried to express everything in the more neutral way, as a sincere question!", "Solution_1": "Having moved to the US early, I have lost a lot of Chinese fluency, but while there are very different accents, it is still often comprehensible. Also, many Chinese can speak both standard Mandarin and their local dialects (possibly similar to Mandarin). Furthermore, Mandarin is becoming more and more standard. I'm not sure about specific numbers, but I think the impression I get from your post is somewhat inaccurate. However, there certainly are many different accents and dialects, so if you wanted the language to be as consistent as English, your number might become something like 550 million. Of course this is just an estimate--someone living in China could give you better information.", "Solution_2": "Well yea duh, China has the largest population in the world, and Mandarin is the most common Chinese dialet. When most people say \"Chinese\", most of the time they refer to \"Mandarin\", or \"Cantonese\". I'm surprised Cantonese didn't appear on the chart.\r\n\r\nI've never heard of the last four types of Chinese dialets on the second chart. ;) So yea I guess Mandarin is still the most widely spoken language in the world, because like almost all Chinese can speak Mandarin, and that's a whole lot of people. It's like the basic language you have to learn in China.", "Solution_3": "i'm surprsied the number for \"chinese\" is only 900,000,000 actually... but neways\r\ni think u should count mandarin as number 1 language because even though there are many dialects, as probability1.01 said they are still based on the mandarin language. when he said they can speak mandarin and their local dialects i think thats sorta true but there are also people whose mandarin just sounds different i dont know what he meant by mandarin and local dialects because the local dialects are mandarin just with different sounds and pronunciations\r\nim no expert on this and sorry for being confusing but id still say mandarin is #1 language because number 2 on the list was 1/3 of number 1 anyways :D", "Solution_4": "Well, i wonder how do you define Madarin?? \r\nI think that Chinese [b]is[/b] the most widely spoken native language. There are only some differences in the pronunciation like british english and amerian english, but they are the same language!", "Solution_5": "[quote=\"shobber\"]Well, i wonder how do you define Madarin?? \nI think that Chinese [b]is[/b] the most widely spoken native language. There are only some differences in the pronunciation like british english and amerian english, but they are the same language![/quote]\r\nYes, but my point is that an American can easely understand a British and viseversa, nevertheless according to some chineses I know, in China it is almost impossible to understand other people if they are from far away of your place!", "Solution_6": "many chinese speaks more than 1 dialect. my dad speaks 3", "Solution_7": "[quote=\"sheepwarrior\"]i'm surprsied the number for \"chinese\" is only 900,000,000 actually... but neways\ni think u should count mandarin as number 1 language because even though there are many dialects, as probability1.01 said they are still based on the mandarin language. when he said they can speak mandarin and their local dialects i think thats sorta true but there are also people whose mandarin just sounds different i dont know what he meant by mandarin and local dialects because the local dialects are mandarin just with different sounds and pronunciations\nim no expert on this and sorry for being confusing but id still say mandarin is #1 language because number 2 on the list was 1/3 of number 1 anyways :D[/quote]\r\nI speak both Mandarin and Shanghainese, and knowing the former is not near sufficient to understand the latter. Shanghainese is probably derived from Mandarin, but it's too different to be counted as Mandarin.", "Solution_8": "[quote=\"shobber\"]Well, i wonder how do you define Madarin?? \nI think that Chinese [b]is[/b] the most widely spoken native language. There are only some differences in the pronunciation like british english and amerian english, but they are the same language![/quote]\r\nWell see in Mandarin there are four accents (1, 2, 3, 4) while in Cantonese there are nine accents.\r\nSo I suppose that's a way to define Mandarin.", "Solution_9": "[quote=\"djimenez\"] Nevertheless chineses that I know (including my advisor) say that Mandarin Speakers that comes from different places of China, do not understand each other if they are speaking (even when they understand each other perfectly if they are writting), but just locally they really can [b]speak[/b] their Mandarin. Therefore, depending on your definition of speakers, it would not be the most widely speaking mother language, but the most widely writing mother language.\n[/quote]\r\nWe can understand each other in Mandarin (offcially called Pu Tong Hua) and we also have accents. But we do not understand their dialects. People from South China have stronger accents, esp from Fujian and Guangdong. I am from Fuzhou, Fujian. And I do not speak Mandarin correctly for all sounds (that is an accent), but they can understand me well.", "Solution_10": "Okay, this stuff with the \"dialects\" being counted as different languages and whatnot is just wrong. It's kind of like saying we should split English into \"American English\" and \"British English\" because the Brits use words like buggy. And furthermore, we should create different categories for \"New York English\", \"Texas English\", \"California English\", and etc to represent every region that employs its own colloquialisms.\r\n\r\nAll of the regional dialects that are under discussion are merely ways of accenting \"Pu Tong Hua\" that have been adopted by those areas. And frankly, no matter how different two dialects of Mandarin are, anyone who has a solid foundation in the language (a decade or more) should be able to be the general gist of another speaker's message. A lot of adults don't know what \"top dawg\" or \"sup\" means, but that doesn't mean they speak a different language. I know northerners that have never heard anyone say \"ya'll\" but that doesn't alienate them from the English language. There's less than a half billion fluent English speakers and we already have different accents. Imagine what would happen if we doubled (or, probably more appropriately, tripled) that number.\r\n\r\nCantonese and Taiwanese (which isn't really a separate language at all, btw) may be arguably a greater departure from mainstream Mandarin, but even those are by no means separate languages.\r\n\r\nImagine what Madrid would think if Mexico City all of a sudden said they speak Mexican instead of Spanish.", "Solution_11": "Anyway, first, let's correct a few misconceptions:\r\n-Mandarin has four tones plus one neutral tone. So that means there are five tones total.\r\n-Cantonese is reported to have six tones. (Personally, I don't know how true that is because although I speak the language fluently, I could never identify only six distinct tones. Some people report that there are nine tones. So I guess the number of tones range from six to nine, depending on whom you ask.) \r\n\r\nI really don't think that understanding people who speak Mandarin with different accents is a problem. That's why it's called \"Pu Tong Hua,\" or literally, the \"Common Language.\" (It's also known as \"Guo Yu\", or the \"National Language,\" but personally, I have some issues with that name. :P )", "Solution_12": "Now, onto the dialect issue. :lol: \r\n\r\nThis is actually quite tricky to answer. Each dialect is distinctly different, and there is actually no guarantee that knowing one will allow you to even partially understand another.\r\n\r\nThis next analogy is a stretch, but I'm going to try it anyway. Think of all the Romantic languages that are in existence today. One could argue that if you spoke Latin, then although you would not be able to converse in each of the Romantic languages, you would be able to understand a lot of it and get the main idea. As you know, this isn't really true, but your knowledge of Latin certainly doesn't hurt.\r\n\r\nThe various Chinese dialects are like this in a sense. Each dialect carries is own character and flair. If you are fluent in more than one dialect, then you'll notice that each time you speak another dialect, you assume a new persona and attitude. What's more, even people who spent their life speaking one dialect of Chinese (so they are quite fluent) may not necessarily understand other dialects. This is most evident in the elderly generations who had their language training before the Cultural Revolution. \r\n\r\nIn the past, the various Chinese dialects were united under a strong leadership, and thus, for political and historical reasons, all Chinese dialects are [i]written[/i] in the same way. Still, they are spoken very differently. For example, in Cantonese, you speak in patois and would usually not speak as you would write.\r\n\r\nToday, the Chinese dialects are united under technology. For example, when movies are filmed today, actors and actresses speak their native tongue. Later, the editing crew would then record the appropriate voice over in the desired language. Personally, this drives me crazy when I watch the films. But *shrugs* what can I do?", "Solution_13": "Hey guys,\r\n\r\nThank you very much for your opinions and comments. I have another question. I was talking to a friend, a chinese girl, one of these days, and talking to here I told her how [b]China[/b] is pronounced in Spanish, and she told me that it sounds like an offensive word in chinese, but she didn't want to tell me what exactly mean. We don't pronounce the [b]i[/b] in China as in english, it is more like the vowel sound in [b]green[/b] or in [b]Teena[/b], you could probably say that we pronounce [b]Cheena[/b]. Could someone explain me what does it mean?", "Solution_14": "[/quote]\r\nI speak both Mandarin and Shanghainese, and knowing the former is not near sufficient to understand the latter. Shanghainese is probably derived from Mandarin, but it's too different to be counted as Mandarin.[/quote]\r\n\r\nShanghainese is not derived from mandarin. i believe, i may be wrong, some of it is derived from european languages. Shanghainese is totally different from mandarin.", "Solution_15": "so where do u actually live....California...???", "Solution_16": "Yeah I live in California. Born in California too.", "Solution_17": "[quote=\"ambierona\"]Yeah it's a Cuban flag. I'm half Cuban and half Chinese, so I keep changing my flag from Cuba to Chinese.. maybe I should put the US flag there.. but I put California as my location so that's cool.[/quote]Cool. Just a thought: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239316#p239316]How do you feel about this?:[/url] \"Bush got 566...\" j/k.\n\"One reason why I wonder what it means to be American is because I'm not Caucasian. ... Now, I'm proud of my ethnic heritage but I'm a little bothered by the fact that no matter how many generations I and my descendants live in the U.S., we may still be considered less than full-blooded Americans.\"\nIt's an interesting remark (and understandable).\n\nI will say that this statement strikes me differently knowing you're (a) Asian--sorry if that sounds reminiscient of the other (linked above) thread--(b) in America. What else would one learn--besides English (over Spanish??) when approaching an age like two?[quote=\"ambierona\"]... I wish I could speak Chinese.. grr I was stupid though and wanted to learn English when I was 2.[/quote]", "Solution_18": "I was wondering if any of you guys read the China's Century NEWSWEEK. Inside it there was an article about how more and more American schools are offering Chinese. Pretty nice...", "Solution_19": "yeah, I have. But I've heard that Chinese is an impossibly difficult language to learn, unlike Spanish, which is ridiculously easy to learn (at least for me). I would still choose Spanish over Chinese if I had the choice. \r\n\r\nIn addition, Chinese isn't that useful yet for us to learn because in our daily experience, we won't use Chinese too often. The only time we'd use is when we're in China. On the other hand, there are 17 million + spanish-speaking people in the U.S., so Spanish is very useful. And if you're on vacation in Mexico, Spain, Latin America...\r\n\r\nHowever, nothing you learn in school is useful. I don't think that's the point of school--if it were, we should go back to the apprentice system of the 1700s. School is to stretch your mind and grow--education for education's sake, so in that case, it'd probably be better to learn Chinese.", "Solution_20": "But it's not about now but the future, soon China will become a major force (in fact it is already) in world economics and politics so knowing Chinese will definitely give you an edge in society.", "Solution_21": "[quote=\"fedwinri\"][quote=\"ambierona\"]Yeah it's a Cuban flag. I'm half Cuban and half Chinese, so I keep changing my flag from Cuba to Chinese.. maybe I should put the US flag there.. but I put California as my location so that's cool.[/quote]Cool. Just a thought: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239316#p239316]How do you feel about this?:[/url] \"Bush got 566...\" j/k.\n\"One reason why I wonder what it means to be American is because I'm not Caucasian. ... Now, I'm proud of my ethnic heritage but I'm a little bothered by the fact that no matter how many generations I and my descendants live in the U.S., we may still be considered less than full-blooded Americans.\"\nIt's an interesting remark (and understandable).\n\nI will say that this statement strikes me differently knowing you're (a) Asian--sorry if that sounds reminiscient of the other (linked above) thread--(b) in America. What else would one learn--besides English (over Spanish??) when approaching an age like two?[quote=\"ambierona\"]... I wish I could speak Chinese.. grr I was stupid though and wanted to learn English when I was 2.[/quote][/quote]\r\n\r\nWell, about that remark thing.. I don't think there are any \"full-blooded Americans.\" So yeah.. the closest you can get is a Native American or Indian or whatever we call that race to be politically correct. And America is a nation of all these mixed races, so I think I'm pretty American.\r\n\r\nUh, I was actually speaking Chinese (Cantonese) and Spanish until I was 2.. because my parents spoke that to me. But then my older brother got into preschool without knowing English, and people made fun of him, he came home crying, and learned English. And I wanted to be just like him, so I learned English and forgot Chinese and Spanish. But it would have been nice to keep up three languages, so I could communicate with my grandparents. (that's what I would use Chinese and Spanish for in the real world, communicating with relatives... but yeah, here in California, Spanish is more useful to know for like going to Mexico or something)", "Solution_22": "[quote=\"tokenadult\"][quote=\"juicybooty911\"]mandarin and taiwanese are actually very very similar[/quote]\n\nEnglish and German are more similar, but I don't see many English speakers counting themselves as speakers of German just because of that similarity.[/quote]\r\n\r\n\r\nI would like to tell you, mandarin and taiwanese may be a little different in accent, but, the grammar is the same. Well, I think the grammar of English and German might be totally different(just guess). \r\n\r\nJust wondering, why you have almost every opinion on the opposite side of China in almost every topic? such as, you insist on to find the difference between Mandarin and taiwanese--a branch of Chinese.\r\n\r\nBesides, I can tell you, I am from south Jiangsu, if you would like difference between Mandarin and other branches of Chinese, hmm, I would like to tell you: personally, I think my native tongue(similar to shanghainese, called Wu) has greater difference between mandarin than taiwanese(although my hometown is nearer to centre China than taiwan) or cartoonese, many people say that it has similar accent with janpanese.", "Solution_23": "Actually the English and German grammar are quite different, rather than similar. They have some common words, but that's it. You might say that the French and English grammars are similar :)", "Solution_24": "[quote=\"tokenadult\"][quote=\"Zratonx\"]no matter wut dialect ppl in China speak they write in the same exact words.[/quote]\n\nThat is a factually incorrect statement. A lot of visitors to Hong Kong who look carefully at local writing in Han characters will see Han characters used there that aren't used anywhere in China where Cantonese is not spoken. The same is true, but less frequent because of the pervasiveness of Mandarin as a second language, in Taiwan. \n\nThere has always been a tendency in the writing of people who are literate at all in Chinese (which was always a small minority of Chinese people until very recently) to write in patterns that reflect the standard language, even if that standard language is not the normal pattern of speech for that speaker. Thus, Cantonese speakers and Taiwanese speakers and Wu speakers are implicitly learning Mandarin when they learn to write in Han characters. But even at that most speakers of Sinitic languages write as they talk--as all human beings do when they write--and that means there are various forms of \"mistaken\" usage (from the point of view of the standard national language) that creep into everyday writing. I have had to deal with that as a translator on many occasions.[/quote]\r\n\r\n\r\nI speak Wu.\r\nI can ensure that Wu and Mandarin only have difference in accent.\r\nGive an example:\r\nEnglish: Hello, what is your name?\r\nMandarin: Ni3 hao3, ni1 shi4 she2 me0 ming2 zi4?\r\nWu: Ni ho , ni si se me min si? (wu doesn't have tones)\r\n\r\nbesides, word \"what\" , in mandarin is she2 me0, in wu is se me, but it could also be accented as \"so ge\" or \"sa ge\"\r\nor \"dia gu\" in Wu.", "Solution_25": "[quote=\"Alexander Lenin\"][quote=\"tokenadult\"][quote=\"juicybooty911\"]mandarin and taiwanese are actually very very similar[/quote]\n\nEnglish and German are more similar, but I don't see many English speakers counting themselves as speakers of German just because of that similarity.[/quote]\n\n\nI would like to tell you, mandarin and taiwanese may be a little different in accent, but, the grammar is the same. Well, I think the grammar of English and German might be totally different(just guess). \n\n[b]Just wondering, why you have almost every opinion on the opposite side of China in almost every topic? such as, you insist on to find the difference between Mandarin and taiwanese--a branch of Chinese.[/b]\n\nBesides, I can tell you, I am from south Jiangsu, if you would like difference between Mandarin and other branches of Chinese, hmm, I would like to tell you: personally, I think my native tongue(similar to shanghainese, called Wu) has greater difference between mandarin than taiwanese(although my hometown is nearer to centre China than taiwan) or cartoonese, many people say that it has similar accent with janpanese.[/quote]\r\n\r\nFirst time poster here,\r\n\r\nI don't think there is any political intentions behind his posts. \r\nYou are reading too much into his posts. \r\n\r\nAs a Taiwanese, i can tell you Taiwanese(Minnan) is in fact very different from Mandarin. \r\nJust look at the origination of each dialect, one (Minnan/Taiwanese) originated from southern coast of china whereas the other (Mandarin) originated from the very northeastern portion of China (Manchuria, bordering north Korea). Heck, mandarin and Manchurians were considered as both barbaric and barbarians until the establishment of the Qing Dynasty.\r\n\r\nAnd take my professor for my Chinese Film and Literature class for example, She is a native taiwanese, as in born in Taiwan. She graduated from NTU(Natiaonal Taiwan University) and received her Ph.D from Stanford. She speaks perfect Mandarin, yet, she told us can neither speak nor understand Taiwanese. \r\n\r\nHeck, just look at me, I was not taught Taiwanese as a kid growing up.\r\nEven though i hear it spoken around me all the time, I can not speak Taiwanese or understand it when spoken to me. \r\n\r\nAnd the Taiwanese \"writting system\" is somewhat different from the standard Chinese writting system.\r\nFor example, i was watching cable the other day, and a Taiwanese song came on. I was reading the lyrics, and the usage of some of the characters made me go \"eh?\"\"", "Solution_26": "manchurian was barbaric until the Qing Dynasty cause they ruled China in the Qing Dynasty, how the heck is mandarin barbaric come on ever since the Zhou Dynasty the Han have been the predominant force and gained total control after the Qin Dynasty.....\r\n\r\nthe character differences r prob just traditional and simplied differences....", "Solution_27": "[quote=\"Zratonx\"]manchurian was barbaric until the Qing Dynasty cause they ruled China in the Qing Dynasty, how the heck is mandarin barbaric come on ever since the Zhou Dynasty the Han have been the predominant force and gained total control after the Qin Dynasty.....\n\nthe character differences r prob just traditional and simplied differences....[/quote]\r\n\r\nMandarin is the language of the Manchurians who like you said, were considered as barbarians until they established the Qing dynasty.\r\n\r\nAnd no, Mandarin was spoken during the Han dynasty. If anything else, the common dialect spoken in the courts of the the Han dynasty is closer to your souther coastal dialects such as fujiannese. \r\n\r\nAnother example, dialect of courts of the Tang dynasty is closer to Cantonese. Some tang poems will not rhyme when read in mandarin, however, it will if you read it using cantonese.", "Solution_28": "thanks very mucgh about the language lesson about mandarin...i really had no clue....\r\n\r\n :blush: \r\n\r\n but still mandarin is spoken but most ppl today so it is...well no longer barbaric...", "Solution_29": "[quote=\"Valentin Vornicu\"]Actually the English and German grammar are quite different, rather than similar. They have some common words, but that's it. You might say that the French and English grammars are similar :)[/quote]I'll say that English actually sounds similar to German in some respects and gets a lot of things from it--enough to be considered non-romantic. It is true, when comparing structure (except for our idea of conjugating verbs--English) and roots of many words, English ties more closely to the Latin than anything else; i.e., yay, Christian church and intercom. Btween France and also Mexico, etc. ~~\r\nNone of that jaunt to take away from your good Chinese discussion forum (only that I have nothing to say for any kind of Chinese--or not-Chinese)." } { "Tag": [ "limit", "integration", "induction", "real analysis", "real analysis solved" ], "Problem": "proove that:\r\na) the sequence x_n=\\sum{k=1,n}(1/(n+k)) is monotonous.\r\nb)there exists a sequence (a_n) \\in {0,1} s.t. \r\nlim{n->oo}\\sum{k=1,n}(a_k/(n+k)).", "Solution_1": "a) x_(n+1) - x_n = -1/2.1/(n+1), x_n is decreasing \r\nx_n = 1/n( \\sum _(k=1 to n-1)1/(1+k/n) + 1/(2n) \r\n \\lim x_n = \\int _[0,1]1/(1+x)dx = ln2", "Solution_2": "I don't think the statement for (b) is complete. We're supposed to show that there is a sequence (an)n>=1 with values 0 or 1 s.t. that limit Kuba mentioned is 1/2.", "Solution_3": "the statement for b) is complete", "Solution_4": "Statement b) is not complete , I just looked over the subjects in 2001 and the problem goes like so:\r\n\r\nProve that there exits a sequence (an)n>=1 with values of 0 or 1 so that\r\n\r\nlim(n-->) (a1/(n+1)+...+an/(n+n))=1/2.\r\n(Radu Gologan)", "Solution_5": "OKay... lets post a solution:\r\n\r\ntry to construct this sequence using induction, so that we have:\r\n\r\n1/2<=sum(k=1,n) ak/(n+k)<=1/2+1/(2n)\r\n\r\nbecause we have 1/2<=1/3+1/4<=1/2+1/4 we can choose a1=1 and a2=1. Now lets consider that all numbers a1,..,an were chosen so that we have: 1/2<=sum(k=1,n) ak/(n+k)<=1/2+1/(2n)\r\nnow we need to consider two cases:\r\ni) if a1/(n+2)+..+an/(2n+1)>=1/2 then we take a_(n+1)=0 \r\nii) if a1/(n+2)+..+an/(2n+1)<1/2 then we take a_(n+1)=1\r\n(see what happens in both cases)\r\n\r\nthen the limit is 1/2.\r\n\r\ncheers! :D :D" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Calculate \r\n\r\n$ \\int{\\frac{t^2}{\\sqrt{t^4\\plus{}t^2\\plus{}1}}}$", "Solution_1": "but can it be computed?? :roll:", "Solution_2": "[quote=\"xxxxtt\"]but can it be computed??[/quote]\r\n\r\nIt's an elliptic integral. So, yes, it can be computed in terms of elliptic integrals E and F.\r\nThe zeros in the denominator are all on the unit circle, so this may be a Crystoffel-type integral which maps the circle to a certain polygon. I'd have to look it up to be sure." } { "Tag": [ "Vieta", "absolute value" ], "Problem": "Find the three roots such that the equation $ x^3\\minus{}x\\plus{}k\\equal{}0\\ (k>0)$ has imaginary root with its absolute value 1.", "Solution_1": "[hide=\"Solution\"]Let $ a\\plus{}bi$, $ a\\minus{}bi$, and $ r$ be the roots. Since the absolute value of the imaginary roots is $ 1$, we know that $ a^{2}\\plus{}b^{2}\\equal{}1$. \n\nThen, we have:\n$ (x\\minus{}(a\\plus{}bi))(x\\minus{}(a\\minus{}bi))(x\\minus{}r)\\equal{}(x^{2}\\minus{}2ax\\plus{}(a^{2}\\plus{}b^{2}))(x\\minus{}r)\\equal{}$\n$ x^{3}\\minus{}(2a\\plus{}r)x^{2}\\plus{}(a^{2}\\plus{}b^{2}\\plus{}2ar)x\\minus{}r(a^{2}\\plus{}b^{2})\\equal{}x^{3}\\minus{}(2a\\plus{}r)x^{2}\\plus{}(1\\plus{}2ar)x\\minus{}r$\n\nComparing to the given equation, we get, $ 2a\\plus{}r\\equal{}0$ and $ 1\\plus{}2ar\\equal{}\\minus{}1$. Solving, we get $ r\\equal{}\\pm \\sqrt{2}, a\\equal{}\\mp \\frac{\\sqrt{2}}{2}$. Since $ k > 0$, we can eliminate $ r\\equal{}\\sqrt{2}$ as a solution. So, we have $ r\\equal{}\\minus{} \\sqrt{2}$, $ a\\equal{}\\frac{\\sqrt{2}}{2}$, and $ b\\equal{}\\frac{\\sqrt{2}}{2}$. The roots are $ \\boxed{\\minus{}\\sqrt{2}, \\frac{\\sqrt{2}}{2}\\plus{}\\frac{\\sqrt{2}}{2}i, \\frac{\\sqrt{2}}{2}\\minus{}\\frac{\\sqrt{2}}{2}i}$.[/hide]", "Solution_2": "That's correct answer.\r\nYou can use directly Vieta's fomula." } { "Tag": [ "inequalities", "integration", "geometry", "calculus", "derivative", "complex analysis", "complex analysis unsolved" ], "Problem": "Suppose $f$ is analytic on a set that contains the closed disk centered at $a$ of radius $R.$ Prove that\r\n\r\n$|f(a)|^2\\le\\frac1{\\pi R^2}\\int_0^R\\int_0^{2\\pi}|f(a+re^{i\\theta}|^2r\\,d\\theta\\,dr$\r\n\r\nThe challenge: can you find at leat two [i]different[/i] proofs of this?", "Solution_1": "$f(z)=p(z)+iq(z)$ where $p,q: \\mathbb{C}\\to \\mathbb{R}$ are harmonic functions. Hence:\r\n$p(a)=\\frac{1}{\\pi R^2} \\int \\int_D p(z)dz$, $p(a)^2 \\leq \\frac{1}{\\pi R^2} \\int \\int_D p(z)^2dz$. \r\nSimilarly:\r\n$q(a)^2 \\leq \\frac{1}{\\pi R^2} \\int \\int_D q(z)^2dz$.\r\nThis is why:\r\n$|f(a)|^2=p(a)^2+q(a)^2\\leq \\frac{1}{\\pi R^2} \\int \\int_D p(z)^2+p(z)^2 dz$,\r\n$|f(a)|^2 \\leq \\frac{1}{\\pi R^2} \\int \\int_D |f(z)|^2 dz$,", "Solution_2": "$D$ the unit open disk of $\\mathbb{C}$\r\n$f: D \\to \\mathbb{C}$ an injective analytic function.\r\nProve that the area $A$ of $f(D)$ is greater than $\\pi |f'(0)|^2$.", "Solution_3": "That should be $\\ge$; we don't want a problem when $f$ is multiplication by a constant.\r\n\r\nThe proof is an easy application of the theorem already discussed and the change-of-variables formula for multiple integrals. $\\int\\int_{f(D)}1\\,dA=\\int\\int_{f(D)}\\det f'(f^{-1}(x,y))\\cdot \\det(f^{-1})'(x,y)\\,dx\\,dy=\\int\\int_D \\det f'(x,y)\\,dx\\,dy$.\r\nThose derivatives are $2\\times 2$ matrices; the determinant of $f'$ in the $\\mathbb{R}^2$ sense is $|f'|^2$ in the complex sense.", "Solution_4": "The Fourier series argument:\r\n\r\n$f(z)=\\sum_{n=0}^{\\infty}c_n(z-a)^n,$ with radius of convergence at least $R.$\r\n\r\nThus, $f(a+re^{i\\theta})=\\sum_{n=0}^{\\infty}c_nr^ne^{in\\theta}$\r\n\r\nSo $\\int_0^{2\\pi}|f(a+re^{i\\theta}|^2\\,d\\theta=\\sum_{n=0}^{\\infty}\\sum_{m=0}^{\\infty} \\int_0^{2\\pi}c_n\\overline{c_m}\\,r^nr^me^{in\\theta}e^{-im\\theta}\\,d\\theta$\r\n\r\n$=\\sum_{n=0}^{\\infty}\\sum_{m=0}^{\\infty}c_n\\overline{c_m}\\,r^{n+m}2\\pi\\delta_{nm} =\\sum_{n=0}^{\\infty}|c_n|^22\\pi r^{2n}$\r\n\r\nSo $\\frac1{\\pi R^2}\\int_0^R\\int_0^{2\\pi}|f(a+re^{i\\theta}|^2r\\,d\\theta\\,dr= \\sum_{n=0}^{\\infty}\\frac{2|c_n|^2}{R^2}\\int_0^Rr^{2n+1}\\,dr$\r\n\r\n$=\\sum_{n=0}^{\\infty}|c_n|^2\\frac{R^{2n}}{n+1}\\ge |c_0|^2=|f(a)|^2.$" } { "Tag": [ "puzzles" ], "Problem": "Mel Colly was found dead in his library. His forehead was resting on the keys of a tape recorder, a pool of blood lay on the desk beside him. From the door, Shadow could see that Mel's right hand covered the gun and that a bullet had entered his right temple. Shadow removed the tape recorder from under Mel's head and pressed the play button. A voice said,\"This is Mel Colly speaking. I am not going to go on with a long winded farewell. I'm tired of life and all its anguish. I'll see you on the other side, wherever that may be.\" There was a gunshot and the sound of his head hitting the tape recorder keys, followed by a click indicating the tape recorder had been shut off. Shadow was certain that someone else had imitated Mel's voice and made this recording to sound like a suicide. Why does Shadow suspect murder?\r\n\r\nFeel free to ask questions.", "Solution_1": "[hide=\"classic...\"]Someone else must have shut the tape recorder off after Mel died.[/hide]", "Solution_2": "[hide=\"Something Else\"]His left hand could've landed on a switch or something that turned it off.[/hide]", "Solution_3": "When his head hit the tape recorder it turned off...", "Solution_4": "was there someone else present in the room at that time?\r\nor...\r\nshadow is teh murderer!! lol, no probably not...\r\nhmm....tricky, tricky, tricky.....", "Solution_5": "[hide]If Mel Colly shut off the tape recorder when he died by hitting it with his forhead then the recording should've left off at the point after he died.\n\nBut when Shadow played the recording it started from the beginning, but if Mel Colly really killed himself Shadow would've had to rewind the recorder to listen to it.\n\nSo another person (murderer) would have had to rewound the tape recorder so that whoever picks it up can hear the fake message.[/hide]\r\n\r\nam i right?", "Solution_6": ":first: :starwars: yes gj" } { "Tag": [ "Putnam", "number theory unsolved", "number theory" ], "Problem": "This problem may have been already posted in this forum.I am sorry if it is the case.\r\nFind all real numbers $ x$ such that for all integer $ n$, $ n^x$ is an integer.", "Solution_1": "$ x$ should be integer or $ n\\equal{}a^{x}$ where $ a$ is integer,and $ x$ is Real number.", "Solution_2": "[url]http://www.mathlinks.ro/viewtopic.php?t=24042[/url] (Putnam problem)\r\nI am just posting a post there, asking some clarification..." } { "Tag": [ "inequalities", "geometry", "geometric transformation", "reflection", "inequalities proposed" ], "Problem": "For $ a,b,c$ positive reals and $ k > 1$ prove that\r\n\r\n$ \\sum \\sqrt[k] {\\frac{a^2\\plus{}b^2}{c^2\\plus{}ab}} \\leq \\frac{a\\plus{}b\\plus{}c}{\\sqrt[3]{abc}}$", "Solution_1": "Dear can_hang2007,\r\n\r\nhave you found a solution for this very nice and very hard inequality?\r\n\r\nThank you very much. :)", "Solution_2": "[quote=\"manlio\"]Dear can_hang2007,\n\nhave you found a solution for this very nice and very hard inequality?\n\nThank you very much. :)[/quote]\r\nThanks Manlio. I haven't seen this problem yet. I'll try.", "Solution_3": "[quote=\"manlio\"]For $ a,b,c$ positive reals and $ k > 1$ prove that\n\n$ \\sum \\sqrt [k] {\\frac {a^2 \\plus{} b^2}{c^2 \\plus{} ab}} \\leq \\frac {a \\plus{} b \\plus{} c}{\\sqrt [3]{abc}}$[/quote]\r\nThe inequality is not true for $ 2 \\ge k \\ge 1$ (for example $ a\\equal{}1.4,b\\equal{}c\\equal{}1$), for $ k \\ge 3$, I got the following solution", "Solution_4": "[quote=\"manlio\"]For $ a,b,c$ positive reals and $ k > 1$ prove that\n\n$ \\sum \\sqrt [k] {\\frac {a^2 \\plus{} b^2}{c^2 \\plus{} ab}} \\leq \\frac {a \\plus{} b \\plus{} c}{\\sqrt [3]{abc}}$[/quote]\r\nDear Manlio,\r\nI don't recall that I posted the problem in this form; more specifically, I don't know where the $ k\\ge 1$ is from. But my original post can be seen here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113060[/url]\r\nSince then, a more or less direct proof of it can be found in Ho Phu Thai's solution to Math. Reflections S27 [url]http://reflections.awesomemath.org/2006_6/2006_6_solutions.pdf[/url]", "Solution_5": "Dear Ductrung,\r\n\r\nsorry for my misprint :blush: ; I copied it from VIF(Vietnam Inequality Forum -- http://batdangthuc.net/forum/showthread.php?t=125.) and I supposed $ k \\geq 1$ but I was wrong. Sorry again.\r\n\r\n\r\nDear Can_hang,\r\n\r\nthank you very much for your solution.", "Solution_6": "Dear can_hang2007,\r\n\r\nin your solution you say: \"....According to my lemma at Mathematical Reflection.\"\r\n\r\nbut I don't remember you published it. Do you refer at ABC theorem?", "Solution_7": "[quote=\"manlio\"]Dear can_hang2007,\n\nin your solution you say: \"....According to my lemma at Mathematical Reflection.\"\n\nbut I don't remember you published it. Do you refer at ABC theorem?[/quote]\r\nI don't know what's ABC theorem at all, the lemma I told is in the article \"On a class three variable inequality\".\r\nAs in the theorem, we have $ \\frac {1}{27}(p \\plus{} q)^2(p \\minus{} 2q) \\le r \\le \\frac {1}{27} (p \\minus{} q)^2(p \\plus{} 2q)$, now if we denote $ x \\equal{} \\frac {p \\minus{} q}{3} \\ge 0, y \\equal{} \\frac {p \\plus{} 2q}{3} \\ge 0$ then $ r\\le x^2y$. Moreover, it's easy for us to check that\r\n\\[ p \\equal{} a \\plus{} b \\plus{} c \\equal{} 2x \\plus{} y\r\n\\]\r\n\r\n\\[ ab \\plus{} bc \\plus{} ca \\equal{} x^2 \\plus{} 2xy \\equal{} \\frac {p^2 \\minus{} q^2}{3}\r\n\\]\r\nThat's what I want to talk about. :)", "Solution_8": "Thank you very much :)", "Solution_9": "[quote=\"can_hang2007\"]I don't know what's ABC theorem at all, the lemma I told is in the article \"On a class three variable inequality\".\nAs in the theorem, we have $ \\frac {1}{27}(p \\plus{} q)^2(p \\minus{} 2q) \\le r \\le \\frac {1}{27} (p \\minus{} q)^2(p \\plus{} 2q)$, now if we denote $ x \\equal{} \\frac {p \\minus{} q}{3} \\ge 0, y \\equal{} \\frac {p \\plus{} 2q}{3} \\ge 0$ then $ r\\le x^2y$. Moreover, it's easy for us to check that\n\\[ p \\equal{} a \\plus{} b \\plus{} c \\equal{} 2x \\plus{} y\n\\]\n\n\\[ ab \\plus{} bc \\plus{} ca \\equal{} x^2 \\plus{} 2xy \\equal{} \\frac {p^2 \\minus{} q^2}{3}\n\\]\nThat's what I want to talk about. :)[/quote]\r\n\r\nSorry if I am so stupid, but I don't understand why, from $ r\\leq x^2y$ you can deduce that it suffices to consider the expression in your proof only for $ b \\equal{} c \\equal{} 1$ :oops:\r\n\r\n\r\n$ \\frac{(\\sum a)^3}{\\sum a^2} \\plus{} 6\\frac{(\\sum a^2)^2}{\\sum a^3 \\plus{} 3abc} \\geq 6\\sum a$", "Solution_10": "[quote=\"manlio\"][quote=\"can_hang2007\"]I don't know what's ABC theorem at all, the lemma I told is in the article \"On a class three variable inequality\".\nAs in the theorem, we have $ \\frac {1}{27}(p \\plus{} q)^2(p \\minus{} 2q) \\le r \\le \\frac {1}{27} (p \\minus{} q)^2(p \\plus{} 2q)$, now if we denote $ x \\equal{} \\frac {p \\minus{} q}{3} \\ge 0, y \\equal{} \\frac {p \\plus{} 2q}{3} \\ge 0$ then $ r\\le x^2y$. Moreover, it's easy for us to check that\n\\[ p \\equal{} a \\plus{} b \\plus{} c \\equal{} 2x \\plus{} y\n\\]\n\n\\[ ab \\plus{} bc \\plus{} ca \\equal{} x^2 \\plus{} 2xy \\equal{} \\frac {p^2 \\minus{} q^2}{3}\n\\]\nThat's what I want to talk about. :)[/quote]\n\nSorry if I am so stupid, but I don't understand why, from $ r\\leq x^2y$ you can deduce that it suffices to consider the expression in your proof only for $ b \\equal{} c \\equal{} 1$ :oops:\n\n\n$ \\frac {(\\sum a)^3}{\\sum a^2} \\plus{} 6\\frac {(\\sum a^2)^2}{\\sum a^3 \\plus{} 3abc} \\geq 6\\sum a$[/quote]\r\nWell, if you change the inequality into $ p,t,r$ ($ t\\equal{}ab\\plus{}bc\\plus{}ca$), then it becomes\r\n\\[ \\frac{p^3}{p^2\\minus{}2t}\\plus{}\\frac{6(p^2\\minus{}2t)^2}{p^3\\minus{}3pt\\plus{}6r} \\ge 6p\r\n\\]\r\nNow,\r\n\\[ \\frac{p^3}{p^2\\minus{}2t}\\plus{}\\frac{6(p^2\\minus{}2t)^2}{p^3\\minus{}3pt\\plus{}6r} \\ge \\frac{p^3}{p^2\\minus{}2t}\\plus{}\\frac{6(p^2\\minus{}2t)^2}{p^3\\minus{}3pt\\plus{}6x^2y}\r\n\\]\r\nIt suffices to prove\r\n\\[ \\frac{p^3}{p^2\\minus{}2t}\\plus{}\\frac{6(p^2\\minus{}2t)^2}{p^3\\minus{}3pt\\plus{}6x^2y} \\ge 6p\r\n\\]\r\nNow, replace $ p,t$ by $ 2x\\plus{}y,x^2\\plus{}2xy$, respectively, it becomes\r\n\\[ \\frac{(x\\plus{}2y)^3}{x^2\\plus{}2y^2} \\plus{}\\frac{6(x^2\\plus{}2y^2)^2}{x^3\\plus{}3xy^2\\plus{}2y^3} \\ge 6(x\\plus{}2y)\r\n\\]\r\nThe inequality is homogenous, then we can assume $ y\\equal{}1$ :)", "Solution_11": "Many thanks again :) \r\n\r\nYour proof is very nice and very interesting :wink:" } { "Tag": [ "function", "floor function", "logarithms", "AMC", "AIME" ], "Problem": "If $\\lfloor {x} \\rfloor$ is the greatest integer less than or equal to $x$, then\r\n\r\n\\[ \\sum^{1024}_{N=1} \\lfloor {\\log_2 N} \\rfloor = \\]\r\n\r\n$\\text{(A)} \\ 8192 \\qquad \\text{(B)} \\ 8204 \\qquad \\text{(C)} \\ 9218 \\qquad \\text{(D)} \\ \\lfloor {\\log_2 (1024!)} \\rfloor \\qquad \\text{(E)} \\ \\text{none of these}$", "Solution_1": "[hide]knowing our powers of 2, $\\log_2 1 = 0$, $\\log_2 2 = 1$, etc., we get 1(0) + 2(1) + 4(2) + 8(3) + 16(4) + .... + 512(9) + 10 = 8204. [/hide]", "Solution_2": "I've never gotten the sum function...Could someone explain it to me?", "Solution_3": "You basically just apply the the thing on the right for each number, starting with the bottom number, to the top number. For example, in this problem, you would add up all the values of ${\\lfloor {\\log_2 N} \\rfloor}$, where N is all the numbers from 1 to 1024.\r\n\r\nSo $\\sum^{5}_{N=1} N = $ would be 15.", "Solution_4": "I posted a thread concerning the use of summation a while ago.\r\n\r\nHere it is:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Summation&t=22820\r\n\r\nHope this helps you on the understanding the $\\sum$!", "Solution_5": "Thanks!! That's helped me to understand a lot more of the sum function. I've seen it a lot before, just all I knew was that it meant sum. There's another greek symbol for product (It is an uppercase $\\pi$: I believe), would it be appropriate to ask what that means here?", "Solution_6": "Do you mean $\\prod$?", "Solution_7": "Yeah, I'm pretty sure that's the sign.", "Solution_8": "$\\prod$ works same as $\\sum$.\r\n\r\nOnly difference is instead of $\\sum^{5}_{i=1} i = 1+2+3+4+5 = 15$, $\\prod^{5}_{i=5} i = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 = 5! = 120$.\r\n\r\nDoes this answer your question?", "Solution_9": "Yes, it does. Thanks for all your help!!!", "Solution_10": "[hide]1*0+2*1+4*2+8*3+...+512*9+10=8204[/hide]", "Solution_11": "[hide]For $N=1$, $\\lfloor \\log_2 N \\rfloor = 0$. \nFor $2 \\le N \\le 3$, $\\lfloor \\log_2 N \\rfloor = 1$.\nFor $4 \\le N \\le 7$, $\\lfloor \\log_2 N \\rfloor = 2$.\n...\nFor $512 \\le N \\le 1023$, $\\lfloor \\log_2 N \\rfloor = 9$.\nFor $N=1024$, $\\lfloor \\log_2 N \\rfloor = 10$.\n\nSo $0 + 1(2) + 2(4) + 3(8) + \\cdots + 9(512) + 10 = 10 + 2 + 8 + 24 + \\cdots + 4608$\n\nDon't know how to evaluate this though. :blush: [/hide]", "Solution_12": "[quote=\"chess64\"]So $0 + 1(2) + 2(4) + 3(8) + \\cdots + 9(512) + 10 = 10 + 2 + 8 + 24 + \\cdots + 4608$\n\nDon't know how to evaluate this though. :blush:[/quote]\r\n\r\nTake out your calculator (or your computer program) :roll:", "Solution_13": "Oh. But no calculators were allowed on AHSME :( Guess you'll have to do it by hand :P", "Solution_14": "[quote=\"chess64\"]Oh. But no calculators were allowed on AHSME :( Guess you'll have to do it by hand :P[/quote]\r\n\r\nThey weren't? :o I guess I was thinking the AMC 12 :P \r\nI cheated :blush:", "Solution_15": "[quote=\"chess64\"]Oh. But no calculators were allowed on AHSME :( Guess you'll have to do it by hand :P[/quote]Yeah, there was a problem like this on AIME one year. The only way is just do it by hand, I think." } { "Tag": [ "AMC", "AMC 10", "AMC 12", "AIME", "MIT", "college", "USA(J)MO" ], "Problem": "I have heard a ton of people talking about the point system for next year's AMC; everyone is giving me different numbers for the amount of points gained when you leave a question blank, so does anyone know how the point system will work next year?", "Solution_1": "6 points for a correct answer.\r\n1.5 points for no answer.\r\n0 points for an incorrect answer.\r\n\r\n25 questions still. Generally, if you're one of those people who purposely does 17 questions on the AMC10 or 11 questions on the AMC12 to qualify, this change is to disrupt you and prevent you from qualifying for the AIME easily like this.", "Solution_2": "Where/when was this announced?", "Solution_3": "This year, 28,000 qualified for AMC12. Among them, 17,000 scored between 100 - 110. I guess you won't see these people in AIME next year with the change.", "Solution_4": "http://en.wikipedia.org/wiki/American_Mathematics_Contest#Rules_and_scoring", "Solution_5": "rofl; wikipedia?\r\n\r\nI think it was announced in the manual(for this year's AMC)\r\n\r\nalso, AMC Director has made posts on AoPS about the change", "Solution_6": "[quote=\"krustyteklown\"]rofl; wikipedia?\n\nI think it was announced in the manual(for this year's AMC)\n\nalso, AMC Director has made posts on AoPS about the change[/quote]\r\n\r\nThis is correct. There was a thread that asked whether 1.5 point scoring is true or not and I think Mr.Dunbar told people that it is true. I think he also mentioned that the main reason for this was to prevent some \"random\" students who would just solve exactly 11 problems and qualify without looking at the later problems (which are much better and exciting). By new scoring, it won't be that easy I think.\r\n\r\nI do not really like wikipedia and never recommend this for any type of serious information as this case. It is because that anyone can edits it and who knows, someone could've put something like \"If you do well on AoPS's Mock AMC E, you'll qualify to AIME next year :D \"", "Solution_7": "ahh, 14 questions. I got 13 questions right this year. :blush:", "Solution_8": "The first I heard about it was in the 2005 awards booklet. As for the information on wikipedia, I'm the one who put it there, and it came from the booklet.", "Solution_9": "[quote=\"Silverfalcon\"]\n...It is because that anyone can edits it...[/quote]\r\n\r\nPoint in case. :-)", "Solution_10": "i didnt think people actually updated those online encyclopedias that are available for anyone to change lol...i guess i was wrong. well, thanks!", "Solution_11": "Wow I actually really like Wikipedia. It gives lots of information about things that other online encyclopedias don't. And usually if you edit in something incorrect, it is changed within seconds. Try it - I've done it once. But whatev, maybe it's just because I'm the only one who doesn't have an encyclopedia at home... :blush:", "Solution_12": "[quote=\"zmli\"]This year, 28,000 qualified for AMC12. Among them, 17,000 scored between 100 - 110. I guess you won't see these people in AIME next year with the change.[/quote]\r\n\r\nI think it makes as much of an impact on the AMC 10. TONS of people qualify with scores like 121.5 (ahem, me last year) and like 123. They probably wouldn't get in now, since you need 19 right/6 blank or better to get in. This probably also explains why the first few questions are SOO easy in both tests...", "Solution_13": "Wikipedia pretty much rules all! \r\n\r\nCheck out its entries for some of the prestigious universities around the country, I found the MIT entry to pretty amusing (and accurate, on the whole)\r\n\r\nAdditionally, Wikipedia is a nice place to get a quick dose of information about famous mathematicians.", "Solution_14": "It's not like Wikipedia isn't nice if you want to look up song lyrics or something but you really shouldn't use it if you're doing formal research; it's not a credible course.", "Solution_15": "[quote=\"hockeyadi23\"][quote=\"zmli\"]This year, 28,000 qualified for AMC12. Among them, 17,000 scored between 100 - 110. I guess you won't see these people in AIME next year with the change.[/quote]\n\nI think it makes as much of an impact on the AMC 10. TONS of people qualify with scores like 121.5 (ahem, me last year) and like 123. They probably wouldn't get in now, since you need 19 right/6 blank or better to get in. This probably also explains why the first few questions are SOO easy in both tests...[/quote]\r\nDoes it really? I suppose it does explain.", "Solution_16": "I am rather happy about the changes, it filters out all the borderline people. Not that I'm great at math or anything, but I did qualify decently this year for AIME (125.5 on AMC 12B) \r\nBut if USAMO # of qualifiers expands it seems natural for all the numbers for AIME and AMC12 to expand too... unless the AMC people want to save money on those who won't be moving on at all, which is a very natural thing to do :)", "Solution_17": "Many of us on this site have been used to a rather elitist outlook on math competitions. AIME qualification seems to come naturally for almost all the users who browse this site on a normal basis, so it doesn't mean all that much anymore. To many people, making the AIME is what making USAMO or MOP is to most of us here. The traditional 12,000 qualifiers, while certainly not a top-level prestige, does make a statement about an individual's problem solving abilities. And that, I think, is a level of achievement that should be preserved. A change in the point system is a great way to do so.\r\n\r\nHeck, even mathletes not enamored by AIME qualification should be glad that this move more fairly calculates a person's true problem solving ability by cancelling some of the USAMO index inflation created by people simply leaving many answers blank.", "Solution_18": "Heh, I'd like to point out that I got 3 on this year's AIME :rotfl: and I'm a junior, so time is running out :o \r\n\r\nBut that's ok, Mith, you're absolutely right!", "Solution_19": "[quote=\"randomdragoon\"][quote=\"hockeyadi23\"][quote=\"zmli\"]This year, 28,000 qualified for AMC12. Among them, 17,000 scored between 100 - 110. I guess you won't see these people in AIME next year with the change.[/quote]\n\nI think it makes as much of an impact on the AMC 10. TONS of people qualify with scores like 121.5 (ahem, me last year) and like 123. They probably wouldn't get in now, since you need 19 right/6 blank or better to get in. This probably also explains why the first few questions are SOO easy in both tests...[/quote]\nDoes it really? I suppose it does explain.[/quote]\r\n\r\nWell a lot of AMC 10-ers who make the AIMEs get most of the first 20, then just bomb the last five, or at least like four of them. Given that it's wicked easy to make a silly mistake on the AMC10 (cough cough number sixteen :wallbash_red: surprisingly the least answered question), we can assume this hits the 16-20 question right range. That leaves a lot of people on the boarderline, forcing at least 19 right (while before 17 right, 8 blank gets in), and that's if you leave the rest blank. A lot of middle schoolers and freshmen thus get knocked out. I don't exactly know why I took the AMC 10 (maybe to get a perfect? :lol:) but I was one of only three in my school, while 45 (!!!) made it on the AMC 12. Now my score was clearly in, but at least one of the other two may not have made it with the new system.", "Solution_20": "I think they're going to lower the difficulty of the questions on the AMC's because of the new scoring system, so scores should be about the same.", "Solution_21": "That sounds like a positive change, IMO......\r\n\r\nI could never do the borderline slip by approach anyhow, since I am almost guarenteed to miss 1 or more problems that I am convinced are right when I'm doing them. Last 2 years I slipped by, since I got 4 questions wrong both times. I am the LORD of stupid mistakes. Something I should think about/work on for next year, since I really can't afford it if I want to do the USAMO next year." } { "Tag": [ "factorial" ], "Problem": "A group of 25 friends were discussing a large positive integer. \"It can be divided by 1,\" said the first friend. \"It can be divided by 2,\" said the second friend. \"And by 3,\" said the third friend. \"And by 4,\" added the fourth friend. This continued until everyone had made such a comment. If exactly two friends were incorrect, and those two friends said consecutive numbers, what was the least possible integer they were discussing?", "Solution_1": "I think that the two numbers were 16 and 17 so the answer is 787386600.", "Solution_2": "Do you multiply this all out by hand, do you need to memorize factorials up to $25$, or would this be a calculator question?", "Solution_3": "no, the two numbers were the two largest prime numbers, 19, 23", "Solution_4": "I'm not sure if we have to memorize the factorials up to 25 (I don't think so) I think we just do this out by hand or of this part of the test we were aloud to use a calc" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "yello ! :D \r\n\r\n$ \\text{(i)}$\r\nnote that signs alternate, and the denominators are composed of [i]odd[/i] numbers that are [u]not[/u] divisible by $ 3$, i.e.\r\n\r\n$ 1 \\minus{} \\dfrac{1}{5^3} \\plus{} \\dfrac{1}{7^3} \\minus{} \\dfrac{1}{11^3} \\plus{} \\dfrac{1}{13^3} \\minus{} \\dfrac{1}{17^3} \\plus{} \\dfrac{1}{19^3} \\minus{} \\dfrac{1}{23^3}\\;\\; \\plus{} \\dfrac{1}{25^3} \\minus{} \\dfrac{1}{29^3}\\;\\; \\plus{} \\;\\; \\minus{} \\;\\;\\ldots$\r\n\r\n$ \\text{(ii)}$\r\nnote that signs alternate, and the denominators are composed of [i]odd[/i] numbers that are [u]not[/u] divisible by $ 7$, i.e.\r\n\r\n$ 1 \\minus{} \\dfrac{1}{3^3} \\plus{} \\dfrac{1}{5^3} \\minus{} \\dfrac{1}{9^3} \\plus{} \\dfrac{1}{11^3} \\minus{} \\dfrac{1}{13^3} \\plus{} \\dfrac{1}{15^3} \\minus{} \\dfrac{1}{17^3}\\;\\; \\plus{} \\dfrac{1}{19^3} \\minus{} \\dfrac{1}{23^3}\\;\\; \\plus{} \\;\\; \\minus{} \\;\\;\\ldots$\r\n\r\n\r\n\r\ndo [b]not[/b] use complex variables or fourier series. those are things that are taught at later stages in an university. instead, try and use techniques that a 7th or 8th grader will be likely to have, i.e. general integral calculus, trigonometry, etc. i know the answers and a rather elementary method for solving both, but i want to see if anyone here knows of any simpler method...", "Solution_1": "[quote=\"Misan\"]that a 7th or 8th grader will be likely to have, i.e. general integral calculus[/quote]\r\n\r\nAgain, a general 7th or 8th grader is not supposed to know calculus. You deal with special students, don't forget. Also, for your genius students complex variables and Fourier series should be elementary, right? Aren't you exagerating a little already?" } { "Tag": [ "ARML", "AMC", "AIME", "HMMT", "MATHCOUNTS" ], "Problem": "ok so tell me how farr youve made it in Mathcounts.....and how many years it took you\r\n\r\nim bored...might as well find out\r\n\r\n\r\nif anyof you are wondering...last year was my first year and i came 6th in state", "Solution_1": "I joined when I was 9, and I competed when I was 10, 11, and 12. I got 29th at nats when I was 12(going on 13 very soon then)", "Solution_2": "I got 18th in state in 6th grade and then 16th at Nationals in 7th grade.", "Solution_3": "I got 19th in chapter in 6th grade and 5th at state in 7th grade (sounds much more of an improvement than it really is).", "Solution_4": "Havn't gone yet going next year :lol: \r\n\r\nI've competed at age 9,10,11 in ARML, PuMAC, AMCs (GO 25 ON 8!), AIME, NYSML, HMMT, BCA math competition, ATL-PAF, AMCNJ, Math Contests, Four-by-Four, Fall Startup event.\r\n\r\nI think.\r\n\r\nGood luck to everyone at MC (Unless, of course, you're facing me in CD)", "Solution_5": "I made Nats ~130th in 8th grade, the first year I'd even heard of Mathcounts.", "Solution_6": "6th: 1st district, 5th state\r\n7th: 1st district, 2nd state, 65th nats\r\n8th: tie 1st school, ___ district, ___ state, ___nats", "Solution_7": "6th: 8th chapter, 22nd state (in Michigan)\r\n7th: (moved to Ohio) 1st chapter, 4th state, 92nd nats", "Solution_8": "6th grade: didn't make school team :( \r\n7th grade: 3rd chapter, 45th state", "Solution_9": "8th grade: 1st state (MN), my first nats, T5th. Good luck to everyone else...\r\n\r\nI wish I could have been smarter earlier in my lifetime... (like the younger people who won last year :P)", "Solution_10": "6th grade: Alternate... :o\r\n7th grade: hasn't happened yet... definitely made the team though...\r\n8th grade: 2010 :P", "Solution_11": "6th grade: Got lucky and was an individual. Rank in Chapter: Not Placed ( I got 9 right in the sprint)\r\n7th Grade: chapter: 13th place and alternate for states\r\n8th grade: Chapter 0th place, state: 0th place, nats: 0th place\r\n\r\n\r\n0= TBA", "Solution_12": "6th grade: 6th chapter. Didn't make state. School sucks.\r\n7th grade: 4th chapter, went to state as an individual. 6th place state.\r\n8th grade: Hoping for 1st chapter and 1st state, top 20 (?) nationals", "Solution_13": "Our school didn't have mathcounts when I was in 6th grade.\r\n7th: This awesome teacher comes, our team makes state as a wild card (although we had the 2nd highest chapter score in the state, we came 2nd in chapter, lost by a quarter of a point :P) and gets 4th place there. Individually, I got 3rd at chapter, and 16th at state.\r\n8th: Our school got 1st chapter, and 4th state again. I got 1st chapter, 1st state, T3 nationals.\r\n\r\nGood luck to everyone who's preparing for it this year!", "Solution_14": "someone really needs to make a thread or forem for all of these surveys...\r\nthis is getting nuts...", "Solution_15": "6th grade, 3rd regional, blown out of water in cd by two people who ended up making nats (they had 1-2-3 play cd in first round!!!). At state that year, I was tied for first after sprint round with a guy who would go on to come in 12th at nats. Then I epic failed target and dropped to 11th!\r\n\r\n7th grade I made nats coming in a tie for 69th. Too many potatoes that morning :rotfl:" } { "Tag": [ "function", "calculus", "derivative", "algebra", "domain", "integration", "Putnam" ], "Problem": "[quote=\"Originally\"]Assume that $ |f(x)| \\leq 1$ and $ |f'(x)| \\leq 1$ for all x on an interval of length at least 2. Show that $ |f''(x)| \\leq 2$ on the interval.[/quote]\r\n\r\n[u]Corrected statement of the problem:[/u] Assume that $ |f(x)| \\leq 1$ and $ |f''(x)| \\leq 1$ for all x on an interval of length at least 2. Show that $ |f'(x)| \\leq 2$ on the interval.", "Solution_1": "This is indeed a challenge! If we start with $ f'(x) \\equal{} \\frac {|x|}{x}$, by smoothing at $ x \\equal{} 0$ we can make $ f''(0)$ arbitrary large.", "Solution_2": "I would assume that the intended problem was that $ |f(x)|\\le1$ and $ |f''(x)|\\le1,$ from which we seek a bound on $ |f'(x)|.$\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123244]This[/url] isn't the same problem but it is related.", "Solution_3": "I did indeed mistype the problem. It is corrected above. Sorry for the mistake.\r\n\r\n[hide=\"Kent:\"]I saw that other thread and I do suspect that the solution involves Taylor's Polynomial/Mean Value Theorem. But since this problem does not specify the derivatives at any particular point in the interval, I'm having trouble pinning it down.[/hide]", "Solution_4": "There was a similar problem in Rudin's Principles of Mathematical Analysis (Ch. 5 #15)\r\n\r\n[hide=\"Solution?\"]We use Taylor's Theorem and we have $ f(x\\plus{}2)\\equal{}f(x)\\plus{}2f'(x)\\plus{}2f''(c)$ where $ c\\in (x, x\\plus{}2)$. We solve for $ f'(x)$ and we have $ f'(x)\\equal{}\\frac{1}{2}[f(x\\plus{}2)\\minus{}f(x)]\\minus{}f''(c)$. We now notice that $ |f'(x)| \\leq \\frac{1}{2}[|f(x\\plus{}2)|\\plus{}|f(x)|]\\plus{}|f''(c)| \\leq 2$ as desired.[/hide]", "Solution_5": "[quote=\"1/(ln x)\"]There was a similar problem in Rudin's Principles of Mathematical Analysis (Ch. 5 #15)\n\n[hide=\"Solution?\"]We use Taylor's Theorem and we have $ f(x \\plus{} 2) \\equal{} f(x) \\plus{} 2f'(x) \\plus{} 2f''(c)$ where $ c\\in (x, x \\plus{} 2)$. We solve for $ f'(x)$ and we have $ f'(x) \\equal{} \\frac {1}{2}[f(x \\plus{} 2) \\minus{} f(x)] \\minus{} f''(c)$. We now notice that $ |f'(x)| \\leq \\frac {1}{2}[|f(x \\plus{} 2)| \\plus{} |f(x)|] \\plus{} |f''(c)| \\leq 2$ as desired.[/hide][/quote]\nI think this works only if $ x \\pm 2$ is in the domain of $ f$.\n\n[hide=\"Proof by Picture\"]WLOG say $ f$ is defined on $ [\\minus{}1,1]$, and let $ s \\in [\\minus{}1,1]$. Using $ |f''| \\le 1$ and the MVT we get $ f'(x) \\ge f'(s) \\minus{} (s\\minus{}x)$ for $ x \\le s$ and $ f'(x) \\ge f'(s) \\minus{} (x\\minus{}s)$ for $ x \\ge s$. Then\n\\[ 2 \\ge f(1) \\minus{} f(\\minus{}1) \\equal{} \\int_{\\minus{}1}^s f'(x)dx \\plus{} \\int_s^1 f'(x)dx \\\\\n\\ge \\int_{\\minus{}1}^s f'(s) \\minus{} (s\\minus{}x) dx \\plus{} \\int_s^1 f'(s) \\minus{} (x\\minus{}s) dx \\\\\n\\equal{} 2f'(s) \\minus{} s^2 \\minus{} 1 \\ge 2(f'(s)\\minus{}1).\\]\nThis shows $ f'(s) \\le 2$, and we can similarly get $ f'(s) \\ge \\minus{}2$.[/hide]", "Solution_6": "When $ x\\geq 0$\r\nWe have \r\n$ f(x)\\minus{}f(0)\\equal{}\\int_{0}^xf'(x)dx\\equal{}xf'(x)\\minus{}\\int_{0}^xxf''(x)dx$\r\n\r\n$ |xf'(x)|\\equal{}|\\int_{0}^xxf''(x)dx\\plus{}f(x)\\minus{}f(0)|$\r\n\r\n $ \\leq |f(x)|\\plus{}|f(0)|\\plus{}|\\int_{0}^xxf''(x)dx|$\r\n \r\n $ \\leq |f(x)|\\plus{}|f(0)|\\plus{}\\int_{0}^xx|f''(x)|dx$\r\n\r\n $ \\leq 1\\plus{}1\\plus{}\\int_{0}^xxdx$\r\n\r\n$ |xf'(x)|\\leq 2\\plus{}\\frac{x^2}{2}$\r\n\r\n$ x^2\\minus{}2x|f'(x)|\\plus{}4\\geq 0$ which is true for all $ x\\in R$\r\n\r\nDiscriminant $ \\leq 0$\r\n\r\n$ 4|f'(x)|^2\\leq 16$\r\n\r\n$ |f'(x)|\\leq 2$\r\n\r\nThis solution was on this website and I had copied it long back.", "Solution_7": "Thats bad notation all the way Pankaj :wink:" } { "Tag": [ "complex numbers", "geometry", "perpendicular bisector" ], "Problem": "Two fixed points $ Z_{1}, Z_{2}$ and a variable point $ Z$ represent the complex numbers $ z_{1},z_{2}, z$ respectively. \r\n\r\nGive the locus of Z if $ |z\\minus{}z_{1}| \\equal{} |z\\minus{}z_{2}|$\r\n\r\nWhy does Z have to be on the perpendicular bisector? Why can't it remain where it is??", "Solution_1": "Well the question stated that $ z$ is a variable point which means it's not fixed........\r\n\r\nSo by definition your locus represents the set point points $ z$ which are equi-distance from $ z_{1}$ and $ z_{2}$. So it is a perpendicular bisector.", "Solution_2": "Can't Z be a variable point at the vertex and moving so that the 2 arms remain equal??\r\n\r\nInstead of being the perendicualr bisector?\r\n\r\n[url=http://img459.imageshack.us/my.php?image=81210114qt5.png][img]http://img459.imageshack.us/img459/3755/81210114qt5.th.png[/img][/url]", "Solution_3": "bos1234...yes but any point at the vertix of this trinagle will lie on the perpendicular bisector...\r\n\r\nProof\r\n\r\nUsing your diagram, let the line $ zx$ be perpendicular to $ z_{1}z_{2}$ with $ x$ lying on the side $ z_{1}z_{2}$.\r\nBy using pythagoras' theorem\r\n\r\n$ {z_{1}z}^{2}\\equal{}{z_{1}x}^{2}\\plus{}{xz}^{2}$ and $ {z_{2}z}^{2}\\equal{}{z_{2}x}^{2}\\plus{}{xz}^{2}$\r\n\r\nBut since $ z_{1}z \\equal{} z_{2}z$ then $ {z_{1}z}^{2}\\equal{}{z_{2}z}^{2}$\r\n\r\n$ \\Rightarrow{z_{1}x}^{2}\\plus{}{xz}^{2}\\equal{}{z_{2}x}^{2}\\plus{}{xz}^{2}$\r\n\r\n$ \\Rightarrow{z_{1}x}^{2}\\equal{}{z_{2}x}^{2}$\r\n\r\n$ \\Rightarrow z_{1}x \\equal{} z_{2}x$\r\n\r\nHence $ x$ is the midpoint of $ z_{1}z_{2}$. But since $ xz$ is perpendicular to $ z_{1}z_{2}$, then $ zx$ is the perpendicular bisector of $ z_{1}z_{2}$.\r\n\r\nHence $ z$ lies on the perpendicular bisector.", "Solution_4": "wonderful proof. They should include this in the books\r\n\r\nNow I understand thoroughly \r\nTHANKS!!" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $x;y;z$ be positive real numbers.Prove that:\r\n$\\sqrt{\\frac{y^{2}z^{2}}{x^{2}(y+z)^{2}}+\\frac{x^{2}y^{2}}{z^{2}(x+y)^{2}}+\\frac{y^{2}}{(x+y)^{2}}+\\frac{y^{2}}{(y+z)^{2}}}$\r\n$+\\sqrt{\\frac{x^{2}z^{2}}{y^{2}(x+z)^{2}}+\\frac{x^{2}y^{2}}{z^{2}(x+y)^{2}}+\\frac{x^{2}}{(x+y)^{2}}+\\frac{x^{2}}{(x+z)^{2}}}$\r\n$+\\sqrt{\\frac{y^{2}z^{2}}{x^{2}(y+z)^{2}}+\\frac{x^{2}z^{2}}{y^{2}(x+z)^{2}}+\\frac{z^{2}}{(y+z)^{2}}+\\frac{z^{2}}{(x+z)^{2}}}\\geq 3$", "Solution_1": "$\\sqrt{\\frac{y^{2}z^{2}}{x^{2}(y+z)^{2}}+\\frac{x^{2}y^{2}}{z^{2}(x+y)^{2}}+\\frac{y^{2}}{(x+y)^{2}}+\\frac{y^{2}}{(y+z)^{2}}}2\\geq\\sum(\\frac{yz}{x(y+z)}+\\frac{xy}{z(x+y)}+\\frac{y}{x+y}+\\frac{y}{y+z})$\r\nFinally,use nesbit's inequality" } { "Tag": [ "videos" ], "Problem": "Has anyone played .hack/infection here? What did you guys think of it??? Being only able to play few video games with my little time, I have found it to be pretty interesting. My only regret was that there was no bonus round where you could kill/hunt grunty's... that would have been pretty tite.\r\n\r\nEnjoy.\r\nIW.OAV", "Solution_1": "I have no clue what it is or where to get it. Can you post some more info? Thanks.\r\n\r\n\r\n\r\n-Isaac", "Solution_2": "[color=white]Same here. What platform is it for? It sounds like Banjo-Kazooie...grunty's?[/color]", "Solution_3": "I've never played it, having spent too much time on this forum. But I think grunty's are referring to grunts, as in common, usual soldiers. Dunno. Just a guess.", "Solution_4": "[color=white]Wow, mathfanatic...you're really gaining on ComplexZeta on posts. Syntax Error isn't going to be happy.[/color]", "Solution_5": "And guess what... every time you post something, I post something in reply...so it's all YOUR fault. :lol: But thanks. #2 right now.", "Solution_6": "[color=white]...\n\nWell, when I post, I get a point as well. So we're helping each other. Seeing that you're not too far away, from ComplexZeta, some faraway day in the future you may be #1.[/color]", "Solution_7": "Oooh, jojo's gunning in on 100 posts and a Poincare Conjecture. Eek! He's coming up FAST too. I'd better watch out.\r\n\r\nP.S. Maybe if I have another \"hyper mouse session\" and post this 5 times, I'll get credit for 5 posts. ", "Solution_8": "Can you help me get posting points too?\r\n\r\n\r\n\r\n-Isaac", "Solution_9": "You guys can find about it here.. it's a freakin' awesome tastic game! \r\n\r\nhttp://www.dothack.com/game/index.html\r\n\r\nBTW it's one of the best RPGs out next to the FF series.", "Solution_10": "Me going away for a couple of weeks ruined my post ranking :( oh well, I'm trying to come back..", "Solution_11": "Gosh...it seemed like yesterday that Mathfanatic passed 210, but now he's in the 240s???\r\n\r\n-interesting_move", "Solution_12": "yah, I need to make a comeback :P\r\n\r\n\r\n\r\n\r\n-Isaac", "Solution_13": "[color=white]He he, I'm catching up, Issac...and I passed IW.OAV a long time ago.[/color]", "Solution_14": "the sequel to infection is already out. and more is to come. i have so many games that are just sitting there to be played on my ps2. like xenosaga. it rocks. but i dunno. i havent been in the mood to play. .hack is also a show on cartoon network.", "Solution_15": "I'm screwed now...\r\n\r\n\r\n\r\n-Isaac", "Solution_16": "welcome back!", "Solution_17": "nope", "Solution_18": "nope what?", "Solution_19": "I think \"nope, I don't play this game\" and I agree.", "Solution_20": "Ahh...I see. in that case, i concur" } { "Tag": [ "Divisibility Theory" ], "Problem": "Determine all triples of positive integers $(a, m, n)$ such that $a^m +1$ divides $(a+1)^n$.", "Solution_1": "[quote=\"Peter\"]Determine all triples of positive integers $ (a, m, n)$ such that $ a^m \\plus{} 1$ divides $ (a \\plus{} 1)^n$.[/quote]\r\nhttp://www.kalva.demon.co.uk/short/soln/sh00n4.html", "Solution_2": "If $ a,m > 1$ and $ (a,m) \\neq (2,3)$, by Zsigmondy's theorem there is a prime $ p$ dividing $ a^m \\plus{} 1$ but not $ a \\plus{} 1$. Thus there are no solutions.\r\nTrivially checking the other cases, we get that all other cases always are solutions, regardless what $ n$ is, the only exception being $ (a,m) \\equal{} (2,3)$ where we need $ n \\geq 2$." } { "Tag": [ "trigonometry" ], "Problem": "Prove that $ x^{\\sin^2 a} \\cdot y^{\\cos^2 a}0, a \\in R$.", "Solution_1": "[quote=\"balan razvan\"]Prove that $ x^{\\sin^2 a} \\cdot y^{\\cos^2 a} < x \\plus{} y$ , $ x,y > 0, a \\in R$.[/quote]\r\nAssume that $ y>x$ ( similarly when $ yx$", "Solution_2": "For $ x\\geq y>0,\\ \\left(\\frac {x}{y}\\right)^{\\sin ^ 2 a} < \\frac {x}{y} < \\frac {x}{y} \\plus{} 1$.", "Solution_3": "Since the maximum value of $ sin^2a$ is 1, $ x^{sin^2a}\\leq x$ and $ y^{sin^2a}\\leq y$.\r\nMultiplying them together we get that the LHS is less than or equal to xy, which is less than x+y because by AM-GM, 2xy<=x+y, and so xy=0 va dashte bashim $ a\\plus{}b\\plus{}c\\equal{}3$ sabet konid\r\n\\[ \\frac{a}{1\\plus{}b^2}\\plus{}\\frac{b}{1\\plus{}c^2}\\plus{}\\frac{c}{1\\plus{}a^2}\\geq \\frac{3}{2}\\]", "Solution_1": "be rahati ba estefade az raveshe lagranj tabe $ L$ ra tarif mikonim:\r\n\\[ \\begin{array}{l} L \\equal{} \\frac {a}{{1 \\plus{} b^2 }} \\plus{} \\frac {b}{{1 \\plus{} c^2 }} \\plus{} \\frac {c}{{1 \\plus{} a^2 }} \\minus{} \\lambda (a \\plus{} b \\plus{} c \\minus{} 3) \\\\\r\n\\Rightarrow \\frac {{\\partial L}}{{\\partial a}} \\equal{} \\frac {1}{{(1 \\plus{} b)^3 }} \\\\\r\n\\frac {{\\partial L}}{{\\partial b}} \\equal{} \\frac {1}{{(1 \\plus{} c)^3 }} \\\\\r\n\\frac {{\\partial L}}{{\\partial c}} \\equal{} \\frac {1}{{(1 \\plus{} a)^3 }} \\\\\r\n\\to a \\equal{} b \\equal{} c \\equal{} 1 \\\\\r\n \\\\\r\n\\end{array}\r\n\\]\r\npas ekstermom tabe\r\n\\[ f \\equal{} \\frac {a}{{1 \\plus{} b^2 }} \\plus{} \\frac {b}{{1 \\plus{} c^2 }} \\plus{} \\frac {c}{{1 \\plus{} a^2 }}\r\n\\]\r\ndar noghteye $ (a,b,c) \\equal{} (1,1,1)$ ast va choon tabe mohadabast pas in ekstermom , minimum ast.va be ezaye in noghte masale hal ast", "Solution_2": "[quote=\"m bahador\"]be rahati ba estefade az raveshe lagranj tabe $ L$ ra tarif mikonim:\n\\[ \\begin{array}{l} L \\equal{} \\frac {a}{{1 \\plus{} b^2 }} \\plus{} \\frac {b}{{1 \\plus{} c^2 }} \\plus{} \\frac {c}{{1 \\plus{} a^2 }} \\minus{} \\lambda (a \\plus{} b \\plus{} c \\minus{} 3) \\\\\n\\Rightarrow \\frac {{\\partial L}}{{\\partial a}} \\equal{} \\frac {1}{{(1 \\plus{} b)^3 }} \\\\\n\\frac {{\\partial L}}{{\\partial b}} \\equal{} \\frac {1}{{(1 \\plus{} c)^3 }} \\\\\n\\frac {{\\partial L}}{{\\partial c}} \\equal{} \\frac {1}{{(1 \\plus{} a)^3 }} \\\\\n\\to a \\equal{} b \\equal{} c \\equal{} 1 \\\\\n \\\\\n\\end{array}\n\\]\npas ekstermom tabe\n\\[ f \\equal{} \\frac {a}{{1 \\plus{} b^2 }} \\plus{} \\frac {b}{{1 \\plus{} c^2 }} \\plus{} \\frac {c}{{1 \\plus{} a^2 }}\n\\]\ndar noghteye $ (a,b,c) \\equal{} (1,1,1)$ ast va choon tabe mohadabast pas in ekstermom , minimum ast.va be ezaye in noghte masale hal ast[/quote]\r\n :huh: \r\nheife soale be in ghashangi nis ke ba in raha kharabesh konim?\r\n\r\n$ x(x \\minus{} 1)^2\\geq 0\\Rightarrow \\frac 1{1 \\plus{} x^2}\\geq\\frac {2 \\minus{} x}2$\r\n\r\nbaraye har $ x\\in\\mathbb{R}^ \\plus{}$,pas darim:\r\n\r\n$ \\frac {a}{1 \\plus{} b^2}\\geq\\frac {a(2 \\minus{} b)}2$\r\n\r\npas darim:\r\n\r\n$ \\sum\\frac {a}{1 \\plus{} b^2}\\geq\\sum\\frac {a(2 \\minus{} b)}2$\r\n\r\npas kafie sabet konim:\r\n\r\n$ \\sum\\frac {a(2 \\minus{} b)}2\\geq\\frac 32\\iff 3\\geq ab \\plus{} bc \\plus{} ca$\r\n\r\nke inam vazehe cherake $ a \\plus{} b \\plus{} c \\equal{} 3$ va $ (a \\plus{} b \\plus{} c)^2\\geq 3(ab \\plus{} bc \\plus{} ca)$", "Solution_3": "are bavar kon ke ghalban ham razi naboodam in raho post konam vali choon ke too emtehanaeem va vaght kafi baraye fek kardan nadaram majboor shodam ,kholase bebakhshid :oops:", "Solution_4": "akh akh gofti emtehana yade parsal oftadam,cheghad hal midad. :D" } { "Tag": [ "function", "logarithms", "algebra", "polynomial", "limit", "induction", "calculus" ], "Problem": "Consider a differentiable function f:R---->R which satisfies $ f^2(\\frac{x}{\\sqrt{2}})\\equal{}f(x)$ for all x belonging to R and f(1)=2\r\nFind f(x)", "Solution_1": "Note that $ f(t)$ on $ \\left[1, x\\right]$ satisfies the hypotheses of the Mean Value Theorem.\r\nSo there exists $ \\theta \\in \\left(1, x\\right)$ such that we have:\r\n\r\n$ \\frac {f(x) \\minus{} f(1)}{x \\minus{} 1} \\equal{} f'(\\theta) \\implies f(x) \\equal{} \\left(x \\minus{} 1 \\right)\\cdot f'(\\theta) \\plus{} f(1)$ *\r\n\r\nThe condition that $ f^{2}\\left(\\frac {x}{\\sqrt {2}}\\right) \\equal{} f(x) \\implies \\sqrt {2} \\cdot f\\left(\\frac {x}{\\sqrt {2}}\\right) \\cdot f'\\left(\\frac {x}{\\sqrt {2}}\\right) \\equal{} f'(x)$\r\n\r\n$ \\implies \\sqrt {2} \\cdot f(1) \\cdot f'(1) \\equal{} f'(\\sqrt {2})$\r\n\r\nNow to find $ f(x)$ explicitly you need to know the values assigned to $ f(1)$ and $ f'(1)$ and then the * will give us the result by setting $ \\theta \\equal{} \\sqrt {2}$.", "Solution_2": "To akech:\r\nNote, that in your formula * you have that $ \\theta \\equal{} \\theta(x)$ depends on $ x$ and,hence, you cannot take $ \\theta \\equal{} \\sqrt {2}$ for arbitrary $ x$ :)", "Solution_3": "Probably the point is that the condition that $ f$ is differentiable [b]at 0[/b] is the crucial thing that rules out all other functions.\r\n\r\nThe obvious solution is $ f(x) \\equal{} 2^{(x^2)}$", "Solution_4": "Can anyone solve it completely ?", "Solution_5": "If I'm not mistaken, there should be some extra conditions on our function or on its derivative(if not, then there exist answers,which are not very ''simple\").But maybe I've done some stupid mistake.I'll check it later :)", "Solution_6": "Nothing else is given :(", "Solution_7": "it's easy to note that $ f(0) \\equal{} 1$ and $ f > 0$ for all $ x$\r\nconsider $ g(x) \\equal{} \\ln f(x)$ then we have $ g(1) \\equal{} \\ln 2$ and $ 2g(x/\\sqrt {2}) \\equal{} g(x)$\r\nso if we suppose that $ g$ is twise differentiable function then it's easy to show that $ g$ is polinomial for example by this way :\r\n$ g''(x/\\sqrt {2}) \\equal{} g''(x)$ so for any point $ z$ we have $ g''(z) \\equal{} g''(z/2^{n/2}) \\to g''(0)$ thus $ g''(z) \\equal{} g''(0)$\r\nso $ g$ is polynomial hence it's equal to $ x^{2} \\ln 2$ since $ g(1) \\equal{} \\ln 2$\r\nI think it's not hard to solve this problem without assumption twise differentiability of $ g$.\r\nfor example consider $ y(x) \\equal{} \\frac {g(x)}{x^{2}}$ then condition $ 2g(x/\\sqrt {2}) \\equal{} g(x)$ becames $ y(x/\\sqrt {2}) \\equal{} y(x)$ thus similary for any point $ z$ we have $ y(z) \\equal{} y(z/2^{n/2}) \\to y(0)$ thus we have $ g(x) \\equal{} y(0)x^{2}$ so $ g(x) \\equal{} x^{2} \\ln 2$\r\nso you just need to explain why $ y(x)$ is continous at point $ 0$ or why there exist limit $ \\lim _{0}\\frac {\\ln f(x)}{x^{2}}$ , does it follows from the differentiability of $ f$ ?", "Solution_8": "No, it does not follow from the differentiability,in general :wink:", "Solution_9": "Note that $ f(x)\\ge 0$ for all $ x$. It is easy to prove using induction that for all naturals $ n$, \r\n$ f(x)\\equal{}f^{2^n}\\left(\\dfrac{x}{(\\sqrt{2})^n}\\right)\\qquad(1)$\r\nwhere $ f^k(x)$ means $ (f(x))^k$. This means \r\n$ \\sqrt[2^{n}]{f(x)}\\equal{}f\\left(\\dfrac{x}{(\\sqrt{2})^n}\\right)$\r\nAs such \r\n$ \\sqrt[2^{n}]{f(1)}\\equal{}f\\left(\\dfrac{1}{(\\sqrt{2})^n}\\right)\\qquad \\Rightarrow\\ \\sqrt[2^{n}]{2}\\equal{}f\\left(\\dfrac{1}{(\\sqrt{2})^n}\\right)$\r\nTaking limit as $ n\\to \\infty$, we obtain $ 1\\equal{}f(0)$.\r\nNext, if for some $ x_0\\neq 0$, $ f(x_0)\\equal{}0$, then that means that at $ x\\equal{}x_0/\\sqrt{2}$ also $ f\\equal{}0$ which further means that at $ x\\equal{}x_0/(\\sqrt{2})^2$, $ f\\equal{}0$ and so on and in general at $ x\\equal{}x_0/(\\sqrt{2})^n$, $ f$ is zero. But again, in the limit, that will imply that $ f(0)\\equal{}0$ --- a contradiction. So $ f$ is not zero for any $ x$. Hence, $ f>0$ for all $ x$. \r\n\r\nBy setting $ F(x)\\equal{}\\log_af(x)$, for some positive $ a$, the given functional equation can be written as \r\n$ F\\left(\\dfrac{x}{\\sqrt{2}}\\right)\\equal{}\\dfrac{1}{2}\\,F(x)$\r\nwhich is the so called $ \\mathrm{Schr\\ddot{o}der}$ equation. Either, from this equation or by taking log of the equation (1), we get that $ F(x)$ satisfies for all $ x$ the following condition for all natural $ n$:\r\n$ F(x)\\equal{}2^n F\\left(\\dfrac{x}{(\\sqrt{2})^n}\\right)$\r\nIt is easy to see that $ F(x)\\equal{}x^2$ satisfies the above equation for all $ x$ (I haven't been able to prove so far the uniqueness of this assertion). As such \r\n$ f(x)\\equal{}a^{x^2}$\r\nBy the condition $ f(1)\\equal{}2$, we get $ a\\equal{}2$. As such, we get \r\n$ f(x)\\equal{}2^{x^2}$", "Solution_10": "Ok.I'll try to briefly explain the idea of my counterexample (maybe somebody explain to me where I'm wrong :maybe: ).\r\nFirst note, that if we have differentialble function $ g: R\\to R$ for which $ g(\\frac {x}{\\sqrt {2}}) \\equal{} \\frac {1}{2}g(x)$ and $ g(1) \\equal{} 1$ then we can take $ f(x) \\equal{} 2^{g(x)}$ and get the function that satisfies our functional equation.\r\nNow choose the 'small' interval $ (a,b)$ for $ a < 1 < b$ and $ a\\sqrt {2} > b$ and define nonconstant function $ h(x)$ that is differentiable on $ \\mathbb {R}$ and $ h(1) \\equal{} 1$, $ h(x) \\equal{} 0$ for $ x\\notin (a,b)$ (this is a classical example). Consider the collection of intervals $ (a(\\sqrt {2})^n, b(\\sqrt {2})^n)$, $ n\\in\\mathbb {Z}$ and define our function $ g$ recursively using our equation $ g(\\frac {x}{\\sqrt {2}}) \\equal{} \\frac {1}{2}g(x)$ and $ h(x) \\equal{} g(x)$ on $ [a,b]$, $ g(0) \\equal{} 0$. Then because our intervals do not intresect each other the only thing that we should check is the differentiability in point $ 0$. But this is obvious because if $ x < \\varepsilon$ then \r\nf(x)=0 if $ x\\ne \\frac{c}{(\\sqrt {2})^n}$ for some $ c\\in (a,b)$ or $ f(x) \\equal{} \\frac {f(c)}{2^n}$ if $ x \\equal{} \\frac {c}{(\\sqrt {2})^n}$. But then $ |\\frac {f(x)}{x}| < 2^{ \\minus{} \\frac {n}{2}}|\\frac {f(c)}{c}|$.This bound provides differentiabiliry in point $ 0$. It's easy to extend oyr function for $ x < 0$ in the same way.", "Solution_11": "You're right, of course, but let's understand the word \"differentable\" as \"has as many derivatives as you wish\" (one is certainly not enough, but, I guess, two or three would do the trick). ;)", "Solution_12": "i didn't understand both of the above solutions.\r\n :(", "Solution_13": "If we suppose that the function $ h(x)\\equal{}log_2 f(x)$ has two dirivatives, and second derivative is continous in point $ 0$ then,of course, everything will be fine.\r\nIndeed, for function $ h(x)$ we have the equation $ h(\\frac{x}{\\sqrt{2}})\\equal{}\\frac{1}{2} h(x)$. After differentiation two times we'll get $ h^{''}(x)\\equal{}h^{''}(\\frac{x}{\\sqrt{2}})$ or $ h(x)\\equal{}lim_{n\\to\\infty}(h(\\frac{x}{2^\\frac{n}{2}}))\\equal{}h(0)$." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $f: [a,b]\\to\\mathbb{R}$ a derivable on $[a,b]$ and continous on $(a,b)$. prove that there is \r\n $c\\in (a,b)$ such that: $(b-a)f(c)=(c-a)f(a)+(b-c)f(c)$ ;)", "Solution_1": "Manifestly that's false ! :D ! Could you make a correct statement ?\r\n :cool:", "Solution_2": "The obvious fix is to make it more symmetric and replace $(b-c)f(c)$ with $(b-c)f(b)$. The problem is, that isn't true either.", "Solution_3": "when $f=constant$ then $\\forall c\\in(a,b) :$ $(b-a)f(c)=(c-a)f(a)+(b-c)f(c)$" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Given a function $f: X \\to Y$\r\nWe have \r\n1. $f(f^{-1}(B)) \\subset B \\textrm{ for } B \\subset Y$ and\r\n2. $A \\subset f^{-1}(f(A)) \\textrm{ for } A \\subset X$. \r\nEquality need not hold in either case.\r\n\r\nWhy? Could someone explain that? thanks", "Solution_1": "Two simple counterxamples. In #1, we need a function that does not map $X$ onto $Y.$ In #2 we need a function that is not one-to-one.\r\n\r\n#1. Define $f: \\mathbb{R}\\mapsto\\mathbb{R}$ by $f(x)=e^x$ and let $B=[-1,1].$\r\n\r\n$f^{-1}(B)=\\{x: e^x\\in[-1,1]\\}=(-\\infty,0].$\r\n\r\n$f(f^{-1}(B))=f((-\\infty,0])=(0,1]$ which is not $B.$\r\n\r\n#2. Define $f: \\mathbb{R}\\mapsto\\mathbb{R}$ by $f(x)=x^2$ and let $A=[1,2].$\r\n\r\n$f(A)=[1,4].$\r\n\r\n$f^{-1}(f(A))=\\{x: x^2\\in[1,4]\\}=[-2,-1]\\cup[1,2].$" } { "Tag": [], "Problem": "Consider the sequence of primes $ p_{1}, p_{2},\\ldots$ such that for each $ i\\geq 2$, either $ p_{i}\\equal{} 2p_{i\\minus{}1}\\minus{}1$ or $ p_{i}\\equal{} 2p_{i\\minus{}i}\\plus{}1$. An example is the sequence $ 2, 5, 11, 23, 47$. Show that such a sequence has a finite number of terms.", "Solution_1": "we suppose that $ (p_{i})$ is infinite.\r\nso, $ \\exists k:\\ p_{k}> 5$ ($ p_{k}\\equiv\\pm 1[6]$).\r\n\r\nif $ p_{k}\\equiv 1[6]$ then $ 3|2p_{k}+1$ so $ p_{k+1}= 2p_{k}-1\\equiv 1[6]$\r\nand we have $ 3|2p_{k+1}+1$ so $ p_{k+2}= 2p_{k+1}-1\\equiv 1[6]$\r\nthen $ \\forall n\\ge k:\\ p_{n+1}= 2p_{n}-1$\r\nso $ \\forall n\\ge k:\\ p_{n}= 2^{n-k}(p_{k}-1)+1$\r\nbut $ p_{k}|2^{p_{k}-1}(p_{k}-1)+1=p_{k+p_{k}-1}$ (impossible)\r\n\r\nlike the same if $ p_{k}\\equiv-1[6]$ then $ \\forall n\\ge k:\\ p_{n}= 2^{n-k}(p_{k}+1)-1$\r\nbut $ p_{k}|2^{p_{k}-1}(p_{k}-1)+1 = p_{k+p_{k}-1}$ (impossible)\r\n\r\nthen $ (p_{i})$ is finite." } { "Tag": [ "function", "algebra", "functional equation", "algebra proposed" ], "Problem": "Find all functions f defined on the positive reals with values in the same set such that f(1+xf(y))=yf(x+y).", "Solution_1": "No solution? Boy, this one is tough!", "Solution_2": "The only significant thing I managed to get is f(1)=1... :D I also got that f takes all values in (0,1], if that is somehow helpful. That's not much, is it? :D :D", "Solution_3": "f(x)= 1/x", "Solution_4": "Well, thank you very much, as if it were extremely difficult to find that f(x)=1/x is a solution. Many IMO problems can be \"solved\" in this way , observing that f(x)=x verifies the conditions. I would give 0 points for such a solution. So, I'm waiting for a solution.", "Solution_5": "My solution doesn't use the fact that f(1)=1 and that f takes all the values from (0,1], but this may be helpful. Nobody says there is only one solution to this problem. In fact, I would be very happy to get another solution.", "Solution_6": "I'm still here and noone solves me! I will start crying if you let me alone!", "Solution_7": "harazi! If you have solution what do you still want? What is the source of this problem?", "Solution_8": "A quick answer:\r\n MORE SOLUTIONS\r\n VRANCEANU-PROCOPIU CONTEST, BACAU, 2003", "Solution_9": "Do you, by any chance, know the grade for which this was proposed?", "Solution_10": "There was a kind of TST for seniors and there was the problem proposed. Of course, nobody managed to finish it and that took my attention.", "Solution_11": "to be more precise: a districtual contest in Romania (of no big importance whatsoever) makes a simulation of TST (at least that's what they call it). Of course it's never such difficult, but this year this problem was difficult enough. But please do not underestimate the Romanian Team :D Andreis solved this problem in 20 minutes, and a nice one too. I will ask him to post it. \r\n\r\nOnly two students from the ones that have participated last year at RoMOP (first 25 in country) went to this contest :)\r\n\r\nSo the problem itself is not that hard :)", "Solution_12": "You say the problem is not that hard. I really don't have the same opinion. Could you please post your solution?", "Solution_13": "And you should not concider that it is easy only because someone solved it in 20 minutes. Just consider the fact that it stayed for quite a long time on this forum with no solution received. If you consider it \"not so difficult\", then why didn't you post any solution?", "Solution_14": "well it's not MY solution, because Andreis solved it. I said above that I will ask him to post it. Unfortunately at this moment he's on his way to Singapore (in the plane) so it might take a while for him to get near a computer. \r\n\r\nAlso do not feel upset: I said that the problem is not that difficult, not that it is easy. You said that at a Romanian TST nobody could solve it, but I have explained you that only 2 (and 2 of the youngest!) components of the big team (25 students) could not solve it, and that is was not a TST. Andreis is one of the first 6 and he solved it pretty quickly. :D", "Solution_15": "I said that it was a kind of romanian TST, not a TST. I'm convinced that the members of the Romanian team would have solved the problem, but I'm really not convinced that more than half of the first 25 people you mentioned would have solved it. I find it very difficult and I gave it to lots of people who couldn't do absolutely anything with it.\r\n And what do you want to say:\"it's not that difficult\". In my opinion, when you say something like this, you must have a solution-otherwise the problem is difficult. Could you post it?", "Solution_16": "I already know Andreis' solution and I will let him post it. It's rather hard to think about another solution :) once you know one. \r\n\r\nYou are probably right about that half :) but then again didn't you solve the problem? If this is a very diffcult problem as you say, how do you call a functional equation that you can't solve? (considering that you already have a solution for this :D) an extremely difficult problem? :) it's a matter of choosing the right adjectives I think. :)", "Solution_17": "It depends on how much time you have spent for finding that solution. Taking into account hat I have managed to solve all the functional equations given in any IMO, except problem 6 proposed by Bulgaria one year, I can say that there aren't so many functional equations given in IMO's that could be called \"extremely difficult\". When I said this one is very difficult, I compared the time spent for finding a solution to it with the time spent for finding solutions to many other problems. And just another thought: If someone solves a problem, that doesn't become automatically easy. Reid Burton or Ciprian Manolescu solved extremely difficult problems and I don't think that you could say: well, if you solved it, the problem is easy.", "Solution_18": "I've found your interesting disscusion only yesterday. I don't think that this problem is very simple (30 min) but I don't think that it's extremelly\r\nhard (30 min). \r\n\r\nSol.\r\nAnswer: f(x)=1/x. \r\n\r\nLet g(x)=1/f(x), then for g we get an equation g(x+y)=y g(1+x/g(y)).\r\nPutting x=a g(y) we get that \r\n\r\n1. g(ag(y)+y)=y g(1+a)\r\n\r\n2. Now we can see, that if g(y_1)=g(y_2) than y_1=y_2.\r\n\r\n2'. putting y=1 in 1 we get, that g(1)=1.\r\n\r\n3. For each t>0 there exist y such that g(y)=t: y=a g(t/g(a+1)) +t/g(a+1) \r\n\r\n4. If y_2>y_1 than g(y_2)>g(y_1). If (y_1-y_2)/(g(y_1)-g(y_2))<0 then\r\nlet a=-(y_1-y_2)/(g(y_1)-g(y_2)) >0. Then a g(y_1)+y_1=a g(y_2)+y_2, then\r\ny_1g(1+a)=y_2g(1+a), so y_1=y_2.\r\n\r\n5. From 4&3 we can get, that g is continious, and we can define th limit\r\ng(0), so we can put a=0 in the equation 1. So g(y)=y.", "Solution_19": "Excellent solution, Fedor. Only now I have realised that the problem is not very difficult. Well, I just hope now that everyone has this feeling at least once in his life. My solution is much more complicated and made me believe that the problem is so difficult: I WAS WRONG! Anyway, I still believe it isn't trivial. I saw far less difficult problems given in important contests. MY solution:\r\n First of all, we prove that f(x)<=1 for x>1 and x>=1 for x<1. This follows easy if we suppose for example that for an y>1 we have f(y)>1. Take x=(y-1)/(f(y)-1).\r\n Thus, f(x+y)=f(1+xf(y))/y<=1/y for all x and y, from where f(x)<=1/x.\r\n Then, yf(x+y)=f(1+xf(y))=1/xf(y)*f(1+f(xf(y)) and so xyf(y)f(x+y)=f(1+f(xf(y)). Take x=1/f(y) and find that f(x)>=(1-k)/(kx) for all x, where k=1-f(1+f(1))>0.\r\n We prove now the injectivity:If f(a)=f(b), then for all x we have af(x+a)=bf(x+b) and so a^nf(na)=b^nf(bn). Let a=(1-k)/(knb) for all n.\r\n Finally, let y>1 and x>0. Then yf(x+y)=f(1+xf(y))=f((1+xf(y)-1/y)+1/y)=yf(1+f(1/y)(1+xf(y)-1/y)). From injectivity we deduce that \r\n x+y=1+f(1/y)(1+xf(y)-1/y) from where f(1/y)f(y)=1 and y=1+f(1/y)-1/yf(1/y). From here we find that for y>1 we have f(y)=1/y and since f(1/x)=1/f(x) we must have f(x)=1/x for x<>1. It's easy to see that f(1)=1.", "Solution_20": "let c>0 then \r\n1 + cf(1+xf(y))=1+cyf(x+y) so f(1 + cf(1+xf(y)))=f(1+cyf(x+y)) so\r\n(1+xf(y))f(c+1+xf(y))=(x+y)f(y(c+1)+x) (1)\r\nwe have: f(1+(1+xf(y))f(c+1+xf(y)))=f(1+(x+y)f(y(c+1)+x)) so\r\n(c+1+xf(y))f(c+2+2xf(y))=(y(c+1)+x)f(y(c+2)+2x) (2)\r\nnow dividing (1) by (2) and taking in (1) x=2x and c=c+1 we have\r\n(c+1+xf(y))(2x+y)=(y(c+1)+x)(1+2xf(y)) or\r\n(1+c)y+(c+1)2x+2x^2f(y)+xyf(y)=(1+c)y+x+2x^2f(x)+2xy(1+c)f(y) or\r\n(c+1)2x+xyf(y)=x+2xy(c+1)f(y) or 2c+1=(2c+1)yf(y) or yf(y)=1 for any y>0", "Solution_21": "hi, Fedor\r\nI don't know you how to get this? can you ......?thank!\r\n\r\n2.Now we can see, that if g(y_1)=g(y_2) than y_1=y_2" } { "Tag": [ "geometry", "3D geometry", "algebra", "function", "domain", "greatest common divisor", "number theory unsolved" ], "Problem": "Find all $x,y\\in Z$ such that\r\n$x^2+2=y^3$.", "Solution_1": "I'm not sure if it was posted before, but I think yes...\r\n\r\nLooking $\\mod 4$ gives us that $x$ is odd.\r\nFactor it as $(x+ \\sqrt{-2})(x-\\sqrt{-2}) = y^3$. Now $\\gcd(x+ \\sqrt{-2} , x-\\sqrt{-2}) =1$, so the numbers are coprime.\r\nThus (using unique factorisation in $\\mathbb{Z}{\\sqrt{-2}]}$) $x+\\sqrt{-2}$ is a cube, so $x+\\sqrt{-2} = (a+\\sqrt{-2}b)^3 = a^3 + 3 \\sqrt{-2} \\cdot a^2b - 6 \\cdot ab^2 - 2 \\sqrt{-2}\\cdot b^3$.\r\nComparing imaginary components gives $1= b \\cdot (3 \\cdot a^2 - 2\\cdot b^2)$, thus $b = \\pm 1 = 3 \\cdot a^2 - 2\\cdot b^2 = 3 \\cdot a^2 -2$, thus $a=\\pm 1 , b=1$, giving the solutions $x=\\pm 5 , y=3$.", "Solution_2": "[quote=\"ZetaX\"]I'm not sure if it was posted before, but I think yes...\n\nLooking $\\mod 4$ gives us that $x$ is odd.\nFactor it as $(x+ \\sqrt{-2})(x-\\sqrt{-2}) = y^3$. Now $\\gcd(x+ \\sqrt{-2} , x-\\sqrt{-2}) =1$, so the numbers are coprime.\nThus (using unique factorisation in $\\mathbb{Z}{\\sqrt{-2}]}$) $x+\\sqrt{-2}$ is a cube, so $x+\\sqrt{-2} = (a+\\sqrt{-2}b)^3 = a^3 + 3 \\sqrt{-2} \\cdot a^2b - 6 \\cdot ab^2 - 2 \\sqrt{-2}\\cdot b^3$.\nComparing imaginary components gives $1= b \\cdot (3 \\cdot a^2 - 2\\cdot b^2)$, thus $b = \\pm 1 = 3 \\cdot a^2 - 2\\cdot b^2 = 3 \\cdot a^2 -2$, thus $a=\\pm 1 , b=1$, giving the solutions $x=\\pm 5 , y=3$.[/quote]\r\nI am sorry $ZetaX$,but I can't understand what means $\\gcd(x+ \\sqrt{-2} , x-\\sqrt{-2}) =1$ :? .", "Solution_3": "As ever, it is the greatest common divisor, but this times in another ring as $\\mathbb{Z}$, here $\\mathbb{Z}[\\sqrt{-2}]$:\r\nWhen $a|b$ (in the ring $R$), then call $a \\leq b$ as a definition. By this, the term \"greatest\" is defined. What remains to show is that there is (up to units) exactly one greatest divisor. That possibly sounds rather obvious, but isn't. You will need a unique factorisation domain (or some other property) to show this.\r\nAt least, the known rule for $\\gcd$'s applies: $\\gcd(a,b) = \\gcd(a,b+ka)$ for all $k$. This follows directly from the definition the same way as for the standard $\\gcd$.\r\n\r\nFor this special case, it is\r\n$(x-\\sqrt{-2}) \\cdot (x+\\sqrt{-2}) =x^2+2=4n^2+4n+3$ (odd) together with\r\n$(x+\\sqrt{-2})+(x-\\sqrt{-2})=2x=4n+2$ (even, coprime as normal integer to the above) gives after using\r\n$c \\cdot (4n^2+4n+3) + d \\cdot (4n+2) =1$ from Bezeout finally\r\n$(c(x-\\sqrt{-2})+d) \\cdot (x+\\sqrt{-2})+d \\cdot(x-\\sqrt{-2})=1$, showing the desired property.\r\nWell, looks, ugly, so another way:\r\n$\\gcd(x+\\sqrt{-2},x-\\sqrt{-2}) = \\gcd(2\\sqrt{-2},x-\\sqrt{-2})$, dividing\r\n$\\gcd(2\\sqrt{-2},2x-2\\sqrt{-2})=\\gcd(2\\sqrt{-2},2x)=2$.\r\nBut $\\sqrt{2}$ doesn't divide $x+\\sqrt{2}$, so the $\\gcd$ is $1$.\r\n\r\nThere is also another definition with ideals:\r\nLet $\\mathfrak{a} , \\mathfrak{b}$ be ideals of the ring $R$, then their $\\gcd$ can be defined as $\\mathfrak{a} + \\mathfrak{b} : = \\{ a+b | a\\in \\mathfrak{a} , \\ b \\in \\mathfrak{b} \\}$.\r\n\r\n\r\nPS: could some moderator please move it to Number Theory\u00bf", "Solution_4": "[quote=\"ZetaX\"]I'm not sure if it was posted before, but I think yes...\n[/quote]\r\nYou are right :) : http://www.mathlinks.ro/Forum/viewtopic.php?t=15408\r\n\r\nOn a side note, I've learnt this technique with Euclidian rings not long ago, and I've to admit that I'm now a fan of it! :love:", "Solution_5": "Thanks $ZetaX$,but I will be glad to learn more about this theory,can you advise any books in the net?" } { "Tag": [ "number theory", "greatest common divisor", "geometry", "3D geometry" ], "Problem": "Hello, can anyone please give me a hint on how I would solve the following problems?\r\n1.P.T. the equation $ x^3 \\plus{} y^3 \\plus{} z^3 \\equal{} 1969^2$ does not have any solutions for $ x,y,z\\in\\mathbb{N}$.\r\n2. Let $ S$ denote the set of all 6-tuples $ (a,b,c,d,e,f)$ of natural numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\plus{} e^2 \\equal{} f^2$. Consider the set \r\n $ T \\equal{} [abcdef: (a,b,c,d,e,f)]\\in S$\r\n Find the gcd of all the members of $ T$\r\n3.Prove that the # of 5-tuples of positive integers $ (a,b,c,d,e)$ satisfying \r\n $ abcde \\equal{} 5(bcde \\plus{} acde \\plus{} abde \\plus{} abce \\plus{} abcd)$\r\nIS AN ODD INTEGER.\r\n4.Find all $ n\\in\\mathbb{N} ; n\\geq3$ such that \r\n $ n_{C_0} \\plus{} n_{C_1} \\plus{} n_{C_2} \\plus{} n_{C_3} | 2^{2001}$\r\n\r\n\r\nPlease help me solve these problems ASAP! :| \r\nThanks!!", "Solution_1": "[hide=\"3\"]The expression is symmetric, so consider possible permutations of any solution.[/hide]", "Solution_2": "I think the first problem can be done by considering mod 7.\r\nThe right hand side is 4 mod 7 whereas any cube can only be 0,1,6 mod 7\r\n\r\nAny permutation of 0,1,6 for the right hand side can never give 4 mod 7.\r\nSo we are done.", "Solution_3": "[hide=\"hint for number four\"]\n\nPascal's Identity: $ \\binom{n}{k} \\plus{} \\binom{n}{k\\plus{}1} \\equal{} \\binom{n\\plus{}1}{k\\plus{}1}$[/hide]", "Solution_4": "[quote=\"befuddlers\"]Any permutation of 0,1,6 for the right hand side can never give 4 mod 7.[/quote]\r\n\r\nExcept $ (6,6,6)$, you mean. ;)", "Solution_5": "Sorry guys mod 7 does not work.\r\n\r\nBut mod 9 does.\r\n\r\nAny cube is only 0,1,-1 mod 9.\r\n\r\nBut $ 1969^2$ is 4 mod 9.\r\n\r\nAny permutation of 0,1,-1 never yields 4 mod 9.", "Solution_6": "I have got quite a bit far in no. [b]2[/b]::\r\nFirst of all, all of $ a, b, c, d, e$ cannot be odd. As a square is congruent to 1(mod8)\r\nHence, if $ a^2\\equiv b^2\\equiv c^2\\equiv d^2\\equiv e^2\\equiv1(mod8)$, then it implies that $ f^2\\equiv5(mod8)$ which, of course, is impossible.\r\nThus at least one of $ a, b, c, d, e$ have to be even.\r\nAlso I have observed that if none of $ a, b, c, d, e$ are divisible by $ 3$,i.e. if \r\n $ a^2 \\equiv b^2\\equiv c^2\\equiv d^2\\equiv e^2\\equiv1(mod 3)$, then it would imply that $ f^2\\equiv5(mod 3)\\equiv2(mod 3)$ which is impossible for any square no. :) \r\n\r\n BUT I DON'T KNOW HOW TO CONTINUE FROM NOW ON!! :( \r\n Can anyone help me? \r\n Thanks", "Solution_7": "I have found no.[b]1[/b]!! :idea: \r\nI read in a number theory book that $ x^3 \\plus{} y^3 \\plus{} z^3\\equiv x \\plus{} y \\plus{} z\\equiv1(mod3)$\r\nWell, if $ x, y, z$ would have been in a complete residue system then it $ \\Rightarrow$ $ x \\plus{} y \\plus{} z\\equiv0(mod3)$ :) \r\nHence we may assume that $ x\\equiv y(mod3)$\r\nWe have three cases(obvious):\r\n[b][u]CASE I:[/u][/b] $ x \\equal{} y \\equal{} 0,z \\equal{} 1 \\Longrightarrow x^3 \\plus{} y^3 \\plus{} z^3\\equiv 1(mod9)$\r\n[b][u]CASE II:[/u][/b] $ x \\equal{} y \\equal{} 1, z \\equal{} 2 \\Longrightarrow x^3 \\plus{} y^3 \\plus{} z^3\\equiv 1(mod9)$\r\n[b][u]CASE III:[/u][/b] $ x \\equal{} y \\equal{} 2, z \\equal{} 0 \\Longrightarrow x^3 \\plus{} y^3 \\plus{} z^3\\equiv 7(mod9)$\r\n\r\nBut, as $ 1962\\equiv \\minus{} 2(mod9)$,\r\n $ \\Longrightarrow 1962^2 \\equiv 4(mod9)$\r\n\r\nBut as observed none of our three cases yield the fact above and hence there are no solutions to the given equation! :!: :idea: :lol:\r\n\r\nBUT, is it correct??? :maybe:", "Solution_8": "hello, I have found the answer to no 2. I continue from the place where I had written how far I got...\r\n[hide=\"answer\"]$ 24$ is the gcd[/hide]\n[hide=\"continuation\"]I found out that the smallest such set a,b,c,d,e,f is:(such that the product $ abcdef$ is smallest)::\n$ (a,b,c,d,e,f)\\equal{}(1,1,1,2,3,4)$ as $ 1^2\\plus{}1^2\\plus{}1^2\\plus{}2^2\\plus{}3^2\\equal{}16\\equal{}4^2$!\nThen with the help of a few examples I saw that in all cases $ 2^3$ is the highest power of $ 2$ dividing $ abcdef$\n[hide=\"request\"] Can you please help me to prove the fact that $ 8$ really is the highest power of $ 2$ dividing $ abcdef$?[/hide]\n[/hide]" } { "Tag": [ "geometry", "inequalities" ], "Problem": "Let be given a triangle $ABC$. Prove that: $R^{2}+r^{2}\\ge\\frac{5S}{3\\sqrt3}$", "Solution_1": "Let F denote the area of the triangle and s the semiperimeter.\r\n\r\nTo prove the given inequality we use these facts:\r\n\r\n1. $\\sqrt(27)R$ $\\geq$ 2s\r\n(which we will use in the form $\\frac{R^{2}}_{4}$ $\\geq$ $\\frac{s^{2}}_{27}$)\r\n\r\n2. $s^{2}$ $\\geq$ $\\sqrt(27)F$\r\n(which we will use in the form $s^{6}$ $\\geq$ $\\sqrt27)^{3}F^{3}$)\r\n\r\n3. rs=F\r\n\r\nNow $\\frac{R^{2}+r^{2}}_{5}$ = $\\frac{1}{5}$($\\frac{R^{2}}_{4}$ +$\\frac{R^{2}}_{4}$ +$\\frac{R^{2}}_{4}$ +$\\frac{R^{2}}_{4}$ +$r^{2}$) $\\geq$ $\\frac{1}{5}$($\\frac{s^{2}}_{27}$+$\\frac{s^{2}}_{27}$+$\\frac{s^{2}}_{27}$+$\\frac{s^{2}}_{27}$+$r^{2}$) using the inequality in 1 above.\r\n $\\geq$ ${\\frac{s^{2}.s^{2}.s^{2}.s^{2}.r^{2}}_{27^{4}}}}^{\\frac{1}{5}}$ by the arithmetic geometric mean inequality.\r\n=${\\frac{s^{6}.F^{2}}_{27^{4}}}^{\\frac{1}{5}}$ after using equation 3\r\n$\\geq$ ${\\frac{\\sqrt27)^{3}F^{5}}_{27^{4}}}^{\\frac{1}{5}}$ after using the inequality 2 above\r\n=$\\frac{F}{\\sqrt27}$\r\n\r\n Concerning inequalities 1. and 2.\r\n\r\nThe first is equivalent to SinA +SinB +SinC $\\leq$ $\\frac{\\sqrt27}{2}$ and the proof is left as an exercise.\r\n\r\nFor the second apply AM-GM to s-a, s-b ,s-c and use F in the form $\\sqrt{s(s-a)(s-b)(s-c)}$", "Solution_2": "[quote=\"April\"][color=darkred]Let be given a triangle $ABC$. Prove that : $\\frac{5}{3\\sqrt3}\\cdot S\\le R^{2}+r^{2}\\ \\ (*)$[/color][/quote]\r\n\r\n[color=darkblue][b]Proof.[/b] Denote $2p=a+b+c\\ .$ Therefore,\n\n$\\left\\{\\begin{array}{c}p\\le\\frac{3\\sqrt 3}{2}\\cdot R\\Longrightarrow S=pr\\le \\frac{3\\sqrt 3}{2}\\cdot Rr\\Longrightarrow\\frac{5}{3\\sqrt 3}\\cdot S\\le \\frac{5}{2}\\cdot Rr\\\\\\\\ 2r\\le R\\Longrightarrow 0\\le (R-2r)(2R-r)\\Longrightarrow\\frac{5}{2}\\cdot Rr\\le R^{2}+r^{2}\\end{array}\\right\\|$ $\\Longrightarrow$ $(*)\\ .$[/color]" } { "Tag": [ "videos" ], "Problem": "Please Post any Video Game Systems you have (even ones not on the poll but still relevant)\r\n\r\nMe:\r\nSuper Nintendo\r\nGameboy Color\r\nGameboy Advance SP\r\nN64\r\nGameCube\r\nWii*\r\nPSP\r\nTI-83+ Siver Edition\r\nTI-89 Titanium\r\nMotorola Cellphone\r\nIpod\r\n\r\nEdit:\r\nIf you have more than one of the systems on the poll\r\nPlease choose the one you like better\r\n\r\nEdit2:\r\nWhat happened to the poll?\r\noh well\r\njust post your systems\r\nand\r\nput an * next to your favorite", "Solution_1": "I chose PC because it's the only one I have on the poll. :D \r\n\r\nMe:\r\nXbox*\r\nPS2\r\nGame Boy COLOR\r\nGame Boy Advance (doesn't work)\r\nHad a Game Boy Advance SP (little brother lost it :mad: )\r\nNintendo 64\r\nNokia Cell Phone\r\niPod (can play breakout and parachute)\r\nTI-84 Plus Silver Edition :D (currently has ChuChu Rocket, Donkey Kong, Dr. Mario, Super Mario Bros., Phantom Star, Phoenix and Tetris)\r\nGetting an Xbox 360 after Nationals...", "Solution_2": "Sega Genisis(got boring)\r\nComputer\r\niPod\r\nTI-84 Plus Silver Edition\r\nmy imagination.", "Solution_3": "comp \r\nipod(IBOY and doom)\r\nPS2\r\nGBA", "Solution_4": "Whoa, u people have so many game systems...\r\nWii*\r\nPS1 (never play)\r\nGameBoy SP (Stupid L button broke)\r\nIPod\r\nTI-84 Silver\r\n\r\nComing Soon:\r\nCell phone\r\nNintendo DS", "Solution_5": "Gameboy Color\r\nGameboy Advance\r\nGameboy SP\r\nNintendo DS\r\nN64\r\nGamecube\r\nWii*\r\nPS2\r\nPC\r\nTI-83\r\nTI-84\r\nIpod\r\nCellphone", "Solution_6": "hey perfect628\r\nand\r\nra5249\r\n\r\ndo you want to pm me your wii number?\r\nI'll tell you mine", "Solution_7": "Xbox 360*\r\nGame Boy\r\n\r\nNo Live :( and I want to buy the Elite.", "Solution_8": "PC*\r\nGBA\r\nIpod Nano (parachute & solitaire)\r\nTI-84+SE (mostly games I build; the most entertaining other than the computer)", "Solution_9": "whats a wii number? where can i find it?", "Solution_10": "[quote=\"ra5249\"]whats a wii number? where can i find it?[/quote]\r\n\r\nPoint at the lower right hand corner of your screen\r\nthere should be a message board type thing there\r\nthere a bunch of buttons in the lower right hand corner\r\nclick on one of those\r\nI can't remember exactly\r\nbut the rest you should be able to find by yourself (or with the manual)", "Solution_11": "Wow. how do you guys have time to play on so many systems?\r\n\r\nPC\r\nTI 84 Plus Silver", "Solution_12": "[quote=\"alkjash\"]Wow. how do you guys have time to play on so many systems?\n\nPC\nTI 84 Plus Silver[/quote]\r\n\r\nWell I play the Super Nintendo when i get nostalgic\r\nNever play Gameboy Color\r\nSometimes on roadtrips or vacations I play the Gameboy Advance SP or even at school sometimes\r\nN64 I gave away recently\r\nGameCube I don't play anymore (the Wii can play GameCube games)\r\nWii I play the most during free time at home\r\nPSP the most when I'm out of my house and sometimes at school (those days that are like free days aka last day before \"holiday\")\r\nI only play tetris on my calculators at school\r\nI rarely (if ever) play on my cellphone\r\nI play on my Ipod when listening to music\r\n\r\nSo Ones I play frequently:\r\nWii\r\nPSP\r\nTI-89 Titanium\r\nIpod\r\n\r\nEdit:\r\nPC once in a while", "Solution_13": "Wow... everyone has a LOT of game consoles. :o" } { "Tag": [ "IMO", "IMO 2008", "Spain", "Countries" ], "Problem": "It seems that IMO'2008 will be huge...\r\nCurrently there are 87 registered countries ( http://www.imo-2008.es/participants.php ).\r\n\r\nAnd countries like Costa Rica, Cuba, Georgia, India, Kazakhstan, Latvia, Macedonia,\r\nMoldova, Romania, Slovakia, Turkmenistan, Vietnam are not registered yet.\r\n\r\nAnd North-Korea will participate again! Good!", "Solution_1": "I guess that is good that they are fostering more international participation and love of math.", "Solution_2": "India is definitely going to be there this year :). We hope that this time will lead to a significant improvement over the average performances of the last few years!", "Solution_3": "89 and growing. (Slovakia and Latvia joined the fun.) I think 93 (the record from last year) is at least going to be equaled (India, Moldova, Romania, Vietnam are not on the list yet), if not broken.", "Solution_4": "Well, you know that not all countries (and regions) registered for the competition usually appear on IMO. That's why we'll have to wait the opening ceremony to say the official number of teams competing. ;)", "Solution_5": "Yeah... I guess I know. So, are you coming to IMO as an observer, hsiljak? (You wrote: \"(...)we'll have to wait the opening ceremony(...)\")", "Solution_6": "No, no, I haven't planned a trip to Spain this year :) I did use the term \"We'll\", but that still doesn't mean I'll go :) Just a figure of speech.", "Solution_7": "Well, to be honest - I expected that. Just wanted to be sure.", "Solution_8": "[quote=\"test20\"]It seems that IMO'2008 will be huge...\nCurrently there are 87 registered countries ( http://www.imo-2008.es/participants.php ).\n\nAnd countries like Costa Rica, Cuba, Georgia, India, Kazakhstan, Latvia, Macedonia,\nMoldova, Romania, Slovakia, Turkmenistan, Vietnam are not registered yet.[/quote]\r\n\r\nNow there are 93 registered countries: http://www.imo-official.org/year_reg.aspx\r\n\r\nCosta Rica, Georgia, Latvia, Romania, Slovakia, Turkmenistan have been \r\nregistered in the meantime.\r\n\r\nCuba, India, Kazakhstan, Macedonia, Moldova, Vietnam are still missing.\r\nIt would be strange to have an IMO without India, Kazakhstan, Macedonia, Moldova, \r\nor Vietnam.\r\nAnd Cuba does not always participate with a full team of six students, but has\r\nparticipated at every IMO since 1971, except for 1975 and 2006.\r\n\r\nSo all in all, Spain may end up with 99 participating countries.", "Solution_9": "[quote=\"test20\"]Now there are 93 registered countries: http://www.imo-official.org/year_reg.aspx\n[/quote]\r\nAnd now they are up to 96 countries.\r\nIndia and Macedonia have registered, and it looks as if Benin will participate \r\nfor the first time in an IMO!\r\n\r\nCuba, Kazakhstan, Moldova, Vietnam are still missing.\r\nI wonder whether this might mean that Kazakhstan has run into problems \r\nwith organizing IMO'2010.", "Solution_10": "[quote=\"test20\"]\nCuba, Kazakhstan, Moldova, Vietnam are still missing.\n[/quote]\r\n\r\nWell, some countries do their TSTs first, and then confirm the participation together with the registration of participants ;)", "Solution_11": "test20: You seem to be from Cuba (the little flag under your name). Will Cuba participate this year?\r\n\r\nIf so, the number of 100 will probably get reached as I'm sure Vietnam, Moldova and Kazakhstan will participate. I couldn't imagine what would need to happen in order for them not to participate.", "Solution_12": "[quote=\"hsiljak\"][quote=\"test20\"]\nCuba, Kazakhstan, Moldova, Vietnam are still missing.\n[/quote]\nWell, some countries do their TSTs first, and then confirm the participation together with the registration of participants ;)[/quote]\r\n\r\nYes, different countries have different ways of doing their TSTs.\r\n\r\nBut: The registration deadline for leaders, deputy leaders and observers of\r\n*ALL* participating countries is April 30, and does not depend on the local TSTs:\r\nhttp://www.imo-2008.es/deadlines.html", "Solution_13": "[quote=\"test20\"]And now they are up to 96 countries.\nIndia and Macedonia have registered, and it looks as if Benin will participate \nfor the first time in an IMO![/quote]\r\n\r\nAs of today, there are 99 registered countries.\r\nCuba and Kazakhstan have not been registered. \r\n\r\nAnd among the registered countries, there are eleven countries that still\r\nhave neither registered team leaders nor deputy leaders nor contestants:\r\n---Benin\r\n---Bosnia and Herzegovina\t\r\n---Bolivia\r\n---Chile\r\n---Guatemala\r\n---Iceland\r\n---Saudi Arabia\r\n---El Salvador\r\n---Turkmenistan\r\n---Uruguay\r\n---Vietnam\r\n\r\n* Benin will send an observer-A. Is this their first appearance at an IMO?\r\n* Guatemala has not shown up in 2006 and 2007. Does anybody know about them?\r\n* Uruguay particpated every year from 1997-2006, but did not come to IMO'2007.\r\nWill they participate in Spain?\r\n* Bolivia, Bosnia and Herzegovina, Chile, El Salvador, Iceland, Saudi Arabia, \r\nTurkmenistan, and Vietnam are regular IMO-countries, and they all were present \r\nat IMO'2007. They are probably just sloppy in international communications, or \r\nperhaps they want to make the life of the Spanish organizers more exciting.", "Solution_14": "El Salvador will participate for sure :) Our TST was held about one week ago, so I guess our country's leaders and contestants will be registered all at the same time.", "Solution_15": "[quote=\"Jutaro\"]El Salvador will participate for sure.[/quote]\r\nGood to know!\r\n\r\nIn the meantime, also Kazakhstan has registered.\r\nAnd for the first time, there will be a team from the United Arab Emirates at an IMO!!", "Solution_16": "IMO'2008 might be the first IMO with contestants from 100 different countries!!!\r\n\r\nCurrently, there are 102 registered countries. \r\nTwo of these registered countries will not send a team, but only an observer (Benin \r\nand Syria; it's their first appearance at an IMO; in 2009 they may send a team).\r\nFurthermore Cuba is not registered yet, but will participate for sure.\r\nAll in all, this yields a total of 101 countries with teams.\r\n\r\nOut of these 101 countries, there are only five countries that still have neither \r\nregistered team leaders nor deputy leaders nor contestants: \r\n---Cuba\r\n---Guatemala\r\n---El Salvador\r\n---Turkmenistan\r\n---Vietnam\r\n\r\nCuba has a team and will participate for sure.\r\nEl Salvador has a team and will participate for sure.\r\nGuatemala has participated at some IMOs, but neither in 2006 nor in 2007.\r\nTurkmenistan has participated at every IMO since 2001.\r\nVietnam has participated at every IMO since 1982.\r\n\r\nA cautiously optimistic prediction would be to have the 96 actively registered countries \r\nplus Cuba, El Salvador, Turkmenistan, Vietnam.\r\n\r\nThis makes 100 countries.", "Solution_17": "[quote=\"test20\"]IMO'2008 might be the first IMO with contestants from 100 different countries!!![/quote]\r\n\r\nCurrently, there are 104 registered countries.\r\nFor three of them (Benin, Syria, Mauritania) it is the first appearance at IMO.\r\nHence, they only send an oberver but no contestants.\r\n\r\nAmong the remaining 101 countries, 96 have registered contestants.\r\nOnly five countries did not register their contestants by the very last (extended) deadline.\r\n--Cuba\r\n--Mozambique\r\n--Romania\r\n--Turkmenistan\r\n--USA\r\n\r\nCuba, Romania, USA have a team.\r\nI guess they are playing some kind of political game, and keep their student data \r\nsecret from the Spanish IMO organisation (although I do not understand the\r\nmotivation for this; how can they get an advantage out of this?)\r\nRomania and USA have also registered observers, but did not settle the \r\ncorresponding fees by the deadline. Very strange.\r\n\r\nWhat about Mozambique and Turkmenistan?\r\nIf there would be just one contestant from one of these countries, Spain would\r\nhave contestants from 100 different countries.", "Solution_18": "I will go by this calculation:\r\n104 countries are registered.\r\n98 countries have registered team leaders. Panama has no registered leader, but it does have 2 registered contestants. That's 99.\r\nIf you are certain that Cuba will participate, we reach 100. With Turkmenistan, the number could rise to 101.", "Solution_19": "[quote=\"test20\"]I guess they are playing some kind of political game, and keep their student data \nsecret from the Spanish IMO organisation (although I do not understand the\nmotivation for this; how can they get an advantage out of this?)\n[/quote]There is no \"political game\". The teams haven't been selected for either Romania or USA. There are still 2 TSTs to go for both countries until the final 6 are selected.", "Solution_20": "Yes , it couldn't be a political game . The is no reason for something like this .\r\n\r\nBTW , Could someone post please the first TST of Romania ? And someone the TST's from USA ?", "Solution_21": "[b]We've got 100![/b]\r\n\r\n99 countries have registered their team leaders + Panama which has registered 2 contestants, but not a team leader.", "Solution_22": "Damn ... More work for me :mad: \r\n\r\n:)", "Solution_23": "Are we certain that Cuba will participate? If so, why are there no registered participants 3 weeks before IMO? What about Mauritania and Mozambique (Mozambique has registered their team leader, though)?", "Solution_24": "[quote=\"mornik\"]Are we certain that Cuba will participate? If so, why are there no registered participants 3 weeks before IMO? What about Mauritania and Mozambique (Mozambique has registered their team leader, though)?[/quote]\r\n\r\nCurrently, there are 104 registered countries.\r\nFor three of them (Benin, Syria, Mauritania) it is the first appearance at IMO.\r\nHence, they will only send an oberver but no contestants.\r\n\r\nAmong the remaining 101 countries, 98 have registered contestants and registered leaders.\r\nOnly three countries did not fully register their participants:\r\n--Cuba (nobody registered at all)\r\n--Mozambique (only 1 registered team leader)\r\n--Panama (only 2 registered contestants)\r\n\r\nCuba has a team (one contestant, one team leader), but they are horribly unreliable in their communications.\r\nMozambique and Panama are totally unclear.", "Solution_25": "Thanks. You are sure that Cuba will participate, though?", "Solution_26": "Finally also Cuba has registered a participant and a team leader!\r\n\r\nAll in all, there are 104 registered countries.\r\nFor three of them (Benin, Syria, Mauritania) it is the first appearance at IMO.\r\nHence, they will only send an observer but no contestants.\r\n\r\nAmong the remaining 101 countries, now 99 have registered contestants and registered leaders.\r\nOnly two countries did not fully register their participants:\r\n--Mozambique (only 1 registered team leader)\r\n--Panama (only 2 registered contestants)\r\n\r\nDoes anybody have any information about Panama or Mozambique?\r\nIf one of them indeed sends contestants, then IMO'2008 would be the\r\nfirst IMO to have contestants from 100 different countries!", "Solution_27": "[quote=\"test20\"]Finally also Cuba has registered a participant and a team leader!\n[/quote]\r\n\r\nHowever, the Cuban contestant seems to be a bit too old for the IMO ;)\r\n\r\nhttp://www.imo-official.org/year_reg_team.aspx?code=CUB", "Solution_28": "[quote=\"michaj\"][quote=\"test20\"]Finally also Cuba has registered a participant and a team leader!\n[/quote]\n\nHowever, the Cuban contestant seems to be a bit too old for the IMO ;)\n\nhttp://www.imo-official.org/year_reg_team.aspx?code=CUB[/quote]\r\n\r\nAnd you have special eyes that can detect ones age ? :maybe:", "Solution_29": "Originally, there was a typo in the page: it said 1817 years old (or 1718, I'm not sure). That's why michaj made this comment (right, michaj?).", "Solution_30": "I think there was also an (Israeli? I'm also not sure.) contestant who was 0 years old. I can't find it now, it was probably changed anyway.", "Solution_31": "[quote=\"hsiljak\"]Originally, there was a typo in the page: it said 1817 years old (or 1718, I'm not sure). That's why michaj made this comment (right, michaj?).[/quote]\r\n\r\nYep, 1817 years it was. An admirable age for a contestant indeed :)" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "For the positive integer n let 4^n+2^n+1 be a prime. Prove that n is represented as n = 3^k with k being a positive integer.", "Solution_1": "It is easy to see, that if k is not 3l, with l -- integer, than \r\nx^{2k}+x^k+1 devided by x^2+x+1", "Solution_2": "Another easy way : if n = 3k+1 then 4 n + 2 n + 1 = 0 mod[7] and is greater than 7.\r\nIf n = 3k+2, then 4 n + 2 n + 1 = 21 = 0 mod[7] and is greater than 7.\r\n\r\nPierre.", "Solution_3": "But what will yuo do if n=3^s k, where k=3m \\pm 1", "Solution_4": "Sorry, I misread the text (3k is not 3^k)...\r\n\r\nPierre." } { "Tag": [ "algorithm" ], "Problem": "What is considered a good time complexity for IOI problems? O(10^6), O(10^7) or what?", "Solution_1": "I don't believe there is such a rule.. it all depends on the task and the time limit given. And by the way, O(10^6) , O(10^7) is equivalent to O(1) :P", "Solution_2": "I'm sorry, I didn't make myself understood. What I mean is: in order for a program whose running time depends on a variable n (n<=10) to run in 1 second, should we aim for an algorithm of O(n^6), O(n^7) or what?", "Solution_3": "Remember New Jersy style guidlines. The relevent portion is [quote]n is usually small[/quote]. A O(1) algorithm with large overhead will not outpreform a O(n) algorithm with next-to-no overhead for small .", "Solution_4": "[quote=\"richard\"]I'm sorry, I didn't make myself understood. What I mean is: in order for a program whose running time depends on a variable n (n<=10) to run in 1 second, should we aim for an algorithm of O(n^6), O(n^7) or what?[/quote]\r\n\r\nI think you can assume that n <= 10 usually means there's an O(2^n) or O(3^n) algorithm.. I don't think I've ever seen a programming problem for which the official solution was O(n^6) or O(n^7). :)" } { "Tag": [ "modular arithmetic" ], "Problem": "Suppose that $ m\\equiv 1\\pmod{b}$. \r\nWhat integer between $ 1$ and $ m\\minus{}1$ is equal to $ b^{\\minus{}1}\\pmod{m}$?", "Solution_1": "Let $ m\\minus{}1\\equal{}kb$ (for some integer $ k$, by the given condition). Then it's obvious that the answer is $ m\\minus{}k$, because $ kb\\equiv\\minus{}1\\mod{m}$." } { "Tag": [ "pigeonhole principle", "AMC", "USA(J)MO", "USAMO" ], "Problem": "Mr. and Mrs. Adams recently attended a party at which there were three other couples. Various handshakes took place. No one shook hands with his or her own spouse, no one shook hands with the same person twice and no one shook his or her own hand. After the handshaking was done, Mr. Adams asked each person, including his own wife, how many hands they had shaken. Each gave a different answer. How many hands did Mrs. Adams shake?\r\n\r\n[hide]Let's see...\nA B C D are the Mr.'s\nAa Bb Cc Dd are the Mrs.'\n\nIf the handshake happens in order:\nA) B C D Bb Cb Db (means A with B, A with C, A with D, etc.)\nB) C D Ab Cb Db\nC) D Ab Bb Db\nD) Ab Bb Cb\nAb) Bb Cb Db\nBb) Cb Db\nCb) Db\nDb) 0\n\nI am not sure why D) and Ab) have to turn to the same number of people for handshakes...\nEach person seems to have to shake hands with other six eventually. But it didn't state in the problem that each person have to shake hands with everyone else...\n[/hide]", "Solution_1": "one question about solution..\r\n\r\n\r\n[hide=\".....??\"]but if A DID shake hands with B, then wouldn't B say i shook hands with A, C, D, Aa, Cc, Dd???\nso technically, everybody shook hands 6 times.... and the total number of handshakes in the room SHOULD be 18... (if i'm not mistaken)...[/hide]", "Solution_2": "Not everyone has to shake hands with everyone else. That would trivialize the problem.", "Solution_3": "We can generalize this. Let there be $ n$ couples.\r\nEach person handshook $ 0,1,..,2n\\minus{}3,2n\\minus{}2$ persons. \r\nSuppose someone handshake $ 2n\\minus{}2$ persons, then this persons husband/wife shakes hands with $ 0$ persons. \r\nSuppose someone handshake $ 2n\\minus{}3$ persons, then this persons husband/wife shakes hands with $ 1$ person.\r\nAnd so on..\r\nSo the handshake becomes $ 2n\\minus{}2\\minus{}r$ persons for say wife and $ r$ persons for husband.\r\nSince each person gave different answer, Mr.Adams and Mrs. Adams shook hands with equal number of hands, i.e. $ r\\equal{}\\frac{2n\\minus{}2}{2}\\equal{}n\\minus{}1$.\r\nHence Mrs.Adams shook hands with $ n\\minus{}1$ in this case.\r\nIn our case $ 2$ persons.", "Solution_4": "[quote=\"dran\"]\nSince each person gave different answer, Mr.Adams and Mrs. Adams shook hands with equal number of hands, i.e. $ r \\equal{} \\frac {2n \\minus{} 2}{2} \\equal{} n \\minus{} 1$.[/quote]\r\n\r\nI got everything else but lost on the above reasoning. Explanation please?", "Solution_5": "[quote=\"jeez123\"]\nI got everything else but lost on the above reasoning. Explanation please?[/quote]\r\n\r\nSince $ 2n\\minus{}2$ is even, there exist number of handshakes which are equal for both the wife and the husband of some couple.\r\nThe problem supposes that 'Each gave a different answer'. This means that Mrs.Adam and Mr.Adam handshook equal number of handshakes. \r\n$ 2n\\minus{}2\\minus{}r$=(number of handshakes of say wife) equals to $ r$=(number of handshakes with husband) when $ r \\equal{} \\frac {2n \\minus{} 2}{2} \\equal{} n \\minus{} 1$.", "Solution_6": "[quote=\"dran\"]\nSuppose someone handshake $ 2n \\minus{} 2$ persons,[b] then this persons husband/wife shakes hands with $ 0$ persons[/b]. \nSuppose someone handshake $ 2n \\minus{} 3$ persons, [b]then this persons husband/wife shakes hands with $ 1$ person[/b].\nAnd so on..\n.[/quote]\r\n\r\nI am almost there, please be patient! \r\n\r\nWhy does the person's husband/wife shook can only shook hands with $ 1$ persons when his/her spouse shook $ 2n\\minus{}3$, and so forth?\r\n\r\nThanks so much!", "Solution_7": "Huh?\r\nHow is it 2?\r\n\r\nHere is a way in which she shakes with 3:\r\n\r\nB.) A C D a c d\r\nC.) A B D a d\r\nD.) A B C a\r\na.) B C D\r\nb.)\r\nc.) B\r\nd.) B C", "Solution_8": "[quote=\"jeez123\"]\nWhy does the person's husband/wife shook can only shook hands with $ 1$ persons when his/her spouse shook $ 2n \\minus{} 3$, and so forth?[/quote]\r\n\r\nSuppose you shake hands with $ 2n \\minus{} 2$ persons. Then you have shaken hand with people who have shaken $ 1,...,2n \\minus{} 3$ hands. Your wife/husband can then have handshaked $ 0$ or $ 2n \\minus{} 2$ persons (since you didn't shake your wife/husband). But all people shake different number of handshakes. Thus your wife/husband shakes hand with $ 0$ people.\r\n\r\nRemove this couple from the counting and the number of handshakes. Apply same argument for $ 2n \\minus{} 3,2n \\minus{} 4,...$ and so on.", "Solution_9": "OH!\r\n\r\nSo the keyword here is that everyone hand-shook different numbers of people. Since there are $ 2n\\minus{}2$ people besides the couple in a room, we have to find out [b]how[/b] every husband and wife pair hand shake (2n - 2, 0) (2n - 3, 1) (2n - 4, 2) so that everyone hand-shook with different number of people. Every husband and wife's combined number of handshake is $ 2n\\minus{}2$.\r\n\r\nThank you for your explanations!", "Solution_10": "[hide=\"super ownage\"]\nThe only possible numbers of hands that people shook are 0,1,2,3,4,5, and 6.\nEach person can shake with everyone except him/herself and his/her spouse.\nBut we need to fit these numbers to 8 people, so by pigeonhole 2 people shook the same number of hands. \n\nLet us suppose WLOG B shook hands 6 times. (We will see why we can do this later.) That means b shook hands 0 times, because if the 0 shaker was any of the other people, we could reach a contradiction.\nThen let us suppose C shook hands 5 times. That means c shook hands 1 time, because if the 1 shaker was A, a, D, or d, then he would have had to shake C and B.\nEtc. \nSo 6 is \"paired\" with 0, 5 is \"paired\" with 1, 4 is \"paired\" with 2, and 3 is \"paired\" with 3. \n\nBecause Mr. Adams is the \"leftover\" in the pigeonholes, then he must have value 3, and that means Mrs. Adams shook 3 times, no matter how we originally named the numbers.\n\nClearly we can do this for [b]any[/b] even number of people.\n[/hide]", "Solution_11": "[quote=\"CircleSquared\"]That means b shook hands 0 times, because if the 0 shaker was any of the other people, we could reach a [b]contradiction[/b].[/quote]\r\n\r\nThe \"Pigeon Hole Principle\" is exactly what I needed to sort myself out. Though, could you go on about the contradiction part? \r\nThanks!", "Solution_12": "B shook hands with everyone but his spouse.\r\nThat means each person except b had to have shaken at least 1 time.\r\nSo if anyone but b shook hands 0 times, we reach a contradiction.", "Solution_13": ":jump: I understood the whole thing now. I am super happy! Thanks!", "Solution_14": "Metamathematical solution:\r\n\r\n[hide] If he instead asked each person how many peoples hands they didn't shake, each person would still give a different answer. So it is essentially the same problem, only since the original problem was solvable so must this one and the answer must be the same, so the wife must have shaken as many hands as she didn't shake. The answer follows.[/hide]", "Solution_15": "[quote=\"nateharman1234\"]Metamathematical solution:\n\n[hide] If he instead asked each person how many peoples hands they didn't shake, each person would still give a different answer. So it is essentially the same problem, only since the original problem was solvable so must this one and the answer must be the same, so the wife must have shaken as many hands as she didn't shake. The answer follows.[/hide][/quote]\r\n\r\nThis is somewhat abusive of the original problem, and would never pass on an proof exam such as the USAMO. How is it a mathematical solution?", "Solution_16": "[quote=\"CircleSquared\"][quote=\"nateharman1234\"]Metamathematical solution:\n\n[hide] If he instead asked each person how many peoples hands they didn't shake, each person would still give a different answer. So it is essentially the same problem, only since the original problem was solvable so must this one and the answer must be the same, so the wife must have shaken as many hands as she didn't shake. The answer follows.[/hide][/quote]\n\nThis is somewhat abusive of the original problem, and would never pass on an proof exam such as the USAMO. How is it a mathematical solution?[/quote]\r\n\r\nI said it was a metamathematical solution, in that I go outside of mathematics by assuming that a unique solution exists, and then by that assumption I determine what it must be. Indeed it would not pass for a proof on USAMO, I just posted it as an interesting way to think about it.", "Solution_17": "How do you know the problem is solvable again? :D" } { "Tag": [ "geometry", "circumcircle", "parallelogram" ], "Problem": "In $\\triangle ABC$, $O$ is the circumcenter and $H$ the orthocenter. $D$ is a point of $BC$ that $OD$ is perpendicular to $BC$. Show that $AH = 2OD$", "Solution_1": "It's a known theorem.\r\nNevertheless is a good idea for Intermediate Topics.....\r\nHave fun!!! :) \r\n\r\n[hide=\"hint\"]Extend $BO$ and Parallelograms... [/hide]", "Solution_2": "This problem was posted by me several days ago ( right in this forum). :) \r\nMy proof is: Construct the circumcircle. And let the intersection of CO and the circle be E. Then we can show that AHBE is a parallelogram. Hence AH=BE. Because OD $\\perp$ BC, EB $\\perp$ BC, and CD=BD, so BE=2OD, hence we can imply AH=2OD.", "Solution_3": "how to prove the parallelogram?", "Solution_4": "[quote=\"indybar\"]how to prove the parallelogram?[/quote]\r\n\r\nThe quadrilateral AHBE is a parallelogram since $AH\\perp BC$, $BE\\perp BC$, $BH\\perp CA$ and $AE\\perp CA$.\r\n\r\nFor another proof of the theorem in question, see http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #4 Theorem 3 (of course, since O is the circumcenter of triangle ABC, the point D, lying on BC and satisfying $OD\\perp BC$, must be the midpoint of the segment BC).\r\n\r\n darij" } { "Tag": [ "integration", "trigonometry", "calculus", "calculus computations" ], "Problem": "Compute $ \\int \\frac{1}{\\sin^4x \\plus{} \\cos^4x}\\,dx$.", "Solution_1": "[hide]\nWe have the identity\n\\[ \\sin^4 x\\plus{}\\cos ^4 x\\equal{}(\\sin^2 x\\plus{}\\cos^2 x)^2\\minus{}2\\sin^2 x\\cos^2 x\\equal{}1\\minus{}\\frac{1}{2}\\sin^2 2x\\equal{}\\frac{3\\plus{}\\cos 4x}{4}\\]\nNow make the substitution $ u\\equal{}\\tan 2x$. We then have $ \\cos 4x\\equal{}\\frac{1\\minus{}u^2}{1\\plus{}u^2}$ and $ \\frac{du}{dx}\\equal{}2(1\\plus{}u^2)$, so that our integral is equal to\n\\[ \\int \\frac{4}{3\\plus{}\\frac{1\\minus{}u^2}{1\\plus{}u^2}}\\cdot \\frac{du}{2(1\\plus{}u^2)}\\equal{}\\int \\frac{du}{2\\plus{}u^2}\\]\nNow use $ \\int_0^y \\frac{dt}{1\\plus{}t^2}\\equal{}\\arctan y$ to get\n\\[ \\int \\frac{du}{2\\plus{}u^2}\\equal{}\\frac{1}{2}\\int \\frac{du}{1\\plus{}\\left( u/\\sqrt{2}\\right)^2}\\equal{}\\frac{\\arctan \\frac{u}{\\sqrt{2}}}{\\sqrt{2}}\\plus{}C\\]\n\\[ \\equal{}\\frac{\\arctan \\frac{\\tan 2x}{\\sqrt{2}}}{\\sqrt{2}}\\plus{}C\\]\nSpecifically, $ \\int_0^x \\frac{dx}{\\sin^4 x\\plus{}\\cos^4 x}\\equal{}\\frac{\\arctan \\frac{\\tan 2x}{\\sqrt{2}}}{\\sqrt{2}}$.\n[/hide]", "Solution_2": "Yes, that's right!", "Solution_3": "[quote=\"Carcul\"]Compute $ \\int \\frac {1}{\\sin^4x \\plus{} \\cos^4x}\\ \\mathrm {dx}$.[/quote]\r\n\r\n$ I\\equiv\\int \\frac {1}{\\sin^4x \\plus{} \\cos^4x}\\ \\mathrm {dx}\\equal{}$ $ \\int \\frac {\\left(\\sin^2x\\plus{}\\cos^2x\\right)^2}{\\sin^4x \\plus{} \\cos^4x}\\ \\mathrm {dx}\\equal{}$ $ \\int\\frac {\\tan^2x\\plus{}1}{\\tan^4x\\plus{}1}\\cdot\\left(\\tan x\\right)'\\ \\mathrm {dx}\\equal{}$ $ F\\left(\\tan x\\right)\\plus{}\\mathcal C$, where $ F(t)\\equal{}\\int\\frac {t^2\\plus{}1}{t^4\\plus{}1}\\ \\mathrm {dt}\\equal{}$ $ \\int\\frac {\\left(t\\minus{}\\frac 1t\\right)'}{\\left(t\\minus{}\\frac 1t\\right)^2\\plus{}2}\\ \\mathrm {dt}\\equal{}$ $ \\frac {1}{\\sqrt 2}\\arctan\\frac {t^2\\minus{}1}{t\\sqrt 2}\\plus{}\\mathcal C$. Therefore, $ I\\equal{}\\frac {1}{\\sqrt 2}\\arctan\\frac {\\tan^2\\minus{}1}{\\sqrt 2\\tan x}\\plus{}\\mathcal C\\equal{}$ $ \\minus{}\\frac {1}{\\sqrt 2}\\arctan\\frac {\\sqrt 2}{\\tan 2x}\\plus{}\\mathcal C$, i.e. $ \\boxed {I\\equal{}\\frac {1}{\\sqrt 2}\\arctan\\frac {\\tan 2x}{\\sqrt 2}\\plus{}\\mathcal C}$ because $ \\arctan x\\plus{}\\arctan \\frac 1x\\equal{}\\frac {\\pi}{2}\\cdot\\mathrm {sign\\ x}\\ .$", "Solution_4": "Yes, that's another approach. We could also use directly the substitution tan x = t." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all polynomials P(x) with integer coefficients such that the set {P(n) | $ n\\in Z$} contains the fibonacci sequence.", "Solution_1": "It's from the first stage of Polish MO, please lock this." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Hello!:)\r\nLet $A,\\ B \\subset X$ and let $\\alpha , \\beta \\in R$. What is a necessary and sufficient condition for $\\alpha* \\chi_A + \\beta* \\chi_B$ being measurable?\r\n($\\chi_A$ - characteristic function $A$)\r\nThaks for help!", "Solution_1": "Three possibilities:\r\n\r\n$A$ and $B$ are both measurable\r\n\r\nOR\r\n\r\n$\\alpha=\\beta$ and $A\\cup B$ is measurable\r\n\r\nOR\r\n\r\n$\\alpha=\\beta=0.$", "Solution_2": "Not exactly.\r\nThe right conditions:\r\n\r\n$(A$ is measurable or $\\alpha=0)$ and $(B$ is measurable or $\\beta=0)$\r\nor\r\n$\\alpha=\\beta$ and both $A\\cup B$ and $A\\cap B$ are measurable\r\nor\r\n$\\alpha=-\\beta$ and both $A\\setminus (A\\cap B)$ and $B\\setminus (A\\cap B)$ are measurable." } { "Tag": [ "binomial coefficients" ], "Problem": "1.If x is a real number not less than 1, which is larger: \u221a(x+1) - \u221ax or \u221ax - \u221a(x-1)?\r\n\r\n2.Find 1*1! + 2*2! + ... + n*n!\r\n\r\n3.What is the remainder on dividing 6^83 + 8^83 by 49? \r\n\r\n4. A nice number equals the product of its proper divisors (positive divisors excluding 1 and the number itself). Find the sum of the first 10 nice numbers.", "Solution_1": "1. [hide]We compare their reciprocals instead.\n\n\n\n(:sqrt:(x + 1) - :sqrt:x)-1 = :sqrt:(x + 1) + :sqrt:x\n\n(:sqrt:x - :sqrt:(x - 1))-1 = :sqrt:x + :sqrt:(x - 1)\n\n\n\nSo (:sqrt:(x + 1) - :sqrt:x)-1 > (:sqrt:x - :sqrt:(x - 1))-1. Thus :sqrt:(x + 1) - :sqrt:x < :sqrt:x - :sqrt:(x - 1).[/hide]\n\n\n\n2. [hide]1 * 1! + 2 * 2! + ... + n * n!\n\n= (2 - 1) * 1! + (3 - 1) * 2! + ... + (n + 1 - 1) * n!\n\n= 2 * 1! - 1! + 3 * 2! - 2! + ... + (n + 1) * n! - n!\n\n= 2! - 1! + 3! - 2! + ... + (n + 1)! - n!\n\n= (n + 1)! - 1.[/hide]\n\n\n\n3. [hide]683 + 883\n\n= (7 - 1)83 + (7 + 1)83\n\n= ((83, 0) * 783 - (83, 1) * 782 + ... - (83, 81) * 72 + (83, 82) * 7 - (83, 83) * 1) + ((83, 0) * 783 + (83, 1) * 782 + ... +(83, 81) * 72 + (83, 82) * 7 + (83, 83) * 1)\n\n\n\nNote that the binomial coefficients are not important here. Then, all the even-powered terms cancel out, so we're left with: (I'll leave out the binomial coefficients except where necessary)\n\n\n\n2 (783 + 781 + ... + 73 + (83, 82) * 7)\n\n\n\nAll terms except the last one are divisible by 49. So this expression :equiv: 2 * (83, 82) * 7 = 2 * 83 * 7 = 35 (mod 49)[/hide].\n\n\n\n4. [hide]I claim that a nice number has exactly 2 proper divisors. To prove this, consider a number n with proper divisors d1, d2, d3, ..., dk. It is obvious that da * dk - a + 1 = n. If k > 2, then the product of the proper divisors > n; if k = 1, the sole proper divisor < n.\n\n\n\nThus k = 2. \n\n\n\nn's prime expansion = p1a1 * p2a2 * ... * pmam.\n\n\n\nThe number of divisors (proper or otherwise) is given by\n\n\n\n(a1 + 1)(a2 + 1) ... (am + 1).\n\n\n\nNow this obviously equals 4 (2 proper divisors, the number 1, and n itself), so there are only two possibilities:\n\n\n\n1) a1 = a2 = 1\n\n2) a1 = 3\n\n\n\nThis means that n is either the product of two distinct primes, or is the third power of some prime. We list some numbers from each case:\n\n\n\n1) 2 * 3 = 6, 2 * 5 = 10, 2 * 7 = 14, 2 * 11 = 22, 2 * 13 = 26, 3 * 5 = 15, 3 * 7 = 21, 3 * 11 = 33;\n\n\n\n2) 23 = 8, 33 = 27.\n\n\n\nSo the 10 smallest ones are 6, 8, 10, 14, 15, 21, 22, 26, 27, 33. They sum to 182.[/hide]", "Solution_2": "[hide]\n\n3. use binomial theorem. (7-1)^83 will have terms that are divisible by 49 except for the last 2, which are 83c1*7*1 +83C0*-1, which is 41 mod 49, and (7+1)^83, which will give 83c1*7+83C0*1, which is 43 mod49. So the sum is 84, which is 35 mod 49.[/hide]" } { "Tag": [], "Problem": "im new in mathlinks please, can somebody post some problems?", "Solution_1": "i dont have any problemz" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Can you prove\r\n\r\n$ \\sum_{k\\equal{}0}^n {n\\choose k}\\, x\\, (x\\plus{}ak)^{k\\minus{}1}\\, (y\\minus{}ak)^{n\\minus{}k}\\equal{}(x\\plus{}y)^n$", "Solution_1": "hello, for the proof see here\r\nhttp://www.combinatorics.org/Volume_3/PDF/v3i2r16.pdf\r\nSonnhard.", "Solution_2": "Thanks for the link.\r\nI reproduced this computer-generated proof. I can only confirm that this is a valid proof.\r\nIt does the trick but it gives me no clue how to derive this proof.\r\nI'm sure Abel had another method to derive his formula." } { "Tag": [ "geometry", "3D geometry", "sphere", "geometry unsolved" ], "Problem": "In $3$-dimensional space there is given a point $O$ and a finite set $A$ of segments with the sum of lengths equal to $1988$. Prove that there exists a plane disjoint from $A$ such that the distance from it to $O$ does not exceed $574$.", "Solution_1": "[quote=\"orl\"]In 3-dimensional space there is given a point $O$ and a finite set $A$ of segments with the sum of lengths equal to 1988. Prove that there exists a plane disjoint from $A$ such that the distance from it to $O$ does not exceed 574.[/quote]\n\nObserve that $574>\\frac{1988}{\\sqrt{12}}$.W.L.O.G., we let $O$ be $(0,0,0)$ For every plane cuts through the sphere $x^2+y^2+z^2=(\\frac{1988}{\\sqrt{12}})^2$, then it must also cut some of the segments. We will prove it by contradiction, that means the sum of the segments must be larger than $1988$.\n\nConsider every plane that parallel to one of $x=0$, $y=0$, $z=0$. For every segment, let its ends are $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. Then it covers $(x_2-x_1)$ length of plane which is parallel to $x=0$. Similarly, it covers $(y_2-y_1)$ length of plane which is parallel to $y=0$ and covers $(z_2-z_1)$ length of plane which is parallel to $z=0$. Note that the length of segment $m$ is $\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Then the total lengths of plane being covered is $\\le \\sqrt{3m}$ by $QM\\ge AM$. As there are totally $3*\\frac{1988}{\\sqrt{3}}$ length of plane being covered. We have $m\\ge 1988$.\nAs $574>\\frac{1988}{\\sqrt{12}}$, there must be a plane disjoint from all the segments such that the distance from it to $O$ does not exceed $\\frac{1988}{\\sqrt{12}}$." } { "Tag": [ "geometry", "circumcircle", "conics", "ellipse", "incenter", "ratio", "parallelogram" ], "Problem": "A quadrilateral $ABCD$ is inscribed in a circle with center $O$. A point $P$ is given on the plane. Let $K$, $L$, $M$, $N$ be the circumcenters of triangles $ABP$, $BCP$, $CDP$, $DPA$, respectively. Prove that the midpoints of the segments $KM$, $LN$, $OP$ are collinear.", "Solution_1": "This is a really nice problem :!:\r\n\r\nThe lines NK, KL, LM, MN are perpendicular bisectors of the segments AP, BP, CP, DP. Let perpendiculars n, k, l, m to AP, BP, CP, DP at A, B, C, D meet at to $K' \\equiv n \\cap k,$ $L' \\equiv k \\cap l,$ $M' \\equiv l \\cap m,$ $L' \\equiv m \\cap n,$ forming a quadrilateral K'L'M'N' centrally similar to KLMN with the similarity center P and similarity coefficient 2. In this similarity, the midpoints U, V of KM, LN go into the midpoints U', V' of K'M', L'N' and the midpoint Z of OP into O. Thus we have to show that O lies on the Newton line U'V' of the quadrilateral K'L'M'N'.\r\n\r\nLet the opposite sides K'L', M'N' meet at S, WLOG on the opposite side of L'M' than K', N', and let the opposite sides L'M', N'K' meet at T, WLOG on the opposite side of K'L' than M', N'. Let $E \\in N'K',\\ F \\in K'L',\\ G \\in L'M',\\ H \\in M'N'$ be the other intersections of (O) with the side lines of the quadrilateral K'L'M'N'. (O) is the circumcircle of the pedal triangle $\\triangle ACD$ of the triangle $\\triangle N'TM'$ with respect to P. Consequently, the 3 normals to $N'T \\equiv N'K',\\ TM' \\equiv L'M', M'N'$ at E, G, H concur at the isogonal conjugate Q of the point P with respect to the $\\triangle N'TM'$ and O is the midpoint of PQ. Likewise, (O) is the circumcircle of the pedal triangle $\\triangle BDA$ of the triangle $\\triangle K'SN'$ with respect to P. Consequently, the 3 normals to $K'S \\equiv K'L',\\ SN' \\equiv M'N',\\ N'K'$ at F, H, E concur at the isogonal conjugate Q' of the point P with respect to the $\\triangle K'SN'$ and O is the midpoint of PQ'. Thus the points $Q \\equiv Q'$ are identical and all 4 normals to N'K', K'L', L'M', M'N' at E, F, G, H concur at Q.\r\n\r\nSince the angles $\\angle PAK',\\ \\angle PBL',\\ \\angle PCM',\\ \\angle PDN'$ and the angles $\\angle QEK',\\ \\angle QFL',\\ \\angle QGM',\\ \\angle QHN'$ are all right, the lines $N'K' \\equiv AE,\\ K'L' \\equiv BF,\\ L'M' \\equiv CG,\\ M'N' \\equiv DH$ are all tangents of an ellipse $\\mathcal E$ with the focal points P, Q and pedal circle (O). Thus the quadrilateral K'L'M'N' is tangential to the ellipse $\\mathcal E$ with the center O. Using a parallel projection, project the ellipse $\\mathcal E$ into the circle (O). The quadrilateral K'L'M'N' tangential to the ellipse $\\mathcal E$ goes into a quadrilateral K''L''M''N'' tangential to the circle (O), the Newton line U'V' of K'L'M'N' goes into the Newton line U''V'' of K''L''M''N'', while the line PQ and the midpoint O of the segment PQ stay in place. Since the Newton line U''V'' of the tangential quadrilateral K''L''M''N'' passes through its incenter O and O stays in place in our parallel projection, the Newton line U'V' of the original quadrilateral K'L'M'N' also passes through O.", "Solution_2": "Why the projection of ellipse $\\mathcal E$ into the circle (O) projects the Newton line U'V' of K'L'M'N' into the Newton line U''V'' of K''L''M''N''??? :huuh:", "Solution_3": "Because parallel projection transforms lines into lines and it preserves ratios of segment lengths on the same line. Therefore, midpoints U', V' of the diagonals K'M', L'N' of K'L'M'N' are projected to the midpoints U'', V'' of the diagonals K''M'', L''N'' of K''L''M''N''.", "Solution_4": "Why does parallel projection preserve ratios of segments on the line?", "Solution_5": "[quote=\"nxxx\"]Why does parallel projection preserve ratios of segments on the line?[/quote]\r\n\r\n You can simply use Thales Thoerem.\r\n\r\n [b]Segments on lines between parallel lines are proportional. [/b]", "Solution_6": "Can you explain it :?: \r\nI don't know how to use Thales Theorem in this problem... :blush:", "Solution_7": "Suppose a line $l$ with points $A,\\ B,\\ C \\in l$ in a plane $\\mathcal P$ is transformed by a parallel projection into a line $l'$ with points $A',\\ B',\\ C' \\in l'$ in a plane $\\mathcal P'.$ If the two planes $\\mathcal P \\parallel \\mathcal P'$ are parallel, then every figure in $\\mathcal P$ is congruent to the projected figure in $\\mathcal P',$ and A'B' = AB, B'C' = BC. If the planes $\\mathcal P,\\ \\mathcal P'$ are not parallel, they intersect in a line $k,$ which stays in place in the parallel projection. If $l \\parallel k,$ then $l' \\parallel k,$ ABB'A', BCC'B' are parallelograms, and again A'B' = AB, B'C' = BC. If $l$ is not parallel to $k,$ they intersect at a point P, which stays in place, and $P \\equiv l \\cap l'.$ The lines $AA' \\parallel BB' \\parallel CC'$ are parallel (rays of the parallel projection), the triangles $\\triangle PAA' \\sim \\triangle PBB' \\sim \\triangle PCC'$ are similar, having equal angles, and $\\frac{PA'}{PA}= \\frac{PB'}{PB}= \\frac{PC'}{PC}.$ If follows that $\\frac{AB}{BC}= \\frac{A'B'}{B'C'}.$ Isn't it simple?", "Solution_8": "[quote=\"yetti\"]\n\nSince the angles $ \\angle PAK',\\ \\angle PBL',\\ \\angle PCM',\\ \\angle PDN'$ and the angles $ \\angle QEK',\\ \\angle QFL',\\ \\angle QGM',\\ \\angle QHN'$ are all right, the lines $ N'K' \\equiv AE,\\ K'L' \\equiv BF,\\ L'M' \\equiv CG,\\ M'N' \\equiv DH$ are all tangents of an ellipse $ \\mathcal E$ with the focal points P, Q and pedal circle (O). Thus the quadrilateral K'L'M'N' is tangential to the ellipse $ \\mathcal E$ with the center O. [/quote]\r\n Why you have M'N'P'K' are tangents of an ellipse with the focal P;Q and pedal circle (O)", "Solution_9": "[quote=\"hoangclub\"]Why you have M'N'P'K' are tangents of an ellipse with the focal P;Q and pedal circle (O)[/quote]\r\n\r\nLet (O) be a circle with radius a, centered at the coordinate origin O, with equation $ x^2 \\plus{} y^2 \\equal{} a^2.$ Let P be arbitrary point inside the circle (O), WLOG on the positive x-axis, so that $ p \\equal{} OP < a.$ Let a line through P with arbitrary slope m, with equation $ y \\equal{} m(x \\minus{} p),$ meet the circle (O) at points A, A'. Substituting the line equation into the circle equation leads to a quadratic equation for the x-coordinate of A:\r\n\r\n$ x^2 \\plus{} m^2(x \\minus{} p)^2 \\equal{} a^2$\r\n\r\nwith solution\r\n\r\n$ x_{A} \\equal{} \\frac {pm^2 \\pm \\sqrt {D}}{1 \\plus{} m^2},\\ \\ D \\equal{} a^2(1 \\plus{} m^2) \\minus{} p^2m^2$\r\n\r\nA line through the point A and perpendicular to PA has equation\r\n\r\n$ y \\minus{} y_{A} \\equal{} y \\minus{} m(x_{A} \\minus{} p) \\equal{} \\minus{} \\frac {1}{m} (x \\minus{} x_{A})$\r\n\r\nSubstituting for $ x_{A}$ and squaring yields\r\n\r\n$ my \\plus{} x \\equal{} x_A(m^2 \\plus{} 1) \\minus{} pm^2 \\equal{} \\pm \\sqrt {D}$\r\n\r\n$ F(x, y, m) \\equal{} (my \\plus{} x)^2 \\minus{} D \\equal{} m^2(y^2 \\minus{} a^2 \\plus{} p^2) \\plus{} 2mxy \\plus{} x^2 \\minus{} a^2 \\equal{} 0$\r\n\r\nThis is equation of a family of lines dependent on the arbitrary parameter m. Equation of their envelope is obtained by eliminating m from the equations $ F(x, y, m) \\equal{} 0$ and\r\n\r\n$ \\frac {\\partial F}{\\partial m} \\equal{} 2m (y^2 \\minus{} a^2 \\plus{} p^2) \\plus{} 2xy \\equal{} 0$\r\n\r\nThe 2nd equation gives $ m \\equal{} \\minus{} \\frac {xy}{y^2 \\minus{} a^2 \\plus{} p^2}.$ Substituting this to the 1st equation, the envelope equation comes out as\r\n\r\n$ (x^2 \\minus{} a^2)(y^2 \\minus{} a^2 \\plus{} p^2) \\minus{} x^2y^2 \\equal{} 0,\\ \\ \\ \\frac {x^2}{a^2} \\plus{} \\frac {y^2}{a^2 \\minus{} p^2} \\equal{} 1$\r\n\r\nwhich is equation of an ellipse $ \\mathcal E$ with center O, semimajor axis a, and focus P. Conversely, locus of the feet of normals from the ellipse focus P to ellipse tangents is the circle (O)." } { "Tag": [ "geometry" ], "Problem": "An eight by eight square has a 2 by 2 square in the middle (where nothing goes in). There are 12 pieces, how do you fit all of the pieces in? (I have really quick \"pictures\" of these). I'm sorry if the pieces and board are not proportionate. You are allowed to flip the pieces, you may not reuse the pieces. The black spot on the board is where no pieces may go into. (for example I can't put a part of piece on/in it)", "Solution_1": "The best way to do this is to make an actual board with pieces and solve it form there. This is just an puzzle, not a math problem.", "Solution_2": "Is the problem asking whether it is [i]possible[/i] to fit all pieces in? Or is it definitely possible?", "Solution_3": "It's possible that all the pieces fit it, I've counted before. And yes I have tried with a real board, almost like a bazillion times... :| :mad: It's really getting me fustrated. Any easier ways to \"solve\" it other than trying so many times? Oh and also, we have to put the pieces on the way it fits.", "Solution_4": "Just counting doesn't ensure it. For example, can you put a 2x2 square on a 1x4 board? They both have 4 squares...", "Solution_5": "Some things to get you started. Obviously the first piece (the plus sign) cannot go into a corner. Otherwise it would block off a square. Is it possible to place the plus sign one spot over from the corner? Which spots can you put the plus sign in, and if you put it in those spots, which pieces must you use around it?\r\n\r\nWhat other pieces have areas \"restricted?\" In other words, there are certain spots that each piece is not allowed to go, otherwise it would block off a square from the other pieces.\r\n\r\nUsing this logic, the answer should come in maybe 20 minutes of playing with it since you start significantly reducing the amount of options that are available.", "Solution_6": "I have a puzzle exactly like this, with the same pieces but the board is different(W, H, and Y). By any chance, are you just asking this on AoPS so you can solve it and get some thing( my parents told me they'd buy anything if I could solve it, sadly, i was really close but haven't yet)\r\n\r\nFor these puzzles, they'll take a lot of time, but sadly, i don't think there is an easier way other than to use common sense.\r\n\r\nThis should kinda belong in the games forum...not sure...", "Solution_7": "Well anyway, I don't need anymore help :D I finished it this afternoon after finishing exactly half of it and then going on :D" } { "Tag": [ "geometry", "trigonometry", "limit", "calculus", "calculus computations" ], "Problem": "limt x--------infinity \r\n\r\n \r\n\r\n [(x+a)(x+b)(x+c)(x+d) ]^1/4 - x\r\n\r\n \r\n\r\n2. (sinx)/x<1 and (tanx/x)>1 prove it\r\n\r\n \r\n\r\n3. limit r---infinity\r\n\r\n \r\n\r\n product (r^3-8)/(r^3+8)\r\n\r\n \r\n\r\n r=1 to infinity", "Solution_1": "Hello\r\nSorry but Do u mean $ \\frac {{(x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d)}^{1/4}}{x}$?\r\nif so [hide=\"solution\"]\nTake $ x^4$ common from num.Answer is 1.\n[/hide]\r\nSolution.", "Solution_2": "Hello\r\nFirst of all $ \\frac {tan(x)}{x} > 1$.Kindly edit it.\r\n[hide=\"2\"]\nConsider a sector $ OAP$ of unit circle with center $ O$.Draw a perpendicular$ PM$ from $ P$ to $ OA$.Extend $ P$ to $ B$ such that angle $ BAM$ is right angle.\n$ OA = OP = 1 unit$\n$ AB = OAtanx$\n$ area of triangle OAP$=$ \\frac {1}{2}OA*OPsinx$\n$ area of sector OAP$=$ \\frac {x}{2\\pi}\\pir^2$\n$ area of triangle OAB$=$ \\frac {1}{2}OA*AB$\n$ area of triangle OAP < area of sector OAP < area of OAB$\n$ sinx < x < tanx$\n$ \\frac {sinx}{x} < 1 < \\frac {tanx}{x}$\n[/hide]\r\nThank u.", "Solution_3": "Hello\r\nKindly start using Latex.\r\nIn question 3 are u asking to find $ lim_{r\\to\\infty}\\frac{r^3\\minus{}8}{r^3\\plus{}8}$?\r\nThank u.", "Solution_4": "Problem 3 is probably meant to be $ \\prod_{n\\equal{}3}^{\\infty}\\frac{n^3\\minus{}8}{n^3\\plus{}8}$.", "Solution_5": "[quote=\"Carcul\"]Problem 3 is probably meant to be $ \\prod_{n \\equal{} 3}^{\\infty}\\frac {n^3 \\minus{} 8}{n^3 \\plus{} 8}$.[/quote]\r\nIf that's the case then just see that $ n^3 \\minus{} 8 \\equal{} (n\\minus{}2)(n\\plus{}1\\plus{}\\sqrt{3}i)(n\\plus{}1\\minus{}\\sqrt{3}i)$ and $ n^3 \\plus{} 8 \\equal{} (n\\plus{}2)(n\\minus{}1\\plus{}\\sqrt{3}i)(n\\minus{}1\\minus{}\\sqrt{3}i)$.\r\n\r\nThen, if i didn't make any mistakes, we have: $ \\prod_{n \\equal{} 3}^{m}\\frac {n^3 \\minus{} 8}{n^3 \\plus{} 8} \\equal{} \\frac{2}{7} \\cdot \\frac{m^4\\plus{}2m^3\\plus{}7m^2\\plus{}6m\\plus{}12}{m^4\\plus{}2m^3\\minus{}m^2\\minus{}2m}$. So the result should be $ \\frac{2}{7}$. :)", "Solution_6": "[quote=\"kabi\"]Hello\n\nFirst of all $ \\frac {tan(x)}{x} > 1$.Kindly edit it.\n[hide=\"2\"]\nConsider a sector $ OAP$ of unit circle with center $ O$.Draw a perpendicular$ PM$ from $ P$ to $ OA$.Extend $ P$ to $ B$ such that angle $ BAM$ is right angle.\n$ OA = OP = 1 unit$\n$ AB = OAtanx$\n$ area of triangle OAP$=$ \\frac {1}{2}OA*OPsinx$\n$ area of sector OAP$=$ \\frac {x}{2\\pi}\\pir^2$\n$ area of triangle OAB$=$ \\frac {1}{2}OA*AB$\n$ area of triangle OAP < area of sector OAP < area of OAB$\n$ sinx < x < tanx$\n$ \\frac {sinx}{x} < 1 < \\frac {tanx}{x}$\n[/hide]\nThank u.[/quote]\r\nsir pl explain the method i am not getting", "Solution_7": "[quote=\"MUN\"]\nsir pl explain the method i am not getting[/quote]\r\nHello \r\nI remember Dr had given a pdf with d same proof.It is much clear.\r\nor \r\ncheck this::\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=242582\r\nHope ur doubt will be clear this time.\r\nThank u.", "Solution_8": "1) $ \\dfrac{1}{4}(a+ b + c + d)$\r\n\r\nInformally:\r\n[hide]\n\\begin{align*}\n\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} - x\n&=& \\left(\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} - x\\right) \\times \n\\dfrac{\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} + x}\n{\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} + x}\n\\\\&=&\n\\dfrac{\\sqrt{(x+a)(x+b)(x+c)(x+d)} - x^2}\n{\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} + x}\n\\times \\dfrac{\\sqrt{(x+a)(x+b)(x+c)(x+d)} + x^2}\n{\\sqrt{(x+a)(x+b)(x+c)(x+d)} + x^2}\n\\\\&=&\n\\dfrac{(x+a)(x+b)(x+c)(x+d) - x^4}\n{\\left(\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} + x\\right)\\left(\\sqrt{(x+a)(x+b)(x+c)\n(x+d)} + x^2\\right)}\n\\end{align*}\n\nConsider the highest powers of $ x$ in the numerator and denominator, which is 3. \n\nCoefficient of $ x^3$ in the numerator is $ (a + b + c + d)$. \n\nand in the denominator, for large (positive) $ x$, \n\\begin{align*}\n\\left(\\sqrt[4]{(x+a)(x+b)(x+c)(x+d)} + x\\right)\\left(\\sqrt{(x+a)(x+b)(x+c)(x+d)} + x^2\\right)\n&\\approx& \\left(\\sqrt[4]{x^4} + x\\right)\\left(\\sqrt{x^4} + x^2\\right)\n\\\\&=& \\left(x + x\\right)\\left(x^2 + x^2\\right)\n\\\\&=& 4x^3\n\\end{align*}\n\n\n[/hide]", "Solution_9": "[quote=\"kabi\"][quote=\"MUN\"]\nsir pl explain the method i am not getting[/quote]\nHello \nI remember Dr had given a pdf with d same proof.It is much clear.\nor \ncheck this::\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=242582\nHope ur doubt will be clear this time.\nThank u.[/quote]\r\nthanks kabi \r\ni want to meet u", "Solution_10": "The statement $ \\frac{\\tan x}{x} > 1$ is wrong, unless some specific interval is given.", "Solution_11": "[quote=\"MUN\"]limt x--------infinity \n\n[(x+a)(x+b)(x+c)(x+d) ]^1/4 - x\n\n[/quote]\r\n$ \\lim_{x \\to \\infty} [(x\\plus{}a)(x\\plus{}b)(x\\plus{}c)(x\\plus{}d) ]^{\\frac{1}{4}} \\minus{} x$\r\n\r\nIs this the question???\r\nOr is it:\r\n$ \\lim_{x \\to \\infty} \\frac{[(x\\plus{}a)(x\\plus{}b)(x\\plus{}c)(x\\plus{}d) ]^{\\frac{1}{4}}}{x}$", "Solution_12": "[quote=\"Rajiv\"][quote=\"MUN\"]limt x--------infinity \n\n[(x+a)(x+b)(x+c)(x+d) ]^1/4 - x\n\n[/quote]\n$ \\lim_{x \\to \\infty} [(x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d) ]^{\\frac {1}{4}} \\minus{} x$\n\nIs this the question???\nOr is it:\n$ \\lim_{x \\to \\infty} \\frac {[(x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d) ]^{\\frac {1}{4}}}{x}$[/quote]\r\nfirst one", "Solution_13": "I was wondering if writing $ \\lim_{x \\to \\infty} [(x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d) ]^{\\frac {1}{4}} \\minus{} x$ (please do not laugh if this is wrong :D ) as:\r\n\r\n$ \\Large \\lim_{x \\to \\infty} \\frac {\\sqrt [4] {\\left(\\frac {1}{x} \\plus{} \\frac {a}{x^2}\\right) \\left(\\frac {1}{x} \\plus{} \\frac {b}{x^2}\\right) \\left(\\frac {1}{x} \\plus{} \\frac {c}{x^2}\\right) \\left(\\frac {1}{x} \\plus{} \\frac {d}{x^2}\\right)} \\minus{} \\frac {1}{x^3}}{\\frac {1}{x^4}}$ could help get an indeterminable form of $ \\left( \\frac {0}{0} \\right)$ so as to be able to use $ \\mathfrak {L'Hopital's}.$\r\nThen applying the $ \\mathfrak {L'Hopital's}$, we get the numerator as : $ \\lim_{x \\to \\infty} \\frac {(\\cdots) \\plus{} \\frac {3}{x^4}}{\\frac { \\minus{} 4}{x^5}}$ where the $ (\\cdots)$ is some ugly term which ultimately equates to $ 0$ as $ x$ is in the denominator of all these terms. Hence we get finally $ \\lim_{x \\to \\infty} \\left(\\frac {\\minus{}3}{4} \\right)x$ and hence we can pretty [b]neatly[/b] conclude the value of $ \\lim_{x \\to \\infty} [(x \\plus{} a)(x \\plus{} b)(x \\plus{} c)(x \\plus{} d) ]^{\\frac {1}{4}} \\minus{} x \\equal{} \\color{red}{\\lim _{x \\to \\infty} \\frac { \\minus{}3x}{4} \\equal{} \\minus{} \\infty}$\r\n\r\n[hide]Did some one out there get $ 4x^3$??? More or less, we have got the same answer, the only noticeable difference being that in the sign... :D [/hide]", "Solution_14": "[hide=\"Solution to 1\"]\nSubstitute $ x \\equal{} \\frac{1}{h}$ which transforms the limit into:\n\nLet $ f(x) \\equal{} [(1\\plus{}ax)(1\\plus{}bx)(1\\plus{}cx)(1\\plus{}dx)]^{\\frac{1}{4}}$\n\n$ \\lim_{h\\to 0^{\\plus{}}} \\frac{f(0 \\plus{} h) \\minus{} f(0)}{h}$\n\n$ \\equal{} f'(0) \\equal{} \\frac{1}{4}(a\\plus{}b\\plus{}c\\plus{}d)$\n\n[/hide]" } { "Tag": [], "Problem": "cum vi s-a parut ?", "Solution_1": "..............", "Solution_2": "Ideile mi-au venit mult mai repede decat la locala (faza pe sector), am fost multumit de subiecte, nici foarte usoare, nici foarte grele. Noroc cu functiile alea si problema 4, ca geometria aia de la 2, mama ei, m-a tocat marunt. :D Nu ma duc la nationala, adevarul este ca nici nu-mi permiteam sa ma gandesc la asa ceva, in ultima luna chiar m-am autodepasit (cu 1000%) si sunt multumit de mine.", "Solution_3": "................", "Solution_4": "vreau sa aud pareri de la cei care au fost clasa a XII-a. cat de banale vi s-au parut ? :P am zis la misto, ca toata lumea zice : vaaai, ce simple ce banaleee subiectele de judeteanaa vaiii. si la nationala, 3 pt :blush: felicitari icx pt autodepasire ( asta e si scopul ascuns al olimpiadei, dincolo de orgoliile prostesti de 2 lei ) si capul sus mychrom, ca esti tanar si mai ai multe astfel de ocazii. persevereaza, si nu te lasa afectat de shobolanismele unora cum m-am lasat eu :roll:", "Solution_5": "[quote]Nu ma duc la nationala, adevarul este ca nici nu-mi permiteam sa ma gandesc la asa ceva, in ultima luna chiar m-am autodepasit (cu 1000%) si sunt multumit de mine.[/quote]\r\n\r\nSi eu sunt in aceeasi situatie, cred.Oricum ma simt foarte multumit :cool: .Felicitari! :) \r\n\r\nSubiectele au fost destul de accesibile, ca sa zic, la 7a , \r\ntotusi nu am rezolvat prima problema care se pare ca era cea mai usoara :huh: .\r\nBafta tuturor celor care merg la nationala si la internationala !\r\n\r\n\r\nPS:In incinta scolii unde s-a tinut olimpiada se amplasasera niste\r\n domni care vindeau culegeri de matematica pentru olimpiada (loc strategic :D ). \r\nAm cumparat cateva , deci tot m-am ales cu ceva :P .", "Solution_6": ".................", "Solution_7": "Cum vi s-au parut subiectele la a IX-a ? .. Eu zic ca au fost palpabile... in comparatie cu unele subiiecte din anii anteriori ?", "Solution_8": "mychrom, eu sunt a IX-a :) \r\n\r\nEram atat de ofticat ca am stat toata saptamana acasa ca sa fac geometrie incat am profitat de standul celor de la GIL din curtea liceului si mi-am luat cartea lui V. Nicula, \"Geometrie plana (sintetica etc)\". Nu mai ratez io sansa asta la anu' :P Hai, felicitari si bafta in continuare celor care au trecut ;) Ceilalti, la treaba pt ONM 2008 :P\r\n\r\n@turcas intr-adevar, mi s-au parut [u]un pic[/u] mai accesibile", "Solution_9": "bine subliniat \"un pic\" . :)) Mi-am luat si eu aceeasi carte ... avand in vedere ca stau mai prost cu geometria :) ... Si pe mine m-a macinat problma 2 , sper sa ma pun la punct cat de cat cu geometria , pana la faza urmatoare :)", "Solution_10": "[quote=\"icx\"]Ideile mi-au venit mult mai repede decat la locala (faza pe sector), am fost multumit de subiecte, nici foarte usoare, nici foarte grele. Noroc cu functiile alea si problema 4, ca geometria aia de la 2, mama ei, m-a tocat marunt. :D Nu ma duc la nationala, adevarul este ca nici nu-mi permiteam sa ma gandesc la asa ceva, in ultima luna chiar m-am autodepasit (cu 1000%) si sunt multumit de mine.[/quote]\r\n\r\n\r\npai cu 3 probleme, eu zic ca treci lejer.", "Solution_11": "Si cu 2 probleme poti sa te califici, dar nu stiu ce sa zic despre punctajele de la celelalte 3 probleme care mi-au ramas, clar nu iau 21 puncte (ca nu-mi faceam griji asa).\r\n\r\nTu ce-ai facut pohoatza?\r\n\r\nBtw cred ca te-am vazut azi pe acolo :P", "Solution_12": "nu stiu ce sa zic, astept rezultatele, caci am o solutie sa zic eu \"dubioasa\" la subiectul 1, sa vedem.\r\n\r\noricum eu anul asta, cred ca o sa se mearga de la 14, cel putin la a 9a, daca nu mai putin, dar din cate am auzit eu, parca asa ceva nu se putea.\r\n\r\n\r\nam aflat niste rezultate din valcea, si am inteles ca la a 9a, cea mai mare nota a fost 6, deci sa fim optimisti.", "Solution_13": "Ma intreb daca se vor posta pe site-ul [url=http://www.cnih.ro]CN \"Iulia Hasdeu\"[/url], cum au facut si cu faza pe sector :maybe: Daca pana maine pe la pranz nu apar pe net, o sa ma duc acolo...", "Solution_14": "...................", "Solution_15": "[quote=\"maky\"]\nedit : in legatura cu link`ul prezentat mai sus, prefer sa am reactia \"no comment\". eventual poate primesc o cutie noua de vopsea drept cadou :maybe:[/quote]\r\n\r\n\r\n :D", "Solution_16": "[quote]http://forum.portal.edu.ro/index.php?showtopic=42605 [/quote]\r\n\r\n\r\nCea mai tare discutie de forum pe care am vazut-o, ma amuza de fiecare data cand o citesc :play_ball:", "Solution_17": ".....................", "Solution_18": "[quote=\"mychrom\"][quote=\"spix\"][quote]http://forum.portal.edu.ro/index.php?showtopic=42605 [/quote]\n\n\nCea mai tare discutie de forum pe care am vazut-o, ma amuza de fiecare data cand o citesc :play_ball:[/quote]\n\nmai e una haioasa, cu \"fraude in olimpiada de matematica\" sau ceva de genu' asta.[/quote]\r\neste cu adevarat amuzanta dar as prefera sa nu fie deloc.Cei neavizati care intra pe acel forum pot crede ca acele lucruri sunt adevarate si imaginea unor oameni de valoare si a olimpiadei in general va avea de suferit.", "Solution_19": "on-topic: mai are cineva rezultate de la alte judete? ce asteptari aveti de la ON din acest an?", "Solution_20": "exista o pozitie oficiala a factorilor de decizie cu privire la elevii calificati la onm 2007?", "Solution_21": "Pot sa va spun ceva rezultate din Satu Mare :\r\na X-a: Tataran Alexandru 21p\r\na XI-a: eu 26p\r\na XII-a: Gavrus Cristian 26p\r\n Marginean Vlad 17p\r\nma astept ca subiectele de la ONM sa fie mult mai dificile,in special la anliza, la judeteana s-au dat doar siruri.", "Solution_22": "Sabin , nu sti cumva ce punctaj a avut loc I la a IX-a din Satu Mare ?", "Solution_23": "sunt curios de rezultatele din slatina/olt de la clasa a 8a, daca poate sa le puna cineva, primele locuri. [maky stie dc :D ]", "Solution_24": "....................", "Solution_25": "la a IX-a parca a castigat andra cu 20-21.", "Solution_26": ".......................", "Solution_27": "La a 9a locu 1 9p :blush: dar nu m-as mira sa il duca la ONM ,n-ar fi prima oara :P", "Solution_28": "nu va mai agitati atata....:)....lasati oricum antionala anu asta e in vacantza....cand is apstele..asha ca mai bine mergeti la stropit..:)).....oricum abia astept si eu sa ajung al pitesti sal vizitez...ca nam fost inca..:D.....", "Solution_29": "nationala a fost mereu in vacantza....\r\noricum.. si eu abia astept sa ajung la piteshti(in sf nu mai zic ploieshti) \\:D/" } { "Tag": [ "Asymptote", "AoPSwiki" ], "Problem": "Bonjour,\r\n\r\nje suis \u00e0 la recherche d'un plugin pour ajouter Asymptote sur un forum et un wiki.\r\n\r\nLe wiki n'est pas du m\u00eame type que AoPSWiki... mais je me dis que je saurais peut-\u00eatre adapter le code qui est utilis\u00e9 sur AoPSWiki... \r\n\r\ndonc si quelqu'un le connait... et s'il consent \u00e0 partager ce savoir pr\u00e9cieux, je le remercie d'avance.\r\n\r\nGa\u00e9tan M.", "Solution_1": "Pour commencer \u00e0 utiliser Asymptote, vous devez d'abord t\u00e9l\u00e9charger et l'installer. Pour afficher les images au format eps que vous produisez, vous aurez \u00e9galement besoin de t\u00e9l\u00e9charger une visionneuse eps comme GSview. (GSview est tr\u00e8s pratique car il permet d'afficher les fichiers EPS et PDF, et avec Ghostscript, Asymptote pouvez facilement imprimer les images au format pdf ainsi.)\n[modifier] Asymptote\n\n\u00a0\u00a0\u00a0\u00a0\u00a0 Pour t\u00e9l\u00e9charger et installer Asymptote sur votre machine Windows:\nAller \u00e0 la FileCluster et cliquez sur le Miroir 1 Lien externe \u00e0 droite. Cela vous am\u00e8nera \u00e0 SourceForge, qui vous aidera \u00e0 t\u00e9l\u00e9charger asymptote-2.15-setup.exe.\nUne fois le t\u00e9l\u00e9chargement termin\u00e9, acc\u00e9dez \u00e0 D: \\ downloads \\ Asymptote, ou partout o\u00f9 vous avez enregistr\u00e9 le fichier, puis double-cliquez sur le fichier exe (asymptote-2.15-setup.exe).. Cela va ouvrir une fen\u00eatre de l'installateur, o\u00f9 vous pouvez choisir le dossier dans lequel sera install\u00e9 Asymptote (ou simplement utiliser la valeur par d\u00e9faut C: \\ Program Files Asymptote \\), puis choisissez si vous souhaitez avoir des raccourcis ajout\u00e9s sur le bureau et le menu D\u00e9marrer. Lorsque vous avez termin\u00e9, cliquez sur Installer.\nAsymptote est pr\u00eat \u00e0 d\u00e9marrer la production d'images, mais vous avez encore besoin d'un moyen de voir ces images. (REMARQUE:.. Si vous avez d\u00e9j\u00e0 un visualiseur capable d'afficher les fichiers EPS et PDF, vous n'avez pas besoin de t\u00e9l\u00e9charger le fichier ci-dessous.)\n\n\nCeci est l'edition anglais. Je regrette, mais mon fran\u00e7ais n'est pas tr\u00e8s bon. En haut est l'edition de Google Translate. J'esp\u00e8re que \u00e7a t'aide.\n\n \n To begin using Asymptote, you must first download and install it. To view the eps-format images you produce, you will also need to download an eps viewer such as GSview. (GSview is convenient because it can display both eps and pdf files, and with GhostScript installed, Asymptote can easily output the images in pdf format as well.)\n[edit] Asymptote\n\n To download and install Asymptote on your Windows machine:\nGo to FileCluster and click on the External Mirror 1 link on the right. This will bring you to SourceForge, which will help you download asymptote-2.15-setup.exe.\nWhen the download is complete, browse to D:\\downloads\\Asymptote, or wherever you saved the file, and double-click on the .exe file (asymptote-2.15-setup.exe). This will open an installer window, where you can choose the folder that Asymptote will be installed to (or simply use the default C:\\Program Files\\Asymptote), and choose whether you wish to have shortcuts added to the desktop and start menu. When you have finished, click Install.\nAsymptote is ready to start producing images, but you still need a way to view these images. (NOTE: if you already have a previewer capable of viewing .eps and .pdf files, you do not need to download the file below.)", "Solution_2": "Aussi, tu dois t\u00e9l\u00e9charger \"GSView\" d'ici: http://downloads.ghostscript.com/public/gs907w32.exe\nLe plugin est d'ici: http://iweb.dl.sourceforge.net/project/asymptote/2.21/asymptote-2.21-setup.exe" } { "Tag": [ "function", "floor function" ], "Problem": "Find the total number of different integer values the function \\[f(x) = \\lfloor x\\rfloor+\\lfloor 2x\\rfloor+\\left\\lfloor \\frac{5x}{3}\\right\\rfloor+\\lfloor 3x\\rfloor+\\lfloor 4x\\rfloor\\] takes for real numbers $x$ with $0 \\leq x \\leq 100$.", "Solution_1": "http://www.kalva.demon.co.uk/apmo/asoln/asol932.html", "Solution_2": "The link doesn't work...can you post the solution?" } { "Tag": [ "function", "modular arithmetic", "algebra unsolved", "algebra" ], "Problem": "Determine all functions $ f$ mapping the set of positive integers to the set of non-negative integers satisfying the following conditions:\r\n(1) $ f(mn) \\equal{} f(m)\\plus{}f(n)$,\r\n(2) $ f(2008) \\equal{} 0$, and\r\n(3) $ f(n) \\equal{} 0$ for all $ n \\equiv 39\\pmod {2008}$.", "Solution_1": "[quote=\"April\"]Determine all functions $ f$ mapping the set of positive integers to the set of non-negative integers satisfying the following conditions:\n(1) $ f(mn) \\equal{} f(m) \\plus{} f(n)$,\n(2) $ f(2008) \\equal{} 0$, and\n(3) $ f(n) \\equal{} 0$ for all $ n \\equiv 39\\pmod {2008}$.[/quote]\r\n\r\n$ f(n) \\ge 0$ for all $ n$ then if $ 2008 \\equal{} mn$ then $ 0 \\equal{} f(2008) \\equal{} f(n) \\plus{} f(m)$ then $ f(n) \\equal{} 0$ for any divisor of $ 2008$\r\n\r\n$ (m,2008) \\equal{} 1$ then exists $ a$ such $ am \\equiv 1 \\pmod {2008}$ then \r\nfor $ n \\equal{} 39a$ we have $ nm \\equiv 39 \\pmod {2008}$ then \r\n$ 0 \\equal{} f(nm) \\equal{} f(n) \\plus{} f(m)$ then $ f(m) \\equal{} 0$ for any integer $ m$ coprime with $ 2008$\r\n\r\nIf $ (m,2008) \\equal{} d$ then $ m \\equal{} da$, $ (a,2008) \\equal{} 1$ and $ d|2008$\r\n$ f(m) \\equal{} f(da) \\equal{} f(d) \\plus{} f(a) \\equal{} 0 \\plus{} 0 \\equal{} 0$\r\nthen $ f(n) \\equal{} 0$ for all $ n$\r\n\r\nSame argument for any natural number instead of $ 2008$ and for any nonzero instead of $ 39$.", "Solution_2": "if (n,2008)=1,then let $m=39n^{-1}$,we get $f(m)+f(n)=0$so$f(m)=0$\nbesides,$0\\equiv f(2008)=f(2)+f(1004)=f(8)+f(251)$\nhence $f(2)=f(251)=0$\nlet $n=2^s251^tr$\nthen since $f(2n)=f(n)=f(251n)$\nhence $f(n)=f(r)=0$\nso for any n,f(n)=0.", "Solution_3": "Consider a prime $p$ such that $(p,2008)=1$\nNow consider the set $2008k+39$ such that $0\\leq k\\leq (p-1)$\nBy php we'll get $p$ divides one of them that implies $f(p)=0$\nNow as $f(2008)=0$ so $f(p)=0$ for all prime $p$.\nNow easily we can see $f(n)=0$ for all $n$." } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "f and g are two periodic functions, and $ \\lim_{x\\to \\plus{} \\infty}[f(x) \\minus{} g(x)] \\equal{} 0$. Prove: f=g.\r\n\r\nProof. Let f's period is $ T_1$, g's $ T_2$. Let $ T_1,T_2 > 0$, then for definite $ x$, we have\r\n$ f(x) \\minus{} g(x) \\equal{} \\lim_{x\\to \\plus{} \\infty}[f(x) \\minus{} g(x)]$\r\n$ \\equal{} \\lim_{n\\to \\plus{} \\infty}\\{[f(x \\plus{} n T_1) \\minus{} g(x \\plus{} n T_1)]$ \r\n$ \\plus{} [g(x \\plus{} nT_1 \\plus{} nT_2) \\minus{} f(x \\plus{} nT_1 \\plus{} nT_2)]$\r\n$ \\plus{} [f(x \\plus{} nT_2) \\minus{} g(x \\plus{} nT_2)]\\}$\r\n$ \\equal{} 0 \\plus{} 0 \\plus{} 0 \\equal{} 0.$\r\n\r\nSo , for $ \\forall x$, we have $ f(x) \\equal{} g(x)$.\r\n\r\nQuestion: why here we can make $ f(x) \\minus{} g(x) \\equal{} \\lim_{x\\to \\plus{} \\infty}[f(x) \\minus{} g(x)]$? thanks", "Solution_1": "[quote=\"water2011\"]\n $ \\lim_{n\\to \\plus{} \\infty}[f(x) \\minus{} g(x)] \\equal{} 0$\n[/quote]\nIt should read $ \\lim_{\\mathbf{x}\\to \\plus{} \\infty}[f(x) \\minus{} g(x)] \\equal{} 0$, otherwise there's nothing to prove.\n[quote]\nQuestion: why here we can make $ f(x) \\minus{} g(x) \\equal{} \\lim_{n\\to \\plus{} \\infty}[f(x) \\minus{} g(x)]$? thanks[/quote]\r\nWhy can we write $ \\lim_{n\\to\\infty} c \\equal{} c$, where $ c$ is a constant?", "Solution_2": "[quote=\"zhoraster\"][quote=\"water2011\"]\n $ \\lim_{n\\to \\plus{} \\infty}[f(x) \\minus{} g(x)] \\equal{} 0$\n[/quote]\nIt should read $ \\lim_{\\mathbf{x}\\to \\plus{} \\infty}[f(x) \\minus{} g(x)] \\equal{} 0$, otherwise there's nothing to prove.\n[quote]\nQuestion: why here we can make $ f(x) \\minus{} g(x) \\equal{} \\lim_{n\\to \\plus{} \\infty}[f(x) \\minus{} g(x)]$? thanks[/quote]\nWhy can we write $ \\lim_{n\\to\\infty} c \\equal{} c$, where $ c$ is a constant?[/quote]\r\n\r\nIt is my typo. I have modified it . Thanks" } { "Tag": [ "LaTeX" ], "Problem": "This expression is OK in the Forum, but when I try to make it into a PDF via TeXnicCentre, it becomes mess. I don't know how to fix it...... Can anyone help me with it?\r\n$ x_{i}^{k}= \\begin{cases}0,\\text{if}~x_{i}^{(k-1)}=x_{i+1}^{(k)},i=1,2,\\cdots,n \\\\ 1 ,\\text{if}~x_{i}^{(k-1)}\\neq x_{i+1}^{(k)},i=1,2,\\cdots,n \\end{cases}$", "Solution_1": "I had the same problem with equation erray. \r\n\r\nPut it in the Center exactly like this:\r\n\r\n\\$ x_{i}^{k}=\r\n\\begin{cases}0,\\text{if}~x_{i}^{(k-1)}=x_{i+1}^{(k)},i=1,2,\\cdots,n \\\\ \r\n1 ,\\text{if}~x_{i}^{(k-1)}\\neq x_{i+1}^{(k)},i=1,2,\\cdots,n \r\n\\end{cases} \\$\r\n\r\nWhere \\$=dollar sign", "Solution_2": "I have done what you said, but it still doesn't work...... :maybe:", "Solution_3": "Eh? I put it in TeXnicCenter, it comes out fine... :huh: :maybe:", "Solution_4": "The cases environment only works if you load the amsmath package. This package is loaded on this forum which is why it works here.", "Solution_5": "[quote=\"stevem\"]The cases environment only works if you load the amsmath package. This package is loaded on this forum which is why it works here.[/quote]\r\nThanks for the information. But where can I find the Amsmath Package? It is not in the LaTeX Packeges Manager.... ...\r\n\r\nEdit: I have fixed it. Thanks for all the help above!" } { "Tag": [ "probability" ], "Problem": "[size=150]Does anyone know where to find solutions for the problems in here:[/size]\r\n\r\n[u]http://www.artofproblemsolving.com/Books/Excerpts.pdf[/u]?\r\n\r\n\r\nI'm having a bit of trouble (esp. with the harder problems) and would appreciate if anyone knows where to find the solutions...", "Solution_1": "Only if you buy the solutions manual. You could also ask it on the forum, where people will help you.", "Solution_2": "[quote=\"Lilac\"][size=150]Does anyone know where to find solutions for the problems in here:[/size]\n\n[u]http://www.artofproblemsolving.com/Books/Excerpts.pdf[/u]?\n\n\nI'm having a bit of trouble (esp. with the harder problems) and would appreciate if anyone knows where to find the solutions...[/quote]\r\nJust so you know, there are hints in the back of the book too.\r\nand as izzy mentioned, the solutions manual. But don't use that all of the time", "Solution_3": "Thanks for the help, everybody...I'll probably buy the Intro to Probability book.\r\n\r\nSorry about being downright clueless ;)" } { "Tag": [ "function", "calculus", "derivative", "real analysis", "integration", "real analysis unsolved" ], "Problem": "Hi,\r\n\r\nI'm wondering if there exists a continuous fonction f:[0,1]->R such that \r\n\r\n$ |f(x)\\minus{}f(y)| \\geq |x\\minus{}y|^{1/2}$\r\n\r\nthanks", "Solution_1": "Note that such a function would have to be nowhere differentiable.\r\n\r\nBrownian motion displays this behavior on average, but not as a worst case.\r\n\r\nIn fact, as a worst case - you can't do this. There is no such function.\r\n\r\nStep 1: Such an $ f$ would have to be monotone. If it were not, then (by the IVT) there would be points $ x\\ne y$ such that $ f(x)\\equal{}f(y),$ which would violate the hypothesis.\r\n\r\nStep 2. WLOG, assume that $ f$ is increasing.\r\n\r\n$ f(1)\\minus{}f(0)\\equal{}\\sum_{k\\equal{}1}^n\\left[f\\left(\\frac kn\\right)\\minus{}f\\left(\\frac{k\\minus{}1}n\\right)\\right]\\ge\\sum_{k\\equal{}1}^n\\sqrt{\\frac1n}\\equal{}\\sqrt{n}.$\r\n\r\nThat cannot hold as $ n\\to\\infty,$ so we cannot construct such a function.", "Solution_2": "Aclually, such function does not exist for every $ \\alpha<1$. There is very beautiful rezult by Lebesque that every monotone function is differentiable a.e. This gives a contradiction( with Kent's observation) :)", "Solution_3": "The concrete estimation I gave - the telescoping sum on a partition - would work in the same way for any $ \\alpha<1,$ giving rise to the estimate $ f(1)\\minus{}f(0)\\ge n^{1\\minus{}\\alpha}.$ But leshik's observation - that monotone functions can't be nowhere differentiable - is also a good point.", "Solution_4": "Thanks for the solution.\r\n\r\nAnd now, do you think there's a continous fonction $ f : R\\minus{}> R^2$ such that\r\n\r\n$ \\parallel{}f(x)\\minus{}f(y)\\parallel{} \\geq \\sqrt{|x\\minus{}y|}$\r\n\r\n?", "Solution_5": "[quote=\"leshik\"]There is very beautiful rezult by Lebesque that every monotone function is differentiable a.e.[/quote]\r\nleshik refers to the [url=http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem]Lebesgue Differentiation Theorem[/url], which helps with the Lebesgue integral analogue to the Fundamental Theorem of Calculus.", "Solution_6": "you can drop the assumption that $ f$ is continuous, and this is still a good exercice. It has already been discussed on the forum ...", "Solution_7": "Are you speaking about the continuous function $ f: R \\rightarrow R^2$ such that\r\n\\[ \\parallel{}f(x) \\minus{} f(y)\\parallel{} \\geq \\sqrt {|x \\minus{} y|}\\]\r\nwhere $ \\parallel{}\\cdot\\parallel{}$ is a norm on $ R^2$\r\n?\r\n\r\nWhere has it been discussed on the forum ?\r\n\r\nthanks" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "A group consist of $ n$ tourists. Among every $ 3$ of them there are $ 2$ of which are not familiar.For every partition of the tourists in 2 buses,you can find 2 tourists that are in the same bus and are familiar with each other.Prove that a tourist familiar to at most $ \\frac{2n}{5}$ tourists.", "Solution_1": "[hide=\"An idea\"]Turn this into a graph. Let each tourist be represented by a vertex, and let two vertices be connected by an edge if and only if their respective tourists are familiar.\n\nThe first condition is equivalent to saying that there are no cycles of length 3.\n\nThe second condition is equivalent to saying that the graph is nonbipartite. This is equivalent to saying there's a cycle of odd length.\n\nSo there exists a cycle of odd length greater than or equal to 5.[/hide]\r\n\r\nI don't know if this helps though...", "Solution_2": "[hide=\"Ignore this\"]It's false. Consider a graph whose vertices are $ a, b, c, d, e, a_{1}, a_{2},\\ldots, a_{n\\minus{}5}$ such that we have edges $ \\{a,b\\},\\{b,c\\},\\{c,d\\},\\{d,e\\},\\{e,a\\}$ and $ \\{a,a_{i}\\}$ for all $ i$. This graph has no triangles and is not bipartite, but $ \\deg(a) \\equal{} n\\minus{}3$ which certainly can be larger than $ \\frac{2n}{5}$ for large enough $ n$. \n\nOne alternative possible statement is that the [i]average[/i] vertex can't have degree larger than $ \\frac{2n}{5}$. Then this is a Turan-type result.[/hide]\r\n\r\nWait, I've misunderstood the question: we almost certainly are supposed to prove that there exists a tourist who knows at most $ \\frac{2n}{5}$ tourists. Or, in other words, show that a graph in which every vertex has degree greater than $ \\frac{2n}{5}$ either contains a triangle or is bipartite.", "Solution_3": "This was an interesting problem. Suppose the opposite, i.e., lets assume that the degree of every vertex is more than $ \\frac{2n}{5}$. Because $ G$ is bipartite, like K81o7 said, it must contain a cycle of odd length. Let $ C_{2k\\plus{}1}$ be the cycle of minimial length. We know $ k\\ge 2$ because $ G$ is triangle-free, and now I will make an argument by contradiction with this odd cycle.\r\n\r\nSuppose that the smallest odd cycle has length greater than or equal to $ 5$. Let the vertices of $ C$ be $ \\{v_{1},v_{2},\\dots, v_{2k\\plus{}1}\\}$ where $ v_{i}$ is connected to $ v_{i\\plus{}1}$ (cyclic notation is used). Notice that if we consider the induced subgraph of the vertices of $ C$, no edges besides the $ 2k\\plus{}1$ explained above can exist, since this would yield a smaller odd cycle. Now, because the length of the cycle is at least $ 5$ and each vertex is connected to exactly two others, each vertex in the cycle must be connected to more than $ \\frac{2}{5}$ of the vertices outside the cycle. Because of this fact and the fact that $ G$ cannot contain any triangles, it is easy to see that $ v_{i}$ and $ v_{i\\plus{}2}$ must both be connected to more than one-fifth of the vertices outside the cycle. But because the cycle is at least length $ 5$, we know that there at least $ 5$ of these pairs $ (v_{i}, v_{i\\plus{}2})$, and thus there are two pairs of vertices in the cycle $ (v_{i}, v_{i\\plus{}2})$ and $ (v_{j},v_{j\\plus{}2})$ that are all connected to some point $ x\\in V$ outside the cycle. This leaves three different cases:\r\n\r\nCase I: $ v_{i}, v_{i\\plus{}1}, v_{i\\plus{}2}, v_{i\\plus{}3}$ are all connected to $ x$.\r\n\r\nThe fact that this is a contradiction is trivial, since there are $ 3$ triangles admitted.\r\n\r\nCase II: $ v_{i}, v_{i\\plus{}2}, v_{i\\plus{}4}$ are all connected to $ x$.\r\n\r\nIn this instance simply consider the new cycle $ v_{i}, x, v_{i\\plus{}4},\\dots, v_{i\\minus{}1}$. This is still an odd cycle, and yet it is shorter in length by $ 2$. But since $ C$ was the shortest odd cycle, we have a contradiction.\r\n\r\nCase III: $ v_{i}, v_{i\\plus{}2}, v_{i\\plus{}j\\plus{}2}, v_{i\\plus{}j\\plus{}4}$ are all connected to $ x$.\r\n\r\nIn this last and final case, we know that either the length from $ v_{i}$ to $ v_{i\\plus{}j\\plus{}4}$ or the length from $ v_{i\\plus{}2}$ to $ v_{i\\plus{}j\\plus{}2}$ must be odd. Then we simply join $ x$ and create a new odd cycle like in Case II, arriving at a contradiction.\r\n\r\nHence, repeating the above argument, we either arrive at a graph with a triangle, or we simply receive an odd cycle of smaller length. Either way, once we obtain an odd cycle of length $ 5$, either Case I or Case II must occur, and this will produce a triangle, which is the desired contradiction. Hence we are finished.\r\n\r\nIt is worth noting that this result is best possible, for if we suppose $ n\\equal{}5k$, then consider the $ k$ disjoint 5-cycles $ P_{1},P_{2},\\dots, P_{k}$ such that $ P_{i}\\equal{}\\{v_{i,1}, v_{i,2},\\dots, v_{i,5}\\}$. Then if the only edges in this graph are the ones in the cycles (i.e. $ v_{i,j}$ is connected to $ v_{i,j\\plus{}1}$) and the edges between $ v_{i,j}$ and $ v_{k,j\\pm 1}$, then each vertex will have degree $ 2k$, the graph will triangle-free and not be bipartite.\r\n\r\nThis problem implores a greater generalization: Given a graph $ G$ that is not $ k$-partite and does not contain a $ m$-clique, what is the greatest degree that the smallest vertex of all possible graphs can have?", "Solution_4": "[quote=\"zgorkster\"]This problem implores a greater generalization: Given a graph $ G$ that is not $ k$-partite and does not contain a $ m$-clique, what is the greatest degree that the smallest vertex of all possible graphs can have?[/quote]\r\n\r\n\"Has girth $ m$\" (i.e., does not contain any cycles of size smaller than $ m$) seems to me the more natural generalization. I suspect asymptotic results for this type of question are known." } { "Tag": [ "algorithm", "search", "geometry", "function", "analytic geometry", "email" ], "Problem": "This thread is for discussion of the USACO November 2006 contest. Discuss results, algorithms, etc., and perhaps we can compile solutions for all $9$ problems while we're waiting on the analysis. I've managed to get together solutions for the gold problems from the USACO discussion forum, but I don't have any silver or bronze solutions yet, so they would be much appreciated (or expansions on the gold solutions would also be cool).\r\n\r\n[hide=\"block\"]\nAt each node, keep track of the two shortest distances to that node found so far. Then proceed with a Dijkstra-like algorithm in which you expand the closest unexpanded node so far (or if the second-shortest path to an already expanded node is the minimal distance, expand that one provided you haven't already). A heap is necessary to run within the time limits.\n[/hide]\n[hide=\"cowfood\"]\nThere are two solutions, you can either just do a straight-up memoization where you recurse to the right, and this is, though slightly counterintuitively, pretty fast because the number of possible configurations is actually bounded above by like either the 12th Fibonacci or Tribonacci number, so the state space for memoization is rather small. However, it is a bit difficult to code because you need a bitmask of size 144 to memoize. \n\nThe second solution involves using a bitmask of the previous row, and constructing a dynamic programming algorithm with that as the state, and you can use various bit-trickery to make it fairly easy to code. \n[/hide]\n[hide=\"plank\"]\nInstead of cutting apart a board, merge the target planks together until you are left with the original board (then merging boards of length $a$ and $b$ costs $a+b$). Merge the two smallest, and repeat until there is only one left. A heap lets you get O(NlogN).\n[/hide]\n[hide=\"badhair\"]We need a solution for this one still.[/hide] *solution needed*\n\n[hide=\"bigsq\"]We need a solution for this one still.[/hide] *solution needed*\n\n[hide=\"rndnum\"]We need a solution for this one still.[/hide] *solution needed*\n\n[hide=\"brand3\"]We need a solution for this one still.[/hide] *solution needed*\n\n[hide=\"scrbl\"]\nAll we need is a way of comparing the input rack with a given word from the dictionary. The easiest way is to sort the input letters (not including #s, but we need to know how many there are), and sort each dictionary word as we read it. Start by looking at the first characters A and B (input/dictionary respectively). If B matches A, go forward in both words; if B>A go forward in the first word; if BA go forward in the first word; if B0; d--) {\n test();\n }\n}[/code]\nThat works in 0.2 seconds on the worst test case for me.[/hide]\nrndnum:\n[hide]Let f(n,x) be how many numbers from 1 to n have x more zeroes than ones. Split those numbers into three groups: {1},{even numbers},{odd numbers}. Those last two groups can be calculated recursively by taking off the last digit and accounting for that with the 'x'. That gives us the recurrence f(n,x) = f([n/2],x-1) + f([(n-1)/2],x+1) + (1 more if x is -1). Since there aren't too many values for x (somewhere between -31 and 31 at worst) and there will only be approximately a logarithmic number of 'n's, we can do this with memoization, storing previous results in a map. You can then calculate the answer as (sum for x>=0 f(end,x)) - (sum for x>=0 f(start-1,x)).That gives me a worst case of 0.008s in C++.[/hide]", "Solution_4": "badhair\r\n\r\n[hide]\nAnother way of phrasing this problem is: for each cow, find the number of cows that can see it.\n\nNow, process the cows in order, and keep a stack of cows that still have the potential to see other cows (in other words, a stack of cows whose lines of vision haven't been blocked by a taller cow yet).\n\nIf the cow you are processing is shorter than the cow on top of the stack, push the current cow on to the stack.\nOtherwise, if it's taller or equal, pop off cows until the current cow shortest than the cow on top of the stack. For each cow you pop off, add the size of the stack to the total, because the size of the stack represents the number of cows that can see the cow you just popped off.\n[/hide]", "Solution_5": "I've edited in solutions to all silver problems above. That just leaves me with one bronze which is stumping me for some reason..", "Solution_6": "brand3 is stumping me too. I can't get the last three test cases done under a second and can't think of a better way to do it other than recursively going through the possibilities.", "Solution_7": "Ah. Well, I just tried it and it passes in 0.916 in C++. So I guess theres no smarter way of doing it (I assumed 22 million was a bit large..)\r\nThus:\r\nbrand3:\r\n[hide]If you use java, give up. Otherwise, just recursively run through all possibilities and stop once you've hit the last one you need.[/hide]", "Solution_8": "Well, I may have only gotten gotten half the test cases for badhair, but my solution was only one line off from being perfect (makes me sad :()\r\n\r\n[hide=\"here is the outline of the solution that is similar to the analysis now up\"]\n\nFirst, keep an array of the heights and a boolean array of which cows can currently see. Also keep a number keeping track of how many can see so you don't have to count the array every time.\n\nLoop through the cows. If a cow is the first cow or shorter than the previous cow, then do nothing, add the number of cows that can see to the sum, then toggle the cow's position in the array because it can now see (also add one to the number seeing).\n\nIf a cow is taller than the previous cow, loop back through the cows until you find one that is taller than the current cow, decrease the number seeing by the number of cows that were blocked and add the number of cows that can still see, then the current cow becomes seeing and the number is incremented.\n\nIf a cow is the same height as the previous cow, then only the previous cow stops seeing, so toggle that and add the number remaining (adding the number was the one line I missed), then add the current cow to the number seeing and the array.\n\nWhen you finish the loop, your sum will be the sum requested (the sum of the number of cows the ith cow can see is the same as the sum of the number of cows that can see the ith cow).\n\nAlso, if you wish:\n[hide=\"full fixed java code of badhair solution outlined above\"]\nIt's slightly less compact than the one in the analysis but it works\n[code]\nimport java.io.*;\nimport java.util.*;\n\nclass badhair {\n\tpublic static void main(String[] args) throws IOException {\n\t\tBufferedReader f = new BufferedReader(new FileReader(\"badhair.in\"));\n\t\tint N = Integer.parseInt(f.readLine());\n\t\tint[] heights = new int[N];\n\t\tint working = 0;\n\t\tboolean[] work = new boolean[N];\n\t\tfor(int i = 0; i-1; j--){\n\t\t\t\t\t\tif(heights[j]<=heights[i]){\n\t\t\t\t\t\t\tif(work[j]){\n\t\t\t\t\t\t\t \tworking--;\n\t\t\t\t\t\t\t\twork[j]=false;\n\t\t\t\t\t\t\t}\n\t\t\t\t\t\t} else {\n\t\t\t\t\t\t\tsum+=working;\n\t\t\t\t\t\t\tbreak;\n\t\t\t\t\t\t}\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t\tworking++;\n\t\t\twork[i]=true;\n\t\t}\n\t\tPrintWriter out = new PrintWriter(\"badhair.out\");\n\t\tout.println(sum);\n\t\tout.close();\n\t}\n}\n[/code]\n\n[/hide]\n[/hide]\n\n[quote=\"TripleM\"]Ah. Well, I just tried it and it passes in 0.916 in C++. So I guess theres no smarter way of doing it (I assumed 22 million was a bit large..)\nThus:\nbrand3:\n[hide]If you use java, give up. Otherwise, just recursively run through all possibilities and stop once you've hit the last one you need.[/hide][/quote]\n\nremember that java gets 2x the runtime\n\n\nAs for bigsq, my current solution (fixed to prevent wrong answers and bad exit statuses giving me only 6 test cases) is looping through all the Js ignoring pairs that can't give smaller areas, but on something like an 80x80 square filled with Js and one empty slot, you're not going to finish in time. Interesting that bigsq is the only one of the nine problems without an analysis up.\n\nalso: my results\n[code]\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15\nbadhair * * * * x x x * x x\nbigsq * * x e * * e e e * * e e e e\nrndnum * x x x x t t t t t\n\n[/code]\r\n\r\nMy five missed badhairs were because of the missing line I explained above (they had consecutive cows with the same height). The wrong answer on bigsq was because my program said OOH! you can go 9 up and 1 right from this J to another J and 9 left and 1 down from the same J to another J! That's a right angle! (draw it and you see it isn't). The bad exit statuses are array out of bounds, mostly - possibly a string index out of bounds which means I essentially didn't include the upper bounds of x\nusing namespace std;\nchar grid[100][100];\nint N;\nint type(int i, int j) {\n if (i<0 || j<0 || i>=N || j>=N || grid[i][j]=='B') return -1; \n if (grid[i][j]=='*') return 0;\n return 1;\n}\nint main() {\n ofstream fout (\"bigsq.out\");\n ifstream fin (\"bigsq.in\");\n fin >> N;\n for (int i=0; i> grid[i][j];\n int best = 0;\n for (int i=0; ibest && l?=(k-i)*(k-i)+(l-j)*(l-j);\n }\n }\n for (int l=N-1; (k-i)*(k-i)+(j-l)*(j-l)>best && l>=0; l--) {\n if (grid[k][l]!='J') continue;\n int x = type(i+j-l,j+k-i);\n int y = type(j+k-l,l+k-i);\n if (x!=-1 && y!=-1 && (x==1 || y==1)) {\n best>?=(k-i)*(k-i)+(l-j)*(l-j);\n }\n }\n \n }\n }\n fout << best << endl; \n return 0;\n}[/code][/hide]", "Solution_10": "hi\r\ncan i go to silver division with this result?\r\n 1 2 3 4 5 6 7 8 9 10\r\nbrand3 * * * * * * * * t t\r\ncowtag * * * * * * * * * *\r\nscrbl * * * * * * * * * *\r\nif i can , how i do it? \r\n :?:", "Solution_11": "[quote=\"TripleM\"][quote=\"Hamster1800\"]\nremember that java gets 2x the runtime[/quote]\nYeah, but the input/output will most likely be far too slow (anything that comes close to running out of time in C++ in the training problems is guaranteed to run out of time in java even with a longer time limit; believe me, I know.)\n\n[quote]As for bigsq, my current solution (fixed to prevent wrong answers and bad exit statuses giving me only 6 test cases) is looping through all the Js ignoring pairs that can't give smaller areas, but on something like an 80x80 square filled with Js and one empty slot, you're not going to finish in time. Interesting that bigsq is the only one of the nine problems without an analysis up.[/quote]\nMy program solves an 80x80 square with one empty slot instantly (0.03 seconds or so), so I guess it depends on your implementation.\nHeres mine, if you want to see it:\n[hide][code]\n#include \nusing namespace std;\nchar grid[100][100];\nint N;\nint type(int i, int j) {\n if (i<0 || j<0 || i>=N || j>=N || grid[i][j]=='B') return -1; \n if (grid[i][j]=='*') return 0;\n return 1;\n}\nint main() {\n ofstream fout (\"bigsq.out\");\n ifstream fin (\"bigsq.in\");\n fin >> N;\n for (int i=0; i> grid[i][j];\n int best = 0;\n for (int i=0; ibest && l?=(k-i)*(k-i)+(l-j)*(l-j);\n }\n }\n for (int l=N-1; (k-i)*(k-i)+(j-l)*(j-l)>best && l>=0; l--) {\n if (grid[k][l]!='J') continue;\n int x = type(i+j-l,j+k-i);\n int y = type(j+k-l,l+k-i);\n if (x!=-1 && y!=-1 && (x==1 || y==1)) {\n best>?=(k-i)*(k-i)+(l-j)*(l-j);\n }\n }\n \n }\n }\n fout << best << endl; \n return 0;\n}[/code][/hide][/quote]\r\n\r\nI'm not quite perfect at reading C++ code, but as I see it, you're looping through the points rather than the Js as I did, so when you get to (0,0) and (80,0), your program essentially terminates. Is this correct? Also, since you're looping through the points. What would be the worst case scenario? the bottom right three being Js and nothing else?\r\n\r\nAlso, could you explain the function of >?= please? I couldn't find it in a web search and it was used in the solution to the hungry cows (OCT06) problem, which I want to know how to do.\r\n\r\nLastly, the brand3 analysis was in java. So obviously if you're working in java, don't give up ;)", "Solution_12": "I am looping through the Js. Let me add a few comments:\r\n[code] for (int i=0; ibest && l?=, its some deprecated operator that probably isn't meant to be used, but I like it. a >? b returns the maximum of a and b, so a >?= b is just like saying a = max(a,b).", "Solution_13": "[quote=\"syboy\"]hi\ncan i go to silver division with this result?\n 1 2 3 4 5 6 7 8 9 10\nbrand3 * * * * * * * * t t\ncowtag * * * * * * * * * *\nscrbl * * * * * * * * * *\nif i can , how i do it? \n :?:[/quote]\r\n\r\nYou will probably automatically be invited. I think you get an email about it, but then in the next competition you will have access to the Silver problems instead.", "Solution_14": "syboy: You will either get an e-mail, or be listed in the results as being allowed to move up to silver or something. If you don't get any e-mail, but the contest results seem to indicate that you should be in silver, just send an e-mail to the Head Coach (Rob Kolstad), making sure to put USACO in the header, and he will probably take care of it. Your score is almost certainly high enough to move up to silver.\r\n\r\nAs for the >?= operator, as mentioned, a>?=b basically sets a to be the max of a and b, and ab?a:b; which isn't nearly as convenient to type.\r\n\r\nmahbub: What country are you from?\r\n\r\nDid anyone actually code up the heap for the blocks problem?", "Solution_15": "[quote=\"TripleM\"]I am looping through the Js. Let me add a few comments:\n[code] for (int i=0; ibest && l?=, its some deprecated operator that probably isn't meant to be used, but I like it. a >? b returns the maximum of a and b, so a >?= b is just like saying a = max(a,b).[/quote]\r\n\r\nAh. There's the difference. We're both looping through the Js, but you loop through the Js much more efficiently (I kept an array of the J positions and looped through that in a very slow manner whereas you go through each point checking if it's a J). I really don't feel like coding a new solution from scratch, but I might port your code to java sometime to see if it runs in time. Also, thanks for explaining what >?= does.\r\n\r\nSyboy: I've got the same opinion as everyone else. With only missing two test cases, you probably have a good chance of being invited to silver. If you really want to be invited and aren't, then send an email asking.", "Solution_16": "[quote=\"JSteinhardt\"]\nmahbub: What country are you from?\n[/quote]\n\nI am from Bangladesh. Small country in south Asia.\n\n[quote=\"JSteinhardt\"]\nDid anyone actually code up the heap for the blocks problem?\n[/quote]\r\n\r\nIn STL there is priority queue. but it works as max heap. but in problem we need min heap. so i inserted the number a as -a. and when popping, i just negated it :lol: So we actually dont need to implement heap if u r using C++.", "Solution_17": "That is a clever way to use the C++ heap. I am curious, though, as to why the solution times out even with a heap. Can you explain?", "Solution_18": "may be there is some mistake in your code.\r\n\r\nyou know, STL priority queue is much slower than self implimented heap. as STL heap got AC so implemented heap also should get AC.", "Solution_19": "ok \r\nthanks\r\n :idea:", "Solution_20": "I missed the qualification round.. so i had to start from bronze.. i submitted 2 and got all ok. but brand3 really hurt me, statement told that it can be upto 22,000,000 ; and i thought.. wow.. may be i need to use dp to find the first element to print... and i just left the contest right then.. in analysis mode, i wrote a simple backtracking with little pruning and that gets all tests okay, and the max was not 22,000,000 the max was 12,000,000 .. I was really heart broken... I still am heart broken...\r\n\r\nAnd you know the funny thing about cowtag.. if you write a nice soln with linked list - okay - you'll get all test case okay.. even if you write a nasty for loop ( 0..n )continuous going until there is nothing left, it will also get all test case okay.. :(", "Solution_21": "The basic rule is: if its bronze, there must be an 'easy' solution ;) I did exactly the same with brand3 for a while, but then realised everything I was trying was far too hard for bronze; lets try brute force.", "Solution_22": "Be careful with that though, while usually this is true, if you are worried about the time limit you might want to at least optimize a little and write tight inner loops.\r\n\r\nAlso, I recall at least three times when a bronze problem was dynamic programming.", "Solution_23": "I didn't do the qualification round either, so I also had to start with bronze. Unfortunately I had to leave in the middle of the contest, but I probably wasn't going to do good on it anyway, probably because I just started learning Java the few days before the contest (I don't normally use any of the languages you have to submit USACO answers in :() and also because I got confused on brand3, trying to figure out how to do an unknown number of nested for loops. Eventually I just did some weird recursion thing...", "Solution_24": "Yeah recursion would be the idea.\r\n\r\nOr you can be like McCutchen and use templates.\r\n\r\nBut that requires C++.\r\n\r\nThen again C++ is far better than JAVA." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "For an even positive integer $n$, let $S$ denote the set of natural numbers $a$, $1?", "Solution_4": "Actually, for us, it's riding a bus for about a day to Penn State. Also, there's a \"traditional\" visit to a certain small food court in the middle of nowhere along the way, where most Georgia ARML members are fond of sending pennies by the hundred down a device which the mall calls a \"hyperbolic funnel\". (The other members of the team and the coaches fail to understand this behavior.) It is also a tradition to go to a certain restaurant afterward, to get yelled at by Penn State staff for getting frisbees in odd places, to stay in a certain hotel on the way, to pull all-nighters playing ultimate frisbee at said hotel on the way back, to eat at Cracker Barrel on the way up, to practice in the hotel rooms the night before we get there, to complain about the freshman dorms we get at Penn State, etc., etc., etc.", "Solution_5": "i got one question\r\n\r\nwhy are you playin ultimate frisbeE??", "Solution_6": "[quote=\"Elemennop\"]i got one question\n\nwhy are you playin ultimate frisbeE??[/quote]\r\n\r\nDude, in general, about half our customs are down the drain. In addition, it's no more fun, because the number of people at UGA is probably going to be quite small (< 600... probably <300).", "Solution_7": "ultimate frisbee is a custom?\r\n\r\nroflwtf?", "Solution_8": "With our group, injuries during Ultimate Frisbee are the custom! :-)", "Solution_9": "So, does *anybody* from GA want to go to the GA ARML site? If we pay more, can we go to Penn State?", "Solution_10": "Yeah, I want to go to UGA and I'll tell you why. As a very successful ARML team, I belive that we have the obligation to help other, less successful, teams, especially in the south. Having a southern location will allow other southern teams more of a chance to compete. It will also allow us to take more than 2 teams if we so choose.\r\n\r\nI'm really going to miss going to Penn State and the ride up. (Not necessarily the ride back! :-) But this is probably better for ARML and the southern teams in the long run.\r\n\r\nRuby had a great suggestion today however. We could leave on Thursday, take a bus to Virginia, and return to UGA on Friday! At least we would get the bus practice time on the way up! :D", "Solution_11": "That would be fun indeed... but then we'd stay up all Thursday night :|.", "Solution_12": "lol that would certainly be interesting....and why would they pick georgia over say...a team from the south that has won recently (as in last year)", "Solution_13": "I feel sad about having to miss that bus ride...there was a bunch of team-building on the way up to Penn State... though I guess what's done is done.", "Solution_14": "[quote=\"captcha000\"]lol that would certainly be interesting....and why would they pick georgia over say...a team from the south that has won recently (as in last year)[/quote]\r\n\r\nNoo that would be bad. As it is, we might still be able to go to Penn State... Personally, I think it should have been in Texas. I mean if they added a site (which they shouldn't have done).", "Solution_15": "We should be able to choose where we go for the competition. That way we may be able to keep traditions.", "Solution_16": "If all of us were willing to pay for the additional cost, then we might convince the coaches to going to Penn State. Another alternative is for all of us to refuse to participate unless we go to Penn State. But, we'll probably give in before they do, especially since there are plenty of people who would still go even if it is only to UGA so that isn't a good idea.", "Solution_17": "[quote=\"Xantos C. Guin\"]If all of us were willing to pay for the additional cost, then we might convince the coaches to going to Penn State. Another alternative is for all of us to refuse to participate unless we go to Penn State. But, we'll probably give in before they do, especially since there are plenty of people who would still go even if it is only to UGA so that isn't a good idea.[/quote]\r\n\r\nLol, but they won't win probably even top 10... anyways... it'll look really corny if Georgia doesn't go to UGA while about 5 other states which are farther away go to UGA. :D", "Solution_18": "who cares how it looks? you guys should do wahtever you want t odo", "Solution_19": "Hmmm guys. We are basically the hosts. It would be really strange for the hosts to go somewhere else!", "Solution_20": "i hate to sound cliche / corny, but of course uga will be no fun with a negative attitude. i may be biased because I've only gone to penn state once and played no ultimate, but i think uga may hold new opportunities...", "Solution_21": "...or maybe you are biased because you won UGA?", "Solution_22": "[quote=\"Criticalline1859\"]...or maybe you are biased because you won UGA?[/quote]\r\n :lol: \r\narml at uga really doesn't have much to do with the uga tournament except that they are at the same place (and i don't really believe that uga gives me good luck)." } { "Tag": [ "calculus", "integration", "floor function", "algebra", "function", "domain", "calculus computations" ], "Problem": "\\[ \\int\\limits_{ \\minus{} 2}^2 {\\left[ {x^2 \\minus{} 1} \\right]dx} \r\n\\]\r\n\r\n$ {\\left[ x \\right]}$=greatest integer less than equal to $ x$", "Solution_1": "To do this integral, we should break it into peices over which the integrand is constant.\r\n\r\n$ x^2\\minus{}1$ ranges from $ 3$ to $ \\minus{}1$ over the interval.\r\n\r\n$ x^2\\minus{}1\\equal{}3 \\implies x\\equal{} \\pm 2$\r\n$ x^2\\minus{}1\\equal{}2 \\implies x \\equal{} \\pm \\sqrt{3}$\r\n$ x^2\\minus{}1\\equal{}1 \\implies x\\equal{} \\pm \\sqrt{2}$\r\n$ x^2\\minus{}1\\equal{}0 \\implies x\\equal{} \\pm 1$\r\n$ x^2\\minus{}1\\equal{}\\minus{}1 \\implies x\\equal{}0$\r\n\r\nSo $ \\lfloor x^2\\minus{}1 \\rfloor$ is $ 2$ for $ x \\in (\\minus{}2,\\minus{}\\sqrt{3}) \\cup (\\sqrt{3},2)$, $ 1$ for $ x \\in (\\minus{}\\sqrt{3},\\minus{}\\sqrt{2})\\cup(\\sqrt{2},\\sqrt{3})$, $ 0$ for $ x \\in (\\minus{}\\sqrt{2},\\minus{}1)\\cup(1, \\sqrt{2})$ and $ \\minus{}1$ for $ (\\minus{}1,1)$. Putting this all together,\r\n\r\n$ \\int_{\\minus{}2}^2 \\lfloor x^2\\minus{}1 \\rfloor dx \\equal{} 2[2(2\\minus{}\\sqrt{3})]\\plus{}1[2(\\sqrt{3}\\minus{}\\sqrt{2})]\\minus{}1(2)$\r\n$ \\equal{} 8\\minus{}4\\sqrt{3} \\plus{} 2\\sqrt{3}\\minus{}2\\sqrt{2}\\minus{}2 \\equal{} 6\\minus{}2\\sqrt{3}\\minus{}2\\sqrt{2}$", "Solution_2": "First question to ask on seeing a symmetric domain, $ \\int_{\\minus{}a}^af(x)\\,dx:$\r\n\r\nIs the function being integrated even, odd or neither? And if neither, can the decomposition into even and odd parts be accomplished painlessly?\r\n\r\nIn this case, the function, $ \\lfloor x^2\\minus{}1\\rfloor,$ is even. It's even because it's some arbitrary thing (the floor function) done to an even function. The the thing we do with the integral of an even function is \"fold it over.\" Write what we want as\r\n\\[ 2\\int_0^2\\lfloor x^2\\minus{}1\\rfloor\\,dx\\]\r\nThe rest proceeds as facis has done, only with the \"fold-over,\" we don't need as many cases." } { "Tag": [], "Problem": "MR.SEIBOLD HAS 6 DAUGHTERS.\r\n\r\nEACH DAUGHTER IS 4 YEARS OLDER THAN THE YOUNGEST SISTER.\r\n\r\nTHE OLDEST DAUGHTER IS 3TIMES AS OLD THAN HER YOUNGER SISTER.\r\n\r\nHOW OLD IS EACH OF THE THE DAUGHTER.\r\n\r\nI AM NOT GETTING IT HOW TO DO. :huh:", "Solution_1": "Solution: I think you mean \"THE OLDEST DAUGHTER IS 3 TIMES AS OLD AS HER [b]YOUNGEST[/b] SISTER\" \r\nSo if we use algebra to do this, it'll be like this:\r\nYoungest daughter's age: $ x$\r\nSecond daughter's age: $ x\\plus{}4$\r\nThird daughter's age: $ x\\plus{}8$\r\nFourth daughter's age: $ x\\plus{}12$\r\nFifth daughter's age: $ x\\plus{}16$\r\nOldest daughter's age: $ x\\plus{}20$\r\n\r\nSince it states that the oldest daughter's age is 3 times the youngest daughter's age, we can set up an equation: $ 3x\\equal{}x\\plus{}20$ So by using simple algebra, we isolate the $ x$, (Subtract $ x$ from both sides) and we get $ 2x\\equal{}20$ Now we divide both sides by $ 2$ and we get $ x\\equal{}10$!!!! \r\nSo our ages would be\"\r\n$ 10$\r\n$ 14$\r\n$ 18$\r\n$ 22$\r\n$ 26$\r\n$ 30$\r\n\r\nHope this helped. :lol:", "Solution_2": "we have x is the youngest one. And x+20 as the oldest. \r\nx+20=3x\r\n20=2x\r\nx=10\r\n10, 14, 18, 22, 26, 30 are the ages of the sisters", "Solution_3": "thank you. now I understood it better." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "incenter", "perpendicular bisector", "geometry unsolved" ], "Problem": ":( :( :(", "Solution_1": "Circumcircle $ (N)$ of $ \\triangle FM_aM_b$ is the 9-point circle of $ \\triangle ABC.$ Construct this circumcircle. $ M_c \\in (N)$ is the unknown midpoint of $ AB.$ Then $ \\overline {FM_a} + \\overline {FM_b} + \\overline {FM_c} = \\pm \\frac {r}{IO} [(b - c) + (c - a) + (a - b)] = 0.$ This is sum of directed segments and it has to be decided, which ones have to be taken positive / negative. For a proof, see\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=100242[/url],\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=85415[/url], etc.\r\n\r\nConstruct 2 concentric circles $ \\mathcal F_1, \\mathcal F_2$ with common center $ F$ and with radii $ r_1 = |FM_a - FM_b|$ and $ r_2 = FM_a + FM_b.$ WLOG, suppose $ FM_a < FM_b,$ as in your sketch. Draw circle with center $ F$ and radius $ FM_a,$ intersecting the ray $ \\overrightarrow{FM_b}$ at $ P_1$ and the opposite ray at $ P_2.$ Reflect $ P_1, P_2$ in the perpendicular bisector of $ FM_b$ into $ Q_1, Q_2.$ Then $ FQ_1, FQ_2$ are radii of the circles $ \\mathcal F_1, \\mathcal F_2.$\r\n\r\nThe smaller circle $ \\mathcal F_1$ always intersects $ (N)$ at 2 points; select the one on the larger of the arcs $ FM_a, FM_b$ of $ (N)$ and label it $ M_c.$ The larger circle $ \\mathcal F_2$ may or may not intersect $ (N)$ at 2 points; if it does, label them $ M_c', M_c''.$ Apart from special cases (such as isosceles $ \\triangle ABC$), this yields either one solution $ M_c$ or 3 solutions $ M_c, M_c', M_c''.$ Consider only one, say $ M_c.$ $ \\triangle M_aM_bM_c$ is medial triangle of $ \\triangle ABC.$ Parallels to $ M_bM_c, M_cM_a, M_aM_b$ through $ M_a, M_b, M_c,$ respectively, pairwise intersect at $ A, B, C.$\r\n\r\nAs a check, construct incenter $ I$ of the $ \\triangle ABC$ and make sure it falls on the segment $ NF.$", "Solution_2": "thanks :lol:" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "If H is a subgroup of finite index in a group G, then H contains a subgroup N which is of finite index and normal in G.", "Solution_1": "I think this has been solved twice (at least :)) on the forum.", "Solution_2": "[quote=\"eugene\"]If H is a subgroup of finite index in a group G, then H contains a subgroup N which is of finite index and normal in G.[/quote]\r\n\r\nWhat does \"finite index\" mean?", "Solution_3": "The set of left cosets of $H$ has finitely many elements.\r\n\r\nP.S. I know you know what it means, liyi, because I've seen you use it :). The index of $H$ in $G$ is $[G:H]$ (you used this notation in some problems you have recently posted).", "Solution_4": "Hmm. Thanks for your reply. Now I know what it refers to... I didn't know that because I'm reading a Chinese book and I don't match the terms in different languages.", "Solution_5": "grobber... could you please give me the link of that proof? thx :) or a hint", "Solution_6": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22796]Here's something[/url], although it's not fully detailed there. There are other threads, I think.", "Solution_7": "Aww well I remember this problem very well. I have been posting it for the last few days but without answers.Only I get the reply that this problem has already been discussed before.Can anyone here give me the link of the proof.", "Solution_8": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=beautiful&t=19724[/url]. Are we all happy now? :)", "Solution_9": "Thanks a lot, grobber.That was very illuminating." } { "Tag": [ "geometry", "Mafia", "algebra", "polynomial", "articles" ], "Problem": "Those (e. g. Santosh and Karthik) who have been at this program, could I get your comments please? Any information would be helpful, including format of classes, daily schedule, free time activities, paths from building to building with the most shade...", "Solution_1": "Governor's Honors ('07 here, wootz) was basically the greatest summer ever. You spend four hours every morning (8:00-12:00) Monday-Saturday in your major, which I'm just assuming is math... you also select a minor area once there, which you'll go to for two-and-a-half hours after lunch Monday-Friday. The rest of the time is basically yours until 10:30, at which time you'll have to be on your hall and your RA will conduct a hall check. They hold random seminars every day at all hours all throughout the week, anything from playing Guitar Hero in a lecture hall to making and eating baklava, playing large group games like Mafia, whatever, and you'll be stuck at VSU with 700 of the most awesome people in Georgia, so you'll always have something to do. The different majors will also hold events and concerts the entire time you're there, which you should not miss out on for any reason. There's no excuse for not seeing the best musicians in the state for free. Some of the classes are mostly lecture format, and some are based more around individual and group problem solving and the like. You'll also take a programming course, different ones depending on your abilities. The professors are awesome and brilliant teachers. You're bound to learn something about something nearly every day.\r\n\r\nIt'll be hot. Shade is scarce, and gnats are plentiful. The food is apparently better than it was in the summers before I was there, but it's still hit or miss. There's always a salad bar, a sandwich bar, and amazing frozen yogurt in addition to the main meal, though, and there's a small store where you can buy snacks and stuff, not to mention the pizza place that's a little pricey, but great food. Your whole hall will share only a handful of showers, so there may be some waiting in line going on. You'll also have to keep your room relatively neat, as your RA will check it often. The staff don't have too many rules, and they're not hard to follow, but they enforce them strictly, and breaking them repeatedly may get you sent home early. Happened to two or three people while I was there, so don't be like them. All of the staff are great people, and most of them have been in your shoes before, participated in GHP themselves, so they know everything that you'll be going through, good or bad. They'll always be there to help out if you ever need anything.\r\n\r\nI could ramble all day about GHP, so I'll cut it out here. Make sure you enjoy it to the fullest, because you'll spend the next year or so wishing you could go back. And by the way, congratulations.", "Solution_2": "[hide]?? HWat?\n?[/hide]", "Solution_3": "Hmm, DannyKim, is it just me, or are all of your posts pointless and spammy and completely nonsensensical? :wink:", "Solution_4": "who is \"Criticalline1859\" and \"DaNnYkIm\"?", "Solution_5": "Hi, Boru! How's your GHP going? (Or is it started yet?)\r\n\r\nI think Criticalline1859 is Andrei. I don't know the other one, but I'm sure he is not from Georgia.", "Solution_6": "Hence his location saying \"Tallahassee, FL\" :P !\r\n\r\nPS: Hi, Boru, Tim, and maybe-Andrei!", "Solution_7": "Same from me!\r\n\r\nI didn't get into GHP because I'm not beast/balanced like that.", "Solution_8": "Yeah, I'm Andrei (my username is 50% reference to math stuff-Riemann Hypothesis, by the way, 30% not thinking of anything else, and 20% a failed attempt at poking fun at other usernames). The second week of GHP is coming to an end, and Boru will not tell me how his GHP is going so I can put it into my post and be efficient (we're in the same computer lab working on a project).", "Solution_9": "Hey Andrei, Tianqi, and Mathgeek. \r\n\r\nHow are you?\r\n\r\nGHP is pretty fun...it is slightly slower than I would like, but we're still learning lots :)\r\n\r\nFor example, I learned that i^2 = -1\r\n\r\n\r\n(Who is JohnJimJoeBob?)", "Solution_10": "[quote=\"llolloll\"]Hey Andrei, Tianqi, and Mathgeek. \n\nHow are you?\n\nGHP is pretty fun...it is slightly slower than I would like, but we're still learning lots :)\n\nFor example, I learned that i^2 = -1\n\n\n(Who is JohnJimJoeBob?)[/quote]\r\n\r\nUm, I'm Sitan. :wink: \r\n\r\nWait, you didn't know that i^2=-1???????????????????????? :rotfl: \r\n\r\nSarcasm is hard to detect, you know...", "Solution_11": "Boru is taking a complex numbers class (mainly about the connections to analytical geomety, I think), hence the joke.", "Solution_12": "I have a sudden urge to spam", "Solution_13": "Sitan, I guess I did know that i^2= -1....\r\n\r\nWho is Iniquitus?\r\n\r\nBy the way, since this is a GHP forum, maybe I should post some GHP problems here?", "Solution_14": "I'm pretty sure Iniquitus is Oliver the |_|83R 1337 beaaaaaaaaaaaaaaaaaast.", "Solution_15": "[quote=\"Arvind_sn\"]It's just you. :wink:[/quote]\r\n\r\nAh... That explains the fact that at least the first 10 pages of your posts from your profile are from GFF. :P \r\n\r\nHmm, this thread is getting quite spammy and non-GHP related. :ninja:", "Solution_16": "Hi guys I'm Eric Chen\r\n\r\nI'm at GHP with Andrei and Boru", "Solution_17": "[quote=\"Arvind wrote\"]He is andrei's long lost twin.[/quote]\r\n?\r\nAnyway, here is something I found funny.\r\n[url]http://en.wikipedia.org/wiki/Proof_by_intimidation[/url]", "Solution_18": "Hey Eric,\r\n\r\nIt's nice to meet you. Since this is a GHP forum, maybe you would like to share some GHP (math) problems?\r\n\r\n\r\nHey Andrei,\r\nThat article was pretty funny.", "Solution_19": "Please choose another name, chenxiyuan. \"Eric\" has already been claimed. Thanks.", "Solution_20": "[quote=\"Criticalline1859\"][quote=\"Arvind wrote\"]He is andrei's long lost twin.[/quote]\n?\nAnyway, here is something I found funny.\n[url]http://en.wikipedia.org/wiki/Proof_by_intimidation[/url][/quote]\r\n\r\nArvind wrote wrote?\r\n\r\nAnyways, that means that half of Karthik's proofs are proof by intimidation. :| :rotfl: \r\n\r\nAlso, @Eric M., yeah I thought he was talking to you for a sec, and I'm like, wait, Eric hasn't posted much in this thread, and then I'm like, oh, yeah, Eric Chen. :P", "Solution_21": "Now I know I shouldn't post at 10:00.\r\nBut seriously, Karthik's ability to see through those long series of steps is pretty impressive.", "Solution_22": "Fortunately for mere mortals, the methods in his proofs by intimidation are mostly unnecessary - he makes problems much harder than they need to be :).", "Solution_23": "Mathematicians have long judged proofs not only by their correctness, but also by their elegance. An \"ugly\" proof will frequently be dismissed purely because it is ugly.", "Solution_24": "Hi Mr. Fulton!!!!!!!!!\r\n\r\nAnyways, ugly proofs would include computer proofs, right? Like the Four Color Theorem's proof? Well, sometimes, the problem is too complicated to call for an elegant proof, so we have no choice.\r\n\r\nUnless if, of course, we're talking about Karthik's proofs. :| :rotfl: jkjkjkjkjkjkjkjk", "Solution_25": "Hi Sitan,\r\n\r\nI don't believe that all computer proofs are necessarily ugly. The four color map proof was certainly a unique proof at the time. I don't know if anyone has given any better proof even 30 years later.\r\n\r\nTom", "Solution_26": "Hi,\r\n\r\nDoes anyone know how to do that problem where we have a 4 digit number and each digit is less than or equal to the digit to its left...well, does anyone know how many such numbers there are? Can this be done without casework?", "Solution_27": "Yes, I'm sure that someone knows how to do that problem, and that someone knows how many numbers there are. :wink: \r\nYes, it can be done without casework. \r\n\r\n[hide=\"hint\"]Try picking which digits you are going to use before arranging them in order.[/hide]", "Solution_28": "[hide]\nFirst note that each group of chosen digits can be arranged in exactly one way. Let $ x_i$ denotes the number of digit $ i$ used. We have\n\\[ x_0\\plus{}...\\plus{}x_9\\equal{}4\n\\]\nBy balls-and-urns, this equation has $ \\binom{13}{4}\\equal{}715$ solutions, so there are $ 715$ such 4-digit numbers. But we also counted $ 0000$, so the answer is $ 714$.\n[/hide]", "Solution_29": "Thanks Tim :)" } { "Tag": [ "\\/closed" ], "Problem": "I seem to have lost my ability to post an avatar. My other account can, but this account has lost the whole avatar control panel. The whole control panel disappeared. Any chance an admin may have removed it? \r\n\r\nThanks.", "Solution_1": "go to your profile and check if you enabled avatars?" } { "Tag": [ "modular arithmetic", "superior algebra", "superior algebra unsolved" ], "Problem": "Hi, please help!\r\n\r\nB(subscript42) = {0,1,7,15,21,22,28,36}. I need to find all atoms of B(subscript42). \r\n\r\nI understand that all of these are (mod42). Don't know what do to next!\r\n\r\nMany thanks in advance!", "Solution_1": "What is an atom?", "Solution_2": "An atom of a Boolean algebra is a nonzero element a that \r\ncannot be written in the form \r\n\r\na = b $ \\vee$ c with\r\n\r\na $ \\neq$ b and\r\n\r\na $ \\neq$ c.", "Solution_3": "Hello again! \r\n\r\nAre there any other ideas as to how solve this?\r\n\r\nMany thanks!", "Solution_4": "Consider divisibility by each of the prime factors of 42. Each such prime will appear in a product if and only if it appears in one of the factors.", "Solution_5": "Thank you!\r\n\r\nSo, the prime factors of 42 are: 2, 3, 7\r\n\r\nForgot to mention that B(subscript42) is a set of idempotents. \r\n\r\nSo, B(subscript42) = {0,1,7,15,21,22,28,36}\r\n\r\n0 divides 2,3,7\r\n1 does not divide any of the prime factors of 42\r\n7 divides 7\r\n15 divides 3\r\n21 divides 3 and 7\r\n22 divides 2\r\n28 divides 2 and 7\r\n36 divides 2 and 3\r\n\r\nSo, which ones are the atoms of B(subscript42)? \r\n\r\nI could not quite understand this: \"Each such prime will appear in a product if and only if it appears in one of the factors\". Does it mean that the only atom is 0?\r\n\r\nThanks!", "Solution_6": "Quite the opposite: 0 is not an atom since $ 7\\cdot 36\\equal{}0$. As 7 is divisible by 7 and 36 is divisible by 2 and 3, their product is divisible by 42 so must be 0.\r\n\r\nIn a similar way, $ 36\\equal{}22\\cdot 15$ is not an atom, but 7 is.", "Solution_7": "Hi, sorry but I am struggling to understand it.\r\n\r\nI have just done all 64 multiplications.\r\n\r\nHow do we define an atom?\r\n\r\n7*36=0 is not an atom because it divides 2,3,7 and its product is divisible by 42\r\n\r\nHow about 7*15=21 i.e. 7/7, 15/3, 21/ (7 and 3), but not 2. So understand it's an atom, as the product is not divisible by 42?\r\n\r\nOr 7*28=28, so 7/7, 28/ (2 and 7), 28 / (2 and 7), but not 3 - is it again an atom?\r\n\r\nOr 7*22=28, so 7/7, 22/2, 28/ (7 and 2), but not 3 - it is an atom. \r\n\r\nAll these products are divisible by either 7,2,3 or all of them.\r\n\r\nBoolean algebra is a new territory for me. So, apologies, that it's taking me a long time to understand it. \r\n\r\nThanks for your help!", "Solution_8": "Umm... [i]you[/i] defined an atom for us, by giving the definition. An atom is a number $ a$ that can't be written as a product of two elements other than $ a$. So:\r\n\r\n- 0 is not an atom because it equals $ 7\\cdot 36$ and neither 7 nor 36 is equal to 0.\r\n\r\n- Saying that $ 7\\cdot 15\\equal{}21$ and that $ 7\\cdot 22\\equal{}28$ means 21 and 28 aren't atoms either, since neither 7 nor 15 equals 21 and neither 7 nor 22 equals 28.\r\n\r\n- Saying that $ 7\\cdot 28\\equal{}28$ doesn't say anything about whether something is an atom. You wrote 28 as a product of two numbers, but one of them is 28. (But we know 28 isn't an atom from the above paragraph.)\r\n\r\nIn this case, I think I should also explain why 7 is an atom. This is more complicated because you can't just use a single equation to verify this.\r\n\r\nYou could go through the entire multiplication table. That's a lot of work, but if you really did multiply every pair of numbers, you can check that the only times you get 7 as a result of multiplying two numbers is when at least one of the numbers is 7.\r\n\r\nHowever, a more elegant way is this: Suppose $ 7\\equal{}bc$ where $ b,c$ are elements of $ B_{42}$. Because $ bc\\equiv 7\\pmod{42}$, $ bc$ is divisible by 7 so either $ b$ or $ c$ is.\r\n\r\nWLOG say $ b$ is divisible by 7, so we want to show $ b\\equal{}7$. If $ b$ were not equal to 7, then $ b$ would be divisible by either 2 or 3 (or both). But then so would their product (i.e. if $ b$ were even then $ bc$ modulo 42 would be even), so we would not get a product of 7, a contradiction. Therefore, $ b\\equal{}7$. So the only factorizations of 7 are the \"trivial\" ones where at least one element equals 7, making 7 an atom." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "find f(n) if :\r\nf(0)=1,f(1)=2 , f(2)=3 , f(4)=5 and f(n+4)=4f(n+3)-6f(n+2)+4f(n+1)-f(n) for n=0 , 1 , 2 , ...", "Solution_1": "i think you meant to say $ f(3)\\equal{}5$, not $ f(4)$.\r\n\r\nthe characteristic polynomial of the recurrence $ t^4\\minus{}4t^3\\plus{}6t^2\\minus{}4t\\plus{}1\\equal{}(t\\minus{}1)^4$, whose only root, 1, has multiplicity 4. the solution thus takes the form $ f(n)\\equal{}c_0\\plus{}c_1n\\plus{}c_2n^2\\plus{}c_3n^3$. to get the $ c_i$, use the initial values. the solution is $ f(n)\\equal{}1\\plus{}\\frac{4}{3}n\\minus{}\\frac{1}{2}n^2\\plus{}\\frac{1}{6}n^3$." } { "Tag": [], "Problem": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03cc\u03bb\u03bf\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03b1\u03bd\u03c4\u03ae\u03c3\u03b5\u03c4\u03b5 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03b7\u03bd \u03c4\u03b7\u03bd \u03b4\u03b7\u03bc\u03bf\u03c3\u03ba\u03cc\u03c0\u03b7\u03c3\u03b7 \u03bc\u03b5 reply \u03b3\u03b9\u03b1 \u03c4\u03bf \u03cc\u03bd\u03bf\u03bc\u03b1 \u03ba\u03b9 \u03c4\u03b7\u03bd \u03bf\u03bc\u03ac\u03b4\u03b1 \u03c3\u03b1\u03c2 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c8\u03ae\u03c6\u03bf \u03c3\u03b1\u03c2 \u03ba\u03b1\u03bb\u03cc \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b3\u03bd\u03c9\u03c1\u03b9\u03c7\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03bf \u03ad\u03bd\u03b1\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03ac\u03bb\u03bb\u03bf \u03ba\u03b9 \u03c4\u03b9 \u03ba\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03bf \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03bf\u03bc\u03ac\u03b4\u03b1 \u03c3\u03b1\u03c2\r\n :rotfl: :rotfl: :D \u03bf\u03ba \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03b5\u03ce \u0395\u03af\u03bc\u03b1\u03b9 \u03bf\u03c0\u03b1\u03b4\u03cc\u03c2 \u03c4\u03bf\u03c5 \u03b4\u03b9\u03ba\u03ad\u03c6\u03b1\u03bb\u03bf\u03c5 \u03b1\u03b5\u03c4\u03bf\u03cd \u03c0\u03b5\u03c1\u03b9\u03bc\u03ad\u03bd\u03c9 \u03b1\u03c0\u03b1\u03bd\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c3\u03b1\u03c2", "Solution_1": "\u03a4\u03bf \u03b5\u03af\u03c7\u03b1 \u03b4\u03b5\u03b9 \u03bd\u03b1 \u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 ... :D\r\n\r\n\u0391\u03b5\u03ba\u03ac\u03c1\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 , \u03b4\u03b5\u03bd \u03c0\u03bf\u03bb\u03c5\u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bc\u03b1\u03b9 \u03cc\u03bc\u03c9\u03c2 :) \r\n\u03a4\u03bf\u03c5 \u03b2\u03bf\u03c1\u03c1\u03ac \u03ae \u03c4\u03bf\u03c5 \u03bd\u03cc\u03c4\u03bf\u03c5 \u0391\u03bc\u03b2\u03c1\u03cc\u03c3\u03b9\u03b5 ??? :P", "Solution_2": "\u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03cc\u03c2 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03c4\u03bf \u03ba\u03cc\u03ba\u03ba\u03b1\u03bb\u03bf...... :D", "Solution_3": "[u][b]PANATHINAIKOS FOOTBALL CLUB[/b][/u]\r\n\r\n[hide]Na lynoume omws kai kamia askisi pou kai pou.......[/hide]", "Solution_4": "\u03a0\u03b1\u03bd\u03b1\u03b8\u03b7\u03bd\u03b1\u03b9\u03ba\u03bf\u03c2\u03c2\u03c2 :omighty: ... :D", "Solution_5": "\u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03bf\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf\u03b9 \u03c3\u03c4\u03b1 \u03c0\u03c1\u03bf\u03c6\u03bf\u03c1\u03b9\u03ba\u03ac \u03c0\u03ad\u03c1\u03b9\u03c3\u03b9 \u03ae\u03c4\u03b1\u03bd [color=green]\u03a0\u0391\u039f[/color]", "Solution_6": "O\u03c0\u03b1 sorry +\u03b5\u03b3\u03c9 \u03c8\u03b7\u03c6\u03b9\u03c3\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b5\u03c3\u03c4\u03b5\u03b9\u03bb\u03b1 \u03bc\u03bd\u03bc.....\u0395,\u03bc\u03b7\u03bd \u03bb\u03b5\u03bc\u03b5 \u03c4\u03b1 \u03b9\u03b4\u03b9\u03b1 ...\u039f\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bf\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03b9\u03bf\u03b9 \u03ae \u03ba\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03bf\u03c4\u03b5\u03c1\u03bf\u03b9...(\u03b2\u03bb,results...)\r\n\r\n :D :D :D", "Solution_7": "\u0395\u03b9\u03bc\u03b1\u03b9 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03bf\u03c2 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03c5\u03bc\u03b1\u03b9 \u03c0\u03bf\u03bb\u03c5 \u03bc\u03b5 \u03c0\u03bf\u03b4\u03bf\u03c3\u03c6\u03b1\u03b9\u03c1\u03bf!\u0398\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03bf\u03c5\u03c3\u03b1 \u03bd\u03b1 \u03be\u03bf\u03b4\u03b5\u03c8\u03c9 \u03c4\u03bf\u03bd \u03c7\u03c1\u03bf\u03bd\u03bf \u03bc\u03bf\u03c5 \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03c4\u03b9\u03b4\u03b1 \u03c3\u03ba\u03b1\u03ba\u03b9 \u03b7 \u03b5\u03bd\u03b1\u03bd \u03b1\u03b3\u03c9\u03bd\u03b1 \u03bc\u03c0\u03b9\u03bb\u03b9\u03b1\u03c1\u03b4\u03bf\u03c5....... :)", "Solution_8": "\u039c\u03b5\u03b3\u03ac\u03bb\u03c9\u03c3\u03b1 \u03c3\u03c4\u03bf \u039a\u03b5\u03c1\u03b1\u03c4\u03c3\u03af\u03bd\u03b9 \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03b5\u03af\u03bc\u03b1\u03b9 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03ba\u03cc\u03c2, \u03b1\u03bb\u03bb\u03ac \u03c0\u03c1\u03ce\u03b7\u03bd, \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bc\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03bf\u03b4\u03cc\u03c3\u03c6\u03b1\u03b9\u03c1\u03bf. \u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03bc\u03b5\u03c4\u03c1\u03ac\u03b5\u03b9 \u03b1\u03c5\u03c4\u03cc, \u03c3\u03c4\u03b7 \u03b4\u03b7\u03bc\u03bf\u03c3\u03ba\u03cc\u03c0\u03b7\u03c3\u03b7.\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_9": "\u039f\u03bb\u03b1 \u03bc\u03b5\u03c4\u03c1\u03ac\u03bd\u03b5 \u03b1\u03b3\u03b1\u03c0\u03ae\u03c4\u03bf\u03af \u03bc\u03bf\u03c5 \u03c6\u03af\u03bb\u03bf\u03b9\r\n\u0396\u03ae\u03c4\u03c9 \u03b7 \u03c4\u03b9\u03bc\u03b7\u03bc\u03ad\u03bd\u03b7 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ae \u03ba\u03b9 \u03bf\u03b9 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03a0\u03b5\u03b9\u03c1\u03b1\u03b9\u03ac \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03c4\u03b9\u03bc\u03b7\u03bc\u03b5\u03bd\u03b7 \u03c6\u03c4\u03c9\u03c7\u03bf\u03bb\u03bf\u03b3\u03b9\u03ac \u03c4\u03bf \u03bc\u03b5\u03b4\u03bf\u03cd\u03bb\u03b9 \u03c4\u03bf\u03c5 \u03bb\u03b1\u03bf\u03cd \u03bc\u03b1\u03c2", "Solution_10": "\u03b6\u03b7\u03c4\u03c9 \u03b7 \u03bc\u03c0\u03b1\u03bb\u03bb\u03b1 \u03b1\u03ba\u03b9 \u03b6\u03b7\u03c4\u03c9 \u03b7 \u0391\u0395\u039a.\u03a0\u0391\u039c\u0395 \u0393\u0399\u0391 \u03a4\u039f \u0394\u0399\u03a0\u039b\u039f \u0391\u039c\u0392\u03a1\u039f\u03a3\u0399\u0395", "Solution_11": "\u039f\u039b\u03a5\u039c\u03a0\u0399\u0391\u039a\u039f\u03a3- \u0391\u0395\u039a 1-0 :lol:", "Solution_12": "[quote=\"silouan\"]\u039f\u039b\u03a5\u039c\u03a0\u0399\u0391\u039a\u039f\u03a3- \u0391\u0395\u039a 1-0 :lol:[/quote]\r\n\r\n\u039d\u03b1\u03b9 \u03b5\u03af\u03b4\u03b1\u03bc\u03b5 \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03bc\u03c0\u03ac\u03bb\u03b1 \u03c0\u03ac\u03bb\u03b9 :D", "Solution_13": "\u039d\u0391\u0399 \u03a6\u0399\u039b\u039f\u0399 \u039c\u039f\u03a5 \u0391\u039b\u039b\u0391 \u03a4\u0395\u03a4\u039f\u0399\u0391 \u039d\u0399\u039a\u0397 \u039c\u0395 \u0391\u03a5\u03a4\u039f\u0393\u039a\u039f\u039b \u039a\u0399 \u03a3\u0391\u03a3 \u03a0\u0391\u0399\u0396\u0391\u039c\u0395 \u039c\u039f\u039d\u039f\u03a4\u0395\u03a1\u039c\u0391 \u039a\u0399 \u039c\u0391\u03a3 \u0391\u0394\u0399\u039a\u0397\u03a3\u0395 \u039f \u0392\u0391\u03a3\u0391\u03a1\u0391\u03a3 \u039d\u0391 \u03a4\u039f \u03a7\u0391\u0399\u03a1\u0395\u03a3\u03a4\u0395 \u03a4\u039f \u03a0\u03a1\u03a9\u03a4\u0391\u0398\u039b\u0397\u039c\u0391 \u03a3\u0391\u03a3 \u039a\u0399 \u03a4\u0397\u039d \u03a3\u039f\u03a5\u03a0\u0395\u03a1 \u039a\u0399\u039b\u039a\u0391 \u0395 \u03a3\u03a5\u0393\u039d\u03a9\u039c\u0397 \u039b\u0399\u0393\u039a\u0391", "Solution_14": "\u03a0\u03bf\u03af\u03bf\u03c2 \u039c\u0391\u0393\u039a\u0391\u03a3 \u03a3\u03c0\u03ac\u03b5\u03b9 \u03c4\u03bf \u03ba\u03b1\u03c4\u03b5\u03c3\u03c4\u03b7\u03bc\u03ad\u03bd\u03bf \u03ba\u03b9 \u03b4\u03b9\u03ac\u03bb\u03b5\u03be\u03b5 \u03c4\u03bf\u03bd \u03c4\u03b9\u03bc\u03b7\u03bc\u03ad\u03bd\u03bf \u0395\u03c1\u03b3\u03bf\u03c4\u03ad\u03bb\u03b7 ????\r\n\u03a6\u03b1\u03bd\u03c4\u03ac\u03b6\u03bf\u03bc\u03b1\u03b9 \u03bf \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2 \u03bf \u039a\u03c1\u03b7\u03c4\u03b9\u03ba\u03cc\u03c2 \u039c\u03c0\u03c1\u03ac\u03b2\u03bf \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c0\u03bf\u03c5 \u03c4\u03bf \u03b5\u03ba\u03b1\u03bd\u03b5 \u03ba\u03b9 \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03c0\u03b5\u03af \u03b4\u03b7\u03bc\u03cc\u03c3\u03b9\u03b1", "Solution_15": "\u0395\u03c1\u03b3\u03bf\u03c4\u03b5\u03bb\u03b5\u03c1\u03b1!!! \u03bc'\u03b1\u03c1\u03b5\u03c3\u03b5\u03b9. \u03ba\u03b1\u03bd\u03b5\u03bd\u03b1\u03c2 \u03a0\u03b1\u03bf\u03ba\u03c4\u03b6\u03b7\u03c2? \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9? \u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03b5 \u03ba\u03b1\u03bb\u03bf \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03b9\u03b4\u03b9 \u03c7\u03b8\u03b5\u03c2 \u03b1\u03bb\u03bb\u03b1 \u03b7 \u03b9\u03c3\u03bf\u03c0\u03b1\u03bb\u03b9\u03b1 \u03b4\u03b5 \u03b2\u03bf\u03bb\u03b5\u03c5\u03b5\u03b9.", "Solution_16": "Pragmati egw ipostirizw ton Ergoteli (kai amesws meta fisika ton OFH- parolo pou mas ta xei kanei mantara apo tote pou efige o Gerald(akis)). Den asxoloumai sxedon katholou me mpala. Eimai omws tis apopsis na ipostirizoume tis omades tou topou mas... Diladi apo poion prepei na perimenei o Ergotelis na ton ipostiriksei? Apo ton Serraio 'h ton Salonikio? (Doksa tw thew iparxei kai o Panserraikos, o Paok, o Aris, h Kalamaria...) \r\n\r\nAlexandros", "Solution_17": "\u03b5\u03c7\u03b5\u03b9 \u03b4\u03b9\u03ba\u03b9\u03bf \u03bf \u0391\u03bb\u03b5\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2 .\u03ba\u03b1\u03b9 \u03b5\u03b3\u03c9 \u03c5\u03c0\u03bf\u03c3\u03c4\u03b7\u03c1\u03b9\u03b6\u03c9 \u03b4\u03c5\u03bf \u03bf\u03bc\u03b1\u03b4\u03b5\u03c2 (\u03b7 \u03bc\u03b9\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf \u03a0\u0391\u039f). \u03b7 \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03c6\u03b1\u03bd\u03b1\u03c4\u03b9\u03ba\u03bf\u03c2 \u03b4\u03b5\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03b9\u03bd\u03b5\u03b9 \u03bf \u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03bf\u03c2. :roll:", "Solution_18": "\u0394\u03b9\u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b7\u03c4\u03b1\u03bd \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc \u03b1\u03c0\u03cc \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bd\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03cc\u03c3\u03bf \u03c0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ad\u03c2", "Solution_19": "\u03c4\u03bf \u03be\u03b5\u03c1\u03c9 \u03c1\u03b5 \u0391\u03bc\u03b2\u03bf\u03c3\u03b9\u03b5.\u03b5\u03c4\u03c3\u03b9 \u03ba\u03b1\u03b9 \u03b1\u03bb\u03bb\u03b9\u03c9\u03c2 \u03b5\u03b9\u03bc\u03b1\u03c3\u03c4\u03b5 \u0392 \u03b5\u03b8\u03bd\u03b9\u03ba\u03b7. :blush:" } { "Tag": [ "inequalities", "geometry", "circumcircle", "angle bisector", "geometry proposed" ], "Problem": "$r,R$ are inradios and circumradius\r\nProve :\r\n$\\frac{R}{r}\\ge \\frac{a}{b}+\\frac{b}{a}$", "Solution_1": "See\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=15808\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=15807\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=49433\r\n\r\n Darij", "Solution_2": "[hide=\" Here is the shortest solution what I know for this nice and yet simple inequality.\"]Denote $l_a,\\ h_a$ the lengths of the angle-bisector and altitude from the vertex $A$.\n$l_a=\\frac{2}{b+c}\\sqrt{bcp(p-a)}\\Longrightarrow l^2_a=\\frac{4bc}{(b+c)^2}\\cdot p(p-a)\\le p(p-a)\\ .$ Therefore,\n$h_a\\le l_a,\\ l_a\\le p(p-a)\\Longrightarrow h^2_b+h^2_c\\le l^2_b+l^2_c\\le p(p-b)+p(p-c)\\Longrightarrow$\n$h^2_b+h^2_c\\le ap\\Longrightarrow 4S^2\\left(\\frac{1}{b^2}+\\frac{1}{c^2}\\right)\\le ap\\Longrightarrow4S^2\\left(\\frac cb+\\frac bc\\right)\\le abcp\\Longrightarrow$\n$4S^2\\left(\\frac cb+\\frac bc\\right)\\le 4pRS\\Longrightarrow \\frac cb+\\frac bc\\le \\frac{Rp}{S}\\Longrightarrow \\frac cb+\\frac bc\\le \\frac Rr\\ .$[/hide]" } { "Tag": [ "geometry", "parallelogram" ], "Problem": "Doua cercuri C1 si C2 sunt tangente intern intr-un punct N astfel incat C2 se gaseste in interiorul lui C1. Punctele C,S,T apartin lui C1 si au proprietatea ca CS si CT sunt tangentela C2 in M si K.Demonstrati ca UCVW este un paralelogram , unde W este al doilea punct de intersectie dintre cercurile circumscrise triunghiurilor UMC si VCK.", "Solution_1": "Ar fi bine sa revezi enuntul.", "Solution_2": ":blush: :oops: \r\nAm uitat sa spun ca U-mij arcului CS si V-mij arcului CT \r\n\r\nMa scuzati!! :oops:", "Solution_3": "O sa formulez problema inversa (obtinuta prin inversiunea problemei initiale in raport cu polul $C$ si puterea $CM^2$). Pentru orice punct $X$, imaginea lui $X$ prin inversiune va fi numita $X'$ (si, in general, imaginea figurii $\\mathcal F$ va fi $\\mathcal F'$). Si o sa mai sar peste o gramada de pasi :).\r\n\r\nSe dau punctul $C$ si cercul $c_2'$ astfel incat $CK',CM'$ sunt tangente la $c_2'$. Se iau $U',V'$ pe o dreapta $c_1'$ tangenta la $c_2'$ astfel incat $S'C=S'U',T'C=T'V'$, unde $S',T'$ sunt intersectiile dreptelor $CM',CK'$ cu $c_1'$ ($U',T'$ sunt de parti opuse ale lui $CS'$, si $V',S'$ sunt de parti opuse ale lui $CT'$). Se mai ia $W'=V'K'\\cap U'M'$.\r\n\r\nSa se arate ca $W'U'C$ si $W'CV'$ sunt asemenea (in ordinea asta a varfurilor).\r\n\r\nE usor de vazut ca putem obtine concluzia daca aratam ca $\\angle U'W'V'=\\pi-\\angle S'CT'\\ (*)$ si ca $C$ se afla pe bisectoarea unghiului $U'W'V'$.\r\n\r\nFie $P$ punctul unde $c_1'$ atinge $c_2'$, si fie $P_u=U'W'\\cap CP,P_v=V'W'\\cap CP$. Se vede ca $P$ este mijlocul lui $U'V'$. $P_uCU',P_vCV'$ sunt isoscele, deci distanta de la $C$ la $U'W'$ este egala cu distanta de la $U'$ la $CP$, si, la fel, distanta de la $C$ la $V'W'$ este egala cu cea de la $V'$ la $CP$. Din moment ce $P$ este mijlocul lui $U'V'$, iese de aici ca distantele de la $C$ la $U'W',V'W'$ sunt egale, deci $C$ e pe bisectoarea lui $\\angle U'W'V'$.\r\n\r\nMai trebuie acum sa obtinem $(*)$, ceea ce este usor, pentru ca unghiurile $\\angle WP_uP_v,\\angle WP_vP_u$ ale lui $WP_uP_v$ sunt egale cu $\\pi-2\\angle PCU'$ si $\\pi-2\\angle PCV'$ respectiv.\r\n\r\nImi pare rau ca nu am demonstrat tot, dar ar fi devenit plictisitor, si oricum, o figura bine facuta ar trebui sa clarifice tot :). Sunt sigur ca iese repede si problema initiala, dar din moment ce primul lucru pe care l-am facut a fost sa o inversez, nu m-am mai chinuit sa o rezolv direct, fara inversiune.", "Solution_4": "Am gasit si eu o rezolvare sintetica\u2026 \r\n :roll: In esenta se demonstreaza ca (U, M, N); (V, K, W); (M, W, K) sunt coliniare si de aici se rezolva usor. \r\nO2K si VO1 sunt perpendiculare pe CT => sunt paralele. (N, O1, O2) coliniare si NO2/NO1 = O2K/O1V => N, K, W coliniare. Analog U, M, V coliniare.\r\nDin patrulaterele UCWM si CVKW inscriptibile => m(MWC) = 180 \u2013 m( MUC) = 180 \u2013 m( NUC) si m(CWK) = 180 \u2013 m( CVK) = 180 \u2013 m(CVN). Dar CVNU inscriptibil => m( CVN) + m( NUC) = 180 => m(MWC) +m(CWK)= 180 => (M, W, K) coliniare.\r\nAcum m(CWU) = m(CMU) = m(SMN). Deoarece SM tangenta la cercul circumscris lui MNK => m(SMN)= m(MKN). m(MKN)=m(WCV) => m(CWU)=m(WCV) => UW\u2551CV. Analog UC\u2551WV => UCVW paralelogram. q.e.d \r\n :D" } { "Tag": [], "Problem": "Solve and write the demonstration outline, and the program outline.\n\n\n\nP=>-Q, -R=>Q, P l- R[/b][/hide][tab]", "Solution_1": "Wow, do I ever not know what's going on there.", "Solution_2": "This belongs in Other Problem Solving Topics. It really is a shame that not very many people know this stuff; I was fortunate enough to have logic classes in 6th and 7th grade, so I actually do know this stuff reasonably well, although it was a long time ago.", "Solution_3": "I recognized the last line as a statement in propositional logic, but I have never heard of a demonstration outline or program outline.", "Solution_4": "I now recognize the symbolic logic -- I have no idea what the first line is asking about, though. Also, I don't know what |- means.", "Solution_5": "The turnstile P=>-Q, -R=>Q, P |- R means you are given P=>-Q, -R=>Q, and P, and you have to prove R. Note that P |- R is subtly different from P => R.", "Solution_6": "I'll follow ComplexZeta's suggestion and move this to Other Problem Solving Topics (although I would otherwise have moved it to Intermediate). It should be accessible to high schoolers with an unusual background in logic, but it is definitely not a Getting Started problem. I have also retitled the thread for clarity.", "Solution_7": "[quote=\"ballerkidd4\"]Solve and write the demonstration outline, and the program outline.\n\nP=>-Q, -R=>Q, P l- R[/quote]\r\n\r\nDemonstration\r\n\r\n1. P by hypothesis\r\n2. P => ~Q by hypothesis\r\n3. ~Q by Modus Ponens on 1 & 2\r\n4. ~R => Q by hypothesis\r\n5. ~Q => R by Law of Contrapositive on 4\r\n6. R by Modus Ponens on 3 and 5.\r\n\r\nQ.E.D.", "Solution_8": "Modus Ponens? Doesn't that mean 'way of placing' in latin? (I take latin :)) What does it mean in math?", "Solution_9": "I don't know Latin, but in logic, it means that if you have A and A => B, then you can infer B. \r\n\r\nBy the way, in gauss's argument, I think we can collapse steps 5 and 6 into one step, by using modus tollens. (\"If you have ~B and A => B, then you can infer ~A.\") Unless we consider going from ~~R to R a separate step." } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $X$ be a compact Hausdorff space with a sequence of closed, connected subspaces $C_1 \\supseteq C_2 \\supseteq C_3 ...$. Is it true that $C = \\cap^\\infty_{n=1} C_n$ is connected?\r\n\r\nIs this the right forum? I'm sorry if it's not.", "Solution_1": "[quote=\"Kalle\"]Let $X$ be a compact Hausdorff space with a sequence of closed, connected subspaces $C_1 \\supseteq C_2 \\supseteq C_3 ...$. Is it true that $C = \\cap^\\infty_{n=1} C_n$ is connected?\n\nIs this the right forum? I'm sorry if it's not.[/quote]\r\n\r\nC is connected.\r\nProof:\r\nIf it is not, we can find open sets $U_1$ and $U_2$ such that ${U_1}\\cup{U_2}\\supseteq{C}$ and ${U_1}\\cap{U_2}=\\emptyset$.\r\nNow the union of countably infinite open sets $U_1$, $U_2$, and $(C_n)^c (n=1,2,...)$ equals X.\r\nSince X is compact, we can choose from them finite sets that still cover X.\r\nObviously, because C must be covered, $U_1$ and $U_2$ must be included in those finite sets.\r\nAnd notice that $(C_1)^c\\subseteq(C_2)^c\\subseteq...$, so $(C_n)^c\\cup{U_1}\\cup{U_2}=X$ holds for some n.\r\nThus ${U_1}\\cup{U_2}\\supseteq{C_n}$, which contradicts with the fact that $C_n$ is connected.", "Solution_2": "Nice proof. But do you really need the Hausdorff property here?", "Solution_3": "may I know how can the existence of $U_1, U_2$ be guaranteed?", "Solution_4": "...", "Solution_5": "[quote=\"AlvaroRecoba\"]\nIf it is not, we can find open sets $U_1$ and $U_2$ such that ${U_1}\\cup{U_2}\\supseteq{C}$ and ${U_1}\\cap{U_2}=\\emptyset$.\n[/quote]\nDo you skip a whole bunch of steps here? I could work your reasoning out like this:\n\nLet $C = A \\cup B$ be a disjunction of $C$, where $A$ and $B$ are open subsets of $C$. $A$ and $B$ must also be closed in $C$, and since $C$ is closed in $X$, $A$ and $B$ must also be closed as subsets of $X$. Since $X$ is compact and Hausdorff it is normal and we can find open, disjoint sets $U_1$ and $U_2$ in $X$ such that $A \\subseteq U_1$, $B \\subseteq U_2$\n\nIs this what you mean or is there a much easier argument since you omitted it?\n\n[quote=\"AlvaroRecoba\"]\nThus ${U_1}\\cup{U_2}\\supseteq{C_n}$, which contradicts with the fact that $C_n$ is connected.[/quote]\r\nI also do not follow this contradiction... Please clarify", "Solution_6": "Sorry I made a terrible mistake here...", "Solution_7": "Kalle is right.\r\n\r\nIf $C$ is not connected, we can find closed sets ${V_1}\\neq{\\emptyset}$ and ${V_2}\\neq{\\emptyset}$ in $C$ such that ${V_1}\\cup{V_2}=C$ and ${V_1}\\cap{V_2}=\\emptyset$. And since $C$ is closed, $V_1$ and $V_2$ are also closed sets in $X$.\r\nBecause compact Hausdorff space has the $T_4$ property, there are open sets ${U_1}\\supseteq{V_1}$, ${U_2}\\supseteq{V_2}$ such that ${U_1}\\cap{U_2}=\\emptyset$.\r\n\r\nAt the end of my proof, we conclude that ${U_1}\\cup{U_2}\\supseteq{C_n}$, then ${U_1}\\cap{C_n}\\neq{\\emptyset}$ and ${U_2}\\cap{C_n}\\neq{\\emptyset}$ are two open sets in ${C_n}$, their union equals ${C_n}$, and their intersection is $\\emptyset$.\r\nThis is impossible since ${C_n}$ is connected.", "Solution_8": "Now it works. Thanks!" } { "Tag": [ "function", "geometry", "geometric transformation", "reflection" ], "Problem": "a shining metal ball with a small black dot on its surface is heated to a very high temperature and then taken to a dark room\r\n\r\na) the spot appears brighter than ball\r\nb) they both appear equally bright\r\nc) the spot appears darker than ball\r\nd) both are invisible in the dark room", "Solution_1": "I guess the whole ball emits about the same spectrum of frequencies, but the black dot emits more energy. Therefore it looks brighter, if observed in a right frequency range. If that's what you mean. (What is a \"dark room\"?)", "Solution_2": "[quote=\"kubus\"]I guess the whole ball emits about the same spectrum of frequencies, but the black dot emits more energy. Therefore it looks brighter, if observed in a right frequency range. If that's what you mean. (What is a \"dark room\"?)[/quote]Now why is that? Could you explain the whole radiation thing a little bit? (I know, we learned that in school - I only wish I could remember any of it :) .)", "Solution_3": "Every surface has its own emissivity - that is a constant between 0 and 1 (0 is a perfect mirror, 1 is a black body)(well, in fact, it is not a single constant, but a function of frequency) that describes what fraction of the Planck-law radiation the body really emits. This emissivity is exactly the same for absorbing radiation. You can prove this by contradiction, because if the emissivity factor could be different for emission and absorbtion, you could violate 2nd law of thermodynamics (I guess;). It depends on the very structure of the material, its interaction with EMG waves, resonance in frequencies, ...\r\n\r\nNow clearly the black dot has higher emissivity than the rest of the ball and it radiates more energy.", "Solution_4": "yup ur right kubus :first: coool soln!", "Solution_5": "Nice sol kubus,didnt see ya for a long time. :lol: \r\nSO u needed to know this emissivity thin for this question,i thought that the metal ball shines a bit in the dark room but the black dot looks more prominent in the whole room :blush: :oops: \r\nIS that right or lame? :?", "Solution_6": ":rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl:(hey might be true too , :lol: :P :lol: :P ) \r\n\r\n\r\n\r\n\r\ndont start overacting again please!", "Solution_7": "However, the \"emissivity thing\" (I don't know how it is called properly either) is given in the problem statement. What was kubus trying to say is that if something reflects for example 75 % light (= is shiny) then it only radiates 25 % of what would a normal object at the same temperature radiate. If something's completely black (= reflects 0 % of light) it radiates 100 %, if something reflects 100 % it won't emit any radiation at all, etc. - it always has to add up to [b]exactly[/b] 100 %. So a shiny ball will only radiate 25 % (or whatever fraction) of what the black dot radiates and therefore will be darker.\r\n\r\n\r\n[quote=\"shadysaysurspammed\"]i thought that the metal ball shines a bit in the dark room but the black dot looks more prominent in the whole room :blush: :oops: \nIS that right or lame? :?[/quote]I'm sorry, that doesn't mean that it also looks [i]brighter[/i] :) ." } { "Tag": [ "geometry", "perimeter", "algebra unsolved", "algebra" ], "Problem": "i'm having a very hard time figuring out the answers to these problems so if anyone can figure it out, please help\r\n\r\nWrite 3 equations in 3 variable for each problem\r\n\r\nThe sum of the digits of a three digit number is 11. The hundreds digt exceeds the sum of the tens digit and the units digit by 1. When the digits are reversed, the new number is 396 less than the original number. Find the number. Write the three equations\r\n\r\nOne side of a triangle is 3 inches longer than another side of the riangle. The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle. If the perimeter is 29 inches, find the length of each side. Write the three equations", "Solution_1": "It's easy, I'll do the first :\r\n\r\n$x+y+z=11$\r\n$x=y+z+1$\r\n$99x-99z=396$ ==> $x-z=4$\r\n\r\nSo, $4=y+1$ and so $y=3$ \r\n\r\nSo, $x+z=8$ ==> $x=6$ ==> $z=2$\r\n\r\nThe number is $632$", "Solution_2": "[quote=\"yvette4848\"]\nOne side of a triangle is 3 inches longer than another side of the riangle. The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle. If the perimeter is 29 inches, find the length of each side. Write the three equations[/quote]\nI wonder if this question should be put in the intermediate forum instead.\n\n[quote=\"yvette4848\"]\nThe sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle.[/quote]\r\nDo you mean 'the sum of the lengths of the two sides is one inch greater than the third side'?\r\n$\r\n\\begin{cases}\r\nx=y+3 \\\\\r\nx+y-1=z \\\\\r\nx+y+z=29\r\n\\end{cases}\r\n$\r\nSubstitute the first and second equations into the third one,\r\n$(y+3)+y+(x+y-1)=29$\r\n$y+3+y+(y+3)+y-1=29$\r\n$4y+5=29$\r\n$y=6$\r\nCorrespondingly, $x=9, z=14$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a, b, c$ non-negative real numbers such that $ abc \\equal{} 1$. Prove or disprove that\r\n\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{}\\frac{a}{b} \\plus{}\\frac{b}{c} \\plus{}\\frac{c}{a} \\plus{} 9 \\ge 2(a \\plus{} b \\plus{} c \\plus{} \\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c})$\r\n\r\n_______\r\n :?: :?:", "Solution_1": "[quote=\"Ligouras\"]Let $ a, b, c$ non-negative real numbers such that $ abc \\equal{} 1$. Prove or disprove that\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{}\\frac{a}{b} \\plus{}\\frac{b}{c} \\plus{}\\frac{c}{a} \\plus{} 9 \\ge 2(a \\plus{} b \\plus{} c \\plus{} \\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c})$\n\n_______\n :?: :?:[/quote]\n_____\n This is the statement correct?\nWhere is the equality?\n____\nMarin Sandu", "Solution_2": "[quote=\"Ligouras\"]Let $ a, b, c$ non-negative real numbers such that $ abc \\equal{} 1$. Prove or disprove that\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{}\\frac{a}{b} \\plus{}\\frac{b}{c} \\plus{}\\frac{c}{a} \\plus{} 9 \\ge 2(a \\plus{} b \\plus{} c \\plus{} \\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c})$\n[/quote]\nLet $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, $c = \\frac{z}{x}$, then WLOG $a = \\min\\{a,b,c\\}$, $b = a + u$, $c = a + v$, $u,v \\geq 0$ and the inequality becomes \\[u^3 v^3 + \n u^2 v^4 + (u^3 v^2 + 5 u^2 v^3 + 2 u v^4) x + (u^4 - 3 u^3 v + \n 12 u^2 v^2 + 5 u v^3 + v^4) x^2 + (2 u^3 + 5 u^2 v + 13 u v^2 + \n 2 v^3) x^3 + (6 u^2 + 9 u v + 6 v^2) x^4 + (6 u + 6 v) x^5 + 3 x^6 \\geq 0\\] which is obvious, e.g. by AM-GM $3x^2u^3v \\leq xu^3v^2 + x^2u^4 + x^3u^2v$ or by checking positivity of $f(u) = 1 + 5 u + 12 u^2 - 3 u^3 + u^4$.", "Solution_3": "[quote=\"Ligouras\"]Let $ a, b, c$ non-negative real numbers such that $ abc \\equal{} 1$. Prove or disprove that\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{}\\frac{a}{b} \\plus{}\\frac{b}{c} \\plus{}\\frac{c}{a} \\plus{} 9 \\ge 2(a \\plus{} b \\plus{} c \\plus{} \\frac{1}{a} \\plus{} \\frac{1}{b} \\plus{} \\frac{1}{c})$\n[/quote]\nBecause $a^2+b^2+c^2+3\\geq a+b+c+ab+ac+bc$ and by Schur $\\frac{a}{b} \\plus{}\\frac{b}{c} \\plus{}\\frac{c}{a}+3\\geq a+b+c+ab+ac+bc$" } { "Tag": [ "AMC" ], "Problem": "Does anyone have access to older (complete) AHSME tests on the internet, specifically those given before 1980?", "Solution_1": "Check Tipton's bookcase; she has several copies of Contest Problem Books 1-4, which contain the AHSME's from 1950 to approximately 1990.", "Solution_2": "What is Tipton's bookcase?", "Solution_3": "lol\r\n\r\nWe go to the same school, and our math teacher is named Tipton.\r\n\r\nSorry that nobody else really has the opportunity to browse her bookcase.", "Solution_4": "Thanks Seva, I am aware of the greatest source of knowledge outside of the internet, but I was wondering if anyone knew of online resources containing them." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "A convex quadriteral $ ABCD$ is given with $ \\angle ABC \\equal{}$ $ \\angle ADC$ and $ M$ and $ N$ are points on $ BC$ and $ CD$ such that $ AM \\perp BC$ and $ AN \\perp CD$. $ K$ is a point where $ MD$ and $ NB$ meet. Prove that $ AK \\perp MN$", "Solution_1": "[quote=\"vedran6\"]A convex quadriteral $ ABCD$ is given with $ \\angle ABC \\equal{}$ $ \\angle ADC$ and $ M$ and $ N$ are points on $ BC$ and $ CD$ such that $ AM \\perp BC$ and $ BN \\perp CD$. $ K$ is a point where $ MD$ and $ NB$ meet. Prove that $ AK \\perp MN$[/quote]\r\nI'm sure that you made a mistake, the condition $ AN\\perp CD$, [b]not[/b] $ BN\\perp CD$", "Solution_2": "[quote=\"mr.danh\"][quote=\"vedran6\"]A convex quadriteral $ ABCD$ is given with $ \\angle ABC \\equal{}$ $ \\angle ADC$ and $ M$ and $ N$ are points on $ BC$ and $ CD$ such that $ AM \\perp BC$ and $ BN \\perp CD$. $ K$ is a point where $ MD$ and $ NB$ meet. Prove that $ AK \\perp MN$[/quote]\nI'm sure that you made a mistake, the condition $ AN\\perp CD$, [b]not[/b] $ BN\\perp CD$[/quote]\r\n you were right, I have editted it :)", "Solution_3": "$ \\triangle ADN\\sim\\triangle AMB \\Rightarrow\\frac{AN}{DN}\\equal{}\\frac{AM}{MB}\\equal{}k$\r\nLet AH denote the altitude of the triangle AMN. Let E in the opposite ray of AH such that $ AE\\equal{}k.MN$\r\nWe have $ \\frac{AE}{MN}\\equal{}\\frac{AN}{DN}, \\hat{EAN}\\equal{}\\hat{MND}$, hence $ \\triangle EAN\\sim\\triangle MND$\r\n$ \\Rightarrow EN\\perp MD$.\r\nSimilar, $ EM\\perp BN$\r\nHence, three altitudes $ MD,NB,EH$ of $ \\triangle EMN$ are concurrent. So, $ AK\\perp MN$." } { "Tag": [ "\\/closed" ], "Problem": "not a \"bug\", rahter a weird cosmet(h?)ical thing.", "Solution_1": "I know, I don't mind it :) It's because of the space in the middle of my username. If you have a simple css solution for it, please post it.", "Solution_2": "Of course I have... as I posted in the other post, it's because phpbb standardly sets spaces around your name.\r\n\r\nIn index_body.tpl (or whatever it's called in the new version), where it would generate \r\n\r\n[color=red] [/color]\r\nValentin Vornicu[color=red] [/color]\r\n\r\n\r\nJust remove the 2 red [color=red] [/color]'s *and* the space right after the first one. That should make it perfectly aligned. :)", "Solution_3": "Fixed :)" } { "Tag": [ "LaTeX", "AoPSwiki" ], "Problem": "Dear moderators,\r\nI am a newbie in using Latex, and I would like to create the document like this... But I dont know how, could you upload the tex file of this document for me to know how to create it? Thank you very much.", "Solution_1": "I'll re-type up three for you here:\r\n\r\nProve that there are infinitely many positive integers $ n$ such that $ n^{2}\\plus{}1$ has a prime divisor greater than $ 2n\\plus{}\\sqrt{2n}$.\r\n\r\nTo type this problem up, there is minimal syntax. The text is just text, and the math is pretty basic. If you scroll over what I typed, it should reveal the syntax.", "Solution_2": "[quote=\"JRav\"]I'll re-type up three for you here:\n\nProve that there are infinitely many positive integers $ n$ such that $ n^{2} \\plus{} 1$ has a prime divisor greater than $ 2n \\plus{} \\sqrt {2n}$.\n\nTo type this problem up, there is minimal syntax. The text is just text, and the math is pretty basic. If you scroll over what I typed, it should reveal the syntax.[/quote]\r\nNo, maybe you dont understand me. I meant how to create the file like this.", "Solution_3": "There is a nice tutorial on basic $ \\text{\\LaTeX}$ on AoPSWiki, which, surprisingly enough, is named [[LaTeX]] and has a link to it (quite unexpectedly, with the same name) clearly visible on both skins :P. Just read it. If you have any particular questions after that, feel free to ask here :)." } { "Tag": [ "function", "trigonometry" ], "Problem": "For $0^\\circ \\leq \\theta <180^\\circ,$ consider the function $f(\\theta)=3\\cos 2\\theta+4\\sin\\theta.$\r\nLet $\\alpha$ be the angle such that $0^\\circ <\\alpha <90^\\circ$ and $f(\\alpha)=3.$\r\n\r\n(1) Find the maximum and minimum value of $f(\\theta).$\r\n(2) Find the value of $\\sin (\\alpha +30^\\circ).$", "Solution_1": "The second one:\r\n[hide]$\\cos(2\\alpha)=1-2\\sin^2\\alpha$ so we have $3-6\\sin^2\\alpha+4\\sin\\alpha$. We want to find such an $\\alpha$ where $f(\\alpha)$ is 3. $3=3-6\\sin^2\\alpha+4\\sin\\alpha$. We cancel the constants and we factor the remains. $0=\\sin\\alpha(-6\\sin\\alpha+4)$. Obviously, $\\alpha=0$ is a solution to the equation so $\\sin(\\alpha+30)=1/2$ is one answer. The other solution has $\\sin\\alpha=2/3$. Therefore, when we add 30 degrees, we have $\\sin\\alpha*\\cos\\30+\\sin30*\\cos\\alpha$. Since $\\sin\\alpha=2/3, \\cos\\alpha=\\sqrt(5)/3$. We have $2/3*\\sqrt(3/2)+1/2*\\sqrt(5)/3=\\frac{2\\sqrt(3)+\\sqrt(5)}{6}$. That's the other solution.[/hide]", "Solution_2": "[quote=\"dengmi\"]The second one:\n[hide]$\\cos(2\\alpha)=1-2\\sin^2\\alpha$ so we have $3-6\\sin^2\\alpha+4\\sin\\alpha$. We want to find such an $\\alpha$ where $f(\\alpha)$ is 3. $3=3-6\\sin^2\\alpha+4\\sin\\alpha$. We cancel the constants and we factor the remains. $0=\\sin\\alpha(-6\\sin\\alpha+4)$. Obviously, $\\alpha=0$ is a solution to the equation so $\\sin(\\alpha+30)=1/2$ is one answer. The other solution has $\\sin\\alpha=2/3$. Therefore, when we add 30 degrees, we have $\\sin\\alpha*\\cos\\30+\\sin30*\\cos\\alpha$. Since $\\sin\\alpha=2/3, \\cos\\alpha=\\sqrt(5)/3$. We have $2/3*\\sqrt(3/2)+1/2*\\sqrt(5)/3=\\frac{2\\sqrt(3)+\\sqrt(5)}{6}$. That's the other solution.[/hide][/quote]\r\n\r\nThe answer is the latter part of your answer. :)" } { "Tag": [ "geometry", "AMC", "AIME", "inequalities", "AMC 12", "USA(J)MO", "USAMO" ], "Problem": "I was wondering how i could determine my level with respect to this forum. Since i never took any contest(s) in high school i am having trouble figuring out where i stand (i will of course try problems in both sections but for curiousities sake). I was hoping there is some sort of test (if not, wouldnt it be a great idea?) to determine where i should be reading most actively. \r\n Anyway i have done 2 problems in USSR Olympiad book with double stars, 3 with single stars, and a few dozen of the regular ones. My geometry is really bad (well in the sense that i cant yet draw lines very well), i have worked out a few of the easier AIME problems (non-geometry) and can do most ASHME problems(particulary inequalities and counting). Other than that i am completely self taught (didnt take any classes or have any type of instruction, besides of course my books) for all classes after algebra I (which i actually failed). Hope this might give you guys a better view so you can tell me what is best.", "Solution_1": "Even though I am not the best to tell because I am not nearly as good as you are but you probably belong to the advanced section if you can solve olympiad type problems. Also, I think there are AIME problems in the amc form and intermediate.\r\n\r\nBy the way don't some ussr problems look even harder then usamo/imo problems, for some reason they look like that to me. :P", "Solution_2": "I would say that you are likely \"Pre-Olympiad\", though don't be discouraged into looked at problems in the \"Advanced\" section.\r\n\r\nIf you can do most AMC-12/AHSME problems and a fair number of AIME problems, then you are certainly well beyond \"Getting Started\" and probably at the high end of \"Intermediate\" if not beyond.\r\n\r\nThere is some degree of randomness to which problems end up in which forums, as it is up to the initial poster to decide in which forum to initially post their problem. If the moderators and/or administrators feel that a problem is significantly too hard for its forum, we will move it to a more appropriate forum.", "Solution_3": "pre-olympiad? really?\r\n\r\ni think im on about the same level as you - i got a 6 on the AIME, then fixed some stupid errors and figured out some more when i got the test back...\r\n\r\n...and i find that the intermediate forum suits me just fine for challeges, and the getting started forum has plenty of easier problems to practice with. there r too many theories and concepts in pre-olympiad that ive never heard of, let alone understand... :(", "Solution_4": "Here's the official description of the Intermediate Problem Solving forum:\r\n\r\n[quote=\"rrusczyk\"]The Intermediate Topics board is for problems and questions that are beyond the basics, but not quite olympiad level. These are problems that might be on the AMC, AIME, Mandelbrot, or ARML. If you're not sure what level your problem or question is, you can generally use the other posts in the board as a guideline. \n\nGenerally, the Intermediate Topics board will include topics of high school math such as trigonometry, polynomials, conditional probability, logarithms, etc., as well as some broader beginning problem solving techniques such as basic problems in induction or the pigeonhole principle. \n[/quote]\r\n\r\nIf you can do \"most\" of the AMC12 problems and \"some\" of the AIME problems, then you're not too far from a USAMO-qualifying score, hence I recommend \"Pre-Olympiad\". But as I said above, placing problems into a forum by difficulty is not an exact science; many times a problem in the \"Intermediate\" forum will be harder than a problem in the \"Pre-Olympiad\" forum. \r\n\r\nGenerally, you should try to work on problems that are as hard as you can manage. Working on easier problems can be good for practice, but you'll only really learn if you try to solve hard problems.", "Solution_5": "I agree with DPatrick. You are definitely past Beginner. It sounds like you will have a shot at USAMO. I suggest that you browse the forums however. There might be something of worth in a forum that you don't usually use.", "Solution_6": "One problem with inspecting the forums for examples is that people are still getting used to the new five-level division of the problem-solving forums, and MOST forums have too many problems that are too hard for their intended level.", "Solution_7": "hello: \r\n\r\n Yeah i hear that alot but the USSR olympiads, but the USSR problem book isnt all hardcore. Some of the problems are even ASHME level, but all of them are interesting (except for one of them that required too much calculation).\r\n\r\nL_Li:\r\n\r\n Nah your probally much better then me, 6 questions on a single AIME is not yet reality for me (although i am looking at some of the older ones so i dont know how those compare). \r\n\r\n I will most likely try to do all the ASHME's (untill i get bored) for the next 3 weeks, its been a long time since i solved problems so i need to get back into the groove. I will post my progress as often as possible perhaps along with problems/solutions i found interesting.[/quote]" } { "Tag": [ "algebra", "polynomial", "quadratics", "function", "algorithm", "Ring Theory", "geometry" ], "Problem": "Here's the question: (rephrased)\r\n\r\nThe remainder when P(x) is divided by x-99 = 19.\r\n---------------ditto----------------- x-19 = 99.\r\nWhat's the remainder when P(x) is divided by (x-19)(x-99) in terms of x?\r\n\r\nFor those of you who want to check, the answer is [hide] 118-x [/hide]\r\n\r\nEDIT!!!: HUGE typo mistake, sorry everybody, the original problem was indeed actually with (x-99) and (x-19) instead of what I wrote originally. Sorry!", "Solution_1": "[quote=\"erdogankerem123\"]Here's the question: (rephrased)\n\nThe remainder when P(x) is divided by x+99 = 19.\n---------------ditto----------------- x+19 = 99.\nWhat's the remainder when P(x) is divided by (x+19)(x+99) in terms of x?\n\nFor those of you who want to check, the answer is [hide] 118-x [/hide][/quote]\r\n\r\nthe remainder will be in form ax+b\r\n\r\n$ P(x)=(x^{2}+118x+1881)q(x)+ax+b$\r\n\r\nso $ P(-99)=19=(0)(q(x))-99x+b$ and\r\n $ P(-19)=99=(0)(q(x))-19x+b$\r\n\r\n$ 19=-99a+b$\r\n$ 99=-19a+b$ subtract the two and you get \r\n\r\n$ -80=-80a$\r\n$ a=1$ therefore $ b=118$ \r\nso the remainder is $ \\boxed{x+118}$?????", "Solution_2": "ie, there are f(x) and g(x) such that\r\n\r\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\r\n\r\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...", "Solution_3": "[quote=\"K81o7\"]ie, there are f(x) and g(x) such that\n\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\n\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...[/quote]\r\n\r\nI got that at first and then I realized my mistake, um did you just estimate and come up with x+118?", "Solution_4": "In everything I say in this post, I'm assuming that every polynomial is monic - that is, the coefficient of its highest order term is $ 1.$\r\n\r\nThe remainder when $ P(x)$ is divided by $ x-a$ is $ P(a).$ That's an extremely important part of the theory of polynomials.\r\n\r\nNow consider dividing $ P(x)$ by $ D(x),$ where $ D$ is a monic polynomial of degree $ k.$ Then $ P(x)=Q(x)D(x)+R(x),$ where $ R(x)$ is the remainder. Certainly, $ R$ has degree $ \\le k-1.$ Now, suppose $ D(a)=0.$ Then $ P(a)=Q(a)\\cdot 0+R(a),$ so $ R(a)=P(a).$\r\n\r\nNow for the problem at hand. The first line tells you that $ P(-99)=19.$ The second line tells you that $ P(-19)=99.$ Finally, we divide by the quadratic polynomial $ (x+19)(x+99).$ From what I said above, the remainder $ R(x)$ is a first degree (or smaller) polynomial, which is to say a linear function, and $ R(-19)=99$ and $ R(-99)=19.$\r\n\r\nAt this point, I think you've misquoted the problem. I don't feel like chasing down the 1999 AHSME right now, but weren't we dividing by $ x-99,\\ x-19,$ and $ (x-19)(x-99)?$", "Solution_5": "[quote=\"sonny\"][quote=\"K81o7\"]ie, there are f(x) and g(x) such that\n\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\n\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...[/quote]\n\nI got that at first and then I realized my mistake, um did you just estimate and come up with x+118?[/quote]\r\n\r\n?\r\nIs not x+118 19 mod x+99 and 99 mod x+19?", "Solution_6": "[quote=\"K81o7\"][quote=\"sonny\"][quote=\"K81o7\"]ie, there are f(x) and g(x) such that\n\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\n\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...[/quote]\n\nI got that at first and then I realized my mistake, um did you just estimate and come up with x+118?[/quote]\n\n?\nIs not x+118 19 mod x+99 and 99 mod x+19?[/quote]\r\n\r\nUh yeah, much like Mr.Merryfield I got confused between the posted and the real.", "Solution_7": "Sorry everybody for misquoting the problem. I get it anyways, though, simply by just using same method, different numbers. Thanks everyone!", "Solution_8": "[quote=\"K81o7\"]ie, there are f(x) and g(x) such that\n\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\n\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...[/quote]\r\n\r\nerr...do you have anything else to add about CRT with polynomials? Is there some more general result that corresponds to a field extension of the integers?", "Solution_9": "[quote=\"Altheman\"][quote=\"K81o7\"]ie, there are f(x) and g(x) such that\n\ng(x)(x+99)+19=P(x)=f(x)(x+19)+99\n\nBy the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x). If I'm not mistaken, though, it should be x+118...[/quote]\n\nerr...do you have anything else to add about CRT with polynomials? Is there some more general result that corresponds to a field extension of the integers?[/quote]\r\n\r\nHmm...polynomials behave a little differently from other field extensions. For example, if p(x) is an irreducible polynomial, then Z[x]/p(x) is not necessarily a field (remember that because the constant terms and x terms and irreconcilable, a polynomial residue class has an infinite number of elements, and cannot be a field unless we take it modulo some prime). On the other hand, the Gaussian integers behave almost identically to the integers. Certain field extensions of the form $ \\mathbb{Z}[\\sqrt{-d}]$ also behave nicely for squarefree integers d, but as you might imagine, the behavior of prime factorization and division algorithm and intimately tied to the norm function, which is tied to the solvability of $ x^{2}+dy^{2}=p$ over primes p, which is tied to $ \\left(\\frac{-d}{p}\\right)$. For example, if you consider the subring of the integers from which you form when you take the norms of all different numbers in the field extension, that itself must have unique prime factorization. Even when there is no unique prime factorization in a field extension, we can use what are called \"ideals\" of the extension. This is a group theory term, and although I can define it, I'm not as experienced in that area, so you may want to ask someone else like, say, JSteinhardt.\r\n\r\nBut try proving things about primes and residue classes with $ \\mathbb{Z}[i]$. Because of the nice behavior of the equation $ x^{2}+y^{2}=p$, this ring behaves nicely.", "Solution_10": "[quote=\"Altheman\"]Is there some more general result that corresponds to a field extension of the integers?[/quote]\r\n\r\nMore general than that, actually:\r\n\r\nhttp://en.wikipedia.org/wiki/Chinese_remainder_theorem#Statement_for_general_rings" } { "Tag": [ "geometry", "rectangle", "perimeter", "geometry unsolved" ], "Problem": "Given an $ a \\times b$ rectangle with $ a > b > 0,$ determine the minimum side of a square that covers the rectangle. (A square covers the rectangle if each point in the rectangle lies inside the square.)", "Solution_1": "It has to be the least perimeter of the square possibly.\n\nsolution $\\pm$\n\nfirst you show all $4$ points of the rectangle are on the boundary of the square.\n\nthen you have the angle between rectangle and square $0$ or $45$ degrees.\n\nso $min (a, \\frac{\\sqrt2} 2 (a+b) )$", "Solution_2": "[hide= Potential Solution?] \nLet $R$ denote the rectangle, and let $S$ denote the square with minimum length that covers $R$ . Let $s$ denote the length of a side of $S$. We claim that $R$ is inscribed in $S$. We also claim that $R $can only be inscribed in the two ways shown below. Let $S_1$ and $S_2$ denote the squares shown on the left-hand and right-hand sides, respectively. For $S_2$ , the sides of $R$ are parallel to the diagonals of $S2$. We can see that $s = a$ if $S = S_1$. \n\n[img] https://lh3.googleusercontent.com/8IBseHsKPR3t0qdwjmS_YCmzdjC7tevE6s-KzyEQKCojYni3RwjDSLTX1sSWPXAqziqrSA=s170 [/img]\n\nIt is easy to see that $s=\\frac{\\sqrt{2}(a+b)}{2}$ if $S=S_2$. Taking the minimum value of $(a,\\frac{\\sqrt{2}(a+b)}{2})$, we conclude that\n\n$s=\\[ \\left\\{\n\\begin{array}{ll}\n a & a<(\\sqrt{2}+1)b \\\\\n \\frac{\\sqrt{2}(a+b)}{2} & a\\geq (\\sqrt{2}+1)b \\\\\n \\end{array} \n\\right. \\]$\n\nNow we prove our claim that these are only two ways. Let $R = ABCD$ and $S = XYZW$. WLOG, we place $XY$ horizontally. By the minimality of $S$, we can assume that at least one vertex, say $A$ , of $R$ lies on one side of $S$ , say $WX$ (see the left diagram). If neither $B$ nor $D$ lies on the sides of $S$, we can then slide $R$ down vertically so that one of them, say $B$, lies on side $XY$ (see the middle figure). If neither $C$ or $D$ lies on the sides of $S$ , then we can apply an enlargement, centered at $X$ with scale less than $1$ , to $S$ such that the image of $S$ still covers $R$ .\nThis violates the minimality of $S$. Therefore, at least one of $C$ and $D$ lies on the sides of $S$, that is, three consecutive vertices of $R$ lie on the sides of $S$ (see the right-hand side figure). WLOG assume that they are $A, B, C$. If any of these three vertices coincide with any of the vertices of $S$, then we clearly have $S = S_1$. Hence we may assume that $A , B, C$ are on sides $WX$, $XY$ and $YZ$, respectively. By symmetry, we may also assume that $AB =a>b=BC$.\n\n[img] https://lh3.googleusercontent.com/lX6YAPjUs7xeaT73NSstZRW_31vZ4-Rui5zM7ZRqOvhw-W_DrjH3AHEOdhmGxa9R7-d9Mw=s170 [/img]\n\n[hide= Note] All rectangular-like shapes are indeed rectangles [/hide]\n\nIf $D$ does not lie on line segment $ZW$, then we can slide $R$ up so both $B$ and $D$ lie in the interior of $S$ (see the left-hand slide) Let $O$ be the center of $R$. We can then rotate $R$ around $O$ with a small angle so that all four vertices lie inside $S$ (see the middle figure shown below). It is easy to see that we can use a smaller square to cover $R$ (by applying an enlargement centered at $O$ with a scale less that $1$) , violating the minimality of $S$. This means our assumption is wrong, and D must lie on side $ZW$ (see the righthand side figure shown below) , which is the case when $S = S_2$. \n\n[img] https://lh3.googleusercontent.com/j6nT3PMMp8rueVTHgUEE_B-m-ZHc99H1uG2L4e6-E_Qfi_AKxCUBaUYFn7HOnP8_2NUp=s170 [/img]\n\n[hide= Note] All rectangular-like shapes are indeed rectangles [/hide]\n\nWe finish our proof that if $S = S_2$, then the sides of $R$ are parallel to the diagonals of $S_2$. By symmetry, it suffices to show that $AX = XB$. It is not difficult to see that $\\angle XAB = \\angle YBC = \\angle ZCD= \\angle WDA$. So triangles $ABX$, $BCY$, $CDZ$ and $DAW$ are similar. Set $AX = ax$ and $XB = ay$. Then $BY = bx$ and $CY = by$. Also, $DW = bx$ and $WA = by$, which gives $by + ax = WA + AX = WX = XY = XB + BY =ay+bx$, implying that $(a-b)x= (a-b)y$, or $x=y$, as desired $\\blacksquare$ [/hide]", "Solution_3": "discussed also [url=https://puzzling.stackexchange.com/questions/109313/a-square-covering-a-rectangle]here[/url] and [url=https://artofproblemsolving.com/community/c6h581772p3437573]here[/url]" } { "Tag": [], "Problem": "Note that 1 + 2 + 3 + 45 + 6 + 78 + 9 = 144. In how many [i]other[/i] ways is it possible to make a total of 144 using only 1, 2, 3, 4, 5, 6, 7, 8, and 9 in that order and addition signs?\r\n\r\nSource: Australian Math Competition, J30 (I30, S26), 2002", "Solution_1": "Here's what I got. I'm new to these boards, by the way. I'm glad to be a part of this.\r\n\r\nMy solution is written out longhand, the important things are bolded to make it easier to read.\r\n\r\nAnswer: 3 (12 + 34 + 5 + 6 + 78 + 9; 12 + 3 + 45 + 67 + 8 + 9; 1 + 2 + 34 + 5 + 6 + 7 + 89)\r\n\r\nSolved by:\r\n\r\nSTEP 1: The sum of the numbers from 1 through 9 is 45, so in order for the sum of the new numbers (with certain numbers promoted to the tens place) to end with a four, [b]the total of the numbers promoted must end in a 1[/b]. [b]No number may be promoted to the 100's place[/b] as this will always result in a number greater than 144. 9 cannot be promoted as there is no number after it.\r\n\r\nSTEP 2: The total of all promoted numbers times ten is less than the total number of the sum of all nine numbers. Thus, the total of all promoted numbers cannot be greater than 14.4. The only two numbers the promoted numbers can now add up to are 1 and 11, as they are less than 14.4 and end with a 1. The only way the total of promoted numbers to equal 1 is when we have 12 + 3 + 4 + 5 + 6 + 7 + 8 + 9, which equals 54, a number less than 144. [b]Only series in which the promoted numbers equal 11 can equal 144.[/b]\r\n\r\nSTEP 3: When we promote a total of 11, we add 110 to the total and subtract 11. This gives us 110 + 45 - 11 = 144. [b]Thus, any series of numbers in which the sum of promoted numbers equals 11 has a total sum of 144[/b].\r\n\r\nSTEP 4: If we promote numbers 5 and greater only, we will not be able to reach a total of promotions that equals 11. This is because 5 + 6 is the only way to get 11 from the remaining numbers, but 6 can't be promoted: to do so would promote 5 to the 100's place, which can't be done. [b]All promotions must contain a number from 1 - 4.[/b]\r\n\r\nSTEP 5: [b]If we promote 1, we have to choose numbers from 3 to 8 to promote to reach a total of 10 (to reach 11.) This can only be done with the non-adjacent pairs 3 & 7 and 4 & 6, giving a total of 2 combinations[/b]. We cannot promote 2 - doing so would require a pair of 4 & 5, which are adjacent and thus cannot both be promoted. [b]If we promote 3, we need to also promote 8[/b]. A total of 8 can not be reached in any other way. [b]This gives a final total of 3 combinations[/b], as the combination using a promoted 4 was given in the problem.", "Solution_2": "Correct.. and welcome to this place, it's great! :)", "Solution_3": "Good job cpp - interesting solution. (in the future, please hide your solution)\r\n\r\nMy solution\r\n\r\n[hide]\n$1 + 2 + 3 \\cdots + 9 = 45$\n\nIf we combine to get a three-digit number, the only possible way to keep the number under 144 is to combine 1-2-3, which gives us $123 + 4 + 5 \\cdots + 9 = 162$, which does not work. Thus, we can only combine numbers into other 2-digit numbers.\n\nWhen we combine, say, a 3 and a 4, we subtract 3 from our sum and add 30, essentially adding $9 * 3$. $144 - 45 = 99$, so the sum of the numbers we combine that are in the tens place must be 11.\n\nIf we combine 1-2, then we can also do 3-4 and 7-8, or we can do 4-5 and 6-7.\n\nIf we combine 2-3, we have no options. We cannot use 3-4 and 6-7, since 3 is already combined. We cannot use 4-5 and 5-6, since they are too close. We cannot use 9 since no number follows it.\n\nIf we combine 3-4, then we can also combine 8-9. (We have already counted 1-2, 3-4, and 7-8).\n\nIf we combine 4-5, we can use 7-8, but this is given in the example.\n\nIf we combine anything greater than 4, either we have already counted it or the numbers are too close. \n\nThus, we have three other options.\n12 + 34 + 5 + 6 + 78 + 9 = 144\n12 + 3 + 45 + 67 + 8 + 9 = 144\n1 + 2 + 34 + 5 + 6 + 7 + 89 = 144\n1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 (example)\n[/hide]", "Solution_4": "Thank you, your solution was very good too. : )" } { "Tag": [ "absolute value", "complex analysis" ], "Problem": "I'm proving something about unitary operators, and at one point I want to show that the eigenvalues have absolute value 1.\r\n\r\nAnyway, I've reached the point where I've devised that $ {\\gamma}^2 \\equal{} \\lambda$ and $ |\\lambda|\\equal{}1$, and I want to show that $ |\\gamma|\\equal{}1$, where $ \\gamma , \\lambda \\in \\mathbb{C}$\r\n\r\nSo, $ {\\gamma}^2 \\equal{} \\lambda$\r\n$ \\Rightarrow |{\\gamma}^2| \\equal{} |\\lambda| \\equal{} 1$\r\n$ \\Rightarrow {|\\gamma|}^2 \\equal{} 1$\r\n$ \\Rightarrow |\\gamma| \\equal{} \\sqrt{1}\\equal{}1$\r\n\r\nIs this correct? It seems very trivial, I'm just concerned about pulling out the power in line 2 to line 3. Thanks.", "Solution_1": "$ |zw|\\equal{}|z|\\,|w|$ for all $ z,w\\in\\mathbb{C}.$\r\n\r\nThat's more than enough to justify the step you're worried about.", "Solution_2": "Simple as that. Thanks." } { "Tag": [ "geometry", "circumcircle", "inradius", "incenter", "IMO Shortlist", "reflection", "Inversion" ], "Problem": "Let $ABC$ be a triangle, $\\Omega$ its incircle and $\\Omega_{a}, \\Omega_{b}, \\Omega_{c}$ three circles orthogonal to $\\Omega$ passing through $(B,C),(A,C)$ and $(A,B)$ respectively. The circles $\\Omega_{a}$ and $\\Omega_{b}$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of the circumcircle of $A'B'C'$ is half the radius of $\\Omega$.", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)", "Solution_2": "I remember posting this a long time ago :).\r\n\r\nConsider the inversion wrt the incircle. It invariates the incircle and $\\Omega_a,_b,_c$, so it turns $X$ into $X'$, for any $X\\in\\{A,B,C\\}$. At the same time, if the incircle touches $BC,CA,AB$ in $D,E,F$, then the inversion turns $A$ into the midpoint of $EF$ and the like, so $A',B',C'$ are the midpoints of $EF,FD,DE$ respectively, meaning that the circumcircle of $A'B'C'$ is the nine-point circle of $D EF$, thus having half its circumradius, which is the inradius of $ABC$.", "Solution_3": "[b]OFFICIAL SOLUTION[/b]\r\n\r\nDenote by $I$ the incenter, by $r$ the inradius, by $D,E,F$ the contacts of the incircle with $BC,CA,AB$ respectively and by $P,Q,R$ the midpoints of the segments $[EF],[FD],[DE]$. We will prove that $\\Omega_{a}$ is the circle $(B,C,Q,R)$. We firstly notice that from the right-angled triangles $IBD$ and $IDQ$ we get $IQ \\cdot IB = ID^{2} = r^{2}$ and in the same way $IR \\cdot IC = r^{2}$, therefore the points $B,C,R,Q$ are on a circle $\\Gamma_{a}$. The points $Q$ and $R$ belong to the segment $(IB)$ and $(IC)$, so $I$ is exterior to the circle $(B,Q,R,C)$ and $I's$ power with respect to the circle $(B,Q,R,C)$. In the same way $\\Gamma_{b}$ is the circle $(C,R,P,A)$ and $\\Gamma_{c}$ is the circle $(A,P,Q,B)$. It follows that $A',B',C'$ coincide with $P,Q,R$ and the required conclusion is not obvious.\r\n\r\n[b]Another Solution[/b]:\r\n\r\nLet $O_{a}$ be the center of $\\Omega_{a}$ and $M$ be the midpoint of $(BC)$. Denote, using oriented segments, $\\overline{M,O_{a}} = x$ (the positive sense on the perpendicular bisector of $(BC)$ being $\\overrightarrow{ID}$. The radius of $\\Omega_{a}$ is $x^{2} + \\frac{a^{2}}{4}$ and\r\n\r\n\\[IO^{2}_{a} = DM^{2} + (\\overline{ID}+ \\overline{MO_{a}})^{2} =\r\n\\frac{a}{2} - \\left(\\frac{p-b}{2}\\right)^{2} + (r+x)^{2}.\\]\r\n\r\nThe condition of $\\Omega$ and $\\Omega_{a}$ being orthogonal is $O_{a}I^{2} = O_{b}B^{2}+r^{2}$, that is\r\n\r\n\\[x^{2} + \\frac{a^{2}}{4} + r^{2} = \\left(\\frac{b-c}{2}\\right)^{2} +\r\n(r+x)^{2} \\Leftrightarrow \\frac{(p-b)(p-c)}{2r} \\Leftrightarrow X\r\n= 2R \\sin \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2}.\\]\r\n\r\nIt follows that\r\n\r\n\\[OO_{a} = \\overline{OM} + \\overline{MO_{a}} = 2R \\sin \\frac{A}{2}\r\n\\cos \\frac{B}{2} \\cos \\frac{C}{2} = R + \\frac{r}{2}.\\]\r\n\r\nTherefore $O_{a},O_{b},O_{c}$ are on the circle of the center $O$ and radius $R + \\frac{r}{2}$. Let $\\{N\\} = O_{a}O_{c} \\cap BC$. The angle $\\widehat{O_{a}OO_{c}}$ has measure $\\pi - b$ (regardless of the position of $B$), so $\\widehat{O_{a}OO_{c}} = \\frac{B}{2}$ and\r\n\r\n\\[MN = x \\tan \\frac{B}{2} = 2R \\sin \\frac{A}{2} \\cos \\frac{B}{2}\r\n\\cos \\frac{C}{2} = \\frac{p-c}{2},BN = \\frac{a}{2} - \\frac{p-c}{2}\r\n= \\frac{p-b}{2}.\\]\r\n\r\nHence $BN = \\frac{BD}{2}$ and, because $MN < BM, N \\in (BM)$. The same holds for the common point of $AB$ and $O_{a}O_{c}$, therefore the reflection of $B$ in $O_{a}O_{c}$ is on $DF$. This proves that $B'$ - the second common point of $\\Omega_{a}$ and $\\Omega_{b}$ - is the midpoint of $(DF)$. The conclusion follows easily.", "Solution_4": "Just for the sake of completeness, the other thread where the problem was discussed is http://www.mathlinks.ro/Forum/viewtopic.php?t=4592 .\r\n\r\n darij", "Solution_5": "Let $r$ be the radius of $\\Omega$, then we have that the power of $I$ with respect to $\\Omega_a, \\Omega_b, \\Omega_c$ is $r^2$. Thus, $I$ is the radical center of $\\Omega_a, \\Omega_b, \\Omega_c$, so $AA', BB', CC'$ concur at $I$. Also, \\[r\\cdot AB = 2[AIB] = IA \\cdot IB \\cdot \\sin \\angle AIB\\] where $[AIB]$ is the area of triangle $AIB$. But \\[\\angle AIB = 180^{\\circ} - \\angle IAB - \\angle IBA = 180^{\\circ} - \\angle IAC - \\angle IBC. \\] By the similarity of $ICA$ and $IA'C'$ and of $ICB$ and $IB'C'$, this equals \\[180^{\\circ} - \\angle IC'A' - \\angle IC'B' = 180^{\\circ} - \\angle A'C'B',\\] so \\[r\\cdot AB = IA \\cdot IB \\cdot \\sin \\angle A'C'B'.\\] By the similarity of $IAB$ and $IB'A'$, \\[A'B' = \\frac{IB' \\cdot AB}{IA} = \\frac{r^2 AB}{IA \\cdot IB}\\] (because the power of $I$ with respect to $\\Omega_a$ is $r^2$). And now, we use the (extended) law of sines: Let $R$ be the circumradius of $A'C'B'$. Then \\[R = \\frac{A'B'}{2 \\sin \\angle A'C'B'}=\\frac{r^2 AB}{2IA \\cdot IB \\sin\\angle A'C'B'}.\\] Since $r\\cdot AB = IA \\cdot IB \\cdot \\sin \\angle A'C'B',$ this is equal to $r/2$ and we are done.", "Solution_6": "This problem also has a solution like 2009 ISL G3's solution.\n\n[b]Solution[/b]\n\nLet $O_a$, $O_b$ and $O_c$ denote the centres of circles $\\Omega_a$, $\\Omega_b$ and $\\Omega_c$, respectively. Let $A_1$ and $A_2$ denote the intersections of $\\Omega_a$ and $\\Omega$. Define $B_1$, $B_2$, $C_1$ and $C_2$ analogously. Let $D, E$ and $F$ denote the intersections of $\\Omega$ with $BC$, $CA$ and $AB$, respectively. Let $P$, $Q$ and $R$ denote the midpoints of $EF$, $DF$ and $DE$, respectively.\n\nBy the definition of orthogonality, lines $O_aA_1$ and $O_aA_2$ are tangent to circle $\\Omega$. Since $O_aB=O_aA_1=O_aA_2=O_aB$, it follows that $O_a$ lies on the radical axes of circles $B$ and $\\Omega$ and circles $C$ and $\\Omega$. Note that these radical axes are the midlines of the triangles $\\triangle{BDF}$ and $\\triangle{CDE}$. By the same argument, $O_b$ lies on the midlines of triangles $\\triangle{CDE}$ and $\\triangle{AEF}$ and $O_c$ lies on the midlines of triangles $\\triangle{BDF}$ and $\\triangle{AEF}$. Reflecting points $A$, $B$ and $C$ in the midlines of triangles $\\triangle{AEF}$, $\\triangle{BDF}$ and $\\triangle{CDE}$ respectively yields that $A'=P$, $B'=Q$ and $C'=R$. Therefore, $\\triangle{A'B'C'}$ is the medial triangle of $\\triangle{DEF}$ and has a circumradius equal to half of that of circle $\\Omega$.", "Solution_7": "Let $D,E,F$ be the points of tangency of $\\Omega$ with $BC,CA,AB$. Do an inversion at $\\Omega$. Notice that $\\Omega_i$ maps to itself $\\forall i \\in \\{ a,b,c \\}$. Then, $ \\{C',C\\} = \\Omega_a \\cap \\Omega_b \\rightarrow \\Omega_a \\cap \\Omega_b = \\{ C', C\\} $. However, $C$ is not its own inverse and thus $C'$ is the inverse of $C$, i.e. the midpoint of $DE$. An analogous thing occurs with $A$ and $B$. Notice\n\n$ R(A'B'C')=R(DEF)/2 = R(\\Omega)/2$\n\nsince $A'B'C'$ is the medial triangle of $DEF$. Done.", "Solution_8": "Inversion wrt $\\Omega$ maps $A,B,C$ to $A',B',C',$ so $A'B'C'$ is the medial triangle of $ABC-\\text{tangential triangle},$ thus done.", "Solution_9": "Take an inversion with respect to $\\Omega$. We have that $\\Omega_A$, $\\Omega_B$, and $\\Omega_C$ stay constant from the orthogonal condition. This means that $(A'B'C')$ is the inverse of $(ABC)$ wrt to $\\Omega$. Let $(ABC)=\\omega$. Let $OI \\cap \\omega = P,Q$ with $P$ and $I$ on opposite sides of $O$. Let $R$ and $S$ be the intersection of $\\omega$ with the line through $I$ perpendicular to $OI$. We have that $PI=\\frac{r^2}{R-\\sqrt{R^2-2Rr}}$, $QI=\\frac{r^2}{R+\\sqrt{R^2-2Rr}}$, and $RI=SI=\\frac{r^2}{\\sqrt{2Rr}}$. Let $X$ be the center of $(PQRS)$. We have that $X$ lies on $PQ$, so $XI=\\frac{r\\sqrt{R^2-2Rr}}{2R}$. This means the radius of the new circle is $$\\sqrt{XI^2+RI^2}=\\sqrt{\\frac{r^2(R^2-2Rr)}{4R^2}+\\frac{r^3}{2R}} = \\sqrt{\\frac{r^2R^2-2Rr^3}{4R^2}+\\frac{2r^3R}{4R^2}} = \\frac{r}{2},$$ as desired so we are done.", "Solution_10": "Let $A_1B_1C_1$ be the contact triangle of $ABC$. Now, we redefine the points $A', B', C'$ so that $A'B'C'$ is the medial triangle of $A_1B_1C_1$.\n\nBecause $A$ and $A'$ are clearly inverses wrt $\\Omega$, we know $A'$ belongs to $\\Omega_b$ and $\\Omega_c$. Using analogous reasoning allows us to recover the original definitions of $B'$ and $C'$ as well. Thus, since $(A'B'C')$ is the Nine-Point Circle of $A_1B_1C_1$, the desired result follows readily. $\\blacksquare$", "Solution_11": "Consider the inversion about $\\Omega$. Since $\\Omega_{a},\\Omega_{b},\\Omega_{c}$ are orthogonal, they are fixed. Now, intersections are preserved, so $A\\to A'$ and $A'\\to A$. Thus, $A'B'C'$ is the medial triangle of the contact triangle. We are done." } { "Tag": [ "algebra", "polynomial", "LaTeX" ], "Problem": "Find, with proof, all triples of real numbers $(a, b, c)$ such that all four roots of the polynomial $f(x) = x^4 +ax^3 +bx^2 +cx+b$ are positive integers. (The four roots need not be distinct.)", "Solution_1": "I did this by hand, so I hope I didn't make any mistakes:\r\n\r\n$\\left(x-r_{1}\\right)\\left(x-r_{2}\\right)\\left(x-r_{3}\\right)\\left(x-r_{4}\\right)=x^{4}-\\left(r_{1}+r_{2}+r_{3}+r_{4}\\right)x^{3}+\\left(r_{1}r_{2}+\\left(r_{1}+r_{2}\\right)\\left(r_{3}+r_{4}\\right)+r_{3}r_{4}\\right)x^{2}-\\left(r_{1}r_{2}\\left(r_{3}+r_{4}\\right)+r_{3}r_{4}\\left(r_{1}+r_{2}\\right)\\right)x+r_{1}r_{2}r_{3}r_{4}$\r\n\r\nThen we can set this equal to $x^{4}+ax^{3}+bx^{2}+cx+b$ to say:\r\n\r\n$a=-\\left(r_{1}+r_{2}+r_{3}+r_{4}\\right)$\r\n\r\n$b=r_{1}r_{2}+\\left(r_{1}+r_{2}\\right)\\left(r_{3}+r_{4}\\right)+r_{3}r_{4}=r_{1}r_{2}r_{3}r_{4}$\r\n\r\n$c=-\\left(r_{1}r_{2}\\left(r_{3}+r_{4}\\right)+r_{3}r_{4}\\left(r_{1}+r_{2}\\right)\\right)$\r\n\r\n...and then I guess it would require some more work to find all values. Is there a quicker way maybe?", "Solution_2": "This solution received a 5, but it is rather heavy on casework and very long.\r\n\r\n[hide]In a quartic equation, the coefficient of the $x^2$ term is the second symmetric sum of the roots, while the constant term is the product of the roots. Since they are equal, we have \n\n|\n\n$pqrs=pq+pr+ps+qr+qs+rs$\n\n|\n\nwhere p, q, r, and s are the four roots and are, as specified, positive integers. If we divide both sides by pqrs we get\n\n|\n\n$1=\\displaystyle\\frac{pq+pr+ps+qr+qs+rs}{pqrs}=\\displaystyle\\frac{1}{rs}+\\displaystyle\\frac{1}{qs}+\\displaystyle\\frac{1}{qr}+\\displaystyle\\frac{1}{ps}+\\displaystyle\\frac{1}{pr}+\\displaystyle\\frac{1}{pq}$\n\n|\n\nWe may assume that $rs\\leq qs, qr, ps, pr, pq$. We can also assume $r\\leq s \\leq q \\leq p$. So, if rs=2, r=1 and s=2. Substituting these values in, we have\n\n|\n\n$1=\\displaystyle\\frac{1}{2}+\\displaystyle\\frac{1}{2q}+\\displaystyle\\frac{1}{q}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{pq}\\Rightarrow \\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{2q}+\\displaystyle\\frac{1}{q}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{pq}$ \n\n|\n\nIt is clearly seen that, because of the $\\displaystyle\\frac{1}{q}$ term, $q>2$ so that we do not have a negative number equal to a sum of positive numbers. When q=3 we have \n\n|\n\n$\\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{6}+\\displaystyle\\frac{1}{3}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{3p}\\Rightarrow 0=\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{3p}$\n\n|\n\nThis is also impossible, so $q>3$. When q=4,\n\n|\n\n$\\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{8}+\\displaystyle\\frac{1}{4}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{4p}\\Rightarrow \\displaystyle\\frac{1}{8}=\\displaystyle\\frac{7}{4p}\\Rightarrow p=14$\n\n|\n\nSince 14 is a positive integer, $(r,s,q,p)=(1,2,4,14)$ is a possible set of roots. If q=5, we have\n\n|\n\n$\\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{10}+\\displaystyle\\frac{1}{5}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{5p}\\Rightarrow \\displaystyle\\frac{1}{5}=\\displaystyle\\frac{17}{10p}\\Rightarrow p=\\displaystyle\\frac{17}{2}$\n\n|\n\nThis is not a positive integer, so $q\\neq 5$. If q=6 we get\n\n|\n\n$\\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{12}+\\displaystyle\\frac{1}{6}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{6p}\\Rightarrow \\displaystyle\\frac{1}{4}=\\displaystyle\\frac{5}{3p} \\Rightarrow p=\\displaystyle\\frac{20}{3}$\n\n|\n\nAgain, we do not have a positive integer. We move on to q=7:\n\n|\n\n$\\displaystyle\\frac{1}{2}=\\displaystyle\\frac{1}{14}+\\displaystyle\\frac{1}{7}+\\displaystyle\\frac{1}{2p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{7p}\\Rightarrow \\displaystyle\\frac{2}{7}=\\displaystyle\\frac{23}{14p}\\Rightarrow p=\\displaystyle\\frac{23}{4}$\n\n|\n\nThis time, not only do we not have an integer, $pq$ and p is not an integer.\n\nWe next consider r=1, s=4, so that\n\n|\n\n$1=\\displaystyle\\frac{1}{4}+\\displaystyle\\frac{1}{4q}+\\displaystyle\\frac{1}{q}+\\displaystyle\\frac{1}{4p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{pq}\\Rightarrow \\displaystyle\\frac{3}{4}=\\displaystyle\\frac{1}{4q}+\\displaystyle\\frac{1}{q}+\\displaystyle\\frac{1}{4p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{pq}$\n\n|\n\nHere, due to the assumption that $s\\leq q$, we must start with q=4:\n\n|\n\n$\\displaystyle\\frac{3}{4}=\\displaystyle\\frac{1}{16}+\\displaystyle\\frac{1}{4}+\\displaystyle\\frac{1}{4p}+\\displaystyle\\frac{1}{p}+\\displaystyle\\frac{1}{4p}\\Rightarrow \\displaystyle\\frac{7}{16}=\\displaystyle\\frac{3}{2p}\\Rightarrow \\displaystyle\\frac{24}{7}$\n\n|\n\nThis is not an integer and already we have $p b$ for $ t > a.$ (Note that $ a$ might depend on $ x.$) Observe also that $ t^2 + x$ is positive and bounded away from zero for $ t > a.$\r\n\r\n$ \\int_0^a\\cos(\\frac {t^3}3 + xt)\\,dt$ is the integral of a continuous function over a compact interval; its convergence is not in doubt. That leaves the integral from $ a$ to $ \\infty.$ For that, we're going to integrate by parts.\r\n\r\n$ \\int_a^{\\infty}\\cos(\\frac {t^3}3 + xt)\\,dt = \\int_a^{\\infty}\\frac1{t^2 + x}\\,\\cos(\\frac {t^3}3 + xt)(t^2 + x)\\,dt$\r\n\r\n$ = \\frac1{t^2 + x}\\,\\sin(\\frac {t^3}3 + xt)\\right_a^{\\infty} + \\int_a^{\\infty}\\frac {2t}{(t^2 + x)^2}\\sin(\\frac {t^3}3 + xt)\\,dt$\r\n\r\nSince $ t^2 + x$ is positive and bounded away from zero on $ [a,\\infty)$ and since the required limit at infinity is zero, the boundary terms evaluate to a finite number. The final integral converges absolutely by comparison to\r\n\r\n$ \\int_a^{\\infty}\\frac {2t}{(t^2 + x)^2}\\,dt.$\r\n\r\nThe statement that this doesn't converge as a Lebesgue integral is nothing more than the statement that the original integral doesn't converge absolutely.\r\n\r\nNext question: is the quantity the integral converges to a differentiable function of $ x?$" } { "Tag": [ "calculus", "derivative", "analytic geometry", "conics", "parabola" ], "Problem": "Dear everyone,\r\n\r\n Can you please help me with the following problem?\r\n\r\na) Find the equations of all lines that are tangent to both y = x^2 + 2 and y = -2 - x^2. Indicate the points where the tangent line intersects with the curve.\r\n\r\nb) A man 2m tall walks at 5/3 meters / s along one edge of a straight road that is 10m wide. On the other side \r\n of the road is a light atop a pole that is 6m tall. How fast is the length of the man's shadow (on the \r\n ground) increasing when he is 15m beyond the point directly across the road from the pole?\r\n\r\n Thank you so much!\r\n\r\nBest,\r\nAngel", "Solution_1": "This belongs in the high school basics forum, or the calculus forum\r\n\r\n1. You have 2 coordinates for $ f(x)$ and two coordinates for $ f'(x)$ so sub them in a make four simultaneous equations\r\n\r\n2. ship A is $ 20 \\minus{} 15t$ miles from the Origin and ship B is $ 15 \\minus{} 20t$ miles\r\nNow just apply Pythagoras and differentiate\r\n\r\nEDIT: did you change your questions???", "Solution_2": "[quote=\"angel_09\"]\n\na) Find the equations of all lines that are tangent to both y = x^2 + 2 and y = -2 - x^2. Indicate the points where the tangent line intersects with the curve.Angel[/quote]\r\n\r\nThe two parabola are symmetric for $ x$ axis and for the origin. Thus we can set the tangent lines as $ y \\equal{} \\pm mx$.\r\nThe condition for which the parabola $ y \\equal{} x^2 \\plus{} 2$ touches to the tangent is $ x^2 \\minus{} mX \\plus{} 2 \\equal{} 0$ has double root.\r\n$ D \\equal{} m^2 \\minus{} 8 \\equal{} 0\\Longleftrightarrow m \\equal{} \\pm 2\\sqrt {2}$, yielding the tangent lines are $ y|\\equal{}\\pm \\sqrt{2}x$ and the tangentpoints are $ (\\pm \\sqrt {2}, 4)$.", "Solution_3": "Hi Ocha,\r\n\r\n I am sorry about that, cos I just realized that I posted the wrong Q.\r\n\r\n I was trying to PM you to say thanks to you. But since I am a new member, it forbids me to do so within the following 2 hrs.\r\n\r\n Sorry about that.\r\n\r\nThanks,\r\nAngel" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $x,y$ be real positive numbers such that $x^2+y^2=2$. Prove that\r\n\\[ (1-x)(1-y)+(1-x-y)^2\\leq1. \\]", "Solution_1": "[quote=\"pvthuan\"]Let $x,y$ be real positive numbers such that $x^2+y^2=2$. Prove that\n\\[ (1-x)(1-y)+(1-x-y)^2\\leq1. \\][/quote]\r\n\r\nequinalent to $(1-x)(1-y)leq 0$ :? \r\n\r\nThen square......\r\n\r\nis it ok??", "Solution_2": "First, we break the square $(1-x-y)^2=1-2(x+y)+x^2+y^2+2xy$, and then multiplying $(1-x)(1-y)$ out to have $1-x-y+xy$. Then the desired inequality reduces to\r\n\\[ 3-3(x+y)+3xy\\leq 0. \\]\r\nPut $s=x+y$, then $s\\leq2$, and $xy=\\tfrac12s^2-1$. The rest is for you." } { "Tag": [ "geometry", "trigonometry" ], "Problem": "Among all the triangles which have a fixed side $l$ and a fixed area $S$, determine for which triangles the product of the altitudes is maximum.", "Solution_1": "[hide]The altitude to the side $l$ is fixed, because the length of that altitude is twice $\\frac{S}{l}$. Hence, the other two altitudes, if the angle opposite $l$ is $\\theta$ and the two other sides are $a$ and $b$, are $a\\sin{\\theta}$ and $a\\sin{theta}$, so the product of those two altitudes is $ab\\sin^2{\\theta}$. However, the area is fixed, and the area is $\\frac{ab\\sin{\\theta}}{2}$. So, the product of the altitudes is $\\frac{2S}{l}2S\\sin{\\theta}=\\frac{4S^2\\sin{\\theta}}{l}$. Thus, we want to maximize $\\sin{\\theta}$ by getting it as close to a right angle as possible.[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$x,y,z\\ge0$. Prove or disprove:\r\n$\\sum\\sqrt{3x^2+3xy+3xz+y^2+z^2-2yz}\\ge3(x+y+z)$", "Solution_1": "[quote=\"Nameless\"]$x,y,z\\ge0$. Prove or disprove:\n$\\sum\\sqrt{3x^2+3xy+3xz+y^2+z^2-2yz}\\ge3(x+y+z)$[/quote]\r\nThis is wrong: $x=0,y=z=1.$ ;)", "Solution_2": "What about this:\r\n$\\sum\\sqrt{4x^2+4xy+4xz+y^2+z^2-2yz}\\ge3(x+y+z)$", "Solution_3": "Noone help me?? :( \r\nPlease help :help:" } { "Tag": [ "binomial coefficients", "algebra", "binomial theorem" ], "Problem": "x^-8 in (x^3 -2x^-2)^4\r\n\r\nx^20 in (x^3 -x^-2)^20\r\n\r\nFind the coefficient.", "Solution_1": "Hello Ms. Sexy,\r\n\r\nThese two problems are at most 'Intermediate' and should be placed in that forum. However, getting back to the two problems:\r\n\r\n$\\left(x^3-{2x^{-2}}\\right)^{4}$ = $x^{12}-8x^7+24x^2-\\frac{32}{x^3}+\\frac{16}{x^8}$\r\n\r\nTherefore, the coefficient of $x^{-8}$ is $16$...\r\n\r\nFor the second problem, you can use the binomial coefficients, but the cofficient of $x^{20}$ will be $125970$...", "Solution_2": "$\\left( {x^3 - x^{ - 2} } \\right)^{20} = \\sum\\limits_{k = 0}^{20} {\\left( \\begin{array}{l}\r\n 20 \\\\ \r\n k \\\\ \r\n \\end{array} \\right) \\cdot \\left( {x^3 } \\right)^{20 - k} \\cdot } \\left( { - 1 \\cdot x^{ - 2} } \\right)^k = $\r\n$ = \\sum\\limits_{k = 0}^{20} {\\left( \\begin{array}{l}\r\n 20 \\\\ \r\n k \\\\ \r\n \\end{array} \\right)\\left( { - 1} \\right)^k x^{60 - 5k} } $\r\nthen:\r\n$60 - 5k = 20 \\Leftrightarrow k = 8$\r\nthen coefficient of $x^{20} $ will be $\\left( \\begin{array}{l}\r\n 20 \\\\ \r\n 8 \\\\ \r\n \\end{array} \\right) = \\frac{{20!}}{{8! \\cdot 12!}}$", "Solution_3": "You should know the binomial theorem before you even attempt these problems. If you do you can see that for the first one $x^{-8}$ will happen on the final term of the expansion. Thus, your coeffecient should be $(-2)^4 = 16$.\r\n\r\nFor the second one you have to find the term where $x$ is raised to the 20th. If the max degree of x is 60, you can set $20 = 60-5r$ r must be 8 so it is the eigth term. 20 nCr 8 = 125970 which should be your final coefficient." } { "Tag": [], "Problem": "On the same set of axes are drawn the graph of $ y\\equal{}ax^2\\plus{}bx\\plus{}c$ and the graph of the equation obtained by replacing $ x$ by $ \\minus{}x$ in the given equation. If $ b \\neq 0$ and $ c \\neq 0$ these two graphs intersect:\r\n\r\n$ \\textbf{(A)}\\ \\text{in two points, one on the x\\minus{}axis and one on the y\\minus{}axis}\\\\\r\n\\textbf{(B)}\\ \\text{in one point located on neither axis} \\\\\r\n\\textbf{(C)}\\ \\text{only at the origin} \\\\\r\n\\textbf{(D)}\\ \\text{in one point on the x\\minus{}axis} \\\\\r\n\\textbf{(E)}\\ \\text{in one point on the y\\minus{}axis}$", "Solution_1": "[hide=\"Solution\"]Let $ f(x)\\equal{}ax^2\\plus{}bx\\plus{}c\\equal{}ax^2\\minus{}bx\\plus{}c\\implies bx\\equal{}\\minus{}bx\\implies 2bx\\equal{}0$.\nBecause $ b\\neq 0$, $ x\\equal{}0$.\n\nSo the answer is $ \\boxed{\\textbf{(E)}}$.[/hide]" } { "Tag": [ "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "5 problems were given at a math contest and it turned out that every problem was solved by at least $ 2/3$ of the contestants. Porve that there is a set of 3 problems solved by at least $ 1/5$ of the contestants.", "Solution_1": "let the number of the contestants $ m$\r\nand $ a_i$ is the number of the problems that $ i$th student solved.\r\n\r\nWant to show: $ \\displaystyle\\sum_{i \\equal{} 1}^{m}\\dbinom{a_i}{3} \\ge 2m$\r\n\r\nit is easy to check for positive integers $ a$ and $ b$, ($ a\\ge b \\plus{} 2$), $ \\dbinom{a}{3} \\plus{} \\dbinom{b}{3} \\ge \\dbinom{a \\minus{} 1}{3} \\plus{} \\dbinom{b \\plus{} 1}{3}$\r\nusing this truth and trivial inequality that $ \\displaystyle\\sum_{i \\equal{} 1}^{m}a_i \\ge \\dfrac{10}{3}m$, \r\nwe can know that the minimum of $ \\displaystyle\\sum_{i \\equal{} 1}^{m}\\dbinom{a_i}{3}$ appears\r\nwhen $ \\dfrac{2}{3}m$ of $ a_i$'s are 3 and $ \\dfrac{1}{3}m$ of $ a_i$'s are 4.\r\nand in this case, $ \\displaystyle\\sum_{i \\equal{} 1}^{m}\\dbinom{a_i}{3}$ is $ 2m$\r\n\r\nSo we proved 'Want to show'......\r\n\r\nBecause there are total 10 pairs of 3 problems and a pair can appear only one time from one contestant,\r\nthere is a pair which are solved by at least $ \\dfrac{2m}{10} \\equal{} \\dfrac{m}{5}$ people.(Because of 'Want to show')\r\n\r\nQED" } { "Tag": [ "LaTeX" ], "Problem": "I looked at my MiKtex package manager for the first time this morning and saw \"Obsolete\" on all but one of my packages. I realized I still have MiKtex 2.4.\r\n\r\nI just installed 2.6, but I don't know how to have TeXnicCenter switch programs from 2.4 to 2.6 without deleting any of my .tex files.\r\n\r\nHelp please?\r\nThanks!", "Solution_1": "Build -> Define Output Profiles and change the [i]Path to the (La)TeX compiler[/i] to MiKTeX 2.6's directory." } { "Tag": [], "Problem": "Sa se constriasca corpuri cu 4, 16, 25 elemente.\r\n\r\n\r\n Edit:", "Solution_1": "Ca sa construiesti un corp cu $p^{n}$ elemente(cu $p$ numar prim) este suficient sa gasesti un polinom ireductibil $f$ de grad $n$ peste $Z_{p}$ si sa consideri corpul $Z_{p}[X]/(f)$. Concret in cazurile cerute de tine putem avea \\[\\begin{array}{l}Z_{2}[X]/(X^{2}+X+1)\\,;\\\\ \\ \\\\ Z_{2}[X]/(X^{4}+X+1)\\,;\\\\ \\ \\\\ Z_{5}[X]/(X^{2}+2) \\,. \\end{array}\\]" } { "Tag": [ "abstract algebra", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "Find all the finite groups $ G$ with the property that any automorphism different from $ 1_{G}$ has an unique fix point.", "Solution_1": "Consider the inner automorphisms, this is: those of type $ x \\mapsto axa^{\\minus{}1}$ for some fixed $ a$.\r\nThis will give you that the group is abelian. Now continue.", "Solution_2": "For $ x\\in G\\setminus Z(G)$ we have the automorphism $ G\\to G,y\\mapsto xyx^{\\minus{}1}$ \r\nwith different fixed points $ 1,x$ and $ \\not\\equal{}\\text{id}_G$. Contradiction!\r\nSo $ G\\equal{}Z(G)$ is abelian, and using the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_finitely_generated_abelian_groups#Classification]classifikation of all finite abelian groups[/url]\r\nwe get $ G\\equal{}\\langle x\\rangle \\times H$ with some $ H\\le G$ and some $ x\\in G$ of order $ \\text{exp}(G)$.\r\nThen for $ H\\not \\equal{}1$ and $ 1\\not\\equal{}y\\in H$ we get the well-defined automorphism\r\n$ G\\to G, x^nz\\mapsto x^ny^nz$ for all $ n\\in\\mathbb{Z},z\\in H$\r\nwith $ x\\mapsto xy\\not\\equal{}x$ and fixed points $ H\\not\\equal{}1$. Contradiction!\r\n\r\nSo $ G\\equal{}\\langle x\\rangle$ is cyclic and for $ |G|>2$ we have the non-trivial automorphism\r\n$ G\\to G,g\\mapsto g^{\\minus{}1}$ with no fixed points $ \\not\\equal{}1$, i.e. no involutions in $ G$, i.e. $ |G|$ is odd.\r\nLet $ p$ be the smallest prime divisor of $ |G|$ and $ y\\in G$ of order $ p$.\r\nThen $ p\\plus{}1\\equal{}2\\cdot {p\\plus{}1\\over 2}$ is a product of primes $ 0$, $ a \\geq 0$, $ L > 0$, solve the boundary value problem\r\n$ ut$ \u2212 $ Du_{xx}$ \u2212 $ au$ = $ 0$, for all $ 0 \\leq x \\leq L$, $ t \\geq 0$\r\n$ u_{x}(0, t) \\equal{} u_{x}(L, t) \\equal{} 0$, for all $ t \\geq 0$\r\n$ u(x, 0) \\equal{} cos(\\pi x/L) \\plus{} cos(2\\pi x/L)$.\r\n\r\nThanks", "Solution_1": "$ u(x,t) \\equal{} T_1(t)\\cos(\\pi x/L) \\plus{} T_2(t)\\cos(2\\pi x/L)$", "Solution_2": "could you please explain how you came up with this answer?\r\nThanks", "Solution_3": "Notation question: should the PDE actually be this?\r\n\\[ u_t \\minus{} Du_{xx} \\minus{} au \\equal{} 0\r\n\\]\r\n(That is, $ u_t$ rather than $ ut$ - that would match with calling this a variant on the diffusion equation.)\r\n\r\nNote that what V.V. said isn't really an answer - it's an intermediate step. He's assumed that you can separate variables and that the $ x$ equation would become $ X''(x) \\plus{} \\lambda X(x) \\equal{} 0,\\ X'(0) \\equal{} X'(L) \\equal{} 0.$\r\n\r\nThat's a well-known eigenfunction equation, and its solutions are $ X(x) \\equal{} \\cos(k\\pi x/L),$ with $ \\lambda \\equal{} \\frac {k^2\\pi^2}{L^2}.$\r\n\r\nYour initial condition is a trigonometric polynomial in terms of precisely that form. Hence, provided separation of variables holds, the solution will have just those terms and no others, each multiplied by its own function of $ t.$\r\n\r\nWhat V.V. has not done yet is to look at the equation you get in $ t.$", "Solution_4": "$ u(x,t) \\equal{} T_1(t)\\cos(\\pi x/L) \\plus{} T_2(t)\\cos(2\\pi x/L)$\r\n\r\n$ u_t\\minus{}Du_{xx}\\minus{}au\\equal{}(\\dot{T}_1\\plus{}(D\\frac{\\pi^2}{L^2}\\minus{}a)T_1(t))\\cos(\\pi x/L)\\plus{}(\\dot{T}_2\\plus{}(D\\frac{4\\pi^2}{L^2}\\minus{}a)T_2(t))\\cos(2\\pi x/L)$.\r\n\r\nHence, $ \\dot{T}_1\\plus{}(D\\frac{\\pi^2}{L^2}\\minus{}a)T_1(t)\\equal{}0$, $ \\dot{T}_2\\plus{}(D\\frac{4\\pi^2}{L^2}\\minus{}a)T_2(t)\\equal{}0$.\r\nFrom initial condition $ T_1(0)\\equal{}1$, $ T_2(0)\\equal{}1$.", "Solution_5": "Is this what i should do to prove that the equation separates?\r\n\r\n$ u(x,t) \\equal{} X(x)T(t)$ \r\n$ u_t$ - $ Du_{xx}$ - $ au$ = $ 0$ \r\n$ XT' \\minus{} DX''T \\minus{} aXT \\equal{} 0$ \r\n$ XT' \\minus{} T(DX'' \\plus{} aX) \\equal{} 0$ \r\n\r\n$ T'/T \\equal{} [DX''/X \\plus{} a) \\equal{} \\minus{}h$ \r\n\r\n$ DX'' \\plus{} (a \\plus{} h)X \\equal{} 0$\r\n\r\nPlease correct me.\r\nThanks", "Solution_6": "Well, it's nice to have some sort of argument (in English) other than just typing a bunch of symbols.\r\n\r\nEssentially yes, you show that you can write it as\r\n\r\n$ \\frac {T'}T \\minus{} D \\frac {X''}X \\minus{} a \\equal{} 0$, so both the parts in $ X$ and $ T$ should be equal to some constant, which you call $ h \\in \\mathbb{R}$ apparently.\r\n\r\nBut why is that true?", "Solution_7": "[quote=\"georgech\"]Any ideas?\n\nAssume a solution of the form $ u(x, t) = X(x)T(t)$ to the modified diffusion equation $ ut$ \u2212 $ Du_{xx}$ \u2212 $ au$ = $ 0$. First show that the equation separates and find the general solution for $ X(x)$ and $ T(t)$. \nNext, assuming that $ D > 0$, $ a \\geq 0$, $ L > 0$, solve the boundary value problem\n$ ut$ \u2212 $ Du_{xx}$ \u2212 $ au$ = $ 0$, for all $ 0 \\leq x \\leq L$, $ t \\geq 0$\n$ u_{x}(0, t) = u_{x}(L, t) = 0$, for all $ t \\geq 0$\n$ u(x, 0) = cos(\\pi x/L) + cos(2\\pi x/L)$.\n\nThanks[/quote]\r\n\r\n$ \\forall \\varphi \\in C^2 ( 0 ;L )$, we have: $ \\int _0 ^L u_t \\, \\varphi\\, \\text{d} x - D \\int _0 ^L u_{xx} \\, \\varphi \\, \\text{d} x - a \\int _0 ^L u \\, \\varphi \\ \\text{d} x = 0$\r\n\r\nwe use the Green formula:\r\n\r\n$ \\int _0 ^L u_t \\, \\varphi\\, \\text{d} x - D \\int _0 ^L u \\, \\varphi _{xx}\\, \\text{d} x - \\left. \\left[ \\varphi \\, u_x - \\varphi _x u \\right] \\right|_{x = 0}^{x = L} - a \\int _0 ^L u \\, \\varphi \\ \\text{d} x = 0 \\;\\; (1)$\r\n\r\nSolve the Sturm-Liouville problem : $ \\left\\{\\begin{array}{l} - \\varphi '' = \\lambda \\varphi \\\\\r\n\\varphi_{x} (0) = \\varphi_{x} (L) = 0 \\end{array}\\right. \\Rightarrow \\left\\{\\begin{array}{l} \\varphi_n(x) = \\cos \\left( \\frac { n \\pi x }{ L} \\right) \\\\\r\n\\lambda _n = \\frac { n^2 \\pi ^2 }{ L^2 } \\end{array}\\right. \\;\\; (2)$\r\n\r\nas we know $ \\overline{\\left < \\left\\{\\varphi _n \\right\\} \\right > }^{L^2} = L^2 (0;L)$ and $ \\left < \\varphi _n , \\varphi _m \\right > = \\frac {L}{2} \\delta _{m n}$. From $ (1)(2)$, we have:\r\n\r\n$ \\frac {\\text{d}}{\\text{d} t } \\left[ \\int _0 ^L u\\, \\varphi _n \\, \\text{d} x \\right] + ( \\lambda _n D - a) \\int _0 ^L u \\, \\varphi _{n}\\, \\text{d} x = 0$\r\n\r\nie , $ \\frac {\\text{d}}{\\text{d} s } \\left\\{\\mathrm{e}^{ ( \\lambda _n D - a) s } \\int _0 ^L u (x,s)\\, \\varphi _n (x) \\, \\text{d} x \\right\\} = 0$\r\n\r\nintegrate $ s \\in (0;t)$\r\n\r\n \\begin{align} \\left < u , \\varphi _n \\right > & = \\int _0 ^L u\\, \\varphi _n \\, \\text{d} x = \\mathrm{e}^{ - ( \\lambda _n D - a) t } \\int_0^L u( x , 0) \\, \\varphi _n ( x ) \\text{d} x \\nonumber \\\\\r\n& = \\mathrm{e}^{ - ( \\lambda _n D - a) t } \\int_0^L \\left[ \\cos \\left( \\frac {\\pi x}{L} \\right) + \\cos \\left( \\frac {2\\pi x}{L} \\right) \\right] \\, \\cos \\left( \\frac { n \\pi x }{ L} \\right) \\text{d} x \\nonumber \\end{align}\r\n\r\nbecause $ \\left < \\varphi _n , \\varphi _m \\right > = \\frac {L}{2} \\delta _{m n}$, we have\r\n\r\n $ \\left < u , \\varphi _n \\right > = \\left\\{\\begin{array}{lll} \\frac {L}{2} \\mathrm{e}^{ - ( \\pi ^2 D L^{ - 2} - a) t } & \\text{if} & n = 1 \\\\\r\n\\frac {L}{2} \\mathrm{e}^{ - (4 \\pi ^2 D L^{ - 2} - a) t } & \\text{if} & n = 2 \\\\\r\n0 & \\text{if} & n > 2 \\end{array} \\right.$\r\n\r\nexpand Fourier $ u(x,t) = \\sum\\limits_{n = 1}^{\\infty } \\frac {\\left < u , \\varphi _n \\right > }{\\left\\| \\varphi _n \\right\\|^2 } \\, \\varphi _n (x)$ \r\n\r\nThe solution is:\r\n\r\n$ u(x,t) = \\mathrm{e}^{ - ( \\pi ^2 D L^{ - 2} - a) t } \\cos \\left( \\frac {\\pi x}{L} \\right) + \\mathrm{e}^{ - (4 \\pi ^2 D L^{ - 2} - a) t } \\cos \\left( \\frac {2\\pi x}{L} \\right)$\r\n\r\ncheer! :lol:", "Solution_8": "Let me combine together the work of the last three posters in a classical separation of variables argument.\r\n\r\nWe look for solutions of the form $ u\\equal{}X(x)T(t).$ Put that into the original equation and divide by $ XT$ to get\r\n\\[ \\frac{T'}{T}\\minus{}a\\equal{}D\\frac{X''}{X}\\equal{}\\lambda\\]\r\nThe last part , the $ \\equal{}\\lambda,$ comes from the fact that the only way a function of $ x$ alone could equal a function of $ t$ alone is for both to be constant.\r\n\r\nThe equation $ \\frac{X''}{X}\\equal{}\\frac{\\lambda}{D},\\ X'(0)\\equal{}X'(L)\\equal{}0$ has the well-known solutions $ X(x)\\equal{}\\cos\\left(\\frac{k\\pi x}{L}\\right)$ for $ k\\equal{}0,1,2,\\dots,$ with $ \\frac{\\lambda}{D}\\equal{}\\minus{}\\frac{k^2\\pi^2}{L^2}$ or $ \\lambda\\equal{}\\minus{}\\frac{k^2\\pi^2D}{L^2}.$\r\n\r\nNow bring this back over to the other equation:\r\n\r\n$ \\frac{T'}{T}\\minus{}a\\equal{}\\minus{}\\frac{k^2\\pi^2D}{L^2}$\r\n\r\n$ T'\\equal{}\\left(a\\minus{}\\frac{k^2\\pi^2D}{L^2}\\right)T$\r\n\r\nThis is the equation of exponential growth or decay, and has solutions\r\n\r\n$ T\\equal{}Ce^{\\left(a\\minus{}\\frac{k^2\\pi^2D}{L^2}\\right)t}.$\r\n\r\nSince the original equation is linear and homogeneous, that means that we can construct solutions from arbitrary superpositions of these functions:\r\n\r\n$ u(x,t)\\equal{}\\sum_{k\\equal{}0}^{\\infty}a_ke^{\\left(a\\minus{}\\frac{k^2\\pi^2D}{L^2}\\right)t}\\cos\\left(\\frac{k\\pi x}L\\right).$\r\n\r\nFrom this $ u(x,0)\\equal{}\\sum_{k\\equal{}0}^{\\infty}a_k\\cos\\left(\\frac{k\\pi x}L\\right).$\r\n\r\nBut we are given that $ u(x,0)\\equal{}\\cos\\left(\\frac{\\pi x}L\\right)\\plus{}\\cos\\left(\\frac{2\\pi x}L\\right).$ That means that we can match the initial condition by setting $ a_1\\equal{}1,a_2\\equal{}1,$ and all other $ a_k\\equal{}0.$ This gives us a solution of\r\n\\[ u(x,t)\\equal{}e^{\\left(a\\minus{}\\frac{\\pi^2D}{L^2}\\right)t}\\cos\\left(\\frac{\\pi x}L\\right)\\plus{}e^{\\left(a\\minus{}\\frac{4\\pi^2D}{L^2}\\right)t}\\cos\\left(\\frac{2\\pi x}L\\right)\\]\r\n\r\nIn order to claim that this is the solution, we need some sort of uniqueness theorem. That is, we've shown that we have a solution that matches the initial condition; a uniqueness theorem would say that there is only one such solution.\r\n\r\nNote that this solution has certain curious properties, depending on the relationship among $ a,D,$ and $ L.$ \r\n\r\nIf $ a<\\frac{\\pi^2D}{L^2},$ then the solution tends to zero as $ t\\to\\infty.$\r\n\r\nIf $ a>\\frac{\\pi^2D}{L^2},$ then the solution grows without bound as $ t\\to\\infty.$\r\n\r\n\"The\" diffusion equation (heat equation) would have $ a\\equal{}0$ and the solutions would tend to zero.", "Solution_9": "hey brethrens,\r\n\r\nSolve the problem $ \\left\\{ \\begin{array}{l} \\frac{\\partial u}{ \\partial t} \\minus{} \\frac{1}{r} \\frac{ \\partial }{ \\partial r} \\left( r \\frac{\\partial u }{ \\partial r}\\right) \\equal{} 0 \\; , \\; 0 \\le r \\le 1 \\; , \\, t \\ge 0 \\\\\\\\ u( r , 0) \\equal{} \\tau _0 \\;(\\text{const}) \\\\\\\\ u ( 1,t) \\equal{} 0 \\end{array}\\right.$", "Solution_10": "I tried to find such a uniqueness theorem showing that this is the only solution but failed.. :( ...May i ask what do you have in mind?\r\nThanks", "Solution_11": "[quote=\"georgech\"]I tried to find such a uniqueness theorem showing that [u][b]this[/b][/u] is the only solution but failed.. :( ...May i ask what do you have in mind?\nThanks[/quote]\r\n\r\nwhat is \"this\" ?" } { "Tag": [ "algorithm", "linear algebra", "matrix" ], "Problem": "Let A be a matrix of mxn.What is the algorithm that shows its elements in this order:the elements of the first column upside down then the elements of the second column from the bottom to the top then the elements of the third column from the top to the bottom and so on.", "Solution_1": "Depends on what you mean by upside down and what you mean by top to bottom. (In terms of array offsets.) But I think you're looking for something like this:\r\n[code]\n// The input is two-dimensional array matrix[m][n].\n\nfor (int i = 0; i < m; i ++) {\n\tfor (int j = 0; j < n; j ++) {\n\t\tprint(matrix[i][j]);\n\t}\n\ti ++;\n\tif (i >= m) break;\n\tfor (int j = n - 1; j >= 0; j --) {\n\t\tprint(matrix[i][j]);\n\t}\n}\n[/code]\r\n\r\nYou may need to change the indexing order depending on which index to the array are rows and which are columns.", "Solution_2": "I'm sorry for not well expressing myself.I mean the output should give the elements of the first column from its top to its bottom, then the elements of the second column from its bottom to its top and so on.While we move through a column i doesn't change but I can't resume this in a single condition for all columns", "Solution_3": "[code]// For an array matrix[][]\n\nfor (int col = 0; col < matrix[].length; col++) { // For each column col:\n if (col % 2 == 0) { // If col is even, print the column top to bottom\n for (int row = 0; row < matrix.length; row++) {\n print(matrix[row][col]);\n }\n } else { // If col is odd, print the column bottom to top\n for (int row = matrix.length - 1; row >= 0; row--) {\n print(matrix[row][col]);\n }\n }\n}[/code]" } { "Tag": [ "number theory", "least common multiple", "geometry", "circumcircle" ], "Problem": "given three points in a plane , construct a quadrilateral for which these points are mid points of three\r\n\r\nSuccessive equal sides.", "Solution_1": "Let K, L, M be the 3 given points. Assume that the problem has been solved and K, L, M are the midpoints of the quadrilateral sides AB = BC = CA, respectively. The triangles $\\triangle KBL, \\triangle LCM$ are both isosceles, hence, the vertices B, C lie on the perpendicular bisectors p, q of the known segments KL, LM. These perpendicular bisectors can be constructed and they intersect in the circumcenter O of the triangle $\\triangle KLM$. Since L is the midpoint of BC, OL is the O-median of the triangle $\\triangle OBC$. This means that the areas of the triangles $\\triangle OBL, \\triangle OCL$ are equal. Let $h_p, h_q$ be the L-altitudes of these 2 triangles, respectively, the distances of the given point L from the known lines p, q. We have\r\n\r\n$h_p \\cdot OB = h_q \\cdot OC,\\ \\ \\frac{h_p}{h_q} = \\frac{OC}{OB}$\r\n\r\nTransfer the segments $h_p, h_q$ to the rays $[OC, [OB,$ of the lines p, q, respectively, so that $C' \\in [OC,\\ B' \\in [OB$ and $OC' = h_p,\\ OB' = h_q$. The triangles $\\triangle OB'C' \\sim \\triangle OBC$ with the common angle $\\angle B'OC' \\equiv \\angle BOC$ are then similar by SAS, which implies that their sides $B'C' \\parallel BC$ are parallel. Draw a line a through the point L parallel to the line B'C' intersecting the lines p, q at the points B, C, respectively, 2 vertices of the desired quadrilateral ABCD. Complete the quadrilateral by drawing the lines BK, CM and constructing the remaining vertices A, D, such that AK = BK, DM = CM." } { "Tag": [ "geometry", "rectangle", "LaTeX", "number theory", "least common multiple", "algebra", "system of equations" ], "Problem": "If someone could post the solutions to the problems on the AMC 10B, that'd be really really nice. Here are a few problems:\r\n\r\n1:\r\n\r\n$What is (-1)^1 + (-1)^2 + ... + (-1)^2006?$\r\n\r\n$(A) -2006$\r\n$(B) -1$\r\n$(C) 0$\r\n$(D) 1$\r\n$(E) 2006$\r\n\r\n2:\r\n\r\n$For real numbers x and y, define x * y = (x+y)(x-y). What is 3 * (4 * 5)?$\r\n\r\n$(A) -72$\r\n$(B) -27$\r\n$(C) -24$\r\n$(D) 24$\r\n$(E) 72$\r\n\r\n3:\r\n\r\n$A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?$\r\n\r\n$(A) 10$\r\n$(B) 14$\r\n$(C) 17$\r\n$(D) 20$\r\n$(E) 24$\r\n\r\nThese are, of course, the easiest ones... here are harder ones :D \r\n\r\n16:\r\n\r\n$Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?$\r\n\r\n$(A) Tuesday$\r\n$(B) Wednesday$\r\n$(C) Thursday$\r\n$(D) Friday$\r\n$(E) Saturday$\r\n\r\n20:\r\n\r\n$In rectangle ABCD, we have A = (6, -22), B = (2006, 178), and D = (8, y), for some integer y. What is the area of rectangle ABCD?$\r\n\r\n$(A) 4000$\r\n$(B) 4040$\r\n$(C) 4400$\r\n$(D) 40000$\r\n$(E) 40400$\r\n\r\n25:\r\n\r\n$Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. \\\"Look, daddy!\\\" she exclaims. \\\"That number is evenly divisible by the age of each of us kids!\\\" \\\"That's right,\\\" replies Mr. Jones, \\\"and the last two digits just happen to be my age.\\\" Which of the following is [i: 988c6147b8][b: 988c6147b8]not[/b: 988c6147b8][/i: 988c6147b8] the age of one of Mr. Jones's children?$\r\n\r\n$(A) 4$\r\n$(B) 5$\r\n$(C) 6$\r\n$(D) 7$\r\n$(E) 8$", "Solution_1": "Much :coool: er $\\text{\\LaTeX}$\r\n\r\n1:\r\n\r\n$\\text{What is }(-1)^1 + (-1)^2 + ... + (-1)^{2006}?$\r\n$\\text{(A) -2006}$\r\n$\\text{(B) -1}$\r\n$\\text{(C) 0}$\r\n$\\text{(D) 1}$\r\n$\\text{(E) 2006}$\r\n\r\n2:\r\n\r\n$\\text{For real numbers }x \\text{ and } y\\text{, define }x * y = (x+y)(x-y)\\text{. What is }3 * (4 * 5)?$\r\n\r\n$\\text{(A) -72}$\r\n$\\text{(B) -27}$\r\n$\\text{(C) -24}$\r\n$\\text{(D) 24}$\r\n$\\text{(E) 72}$\r\n\r\n3:\r\n\r\n$\\text{A football game was played between two teams, the Cougars and the Panthers.}$ $\\text{The two teams scored a total of 34 points, and the Cougars won}$ $\\text{by a margin of 14 points. How many points did the Panthers score?}$\r\n\r\n$\\text{(A) 10}$\r\n$\\text{(B) 14}$\r\n$\\text{(C) 17}$\r\n$\\text{(D) 20}$\r\n$\\text{(E) 24}$\r\n\r\n16:\r\n\r\n$\\text{Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?}$\r\n\r\n$\\text{(A) Tuesday}$\r\n$\\text{(B) Wednesday}$\r\n$\\text{(C) Thursday}$\r\n$\\text{(D) Friday}$\r\n$\\text{(E) Saturday}$\r\n\r\n20:\r\n\r\nIn rectangle$ABCD$, we have $A = (6, -22),\\ B = (2006, 178)$, and $D = (8, y)$ for some integer $y$. What is the area of rectangle $ABCD?$\r\n\r\n$\\text{(A) 4000}$\r\n$\\text{(B) 4040}$\r\n$\\text{(C) 4400}$\r\n$\\text{(D) 40000}$\r\n$\\text{(E) 40400}$\r\n\r\n25:\r\n\r\n$\\text{Mr. Jones has eight children of different ages. On a family trip his oldest child,}$ $\\text{who is 9, spots a license plate with a 4-digit number in which each of two}$ $\\text{digits appears two times. \\\"Look, daddy!\\\" she exclaims.}$ $\\text{\\\"That number is evenly divisible by the age of each of us kids!\\\"}$ $\\text{\\\"That's right,\\\" replies Mr. Jones,}$ $\\text{ \\\"and the last two digits just happen to be my age.\\\" Which of the following is the age of one of Mr. Jones's children?}$\r\n\r\n$\\text{(A) 4}$\r\n$\\text{(B) 5}$\r\n$\\text{(C) 6}$\r\n$\\text{(D) 7}$\r\n$\\text{(E) 8}$\r\n\r\n\r\nOK my best solutions:\r\n[hide=\"numero uno\"]\nFor the odd powers, it equals -1\nFor the even powers, it equals 1\nThus, we can group the powers into 1003 groups of 1 odd 1 even.\nThe sum of 1 odd and 1 even is $1+-1=0$\n$0\\cdot1003=0$, the answer[/hide]\n[hide=\"numero dos\"]\n$4*5=(4+5)(4-5)=9\\cdot-1=-9$\n$3*-9=(3+(-9))(3-(-9))=-6\\cdot12=-72$[/hide]\n[hide=\"numero tres\"]\nIf $x$ is the Cougar's score and $y$ is the Panther's score, you can set up the system of equations:\n$x+y=34$\n$x-y=14$\nSolving gives $(24,10)$, so the score of the Panthers is 10[/hide]\n[hide=\"numero diez y seis\"] \nYou have to first sum the number of days between the leap years. If you sum correctly you get\nA number that is $6\\mod{7}$(im too lazy to put the number here), so Sunday+6 days=Saturday[/hide]\n[hide=\"numero veinte\"]\nI didn't get it :blush: [/hide]\n[hide=\"numero veinte y cinco\"]\nClearly only one of the ages 8-1 is not included.\nIt can't be 1, obviously\nIf it is 2, then 4,6,8 can't be there\nIf it is 3, then 6 can't be there\nIf it is 4, then 8 can't be there\nIt can be 5\nIf it is 6, then 2,3,4,8 can't be there\nIt can be 7\nIt can be 8\nLet's try 5 first.\nThe lcm of the remaining numbers is \n504\nThus the license plate must be a multiple of 504, but not divisible by 5.\nIf you keep pressing +504 on your calculator, you will get to 5544.\nIt fits all of the requirements, and the dad is 44 (makes sense).\nThus the answer is 5[/hide]\r\n\r\nto Karth: can you post more problems so I can see if I was right or not (I don't have a copy of the test and am not waiting 3 weeks to get the results; Ill even do solutions to the ones I know) :nhl: :moose: :football: :nhl: :moose: :football: :nhl: :moose:" } { "Tag": [ "function", "calculus", "derivative", "calculus computations" ], "Problem": "If $ \\alpha$ is a real number and $ \\alpha x \\minus{} \\frac{x^{3}}{1\\plus{}x^{2}}$ is an increasing function for all values of $ x$, show that $ \\alpha \\ge \\frac{9}{8}$.", "Solution_1": "If $ f(x)$ is an increasing function, then $ \\forall_x,f'(x)\\geq 0$ .. We obtain something like $ \\forall_x, Ax^4\\plus{}Bx^2\\plus{}C \\geq 0$ and so on .. :cool:", "Solution_2": "[quote=\"AndrewTom\"]If $ \\alpha$ is a real number and $ \\alpha x \\minus{} \\frac {x^{3}}{1 \\plus{} x^{2}}$ is an increasing function for all values of $ x$, show that $ \\alpha \\ge \\frac {9}{8}$.[/quote]\r\n[hide=\"Solution?\"]Since it is an increasing function for all x, this means as x approaches infinity, we obtain larger and larger values, thus this is also concave up. Thus we must have that $ f'(x)\\ge0$, and $ f''(x)\\ge0$.\n$ f'(x)\\equal{}\\alpha\\minus{}\\frac{x^{4}\\plus{}3x^{2}}{(x^{2}\\plus{}1)^{2}}\\ge0$\n$ f''(x)\\equal{}\\frac{\\minus{}2x(x^{2}\\minus{}3)}{(x^{2}\\plus{}1)^{3}}\\ge0$\nFrom the 2nd equation we get that x must be in the interval $ (\\minus{}\\infty,\\minus{}\\sqrt{3}]U[0,\\sqrt{3}]$\nTo find the lowest possible value of $ \\alpha$, we plug in the max value of x into the first equation. Plugging in $ x\\equal{}\\sqrt{3}$ into the first derivative results in $ \\alpha\\ge\\frac{9}{8}$[/hide]", "Solution_3": "[quote=\"ckck\"]\n...\n$ f'(x) \\equal{} \\alpha \\minus{} \\frac {x^{4} \\plus{} 3x^{2}}{(x^{2} \\plus{} 1)^{2}}\\ge0$\n$ f''(x) \\equal{} \\frac { \\minus{} 2x(x^{2} \\minus{} 3)}{(x^{2} \\plus{} 1)^{3}}\\ge0$ \n...\n[/quote]\r\nwarning , $ f''(x) \\equal{} \\frac { 2x(x^{2} \\minus{} 3)}{(x^{2} \\plus{} 1)^{3}}$ . \r\n$ \\minus{}\\infty < x < \\minus{}\\sqrt 3 \\Longrightarrow f''(x)<0 \\Longrightarrow f'(x) \\ decreasing$\r\n$ \\minus{}\\sqrt 3 < x < 0 \\Longrightarrow f''(x)>0 \\Longrightarrow f'(x) \\ increasing$\r\n$ 0< x < \\sqrt 3 \\Longrightarrow f''(x)>0 \\Longrightarrow f'(x) \\ decreasing$\r\n$ \\sqrt 3 < x < \\minus{} \\Longrightarrow <\\infty f''(x)<0 \\Longrightarrow f'(x) \\ increasing$\r\nThe minimum,maximum of $ f'$ are reached for $ x\\equal{}\\pm\\sqrt 3, x\\equal{}0$ so : $ 0\\leq \\alpha\\minus{}\\frac 98\\equal{}f'(\\pm\\sqrt 3) \\leq f'(x) \\leq f'(0)\\equal{}\\alpha$\r\n :cool:", "Solution_4": "$ f(x) \\equal{} \\alpha x \\minus{} \\frac {x^3}{1 \\plus{} x^2} \\equal{} (\\alpha \\minus{} 1)x \\plus{} \\frac {x}{1 \\plus{} x^2}$, \r\n\r\n$ f'(x) \\equal{} \\alpha \\minus{} 1 \\minus{} \\frac {x^2 \\minus{} 1}{(1 \\plus{} x^2)^2}\\geq 0,\\ \\forall x\\in{\\mathbb{R}}$\r\n\r\n$ \\Longleftrightarrow \\frac {x^2 \\minus{} 1}{(x^2 \\plus{} 1)^2}\\leq \\alpha \\minus{} 1, \\ \\forall x\\in{\\mathbb{R}}$\r\n\r\nLet $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{(x^2 \\plus{} 1)^2} \\equal{} \\minus{} \\left(\\frac {1}{x^2 \\plus{} 1} \\minus{} \\frac {1}{4}\\right)^2 \\plus{} \\frac {1}{8}$\r\n\r\nSince $ 0 < \\frac {1}{x^2 \\plus{} 1}\\leq 1\\ \\therefore g_{max} \\equal{} g(\\pm 3) \\equal{} \\frac {1}{8}$\r\n\r\n$ \\therefore \\frac 18\\leq \\alpha \\minus{} 1$, yielding $ \\alpha \\geq \\frac 98$", "Solution_5": "Thanks. I had:\r\n\r\n$ f'(x) \\equal{} \\alpha \\minus{} \\frac{3x^{2}\\plus{}x^{4}}{(1\\plus{}x^{2})^{2}}$\r\n\r\n$ f'(x) \\ge 0$ if $ \\alpha \\ge y\\equal{}\\frac{3x^{2}\\minus{}x^{4}}{1\\plus{}2x^{2}\\plus{}x^{4}} \\equal{}1 \\plus{} \\frac{x^{2}\\minus{}1}{1\\plus{}2x^{2}\\plus{}x^{4}} \\equal{} 1 \\plus{} \\frac{1}{1\\plus{}x^{2}}\\minus{}\\frac{2}{(1\\plus{}x^{2})^{2}}$\r\n\r\n$ \\frac{dy}{dx} \\equal{} \\minus{}\\frac{2x}{(1\\plus{}x^{2})^{2}} \\plus{} \\frac{8x}{(1\\plus{}x^{2})^{3}}$\r\n\r\n$ \\frac{dy}{dx} \\equal{} 0$ for $ 1\\plus{}x^{2} \\equal{} 4$, that is, $ x^{2}\\equal{}3$.\r\n\r\nDifferentiating again, we see that this gives a maximum and the maximum value is $ \\frac{9}{8}$." } { "Tag": [ "geometry", "conics", "projective geometry", "geometry unsolved" ], "Problem": "the incircle of triangle ABC is tangenc to BC at Z.AZ intersect incircle at P.BP and PC intersect incircle at Y and X.prove that AZ,CY,BX are concur.", "Solution_1": "in few minutes later i become crazy because of this question hurry up please please :(", "Solution_2": "where are the lovers of geometry.i can conclude nobody can solve my problems.", "Solution_3": "The second post 9 (!!!) minutes after the first one.\r\nAre you stupid or what?", "Solution_4": "yes i become stupid because of this quastion :( :(", "Solution_5": "[b]More generally:[/b]\r\n$ \\mbox{Let Z the tangency point of BC an inscribed conic in the triangle ABC, and AZ intersect inconic at P. } \\\\\r\n\\mbox{BP and PC intersect the conic at Y and X. Prove that AZ,CY,BX are concur.}$ \r\n(view Figure!)", "Solution_6": "ok,why?(can you prove the main quastion with triangle geometry)", "Solution_7": "i realy conufsed.a realy nice and hard problem.but ithink you can solve it.help help help help help :( :(", "Solution_8": "you are salak", "Solution_9": "Let $ E,F$ be the touch points of the incircle (with center $ I$) with $ AC,AB$\r\nthen by using similarities (or directly the well-known theorem) we get $ FY . PZ \\equal{} ZY . PF$ and similarly $ EX.PZ \\equal{} EP.ZX$\r\nnow we get $ FY.PZ.EP.ZX \\equal{} ZY.PF.EX.PZ$ or in other words $ FY.EP.ZX \\equal{} ZY.PF.EX$ which is in fact the sufficient and needed condition for which $ ZP,EY,FX$ are concurrent(use c\u00e9va's theorem sinus form for proving this).\r\nlet the tangent at point $ P$ to $ (I)$ intersect $ BC$ at $ T$ then by the well-known theorem(of pole and polar) we have the concurrency of $ EF,XY,PP,BC$ at point $ T$.($ PP$ is the tangent)\r\nalso we know that $ (TZBC) \\equal{} \\minus{} 1$ and since $ T,Y,X$ are collinear it implies the concurrency of $ BX,CY,PZ$ and we are done.\r\nalso note that we can prove $ FY,EX,AZ$ are concurrent (since we know the concurrency of $ BX,CY,PZ$ we can prove this by Menelaus's theorem).\r\nLet this point be $ S \\equal{} (FY,EX,AZ)$ then by pascal's theorem we can prove that $ (SPAZ) \\equal{} \\minus{} 1$.", "Solution_10": "you are very good in geometry." } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector is to be painted with one of four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.)", "Solution_1": "There is orginally $ 4!$ ways but you must divide by 4 to compensate for rotation. So the answer is $ 3!\\equal{}6$", "Solution_2": "The answer is 70 (I know from reviewing) and remember, you can have colors repeated.", "Solution_3": "Can someone elaborate on how the answer is 70?", "Solution_4": "We use casework! (4,0,0,0) denotes there are 4 of one color, 0 of another, 0 of 3rd, etc.\r\n\r\n(4,0,0,0)=4 ways (one for each color)\r\n\r\n(3,1,0,0)=4 ways for the 3 and 3 ways for the 1. 4x3=12 ways.\r\n\r\n(2,2,0,0)=4c2=6 ways to pick colors. Either the same colored parts must touch or be opposite, for 2 ways of placing onjce you have selected colors. Thus 6x2=12\r\n\r\n(2,1,1,0)=4 ways to pick the 2, 3c2=3 ways to pick the 1s. 12 ways to pick colors. There are normally 6 ways to arrange table, but we must divide by 2 as two colored parts are indistinct. Thus 12x3=36 ways.\r\n\r\n(1,1,1,1)=1 way to pick colors, 6 ways to arrange (4!/4) so 6 ways.\r\n\r\nAdding, 4+12+12+36+6=70" } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "I think this has been discussed before, but never actually posted in it's own topic, as it deserves :).\r\n\r\nLet $A=(a_{ij})$ be a complex matrix. Prove that:\r\n\r\n(a) The eigenvalues of $A$ lie in the region $\\bigcup_{i=1}^nD_i$, where $D_i=\\{z\\in\\mathbb C\\ |\\ |z-a_{ii}|\\le\\sum_{j\\ne i}|a_{ij}|\\}$;\r\n(b) If the union of $k$ of the disks $D_i$ is disjoint from the other $n-k$ disks, then these $k$ disks contain exactly $k$ eigenvalues of $A$.", "Solution_1": "The numerical analysts that I know declare this to be their favorite theorem about matrices.", "Solution_2": "a)Let $\\lambda$ an eigenvalue.Take a vector $X$(with $||X||=1$) such that $AX=\\lambda X$.There is $i_0$ such that $x_{i_0}=1$.\r\n$AX=\\lambda X \\rightarrow$$\\sum_{j=1}^n a_{i_0j}x_j=\\lambda x_{i_0}\\rightarrow$ \r\n$\\sum_{j\\neq i_0} a_{i_0j}x_j=(\\lambda-a_{i_0i_0})x_{i_0}$.So $|\\lambda-a_{i_0i_0}|\\leq \\sum_{j\\neq i_0} |a_{ij}|$ and $\\lambda \\in D_{i_0}$", "Solution_3": "[quote=\"killer\"]... with $||X||=1$ ... There is $i_0$ such that $x_{i_0}=1.$ [/quote]\r\nJust a clarification: killer is clearly using the $\\infty$ (or sup) norm:\r\n\r\n$||x||_{\\infty}=\\max_{1\\le i\\le n} |x_i|$" } { "Tag": [ "LaTeX" ], "Problem": "From a group of boys and girls, 15 girls leave. There are then left two boys for each girl. After 45 boys leave, there are then 5 girls for each boy. The number of girls in the beginning was:\r\n\r\n(A)40 (B) 43 (C) 29 (D) 50 (E) 60", "Solution_1": "HI\r\nthe number of girls should be 40 \r\nlet the no of boys be 'b'\r\nlet the no of girls be 'g'\r\n\r\nFrom a group of boys and girls, 15 girls leave.\r\n\r\nb/g-15=2\r\nb=2g-30\r\n\r\nAfter 45 boys leave, there are then 5 girls for each boy.\r\nb-45/g-15=1/5\r\nb-45=g-15/5\r\n5b-225=g-15\r\n5b=g-15+225\r\nb=g+210/5\r\n\r\nequating the two\r\n\r\n2g-30=g+210/5\r\n10g-150=g+210\r\n9g=360\r\n[b]g=40[/b]\r\n\r\nsorry for not using latex i am not acquainted with it" } { "Tag": [ "algebra", "polynomial" ], "Problem": "factor the sum\r\n\r\n$x^7+y^7$", "Solution_1": "x^7+y^7=(x^7/3)^3+(y^7/3)^3\r\n\r\nUsing a^3+b^3 formula...\r\n\r\n(x^7/3+y^7/3)(x^28/3-x^14/3*y^14/3+y^28/3)", "Solution_2": "7 is prime it generates cyclotomic polynomial\r\n\r\n$x^7+y^7=(x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)$", "Solution_3": "One question:\r\n\r\nWhat if the exponent isn't prime? For instance, how would you factor $x^6+y^6$? :huh:", "Solution_4": "[quote=\"prototypex24\"]One question:\n\nWhat if the exponent isn't prime? For instance, how would you factor $x^6+y^6$? :huh:[/quote]\r\n\r\n$(x^2)^3+(y^2)^3=(x^2+y^2)(x^4-x^2y^2+y^4)$", "Solution_5": "Oh thanks, so I guess all composite powers can be expressed that way.\r\n\r\nEDIT: How about $x^4+y^4$? $(x^2)^2+(y^2)^2$ doesn't help much... :|", "Solution_6": "[quote=\"prototypex24\"]Oh thanks, so I guess all composite powers can be expressed that way.\n\nEDIT: How about $x^4+y^4$? $(x^2)^2+(y^2)^2$ doesn't help much... :|[/quote]\r\n\r\nthat one cannot be factorise .\r\n\r\nif $x^n+y^n$ where $n\\in\\mathbb{N}$ . Then if $n$ has odd prime , it can be factorised.", "Solution_7": "Ok, thanks a lot ;)", "Solution_8": "Well, you could factor $x^4+y^4$ this way: \r\n\r\n$x^4+y^4=(x^2+y^2)^2-2(xy)^2=(x+y+\\sqrt{2}xy)(x+y-\\sqrt{2}xy)$\r\n\r\nWhere I used the difference of squares.", "Solution_9": "i am pretty sure that a^n+b^n can always be factored if n is odd greater than 1, being prime does not matter", "Solution_10": "is this taught in Algebra II/trig or is it taught in algebra I?", "Solution_11": "i don't know if this is taught, but you can easily show that it is true with algebra 2 concepts,\r\n\r\nlet $f(x)=x^7+a^7$ clearly $-a$ is a root, therefore $(x+a)$ is a factor of $f(x)$ by the factor theorem", "Solution_12": "But the hard part is how would you show that $x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6$ has no more polynomial factors with integer coefficients... maybe rational root?" } { "Tag": [ "geometry", "perimeter", "inequalities", "incenter", "trigonometry", "inequalities unsolved" ], "Problem": "Denote by p the semi-perimeter of a triangle ABC, and by L the sum of the\r\nlengths of the interior angle bisectors of the triangle ABC. Prove that the\r\nfollowing inequality takes place\r\np\u221a3\u2265L", "Solution_1": "$(l_a+l_b+l_c)^2\\leq3(l_a^2+l_b^2+l_c^2)\\leq3\\sum p(p-a)=3p^2$", "Solution_2": "[quote=\"Maverick\"]$(l_a+l_b+l_c)^2\\leq3(l_a^2+l_b^2+l_c^2)\\leq3\\sum p(p-a)=3p^2$[/quote]\r\n\r\nI don't understand your second assertion. Any help please? ;)", "Solution_3": "$l_a^2 \\leq p(p-a)$ this one is easy to prove", "Solution_4": "[quote=\"Iris Aliaj\"]Denote by $s$ the semi-perimeter of a triangle $ABC$, and by $w_{a},w_{b},w_{c}$ the lengths of the interior angle bisectors of the triangle $ABC$. Prove that the following inequality takes place $w_{a}+w_{b}+w_{c}\\leq s\\sqrt{3}$.[/quote]\r\nIn fact, we have the following inequality chain :\r\n\\[w_{a}+w_{b}+w_{c}\\leq\\frac{3}{2}(A\\Gamma+B\\Gamma+C\\Gamma)\\leq\\frac{3}{2}(AI+BI+CI)\\leq s\\sqrt{3}, \\]\r\nwhere $I$ be the incentre and $\\Gamma$ be the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33335]Gergonne Point[/url].", "Solution_5": "I remember having seen this \r\n$l_{a}= \\frac{2bc}{b^{2}+c^{2}}\\sqrt{p(p-a)}$\r\nIs it true, and how to prove it? :D", "Solution_6": "[quote=\"bibobeo\"]I remember having seen this \n$l_{a}= \\frac{2bc}{b^{2}+c^{2}}\\sqrt{p(p-a)}$\nIs it true, and how to prove it? :D[/quote]\r\nActually\r\n$l_{a}=\\frac{2\\sqrt{bc}}{b+c}\\sqrt{p(p-a)}$\r\nBy Stewart it's not hard.", "Solution_7": ":D Sorry, I remember it is $l_{a}=\\frac{2bc}{b^{2}+c^{2}}cos\\frac{A}{2}$ (I use Sin law to prove this)\r\nbtw. who is Stewart?\r\nI use $AD^{2}=AB.AC-DB.DC$ for the first one", "Solution_8": "if $P$ is on the side $BC$ then\r\n$AP^{2}=AB^{2}\\cdot\\frac{PC}{BC}+AC^{2}\\cdot\\frac{BP}{BC}-BP\\cdot CP$", "Solution_9": ":blush: How to prove Stewart's?", "Solution_10": "[quote=\"shalex\"]if $P$ is on the side $BC$ then\n$AP^{2}=AB^{2}\\cdot\\frac{PC}{BC}+AC^{2}\\cdot\\frac{BP}{BC}-BP\\cdot CP$[/quote]\r\n\r\nyou forgot something", "Solution_11": "I think Shalex is right. I've checked it with $m_{a}$, $l_{a}$ and $h_{a}$ with $a=5, b=4,c=3$", "Solution_12": "oh,yes. it`s ok. I was thinking at stewart without proportion (\"/\")", "Solution_13": "but ... how to prove it? :blush:", "Solution_14": "[quote=\"bibobeo\"]but ... how to prove it? :blush:[/quote]\r\njust denote that $\\angle APB=\\theta$ and\r\n$\\frac{AP^{2}+BP^{2}-AB^{2}}{2AP\\cdot BP}=\\cos\\theta$\r\n$\\frac{AP^{2}+CP^{2}-AC^{2}}{2AP\\cdot CP}=\\cos(\\pi-\\theta)=-\\cos\\theta$\r\nSum them up with some computations we reach to the conclusion :wink: .", "Solution_15": "I use similar triangles and Ptoleme Theorem. But it's quite long.\r\nYou have very nice solution Shalex :lol:" } { "Tag": [ "geometry", "circumcircle", "cyclic quadrilateral", "power of a point" ], "Problem": "In $\\triangle ABC$, let $D$ a point on $BC$ so that $BD:DC=5:4$. Also, $\\angle BAD=40^\\circ$, $\\angle CAD=70^\\circ$, and $AD=6$. Find $AB$.\r\n\r\nI know the answer, but I don't know how to solve it. I heard it's from a contest for elementary school kids, so you can't use any special formula :( Any idea?", "Solution_1": "The angle $\\angle CAB$ is equal to \r\n\r\n$\\angle CAB = \\angle CAD + \\angle BAD = 70^\\circ + 40^\\circ = 110^\\circ$\r\n\r\nLet $\\triangle PBC$ be a triangle having the common side BC with the triangle $\\triangle ABC$, the vertex P lying on the opposite side of the line BC than the vertex A, such that the angles $\\angle PBC = 70^\\circ,\\ \\angle PCB = 40^\\circ$. The remaining angle of this triangle is then\r\n\r\n$\\angle BPC = 180^\\circ - (\\angle PBC + \\angle PCB) = 180^\\circ - (70^\\circ + 40^\\circ) = 180^\\circ - 110^\\circ = 70^\\circ$\r\n\r\nSince the angles\r\n\r\n$\\angle CAB + \\angle BPC = 110^\\circ + 70^\\circ = 180^\\circ$\r\n\r\nadd up to a straight angle, the quadrilateral ABPC is cyclic. Let D' be the intersection of its diagonals BC, AP. Since $BD' \\cdot CD' = AD' \\cdot PD'$ is the power of the point D' to the circumcircle of this quadrilateral, we get\r\n\r\n$\\frac{BD'}{PD'} = \\frac{AD'}{CD'}\\ \\Rightarrow \\triangle BPD' \\sim \\triangle ACD'\\ \\Rightarrow \\angle CAD' = \\angle PBD' \\equiv \\angle PBC = 70^\\circ$\r\n\r\n$\\frac{BD'}{AD'} = \\frac{PD'}{CD'}\\ \\Rightarrow \\triangle BAD' \\sim \\triangle PCD'\\ \\Rightarrow \\angle BAD' = \\angle PCD' \\equiv \\angle PCB = 40^\\circ$\r\n\r\n(This certainly can be done without using the power of a point to a circle.) Consequently, the points $D' \\equiv D$ are identical, i.e., the point D is the diagonal intersection of the cyclic quadrilateral ABPD. The internal angles $\\angle PBC = \\angle BPC = 70^\\circ$ of the triangle $\\triangle PBC$ are equal, hence, this triangle is isosceles and $BC = PC$. Using one more time similarity of the triangles $\\triangle BAD \\sim \\triangle PCD$, we get\r\n\r\n$\\frac{AB}{AD} = \\frac{CP}{CD} = \\frac{CB}{CD} = \\frac{BD + CD}{CD} = \\frac{BD}{CD} + 1 = \\frac 5 4 + 1 = \\frac 9 4$\r\n\r\n$AB = \\frac 9 4\\ AD = \\frac 9 4 \\cdot 6 = \\frac{54}{4} = 13. 5$\r\n\r\nYou can pull out the same trick with any other pair of angles $\\beta, \\gamma$, such that $\\beta + 2\\gamma = 180^\\circ$. In the posted problem, $\\beta = 40^\\circ, \\gamma = 70^\\circ$, $\\beta + 2 \\gamma = 40^\\circ + 2 \\cdot 70^\\circ = 180^\\circ$. Angles with an integer number of degrees are constructible iff the number of degrees is divisible by 3. Hence, we may choose the non-constructible angle $\\gamma = 53^\\circ$, in order to make the problem look difficult. If we add the angle $\\beta = 180^\\circ - 2\\gamma = 74^\\circ$, the problem can be solved in exactly the same way.", "Solution_2": "I think I got an alternate solution,\r\n[hide=\"Solution\"]\nwe know that the areas $\\triangle{ABD}:\\triangle{ADC}=5:2$\nwe also know that area of $\\triangle{ABD}=\\frac{1}{2}*AB*AD=3AB*sin40^\\circ$\nSimilarily for the other triangle,\nArea of $\\triangle{ADC}=3AC*sin70^\\circ$\nEquating the $2$ areas to $5:2$ we get,\n$\\frac{5sin70^\\circ}{2sin40^\\circ}=\\frac{AB}{AC}$\nWe also know that $\\text{area of}\\triangle{ABD}+\\text{area of}\\triangle{ADC}=\\text{area of}\\triangle{ABC}$\nwhich gives,\n$\\frac{AB*6sin40^\\circ}{2}+\\frac{AC*6sin70^\\circ}{2}=\\frac{AB*ACsin110^\\circ}{2}$\nusing the 2 equations we can find the value of $AB$\n[/hide]", "Solution_3": "Wow, nice solutions :coolspeak: :10:", "Solution_4": "another solution the short one.", "Solution_5": "[quote=\"gunes\"]another solution the short one.[/quote]\r\nOh gosh! Your solution is very quick :o I knew this property and yet I didn't notice that! :ewpu:", "Solution_6": "What does bisector theorem state ? :blush: Anyone mind to tell", "Solution_7": "yes what does it state, Ive never heard of it :blush:", "Solution_8": "bisector teorems are these." } { "Tag": [ "geometry", "parallelogram", "geometry unsolved" ], "Problem": "Let ABCD a convex quadilateral. a, b, c, d are the angles of A, B, C, D respectively. AC and BD intersect in O. \r\nProve that ABCD is a parallelogram if and only if \r\nOA(sin a)= OC(sin c) and OB(sin b)= OD(sin d)\r\n\r\nIm wait your ingenuos solutions!!", "Solution_1": "I am not a very good geometry solver, nor a good explainer.\r\nOA(sin a)= OC(sin c) and OB(sin b)= OD(sin d) \r\nI don't think a // gram must possess this property in accord with my knowledge. \r\nIf I am wrong, please tell me.", "Solution_2": "If $ABCD$ is a parallelogram, then the result is straightforward. So, let's consider the converse part of the problem.\n\nAssume the condition satisfies, then we have $OA/OC = (DAB)/(DCB) = \\frac{\\sqrt{AB \\cdot AD}}{ \\sqrt{CB \\cdot CD}}$. Analogously, $OB/OD = \\frac{\\sqrt{BA \\cdot BC}}{ \\sqrt{DA \\cdot DC}}$. Therefore, $(AOB)/(DOC) = \\frac{OA\\cdot OB}{OC \\cdot OD} = AB/CD$. Let $X,Y,Z,W$ be the perpendicular feet from O to segments $AB, BC, CD, DA$, respectively. From the last relation, then we have that $OX = OZ$ and $OY = OW$. In addition, by the sine law, we have that $WX = AO \\cdot sin(a) = OC \\cdot sin(c) = YZ$. Analogously, $XY = ZW$. Thus, $XYZW$ is a parallelogram, which implies that $X,O,Z$ and $Y,O,W$ are collinear. Finally, it is easy to conclude that $AD || BC$ and $AB || CD$." } { "Tag": [ "floor function", "combinatorics unsolved", "combinatorics" ], "Problem": "On a soccer tournament with $n\\ge 3$ teams taking part, several matches are played in such a way that among any three teams, some two play a match.\n$(a)$ If $n=7$, find the smallest number of matches that must be played.\n$(b)$ Find the smallest number of matches in terms of $n$.", "Solution_1": "We draw a graph such that any line between vertices A and B we color in blue iff A played with B, and in red iff A didn't play with B. Then there no exits triangle ABC such AB, BC and AC are all red. This is well known problem, the maximum of red edges is $ \\left\\lfloor \\frac{n^{2}}{4} \\right\\rfloor $. So the answer is $ \\frac {n(n-1)}{2}-\\left\\lfloor \\frac{n^{2}}{4} \\right\\rfloor$." } { "Tag": [ "symmetry" ], "Problem": "Each square of the three by three grid is painted so that the whole picture has at least the two lines of symmetry indicated. Each grid square is painted one solid color. What is the maximum number of colors that could have been used?\n[asy]size(100);\ndraw((0,0)--(0,3)--(3,3)--(3,0)--cycle);\ndraw((1,0)--(1,3));\ndraw((2,0)--(2,3));\ndraw((0,1)--(3,1));\ndraw((0,2)--(3,2));\ndraw((-0.2,-0.2)--(3.2,3.2),dashed);\ndot((0,0));dot((3,3)); dot((1.5,0)); dot((1.5,3));\ndraw((1.5,-0.3)--(1.5,3.3), dashed);[/asy]", "Solution_1": "Label the squares 1,2,3,4,5,6,7,8,9 from top left to bottom right.\r\n1,3,7, and 9 have to be the same color.\r\n2,4, 6, and 8 have to be the same color.\r\n5 can be any color.\r\nThus, the maximum number of colors that can be used is $ \\boxed{3}$.", "Solution_2": "What if you do this?", "Solution_3": "[quote=bluehurricane]What if you do this?[/quote]\n\nNo, it isn't diagonally symmetrical :)", "Solution_4": "Oh, OK. Thanks for telling me, mossie!", "Solution_5": "[quote=bluehurricane]Oh, OK. Thanks for telling me, mossie![/quote]\n\nNo problem! :-)", "Solution_6": "[hide=Solution]\nThere are three parts of the square. Corner, middle edge, and center. Therefore, the max is $\\boxed{3}.$\n[/hide]", "Solution_7": " :wacko: :wacko: " } { "Tag": [ "calculus", "integration", "function", "logarithms", "calculus computations" ], "Problem": "Define the function $ f(t)\\equal{}\\int_0^1 (|e^x\\minus{}t|\\plus{}|e^{2x}\\minus{}t|)dx$. Find the minimum value of $ f(t)$ for $ 1\\leq t\\leq e$.", "Solution_1": "$ f(t) \\equal{} \\int_0^1 (|e^x \\minus{} t| \\plus{} |e^{2x} \\minus{} t|)dx \\geq \\int_0^1 (|e^x \\plus{} e^{2x} \\minus{} 2t|)dx \\geq |\\int_0^1 (e^x \\plus{} e^{2x} \\minus{} 2t)dx| \\equal{} |e \\minus{} 1 \\plus{} \\frac {e^{2}}{2} \\minus{} \\frac {1}{2} \\minus{} 2t| \\geq \\frac {1}{2}|t \\minus{} 3\\parallel{}t \\plus{} 1| \\geq 2$.", "Solution_2": "Let $ u \\equal{} e^x$, $ du \\equal{} e^x \\, dx \\equal{} u \\, dx$, from which it follows that\r\n\r\n$ f(t) \\equal{} \\int_{x \\equal{} 0}^1 |e^x \\minus{} t| \\plus{} |e^{2x} \\minus{} t| \\, dx \\equal{} \\int_{u \\equal{} 1}^e \\frac {|u \\minus{} t| \\plus{} |u^2 \\minus{} t|}{u} \\, du \\equal{} \\int_{u \\equal{} 1}^e g(u;t) \\, du$,\r\n\r\nwhere $ g(u;t) \\equal{} \\left|1 \\minus{} \\frac {t}{u}\\right| \\plus{} \\left|u \\minus{} \\frac {t}{u}\\right|$.\r\n\r\nThen since $ 1 < t < e$, for $ 1 < u < \\sqrt {t} < t$,\r\n\r\n$ g(u;t) \\equal{} \\frac {2t}{u} \\minus{} 1 \\minus{} u$,\r\n\r\nand for $ \\sqrt {t} < u < t$,\r\n\r\n$ g(u;t) \\equal{} u \\minus{} 1$,\r\n\r\nand for $ t < u < e$,\r\n\r\n$ g(u;t) \\equal{} 1 \\plus{} u \\minus{} \\frac {2t}{u}$.\r\n\r\nIt follows that we must evaluate the integral over each of these three subintervals:\r\n\r\n$ f(t) \\equal{} \\int_{u \\equal{} 1}^{\\sqrt {t}} \\frac {2t}{u} \\minus{} 1 \\minus{} u \\, du \\plus{} \\int_{u \\equal{} \\sqrt {t}}^t u \\minus{} 1 \\, du \\plus{} \\int_{u \\equal{} t}^1 1 \\plus{} u \\minus{} \\frac {2t}{u} \\, du$\r\n\r\n$ \\equal{} \\left[2t \\log u \\minus{} u \\minus{} \\frac {u^2}{2}\\right]_{u \\equal{} 1}^{\\sqrt {t}} \\plus{} \\left[\\frac {u^2}{2} \\minus{} u\\right]_{u \\equal{} \\sqrt {t}}^t \\plus{} \\left[u \\plus{} \\frac {u^2}{2} \\minus{} 2t \\log u\\right]_{u \\equal{} t}^e$\r\n\r\n$ \\equal{} t \\log t \\minus{} \\sqrt {t} \\minus{} \\frac {t}{2} \\plus{} \\frac {3}{2} \\plus{} \\frac {t^2}{2} \\minus{} t \\minus{} \\frac {t}{2} \\plus{} \\sqrt {t} \\plus{} e \\plus{} \\frac {e^2}{2} \\minus{} 3t \\minus{} \\frac {t^2}{2} \\plus{} 2 t \\log t$\r\n\r\n$ \\equal{} 3t \\log t \\minus{} 5t \\plus{} \\frac {3 \\plus{} 2e \\plus{} e^2}{2}$.\r\n\r\nThis function is minimized at a critical point or endpoint, so we compute\r\n\r\n$ f'(t) \\equal{} 3 \\log t \\minus{} 2 \\equal{} 0$\r\n\r\nfrom which we have $ t \\equal{} e^{2/3}$. Evaluating,\r\n\r\n$ f(1) \\equal{} (e^2 \\plus{} 2e \\minus{} 7)/2$,\r\n\r\n$ f(e^{2/3}) \\equal{} (e^2 \\plus{} 2e \\minus{} 6e^{2/3} \\plus{} 3)/2$,\r\n\r\n$ f(e) \\equal{} (e^2 \\minus{} 2e \\plus{} 3)/2$,\r\n\r\nand the minimum is attained for $ t \\equal{} e^{3/2}$. If we choose, we can verify this algebraically by computing the differences\r\n\r\n$ f(1) \\minus{} f(e^{2/3}) \\equal{} 3e^{2/3} \\minus{} 5 > 3e^{2/3} \\minus{} 2e \\equal{} f(e) \\minus{} f(e^{2/3})$.\r\n\r\nWe wish to show that $ e^{\\minus{}1/3} > 2/3$. Expanding as a power series, we find\r\n\r\n$ e^{\\minus{}1/3} > 1 \\minus{} (1/3) \\plus{} (1/18)$,\r\n\r\nfrom which it follows that $ 3e^{2/3} \\minus{} 2e > 0$, and the minimality of $ f(e^{2/3})$ is proven.", "Solution_3": "[quote=\"akech\"]$ |e \\minus{} 1 \\plus{} \\frac {e^{2}}{2} \\minus{} \\frac {1}{2} \\minus{} 2t| \\geq \\frac {1}{2}|t \\minus{} 3\\parallel{}t \\plus{} 1|$.[/quote]\r\n\r\nThat cannot be correct, since when $ t \\equal{} (e^2 \\plus{} 2e \\minus{} 3)/4 \\approx 2.456 < e$, the LHS is zero.", "Solution_4": "Oops, my bad.", "Solution_5": "[quote=\"1605272\"]$ f(t) = \\int_0^1 (|e^x - t| + |e^{2x} - t|)dx$\n$ = \\int_0^{\\frac {1}{2}\\ln t }[(t - e^x ) + (t - e^{2x})]dx$\n$ + \\int_{\\frac {1}{2}\\ln t}^{\\ln t} [(t - e^x ) + (e^{2x} - t)]dx$\n$ + \\int_{\\ln t}^{1}[(e^x - t) + (e^{2x} - t)]dx$\n$ = f_1 (t) + f_2( t) + f_3 (t)$\n\n$ f_1^'(t) = \\ln t - \\frac {\\sqrt {t}}{2t} + \\frac {1}{2}$ $ f_2^'(t) = t + \\frac {\\sqrt {t}}{2t} - \\frac {3}{2}$ $ f_3^'(t) = - t + 2\\ln t - 1$\n\n$ f'(t) = 3\\ln t - 2$\n\nLet$ f'(t) = 3\\ln t - 2 = 0$,we have $ t = \\sqrt [3]{e^2}$.\n\nSo $ 1 < t < \\sqrt [3]{e^2},f'(x) < 0,f(x)$monotonically decreasing,${ \\sqrt [3]{e^2} < t < e},f'(x) > 0,f(x)$ monotonically increasing.\n\nSo we have the minimum value of the function $ f(x)$ is\n\n$ min_{1 < t < e}f(x) = f(\\sqrt [3]{e^2}) = \\int_0^1 (|e^x - \\sqrt [3]{e^2}| + |e^{2x} - \\sqrt [3]{e^2}|)dx$\n$ = \\int_0^{\\frac {1}{3}}[(\\sqrt [3]{e^2} - e^x ) + (\\sqrt [3]{e^2} - e^{2x})]dx$\n$ + \\int_{\\frac {1}{3}}^{\\frac {2}{3}} [(\\sqrt [3]{e^2} - e^x ) + (e^{2x} - \\sqrt [3]{e^2})]dx$\n$ + \\int_{\\frac {2}{3}}^{1}[(e^x - \\sqrt [3]{e^2}) + (e^{2x} - \\sqrt [3]{e^2})]dx$\n$ = e^2 + e - \\frac {7}{2}\\sqrt [3]{e^2} + \\frac {3}{2}$[/quote]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Happy New Year everyone! Try this 'evil' problem:\r\n\r\nProblem. A positive integer $m$ is called 'evil' if, for some $t \\in \\mathbb{N}$, the sum of the first $t$ decimal digits of $m$ equals $666$. Prove or disprove: for every $n \\in \\mathbb{N}$, there is a $k \\in \\mathbb{N}$ such that each of the numbers $k,2k,\\ldots,nk$ is 'evil.'\r\n\r\nOn that note, has anyone ever wondered which is the most useless entry in MathWorld? http://mathworld.wolfram.com/EvilNumber.html :D \r\n\r\nCheers! :D", "Solution_1": "[quote=\"vess\"]On that note, has anyone ever wondered which is the most useless entry in MathWorld? http://mathworld.wolfram.com/EvilNumber.html :D [/quote]\r\n\r\nI disagree. What about http://mathworld.wolfram.com/LeviathanNumber.html ?\r\n\r\n Darij", "Solution_2": "Hum at fist when i link to the links you wrote there are just few words like \r\n[b]Access Denied to IP Address 203.160.1.67[/b]\r\n---------\r\nThe second plz tell me the sum from the left or right?\r\n\r\n--------\r\n\r\nThe last , happy new year :D", "Solution_3": "[quote=\"pluricomplex\"]\nThe second plz tell me the sum from the left or right?\n[/quote]\r\n\r\nFrom the left: a positive integer is 'evil' if a block of its decimal digits, from left to right, sums to $666$. If we were summing from the right (the last few digits), the answer would (almost trivially) be 'no.'", "Solution_4": "[quote=\"darij grinberg\"][quote=\"vess\"]On that note, has anyone ever wondered which is the most useless entry in MathWorld? http://mathworld.wolfram.com/EvilNumber.html :D [/quote]\n\nI disagree. What about http://mathworld.wolfram.com/LeviathanNumber.html ?\n\n Darij[/quote]\r\n\r\nThere are many more of it like Apocalypse Number and similar :D" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all continuous functions $f: \\mathbb R \\to \\mathbb R$ such that\n\\[ f(2x - y) = 2f(x) - f(y), \\quad \\forall x,y\\in \\mathbb R.\\]", "Solution_1": "[quote=\"TRAN THAI HUNG\"]Find f(x) continuous in R and\n$ {\\rm{f(2x \\minus{} y) \\equal{} 2f(x) \\minus{} f(y)}}\\forall {\\rm{x,y}} \\in {\\rm{R}}$:blush:[/quote]\r\nLet $ P(x,y)$ be the assertion $ f(2x \\minus{} y) \\equal{} 2f(x) \\minus{} f(y)$\r\n\r\nIf $ f(x)$ is solution, $ f(x) \\minus{} f(0)$ is too. So we'll only look for solutions such that $ f(0) \\equal{} 0$\r\n\r\n$ P(x,0)$ $ \\implies$ $ f(2x) \\equal{} 2f(x)$\r\n$ P(0,y)$ $ \\implies$ $ f( \\minus{} y) \\equal{} \\minus{} f(y)$\r\n\r\nSo $ 2f(x) \\minus{} f(y) \\equal{} f(2x) \\plus{} f( \\minus{} y)$ and so $ f(2x \\plus{} ( \\minus{} y)) \\equal{} f(2x) \\plus{} f( \\minus{} y)$ and so $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$ and so $ f(x) \\equal{} ax$ since $ f(x)$ is continuous (Cauchy).\r\n\r\nSo $ f(x) \\equal{} ax \\plus{} b$ and it's easy to check back that these functions indeed are solutions." } { "Tag": [ "vector", "inequalities", "geometry unsolved", "geometry" ], "Problem": "Given a set of $n>2$ planar vectors. A vector from this set is called [i]long[/i], if its length is not less than the length of the sum of other vectors in this set. Prove that if each vector is long, then the sum of all vectors equals to zero.\n[i]N. Agakhanov[/i]", "Solution_1": "Let $v_{i}=(x_{i}, y_{i})$, $1\\leq i \\leq n$, the long vectors and $S=(X,Y)$ their sum. \r\nFor each $i$ we have: \\[|v_{i}| \\geq |S-v_{i}| \\Longleftrightarrow x_{i}^{2}+y_{i}^{2}\\geq (X-x_{i})^{2}+(Y-y_{i})^{2}\\] \\[\\Longleftrightarrow 2Xx_{i}+2Yy_{i}\\geq X^{2}+Y^{2}\\] Adding these $n$ inequalities: \\[2X(x_{1}+x_{2}+\\ldots+x_{n})+2Y(y_{1}+y_{2}+\\ldots+y_{n})=2X^{2}+2Y^{2}\\geq n(X^{2}+Y^{2}).\\] If $S$ is not the zero vector then $2\\geq n$, a contradiction. So $S$ is the zero vector.\r\n\r\n$Tipe$", "Solution_2": "The result is true in any dimension $ d$, for $ n > 2$ vectors $ v_i$ of sum $ \\sigma \\equal{} \\sum_{i\\equal{}1}^n v_i$.\r\nIf $ | v_i | \\geq | \\sigma \\minus{} v_i |$ for all $ 1 \\leq i \\leq n$, then $ \\sum_{i\\equal{}1}^n | v_i |^2 \\geq \\sum_{i\\equal{}1}^n | \\sigma \\minus{} v_i |^2 \\equal{} \\sum_{i\\equal{}1}^n < \\sigma \\minus{} v_i, \\sigma \\minus{} v_i > \\ \\equal{} \\sum_{i\\equal{}1}^n (| \\sigma |^2 \\minus{} 2 < \\sigma, v_i > \\plus{} \\ | v_i |^2) \\equal{}$ $ n| \\sigma |^2 \\minus{} 2< \\sigma, \\sum_{i\\equal{}1}^n v_i > \\plus{} \\sum_{i\\equal{}1}^n | v_i |^2 \\equal{} (n \\minus{} 2)| \\sigma |^2 \\plus{} \\sum_{i\\equal{}1}^n | v_i |^2$, hence $ (n \\minus{} 2)| \\sigma |^2 \\leq 0$, whence $ \\sigma \\equal{} 0$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x_1,x_2,x_3,...,x_n>0$ such that:$\\sum_{i=1}^n\\frac{1}{1+x_i}=1$.Find the maximum :\r\n\\[\\sum_{i=1}^n\\frac{x_i}{4+x_i^2}\\]", "Solution_1": "CWMO 03 no.7", "Solution_2": "[quote=\"siuhochung\"]CWMO 03 no.7[/quote]\r\nYes,I have known it but in CWMO 03 no.7 n=5 and my problem is genaral form of no.7", "Solution_3": "everybody can solve thi pro?", "Solution_4": ":D i think this pro isn't easy\r\n\r\nit've proposed for test in HANOI-VMO" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Find the minimum value of \r\n\r\nf(x,y,z) = x^3 + y^3 + z^3 + 6xyz \r\n\r\nfor x,y,z >= 0 and x+y+z=1.\r\n\r\nthere's something that i'm not getting", "Solution_1": "Find the minimum value of \r\n$ f(x,y,z) \\equal{} x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} 6xyz$ for $ x,y,z\\geq0$ and $ x\\plus{}y\\plus{}z\\equal{}1$", "Solution_2": "[quote=\"kritikos\"]Find the minimum value of \n$ f(x,y,z) \\equal{} x^3 \\plus{} y^3 \\plus{} z^3 \\plus{} 6xyz$ for $ x,y,z\\geq0$ and $ x \\plus{} y \\plus{} z \\equal{} 1$[/quote]\r\nBy Schur inequality, we have:\r\n$ x^3\\plus{}y^3\\plus{}z^3\\plus{}3xyz \\ge xy(x\\plus{}y)\\plus{}yz(y\\plus{}z)\\plus{}zx(z\\plus{}x)$\r\nIt follows that\r\n$ 4(x^3\\plus{}y^3\\plus{}z^3)\\plus{}13xyz \\ge x^3\\plus{}y^3\\plus{}z^3\\plus{}3\\left(xy(x\\plus{}y)\\plus{}yz(y\\plus{}z)\\plus{}zx(z\\plus{}x)\\right)\\plus{}6xyz\\equal{}(x\\plus{}y\\plus{}z)^3\\equal{}1$\r\nSo $ 4f(x,y,z)\\equal{}4(x^3\\plus{}y^3\\plus{}z^3)\\plus{}24xyz \\ge 4(x^3\\plus{}y^3\\plus{}z^3)\\plus{}13xyz \\ge 1$\r\nEquality holds for $ x\\equal{}y\\frac{1}{2},z\\equal{}0$ and permutation.", "Solution_3": "if you put x=y=1/2 and z=0 then you have :\r\nf = 1/4 that is smaller than 1/3.!!!!!!\r\n\r\nalso i found a similiar solution.\r\n\r\npls help", "Solution_4": "$ x^2\\plus{}y^2\\plus{}z^2\\ge \\frac{(x\\plus{}y\\plus{}z)^2}{3}$ I think here u have a mistake because we have that $ x^2\\plus{}y^2\\plus{}z^2\\geq{xy\\plus{}yz\\plus{}xz}$ but $ \\frac{(x\\plus{}y\\plus{}z)^2}{3}\\geq{xy\\plus{}yz\\plus{}xz}$", "Solution_5": "it also can be reduced in \r\n\r\nf = 1 - 3 [(x^2+y^2+z^2)-(x^3+y^3+z^3)]..", "Solution_6": "I was also trying reducing substituting z=1-(x+y).\r\nThen finding some expresssion that we can put t=x+y ... and finally study f(t) for t in [0,1]\r\n\r\nCAN SOMEONE RESOLVE IT?", "Solution_7": "[quote=\"exalibur\"]if you put x=y=1/2 and z=0 then you have :\nf = 1/4 that is smaller than 1/3.!!!!!!\n\nalso i found a similiar solution.\n\npls help[/quote]\r\nSorry, I have some mistakes. I've edited my post.", "Solution_8": "we have ${ \\{x^3}$+${ \\{y^3}$+${ \\{z^3}$+6xyz=x\u00b2+y\u00b2+z\u00b2-xy-yz-zx+6xyz\r\n =1+6xyz-3(xy+yz+zx);\r\nwe have (x+y-z)(y+z-x)(x+z-y)\u2265xyz;\r\nit follows that (1-2x)(1-2y)(1-2z)\u2265xyz;\r\n we attain 1+4(xy+yz+zx)\u22659xyz + 2 \r\n nextly, 4(xy+yz+zx - 2xyz)+1\u2265xyz+2\u22652\r\nsince we have minimax ... \r\n${ \\{(x + y + z)^3}$+9xyz \u22654(x+y+z)(xz+yz+xy) since that we have 1+9xyz\u22654(xz+yz+zx)\r\nit follows that 1+4(2xyz-zx-xy-zy)\u2265-xyz\u2265$ \\frac { - 1}{27}$\r\nsince we have minimum!" } { "Tag": [ "geometry" ], "Problem": "Here's a fun test I found. See how you do, and announce your results here. \r\nIt wouldn't let me make an extension(quite annoying) so here's the test:\r\n\r\nGeography Quiz\r\nUS Geography\r\n\r\n1. Which US State is home to Hot Springs National Park?\r\n \t\tA. Louisiana\r\n\t\tB. Arkansas\r\n\t\tC. Wyoming\r\n\t\tD. California\r\n\r\n2. The bloodiest site of the Civil War, the Antietam battleground, is located in which US State?\r\n\t\tA. Massachusetts\r\n\t\tB. Kentucky\r\n\t\tC. Maryland\r\n\t\tD. Ohio\r\n\r\n3. In terms of area, what is the smallest state in the US?\r\n\t\tA. Delaware\r\n\t\tB. Vermont\r\n\t\tC. Maine\r\n\t\tD. Rhode Island\r\n\r\n4. What state was the first to grant women suffrage?\r\n\t\tA. Colorado\r\n\t\tB. Montana\r\n\t\tC. Wyoming\r\n\t\tD. New Mexico\r\n\r\n5. Which state is nicknamed the \"Sooner State\"?\r\n\t\tA. Texas\r\n\t\tB. Oklahoma\r\n\t\tC. Kansas\r\n\t\tD. Alabama\r\n\r\n6. In which state was the cereal company Kellogg's founded?\r\n\t\tA. Wisconsin\r\n\t\tB. Minnesota\r\n\t\tC. North Dakota\r\n\t\tD. Michigan\r\n\r\n7. Which of the following former Presidents was born in Arkansas?\r\n\t\tA. Ronald Reagan\r\n\t\tB. Jimmy Carter\r\n\t\tC. Bill Clinton\r\n\t\tD. George W. Bush\r\n\r\n8. Where did the Klondike Gold Rush happen?\r\n\t\tA. Colorado\r\n\t\tB. Alaska\r\n\t\tC. California\r\n\t\tD. Montana\r\n\r\n9. The first U.S. City, St. Augustine, was founded in 1519 by whom?\r\n\t\tA. The Portuguese\r\n\t\tB. The English\r\n\t\tC. The Spanish\r\n\t\tD. The Italian\r\n\r\n10. Which are the only two states where gambling is illegal?\r\n\t\tA. Alaska and Hawaii\r\n\t\tB. Utah and Hawaii\r\n\t\tC. Alaska and Utah\r\n\t\tD. Wyoming and Utah\r\n\r\n11. Where is the center of the continental United States located?\r\n\t\tA. Nebraska\r\n\t\tB. Colorado\r\n\t\tC. Oklahoma\r\n\t\tD. Kansas\r\n\r\nWorld Geography\r\n12. Which of the following cities is located in Western Australia?\r\n\t\tA. Melbourne\r\n\t\tB. Sydney\r\n\t\tC. Canberra\r\n\t\tD. Perth\r\n\r\n13. Which middle eastern city has a capital of Ashgabat?\r\n\t\tA. Afghanistan\r\n\t\tB. Pakistan\r\n\t\tC. Tajikistan\r\n\t\tD. Turkmenistan\r\n\r\n14. What is the most densely populated Asian country?\r\n\t\tA. Brunei\r\n\t\tB. Singapore\r\n\t\tC. Taiwan\r\n\t\tD. Qatar\r\n\r\n15. Which of the following countries is located on the Iberian Peninsula?\r\n\t\tA. Italy\r\n\t\tB. Portugal\r\n\t\tC. Ireland\r\n\t\tD. Greece\r\n\r\n16. In which African country is English the official spoken language?\r\n\t\tA. Mozambique\r\n\t\tB. Namibia\r\n\t\tC. Egypt\r\n\t\tD. Sudan\r\n\r\n17. Which country has the highest per capita GDP of any country?\r\n\t\tA. Luxembourg\r\n\t\tB. Singapore\r\n\t\tC. Qatar\r\n\t\tD. Switzerland\r\n\r\n18. Which of the following countries is completely enclosed inside another country?\r\n\t\tA. Tunisia\r\n\t\tB. Sierra Leone\r\n\t\tC. Lesotho\r\n\t\tD. Swaziland\r\n\r\n19. In which continent is Mount Kilimanjaro located?\r\n\t\tA. Africa\r\n\t\tB. Europe\r\n\t\tC. Asia\r\n\t\tD. South America\r\n\r\n20. Open Ended: What is the largest country that doesn't border an ocean?\r\n\r\n\r\nAnswers:\r\n1. B\r\n2. C\r\n3. D\r\n4. C\r\n5. B\r\n6. D\r\n7. C\r\n8. B\r\n9. C\r\n10. B\r\n11. D\r\n12. D\r\n13. D\r\n14. B\r\n15. B\r\n16. B\r\n17. C\r\n18. C\r\n19. A\r\n20. Kazakhstan \r\n\r\n\r\nHere's the grading scale:\r\n20-Geography Expert\r\n19-Genius\r\n16-18 - Smart Alec\r\n13-15 - Geography Whiz\r\n11-12 - Pretty Good\r\n8-10 - Average\r\n5-7 - Not too good\r\n4 - Poor\r\n2-3 - Dreadful\r\n1 - Disgraceful\r\n0 - Know-Nothing", "Solution_1": "Come on, why isn't anyone taking it?", "Solution_2": "There were probably quite a few people who took it but simply didn't bother to post.", "Solution_3": "alright, ill take it :D \r\n[hide=\" :maybe: \"]\n1.B\n2.C\n3.D\n4. C \n5. B \n6. D \n7. C \n8. B \n9. C \n10. B \n11. D \n12. D \n13. D \n14. B \n15. B \n16.a\n17.c\n18.a\n19.a\n20.idk[/hide]", "Solution_4": "Question 13 says \"city\" instead of \"country\"\r\n\r\nQuestion 15 has two answers (both Spain and Portugal are on the Iberian Peninsula, the other country is Andorra)\r\n\r\nAlso, a lot of it is more of a history quiz than a geography quiz", "Solution_5": "Sorry about 15, but on number 13, the question said \r\n\"Which of the following\" meaning there could be more than 1 on the Iberian peninsula, but only one is one of the 4 answer choices.", "Solution_6": "I got 19." } { "Tag": [ "induction", "inequalities", "inequalities solved" ], "Problem": "If $x_{1},x_{2}, \\ldots ,x_{n} \\geq 1 $ then we have $ \\sum_{k=1}^{n} \\frac{1}{1+x_{k}} \\geq \\frac{n}{1+ \\sqrt[n]{x_{1}x_{2} \\ldots x_{n}}}$.\r\n\r\nmade by Berkolaiko L.K.", "Solution_1": "It is Old problem!! :) :) :) \r\nWe prove by induction on n=2^k\r\nor use Jensen ineq :consider f(x)=1/(1+e^x)", "Solution_2": "1)\r\n Let f(x) = 1/ (1+e^x)\r\n f'(x) = -e^x /(1+e^x)^2\r\n f\"(x)= -[e^x(1+e^x)^2 - e^x .2.(1+e^x).e^x]/ (1+e^x)^4 \r\n = -[(1+e^x).e(x)[1+e^x-2e^x)]/ M^2\r\n =-[(1+e^x).e^x.(1-e^x) ]/M^2\r\n --> f''(x)>0 for all x >0 \r\n 2) \r\n Let a_i= lnx_i --> a_i>0 fot all i =1..n\r\n (*) <-> [sum f(a_k) ] /n >= f( sum(a_k)/n)\r\n \r\n ( Jensen 's inequality)" } { "Tag": [ "trigonometry", "analytic geometry", "graphing lines", "slope", "logarithms", "trig identities", "calculus" ], "Problem": "Does $ \\frac{e^{\\left(\\displaystyle\\sum_{k\\equal{}1}^n \\tan^{\\minus{}1}\\frac{1}{k!}\\right)}}{n}$ converge as $ n\\rightarrow\\infty$? If so, what is its limit?", "Solution_1": "Let's think about this one little piece at a time -- inside the summation, we have $ \\arctan \\frac{1}{k!}$. About how big is $ \\frac{1}{k!}$? So, about how big is $ \\arctan \\frac{1}{k!}$? Then about how big is the whole summation?", "Solution_2": "If the summation in the exponent converges to a finite number (and it does), then the limit will converge to 0. The fun bit is proving that the sum converges.", "Solution_3": "$ \\arctan x0$, so $ \\sum_{k\\equal{}1}^\\infty \\arctan \\frac{1}{k!}<\\sum_{k\\equal{}1}^\\infty \\frac{1}{k!}\\equal{}e\\minus{}1$.", "Solution_4": "Um, I guess I typed the wrong problem. It should be\r\n\r\n$ \\frac{e^{\\sum_{k\\equal{}1}^{n} \\tan^{\\minus{}1} \\frac{1}{k}}}{n}$.\r\n\r\nAnyway, the numerator forms a line with approximate slope 1.35, and a $ y$ intercept of about .71. I don't know, I can only go up to about 100 on my Ti-84 before it gets too long to wait. On Mathematica it went to 100000 and still the slope wasn't 0.", "Solution_5": "Claim: $ \\sum_{k \\equal{} 1}^n\\tan^{ \\minus{} 1}\\frac1k \\equal{} \\ln n \\plus{} O(1).$\r\n\r\nI make that claim by the limit comparison of this series to the harmonic series.\r\n\r\nFrom that $ e^{\\sum_{k \\equal{} 1}^n\\tan^{ \\minus{} 1}\\frac1k} \\equal{} n\\cdot O(1).$\r\n\r\nDivide by $ n$ and you get something bounded. \r\n\r\nBy a slight revision of the argument, I'm willing to claim that $ \\sum_{k \\equal{} 1}^n\\tan^{ \\minus{} 1}\\frac1k \\minus{} \\ln n$ is a bounded decreasing (positive) sequence, and hence has a limit, which means that the original problem has a finite limit.\r\n\r\nBy extrapolation from a modest number of terms, I estimate the limit as $ \\approxeq 1.352074826.$ That's the slope in what kidperson3 was saying with his last post, and the slope is what matters.\r\n\r\nI don't yet know whether this can be expressed in closed form; there's at least some chance, since there are interesting trig identities to be tried with arctangents.\r\n\r\nAs an aside: my favorite tool for this sort of thing is not a calculator and it's not Mathmatica/Maple/etc. It's a spreadsheet." } { "Tag": [ "Alcumus", "videos", "Support" ], "Problem": "i started alcumus right?\r\nso i started to earn karma one day and then the next day i had even less???\r\ndo you have an answer??\r\nif so please post it\r\nalso the problems keep repeating them selves", "Solution_1": "Karma is like money, you earn it then you spend it. For each problem you answer, you lose 5 karma. For each video you watch, you lose 5 karma. For each rating you give a problem, you earn 5 karma. For each video you rate, you earn 5 karma. For each comment you submit on a video, you earn 5 karma. Problems and videos can't be rerated, but you can submit more comments.", "Solution_2": "but how do you earn karma\r\ni don't know how to rate\r\nand i think sending error reports work right?", "Solution_3": "sending error reports costs karma (i think it's about 5?), but if your error report is valid, then your receive karma (15 or 20?)\r\n\r\nto rate a problem, you go to the bottom of the solutions screen\r\nthere will be three categories in which you can rate\r\nrate the categories by choosing the number of stars to give to each category\r\n\r\nto rate videos, i believe the rating stars are right underneath the video\r\njust click the number of stars you think the video deserves", "Solution_4": "It costs 5. You recieve 25.\r\n\r\nhttp://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints\r\n\r\n :)", "Solution_5": "You probably used your karma to answer questions, and never earned any again." } { "Tag": [ "geometry", "geometric transformation", "rotation", "reflection", "ratio", "3D geometry", "sphere" ], "Problem": "Hey.\r\n\r\nYou can add two segments/squares/cubes/spheres by putting them together. (same goes to subtraction) Quantification is not required, right?\r\n\r\nWhat does it mean to multiply/divide two segments, w/o assigning a measuring unit?\r\n\r\nI really, really hope this makes sense. :ewpu:", "Solution_1": "Multiplication is really just addition so if you can add two things, you can certainly multiply them. After all, $ 2\\plus{}2\\plus{}2$ is the same thing as $ 3\\times 2$.\r\n\r\nThis is assuming you mean the standard multiplication defined on $ \\mathbb{R}$. In abstract groups, you can define some new binary operation and call it multiplication (or addition), and it doesn't necessarily send $ (a,b)\\mapsto ab$ (or $ a\\plus{}b$). In this case, you don't need the \"quantification\" you're thinking of. For example, in the symmetric group $ S_{n}$, the operation is composition. In the dihedral group $ D_{2n}$, the operation is rotation and reflection (in a group, there is some binary operation $ \\circ$ which is called \"multiplication,\" although as I just pointed out, there may be very little actual multiplication going on).", "Solution_2": "Holy cow I'm humble, I don't know much math. :surrender: \r\nI tried to understand 2nd paragraph, but I have a hunch it's not what I really meant. :D \r\n\r\nI mean, if you don't establish a unit length, you can't really multiply two segments. but you can add them, you know, \"sticking\" them together.\r\n\r\nLet's put it this way:\r\n[img]http://img132.imageshack.us/img132/7519/quantification.jpg[/img]", "Solution_3": "Multiplying two segments the standard way results in a measurement of area.\r\nDividing two segments results in a pure number, or a ratio, or whatever.\r\nMultipliying a segment by a sphere results in a four-dimensional measurement of four-dimensional volume.\r\nIt's mainly just the problem of the dimensions: $ 2 \\text{ km} \\times 5 \\text{ km} \\equal{} 10 \\text{ km}^2$.\r\n\r\nThe second paragraph oversimplified: you can define \"multiplication\" your way, even if it's something like $ x \\times y \\equal{} \\sin \\ln^* (xy)^{5^{(x\\plus{}xy\\plus{}y\\plus{}2010\\pi)!!!}}$. However the multiplication should be associative, commutative, have an identity element, and satisfy a bunch of other properties.\r\n\r\nRepeat: that is an oversimplification.", "Solution_4": "[quote=\"math_explorer\"]Multiplying two segments the standard way results in a measurement of area.\nDividing two segments results in a pure number, or a ratio, or whatever.\nMultipliying a segment by a sphere results in a four-dimensional measurement of four-dimensional volume.\nIt's mainly just the problem of the dimensions: $ 2 \\text{ km} \\times 5 \\text{ km} \\equal{} 10 \\text{ km}^2$.\n[/quote]\r\n\r\nThanks. So that means, segment * segment cannot \"return\" segment, but area?\r\n\r\nIf so, what about sin(x+y) = sin(x)cos(y) - cos(x)sin(y)?\r\nsin(x+y) is represented by a segment, but RHS is square subtraction, so to speak.", "Solution_5": "Unfortunately sin(x) does not return a length, but a ratio of lengths, or a pure number. Or you can think of it as a length, but the unit is still implicit.\r\n\r\nsin(x) is usually defined in one of three ways:\r\n1. of a right triangle with one angle x, the opposite side (to the x angle) over the hypotenuse. Here it is length / length.\r\n2. in a unit circle, the y-coordinate of the point at angle x. This is a length, but relative to the \"unit\" of the unit circle is defined;\r\n3. the sum of a certain purely numerical series (which I'm too lazy to look up, as usual)." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "ratio", "trigonometry", "calculus", "derivative" ], "Problem": "A hollow cylinder of radius $R$ and of mass $M$ is rolling down a slope with an incline angle $\\varphi$. Inside it there is another but solid cylinder of radius $r$ and mass $m$ rolling so that it maintains the same position with respect to the hollow cylinder and the line connecting the axises of the two cylinders (points $O$ and $C$ on the diagram) is always in angle $\\psi$ with the vertical. There is pure rolling everywhere. Find the ratio $\\frac{M}{m}$ in terms of the given parameters.", "Solution_1": "I give up. [b]Thaakisfox[/b], would you please post an answer to this one?", "Solution_2": "I got $\\eta = \\frac{M}{m}=\\frac{\\sin\\varphi-((\\frac{r}{R})^{2}+\\frac{4}{3}(1-\\frac{r}{R}))\\sin\\psi}{\\sin\\varphi-\\frac{4}{3}\\sin\\psi}$", "Solution_3": "My result is:\r\n\r\n$\\frac{M}{m}=\\frac{\\tan\\varphi(1-\\cos(\\varphi-\\psi))+3\\sin(\\varphi-\\psi)}{\\tan\\varphi(2\\cos(\\varphi-\\psi)-1)-4\\sin(\\varphi-\\psi)}$", "Solution_4": "This is exactly why I gave up; a tiny mistake could lead to a lot of useless work, and so far either [b]Astronum[/b] or [b]golduck[/b] have squandered their time -- or maybe both :roll:", "Solution_5": "Yeah, I agree with you [b]Djole[/b] - I tried to solve that problem three times, each time I got a different result :) The one above is the most probable but I still don't believe that it's correct. :roll:", "Solution_6": "That's a good way to solve problems, isn't it -- you do it a lot of times and then calculate the mean solution :D You could even estimate the error you made ;) \r\n\r\nPS: Sorry for the off-topic...", "Solution_7": "I checked my solution and now I got a different answer. :lol: \r\n\r\n$\\frac{M}{m}=\\frac{\\sin2\\varphi+\\cos\\psi\\sin\\varphi-\\tan\\psi(3(\\frac{r}{R})^{2}-4\\frac{r}{R}+2-\\sin\\psi\\sin\\varphi-2\\sin^{2}\\varphi)}{\\sin2\\varphi+\\cos\\psi\\sin\\varphi+\\tan\\psi(\\sin\\psi\\sin\\varphi+2\\sin^{2}\\varphi-2)}$", "Solution_8": "Hello everyone, my exams are finished so im here again. :D \r\nI see you guys have tried a lot to solve this problem :D \r\nBut there is a trick in it.\r\n[hide=\"hint\"]Can this arrangement really take place? :wink: [/hide]", "Solution_9": "I can't find out why it couldn't take place :?: :(", "Solution_10": "Hmm [b]Thaakisfox[/b] could you post a solution to this one :?:", "Solution_11": "Since we have problem where only rolling is involved, then the problem is of mechanics and we'll try to solve the problem with Lagrangian expressed in terms of generalized coordinates $ \\psi$ and $ x$, and their derivatives; $ x$ is the length of the line segment between horizontal ground and the center of hollow cylinder, when we draw parallel to inclined plane:\r\nFirstly, we can easily see that we can divide Lagrangian into two parts corresponding to the Lagrangians of the two cylinders we have:\r\n$ L_{1}= M\\dot{x}^{2}-Mgx\\sin{\\phi}$\r\nThe Lagrangian of the second is pretty large with respect to the previous one\r\n$ L_{2}= \\frac{1}{2}m((R-r)^{2}\\dot{\\psi}^{2}+\\dot{x}^{2}-2(R-r)\\dot{x}\\dot{\\psi}\\cos(\\psi-\\phi))+$\r\n$ +\\frac{1}{4}m(\\dot{x}^{2}+(R-r)^{2}\\dot{\\psi}^{2}-2\\dot{x}\\dot{\\psi}(R-r))-mg(x\\sin{\\phi}-(R-r)\\cos{\\psi})$.\r\nNow let's write the Lagrangian for the system in total:\r\n$ L= (M+\\frac{3}{4}m)\\dot{x}^{2}+\\frac{3}{4}m(R-r)^{2}\\dot{\\psi}^{2}-m(R-r)(\\cos{(\\phi-\\psi)}+\\frac{1}{2})\\dot{x}\\dot{\\psi}-$\r\n$ -(M+m)gx\\sin{\\phi}+mg(R-r)\\cos \\psi$\r\nNow if we use Lagrange's formula we can easily get the equations of motion:\r\n$ \\frac{d}{dt}(\\frac{\\partial L}{\\partial \\dot{x}})=\\frac{\\partial L}{\\partial x}$\r\n$ \\frac{d}{dt}(\\frac{\\partial L}{\\partial \\dot{x}})=\\frac{d}{dt}((2M+\\frac{3}{2}m)\\dot{x}-m(R-r)(\\cos(\\phi-\\psi)+\\frac{1}{2})\\dot{\\psi})=$\r\n$ = (2M+\\frac{3}{2}m)\\ddot{x}-m(R-r)\\sin(\\phi-\\psi)\\dot{\\psi}^{2}-m(R-r)(\\cos(\\phi-\\psi)+\\frac{1}{2})\\ddot{\\psi}$\r\n$ \\frac{\\partial L}{\\partial x}=-(M+m)g\\sin{\\phi}$\r\nTherfore,\r\n$ (2M+\\frac{3}{2}m)\\ddot{x}-m(R-r)\\sin(\\phi-\\psi)\\dot{\\psi}^{2}-m(R-r)(\\cos(\\phi-\\psi)+\\frac{1}{2})\\ddot{\\psi}=$ \r\n$ =-(M+m)g\\sin{\\phi}$\r\nSince, we examine a particular solution for the system $ \\dot{\\psi}=0$, then our equation reduces to \r\n$ (2M+\\frac{3}{2}m)\\ddot{x}=-(M+m)g\\sin\\phi$\r\nNow comes the second Lagrange equation for the system:\r\n$ \\frac{d}{dt}(\\frac{\\partial L}{\\partial \\dot{\\psi}})=\\frac{\\partial L}{\\partial \\psi}$\r\n$ \\frac{d}{dt}(\\frac{\\partial L}{\\partial \\dot{\\psi}})=\\frac{d}{dt}(\\frac{3}{2}m(R-r)^{2}\\dot{\\psi}-m(R-r)(\\cos(\\phi-\\psi)+\\frac{1}{2})\\dot{x})=$\r\n$ =\\frac{3}{2}m(R-r)^{2}\\ddot{\\psi}-m(R-r)\\sin(\\phi-\\psi)\\dot{\\psi}\\dot{x}-m(R-r)(\\cos(\\phi-\\psi)+\\frac{1}{2})\\ddot{x}$\r\n$ \\frac{\\partial L}{\\partial \\psi}=-m(R-r)\\sin(\\phi-\\psi)\\dot{x}\\dot{\\psi}-mg(R-r)\\sin\\psi$\r\nAgain, making $ {\\dot{\\psi}=0}$, we get\r\n$ -(\\cos(\\phi-\\psi)+\\frac{1}{2})\\ddot{x}=-g\\sin\\psi$\r\nCombining the two equations we got, we'll achieve\r\n$ \\frac{(M+m)g\\sin{\\phi}}{2M+\\frac{3}{2}m}=-\\frac{g\\sin{\\psi}}{\\cos(\\phi-\\psi)+\\frac{1}{2}}$\r\n$ \\frac{2M+\\frac{3}{2}m}{(M+m)}=-\\frac{\\sin\\phi(\\cos(\\psi-\\phi)+\\frac{1}{2})}{\\sin{\\psi}}=$\r\n$ =-\\frac{1}{2}(\\frac{\\sin\\psi+\\sin(2\\phi-\\psi)+\\sin\\phi}{\\sin\\psi})=-\\frac{1}{2}(1+\\frac{\\sin(2\\phi-\\psi)+\\sin\\phi}{\\sin\\psi})=$\r\n$ =-\\frac{1}{2}(1+\\frac{2\\sin(3\\phi-\\psi)\\cos(\\phi-\\psi)}{\\sin\\psi})$\r\nCan anybody check the $ crap$ I've just made here..." } { "Tag": [ "inequalities" ], "Problem": "Let $n$ be a natural number. Define $t(n)$ as the number of positive divisors of $n$ (including $1$ and $n$) en define $\\sigma(n)$ as the sum of these numbers. Show that \r\n\r\n\\[\\sigma(n) \\geq \\sqrt{n}. t(n)\\]", "Solution_1": "[hide]\nLet the divisors be $1 = d_{0}< d_{1}< \\cdots < d_{k}= n$. Clearly $d_{0}d_{k}= n$ and in general $d_{i}d_{k-i}= n$. Then we clearly have\n\n$d_{0}d_{1}\\cdots d_{k}= n^{\\frac{k+1}{2}}$.\n\nBy AM-GM, we know\n\n$\\frac{d_{0}+d_{1}+\\cdots+d_{k}}{k+1}\\ge \\sqrt[k+1]{d_{0}d_{1}\\cdots d_{k}}= \\sqrt[k+1]{n^{\\frac{k+1}{2}}}= \\sqrt{n}$.\n\nBut $\\sigma(n) = d_{0}+d_{1}+\\cdots+d_{k}$ and $\\tau(n) = k+1$ so\n\n$\\frac{\\sigma(n)}{\\tau(n)}\\ge \\sqrt{n}\\Rightarrow \\sigma(n) \\ge \\tau(n) \\sqrt{n}$.[/hide]" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "LET a,b,c>0\r\nPROVE THAT:\r\nabc+2(a^2+b^2+c^2)>=8+5(a+b+c) :wink:", "Solution_1": "[quote=\"andrew-mathmaster\"]LET a,b,c>0\nPROVE THAT:\nabc+2(a^2+b^2+c^2)>=8+5(a+b+c) :wink:[/quote]\r\nIt is wrong: try $a=b=c\\to 0$.", "Solution_2": "I am sorry \r\nabc+2(a^2+b^2+c^2)+8>=5(a+b+c)[/hide]" } { "Tag": [ "email", "\\/closed" ], "Problem": "hy all i have some freind who have as email adress \r\nxxxxx@gmail.fr but haw do they make it because i want to have one :roll:", "Solution_1": "I signed up for gmail here:\r\n\r\nhttps://www.google.com/accounts/SmsMailSignup1" } { "Tag": [ "inequalities", "inequalities open" ], "Problem": "As the title say, who can prove Nesbitt's inequality for $7$ numbers? Does an elementary solution exist?", "Solution_1": "Can anybody show me any information about an elementary solution of Nesbitt inequality for $n=7$?" } { "Tag": [], "Problem": "The sum of three numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all three?\r\n\r\n$\\left(A\\right) 28$\r\n$\\left(B\\right) 40$\r\n$\\left(C\\right) 100$\r\n$\\left(D\\right) 400$\r\n$\\left(E\\right) 800$", "Solution_1": "k i saw a lot of unsolved one's so i guess im done for now cuz i'm bored..\r\n\r\n[hide]\n\n1. Let $a+b+c=20$\n\n$a=4(b+c)\\implies (b+c)=\\frac{a}{4}$ sub into 1\n\n$a+\\frac{a}{4}=20\\implies a=16$\n\nsub $b=7c$ and $a=16$ into 1\n\n$\\implies 8c=4\\implies c=\\frac{1}{2}\\implies b=\\frac{7}{2}$\n\n$abc=\\frac{1}{2}\\cdot\\frac{7}{2}\\cdot 16=\\boxed{28}$\n\n[/hide]" } { "Tag": [ "inequalities", "logarithms", "inequalities unsolved" ], "Problem": "If $ a,b > 0$ and $ x,y \\ge 0$ are non-negative real numbers such that $ x \\plus{} y \\equal{} 1$ , prove\r\n\\[ ax \\plus{} by\\ge a^xb^y\\]\r\n\r\nEdited: arqady's right. Inequality is defined for real exponents, and $ a,b >0$ . So how to get through with it ?", "Solution_1": "[quote=\"Obel1x\"]If $ a,b \\ge 0$ and $ x,y$ are non-negative numbers such that $ x \\plus{} y \\equal{} 1$ , prove\n\\[ ax \\plus{} by\\ge a^xb^y\\]\n[/quote]\r\n\r\n\r\n $ ab \\equal{} 0$ obvious true\r\n\r\n$ ab > 0$ \r\n\r\n$ ax \\plus{} by\\ge a^xb^y \\Longleftrightarrow x\\ln{a} \\plus{} y\\ln{b}\\le \\ln{(ax \\plus{} by)}$\r\n\r\ntrue due to Jensen\u3002", "Solution_2": "I wonder we can use Bernoulli inequality.", "Solution_3": "[quote=\"hedeng123\"][quote=\"Obel1x\"]If $ a,b \\ge 0$ and $ x,y$ are non-negative numbers such that $ x \\plus{} y \\equal{} 1$ , prove\n\\[ ax \\plus{} by\\ge a^xb^y\\]\n[/quote]\n\n\n $ ab \\equal{} 0$ obvious true\n\n[/quote]\r\nI think, if $ ab\\equal{}0$ it's obviously wrong.\r\nIndeed, $ a^x$ defined for real numbers $ a$ and $ x$ only for $ a>0.$ :wink:", "Solution_4": "you are right. :D" } { "Tag": [ "geometry", "3D geometry", "sphere" ], "Problem": "I just have a little question, because I can't picture this in my head.\r\n\r\nSay you had three cylinders such that they are each the same size and shape, and the diameter of the cylinders is equal to the height of the cylinders. Say you put the cylinders together at different orientations, such that one is facing front-back, one is facing left-right, and one facing up-down, and they all share the same center. Is the intersection of their three volumes a sphere? If not, what is it?", "Solution_1": "No, it's not a sphere. Check this out for some more info: [url]http://mathworld.wolfram.com/SteinmetzSolid.html[/url]\r\nIt's like two-thirds of the way down the page.", "Solution_2": "Thank you for your help :lol: ." } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "We have $A,B \\in M_{4}(R)$ and $A-B=AB.$\r\nIf it exists $m \\in N^*$ so that:$ A^m-B^m $ is invertible, than prove that A and B are invertible.", "Solution_1": "Since $A-B-AB+I_n=I_n$ (we don't need the fact that they're in $M_4(\\mathbb R)$; any $M_n(\\mathbb R)$ works), we have $()()=I_n$, so $()()=I_n$, meaning that $A,B$ commute. This means that $A^m-B^m=(A-B)(A^{m-1}+\\ldots+B^{m-1})=AB(A^{m-1}+\\ldots+B^{m-1})$, and if this is invertible, then $A,B$ must be invertible." } { "Tag": [ "AMC", "AIME", "geometry", "AIME I" ], "Problem": "hey.. does anyone know how to sign up for other olimpiads other than math?\r\nlike how do i take the physics and bio equivalent of the AMC or AIME?\r\nI dont think my teachers offer it or know much about it.. which is why i missed out last year... although my skool is very involved in the chem olimpiad..", "Solution_1": "For Bio: \r\n\r\n http://www.cee.org/usabo/usabo_how_to_enter.shtml\r\n\r\nFor Physics:\r\n\r\n http://www.aapt.org/Contests/olympiad.cfm", "Solution_2": "For Computer programming:\r\n\r\n http://www.usaco.org", "Solution_3": "Wait...let me get this straight...for biology, all you have to do, is just be like, \"Hey Miss, nominate me for this USABO!\" and then you're in? Wow, that's so much more easier the the USAMO. :-P", "Solution_4": "It's not quite the same. Your teacher can nominate you for the Open Exam which is kind of like taking the Amc 12/10. I believe the teacher can nominate as many people as possible. So it s not the same as taking the USAMO.", "Solution_5": "What about chemistry? To me, chem way cooler than biology.", "Solution_6": "Look on this page for all kinds of competitions [url]http://cty.jhu.edu/imagine/linkB.htm[/url]. It has the olympiads under the math/science section somewhere.\r\n\r\nI took the Chemistry this year. I made it past the first round but did horrible in the second. I was happy with my results considering that I had only been in chemistry for less than half of a year.", "Solution_7": "Chemistry is a bit shady. You're actually competing regionally rather than nationally for the first exam. For example, out of 60 questions, in some areas you may need 40ish questions correct to move on to the national exam whereas in other areas only the high fifties will allow you to take the next test. Read Zumdahl, Chang, or some other chem textbook front to back and you should be set.", "Solution_8": "oh god.. finally.. i was looking for this post for so long.. forgot where i put it.. thanks", "Solution_9": "oh man.. jus did a few problems on physics test.. i am getting rusty only after a few month.. guess i'll have to study for it before i take it.. anyways.. anyone know what the curves are for both bio and physics?" } { "Tag": [], "Problem": "A negatively charged particle is caused to move between two electrically charged plates, as illustrated in Figure 2.7. If the mass of this particle is creased while the speed of the particles remains the same, would you expect the bending to increase, decrease or stay the same, and why?\r\n\r\nThanks in advance!", "Solution_1": "do you mean increased or decreased?\r\n\r\nIf increased, then ur increasing the particles kinetic energy, hence it will have less bending as the object will need more work acted upon it to change its course... The opposite for decrease." } { "Tag": [ "limit", "induction", "inequalities", "function", "integration", "calculus", "calculus computations" ], "Problem": "Solve an equation $\\ds\\lim_{n\\to\\infty}\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\ldots+\\sqrt{x^{n}}}}}=2.$\r\n(A. Kukush)", "Solution_1": "I have managed to prove the following:\r\n1) There exists precisely one $x$ in $\\mathbb{R}^{+}$ that solves the equation (it is also clear that this solution is not in $[0,1]\\cup[6,\\infty)$ through easy calculations, not in say $[0,3.99]\\cup[4.01,\\infty)$ by numerical calculations with computer).\r\n\r\n2. For any $m$ and $x > 1$, $\\ds\\lim_{n\\to\\infty}\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\ldots+\\sqrt{x^{n}}}}}< \\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\ldots+\\sqrt{x^{m}+\\frac{x}{2}\\left(1+\\sqrt{1+4x^{m-1}}\\right) }}}}}$ (and in the proof you also see that the larger $m$ the closer the RHS is to th LHS).\r\n\r\nI am also certain that the solution is $x=4$ from numerical evidence, but I can't prove it.", "Solution_2": "kalle your intuition is good probably $x=4$ is the only solution.\r\nJean_No\u00ebl used mapple he told he found $4$.\r\n\r\nFind the proof", "Solution_3": "Do you know a proof? I have spent a lot of time on this problem, and it seems futile. So I will not think about it any more. The approaches I've tried are: \r\n\r\n1) To consider the the equation you get for each $n$ when you remove the limit sign. By intermediate value theorem you have that for each $n$ there is precisely one solution $a_{n}$. It is clear that $a_{n}$ converges... from some reasonings around this you get my statement \"1)\" in my previous post. I have tried to prove by induction that $4-\\frac{1}{2^{n}}< a_{n}< 4+\\frac{1}{2^{n}}\\quad n \\ge 2$, but I can't do it.\r\n\r\n2) I have tried to calculate the limit directly for each x, mainly focusing on finding good inequalities (which may also help in 1). I think the one I posted is the best one I got. It comes from that I managed to show that:\r\n$\\sqrt{ax^{2}+\\sqrt{ax^{4}+\\sqrt{ax^{8}+\\cdots}}}= \\frac{x}{2}\\left( 1+\\sqrt{1+4a}\\right)$.\r\n\r\nI am not getting further than that.", "Solution_4": "How do you get this equality?\r\n\r\n$\\sqrt{ax^{2}+\\sqrt{ax^{4}+\\sqrt{ax^{8}+\\cdots}}}= \\frac{x}{2}\\left( 1+\\sqrt{1+4a}\\right)$.", "Solution_5": "$\\sqrt{ax^{2}+\\sqrt{ax^{4}+\\sqrt{ax^{8}+\\cdots}}}= x\\sqrt{a+\\sqrt{a+\\sqrt{a+\\cdots}}}$. Presuming convergence,\r\n$S = \\sqrt{a+\\sqrt{a+\\sqrt{a+\\cdots}}}\\implies S^{2}= a+S$ whence the result.\r\n\r\nI don't think this is going to be helpful here, unfortunately -- there's no obvious way to go from powers of 2 to all the integers in such an expression.", "Solution_6": "Well, my idea was to take a certain number of terms in the original nested radical and then approximate the rest of the terms with such a series. I don't think the problem can be solved by finding a clever equality, I believe it must be solved with inequalities.", "Solution_7": "Hey,I think I got it,using numerator-rational.\r\nwe have:\r\n$2-\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}=\\frac{3-\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}{2+\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}}=\\frac{5-\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}{\\left(2+\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}\\right)\\left(3+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}\\right)}$\r\n$=\\cdots=\\frac{2^{n+1}+1}{\\left(2+\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}\\right)\\left(3+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}\\right)\\cdots \\left(2^{n}+1+\\sqrt{4^{n}}\\right)}$\r\nthen it is easy to prove the limit is 0.\r\nSo we have proved the limit of the left side is 2,when x=4.\r\nThen by proving the limit of the left side is strickly increasing by x.We sloved the whole problem. :) \r\nAm I right?", "Solution_8": "Yes, very nice!", "Solution_9": "Bravo zhaobin", "Solution_10": "The above posts culminating with the alleged proof I have a complaint with inasmuch as they it is like working on a multiple choice math question and eliminating choices and then plugging in one of the remaining values to see that it appears to be working.\r\n\r\nNone of the posts have taken a constructivist approach to actually solve the equation outright. If this value of 4 \"weren't \r\neven on the map,\" what would you solvers do instead?", "Solution_11": "I have found a different solution, which is unfortunately not nearly as simple as zhaobin's. Actually, my solution doesn't quite work. I am almost certain it is correct up to and including the point where I conclude $\\lim_{n\\to\\infty}a_{n}= 4$. However, the argument following is rubbish. I must've been thinking that the limit is a continuous function of $x$ or something similar. Of course it sucks to realize that you're wrong. Could anyone help me by finding an argument for the last bit?\r\n \r\nConsider the equation $\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\ldots+\\sqrt{x^{n}}}}}=2$ for every integer $n$. The LHS expression is strictly increasing in $[0,\\infty)$, assuming all values in $[1,\\infty)$, according to the intermediate value theorem. Therefore the equation has precisely one solution $a_{n}$. It is clear that $(a_{n})_{n=1}^\\infty$ is a strictly decreasing bounded sequence, and therefore converges to a limit $A$.\r\n\r\nNow, I hope you can understand the upcoming notation. It might require you to write the following out yourself. By squaring and rearranging the equation many times we find that it's equivalent to\r\n$\\sqrt{x^{n}}= \\left( \\cdots \\left( \\left(9-x\\right)^{2}-x^{2}\\right)^{2}-\\cdots-x^{n-2}\\right)^{2}-x^{n-1}$, (**)\r\nBy an inductive argument you can see that for $x=4$ the RHS in (**) is equal to $2^{n}+1$. From this you see that \r\n $\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}< 2$\r\nand thus $4 < a_{n}$ for every $n$.\r\n\r\nBy another inductive reasoning the RHS in (**) is always positive for $x \\in [0,a_{n}]$. This statement is truthfully a little bit work to show, but it's not hard. Since it is true for every $n$, it implies that you can't find a bracket \"inside\" the RHS that is negative in $[0,a_{n}]$ (remember, $a_{n}$ is decreasing). This in turn implies that the RHS is strictly decreasing (with increasing $x$) in $[0,a_{n}]$.\r\n\r\nNow introduce $f_{n}(x) = \\frac{\\sqrt{x^{n}}}{\\left( \\cdots \\left( \\left(9-x\\right)^{2}-x^{2}\\right)^{2}-\\cdots-x^{n-2}\\right)^{2}-x^{n-1}}$\r\n\r\nOur equation is $f_{n}(x) = 1$. By what I said above we have that $f_{n}(4) = \\frac{2^{n}}{2^{n}+1}= 1-\\frac{1}{2^{n}+1}\\to 1$ as $n\\to\\infty$. From the discussion about monotony, we realize that $f_{n}$ is strictly increasing in $[4,a_{n}]$ for every $n$.\r\n\r\nI claim that $A = 4$. We will prove it by contradiction: Assume that $A > 4$. Then we can choose a large $n$ so that $f_{n}(4) \\approx 1$. Then since $f_{n}$ is stricly increasing in $[4,a_{n}]$ and $f_{n}(a_{n}) = 1$ it follows that for every $b \\in [4,A]$, $\\lim_{n\\to\\infty}f_{n}(b) = 1$. In fact, we have $f_{n}\\to 1$ uniformly in $[4,A]$! This would have the consequence that $\\lim_{n\\to \\infty}\\int_{4}^{A}f_{n}dx = A-4$. However,\r\n$\\int_{4}^{A}f_{n}dx > \\int_{4}^{A}\\frac{\\sqrt{x^{n}}}{2^{n}+1}dx$.\r\nA pure calculation shows that the right hand side tends to $\\infty$ as $n$ gets large. So we have found a contradiction. \r\n\r\nThat is, we have now shown that $\\lim_{n\\to\\infty}a_{n}= 4$.\r\n\r\nFinally, we show that\r\n$\\ds\\lim_{n\\to\\infty}\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}=2$\r\nSuppose the opposite, then\r\n$\\ds\\lim_{n\\to\\infty}\\sqrt{1+\\sqrt{4+\\sqrt{4^{2}+\\ldots+\\sqrt{4^{n}}}}}<2$\r\nThen, since the value of the limit strictly increases with $x$, we can find a $c>4$ such that \r\n$\\ds\\lim_{n\\to\\infty}\\sqrt{1+\\sqrt{c+\\sqrt{c^{2}+\\ldots+\\sqrt{c^{n}}}}}<2$\r\nBut this is absurd because eventually $a_{n}< c$.", "Solution_12": "Sorry for bumping this thread, but as it is I don't think people have seen that I would like some help/discussion. As it stands I have defined a sequence $a_{n}$ which for each $n$ is value of the solution to the equation:\r\n$\\sqrt{1+\\sqrt{x+\\cdots+\\sqrt{x^{n}}}}= 2$.\r\nFurthermore, I have shown that $\\lim_{n \\to \\infty}a_{n}= 4$.\r\n\r\nHowever, although I have included a bogus argument, I don't know how to show that this implies that $x=4$ solves the initial equation. Actually, I don't even know how to show that the solution is unique. \r\n\r\nDoes anyone have any ideas? I'd love some help.", "Solution_13": "[quote=\"Kalle\"]\n\nDoes anyone have any ideas? I'd love some help.[/quote]\r\nI think we can prove that the limit of the left side is strick increasing by $x$.\r\nlet $a(n,x)=\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}}$\r\nFirst step:Prove that $a(n,x)$ has limit when n go infinite for every x.\r\nthis is because:$x<1$,obvious.\r\n$x>1$,$a(n,x)y$ we have:\r\n$\\lim_{n\\to \\infty}\\sqrt{x^{2}+\\sqrt{x^{3}+\\cdots+\\sqrt{x^{n}}}}\\ge \\lim_{n\\to \\infty}\\sqrt{y^{2}+\\sqrt{y^{3}+\\cdots+\\sqrt{y^{n}}}}$\r\nthus:$\\lim_{n\\to \\infty}a(n,x) >\\lim_{n\\to \\infty}a(n,y)$(do it yourself :) ) \r\nNow we proved limit of the left side is strick increasing by $x$.\r\nThe result follows.This is my approach.\r\n\r\nBut a guy in Chinese forum proved:\r\n$\\lim_{n\\to \\infty}\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}}=\\sqrt{x}+\\frac{4-x}{2+\\sqrt{x}}$\r\nbut his proof is only write:\r\n$\\lim_{n\\to \\infty}\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}}-\\sqrt{x}=\\lim_{n\\to \\infty}\\frac{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}-x}{\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}}+\\sqrt{x}}=\\frac{4-x}{2+\\sqrt{x}}$\r\nI can't understand his proof,And I also doubt it...", "Solution_14": "Maybe he meant something different? \r\n\r\nWe have that\r\n$\\sqrt{x}+\\frac{4-x}{2+\\sqrt{x}}= 2$ for every $x$.", "Solution_15": "[quote=\"Kalle\"]Maybe he meant something different? \n\nWe have that\n$\\sqrt{x}+\\frac{4-x}{2+\\sqrt{x}}= 2$ for every $x$.[/quote]\r\nbut he said he is wrong...", "Solution_16": "[quote=\"zhaobin\"]\n\n$\\lim_{n\\to \\infty}\\sqrt{1+\\sqrt{x+\\sqrt{x^{2}+\\cdots+\\sqrt{x^{n}}}}}=\\sqrt{x}+\\frac{4-x}{2+\\sqrt{x}}$\n[/quote]\r\n\r\nthis is not correct for $x=1$ \r\n\r\n$\\lim_{n\\to \\infty}\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots+\\sqrt{1}}}}=L$ positive\r\nwith $L^{2}=1+L$ \r\n\r\n$L=\\frac{1+\\sqrt{5}}{2}$\r\n\r\n\r\nbut when you plug $x=1$ in $\\sqrt{x}+\\frac{4-x}{2+\\sqrt{x}}$\r\nwe get 1++1=2", "Solution_17": "Ok,I also think it is impossible...\r\nBut now the whole problem is also solved.\r\nDo you agree?", "Solution_18": "Which is the official solution for this problem?" } { "Tag": [ "calculus", "real analysis", "real analysis unsolved" ], "Problem": "This problem is from our schools exams .\r\n\r\nLet $ f: [a,b] \\rightarrow R$ continuous to $ [a,b]$.$ f''$ is also continuous to $ (a,b)$ .If $ f(a)=f(b)=0$, and exists real numbers $ c,d\\in (a,b)$ such as$ f(c)f(d)<0$ proove \r\n\r\ni) $ f(x)=0$ has at least one real root to $ [a,b]$ \r\n\r\nii) exists$ \\xi_{1}, \\xi_{2}\\in (a,b)$ such as $ f''(\\xi_{1})>0$ $ f''(\\xi_{2})<0$", "Solution_1": "(i) Since $ f(c)f(d)<0$, we must have $ f(c)$ and $ f(d)$ be of opposite sign. Intermediate Value theorem guarantees a value between $ c$ and $ d$, say $ x$, such that $ f(x)=0$\r\n\r\n(ii) Note that since $ f(a)=f(x)=f(b)=0$, we can not have $ f''(x)=0 \\hspace{1 mm}\\forall x$. But we also can not have $ f''(x)>0 \\hspace{1 mm}\\forall x$ or $ f''(x)<0 \\hspace{1 mm}\\forall x$ since $ f$ has two interior critical points which follows from Rolle's theorem. Therefore the result follows." } { "Tag": [ "MATHCOUNTS" ], "Problem": "29. The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string?\r\n\r\nGracias. :D", "Solution_1": "Huh? It's not possible is it? Cause if you have:\r\n$ 1, 9...$ What's after the 9??????", "Solution_2": "... Wrong problem, wrong thread, jonathan...\r\n\r\nThe numbers possible are:\r\n19\r\n38\r\n57\r\n76\r\n95\r\n\r\n31\r\n62\r\n93\r\n\r\nAs you can see, 38 can be eliminated (as there is no two-digit number with 8 to follow up) and 62 (no leading 2).\r\n\r\nThe numbers possible are:\r\n19\r\n57\r\n76\r\n95\r\n\r\n31\r\n93\r\n\r\nNow, 57,76,and 95 will be eliminated.\r\n\r\n19\r\n31\r\n93\r\n\r\nThus it cycles, 193193193... and the 2002 = 1 (mod 3) digit is [b]1[/b]\r\n\r\nNotice that there exists no configurability, and there exists only one string.", "Solution_3": "Uh.. You might want to see our gmail chat. I edited the names:\r\n\r\nmr.k_74: you better want to change ur post now\r\ni edited my post\r\n\r\njonathanchou711: oh..\r\nok\r\n\r\nmr.k_74: me bad\r\n\r\njonathanchou711: that ok\r\n\r\nSo I was we were both referring to the Mathcounts Marathon place, but we both edited. Apparently, you were in the middle of typing urs when I edited.", "Solution_4": "The answer isn't $ 1$...after the last $ 3$ shows up, the final digit can be anything that the three allows it to be, disregarding the cycles (since there's nothing after it), and the answer is $ 193193\\ldots193\\boxed8$.", "Solution_5": "OMG!!! Yeah!! 8!!! Wow. Nice." } { "Tag": [ "modular arithmetic", "search", "algorithm", "Euler", "geometry", "3D geometry", "number theory" ], "Problem": "Mr X chooses an integer x={0,1,2,...,1924992000}. He squares it, divides it by 4000000007 and gives us the remainder, so\r\n(x 2) mod 4000000007 = k\r\n\r\nThe question is:\r\nIf we know k - how to find x?", "Solution_1": "This is not possible, since a value of $k$ can correspondence with different values of $x$. For example $4\\equiv2^2\\equiv4000000005^2\\pmod{400000007}$.", "Solution_2": "It has to be possible and I know it is - just can't find the right way to solution....\r\n\r\nAnd btw:\r\n1924992000<4000000005, so we don't need to worry about integers larger then 1924992000....", "Solution_3": "If someone gave $k=4000000006$ or $k=5$ you would be out of luck, as there are no solutions .. on the other hand $k=2$ would be ok. and you should get your $x$..:)\r\n\r\n(For example you can try $ x = 2529181847$ and you would get $x^2 \\equiv 2 \\bmod 4000000007$ )\r\n( Another value of x (if you wanted it to be less than 2000000000) is $x= 1470818160 $ which will give $x^2 \\equiv 2 \\bmod 4000000007$)\r\n\r\n\r\n\r\n(The original question is \"given $k$ how do take a \"square-root to find $x$\" ? (all calculations are mod 4000000007)\r\n(and within a reasonable amout of time :) (Hint: It is lucky that 4000000007 is prime and of the form 4m+3) \r\n\r\nImportant Note: Between 0 and 4000000006 there are always two values of x which will satisfy any given k you must restrict 0<= x <= 2000000003 to get an unique value of x)", "Solution_4": "I understand that we have to take such x that remainder mod 4000000007 will be unique. But x is an integer from set {1,2,3,4,...,1924992000} and 2529181847>1924992000, so if x is from the given set - 1470818160 is ok, but 2529181847 is out of it - it's just too big. And of course - as 2000000003>1924992000, there has to be only one value that satisfy given k - I'm just wondering how to find x, when we are given k (the remainder) and all we know is that x has unique value...\r\n\r\nCan you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any \"mathematical\" way of getting x when we know k?", "Solution_5": "First: IMO this problem is probably more suitable for the \u201cPre-Olympiad? Forum.\r\n\r\nAny way here is one method to do this:\r\n\r\nAll calculations below are $\\bmod 4000000007$ and $x$ or $k$ etc are not $0$.\r\nWe are given $k=x^2$ \r\nBut, since $4000000007$ is prime, FLT ===> : $x^{4000000006} \\equiv 1$ \r\nOr $k^{2000000003} \\equiv x^{4000000006} \\equiv 1$ \r\nLet $y \\equiv k^{1000000002}$\r\nThen $y^2 \\equiv k^{2000000004} \\equiv k$\r\nSo $y \\equiv +x \\ or \\ y \\equiv -x $ You take the value between 1 and 2000000003. \r\n\r\n[hide=\"Method works when number is prime and 3 mod 4\"]Note if $4000000007$ was not of the form $4n+3$ the above method (try it and see why, use $13$ for example) will not work, and method would be a little different. \n[/hide] \n\nAlso finding $k^{1000000002}$ is not really that hard to compute. [hide]You go by computing 10th power, eg say $a \\equiv k^{10}$ then compute $a^{10}$ etc.. so you have to do only 8-9 such computations (and not billions).\n[/hide] \r\nHope this is helpful.", "Solution_6": "[quote=\"tytus\"]Can you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any \"mathematical\" way of getting x when we know k?[/quote]\r\n\r\nWhat you are trying to do is a special (very simple) case of a standard cryptographic algorithm. [You might search for \"Berlekamp's Algorithm\"]\r\n\r\nLet $m$ be the plaintext (an integer in $[0,1924902000]$). Let $c$ be the ciphertext (an integer in $[0, 4000000007)$.)\r\n\r\nYou have told us that $c=m^2 \\mod 4000000007$.\r\nIs there a number, $d$, such that $m=c^d \\mod 4000000007?$\r\n\r\nCombining these equations:\r\n$m=(m^2)^d \\mod 4000000007$\r\n$m=m^{2d} \\mod 4000000007$\r\n\r\nIn general:\r\n$m=m^{1+\\phi(n)} \\mod n$\r\n\r\nWhere $\\phi$ is the Euler Totient Function. In the special case, where $n$ is a prime, we have $\\phi(n)=n-1$. [This is equivalent to Fermat's Little Theorem, as Gyan previously noted.]\r\n\r\nSo, all we are left is to find:\r\n$2d=1+j\\phi(n)$, where $j$ is a positive integer.\r\n\r\n$2d=1+j(n-1)$\r\n$2d=1+j(4000000007-1)$\r\n$2d=1+j(4000000006)$\r\n\r\nOops. There is no value of $j$, such that $2d$ will be odd. In a roundabout way, this explains why there is not a unique square root, modulo an odd prime.\r\n\r\nIf $n$ was not prime, then $\\phi(n)$ would have a more complex form, and it would be necessary to factor $n$ to find $\\phi(n)$. [Choosing $n$ to be the product of two very large randomly selected primes is the key to the RSA algorithm's strength.]\r\n\r\nIf you had cubed the number and asked for the cube-root, modulo 4000000007, then we could have found a unique value of $1\\le d<4000000007.$ [In fact, $3\\times 2666666671=1+2\\times 4000000006$.]\r\n\r\nGyan has shown how to obtain the two square roots, under specified conditions. Between my explanation and his, does this answer your questions?", "Solution_7": "[quote=\"tytus\"]I understand that we have to take such x that remainder mod 4000000007 will be unique. But x is an integer from set {1,2,3,4,...,1924992000} and 2529181847>1924992000, so if x is from the given set - 1470818160 is ok, but 2529181847 is out of it - it's just too big. ..\n[/quote]\n\nYes, you are right, and I was just pointing that out in my first message, of course, $1470818160$ is nothing but $(4000000007 - 2529181847)$ so once you restrict the range, the value is unique.\nIn my second post I posted one method to find that value.\n\n[quote] \nCan you please explain me why it's so important that 4000000007 is prime of the form 4m+3?? Is there any \"mathematical\" way of getting x when we know k?[/quote]\r\nWhen the modulo mumber is prime of the form 4k+3 (Or a composite number with all factors of the form 4m+3) Things are a little easier, becuase then you can write x as a power of k. (The reason for this a little complex) \r\n\r\nFor numbers not of the type 4k+3, one may use a little different method but it is still not really that difficult . BTW Note that [b]not [/b]every number will have a \"square-root\".", "Solution_8": "OK - now I understand little bit more, but can you please show me how does it work?? For example: k=1749870067 and I know that then x=1087143639. But what is the way to get x??", "Solution_9": "[quote=\"tytus\"]OK - now I understand little bit more, but can you please show me how does it work?? For example: k=1749870067 and I know that then x=1087143639. But what is the way to get x??[/quote]\n\n[b]That's what was explained before:[/b]\nUsing a calculator (Maple) for example: (Time taken : only a few seconds)\n\n[quote]> n:=4000000007;k:=1749870067; \n\n[color=blue](To avoid overflow we take only 1000 power of k at a time .. we have to find k^1000000002)[/color]\n\n> y:=k^1000 mod n; ====> y := 3603893674 ([color=blue] Done to avoid overflow) [/color]\n> y:=y^1000 mod n; =====> y := 1787853990 [color=blue] ( k^1000000)[/color]\n> y:=y^1000 mod n; =====> y := 3398870423 [color=blue] (k^1000000000)[/color]\n\n> y:=y*k^2 mod n; ======. y := 2912856368 [color=blue] (k^1000000002)[/color]\n\n> x:=n-y; [color=red] [size=150] x := 1087143639[/size] [/color] [color=blue](Since 2912856...> 2000000002 we subtract from n) [/color]\n\n[/quote]\r\n\r\nAdded later: Just curious, what is your back ground as far as number theory is concerned? I have seen other questions (Like how to decompose a prime into sum of two squares, and how to find a prime etc... ) Looks like you are interested in computing methods? ..." } { "Tag": [ "inequalities", "search", "geometry", "inradius", "circumcircle", "trigonometry", "Cauchy Inequality" ], "Problem": "They may have been solved ...but ...I need your solutions...", "Solution_1": "[hide=\"6\"]\nmultiply both sides by $ xyz$,we have to show that:\n\n$ x^4 \\plus{} y^4 \\plus{} z^4\\geq xyz(x \\plus{} y \\plus{} z)$\n\nnow note that by cauchy inequality we have:\n\n$ x^4 \\plus{} y^4 \\plus{} z^4\\geq \\frac {(x^2 \\plus{} y^2 \\plus{} z^2)^2}3\\geq \\frac {(x \\plus{} y \\plus{} z)^4}{27}$\n\nso our inequality is equivalent to:\n\n$ \\frac {(x \\plus{} y \\plus{} z)^4}{27}\\geq xyz(x \\plus{} y \\plus{} z)$\n\n$ \\iff \\frac {x \\plus{} y \\plus{} z}3\\geq \\sqrt [3]{xyz}$\n\nwhich is true by AM-GM...[/hide]", "Solution_2": "The seventh is incorrect,i think it should be:\r\n$ 9(a^3\\plus{}b^3\\plus{}c^3)\\geq (a\\plus{}b\\plus{}c)^3$.\r\nFor a proof,look here:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=171643\r\nThe third is Iran 1996,\r\nSee here:\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=209710799&t=3547\r\nAnd i think that it should be:\r\n$ \\sum \\frac{a}{b\\plus{}c}<2$,\r\ninstead of \r\n$ \\sum \\frac{a}{b\\plus{}c}\\leq 2$.", "Solution_3": "i'm sending some ques.s more!", "Solution_4": "19. \r\n\r\n\r\n$ \\frac {1}{1 \\minus{} x^{2}} \\plus{} \\frac {1}{1 \\minus{} y^{2}}\\geq 2\\sqrt {\\frac {1}{(1 \\minus{} x^{2})(1 \\minus{} y^{2})}} \\equal{} \\frac {2}{\\sqrt {(1 \\minus{} x^{2})(1 \\minus{} y^{2})}}\\geq\\frac {2}{\\frac {1 \\minus{} x^{2} \\plus{} 1 \\minus{} y^{2}}{2}} \\equal{} \\frac {4}{2 \\minus{} (x^{2} \\plus{} y^{2})}\\geq\\frac {4}{2 \\minus{} 2xy} \\equal{} \\frac {2}{1 \\minus{} xy}$\r\n\r\n$ x \\equal{} y$", "Solution_5": "17.\r\n\r\n$ (a\\plus{}b\\plus{}c)\\left(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}\\right)\\geq 9,$ using Jensen's inequality gives\r\n\r\n$ \\frac{\\left(a\\plus{}\\frac{1}{a}\\right)^2\\plus{}\\left(b\\plus{}\\frac{1}{b}\\right)^2\\plus{}\\left(c\\plus{}\\frac{1}{c}\\right)^2}{3}\\geq \\left(\\frac{a\\plus{}\\frac{1}{a}\\plus{}b\\plus{}\\frac{1}{b}\\plus{}c\\plus{}\\frac{1}{c}}{3}\\right)^2$\r\n\r\n$ \\equal{}\\left(\\frac{1\\plus{}\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}}{3}\\right)^2\\geq\\left(\\frac{10}{3}\\right)^2$ yielding $ \\left(a\\plus{}\\frac{1}{a}\\right)^2\\plus{}\\left(b\\plus{}\\frac{1}{b}\\right)^2\\plus{}\\left(c\\plus{}\\frac{1}{c}\\right)^2\\geq \\frac{100}{3}$", "Solution_6": "[quote=\"kim77\"]i'm sending some ques.s more![/quote]\r\nFor problem 17.\r\n[hide=\"Generalisation holds: your case is for n=3.\"]\n[u]Problem:[/u]\n$ \\left(a_1 \\plus{} \\frac 1{a_1}\\right)^2 \\plus{} \\left(a_2 \\plus{} \\frac 1{a_2}\\right)^2 \\plus{} \\ldots \\plus{} \\left(a_n \\plus{} \\frac 1{a_n}\\right)^2\\ge K$\nFind the best K\n\n[b][i]Claim:[/i][/b]$ K \\equal{} \\frac {(n^2 \\plus{} 1)^2}{n}$\n[b][i][u]Proof:[/u][/i][/b]\nDenote LHS of your inequality by I,\nAM-HM says: \n$ (\\sum a_i)(\\sum \\frac1{a_i}) \\geq n^2 \\Rightarrow (1 \\plus{} \\sum \\frac1{a_i})^2 \\geq (n^2 \\plus{} 1)^2$-------------(*)\nApplying CS:\n$ I \\geq \\frac {{(a_1 \\plus{} \\frac 1{a_1} \\plus{} a_2 \\plus{} \\frac 1{a_2} \\plus{} \\ldots a_n \\plus{} \\frac 1{a_n}})^2}{1 \\plus{} 1 \\plus{} 1 \\ldots \\plus{} 1}$\nBut $ ({a_1 \\plus{} \\frac 1{a_1} \\plus{} a_2 \\plus{} \\frac 1{a_2} \\plus{} \\ldots a_n \\plus{} \\frac 1{a_n}} \\equal{} 1 \\plus{} \\sum \\frac1{a_i}$\nSo $ I \\geq \\frac {{(1 \\plus{} \\sum \\frac1{a_i})}^2}{n}$.\nBut by (*), $ \\frac {{(1 \\plus{} \\sum \\frac1{a_i})}^2}{n}\\geq \\frac {(n^2 \\plus{} 1)^2}{n}$\nThus $ I \\geq \\frac {(n^2 \\plus{} 1)^2}{n}$\nEquality for all $ a_i$s equal.\nK cant get any better than that because we showed equality achieving ;)[/hide]\n\n15 follows from AM-GM, applied to both terms....\n[hide]$ (\\Sigma{a^2b})(\\Sigma{ab^2}) \\geq 9{a^6b^6c^6}^{\\frac13}$\n[/hide]\n\n[hide=\"Problem 16 By AM_GM,\"]\n$ abc \\leq (\\frac {a \\plus{} 1}{2})^2 \\cdot (\\frac {b \\plus{} 1}{2})^2 \\cdot (\\frac {c \\plus{} 1}{2})^2 \\equal{} \\frac {8^2}{4^3} \\equal{} 1$[/hide]", "Solution_7": "2. For $ 0 < x < 1,\\ \\frac {1}{1 \\minus{} \\sqrt {x}}\\geq 4x \\plus{} 1\\Longleftrightarrow \\sqrt {x}(2\\sqrt {x} \\minus{} 1)^2\\geq 0,$\r\n\r\n$ \\frac {1 \\plus{} \\sqrt {a}}{1 \\minus{} a} \\plus{} \\frac {1 \\plus{} \\sqrt {b}}{1 \\minus{} b} \\plus{} \\frac {1 \\plus{} \\sqrt {c}}{1 \\minus{} c} \\plus{} \\frac {1 \\plus{} \\sqrt {d}}{1 \\minus{} d} \\equal{} \\frac {1}{1 \\minus{} \\sqrt {a}} \\plus{} \\frac {1}{1 \\minus{} \\sqrt {b}} \\plus{} \\frac {1}{1 \\minus{} \\sqrt {c}} \\plus{} \\frac {1}{1 \\minus{} \\sqrt {d}}$\r\n\r\n$ \\geq 4(a \\plus{} b \\plus{} c \\plus{} d) \\plus{} 1\\cdot 4 \\equal{} 8.$", "Solution_8": "[hide=\"For problem 10, \"]$ 4(a^3 \\plus{} b^3) \\geq (a\\plus{}b)^3 \\Leftrightarrow a^3 \\plus{} b^3 \\geq a^2b \\plus{} b^2a$\nIt follows from Muirhead, since [3,0] majorises [2,1][/hide]", "Solution_9": "[quote=\"isomorphism\"][hide=\"For problem 10, \"]$ 4(a^3 \\plus{} b^3) \\geq (a \\plus{} b)^3 \\Leftrightarrow a^3 \\plus{} b^3 \\geq a^2b \\plus{} b^2a$\nIt follows from Muirhead, since [3,0] majorises [2,1][/hide][/quote]\r\n\r\nJust using Jensen's inequality kills in 10 seconds.:D", "Solution_10": "[quote=\"kunny\"][quote=\"isomorphism\"][hide=\"For problem 10, \"]$ 4(a^3 \\plus{} b^3) \\geq (a \\plus{} b)^3 \\Leftrightarrow a^3 \\plus{} b^3 \\geq a^2b \\plus{} b^2a$\nIt follows from Muirhead, since [3,0] majorises [2,1][/hide][/quote]\n\nJust using Jensen's inequality kills in 10 seconds.:D[/quote]\r\nOr using Holder's inequality kills it in 5 second :wink:", "Solution_11": "[quote=\"Erken\"][quote=\"kunny\"][quote=\"isomorphism\"][hide=\"For problem 10, \"]$ 4(a^3 \\plus{} b^3) \\geq (a \\plus{} b)^3 \\Leftrightarrow a^3 \\plus{} b^3 \\geq a^2b \\plus{} b^2a$\nIt follows from Muirhead, since [3,0] majorises [2,1][/hide][/quote]\n\nJust using Jensen's inequality kills in 10 seconds.:D[/quote]\nOr using Holder's inequality kills it in 5 second :wink:[/quote]\r\n\r\nRegrettably Holder's inequality beyond my ability. :lol:", "Solution_12": "13. \r\n\r\n$ A\\equal{}(\\frac{a\\plus{}b}{c})^{m}\\plus{}2(\\frac{b\\plus{}c}{a})^{m/2}\\plus{}3(\\frac{a\\plus{}c}{b})^{m/3}\\equal{}(\\frac{a\\plus{}b}{c})^{m}\\plus{}(\\frac{b\\plus{}c}{a})^{m/2}\\plus{}(\\frac{b\\plus{}c}{a})^{m/2}\\plus{}(\\frac{a\\plus{}c}{b})^{m/3}\\plus{}(\\frac{a\\plus{}c}{b})^{m/3}\\plus{}(\\frac{a\\plus{}c}{b})^{m/3}\\geq6\\sqrt[6]{(\\frac{(a\\plus{}b)(a\\plus{}c)(b\\plus{}c)}{abc})^{m}}\\geq6\\sqrt[6]{(\\frac{(2\\sqrt{ab})(2\\sqrt{ac})(2\\sqrt{bc})}{abc})^{m}}\\equal{}6\\sqrt[6]{(2^{3})^{m}}\\equal{}6\\sqrt{2^{m}}$\r\n\r\n$ a\\equal{}b\\equal{}c$ --->$ A\\equal{}2^{m}\\plus{}2*2^{m/2}\\plus{}3*2^{m/3}>6*2^{m}$ ----> $ A>6\\sqrt{2^{m}}$", "Solution_13": "12. Let $ 2[ABC]\\equal{}ah_a \\equal{} bh_b \\equal{} ch_c \\equal{} abc\\cdot k\\ (k > 0)\\Longrightarrow h_a \\equal{} bck,\\ h_b \\equal{} cak,\\ h_c \\equal{} abk.$ \r\n\r\n$ \\therefore h_c > \\frac {h_ah_b}{h_a \\plus{} h_b}\\Longleftrightarrow a \\plus{} b > c.$", "Solution_14": "5.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=165510", "Solution_15": "3. $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a},z \\equal{} \\frac {2c}{a \\plus{} b}$\r\nIt's remain to prove\r\n\\[ \\sum_{cyc}\\frac {a}{b \\plus{} c} \\geq \\sum_{cyc}\\frac {2ab}{(a \\plus{} c)(b \\plus{} c)}\r\n\\]\r\n\r\n\\[ \\leftrightarrow \\sum_{cyc}\\frac {a}{b \\plus{} c} \\minus{} \\sum_{cyc}\\frac {2ab}{(a \\plus{} c)(b \\plus{} c)} \\geq 0\r\n\\]\r\n\r\n\\[ \\leftrightarrow \\frac {\\sum_{cyc}a^3 \\plus{} 3abc \\minus{} \\sum_{sym}a^2b}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq 0\r\n\\]\r\nbut it's obviously true by schur.", "Solution_16": "[quote=\"apollo\"]4. $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a},z \\equal{} \\frac {2c}{a \\plus{} b}$\n [/quote]\r\nI think this is proof for 3rd problem,not 4th. :)", "Solution_17": "[quote=\"Erken\"][quote=\"apollo\"]4. $ x \\equal{} \\frac {2a}{b \\plus{} c},y \\equal{} \\frac {2b}{c \\plus{} a},z \\equal{} \\frac {2c}{a \\plus{} b}$\n [/quote]\nI think this is proof for 3rd problem,not 4th. :)[/quote]\r\nSorry,I've edited it. :blush:", "Solution_18": "14. It's equivalent to prove: $ 3(x^3\\plus{}y^3\\plus{}z^3) \\geq (x^2\\plus{}y^2\\plus{}z^2)(x\\plus{}y\\plus{}z)$\r\n$ \\leftrightarrow 2(x^3\\plus{}y^3\\plus{}z^3) \\geq \\sum_{sym}x^2y$\r\nbut it's obviously true by Am-Gm.", "Solution_19": "18. It's obviously true by schur inequality;\r\n$ x^3\\plus{}y^3\\plus{}z^3\\plus{}3xyz \\geq \\sum_{cyc}x^2(y\\plus{}z)$\r\n$ \\rightarrow x^6\\plus{}y^6\\plus{}z^6\\plus{}3x^2y^2z^2 \\geq \\sum_{cyc}x^4(y^2\\plus{}z^2)$", "Solution_20": "11.\r\nI think it should be:\r\nLet $ x,y,z,t \\in R , x\\plus{}y\\plus{}z\\plus{}t\\equal{}0 , x^2\\plus{}y^2\\plus{}z^2\\plus{}t^2\\equal{}1$\r\nProve that $ \\minus{}1 \\leq xy\\plus{}yz\\plus{}zt\\plus{}tx \\leq 0$ :D", "Solution_21": "[hide=\"20\"]\nnote that $ 3 \\minus{} 2a \\equal{} b \\plus{} c \\minus{} a$ and so on,so we have:\n\n$ LHS \\equal{} (b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)$\n\nnow note that we have $ a^2 \\plus{} b^2 \\plus{} c^2 > 0$ and $ RHS \\equal{} 3a^2b^2c^2 > 0$ so if $ (b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c) < 0$ then the inequality becomes trivial because $ LHS < 0 < RHS$ so assume that $ (b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c) > 0$,so we have two possibilities:\n\n1.\ntwo of the numbers $ b \\plus{} c \\minus{} a,c \\plus{} a \\minus{} b,a \\plus{} b \\minus{} c$ were negative,for example $ b \\plus{} c \\minus{} a < 0$ and $ c \\plus{} a \\minus{} b < 0$...\nadding these two inequalities we get that $ c < 0$ which is a contradiction.\nso this case never happens...\n\n2.\nall of the numbers $ b \\plus{} c \\minus{} a,c \\plus{} a \\minus{} b,a \\plus{} b \\minus{} c$ are positive i.e. $ a \\plus{} b > c , b \\plus{} c > a$ and $ c \\plus{} a > b$ which means there exists a triangle $ \\triangle ABC$ with side lengths $ a,b,c$...\nnow let $ p,r,R,S$ be the semiperimeter,inradius,circumradius and area of this triangle we have:\n\n$ 3 \\minus{} 2a \\equal{} b \\plus{} c \\minus{} a \\equal{} 2(p \\minus{} a)$ so we have:\n\n$ LHS \\equal{} 8(p \\minus{} a)(p \\minus{} b)(p \\minus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)$\n\nnow from a well-known theorem we have:\n\n$ S \\equal{} \\sqrt {p(p \\minus{} a)(p \\minus{} b)(p \\minus{} c)} \\equal{} \\frac {abc}{4R} \\equal{} pr$\n\nso we'll have:\n\n$ (p \\minus{} a)(p \\minus{} b)(p \\minus{} c) \\equal{} pr^2$\n\nand\n\n$ abc \\equal{} 4Rrp$\n\nso our inequality reduses to show that:\n\n$ 8pr^2(a^2 \\plus{} b^2 \\plus{} c^2)\\leq 48R^2r^2p^2$\n\n$ \\iff a^2 \\plus{} b^2 \\plus{} c^2\\leq 6R^2p$\n\nnow note that $ p \\equal{} \\frac {a \\plus{} b \\plus{} c}2 \\equal{} \\frac 32$ so we have to prove that:\n\n$ a^2 \\plus{} b^2 \\plus{} c^2\\leq 9R^2$\n\nnow by $ \\sin$ law in $ \\triangle ABC$ we have:\n\n$ \\frac {a}{\\sin A} \\equal{} \\frac {b}{\\sin B} \\equal{} \\frac {c}{\\sin C} \\equal{} 2R$\n\nso we have:\n\n$ a \\equal{} 2R\\sin A , b \\equal{} 2R\\sin B , c \\equal{} 2R\\sin C$\n\nso our inequality is equivalent to:\n\n$ 4R^2(\\sin^2 A \\plus{} \\sin^2 B \\plus{} \\sin^2 C)\\leq 9R^2$\n\n$ \\iff \\sin^2 A \\plus{} \\sin^2 B \\plus{} \\sin^2 C\\leq \\frac 94$\n\nwhich is a well-known inequality...[/hide]", "Solution_22": "8)\r\nSeen it hundreds of times. \r\nGiven the inequality between the arithmetical and harmonical mean of numbers (a+b),(b+c),(c+a) we end in a result:\r\n$ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\plus{}3\\equal{}(a\\plus{}b\\plus{}c)(\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\plus{}\\frac{1}{a\\plus{}b})\\equal{}\r\n\\frac{1}{2}((a\\plus{}b)\\plus{}(b\\plus{}c)\\plus{}(c\\plus{}a))(\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\plus{}\\frac{1}{a\\plus{}b})\\geq \\frac{9}{2}$\r\nWhich is equivalent to our first inequality\r\nSecond part:\r\nWe know, that $ a f(a) \\leq f(b)\\equal{}\\frac{1}{2b\\plus{}k}\\plus{}\\frac{1}{b\\plus{}c\\plus{}k}\\plus{}\\frac{1}{b\\plus{}c\\plus{}k}$\r\n\r\nThen $ f'(b)\\equal{}\\frac{\\minus{}1}{(2b\\plus{}k)^2}\\plus{}\\frac{\\minus{}1}{(b\\plus{}c\\plus{}k)^2} \\leq 0$\r\n\r\n$ \\minus{}>f(b)\\leq f(c)\\equal{} \\frac{1}{2c\\plus{}k}\\plus{}\\frac{1}{2c\\plus{}k}\\equal{}\\frac{2}{2c\\plus{}k}$\r\n\r\nWe see, $ f(c)$ is decrease funcion by $ c$ -> $ c\\minus{}>0$ then $ f(c)$ is max!", "Solution_2": "[quote=\"vipCD_A1\"][quote=\"STARS\"]Let $ a, b, c$ be positive real numbers such that $ abc \\equal{} 1$, k is given positive real number. Find the maximum value of\n\\[ P \\equal{} \\frac {1}{a \\plus{} b \\plus{} k} \\plus{} \\frac {1}{b \\plus{} c \\plus{} k} \\plus{} \\frac {1}{c \\plus{} a \\plus{} k}\n\\]\n[/quote]\n\nAssume $ a \\geq b \\geq c$\n\n$ P \\equal{} f(a)$. We have:\n\n$ f'(a) \\equal{} \\frac { \\minus{} 1}{(a \\plus{} b \\plus{} k)^2} \\plus{} \\frac { \\minus{} 1}{(a \\plus{} c \\plus{} k)^2} \\leq 0$\n\n$ \\minus{} > f(a) \\leq f(b) \\equal{} \\frac {1}{2b \\plus{} k} \\plus{} \\frac {1}{b \\plus{} c \\plus{} k} \\plus{} \\frac {1}{b \\plus{} c \\plus{} k}$\n\nThen $ f'(b) \\equal{} \\frac { \\minus{} 1}{(2b \\plus{} k)^2} \\plus{} \\frac { \\minus{} 1}{(b \\plus{} c \\plus{} k)^2} \\leq 0$\n\n$ \\minus{} > f(b)\\leq f(c) \\equal{} \\frac {1}{2c \\plus{} k} \\plus{} \\frac {1}{2c \\plus{} k} \\equal{} \\frac {2}{2c \\plus{} k}$\n\nWe see, $ f(c)$ is decrease funcion by $ c$ -> $ c \\minus{} > 0$ then $ f(c)$ is max![/quote]\r\n\r\nHave you got any simpler solution?", "Solution_3": "[quote=\"vipCD_A1\"]$ P \\equal{} f(a)$. We have:\n\n$ f'(a) \\equal{} \\frac { \\minus{} 1}{(a \\plus{} b \\plus{} k)^2} \\plus{} \\frac { \\minus{} 1}{(a \\plus{} c \\plus{} k)^2} \\leq 0$\n[/quote]\r\nWrong. You forgot the condition $ abc \\equal{} 1$. You can't calculate $ f'(a)$ like that in the first place because $ b,c$ are dependent on $ a$.", "Solution_4": "No one can prove it?\r\n\r\nI am waiting the nice proof.", "Solution_5": "[quote=\"STARS\"]Let $ a, b, c$ be positive real numbers such that $ abc \\equal{} 1$, k is given positive real number. Find the maximum value of\n\\[ P \\equal{} \\frac {1}{a \\plus{} b \\plus{} k} \\plus{} \\frac {1}{b \\plus{} c \\plus{} k} \\plus{} \\frac {1}{c \\plus{} a \\plus{} k}\n\\]\n[/quote]\r\n\r\nMy proof:\r\n\r\n[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/1234.jpg[/img]" } { "Tag": [ "inequalities", "linear algebra", "matrix", "group theory", "abstract algebra", "analytic geometry", "geometry" ], "Problem": "We say a triple of real numbers $ (a_1,a_2,a_3)$ is [b]better[/b] than another triple $ (b_1,b_2,b_3)$ when exactly two out of the three following inequalities hold: $ a_1 > b_1$, $ a_2 > b_2$, $ a_3 > b_3$. We call a triple of real numbers [b]special[/b] when they are nonnegative and their sum is $ 1$.\r\n\r\nFor which natural numbers $ n$ does there exist a collection $ S$ of special triples, with $ |S| \\equal{} n$, such that any special triple is bettered by at least one element of $ S$?", "Solution_1": "[quote=\"flamingspinach\"]We say a triple of real numbers $ (a_1,a_2,a_3)$ is [b]better[/b] than another triple $ (b_1,b_2,b_3)$ when exactly two out of the three following inequalities hold: $ a_1 > b_1$, $ a_2 > b_2$, $ a_3 > b_3$. We call a triple of real numbers [b]special[/b] when they are nonnegative and their sum is $ 1$.\n\nFor which natural numbers $ n$ does there exist a collection $ S$ of size $ n$ such that any special triple is bettered by at least one element of $ S$?[/quote]\r\n\r\nI hope the collection $ S$ is supposed to consist of special triples (else, just take $ n \\equal{} 1$ and $ S \\equal{} \\left\\{\\left(2,2,0\\right)\\right\\}$).\r\n\r\n[hide=\"Wrong stuff\"]In this case, the minimal possible $ n$ is $ n \\equal{} 3$ with $ S \\equal{} \\left\\{\\left(0,1,1\\right),\\left(1,0,1\\right),\\left(0,1,1\\right)\\right\\}$. Of course, any $ n\\geq 3$ is therefore also possible. The impossibility of $ n\\leq 2$ is easy to see (if $ n \\equal{} 2$ and $ S \\equal{} \\left\\{\\left(a,b,c\\right),\\left(d,e,f\\right)\\right\\}$, then $ \\left(a,b,c\\right)$ must be better than $ \\left(d,e,f\\right)$ and $ \\left(d,e,f\\right)$ must be better than $ \\left(a,b,c\\right)$ (because no triple is better than itself), what is absurd).[/hide]\r\n\r\nStrangely enough, we have got pretty easy problems on places #4-6 this year - at least, this one, the one with the primes matrix and the one with the permutation distances. Then again, the subgroup problem seems to counterpoise this.\r\n\r\n darij", "Solution_2": "If $ S$ is supposed to consist of special triples, how can you include $ (1,0,1)$?", "Solution_3": "[quote=\"darij grinberg\"]I hope the collection $ S$ is supposed to consist of special triples[/quote]\n\nAbsolutely right, sorry. Fixed.\n\n[quote=\"darij grinberg\"]the minimal possible $ n$ is $ n \\equal{} 3$ with $ S \\equal{} \\left\\{\\left(0,1,1\\right),\\left(1,0,1\\right),\\left(0,1,1\\right)\\right\\}$.[/quote]\r\n\r\nThose triples are not special, as kevinatcausa pointed out. If you meant to scale \"specialness\" to a sum of $ 2$, that still doesn't work, as each triple in your $ S$ is itself a counterexample to the statement (though your $ S$ works for all special triples outside $ S$).", "Solution_4": "It's best to view the triples as barycentric coordinates and then use a geometry program to visualise the three parallelogramms defined by each point. It is not hard to see that you can cover the triangle with three of them, for example the midpoints of the sides, but only with weak inequalities, so you have to add a fourth point." } { "Tag": [], "Problem": "There are 6 sentences\r\nfind 1 right sentence \r\n1: only 1 false sentence\r\n2: only 2 false sentences\r\n3: only 3 false sentences\r\n4: only 4 false sentences\r\n5: only 5 false sentences\r\n6: only 6 false sentences", "Solution_1": "[hide]#5\n\nsince only one may be true( no two can both be true), all the others must be false.\n6 statements total-1 true statement=5 false statements,\nwhich is what #5 states.[/hide]", "Solution_2": "[quote=\"acdef\"]There are 6 sentences\nfind 1 right sentence \n1: only 1 false sentence\n2: only 2 false sentences\n3: only 3 false sentences\n4: only 4 false sentences\n5: only 5 false sentences\n6: only 6 false sentences[/quote]\r\n[hide=\"answer\"]since sentences 1~4 contradict what they say, and sentence number 6 says all false sentences yet it's supposed to be tru, the answer would be number 5.[/hide]" } { "Tag": [ "LaTeX", "inequalities unsolved", "inequalities" ], "Problem": "Let x,y,z be positive integer such that $x^{2}$ + $y^{2}$ + $z^{2}$ = 3. Prove that\r\nx+y+z +$\\frac{3}{2}$($\\frac{1}{x}$+$\\frac{1}{y}$+$\\frac{1}{z}$)\\leq$\\frac{15}{2}$", "Solution_1": "[quote=\"Thao\"]Let x,y,z be positive integer such that $x^{2}$ + $y^{2}$ + $z^{2}$ = 3. Prove that\r\nx+y+z +$\\frac{3}{2}$($\\frac{1}{x}$+$\\frac{1}{y}$+$\\frac{1}{z}$)\\leq$\\frac{15}{2}$\r\nIt's wrong. Try $x\\rightarrow0^{+}.$ :wink:", "Solution_2": "x+y+z+3/2*(1/x+1/y+1/z)>=15/2 (x,y,z>0 ,x^2+y^2+z^2=3) (*)\r\n\r\n<==>\r\n \r\n((x+y+z+3/2*(1/x+1/y+1/z)*(x^2+y^2+z^2)/3)^2-(15/2)^2*(x^2+y^2+z^2)/3)>=0\r\n\r\n<==>\r\n\r\nF[8][0,3]+4*F[8][1,1]+10*F[8][1,2]+4*F[8][0,5]+20*F[8][1,3]+2*F[8][2,1]>=0,\r\n\r\nin which\r\n\r\nF[8][0,3]=(x+y+z)^2*(x-y)^2*(y-z)^2*(z-x)^2>=0;\r\nF[8][1,1]=x*y*z*sum(x^3*(x-y)*(x-z))>=0;\r\nF[8][1,2]=x*y*z*sum(x^2*(y+z)*(x-y)*(x-z))>=0;\r\nF[8][0,5]=sum(y^3*z^3*(x-y)*(x-z))>=0;\r\nF[8][1,3]=x*y*z*sum(y*z*(y+z)*(x-y)*(x-z))>=0;\r\nF[8][2,1]=(x*y*z)^2*sum((x-y)*(x-z))>=0.", "Solution_3": "Just use\r\n\\[x+\\frac{3}{2x}\\geq \\frac{5}{2}+\\frac{1}{4}(x^{2}-1)\\]\r\nwhich reduces to $(x+6)(x-1)^{2}\\geq 0$.", "Solution_4": "$x+y+z+\\frac{1}{2}\\cdot\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)\\geq\\frac{9}{2}$ with same conditions true too.", "Solution_5": "Let x,y,z>0,x^2+y^2+z^2=1,k>=7/10, we have\r\n\r\n x+y+z+k*(1/x+1/y+1/z)>=3+3*k,\r\n\r\nit is equivalent to\r\n \r\n k^2*F[8][0,3]+4*k^2*F[8][1,1]+(6*k+6*k^2)*F[8][1,2]+4*k^2*F[8][0,5]+(12*k+12*k^2)*F[8][1,3]\r\n\r\n +(-18+10*k^2)*F[8][2,1]>=0.", "Solution_6": "[quote=\"fjwxcsl\"]Let x,y,z>0,x^2+y^2+z^2=1,k>=7/10, we have\n\n x+y+z+k*(1/x+1/y+1/z)>=3+3*k,\n\nit is equivalent to\n \n k^2*F[8][0,3]+4*k^2*F[8][1,1]+(6*k+6*k^2)*F[8][1,2]+4*k^2*F[8][0,5]+(12*k+12*k^2)*F[8][1,3]\n\n +(-18+10*k^2)*F[8][2,1]>=0.[/quote]\r\nwhy don't you learn about latex? It's very useful for you to write maths" } { "Tag": [ "function", "calculus", "derivative", "integration", "complex analysis", "complex analysis unsolved" ], "Problem": "Thanks all for considering a lot of my questions. \r\n\r\nNow my question is: Consider the punctured disk $ D\\equal{}\\{ z \\in \\mathbb C | 0 <|z| <1\\}$. Suppose $ f : D \\to \\mathbb C$ is an analytic function such that \r\n\\[ |f''(z)| \\leq \\frac{2}{|z|^2}\\]\r\nfor all $ z \\in D$. Is it true that $ f$ has a removable singular point at $ z\\equal{}0$? \r\n\r\nI believe (but I do not know why) that the second derivative can be replaced by an $ \\mathbb Z\\ni m \\geq 1$ and therefore $ \\frac{2}{|z|^2}$ should be $ \\frac{2}{|z|^m}$. Thank you.", "Solution_1": "Write down the Laurent expansion of f, differentiate it twice...", "Solution_2": "Oh, it seems that $ f(z) \\equiv 0$ satisfies all assumptions. So $ f$ has no removable singularity point at $ z\\equal{}0$, right?", "Solution_3": "[quote=\"Kalle\"]Write down the Laurent expansion of f, differentiate it twice...[/quote]\r\n\r\nAssume \r\n\\[ f\\left( z \\right) \\equal{} \\sum\\limits_{n \\equal{} \\minus{} \\infty }^{ \\plus{} \\infty } {{a_n}{z^n}}\\]\r\nwhere \r\n\\[ {a_n} \\equal{} \\frac{1}\r\n{{2\\pi i}}\\int\\limits_{\\left| z \\right| \\equal{} r < 1} {\\frac{{f\\left( z \\right)dz}}\r\n{{{z^{n \\plus{} 1}}}}} .\\]\r\nThen from\r\n\\[ f''\\left( z \\right) \\equal{} \\sum\\limits_{n \\equal{} \\minus{} \\infty }^{ \\plus{} \\infty } {\\underbrace {\\left( {n \\plus{} 2} \\right)\\left( {n \\plus{} 1} \\right){a_{n \\plus{} 2}}}_{{b_n}}{z^n}}\\]\r\nwe get\r\n\\[ {b_n} \\equal{} \\frac{1}\r\n{{2\\pi i}}\\int\\limits_{\\left| z \\right| \\equal{} r < 1} {\\frac{{f''\\left( z \\right)dz}}\r\n{{{z^{n \\plus{} 1}}}}} .\\]\r\nThus,\r\n\\[ \\left| {{b_n}} \\right| \\leqslant \\frac{1}\r\n{{2\\pi }}\\int\\limits_{\\left| z \\right| \\equal{} r < 1} {\\left| {\\frac{{f''\\left( z \\right)}}\r\n{{{z^{n \\plus{} 1}}}}} \\right|} dz \\leqslant \\frac{1}\r\n{\\pi }\\int\\limits_{\\left| z \\right| \\equal{} r < 1} {\\left| {\\frac{1}\r\n{{{z^{n \\plus{} 3}}}}} \\right|} dz \\leqslant \\frac{1}\r\n{{\\pi {r^{n \\plus{} 3}}}}2\\pi r \\equal{} \\frac{2}\r\n{{{r^{n \\plus{} 2}}}}.\\]\r\nWhen $ n \\leq \\minus{}3$, let $ r \\to 0$ we see that $ z\\equal{}0$ is a removable singularity of $ f$ since $ b_n\\equal{}0$ ($ n \\leq \\minus{}3$) implies $ a_n\\equal{}0$ ($ n \\leq \\minus{}1$)." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Point $O$ is the circumcenter of the triangle $ABC$ . Denote $D$ the intersection of $AO$ with the segment $(BC)$. If $OD=BD=\\frac13BC$ detemine the angles of $ABC$.", "Solution_1": "Connect O and the midpoint M of CD. Observe that ODM is a equilateral. And $\\angle{OBC}=\\angle{OCB}=30^\\circ$, the result follows..." } { "Tag": [ "geometry open", "geometry" ], "Problem": "In my zeal to solve this problem: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36502]www.mathlinks.ro/Forum/viewtopic.php?t=36502[/url], I misread it (not an untypical move for myself), and I thought it said to prove that AK, BM, and CN are concurrent at a point P.\r\n\r\nThis seems to be true; furthermore, the point of concurrency also appears to lie on IO!\r\n\r\nCan anyone prove this?", "Solution_1": "I found a proof!\r\n\r\nIn case anyone is interested, it appears in \"The Intouch Triangle and the OI-line\" by Eric Danneels, [url=http://forumgeom.fau.edu/FG2004volume4/FG200416.pdf]forumgeom.fau.edu/FG2004volume4/FG200416.pdf[/url], p. 128.", "Solution_2": "We know $KD,ME,NF$ are bisectors of $\\angle BKC,\\angle CMA,\\angle ANB$. \r\nNow, without signs: $BK/KC=BD/DC, CM/MA=CE/EA , AN/NB=AF/FB$, so:\r\n$(BK/KC)(CM/MA)(AN/NB)=(BD/DC)(CE/EA)(AF/FB)=1$\r\nDue to $AD,BE,CF$ are concurrent in $ABC$ triangle. Hence $AK,BM,CN$ are concurrent by Ceva circular form. But I have not the synthetic proof for this point of concurrency is on $IO$." } { "Tag": [ "function", "induction", "Support", "number theory theorems", "number theory" ], "Problem": "While trying to solve a textbook problem (Apostol, chap.3, prob.11(b)) using an unexpected method (multiplicative induction), I felt that the following identity must hold:\r\n\r\n$\\sigma(mp)-\\sigma(m/p)=\\sigma(m)p$\r\n\r\nwhere $p$ is a prime, $p|m$ and $\\sigma(n)$ is the sum of the divisors of $n$. The following trials support this conjecture:\r\n\r\nLet $m=6$, $p=2$. Then $\\sigma(12)-\\sigma(3)=\\sigma(6).2$ or $1+2+3+4+6+12-1-3=(1+2+3+6).2$ or $24=12.2$.\r\n\r\nLet $m=6$, $p=3$. Then $\\sigma(18)-\\sigma(2)=\\sigma(6).3$ or $3+6+9+18=(1+2+3+6).3$ or $36=12.3$.\r\n\r\nCan anyone prove (or disprove) this identity?\r\n\r\nSincerely Yours,\r\nMurat Aygen", "Solution_1": "Hello Murat abi how are you kostakoz@hotmail.com for you...\r\nAdd me please do you remember me in the IMO 2005 Mexico, Merida, Yucatan \r\nsonra konusacagis inshallah... :) \r\n\r\nDavron", "Solution_2": "Hello Davron. Nice to meet you. But I afraid you are confusing me with some other Murat. I have never been a member of an Olympiad team. If you want to correspond with me, my account is aygenmurat2001@yahoo.com\r\n\r\nBest regards,\r\nMurat Aygen", "Solution_3": "The conjecture is nice, though infinitely often wrong :(\r\n\r\nFor instance, take $m = p^2$ (where $p$ prime as you mentioned). Then\r\n\\begin{eqnarray*} \\sigma ( mp ) - \\sigma ( m / p ) & = & p \\sigma ( m )\\\\ \\sigma ( p^3 ) - \\sigma ( p ) & = & p \\sigma ( p^2 )\\\\ 1 + p + p^2 + p^3 - p - 1 & = & p \\cdot ( 1 + p + p^2 )\\\\ p^2 + p^3 & = & p + p^2 + p^3\\\\ 0 & = & p > 0 \\end{eqnarray*}\r\nWhich is clearly a contradiction", "Solution_4": "Sorry. I soon discovered that the proof follows easily if one replaces the hypothesis $p|m$ with the stronger $p||m$. Then by the \"multiplicative\" property of the $\\sigma$ function, one has:\r\n\r\n$\\sigma(mp)-\\sigma(m/p)=\\sigma((m/p)p^{2})-\\sigma(m/p)= \\sigma(m/p)\\sigma(p^{2})-\\sigma(m/p)=$\r\n$\\sigma(m/p)(\\sigma(p^{2})-1)= \\sigma(m/p)(p+1)p=\\sigma(m/p)\\sigma(p)p=\\sigma(m)p$\r\n\r\nThank you very much,\r\nMurat", "Solution_5": "Yes, exactly, the problem consisted in the squares of $p$ that could occur in $n$.\r\nAnyway now it's solved :)" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let a convex hexagon ABCDEF. M,N,P,Q,R,S are the midpoints of AB,BC,CD,DE,EF,FA respectively. Prove that MQ,NR,PS are concurrent if and only if triangles ACE and BDF have the same area.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17902[/url]" } { "Tag": [ "inequalities", "calculus", "vector", "Functional Analysis", "inequalities unsolved" ], "Problem": "Let a,b,c be sides of a triangle.\r\nThe following equation is held: $ a\\plus{}b\\plus{}c\\equal{}3$\r\nFin the maximum value of:\r\n$ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}\\frac{4abc}{3}$ :lol: :wink:", "Solution_1": "Are you familiar with multivariable calculus? This looks like an excellent problem for which to use Lagrange Multipliers (maybe that's not the solution you're hoping for but that is easier for me to do).", "Solution_2": "Well unfortunately a \"normal\" functional analysis based on Lagrange's multipliers does not work. It gives us (from what I remember) sth like $ a\\equal{}b\\equal{}\\frac{1}{4} \\\\ c\\equal{}2,5$, but it does not satisfy the condition of triangle inequality. :wink:", "Solution_3": "no you are not using lagrange multipliers correctly then\r\n\r\nyou have to check boundaries too\r\n\r\nyou can also say that $ a\\equal{}y\\plus{}z$, etc and do it that way where $ x,y,z$ are non-negative reals", "Solution_4": "[quote=\"polskimisiek\"]Let a,b,c be sides of a triangle.\nThe following equation is held: $ a \\plus{} b \\plus{} c \\equal{} 3$\nFin the maximum value of:\n$ a^{2} \\plus{} b^{2} \\plus{} c^{2} \\plus{} \\frac {4abc}{3}$ :lol: :wink:[/quote]\r\n\r\n\r\n$ a\\equal{}z\\plus{}y$ , $ b\\equal{}x\\plus{}z$ , $ c\\equal{}y\\plus{}x$ ------------------->$ x\\plus{}y\\plus{}z\\equal{}\\frac{3}{2}$\r\n$ A\\equal{}a^{2} \\plus{} b^{2} \\plus{} c^{2} \\plus{} \\frac {4abc}{3}\\equal{}(z\\plus{}y)^{2}\\plus{}(x\\plus{}z)^{2}\\plus{}(x\\plus{}y)^{2}\\plus{}\\frac{4(x\\plus{}y)(z\\plus{}y)(x\\plus{}z)}{3}\\equal{}(\\frac{3}{2}\\minus{}x)^{2}\\plus{}(\\frac{3}{2}\\minus{}y)^{2}\\plus{}(\\frac{3}{2}\\minus{}z)^{2}\\plus{}\\frac{4(\\frac{3}{2}\\minus{}x)(\\frac{3}{2}\\minus{}y)(\\frac{3}{2}\\minus{}z)}{3}\\equal{}$\r\n\r\n$ \\frac{9}{4}\\minus{}3x\\plus{}x^{2}\\plus{}\\frac{9}{4}\\minus{}3y\\plus{}y^{2}\\plus{}\\frac{9}{4}\\minus{}3z\\plus{}z^{2}\\plus{}\\frac{4}{3}((\\frac{3}{2})^{3}\\minus{}(\\frac{3}{2})^{2}(x\\plus{}y\\plus{}z)\\plus{}\\frac{3}{2}(xy\\plus{}yz\\plus{}zx)\\minus{}xyz)\\equal{}\\frac{9}{2}\\minus{}\\frac{4}{3}xyz\\leq\\frac{9}{2}$ ----> \r\n$ x\\equal{}o$ or $ y\\equal{}0$ or $ z\\equal{}0$ -----> there isn't max ---> $ A<\\frac{9}{2}$", "Solution_5": "Why are you substituting for $ a$, $ b$, and $ c$? Why would it not suffice to simply treat those as the variables and then use Lagrange multipliers?", "Solution_6": "[color=green]Because a,b,c are three sides of a triangle .:D[/color]", "Solution_7": "Ah, then $ x$, $ y$, and $ z$ are their vector components.", "Solution_8": "We have:\r\n$ S\\equal{}\\sqrt{p(p\\minus{}a)(p\\minus{}b)(p\\minus{}c)}\\equal{}\\sqrt{\\frac{3}{2}.(\\frac{3}{2}\\minus{}a)(\\frac{3}{2}\\minus{}b)(\\frac{3}{2}\\minus{}c)}$\r\nSo $ 16S^2\\equal{}3(3\\minus{}a)(3\\minus{}b)(3\\minus{}c)\\equal{}81\\minus{}54(a\\plus{}b\\plus{}c)\\plus{}36(ab\\plus{}bc\\plus{}ca)\\minus{}24abc$\r\n$ \\equal{}\\minus{}81\\plus{}36(ab\\plus{}bc\\plus{}ca)\\minus{}24abc$\r\nWe have $ (a\\plus{}b\\plus{}c)^2\\equal{}a^2\\plus{}b^2\\plus{}c^2\\plus{}2(ab\\plus{}bc\\plus{}ca)$\r\nso $ 36(ab\\plus{}bc\\plus{}ca)\\equal{}162\\minus{}18(a^2\\plus{}b^2\\plus{}c^2)$\r\nthen $ 16S^2\\equal{}81\\minus{}18(a^2\\plus{}b^2\\plus{}c^2)\\minus{}24abc>0$\r\nSo $ a^2\\plus{}b^2\\plus{}c^2\\plus{}\\frac{4}{3}abc <\\frac{9}{2}$", "Solution_9": "It's here, with many solutions :P \r\nhttp://www.mathlinks.ro/viewtopic.php?t=147090" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Who can tell me whether the following expression can be simplified or not?\r\n\r\nNote that H is an m x n matrix, G is a k x n matrix, H^T and G^T are respectively the transposes of H and G, I is the n x n identity matrix and O is the n x n zero matrix.", "Solution_1": "I suppose you can provide some additional information due to the usage of matrix inverses (which not necessarily exist)", "Solution_2": "The matrix inverses are assumed to exist. \r\n\r\nI think the conditions for the existence of the matrix inverses, e.g., H and G are of full row rank, m0$. Whenever the line hits a side of the triangle, it will be \"reflected\" and keeps going until it hits another side and becomes reflected again and so on until it reaches a vertice of the rectangle.\r\n\r\n1. Find a formula in terms of $m$ and $n$ for the total number of reflections the line goes through until it hits a vertex.\r\n2. Which vertices can the line eventually reach?\r\n3. For what values of $m$ and $n$ will the line end up at $A$? $B$? $C$? $D$?", "Solution_1": "[b]Examples[/b]:\r\n-For $(m,n)=(1,1)$, the line will end up at $(1,1)$, with no reflections.\r\n-For $(m,n)=(3,2)$, the line will first hit $(2,2)$, bounces off and then hits $(3,1)$, bounces off and then hits $(2,0)$, bounces off and ends up at the vertex $(0,2)$. So there will be $3$ reflections, and the line will end up at $D=(0,n)=(0,2)$.\r\n\r\nCan anyone solve the problem now? Also, I noticed some typos in my original statment of the problem, so I fixed them." } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "function", "algebra proposed" ], "Problem": "Let $\\{W_n\\}$ be a polynomial sequence such that $W_1(x)=1-x$, $W_2(x)=x^2-3x-7$ and\r\n$W_n(x)=(n-x)W_{n-1}(x)-(n+1)^2W_{n-2}(x)$ for $n \\geq 3$.\r\n\r\nProve that for all $n$ $W_n(x)$ has exactly $n$ real roots.", "Solution_1": "We use the Roll's theorem?", "Solution_2": "I think I have excellent proof, but I am not sure with my computations. I consider the symmetric matrix which is also tridiagonal having on the main diagonal $ 1-x,2-x,...,n-x$ and on the second-main diagonal (above and under the main diagonal) the numbers 2,3,...,n+1. Then it seems that $W_n$ is the characteristic polynomial of this matrix and since it is symmetric, it must have all eigenvalues real numbers. Thus conclusion.", "Solution_3": "[quote=\"Rafal\"]Let $\\{W_n\\}$ be a polynomial sequence such that $W_1(x)=1-x$, $W_2(x)=x^2-3x-7$ and\n$W_n(x)=(n-x)W_{n-1}(x)-(n+1)^2W_{n-2}(x)$ for $n \\geq 3$. Prove that for all $n$ $W_n(x)$ has exactly $n$ real roots.[/quote]\r\n Suppose that the reader known basic facts about [b]orthogonal polynomials[/b] (see for instance the monograph [1] and the Szego's bible [8]. Between other important properties, a polynomial sequence $ (P_n)_{n=0}^{\\infty},\\; P_0(x)=1, $ which is orthogonal verifies :\r\n[b] i)[/b] For each $ n\\ge 1$ all roots of $P_n(x) $ are real and distinct, located in the interior of the orthogonality interval ;\r\n[b]ii)[/b] The terms of $ (P_n)_{n=0}^{\\infty} $ satisfies a [b]three-term recurrence relation [/b] of the form\r\n\\[(1)\\; \r\n\\begin{array}{|c|}\r\n\\hline P_n(x)=(x-c_n)P_{n-1}(x)- \\lambda_n P_{n-2}(x)\\\\\r\n\\hline\r\n\\end{array}\\; ,\\; n\\in {\\mathbb N}^* ,\\; P_{-1}{x}:=0, P_0(x)=1,\r\n\\]\r\nwith $ c_n\\in {\\mathbb R} $ and $\\; \\; (*)\\; \\; \\; \\begin{array}{|c|} \\hline \\lambda_{n+1}>0 \\\\ \\hline\r\n\\end{array} $ (for $ n>0 $ ). \r\nFortunately there is known an inverse result, namely the so-called [b]Favard's Theorem--(1935,[3] )[/b] (see also [2],[4]--[7] and [9]).\r\n This theorem asserts that if $ (P_n)_{n=0}^{\\infty} $ is a polynomial sequence whose which satisfies (1) and moreover (*) holds, then $ (P_n)_{n=0}^{\\infty} $ is orthogonal with respect to a certain scalar product.\r\nBut all orthogonal polynomials have only real roots !\r\n Therefore:\r\n[b]Proposition.[/b][i] Suppose that $ (P_n)_{n=0}^{\\infty} $ is a polynomial sequence which satisfies a three-term recurrence relation like (1). If (*) is verified, then for each $ n , n\\ge 1, $ the polynomial function $ P_n(x) $ has all real roots.[/i]\r\n\r\n [b]Regarding your posted question:[/b]\r\n Observe that $ P_n(x):=W_n(-x) $ satisfies the three term recurrence \r\n\\[ P_n(x)=(x-c_n)P_{n-1}(x)- \\lambda_n P_{n-2}(x) , n\\in \\{2,3,...\\},\\]\r\nwhere $c_n:=-n $ and $\\lambda_{n+1}:=(n+2)^2>0 $ for $ n>0 .$ Also $ P_{-1}(x)=0, P_0(x)=1 , P_1(x)=x+1, P_2(x)=x^2+3x-7.$ \r\n\r\n[b]References:[/b]]\r\n[1] T.S.Chihara , [i] An Introduction to Orthogonal Polynomials [/i], Gordon and Breach Science Publishers , New York,1978. \r\n[2] T.S. Chihara , [i] The three term recurrence relation and spectral properties of orthogonal polynomials[/i], in \"Orthogonal Polynomials\" (edited by P.Nevai)Kluwer Academic Publishers,NATO ASI Series, 1990,pp.99-114.\r\n[3] J.Favard , [i] Sur les polynomes de Tchebicheff [/i] , C.R.Acad.Sci.Paris 200 (1935) 2052--2053.\r\n[4] O. Perron , [i]Die Lehre von der Kettenbruechen [/i], third edition, Leipzig, 1957.\r\n[5]J. Sherman , [i] On the numerators of the convergent of the Stieltjes continued fractions [/i], Trans.Amer.Math.Soc., 35(1933) 64--87.\r\n[6] T.J. Stieltejs , [i] Recherches sur les fractions continues [/i] , Annales de la Faculte des SCiences de Toulouse 8(1984) , J 1--122, 9(1985) A 1--47 , Oeuvres , vol. 2 , 398--566.\r\n[7] M.H. Stone , [i] Linear Transformations in Hilbert Spaces and Their Applications to Analysis [/i], Colloquium Publications no. 15, Amer.Math.Soc., New York 1932.\r\n [8] G. Szego , [i] Orthogonal Polynomials[/i], Colloqium Publications n0.23,Amer.Math.Soc.,Providence , fourth edition, 1975. \r\n[9] A. Winter , [i] Spektraltheorie der unendlichen Matrizen [/i] , second edition, Leipzig,1929.", "Solution_4": "I think that my solution works also to prove the theorem presented by Alexandru Lupas.", "Solution_5": "first prize for harazi :)... ( find this matrix ;) )\r\n\r\nI just wondered if there is an elementary proof, but there is none..." } { "Tag": [], "Problem": "[i]\u039b\u03bf\u03b9\u03c0\u03cc\u03bd , \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03b1\u03b9\u03b3\u03bd\u03af\u03b4\u03b9 - puzzle \u03c4\u03bf \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bd \u03c3\u03c4\u03b1 \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03c4\u03bf \u03ad\u03b4\u03c9\u03c3\u03b5 \u03ad\u03bd\u03b1\u03c2 \u03c3\u03c5\u03bd\u03ac\u03b4\u03b5\u03bb\u03c6\u03bf\u03c2 \u03c6\u03b9\u03bb\u03cc\u03bb\u03bf\u03b3\u03bf\u03c2:[/i]\r\n\r\n\u03a3\u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03b4\u03b9\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf\u03c0\u03bf\u03b8\u03b5\u03c4\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2 5+4+3+2+1 = 15 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2, \u03cc\u03c0\u03c9\u03c2 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9: \r\n\r\n 1\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae [color=red]|||||[/color]\r\n\r\n 2\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae [color=red]||||[/color]\r\n\r\n 3\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae [color=red]|||[/color]\r\n\r\n 4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae [color=red]||[/color]\r\n\r\n 5\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae [color=red]|[/color]\r\n\r\n\u03a4\u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1 \u03b1\u03c0\u03cc \u03b4\u03cd\u03bf \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac(\u03c0\u03b1\u03af\u03ba\u03c4\u03b5\u03c2) \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03b2\u03ae\u03bd\u03b5\u03b9 \u03b5\u03bd\u03b1\u03bb\u03bb\u03ac\u03be \u03cc\u03c3\u03b5\u03c2 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b5\u03c0\u03b9\u03b8\u03c5\u03bc\u03b5\u03af , \u03b1\u03c0\u03cc \u03cc\u03c0\u03bf\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae (1 \u03ad\u03c9\u03c2 5)\u03b8\u03ad\u03bb\u03b5\u03b9 .\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03cc\u03bc\u03c9\u03c2 \u03c0\u03c7 \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03c1\u03cc\u03bd\u03c9\u03c2 2 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd 1\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03ba\u03b1\u03b9 1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c4\u03c1\u03af\u03c4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae , \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 .\u0395\u03c0\u03af\u03c3\u03b7\u03bd \u03b1\u03bd \u03c0\u03c7 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c4\u03b7 \u03bc\u03b5\u03c3\u03b1\u03af\u03b1 , \u03bf \u03ac\u03bb\u03bb\u03bf\u03c2 \u03b4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03c1\u03cc\u03bd\u03c9\u03c2 \u03c4\u03b9\u03c2 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 4 .\u03a7\u03ac\u03bd\u03b5\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03b4\u03af \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03b1\u03bd\u03b1\u03b3\u03ba\u03b1\u03c3\u03c4\u03b5\u03af \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae.\r\n \u03a0\u03bf\u03b9\u03bf \u03b1\u03c0\u03cc \u03c4\u03b1 \u03b4\u03cd\u03bf \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03ad\u03c7\u03b5\u03b9 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bd\u03af\u03ba\u03b7\u03c2 , \u03b1\u03bd \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bd \u03b5\u03bd\u03b1\u03bb\u03bb\u03ac\u03be, \u03ba\u03b1\u03b9 \u03c0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ae;", "Solution_1": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1!\r\n\u03a4\u03bf \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c5\u03b6\u03b7\u03c4\u03ae\u03c3\u03b5\u03b9 \u03be\u03b1\u03bd\u03ac \u03c3\u03c4\u03bf \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112203]\u03b5\u03b4\u03ce[/url]\r\n\r\n\u039c\u03b5 \u03c4\u03b7 \u03bc\u03cc\u03bd\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9.\r\n\r\n\u038a\u03c3\u03c9\u03c2 \u03b1\u03c5\u03c4\u03ae \u03b7 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03ad\u03c1\u03b5\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03ba\u03b1\u03c4\u03ac \u03c0\u03bf\u03bb\u03cd \u03c4\u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae.\r\n\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03bd\u03b8\u03b1\u03c1\u03c1\u03c5\u03bd\u03c4\u03b9\u03ba\u03cc \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03cc\u03c2 \u03cc\u03c4\u03b9 \u03ad\u03bd\u03b1 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c0\u03b1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03bf :)", "Solution_2": "\u0386\u03bd\u03bf\u03b9\u03be\u03b1 \u03c4\u03bf mathematical circles \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b3\u03b5\u03bd\u03b9\u03ba\u03cc \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Nim \u03c4\u03bf \u03ad\u03c7\u03b5\u03b9 \u03bb\u03c5\u03bc\u03ad\u03bd\u03bf \u03bc\u03b5 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03b4\u03c5\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd \u03c3\u03c5\u03c3\u03c4\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 , \u03b1\u03bd \u03ba\u03b1\u03b9 \u03b4\u03b5 \u03bc\u03b5\u03bb\u03ad\u03c4\u03b7\u03c3\u03b1 \u03c3\u03b5 \u03b2\u03ac\u03b8\u03bf\u03c2 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 , \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03c7\u03b1 \u03bc\u03ac\u03b8\u03b7\u03bc\u03b1. \u0398\u03ad\u03bb\u03c9 \u03cc\u03bc\u03c9\u03c2 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03bf \u03c3\u03c5\u03bd\u03ac\u03b4\u03b5\u03bb\u03c6\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03bc\u03bf\u03c5 \u03c4\u03bf \u03ad\u03b4\u03b5\u03b9\u03be\u03b5 (\u03c4\u03bf\u03c5 \u03c4\u03bf \u03b5\u03af\u03c0\u03b1\u03bd \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c4\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03c0\u03ac\u03bd\u03b5 \u03c3\u03b5 \u03ac\u03bb\u03bb\u03bf \u03c3\u03c7\u03bf\u03bb\u03b5\u03af\u03bf) \u03ad\u03c0\u03b1\u03b9\u03b6\u03b5 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03bc\u03b5 \u03ba\u03ad\u03c1\u03b4\u03b9\u03b6\u03b5.\u03a0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b7\u03c3\u03b1 \u03bc\u03b9\u03b1 \u03c6\u03bf\u03c1\u03ac \u03bd\u03b1 \u03c0\u03b1\u03af\u03be\u03c9 \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 (\u03c3\u03c4\u03bf \u03b4\u03b9\u03b1\u03bb\u03b5\u03b9\u03bc\u03bc\u03b1 !) , \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce\u03bd\u03c4\u03b1\u03c2 \u03bd\u03b1 \u03c4\u03bf\u03c5 \u03b1\u03c6\u03ae\u03bd\u03c9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ce\u03bd , \u03b1\u03bb\u03bb\u03ac \u03c0\u03ac\u03bb\u03b9 \u03ad\u03c7\u03b1\u03c3\u03b1.\u0398\u03ad\u03bb\u03b7\u03c3\u03b5 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9 \u03c4\u03bf \u03ba\u03cc\u03bb\u03c0\u03bf , \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03c4\u03bf\u03bd \u03ac\u03c6\u03b7\u03c3\u03b1 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03bc\u03b5\u03bb\u03ad\u03c4\u03ae\u03c3\u03c9.\u0392\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03bc\u03b5 '' \u03b4\u03b9\u03b1\u03b2\u03b5\u03b2\u03b1\u03af\u03c9\u03c3\u03b5 '' \u03cc\u03c4\u03b9 \u03b4\u03b5 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b1\u03b9\u03b3\u03bd\u03af\u03b4\u03b9 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7. \u03a4\u03bf\u03c5 \u03c7\u03b1\u03bc\u03bf\u03b3\u03ad\u03bb\u03b1\u03c3\u03b1 ,\u03b3\u03b9\u03b1\u03c4\u03af \u03bc\u03b5 \u03c4\u03b1 \u03bb\u03cc\u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03be\u03ad\u03c6\u03c1\u03b1\u03b6\u03b5 \u03c4\u03b7\u03bd \u03ba\u03bf\u03b9\u03bd\u03ae \u03ac\u03c0\u03bf\u03c8\u03b7 \u03cc\u03c4\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03b5\u03be\u03b9\u03c3\u03ce\u03c3\u03b5\u03b9\u03c2 , \u03c3\u03cd\u03bc\u03b2\u03bf\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 , \u03b1\u03bb\u03bb\u03ac \u03c4\u03bf\u03c5 \u03b5\u03be\u03ae\u03b3\u03b7\u03c3\u03b1 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03bc\u03bf\u03c1\u03c6\u03ae\u03c2 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03c4\u03b7 \u03b6\u03c9\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac , \u03b4\u03b9\u03cc\u03c4\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03b7 \u03bb\u03bf\u03b3\u03b9\u03ba\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac.\u0398\u03b1 \u03c4\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ae\u03c3\u03c9 \u03b1\u03cd\u03c1\u03b9\u03bf \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9 \u03c4\u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03c9 \u03b5\u03ba \u03c4\u03c9\u03bd \u03c5\u03c3\u03c4\u03ad\u03c1\u03c9\u03bd \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c9 \u03c4\u03b7\u03bd \u03bf\u03c1\u03b8\u03cc\u03c4\u03b7\u03c4\u03ac \u03c4\u03b7\u03c2 !\r\n '' ... \u03b3\u03b7\u03c1\u03ac\u03c3\u03ba\u03c9 \u03b1\u03b5\u03af \u03b4\u03b9\u03b4\u03b1\u03c3\u03ba\u03cc\u03bc\u03b5\u03bd\u03bf\u03c2 '' \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c0\u03b9\u03c3\u03c4\u03ad\u03c8\u03c4\u03b5 \u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03bb\u03cc(\u03c4\u03bf \u03b4\u03b9\u03b4\u03b1\u03c3\u03ba\u03cc\u03bc\u03b5\u03bd\u03bf\u03c2 \u03b5\u03bd\u03bd\u03bf\u03ce !).\r\n \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_3": "\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03ad \u039c\u03c0\u03ac\u03bc\u03c0\u03b7 \u03ba\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1.\r\n\r\n\u03a0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03bd\u03b1 \u03bc\u03b5\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 ''\u03c3\u03c4\u03ac\u03c3\u03b9\u03bc\u03bf\u03b9'' \u03c3\u03c4\u03b7\u03bd \u03b7\u03bb\u03b9\u03ba\u03af\u03b1 \u03b3\u03b9\u03b1 \u03bc\u03b5\u03c1\u03b9\u03ba\u03ac \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1. \u0391\u03c2 \u03c0\u03c1\u03bf\u03c7\u03c9\u03c1\u03ac\u03b5\u03b9 \u03c4\u03bf \u03b7\u03bc\u03b5\u03c1\u03bf\u03bb\u03cc\u03b3\u03b9\u03bf, \u03c7\u03c9\u03c1\u03af\u03c2 \u03b5\u03bc\u03ac\u03c2...\r\n\r\n\u0395\u03bc\u03b5\u03af\u03c2 ''\u03b1\u03b5\u03af \u03b8\u03b1 \u03b4\u03b9\u03b4\u03b1\u03c3\u03ba\u03cc\u03bc\u03b1\u03c3\u03c4\u03b5'' (\u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b3\u03b5\u03c1\u03bd\u03ac\u03bc\u03b5).\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_4": "[quote=\"vittasko\"]\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03ad \u039c\u03c0\u03ac\u03bc\u03c0\u03b7 \u03ba\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1.\n\n\u03a0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03bd\u03b1 \u03bc\u03b5\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 ''\u03c3\u03c4\u03ac\u03c3\u03b9\u03bc\u03bf\u03b9'' \u03c3\u03c4\u03b7\u03bd \u03b7\u03bb\u03b9\u03ba\u03af\u03b1 \u03b3\u03b9\u03b1 \u03bc\u03b5\u03c1\u03b9\u03ba\u03ac \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1. \u0391\u03c2 \u03c0\u03c1\u03bf\u03c7\u03c9\u03c1\u03ac\u03b5\u03b9 \u03c4\u03bf \u03b7\u03bc\u03b5\u03c1\u03bf\u03bb\u03cc\u03b3\u03b9\u03bf, \u03c7\u03c9\u03c1\u03af\u03c2 \u03b5\u03bc\u03ac\u03c2...\n\n\u0395\u03bc\u03b5\u03af\u03c2 ''\u03b1\u03b5\u03af \u03b8\u03b1 \u03b4\u03b9\u03b4\u03b1\u03c3\u03ba\u03cc\u03bc\u03b1\u03c3\u03c4\u03b5'' (\u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b3\u03b5\u03c1\u03bd\u03ac\u03bc\u03b5).\n\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.[/quote]\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1 ,\u03ad\u03c4\u03c3\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 ! \r\n\u03b1\u03c2 \u03b1\u03c6\u03b5\u03b8\u03bf\u03cd\u03bc\u03b5 \u03c3\u03c4'\u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1 \u03c3\u03c4\u03b1 \u03b1\u03c0\u03b1\u03bb\u03ac \u03c7\u03ad\u03c1\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c5 \u03bc\u03ad\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bd\u03ad\u03bf\u03b9 \u03c3\u03c4\u03b7 \u03c3\u03ba\u03ad\u03c8\u03b7 \u03ba\u03b1\u03b9 \u03b1\u03c2 \u03c3\u03ba\u03b5\u03c6\u03c4\u03bf\u03cd\u03bc\u03b5 \u03bc\u03b5 \u03b3\u03bb\u03c5\u03ba\u03b5\u03b9\u03ac \u03bd\u03bf\u03c3\u03c4\u03b1\u03bb\u03b3\u03af\u03b1 \u03c4\u03b1 \u03bb\u03cc\u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03bf\u03b9\u03b7\u03c4\u03ae : \r\n\r\n[color=red]'' Ach , dass sie ewig gru''nen bliebe , \ndie scho''ne Zeit der jungen Liebe !!!'' [/color]\r\n\r\n\u0397 \u03b1\u03b3\u03ac\u03c0\u03b7( Liebe ) \u03c4\u03bf\u03c5 \u03c0\u03bf\u03b9\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b1\u03bb\u03b7\u03b8\u03b9\u03bd\u03cc\u03c2 \u03ad\u03c1\u03c9\u03c4\u03b1\u03c2 : \r\n\u03b7 \u03b3\u03bd\u03ce\u03c3\u03b7 , \u03bf \u03ba\u03cc\u03c3\u03bc\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03bb\u03bb\u03ac\u03be\u03b5\u03b9 , \u03b7 \u03b3\u03c5\u03bd\u03b1\u03af\u03ba\u03b1 \u03c0\u03bf\u03c5 \u03b3\u03ad\u03c1\u03bd\u03b5\u03b9 \u03c4\u03c1\u03c5\u03c6\u03b5\u03c1\u03ac \u03c3\u03c4\u03bf\u03bd \u03ce\u03bc\u03bf \u03c4\u03bf\u03c5 \u03ac\u03bd\u03b4\u03c1\u03b1 , \u03c4\u03b1 \u03c4\u03b1\u03be\u03af\u03b4\u03b9\u03b1 \u03ba\u03b1\u03b9 ... \u03c4\u03cc\u03c3\u03b1 \u03ac\u03bb\u03bb\u03b1. \u039c\u03b1\u03ba\u03ac\u03c1\u03b9 \u03bd\u03b1 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\u03c0\u03c1\u03cc\u03c3\u03c9\u03c0\u03bf.\u0393\u03b9 \u03b1\u03c5\u03c4\u03cc \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03c0\u03af\u03c4\u03b9 \u03b7 \u03bc\u03b9\u03ba\u03c1\u03ae \u03bf\u03b8\u03cc\u03bd\u03b7 \u03c0\u03b1\u03c1\u03b1\u03bc\u03ad\u03bd\u03b5\u03b9 \u03b7 ...\u03ac\u03c3\u03b2\u03b5\u03c3\u03c4\u03b7 \u03c6\u03bb\u03cc\u03b3\u03b1.\u039a\u03b1\u03b9 \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c8\u03ac\u03c7\u03bd\u03bf\u03c5\u03bd \u03b1\u03c0\u03b5\u03bb\u03c0\u03b9\u03c3\u03bc\u03ad\u03bd\u03b1 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03b1\u03c0\u03ac\u03bd\u03b8\u03c1\u03c9\u03c0\u03b5\u03c2 \u03c0\u03cc\u03bb\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf\u03c5\u03c2 \u03c6\u03c4\u03b9\u03ac\u03be\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03c5\u03bd \u03c4\u03b7 \u03c7\u03b1\u03c1\u03ac.\u0397 \u03b5\u03c5\u03c4\u03c5\u03c7\u03af\u03b1 \u03c0\u03bf\u03c5 \u03bc\u03b1\u03c2 \u03c7\u03ac\u03c1\u03b9\u03b6\u03b5 \u03ba\u03ac\u03c0\u03bf\u03c4\u03b5 \u03c4\u03bf \u03be\u03b5\u03c6\u03cd\u03bb\u03bb\u03b9\u03c3\u03bc\u03b1 \u03b5\u03bd\u03cc\u03c2 \u03b2\u03b9\u03b2\u03bb\u03af\u03bf\u03c5 - \u03b1\u03ba\u03cc\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac - \u03c4\u03ce\u03c1\u03b1 \u03b4\u03b5 \u03c3\u03c5\u03b3\u03ba\u03b9\u03bd\u03b5\u03af \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1.\r\n \u03a3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03ba\u03b1\u03bc\u03b9\u03ac \u03c6\u03bf\u03c1\u03ac \u03bc\u03ae\u03c0\u03c9\u03c2 \u03ba\u03bb\u03b5\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03b9\u03b4\u03b5\u03b9\u03ba\u03c4\u03b9\u03ba\u03ac \u03c4\u03b1 \u03bc\u03ac\u03c4\u03b9\u03b1 \u03bc\u03c0\u03c1\u03bf\u03c3\u03c4\u03ac \u03c3\u03b5 \u03ad\u03bd\u03b1\u03bd \u03c4\u03c1\u03bf\u03bc\u03b5\u03c1\u03cc \u03ba\u03cc\u03c3\u03bc\u03bf \u03ba\u03b1\u03b9 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03bf\u03b9\u03bf\u03cd\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03b5\u03c5\u03c4\u03c5\u03c7\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5\u03c2. \u039c\u03ae\u03c0\u03c9\u03c2 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03c0\u03b9\u03bf \u03c4\u03af\u03bc\u03b9\u03bf \u03bd\u03b1 \u03ba\u03c1\u03b1\u03c5\u03b3\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 ;\r\n\u03a6\u03af\u03bb\u03b5 , \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c0\u03ac\u03bb\u03b9 \u03c3\u03c4\u03b1 \u03c3\u03bf\u03b2\u03b1\u03c1\u03ac \u03c4\u03b7 \u03b6\u03c9\u03ae !!! \r\n\r\n\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2\r\n\r\n [i]\u03a3\u03c5\u03b3\u03bd\u03ce\u03bc\u03b7 \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03ba\u03c1\u03bf\u03cd\u03c2 \u03ae \u03bd\u03ad\u03bf\u03c5\u03c2 \u03c6\u03af\u03bb\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5 forum.\u0391\u03bb\u03bb\u03ac \u03ba\u03b1\u03bc\u03b9\u03ac \u03c6\u03bf\u03c1\u03ac \u03bf\u03b9 \u03b3\u03b5\u03bd\u03b5\u03ad\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03af\u03bd\u03bf\u03c5\u03bd \u03b7 \u03bc\u03b9\u03b1 \u03c3\u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03c4\u03b7\u03bd \u03b5\u03c5\u03ba\u03b1\u03b9\u03c1\u03af\u03b1 \u03bd\u03b1 \u03b3\u03bd\u03c9\u03c1\u03b9\u03c3\u03c4\u03bf\u03cd\u03bd![/i]", "Solution_5": "[quote=\"stergiu\"]\u03a3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03ba\u03b1\u03bc\u03b9\u03ac \u03c6\u03bf\u03c1\u03ac \u03bc\u03ae\u03c0\u03c9\u03c2 \u03ba\u03bb\u03b5\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03b9\u03b4\u03b5\u03b9\u03ba\u03c4\u03b9\u03ba\u03ac \u03c4\u03b1 \u03bc\u03ac\u03c4\u03b9\u03b1 \u03bc\u03c0\u03c1\u03bf\u03c3\u03c4\u03ac \u03c3\u03b5 \u03ad\u03bd\u03b1\u03bd \u03c4\u03c1\u03bf\u03bc\u03b5\u03c1\u03cc \u03ba\u03cc\u03c3\u03bc\u03bf \u03ba\u03b1\u03b9 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03bf\u03b9\u03bf\u03cd\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03b5\u03c5\u03c4\u03c5\u03c7\u03b9\u03c3\u03bc\u03ad\u03bd\u03bf\u03c5\u03c2.[/quote]\u03a0\u03bf\u03bb\u03cd \u03c6\u03bf\u03b2\u03ac\u03bc\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03af\u03b1 \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03b1\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1.\r\n\r\n[hide=\"\u039b\u03bf\u03b9\u03c0\u03cc\u03bd... \u03cc\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c3\u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1\"]\n\u0391\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf \u03bc\u03af\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 $x$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1:\n$1a)$ \u0391\u03bd $x=1$, \u03c7\u03ac\u03bd\u03c9\n$1b)$ \u0391\u03bd $x>1$, \u03b1\u03c6\u03ae\u03bd\u03c9 \u03bc\u03cc\u03bd\u03bf $1$\n$\\hline$\n\n\u0391\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 $x,y$ \u03cc\u03c0\u03bf\u03c5 $x\\geq y$\n$2a)$ \u0391\u03bd $y=1$ \u03b1\u03c6\u03b1\u03b9\u03c1\u03ce \u03cc\u03bb\u03b1 \u03c4\u03b1 $x$. \u0394\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac:\n$2b)$ \u0391\u03bd $x>y$, \u03c4\u03cc\u03c4\u03b5 \u03c3\u03c4\u03b7 \u03c3\u03c4\u03ae\u03bb\u03b7 \u03bc\u03b5 \u03c4\u03b1 $x$ \u03b1\u03c6\u03ae\u03bd\u03c9 \u03bc\u03cc\u03bd\u03bf $y$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1. \n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b4\u03af\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf \u03b4\u03cd\u03bf \u03c3\u03c4\u03ae\u03bb\u03b5\u03c2 \u03bc\u03b5 \u03af\u03c3\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c3\u03c0\u03af\u03c1\u03c4\u03c9\u03bd. \u0391\u03c0\u03cc \u03ba\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1, \u03cc,\u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2, \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac. \u0395\u03c0\u03b1\u03bd\u03b1\u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b2\u03ae\u03bc\u03b1 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 $(a)$\n$2c)$ \u0391\u03bd $x=y$, \u03c3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b1 \u03b8\u03b1 \u03c7\u03ac\u03c3\u03c9.\n$\\hline$\n\n\u0391\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 3 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 $x,y,z$ \u03cc\u03c0\u03bf\u03c5 $x\\geq y\\geq z$\n$3a)$ \u0391\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03b4\u03cd\u03bf \u03af\u03c3\u03b1 \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ce \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03c0\u03bf\u03c5 \u03b1\u03c0\u03bf\u03bc\u03ad\u03bd\u03b5\u03b9. \u03a0\u03c7 \u03b1\u03bd $y=z$ \u03b1\u03c6\u03b1\u03b9\u03c1\u03ce \u03cc\u03bb\u03b1 \u03c4\u03b1 $x$. \u0388\u03c4\u03c3\u03b9 \u03c0\u03ac\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b9\u03c2 2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2. (\u03b5\u03b4\u03ce \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7, \u03cc\u03c0\u03bf\u03c5 \u03c4\u03b1 \u03b4\u03cd\u03bf \u03af\u03c3\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b1 \u03bc\u03b5 1, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ba\u03ac\u03bd\u03c9 \u03af\u03c3\u03b7 \u03bc\u03b5 1)\n\n[color=red]EDIT:[/color]\n\u0391\u03c0\u03bf \u03b4\u03c9 \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bf\u03b4\u03b7\u03b3\u03b5\u03af \u03c3\u03c4\u03b7 \u03bd\u03af\u03ba\u03b7. \u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c4\u03bf \u03c3\u03ba\u03b5\u03c0\u03c4\u03b9\u03ba\u03cc \u03c3\u03b5 \u03b1\u03c5\u03c4\u03bf\u03cd \u03c4\u03bf\u03c5 \u03b5\u03af\u03b4\u03bf\u03c5\u03c2 \u03c4\u03b1 \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1:\n\n\u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03ba\u03ac\u03b8\u03b5 \u03bc\u03b1\u03c2 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b9\u03b1\u03c4\u03b7\u03c1\u03b5\u03af\u03c4\u03b1\u03b9 \u03bc\u03af\u03b1 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7. \u039a\u03b1\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03bd\u03b1 \u03bc\u03b7 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bc\u03b5 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03c4\u03c1\u03cc\u03c0\u03bf \u03bd\u03b1 \u03b5\u03c0\u03b9\u03c4\u03cd\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7.\n\u039c' \u03ac\u03bb\u03bb\u03b1 \u03bb\u03cc\u03b3\u03b9\u03b1, \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03ba\u03b9\u03bd\u03b7\u03b8\u03bf\u03cd\u03bc\u03b5 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03c4\u03c1\u03b9\u03ac\u03b4\u03b5\u03c2 \u03c0\u03b1\u03c4\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c3\u03b5 \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03b1 \"\u03bc\u03b1\u03be\u03b9\u03bb\u03b1\u03c1\u03ac\u03ba\u03b9\u03b1\"\n\n\n$\\hline$\n\u03a3\u03c7\u03b5\u03c4\u03b9\u03ba\u03cc \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ad\u03bd\u03b1 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9.\n[i][color=brown]\u039a\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03c0\u03b1\u03af\u03ba\u03c4\u03b5\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bb\u03ad\u03b5\u03b9 \u03ad\u03bd\u03b1\u03bd (\u03b1\u03ba\u03ad\u03c1\u03b1\u03b9\u03bf) \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc \u03b1\u03c0\u03cc \u03c4\u03bf 1 \u03c9\u03c2 \u03c4\u03bf 10\n\u039a\u03b1\u03c4\u03b1\u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03b5\u03b9\u03c0\u03c9\u03b8\u03b5\u03af.\n\u039d\u03b9\u03ba\u03b7\u03c4\u03ae\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c6\u03c4\u03ac\u03c3\u03b5\u03b9 \u03c3\u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 100[/color].[/i]\n\n\u03a3'\u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03ad\u03c7\u03b5\u03b9 \u03bf \u03c0\u03c1\u03c9\u03c4\u03bf\u03c2 \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03b5\u03b9 \u03b5\u03bd\u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03bf\u03c5\u03c2 \u03c3\u03c4\u03cc\u03c7\u03bf\u03c5\u03c2-\u03bc\u03b1\u03be\u03b9\u03bb\u03b1\u03c1\u03ac\u03ba\u03b9\u03b1\n\n[/hide]", "Solution_6": "[quote=\"pontios\"]\n\u0391\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03bc\u03b5\u03af\u03bd\u03b5\u03b9 2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 $x,y$ \u03cc\u03c0\u03bf\u03c5 $x\\geq y$\n$2a)$ \u0391\u03bd $y=1$ \u03b1\u03c6\u03b1\u03b9\u03c1\u03ce \u03cc\u03bb\u03b1 \u03c4\u03b1 $x$. \u0394\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac:\n$2b)$ \u0391\u03bd $x>y$, \u03c4\u03cc\u03c4\u03b5 \u03c3\u03c4\u03b7 \u03c3\u03c4\u03ae\u03bb\u03b7 \u03bc\u03b5 \u03c4\u03b1 $x$ \u03b1\u03c6\u03ae\u03bd\u03c9 \u03bc\u03cc\u03bd\u03bf $y$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1.\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae \u03b4\u03af\u03bd\u03c9 \u03c3\u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf \u03b4\u03cd\u03bf \u03c3\u03c4\u03ae\u03bb\u03b5\u03c2 \u03bc\u03b5 \u03af\u03c3\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c3\u03c0\u03af\u03c1\u03c4\u03c9\u03bd. \u0391\u03c0\u03cc \u03ba\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1, \u03cc,\u03c4\u03b9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2, \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03c3\u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac. \u0395\u03c0\u03b1\u03bd\u03b1\u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03b2\u03ae\u03bc\u03b1 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 $(a)$\n$2c)$ \u0391\u03bd $x=y$, \u03c3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03b1 \u03b8\u03b1 \u03c7\u03ac\u03c3\u03c9.\n[/quote]\r\n\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b1\u03c5\u03c4\u03ae \u03b4\u03b5\u03bd \u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03b5\u03c4\u03b5 \u03c5\u03c0\u03cc\u03c8\u03b9\u03bd \u03c3\u03b1\u03c2 \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03cc\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b5\u03bd\u03ac \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd \u03b3\u03c1\u03b1\u03bc\u03bc\u03ce\u03bd, \u03c0.\u03c7. \u03b1\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7:\r\n\r\n1\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac |...|...|\r\n2\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac | | | |\r\n\r\n(\u03bc\u03b5 \"...\" \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03c9 \u03c4\u03bf \u03ba\u03b5\u03bd\u03cc \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03b4\u03c5\u03bf \u03b3\u03c1\u03b1\u03bc\u03bc\u03ce\u03bd)\r\n\r\n\u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03c0\u03bf\u03c5 \u03bb\u03ad\u03c4\u03b5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd 2\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac,\u03b1\u03bb\u03bb\u03ac \u03b1\u03bd \u03c4\u03bf \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03b8\u03b1 \u03c4\u03b9\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03b5\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7 2\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c7\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5, \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b4\u03c5\u03bf \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd 1\u03b7 \u03b1\u03bd \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2\".\r\n\u0397 \u03bb\u03cd\u03c3\u03b7 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c3\u03b5 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c4\u03b7\u03c2 2\u03b7\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ac\u03c2.\r\n\r\n\u0391\u03c5\u03c4\u03ae \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03ba\u03c5\u03c1\u03b9\u03cc\u03c4\u03b5\u03c1\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim ,\u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03b1\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03bd\u03b1 \u03c7\u03ac\u03bd\u03b5\u03b9 \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2, \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bc\u03ad\u03c7\u03c1\u03b9 \u03ba\u03b1\u03b9 \"\u03bb\u03af\u03b3\u03bf \u03c0\u03c1\u03b9\u03bd\" \u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd xor \u03cc\u03c0\u03c9\u03c2 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c3\u03c4\u03bf \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim (...,\u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 \u03bd\u03b9\u03ba\u03ac\u03b5\u03b9).", "Solution_7": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03b3\u03bd\u03af\u03b4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd , \u03c0\u03b1\u03c1\u03ac \u03bc\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03b5\u03b9\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b5\u03c1\u03b5\u03c5\u03bd\u03b7\u03c4\u03ad\u03c2, \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c7\u03b5\u03b9 \u03c3\u03bf\u03b2\u03b1\u03c1\u03ac \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1.\u039f \u03c3\u03c5\u03bd\u03ac\u03b4\u03b5\u03bb\u03c6\u03bf\u03c2 \u03bc\u03bf\u03c5 \u03b1\u03c0\u03bf\u03ba\u03ac\u03bb\u03c5\u03c8\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03bc\u03b9\u03ac \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b4\u03b5 \u03b4\u03b9\u03b1\u03b8\u03ad\u03c4\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c4\u03cd\u03c7\u03b7 , \u03bb\u03cc\u03b3\u03c9 \u03b5\u03be\u03ac\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2 !!!\u0391\u03c2 \u03bc\u03b7\u03bd \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03bf\u03cd\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03ac\u03bb\u03bb\u03bf \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc , \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c0\u03bf\u03bb\u03bb\u03ac \u03ba\u03b1\u03bb\u03ac \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03bd\u03b1 \u03bb\u03cd\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03bf \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03ba\u03c5\u03bb\u03bb\u03ac\u03b5\u03b9 \u03b3\u03c1\u03ae\u03b3\u03bf\u03c1\u03b1 .\r\n \u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03cc\u03bc\u03c9\u03c2 \u03bd\u03b1 \u03bf\u03bc\u03bf\u03bb\u03bf\u03b3\u03ae\u03c3\u03c9 \u03cc\u03c4\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 N\u03b9m \u03ad\u03b3\u03b9\u03bd\u03b1 \u03c3\u03bf\u03c6\u03cc\u03c4\u03b5\u03c1\u03bf\u03c2(\u03cc\u03c7\u03b9 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03bf\u03c6\u03cc\u03c2 , \u03b1\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1\u03c5\u03c4\u03cc \u03b4\u03b5 \u03c6\u03c4\u03b1\u03af\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03bb\u03bb\u03ac 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\"\u03bb\u03af\u03b3\u03bf \u03c0\u03c1\u03b9\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2\", \u03cc\u03c0\u03c9\u03c2 \u03bb\u03ad\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf stedes.[/color]\n\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03b5\u03c1\u03b4\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03b1\u03c4\u03ac\u03bc\u03b5 \u03c3'\u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \"\u03bc\u03b1\u03be\u03b9\u03bb\u03b1\u03c1\u03ac\u03ba\u03b9\u03b1\" \u03cc\u03c0\u03c9\u03c2 \u03ad\u03c7\u03c9 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03c3\u03b5 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03c5\u03bc\u03ad\u03bd\u03bf post\n\n* \u0388\u03c7\u03c9 \u03b2\u03c1\u03b5\u03b9 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03ba\u03b1\u03b9 \u03bd=4 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2. \u0393\u03b9\u03b1 5 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03b1\u03c1\u03ba\u03b5\u03c4\u03ad\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03bc\u03b5 XOR-\u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03bc\u03b7\u03b4\u03ad\u03bd, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5. \u0391\u03c0\u03bb\u03ce\u03c2, \u03b1\u03bd \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c0\u03c1\u03ce\u03c4\u03bf\u03b9 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc \u03c3\u03c0\u03af\u03c1\u03c4\u03bf \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac 1 (\u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03c0\u03ac\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd [b]2345[/b]). \u03a3\u03c4\u03b7 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1, \u03cc,\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03c0\u03ac\u03bc\u03b5 \u03c3\u03b5 \u03bc\u03af\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03ac\u03bb\u03bb\u03b5\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03ae\u03c4\u03c4\u03b1\u03c2.\n-----------------------------------------------------------------------------------------------------------------------\n\n[/hide][hide=\"\u0397 \u03c0\u03c1\u03ac\u03be\u03b7 XOR (\u03b3\u03b9\u03b1 \u03cc\u03c0\u03bf\u03b9\u03bf\u03bd \u03b4\u03b5\u03bd \u03c4\u03b7 \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03b5\u03b9 \u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03ac\u03be\u03b7 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03ce\u03bd)\"]\n\u0388\u03c3\u03c4\u03c9 \u03b4\u03cd\u03bf \u03b4\u03c5\u03b1\u03b4\u03b9\u03ba\u03ac \u03c8\u03b7\u03c6\u03af\u03b1 $x,y$. \u0397 \u03c0\u03c1\u03ac\u03be\u03b7 XOR \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 $x\\otimes y = \\{\\begin{array}{cc}0, & a\\nu\\ x=y \\\\ 1, & a\\nu\\ x\\neq y \\end{array}$\n\u0397 \u03c0\u03c1\u03ac\u03be\u03b7 XOR \u03b5\u03c0\u03b5\u03ba\u03c4\u03b5\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03b4\u03c5\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03bd\u03cc\u03c2 \u03c8\u03b7\u03c6\u03af\u03b1. \u0391\u03c0\u03bb\u03ce\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf XOR \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03c8\u03b7\u03c6\u03af\u03c9\u03bd \u03af\u03c3\u03b7\u03c2 \u03c4\u03ac\u03be\u03b7\u03c2\n\n\u0388\u03c4\u03c3\u03b9, \u03b1\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 $5=(101)_{2}$ \u03ba\u03b1\u03b9 $4=(100)_{2}$, \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $101\\otimes 100 = 001$,\n\u03b1\u03c6\u03bf\u03cd\n$1\\otimes 1 = 0$\n$0\\otimes 0 = 0$\n$1\\otimes 0 = 1$\n\n\u039f\u03bc\u03bf\u03af\u03c9\u03c2 \u03b5\u03c0\u03b5\u03ba\u03c4\u03b5\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf\u03c5\u03c2 \u03c4\u03c9\u03bd \u03b4\u03cd\u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03ce\u03bd\n$x\\otimes y\\otimes z = (x\\otimes y)\\otimes z$\u0384\n\n\u0388\u03c4\u03c3\u03b9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03b7 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 XOR \u03af\u03c3\u03bf \u03bc\u03b5\n$1\\otimes 2\\otimes 3\\otimes 4\\otimes 5=1$\n-----------------------------------------------------------------------------------------------------------------------\n\n[/hide]\r\n\r\n[color=brown]\u03a5\u0393\n\u038c\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bb\u03cc\u03b3\u03bf\u03c2 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03b5\u03af \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1, \u03b1\u03c2 \u03bc\u03b7\u03bd \u03c4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9.\n\u0391\u03bd \u03cc\u03bc\u03c9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b8\u03ad\u03bb\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03b1\u03c0\u03b5\u03c5\u03b8\u03cd\u03bd\u03b5\u03b9 \u03b5\u03c1\u03c9\u03c4\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03ac \u03bc\u03b5 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03c9, \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03b4\u03b9\u03ac\u03b8\u03b5\u03c3\u03ae \u03cc\u03bb\u03c9\u03bd :)[/color]", "Solution_9": "\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03ad \u03c6\u03af\u03bb\u03b5 pontios , \r\n\r\n \u03b8\u03b1 \u03bc\u03b5\u03bb\u03b5\u03c4\u03ae\u03c3\u03c9 \u03c4\u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03c3\u03bf\u03c5 \u03bc\u03b5 \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03ae.\u03a3\u03b5 \u03c3\u03c5\u03b3\u03c7\u03b1\u03af\u03c1\u03c9 \u03c9\u03c3\u03c4\u03cc\u03c3\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03bc\u03bf\u03bd\u03ae \u03c3\u03bf\u03c5 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 \u03bb\u03cd\u03c3\u03b7.\u0395\u03bc\u03ad\u03bd\u03b1, \u03cc\u03c4\u03b1\u03bd \u03bc\u03bf\u03c5 \u03b5\u03af\u03c0\u03b5 \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b7 \u03bc\u03ad\u03c1\u03b1 \u03bf \u03c3\u03c5\u03bd\u03ac\u03b4\u03b5\u03bb\u03c6\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03b3\u03c5\u03b9\u03bf \u03c4\u03bf\u03c5 \u03bc\u03b9\u03b1 \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9 \u03bf \u03ad\u03bd\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03bf \u03ac\u03bb\u03bb\u03bf\u03c2 \u03bc\u03bf\u03c5 ..\u03ba\u03cc\u03c0\u03b7\u03ba\u03b1\u03bd \u03c4\u03b1 \u03c0\u03cc\u03b4\u03b9\u03b1 , \u03b3\u03b9\u03b1\u03c4\u03af \u03c3\u03ba\u03ad\u03c6\u03c4\u03b7\u03ba\u03b1 \u03cc\u03c4\u03b9 \u03c0\u03ae\u03b3\u03b5 \u03c4\u03cc\u03c3\u03bf\u03c2 \u03ba\u03cc\u03c0\u03bf\u03c2 \u03c7\u03b1\u03bc\u03ad\u03bd\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03ba\u03c4\u03cc\u03c2 \u03b1\u03c5\u03c4\u03bf\u03cd \u03ad\u03b2\u03b1\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03c3\u03b1\u03c2 \u03c3\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c0\u03ad\u03c4\u03b5\u03b9\u03b1.\u038c\u03bc\u03c9\u03c2 , \u03cc\u03c0\u03c9\u03c2 \u03b4\u03b9\u03b1\u03b2\u03ac\u03b6\u03c9, \u03ad\u03c7\u03b5\u03b9\u03c2 \u03b2\u03c1\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 , ...\u03bf\u03c5\u03b4\u03ad\u03bd \u03ba\u03b1\u03ba\u03cc\u03bd \u03b1\u03bc\u03b5\u03b9\u03b3\u03ad\u03c2 \u03ba\u03b1\u03bb\u03bf\u03cd !\r\n \u0394\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 '' \u03bf\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2 '' , \u03ce\u03c3\u03c4\u03b5 \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bd\u03b1 \u03b1\u03c0\u03bb\u03bf\u03c5\u03c3\u03c4\u03b5\u03c5\u03c4\u03b5\u03af.\u0391\u03bd \u03bd\u03bf\u03bc\u03af\u03b6\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03b3\u03bd\u03af\u03b4\u03b9 \u03b2\u03b5\u03bb\u03c4\u03b9\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 , \u03ba\u03ac\u03bd\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03ac\u03c1\u03bc\u03bf\u03c3\u03b5 \u03c4\u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae.\r\n [u]\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2 [/u]", "Solution_10": "[quote=\"pontios\"]\n[quote=\"stedes\"]\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b1\u03c5\u03c4\u03ae \u03b4\u03b5\u03bd \u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03b5\u03c4\u03b5 \u03c5\u03c0\u03cc\u03c8\u03b9\u03bd \u03c3\u03b1\u03c2 \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03cc\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b5\u03bd\u03ac \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd \u03b3\u03c1\u03b1\u03bc\u03bc\u03ce\u03bd\n[/quote]\n\u03a3\u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c3\u03b7\u03bc\u03b1\u03c3\u03af\u03b1 \u03c4\u03b1 \u03ba\u03b5\u03bd\u03ac. \u0393\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03ce\u03c0\u03b9\u03c3\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b1\u03c1\u03cc\u03bd \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03ba\u03b5\u03c0\u03c4\u03b9\u03ba\u03cc (\u03b1\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bb\u03b7\u03c6\u03b8\u03bf\u03cd\u03bd \u03c5\u03c0\u03cc\u03c8\u03b9\u03bd \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03ba\u03b5\u03bd\u03ac, \u03c4\u03cc\u03c4\u03b5 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c7\u03ac\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03bf\u03bc\u03bf\u03c1\u03c6\u03b9\u03ac \u03c4\u03bf\u03c5).\n\u0395\u03be\u03ac\u03bb\u03bb\u03bf\u03c5, \u03b1\u03bd \u03c0\u03c7 \u03c3\u03b2\u03ae\u03c3\u03c9 \u03c4\u03b7 2\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae, \u03c4\u03cc\u03c4\u03b5 \u03b7 1\u03b7 \u03ba\u03b1\u03b9 3\u03b7 \u03b3\u03af\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2.\n\u0388\u03c4\u03c3\u03b9 \u03c4\u03bf \"\u03cc\u03c3\u03b5\u03c2 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b5\u03c0\u03b9\u03b8\u03c5\u03bc\u03b5\u03af\" \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03b5\u03b9 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b1\u03bd\u03c4\u03af \u03bd\u03b1 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b5\u03af. \n[/quote]\n\u0395\u03b3\u03ce \u03b1\u03c0\u03bb\u03ac \u03c4\u03bf \u03b1\u03bd\u03ad\u03c6\u03b5\u03c1\u03b1 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c4\u03bf \u03bf\u03c1\u03af\u03b6\u03b5\u03b9 \u03ad\u03c4\u03c3\u03b9:\n[quote=\"stergiou\"]\n\u0395\u03c0\u03af\u03c3\u03b7\u03bd \u03b1\u03bd \u03c0\u03c7 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c4\u03b7 \u03bc\u03b5\u03c3\u03b1\u03af\u03b1 , \u03bf \u03ac\u03bb\u03bb\u03bf\u03c2 \u03b4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03c1\u03cc\u03bd\u03c9\u03c2 \u03c4\u03b9\u03c2 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 4\n[/quote]\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b7 \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \"\u03bf\u03b9 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ad\u03c2\" ,\u03b5\u03af\u03c4\u03b5 \u03c4\u03bf \u03bf\u03bc\u03bf\u03c1\u03c6\u03b1\u03af\u03bd\u03b5\u03b9 \u03b5\u03af\u03c4\u03b5 \u03c4\u03bf \u03c7\u03b1\u03bb\u03ac\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1, \u03c4\u03bf \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c0\u03b9\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf.\r\n\r\n\r\n\r\n\u038c\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03ce\u03c1\u03b1 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c7\u03c9\u03c1\u03af\u03c2 \u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd, \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b8\u03b5\u03c9\u03c1\u03ce \u03c0\u03c9\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2:\r\n\r\n$1)$\u0391\u03bd \u03c3\u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $n\\geq 1$ \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b7 \u03bc\u03af\u03b1 \u03ad\u03c7\u03b5\u03b9 $x>1$ \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03ba\u03b1\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 (\u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd) \u03ad\u03c7\u03bf\u03c5\u03bd \u03b1\u03c0\u03cc \u03bc\u03af\u03b1 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03b7 \u03ba\u03ac\u03b8\u03b5\u03bc\u03b9\u03b1,\u03c4\u03cc\u03c4\u03b5 \u03b1\u03bd $n$ \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 \u03c4\u03b9\u03c2 $x$ \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2, \u03b5\u03bd\u03ce \u03b1\u03bd $n$ \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c6\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 \u03c4\u03b9\u03c2 $x$ \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03c4\u03b9\u03c2 $x-1$ \u03ba\u03cc\u03ba\u03ba\u03b9\u03bd\u03b5\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2.\r\n\r\n$2)$\u03a3\u03b5 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03c9\u03c0\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03cc\u03c0\u03c9\u03c2 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c3\u03c4\u03bf \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim (\u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9).\r\n\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2...", "Solution_11": "\u0388\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf, \u03a3\u03c4\u03ad\u03bb\u03b9\u03bf. \u039f\u03c6\u03b5\u03af\u03bb\u03c9 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03b4\u03b5\u03c7\u03c4\u03ce \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c0\u03c1\u03cc\u03c3\u03b5\u03be\u03b1 \u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7:\r\n[quote=\"stergiu\"]\u0395\u03c0\u03af\u03c3\u03b7\u03bd \u03b1\u03bd \u03c0\u03c7 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ae \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c4\u03b7 \u03bc\u03b5\u03c3\u03b1\u03af\u03b1 , \u03bf \u03ac\u03bb\u03bb\u03bf\u03c2 \u03b4\u03b5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03c1\u03cc\u03bd\u03c9\u03c2 \u03c4\u03b9\u03c2 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 4 .[/quote]\n\n\u0395\u03c0\u03b5\u03b9\u03b4\u03ae \u03c0\u03af\u03c3\u03c4\u03b5\u03c8\u03b1 \u03cc\u03c4\u03b9 \u03ae\u03c4\u03b1\u03bd \u03c4\u03bf \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c4\u03bf\u03c5 Nim \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03bd \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ad\u03b2\u03bb\u03b5\u03c8\u03b1 :D \n\u0393\u03b9 \u03b1\u03c5\u03c4\u03cc \u03ba\u03b1\u03b9 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03cc\u03c4\u03b9 \u03b7 \u03bc\u03cc\u03bd\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03c7\u03ac\u03bd\u03b5\u03b9 \u03bf \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2\n\u0395\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03af\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c1\u03b5\u03c5\u03bd\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c3\u03b5 \u03b4\u03cd\u03bf \u03bc\u03bf\u03c1\u03c6\u03ad\u03c2, \u03bc\u03af\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03ba\u03b1\u03b9 \u03bc\u03af\u03b1 \u03c7\u03c9\u03c1\u03af\u03c2.\n\n[b][color=red]EDIT[/color][/b]\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03b1\u03bd\u03b1\u03b9\u03c1\u03ce \u03c4\u03b7 \u03c6\u03c1\u03ac\u03c3\u03b7 \u03bc\u03bf\u03c5 [quote]\u03b1\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03bb\u03b7\u03c6\u03b8\u03bf\u03cd\u03bd \u03c5\u03c0\u03cc\u03c8\u03b9\u03bd \u03ba\u03b1\u03b9 \u03c4\u03b1 \u03ba\u03b5\u03bd\u03ac, \u03c4\u03cc\u03c4\u03b5 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c7\u03ac\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd \u03bf\u03bc\u03bf\u03c1\u03c6\u03b9\u03ac \u03c4\u03bf\u03c5[/quote]\r\n\r\n\u039a\u03ac\u03b8\u03b5 \u03ac\u03bb\u03bb\u03bf!\r\n\u0395\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03ba\u03bf\u03bd\u03c4\u03b5\u03cd\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c7\u03ac\u03c3\u03b5\u03b9\u03c2, \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c4\u03b7\u03bd \u03b5\u03bd\u03b1\u03bb\u03bb\u03b1\u03ba\u03c4\u03b9\u03ba\u03ae \u03b5\u03c0\u03b9\u03bb\u03bf\u03b3\u03ae \u03bd\u03b1 \"\u03c3\u03c0\u03ac\u03c3\u03b5\u03b9\u03c2\" \u03c0\u03c7 \u03c4\u03b7\u03bd 5\u03ac\u03b4\u03b1 \u03c3\u03b5 2 \u03b4\u03c5\u03ac\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03b1\u03c0\u03bf\u03ba\u03c4\u03ac \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd.\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b9\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03b3\u03b9\u03b1\u03c4\u03af \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \"\u03c3\u03b5\u03b9\u03c1\u03ad\u03c2\" \u03b5\u03ba\u03b5\u03af \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ae\u03c1\u03c7\u03b1\u03bd\r\n\r\n\r\n\u039b\u03bf\u03c5\u03ba\u03ac\u03c2", "Solution_12": "\u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03bc\u03b5 \u03cc\u03c3\u03b1 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2:\r\n\r\n[hide=\"\u03a0\u03c1\u03ce\u03c4\u03b7 \u03b5\u03ba\u03b4\u03bf\u03c7\u03ae - \u03c7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bd \u03c1\u03cc\u03bb\u03bf \u03c4\u03b1 \u03ba\u03b5\u03bd\u03ac\"]\n\u03a7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03c4\u03bf\u03bd \u03cc\u03c1\u03bf \"\u03c3\u03c0\u03af\u03c1\u03c4\u03b1\" \u03b1\u03bd\u03c4\u03af \u03b3\u03b9\u03b1 \"\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2\" \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7 \u03c3\u03c5\u03b3\u03c7\u03ad\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03c2 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \n\n\u03a7\u03c9\u03c1\u03af\u03b6\u03c9 [u]\u03cc\u03bb\u03b5\u03c2[/u] \u03c4\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 [b]\u0397[/b] \u03ba\u03b1\u03b9 [b]\u039d[/b]:\n\n$H = \\{(x_{1},x_{2},x_{3},x_{4},x_{5})\\ |\\ x_{1}\\otimes x_{2}\\otimes x_{3}\\otimes x_{4}\\otimes x_{5}=0\\}$\n$N = \\{(x_{1},x_{2},x_{3},x_{4},x_{5})\\ |\\ x_{1}\\otimes x_{2}\\otimes x_{3}\\otimes x_{4}\\otimes x_{5}\\not =0\\}$\n\n\u0394\u03b7\u03bb\u03b1\u03b4\u03ae, \u03b1\u03bd \u03c4\u03bf XOR \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 0, \u03c4\u03cc\u03c4\u03b5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u0397[/b], \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u039d[/b]. \u03a3\u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b1\u03c5\u03c4\u03cc \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03cc\u03bb\u03bf\u03c5 [b]\u0397[/b] \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03b9\u03c3\u03bf\u03c1\u03c1\u03bf\u03c0\u03b7\u03bc\u03ad\u03bd\u03b7\", \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03b1\u03bd \u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2 \u03c3\u03c4\u03bf \u03b4\u03c5\u03b1\u03b4\u03b9\u03ba\u03cc \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ac\u03c1\u03c4\u03b9\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 1\u03ac\u03b4\u03c9\u03bd, \u03af\u03c3\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 2\u03ac\u03b4\u03c9\u03bd, 4\u03ac\u03b4\u03c9\u03bd \u03ba\u03bb\u03c0\n\n[url=http://www.cut-the-knot.org/nim_theory.shtml]\u0395\u03b4\u03ce[/url] \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ad\u03c4\u03bf\u03b9\u03bc\u03b7 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim \u03cc\u03c4\u03b9 \n[quote]-[color=blue] \u03ba\u03ac\u03b8\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u0397[/b] \u03bf\u03b4\u03b7\u03b3\u03b5\u03af \u03bc\u03cc\u03bd\u03bf \u03c3\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u039d[/b][/color], \u03b5\u03bd\u03ce\n- [color=blue]\u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u039d[/b] \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03bd\u03b1 \u03bf\u03b4\u03b7\u03b3\u03b5\u03af \u03c3\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u0397[/b][/color][/quote]\n\n\u0391\u03c1\u03c7\u03b9\u03ba\u03ac \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]12345[/b], \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae [b]\u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bb\u03b1[/b] [b]1[/b]. \n\u03a4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c7\u03c9\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 2 \u03c6\u03ac\u03c3\u03b5\u03b9\u03c2, \u03c4\u03b7\u03bd [color=red]\u03b1\u03c1\u03c7\u03b9\u03ba\u03ae[/color] (\u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 [u]\u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9[/u] \u03cc\u03bb\u03b1 [b]1[/b]) \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd [color=red]\u03c4\u03b5\u03bb\u03b9\u03ba\u03ae[/color] (\u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 [u]\u03b5\u03af\u03bd\u03b1\u03b9[/u] \u03cc\u03bb\u03b1 [b]1[/b])\n\u0391\u03c2 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [b]\u03a4[/b] \u03cc\u03bb\u03c9\u03bd \u03c4\u03c9\u03bd \u03c4\u03b5\u03bb\u03b9\u03ba\u03ce\u03bd \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae $T = \\{1,11,111,1111,11111\\}$.\n\n\n\u0391\u03c1\u03c7\u03b9\u03ba\u03ac, \u03c3\u03b5 \u03ba\u03ac\u03b8\u03b5 \u03bc\u03b1\u03c2 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c6\u03ad\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \u03c3\u03b5 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u0397[/b]. \u039f \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03b4\u03b5 \u03b8\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03c6\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 \u03c3\u03b5 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7, \u03b1\u03c6\u03bf\u03cd \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 $H \\to H$.\n\n\u0397 \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c6\u03ac\u03c3\u03b7 \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03c3\u03b5\u03b9 \u03bc\u03b5 \u03b4\u03b9\u03ba\u03ae \u03bc\u03b1\u03c2 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7. \u0398\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03bc\u03af\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 [b]1,111,11111[/b]. \u0391\u03c0\u03bf \u03ba\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03ad\u03c1\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b1\u03c6\u03ad\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03ba\u03b5\u03c1\u03b4\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9.\n\n\n\n\u0386\u03c1\u03b1 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd, \u03c4\u03bf \u03bc\u03cc\u03bd\u03bf \u03c0\u03bf\u03c5 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03bc\u03cc\u03bd\u03b5\u03c2 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 (\u03c4\u03b7\u03c2 \u03c0\u03c1\u03ce\u03c4\u03b7\u03c2 \u03c6\u03ac\u03c3\u03b7\u03c2) \u03c0\u03bf\u03c5 \u03bf\u03b4\u03b7\u03b3\u03bf\u03cd\u03bd \u03c3\u03b5 \u03bc\u03af\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 [b]1,111,11111[/b] \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 [b]\u039d[/b] \n\n\n[color=brown]\u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03b1\u03c2 \u03c0\u03ac\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]1[/b].\n\u03a5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd 2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03b9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c6\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b5\u03ba\u03b5\u03af.\n\u039f \u03c0\u03c1\u03ce\u03c4\u03bf\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03be\u03b5\u03ba\u03b9\u03bd\u03ce\u03bd\u03c4\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u03b11[/b], \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae 2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2, \u03b7 \u03bc\u03af\u03b1 \u03bc\u03b5 1 \u03c3\u03c0\u03af\u03c1\u03c4\u03bf \u03ba\u03b1\u03b9 \u03b7 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03b5 \u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 (\u03b5\u03c6\u03cc\u03c3\u03bf\u03bd \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03b1\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ae, \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1>1)\n\u039f \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]\u03b2[/b], \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bc\u03af\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 \u03b2>1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1.\n\n\u0391\u03c5\u03c4\u03ad\u03c2 \u03bf\u03b9 \u03b4\u03cd\u03bf \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bd \u03bd\u03b1 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [b]\u0397[/b]. \u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [b]\u0397[/b] \u03cc\u03c0\u03c9\u03c2 \u03bf\u03c1\u03af\u03c3\u03c4\u03b7\u03ba\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03b9\u03c3\u03bf\u03c1\u03c1\u03bf\u03c0\u03b7\u03bc\u03ad\u03bd\u03bf\".\n\u039a\u03b1\u03b9 \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c6\u03b1\u03bd\u03b5\u03c1\u03cc \u03bf\u03b9 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 [b]\u03b11[/b] \u03ba\u03b1\u03b9 [b]\u03b2[/b] \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c1\u03c1\u03bf\u03c0\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2, \u03b5\u03c0\u03bf\u03bc\u03ad\u03bd\u03c9\u03c2 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf [b]\u039d[/b]\n\u03a0\u03c7 \u03b1\u03bd \u03b1=5, \u03c4\u03cc\u03c4\u03b5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \n\u03b1=1+4\n1=1, \n\u03c3\u03c4\u03bf \u03c3\u03cd\u03bd\u03bf\u03bb\u03bf \u03b4\u03cd\u03bf \u03bc\u03bf\u03bd\u03ac\u03b4\u03b5\u03c2 (\u03c0\u03bf\u03c5 \u03b1\u03bb\u03bb\u03b7\u03bb\u03bf\u03b5\u03be\u03bf\u03c5\u03b4\u03b5\u03c4\u03b5\u03c1\u03ce\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9)\n\u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03bc\u03af\u03b1 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b4\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03b5\u03cd\u03b5\u03b9 \n\n\n\u03a4\u03ce\u03c1\u03b1 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b7 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 [b]111[/b] \u03b4\u03b5 \u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b1\u03c6\u03bf\u03cd \u03c0\u03c1\u03bf\u03ae\u03bb\u03b8\u03b5 \u03b5\u03af\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd [b]\u03b1111[/b] \u03ae \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd [b]\u03b211[/b]\n\u039f\u03bc\u03bf\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03b7 [b]11111[/b] \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c0\u03c1\u03bf\u03ae\u03bb\u03b8\u03b5 \u03bc\u03cc\u03bd\u03bf \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd [b]\u03b21111[/b] [/color]\n[/hide]", "Solution_13": "\u039b\u03bf\u03c5\u03ba\u03ac \u03c3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03ae\u03c1\u03b9\u03b1, \u03ae\u03c3\u03bf\u03c5\u03bd \u03ba\u03b1\u03c4\u03b1\u03c4\u03bf\u03c0\u03b9\u03c3\u03c4\u03b9\u03ba\u03cc\u03c4\u03b1\u03c4\u03bf\u03c2! :) \r\n\u0395\u03b3\u03ce \u03b1\u03c3\u03c7\u03bf\u03bb\u03ae\u03b8\u03b7\u03ba\u03b1 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c4\u03bf\u03c5 \u03c0\u03b1\u03b9\u03c7\u03bd\u03b9\u03b4\u03b9\u03bf\u03cd \u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \u03cc\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03bf\u03c5\u03bc\u03b5 \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \"\u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac\"*. \u039a\u03b1\u03c4\u03ad\u03bb\u03b7\u03be\u03b1 \u03c3\u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03c3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1.\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c3\u03c5\u03bc\u03b5\u03c1\u03b9\u03bb\u03ac\u03b2\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \"\u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac\" ,\u03b7 \u03bd\u03b9\u03ba\u03b7\u03c6\u03cc\u03c1\u03b1 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c3\u03c4\u03bf \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf post, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae:\r\n\r\n[b]\u03a3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae[/b]\r\n$1)$\u0391\u03bd \u03c3\u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd $n\\geq 1$ \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b9\u03c2 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b7 \u03bc\u03af\u03b1 \u03ad\u03c7\u03b5\u03b9 $x>1$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 (\u03b1\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd) \u03ad\u03c7\u03bf\u03c5\u03bd \u03b1\u03c0\u03cc \u03ad\u03bd\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03bf \u03b7 \u03ba\u03ac\u03b8\u03b5\u03bc\u03b9\u03b1,\u03c4\u03cc\u03c4\u03b5 \u03b1\u03bd $n$ \u03ac\u03c1\u03c4\u03b9\u03bf\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 \u03c4\u03b1 $x$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1, \u03b5\u03bd\u03ce \u03b1\u03bd $n$ \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc\u03c2 \u03c4\u03cc\u03c4\u03b5 \u03b1\u03c6\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5 \u03c4\u03b1 $x$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03c4\u03b1 $x-1$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1.\r\n$2)$\u03a3\u03b5 \u03bf\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03c9\u03c0\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03cc\u03c0\u03c9\u03c2 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c3\u03c4\u03bf \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim (\u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9).\r\n\r\n[b][u]\u0391\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7[/u][/b]** (\u03cc\u03c4\u03b9 \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03bb\u03b5\u03b9\u03c4\u03bf\u03c5\u03c1\u03b3\u03b5\u03af)\r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $n$ \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2 \u03bc\u03b5 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03bc\u03b5 $a_{1},a_{2},...,a_{n}$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03bc\u03b9\u03ac \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1.\r\n\u03a3\u03ba\u03bf\u03c0\u03cc\u03c2 \u03bc\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf \u03bd\u03b1 \u03c0\u03b1\u03af\u03be\u03b5\u03b9 \u03c3\u03b5 \u03b8\u03ad\u03c3\u03b7 \u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $a_{1}XORa_{2}XOR...XORa_{n}=0$.\r\n\u03a3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03b5\u03b9 \u03c3\u03c4\u03b7 \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03bc\u03cc\u03bd\u03bf \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac \"\u03b5\u03be\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 (\u03b1\u03c0\u03cc \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim) \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $a'_{1}XORa'_{2}XOR...XORa'_{k}=0$ ,\u03cc\u03c0\u03bf\u03c5 $a'_{1},a'_{2},...,a'_{k}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03c3\u03c0\u03af\u03c1\u03c4\u03c9\u03bd \u03c0\u03bf\u03c5 \u03ad\u03c7\u03b5\u03b9 \u03b7 \u03ba\u03ac\u03b8\u03b5 \u03c3\u03b5\u03b9\u03c1\u03ac \u03bc\u03b5\u03c4\u03ac \u03c4\u03b7\u03bd \u03ba\u03af\u03bd\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03b1\u03bd\u03c4\u03b9\u03c0\u03ac\u03bb\u03bf\u03c5.\r\n\u039f\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bb\u03b5\u03b9\u03c4\u03bf\u03c5\u03c1\u03b3\u03b5\u03af, \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9 \u03bf\u03cd\u03c4\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b1\u03c6\u03b1\u03b9\u03c1\u03b5\u03af \u03bc\u03cc\u03bd\u03bf \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac \"\u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03ba\u03ac\u03b8\u03b5 \u03c3\u03b5\u03b9\u03c1\u03ac.\r\n[color=darkblue][u]\u03a0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7[/u]\n\u039a\u03ac\u03b8\u03b5 \u03c6\u03bf\u03c1\u03ac \u03c0\u03bf\u03c5 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03b8\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03b5\u03af \u03bc\u03cc\u03bd\u03bf \"\u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03c3\u03b5\u03b9\u03c1\u03ac ,\u03b8\u03b1 \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03b8\u03b5\u03af \u03b4\u03cd\u03bf \u03bd\u03ad\u03b5\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2, \u03b7 \u03bc\u03af\u03b1 \u03bc\u03b5 \u03cc\u03c3\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03ac \u03c4\u03c9\u03bd \u03c3\u03c0\u03af\u03c1\u03c4\u03c9\u03bd \u03c0\u03bf\u03c5 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03b8\u03b7\u03ba\u03b1\u03bd \u03ba\u03b1\u03b9 \u03b7 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03b5 \u03cc\u03c3\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b4\u03b5\u03be\u03b9\u03ac \u03c4\u03bf\u03c5\u03c2.[/color]\r\n\u0388\u03c3\u03c4\u03c9 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03cc\u03c4\u03b9 $a_{1}XORa_{2}XOR...XORa_{n}=0$ $(1)$ \u03ba\u03b1\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03b5\u03af \u03bc\u03cc\u03bd\u03bf \"\u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd $1$\u03b7 \u03c3\u03b5\u03b9\u03c1\u03ac. \u039b\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7\u03c2 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bd \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03b7\u03b8\u03b5\u03af \u03b4\u03cd\u03bf \u03bd\u03ad\u03b5\u03c2 \u03c3\u03b5\u03b9\u03c1\u03ad\u03c2, \u03b7 \u03bc\u03af\u03b1 \u03bc\u03b5 $x$ \u03ba\u03b1\u03b9 \u03b7 \u03ac\u03bb\u03bb\u03b7 \u03bc\u03b5 $y$ \u03c3\u03c0\u03af\u03c1\u03c4\u03b1. \u0398\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $a_{1}>x+y$ $(2)$ (\u03b1\u03c6\u03bf\u03cd \u03b1\u03bd\u03b1\u03b3\u03ba\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03b8\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1). \u0391\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $xXORyXORa_{2}XOR...XORa_{n}=0$ $(3)$. \r\n\u0388\u03c3\u03c4\u03c9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03cc\u03c4\u03b5 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 $(1)$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $a_{2}XOR...XORa_{n}=a_{1}$ \u03ba\u03b1\u03b9 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c3\u03c4\u03b7\u03bd $(3)$ \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 $xXORyXORa_{1}=0 \\Rightarrow xXORy=a_{1}$ \u03ba\u03b1\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 $(1)$ $xXORy>x+y$ (\u03ac\u03c4\u03bf\u03c0\u03bf, \u03b1\u03c6\u03bf\u03cd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $x+y\\geq xXORy$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $x,y \\in \\mathbb N$).\r\n\r\n\u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b1\u03bc\u03b5 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03cc\u03c4\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 \u03bd\u03b9\u03ba\u03b7\u03c6\u03cc\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03c3\u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b4\u03b9\u03ba\u03b1\u03af\u03c9\u03bc\u03b1 \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \"\u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c0\u03cc \u03bc\u03af\u03b1 \u03bc\u03cc\u03bd\u03bf \u03c3\u03b5\u03b9\u03c1\u03ac \u03ba\u03b1\u03b9 \u03ba\u03b5\u03c1\u03b4\u03af\u03b6\u03b5\u03b9 \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c0\u03c1\u03bf\u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 (\u03b1\u03c6\u03bf\u03cd \u03c0\u03bb\u03ad\u03bf\u03bd \u03b2\u03bb\u03ad\u03c0\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03b9 \u03b1\u03c5\u03c4\u03cc \u03b2\u03b1\u03c3\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03ba\u03bb\u03b1\u03c3\u03b9\u03ba\u03cc \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 Nim, \u03c7\u03c9\u03c1\u03af\u03c2 \u03b9\u03b4\u03b9\u03b1\u03af\u03c4\u03b5\u03c1\u03b5\u03c2 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ad\u03c2).\r\n\r\n\r\n*\u03a9\u03c2 \"\u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac\" \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03bc\u03b5 \u03b4\u03cd\u03bf \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b4\u03af\u03c0\u03bb\u03b1 \u03c3\u03c4\u03bf \u03ac\u03bb\u03bb\u03bf (\u03c7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b5\u03bc\u03b2\u03ac\u03bb\u03bb\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ac\u03bb\u03bb\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1) \u03ba\u03b1\u03b9 \u03b1\u03bd\u03ac\u03bc\u03b5\u03c3\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03b1\u03c5\u03c4\u03ac \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03c5\u03bc\u03ad\u03bd\u03c9\u03c2 \u03b1\u03c6\u03b1\u03b9\u03c1\u03b5\u03b8\u03b5\u03af \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1. \r\n\r\n**\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03ae \u03bc\u03bf\u03c5 \u03b8\u03b5\u03c9\u03c1\u03ce \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03bf \u03cc\u03c4\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c4\u03bf\u03c5 Nim, \u03c3\u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03bf \u03c0\u03b1\u03af\u03ba\u03c4\u03b7\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03b6\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03bf\u03c2 \u03c7\u03ac\u03bd\u03b5\u03b9, \u03ad\u03c7\u03b5\u03b9 \u03c3\u03b1\u03bd \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bd\u03af\u03ba\u03b7\u03c2 \u03c4\u03b7\u03bd \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03c0\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 (\u03bc\u03b9\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03bf \u039b\u03bf\u03c5\u03ba\u03ac\u03c2 \u03ad\u03b4\u03c9\u03c3\u03b5 \u03bc\u03b9\u03b1 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03c9\u03c2 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03af\u03c0\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b7 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03c3\u03c4\u03b7\u03bd \u03b3\u03b5\u03bd\u03b9\u03ba\u03ae \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae\u03c2 \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ae\u03c2). \u03a4\u03bf \u03bc\u03cc\u03bd\u03bf \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03bf\u03c5 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b7 \u03ba\u03b9 \u03b1\u03bd \u03c0\u03c1\u03bf\u03c3\u03c4\u03b5\u03b8\u03b5\u03af \u03c3\u03c4\u03bf \u03c0\u03b1\u03b9\u03c7\u03bd\u03af\u03b4\u03b9 \u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 \"\u03c4\u03b1 \u03c3\u03c0\u03af\u03c1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03c3\u03b2\u03ae\u03bd\u03bf\u03c5\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b4\u03bf\u03c7\u03b9\u03ba\u03ac\" \u03c4\u03b1 \u03b4\u03b5\u03b4\u03bf\u03bc\u03ad\u03bd\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bd\u03b9\u03ba\u03b7\u03c6\u03cc\u03c1\u03b1 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03b4\u03b5\u03bd \u03b1\u03bb\u03bb\u03ac\u03b6\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 \u03b7 \u03c3\u03c4\u03c1\u03b1\u03c4\u03b7\u03b3\u03b9\u03ba\u03ae \u03bc\u03b1\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bc\u03ad\u03bd\u03b5\u03b9 \u03b7 \u03af\u03b4\u03b9\u03b1.\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2...", "Solution_14": "\u0386\u03c8\u03bf\u03b3\u03b1! :) \r\n[quote=\"stedes\"]$... \\Rightarrow x\\text{XOR}y=a_{1}$ \u03ba\u03b1\u03b9 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 $(1)$ $xXORy>x+y$[/quote](\u03c0\u03c1\u03bf\u03c2 \u03b1\u03c0\u03bf\u03c6\u03c5\u03b3\u03ae \u03c0\u03b1\u03c1\u03b1\u03be\u03b7\u03b3\u03ae\u03c3\u03b5\u03c9\u03bd, \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 $(2)$\r\n\r\n\r\n\u0398\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03bf \u03b1\u03bd\u03c4\u03af\u03c0\u03b1\u03bb\u03bf\u03c2 \u03b4\u03b5 \u03b8\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 $(a_{1},a_{2}, ..., a_{n})$ \u03bd\u03b1 \u03c0\u03ac\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \r\n$a_{i}=1\\ \\forall i$ \u03bc\u03b5 $\\sum{a_{i}}$ \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc. \r\n\r\n\u0391\u03c5\u03c4\u03cc \u03cc\u03bc\u03c9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf, \u03b1\u03c6\u03bf\u03cd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c0\u03ac\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 $(1,1,...,1)$ \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c7\u03c9\u03c1\u03b9\u03c3\u03bc\u03cc, \u03bf \u03bc\u03cc\u03bd\u03bf\u03c2 \u03c4\u03c1\u03cc\u03c0\u03bf\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 $(x,1,...,1)$ \u03cc\u03c0\u03bf\u03c5 $x>2$\r\n\r\n\u039b\u03ad\u03b3\u03bf\u03bd\u03c4\u03b1\u03c2 \"\u03b4\u03b9\u03b1\u03c7\u03c9\u03c1\u03b9\u03c3\u03bc\u03cc\" \u03b5\u03bd\u03bd\u03bf\u03ce \u03bd\u03b1 \u03c3\u03c0\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7 x-\u03ac\u03b4\u03b1 \u03c3\u03b5 \u03b4\u03cd\u03bf \u03bc\u03bf\u03bd\u03ac\u03b4\u03b5\u03c2, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bd\u03b1 \u03b1\u03c6\u03b1\u03b9\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac.\r\n\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 \u03cc\u03bc\u03c9\u03c2 \u03cc\u03c4\u03b9 \u03b7 \u03ba\u03b1\u03c4\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 $(x,1,...,1)$ \u03b4\u03b5\u03bd \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af \u03c4\u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 $x\\otimes 1 \\otimes...\\otimes 1=0$ \u03b1\u03c6\u03bf\u03cd \u03c4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 $x<2$", "Solution_15": "\u039d\u03b1\u03b9, \u03b5\u03bd\u03bd\u03bf\u03cd\u03c3\u03b1 \u03bb\u03cc\u03b3\u03c9 \u03c4\u03b7\u03c2 (2) (...\u03c8\u03b9\u03bb\u03cc \u03c4\u03bf \u03ba\u03b1\u03ba\u03cc). :P \r\n\r\nEdit: \u03a7\u03b1,\u03b5\u03af\u03c3\u03b1\u03b9 \u03b3\u03c1\u03ae\u03b3\u03bf\u03c1\u03bf\u03c2! \u03a0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b5\u03c2 \u03c4\u03bf x=0 \u03c0\u03c1\u03b9\u03bd \u03c0\u03c1\u03bf\u03bb\u03ac\u03b2\u03c9 \u03bd\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03c9. :lol:", "Solution_16": ":) \r\n\u0395\u03bd \u03c4\u03c9 \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03b4\u03b5\u03bd \u03c0\u03c1\u03bf\u03bb\u03ac\u03b2\u03b1\u03b9\u03bd\u03b1 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03ce \u03b1\u03bd \u03c4\u03bf \u03c3\u03c9\u03c3\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 0 \u03ae 1 \u03ba\u03b9 \u03ad\u03c4\u03c3\u03b9 \u03ad\u03b2\u03b1\u03bb\u03b1 $x<2$ :D\r\n\r\n\u0395\u03bd\u03c4\u03ac\u03be\u03b5\u03b9, \u03b1\u03c6\u03bf\u03cd \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03bf \u03c4\u03ad\u03bb\u03bf\u03c2 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $2k+1$ \u03bc\u03bf\u03bd\u03ac\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03bf\u03b9 \u03b4\u03cd\u03bf \u03b1\u03c0\u0384\u03b1\u03c5\u03c4\u03ad\u03c2 \u03c0\u03c1\u03bf\u03ad\u03ba\u03c5\u03c8\u03b1\u03bd \u03b1\u03c0\u03cc \u03c4\u03bf $x$, \u03c3\u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 $(x,1,...,1)$ \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 $2k-1$ \u03bc\u03bf\u03bd\u03ac\u03b4\u03b5\u03c2, \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b4\u03af\u03bd\u03bf\u03c5\u03bd XOR \u03af\u03c3\u03bf \u03bc\u03b5 1", "Solution_17": ":lol: \r\n\u03a4\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03c4\u03c9\u03bd \u03bc\u03bf\u03bd\u03ac\u03b4\u03c9\u03bd \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03cc \u03bc\u03b5\u03c4\u03ac \u03c4\u03b7 \u03b4\u03b9\u03ac\u03c3\u03c0\u03b1\u03c3\u03b7, \u03ac\u03c1\u03b1 \u03c0\u03c1\u03b9\u03bd \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ae \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ac\u03c1\u03c4\u03b9\u03bf \u03c0\u03bb\u03ae\u03b8\u03bf\u03c2 \u03cc\u03c1\u03c9\u03bd \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 $x=1$.\r\n\r\nEdit: \u03a0\u03ac\u03bb\u03b9 \u03bc\u03b5 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5\u03c2...! :|" } { "Tag": [], "Problem": "The Fibonacci sequence is $F_1 = F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \\ge 3$, while the \"Galileo numbers\" satisfy $G_n = G_{n-1} + G_{n-2} + 1$ with $G_1 = G_2 = 1$. If $F_{203} = a$ and $F_{204} = b$, find $G_{205}$ in terms of $a$ and $b$.", "Solution_1": "[hide=\"solution\"]\n\nlisting out the Fibonacci sequence:\n\n1,1,2,3,5,8,13, ..\n\nlisting out the \"Galileo sequence\":\n\n1,1,3,5,9,15,....\n\nafter generating and testing out many formulas, we find that $\\displaystyle G_n = F_{n-1} + F_n + (n-4)$\n\nso, $\\displaystyle G_{205} = F_{204} + F_{205} + 201$\n$\\displaystyle = F_{204} + (F_{203} + F_{204}) + 201$\n$\\displaystyle = b + (a+b) + 201$\n$\\displaystyle = a + 2b + 201$\n[/hide]" } { "Tag": [ "algebra", "polynomial" ], "Problem": "The equation x^5-6x^4+13x^3-14x^2+12x-8=0 has\r\n\r\na. all real roots\r\nb. all complex, non-real roots\r\nc. two complex, non-real roots\r\nd. two different positive real roots\r\ne. two differnet negative roots\r\n\r\nuh.. I found this question on an old Indiana math contest (ICTM) and I was wondering is there a shortcut to this..because Im kinda stuck (like rules for the roots). Thanks", "Solution_1": "We can try to factorize it, we find that\r\n$x^{5}-6x^{4}+13x^{3}-14x^{2}+12x-8=(x-2)(x^{4}-4x^{3}+5x^{2}-4x+4)=$\r\n$(x-2)^{2}(x^{3}-2x^{2}+x-2)=(x-2)^{3}(x^{2}+1).$\r\n\r\nTherefore the roots are $2$, $i$, and $-i$, so the polynomial has two complex roots.", "Solution_2": "thanks\r\nhow long did you take to factor that because i could never factor that..", "Solution_3": "A polynomial of degree 5 must have at least one real root, so you can eliminate choice (B). \r\n\r\nUse Descartes' Rule of Signs to eliminate choices (D) and (E).", "Solution_4": "you can consider a polynom\r\n\r\nP(x)=x^5-6x^4+13x^3-14x^2+12x-8\r\n\r\nand study its variation\r\ntherefor you can find the number of real roots\r\n\r\nand the rest roots are in C (fundamental theorem of algebra.)" } { "Tag": [ "absolute value", "number theory proposed", "number theory" ], "Problem": "Find all pairs of positive integers $ (a,b)$ such that the set of positive integers cannot be partitioned into three pairwise disjoint nonempty subsets such that none of them has two numbers with absolute value of their difference equal to $ a$ or $ b$ or $ a\\plus{}b$", "Solution_1": "Anybody solvve it?" } { "Tag": [ "function", "LaTeX", "algebra unsolved", "algebra" ], "Problem": "Find all functions f:R+ to R+ such that :\r\n(x+y)f(f(x)y)=x^2f(f(x)+f(y)) for all x,y>0.", "Solution_1": "[quote=\"hasan\"]Find all functions f:R+ to R+ such that :\n(x+y)f(f(x)y)=x^2f(f(x)+f(y)) for all x,y>0.[/quote]\n\nI think it is better this way:\n\n[quote]Find all functions $f: R_{+}\\to R_{+}$ such that :\n$(x+y)f(f(x)y)=x^{2}f(f(x)+f(y))$ for all $x,y>0$.[/quote]\r\n\r\nP.S. Learn $LaTeX$! :wink:", "Solution_2": "(*) $(x+y)f(f(x)y)=x^{2}f(f(x)+f(y))$\r\n\r\nIf $f(x)=f(z)$ for $x,z\\in\\mathbb{R}_{+}$, then \r\n${x^{2}\\over x+y}={f(f(x)y)\\over f(f(x)+f(y))}={f(f(z)y)\\over f(f(z)+f(y))}={z^{2}\\over z+y}$ for all $y\\in\\mathbb{R}_{+}$\r\nand $y\\to 0$ gives $x=z$.\r\nSo $f$ is injective, and then $x=\\lambda: ={1+\\sqrt{5}\\over 2}$ and $y=1$ in (*) gives\r\n$(\\lambda+1)f(f(\\lambda))=\\lambda^{2}f(f(\\lambda)+f(1))$ with $\\lambda^{2}=\\lambda+1\\not =0$.\r\nSo $f(f(\\lambda))=f(f(\\lambda)+f(1))$ and $f(\\lambda)=f(\\lambda)+f(1)$ and $f(1)=0$. Contradiction!\r\nSo there is no solution of (*)." } { "Tag": [], "Problem": "I was wondering how many AoPSers play the game. Here are the rules:\r\n\r\nRULE 1: Everyone is playing The Game, and everyone is always playing The Game. \r\nRULE 2: Whenever you remember that you are playing The Game, or think about The Game, you lose. \r\nRULE 3: Loss must be announced.\r\n\r\nBy the way, you just lost The Game!", "Solution_1": "I like cheese, cheesy cheese. I'm just think about cheese. Cheese chees cheese. DONT THINK ABOUT ANYTHING ELSE BESIDES CHEESE MEW! DONT DO IT!", "Solution_2": "i like pie. pie is good. pie pie pie pie pie. all i think about is pie. bluebery pie, apple pie, creme pie, blackberry pie, chess pie, pie pie pie pie. i refuse to think about anything other than pie. pie pie pie pie pie.", "Solution_3": "darn it i just lost the game. ahhhhh i forgot about it for the WHOLE SUMMER!!!! UR SO MEAN!!!!!!!!! me cries. oh and u have 5 mins after starting to play the game to forget you're playing the game (or at least that's how i play it)", "Solution_4": "I LOSEEEEEEEEEEEEEEEEEEEEEE\r\n\r\nit's 30 minutes.\r\nit's a 30 minute grace period.\r\nthere is a way you can win.\r\nit's if you die within the 30 minute period or you die but you don't think of the game.\r\nthere are people who have won the game.\r\n\r\ni learned about THE GAME at CTY.\r\nCTY = amazing.", "Solution_5": "Everyone has a different grace period. I prefer 30 seconds. It's funner. And even if you remember that you are playing in the grace period, you don't win. The game was designed to be impossible to win.", "Solution_6": "Donuts...I love donuts, donuts are amazing why doesnt anyone love donuts as much as me. I REFUSE TO THINK ABOUT ANYTHING BESIDES DONUTS!!!!!!!!!!ARGHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH", "Solution_7": "Hmm.\r\n\r\ni think this was an old one but still funny now this topic is on.\r\n\r\n[img]http://imgs.xkcd.com/comics/anti_mind_virus.png[/img]", "Solution_8": "MOO MOO MOO MOO MOO MOO MOO MOO MOO, FOO! ME MOO TO THE MOON! ME MOOS TO MOOSIC!\r\n\r\nI thought that you won if you said moo without thinking of the game while saying it.", "Solution_9": "Rule 1 makes the poll unnecessary." } { "Tag": [], "Problem": "This is '[b]Teh Challenge[/b]' modded by yours truly,\r\n I wanted to try this type of thing. \r\nI got an inspiration from a whole bunch of games on this forum, (The Fight Against the Dark Lord, Beat the Mod, Big Brother, Mafia.... etc.)\r\nSooo.. like I'll explain the game. \r\n\r\nThere will be [b][i][u]TEAMS[/u][/i][/b] (guess where I got this from)\r\nGot that?\r\nI'll be the one to assign them :)\r\nI will send you a very, very detailed PM when you sign up.\r\nIt will contain what team your on, and your [b][i][u]role[/u][/i][/b] (guess where I got that from)\r\na quicktopic for each team, where you and your teammates will discuss strategies and such. \r\n\r\nExample:\r\n\r\nHELLO! [insert username here]\r\nThank you for signing up for my game!\r\nNow to get on with the PM stuff... eh...\r\nYou are \"[insert role here]\", you [details of role here]\r\nand you are on team [insert team name here] and your quicktopic is quicktopic.com/[insert unique URL here]\r\nIf you don't have an account I suggest you make one.\r\nYour team members are;\r\n[insert team members here]\r\nGood Luck! Be ready for a whole bunch of surprises :)\r\n\r\nEND EXAMPLE\r\n\r\nSo like that.\r\nThere will be a challenge, like... when I get the whole like PM stuff out. (guessy, guessy...)\r\nEach challenge will last at a maximum of [b]4 days[/b].\r\nEach team will have a leader that will send their PM to MOI! (if leader absent tell MOI!)\r\nIf a team sends in their PM late, or not at at all they will automatically be up for eviction/elimination.\r\nGot that?\r\nIf ALL or more than 1 team get's there PM late, I will first then ask if team is active, and/or use the power of random.org.\r\n(I might add extension dates, we'll see)\r\nThen the teammates will send me a PM on who they want to eliminate, blah blah blah..\r\nand stuff.\r\n\r\nNOW ON WITH [size=150]TEH [color=orange]RULEZ![/color][/size]\r\n\r\n[b]Do not reveal your role (to players/teammates) unless I tell you too.\nDo not complain if you have been 'bye-bye''d PMing me\nDo not complain if challenge is hard and/or too meewhee-ish PMing me\nDo NOT keep contact with other players (if not on team) if discussing this game.\nBe active as much as possible.\nJust so I know you have read the rules, say \"Rattatatat\" in your sign-up post.\n[u]I trust you people, do not break that trust[/u]\n[/b]\r\nMore rules will be added on later if needed.\r\nSo YAY! Now we begin sign-ups \r\n\r\n1. Chompy ( the bored person) \r\n2. KING1\r\n3. vahalla\r\n4. jjx1\r\n5. ernie\r\n6. jjfun1\r\n7. Bugi\r\n8. Westiepaw\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.\r\n15.\r\n16.", "Solution_1": "bored says raattatatat ( :| wait...boredom talks :huh: )...and also that:\r\n\r\n1. Chompy ( the bored person)\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.\r\n15.\r\n16.", "Solution_2": "Ha nice. Boredom is very smart.\r\n:)\r\nEDIT: Changed Name to attract more attention :D", "Solution_3": "the title of this thread made me lose the game.", "Solution_4": "Hmm, it attracts attention though :D\r\nWould you like to join?", "Solution_5": "Sounds fun! I join! Yay! :)\r\n\r\n\r\nEDIT: lol meewhee you are hilarious :P Rattatatat", "Solution_6": "Eh hem?\r\nYour missing something...", "Solution_7": "I join.\r\n\r\nI'm not going to say rattatat.... oh wait I just did.", "Solution_8": "Do, do, do...\r\nHumm... 13 spots left :P\r\nmaybe I'll reduce the size to 12... We'll see...", "Solution_9": "I\u00b4ll join", "Solution_10": "Eh hem....\r\nYou forgot something... Re-read the rules :)", "Solution_11": "I join. Rattatatat", "Solution_12": "I join. [b]\"Rattatatat\" in your sign-up post.[/b]", "Solution_13": "Mind if I change the title a bit? (first it makes me lose the game but that barely matters. Second, the title might describe the game a bit better)", "Solution_14": "Sure dragon :),\r\nI can't like.. think of a name. :( \r\nSo I keep having massive name changes.", "Solution_15": "Guys, there's a reason there's a QT. \r\nPlease check it often and post!!!\r\nThanks.", "Solution_16": "Um, its past 4:00 PM PST.", "Solution_17": "Yeah. Results posted ASAMGTHMC (As soon as meewhee gets to her main computer :P)", "Solution_18": "My group is dead.\r\nBleh.", "Solution_19": "I can put you in a new group if members inactive... he. he. he. he.", "Solution_20": "The game is dead now.\r\nBleh.", "Solution_21": "I is active. :)", "Solution_22": "When are the results gonna be posted?", "Solution_23": "Am I considered inactive now if i haven't been on for 2 days? :(", "Solution_24": "I don't think so.", "Solution_25": "[quote=\"polkadotman1\"]Am I considered inactive now if i haven't been on for 2 days? :([/quote]\r\n\r\nYes.\r\nSigh.\r\nREQUEST REPLACEMENTS.....", "Solution_26": "Go PM meewhee.", "Solution_27": "*sigh*\r\n\r\n\r\nthis topic will not survive much longer.\r\n\r\n\r\nI might be inactive for a LONG period of time so.....\r\n\r\n\r\n$ \\text{\\tiny request replacement}$", "Solution_28": "Oh sorry. I'm in Canada soo... I can like post the results when I get home tomorrow. SORRY FOR INACTIVITY!", "Solution_29": "cool!\r\n\r\n\r\nbut i still have to $ \\text{\\tiny\\color{yellow}request replacement}$" } { "Tag": [ "counting", "distinguishability" ], "Problem": "How many different positive, four-digit integers can be formed\nusing the digits 2, 2, 9 and 9?", "Solution_1": "Since there are $ 4$ digits, and there are $ 2$ pairs of digits that are indistinguishable, \r\n$ \\frac{4!}{2!2!}\\equal{}\\boxed{6}$.", "Solution_2": "[hide=\"ANSWER\"]1.we know the following facts\n\n I.there are the #'s 2,2,9,9\n\n II.there are 4! ways to arrange them \n\n III.2 pairs of 2 #'s are the same \n\n2.so we get the equation \n\n 4!/2!2!=6\n\n3.[hide]6[/hide][/hide]\n\n[= I HOPE IT'S EXPLAINED WELL !" } { "Tag": [ "pigeonhole principle", "induction" ], "Problem": "There is a finite amount of people at a party ($n>1$). Show that there are at least two people who know the same number of people. (Here we take it to mean that \"knowing\" is a mutal relationship).\r\n\r\nI think I solved the problem. But my problem is that when my book did examples involving the Pigeonhole Principle it divided the problem into Pigeons and Holes. It explicity stated what the Pigeons are and what the Holes are. I managed to solve it I just dont know how to explicity state what the Pigeons and Holes are.\r\n\r\n[hide]\nI drew a picture to help visualize the situation. Say we have $P_{1},P_{2},P_{3},P_{4}$ people. Draw lines between them any way you like (with no arrows because knowing is mutal). Then at each vertex place a number between $0,n-1$ which is the number of people he knows. And do that at each vertex. Now divide the problem into 2 cases:\n\n1)If everyone knows at least one person, i.e. nobody is isolated in the graph. Then we are assigning $n-1$ numbers, i.e. $\\{1,2,...,n-1\\}$ among $n$ people and so by Pigeonhole Principle says at least two know each other.\n\n2)If exactly one person doesnt know anybody (if exactly 2 or more then proof is ocomplete) i.e. we have an isolated point. Since knowing it mutal that point has nothing to do with the graph, i.e. the knowing of people amongst themselves. So we can as if remove it from the picture. Now we have $n-1$ people which by induction (which I didnt state in the begining that this is an iduction proof bu you get the idea) is true.\n\nSo in both cases it works\n[/hide]\r\n\r\nDoes it help a problem solver to know explicity state what the holes and pigeons are? Because I dont know how to put it that way.", "Solution_1": "[hide=\"Explicit statements\"] Assume WLOG that every person knows at least one person, or else they are irrelevant. The $n$ people are the pigeons. The $n-1$ holes are for people who know $1, 2, 3, ... n-1$ people, respectively (since it is not possible to know more than $n-1$ people at a party with $n$ people). Since there are more pigeons than holes, at least one hole contains two pigeons. Then at least two people know the same number of people. [/hide]", "Solution_2": "Your proof is alright. \r\n\r\nt0rajir0u already explained the pigeonhole stuff.\r\n\r\nOne problem with the proof...no base case.", "Solution_3": "Let position numbers $0,1,\\hdots, n-1$ represent the $n$ positions for the number of acquaintances. Clearly, positions $0,n-1$ cannot exist simultaneously. Therefore, we can have a maximum of $n-1$ positions. Since we have $n$ people, there will exist at least one position with at least $2$ people, as required." } { "Tag": [ "algebra", "polynomial", "quadratics", "linear algebra", "matrix" ], "Problem": "Can anyone help me with this???\r\n\r\nThe Question is:-\r\n\r\nLet A be the mx [0 0 0 -1\r\n 0 1 sqrt(2) 0\r\n 0 sprt(2) 1 0\r\n -1 0 0 -1]\r\n[b]Find all EigenValues of A? For one of these Eigenvalues, Find a corres-unit eigenvactor?[/b]\r\n\r\nthe hint is:- the charactristic polynomial is the product of 2 quadratics. so shuffle the rows and columns first and find the produdt of 2 quadratics and expand by cofactors.", "Solution_1": "maybe this will help you\r\nchange the order of your basis\r\nso now you have the matrix with as basis e1,e2,e3,e4\r\n\r\nnow use the basis e1,e4,e2,e3\r\nnow you get two nice blocks\r\n\r\nwhen now calculating the charpoly, remember this fact about block matrices\r\nA 0\r\n0 B\r\n\r\nhas determinant = det(A) *det(B)" } { "Tag": [ "\\/closed" ], "Problem": "Something I noticed during the Math Jams is that sometimes I need to scroll up to find the original problem and it gets tiresome sometimes. Is there a way we could add a little box inside the virtual classroom to display the problem?", "Solution_1": "I agree. The problem can pop-up or something, if it can that is.", "Solution_2": "We often put problems as images during the classes and Math Jams; we'll be more assiduous about this in the future (the images appear in a browser window, which you can flip to rather than scrolling to the question).", "Solution_3": "Thats right...I forgot about that. thanks" } { "Tag": [ "geometry", "rectangle" ], "Problem": "All angles are right angles. How many rectangles are in this diagram?\n[asy]for(int i = 0; i < 4; ++i){\ndraw((0,(i)*(i+1)/2)--(-7,(i)*(i+1)/2));\ndraw((1-2^i,0)--(1-2^i,6));\n}[/asy]", "Solution_1": "Hmmm... I got $ \\boxed{39}$...", "Solution_2": "We can see that two points on the left and two points on the top will determine a unique rectangle. Thus, our answer is $ \\dbinom42\\dbinom42\\equal{}6\\cdot6\\equal{}\\boxed{36}$.", "Solution_3": "Oh, I remember what I did. I counted the vertical $ 1 \\times 3$ rectangles twice.", "Solution_4": "Or you can count how many of each size rectangle there is\n1x1-9\n1x2-12\n1x3-6\n2x2-4\n2x3-4\n3x3-1\nAdd them all together and you get 36" } { "Tag": [ "summer program", "Mathcamp", "inequalities unsolved", "inequalities" ], "Problem": "If $ a,b,c$ are 3 sides of acute triangle. Prove that:\r\n\r\n$ (a\\plus{}b\\plus{}c)(a^{2}\\plus{}b^{2}\\plus{}c^{2})(a^{3}\\plus{}b^{3}\\plus{}c^{3})\\ge 4(a^{6}\\plus{}b^{6}\\plus{}c^{6})$\r\n :)", "Solution_1": "we have $ x \\equal{} a\\plus{}b\\plus{}c\\geq 2a,2b,2c$ and $ y \\equal{} a^{2}\\plus{}b^{2}\\plus{}c^{2}\\geq 2a^{2},2b^{2},2c^{2}$ because they are sides of an acute triangle.\r\nso $ (a^{3}\\plus{}b^{3}\\plus{}c^{3})xy \\equal{} a^{3}xy\\plus{}b^{3}xy\\plus{}c^{3}xy > 4(a^{6}\\plus{}b^{6}\\plus{}c^{6})$ :wink:" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello, \r\n\r\nthis problem should be an easy exercise but I'm afraid I'm missing something :blush: \r\n\r\nLet $ A_n$ be the alternating group on $ \\{1,\\ldots,n\\}$\r\nLet $ \\Omega$ be the set of non-ordered triples in that last set.\r\n\r\nConsider the action of $ A_n$ on $ \\Omega$. Show that it is primitive.\r\n\r\nNow the stabilizer of a random triple is $ A_n\\cap S_3\\times S_{n\\minus{}3}$, so maybe I should try to prove that this is a maximal subgroup of $ A_n$, but that doesn't seem to work either. :huh: \r\n\r\nCan anyone help? Thanks!", "Solution_1": "You can map a pair of triples with exactly $ k$ common elements ($ 0\\le k\\le 2$) into $ \\{1,2,3\\}$ and $ \\{4 \\minus{} k,5 \\minus{} k,6 \\minus{} k\\}$ (e.g. do it by an element of $ S_n$ and then tweak it to make it even). So, if a nontrivial equivalence relation is preserved, then there is $ k\\in\\{0,1,2\\}$ such that any two triples with exactly $ k$ common elements are equivalent. But then all triples are equivalent.\r\n\r\n(You may need to consider small $ n$ separately. In fact, the statement is false for $ n\\equal{}6$.)" } { "Tag": [ "probability", "number theory", "relatively prime" ], "Problem": "1. How many subsets of $\\left\\{1, 2, 3, 4, 5, 6, 7\\right\\}$ are there such that the sum of their elements is odd? \r\n\r\n2. What is the sum of the numbers less than $1000$ and relatively prime with it?\r\n\r\n3. The flu can be avoided at 95% taking the vaccine or, indipendentely at 25% staying in the country.\r\n What is the probability to avoid the flu if I live in the country and I take the vaccine?", "Solution_1": "[hide=\"Problem 1\"]\nThere are 4 odds and 3 evens. We now ask the question, how many ways are there of choosing odd subsets?\n\nso either we choose 3 ones or we choose 1 one. There are 4 ways of choosing 3 ones and 4 ways of choosign 1 one.\n\nWe can either take a zero or not take a zero: $2^3$ combinations.\n\nTherefore there are $2(4 \\times 2^3)=64$ combinations.[/hide]\r\n\r\n[color=green][size=75]mod edit: hide your solutions[/size][/color]", "Solution_2": "[hide=\"Problem 2\"]We can find the sum of the numbers that share at least one common factor. \n\n2: $\\sum_{n=1}^{499} 2n = 499 \\cdot 500 = 249500$\n5: $\\sum_{n=1}^{199} 5n = \\frac{199(5+995)}{2}=99500$\n10: $\\sum_{n=1}^{99} 10n = \\frac{99(10+990)}{2}=100 \\cdot 1000=49500$ (both)\n\nSo total: $249500+99500-49500=299500$. The total sum of all numbers $<999$ is $\\frac{(1+999)999}{2}=500 \\cdot 999 = 499500$. The difference: $200000$.[/hide]" } { "Tag": [ "linear algebra", "superior algebra", "superior algebra theorems" ], "Problem": "I am a high school studnet interested in studying advanced algebra. I recently read about Maclane's authoritative book \" A survey of modern Algebra\" . I also came across\" Alegbra \" by the same author. Which one covers more topics and which one is more comprehensive and better to comprehend? Also, can you recommend other such books in Advanced Algebra???", "Solution_1": "Do you know any linear algebra or basic group/ring/field theory? If the answer to that is no, then you probably want an easier textbook to begin with. I don't own it, but I've glanced through Herstein's Abstract Algebra a couple of times, and it seems nice for an introductory book (and the exercises are ordered from easy to hard, which is also good). It may be suitable for your needs." } { "Tag": [ "AMC" ], "Problem": "If anyone has a scanned copy of either, could you send it to me so I can type it up for the Contest section? I'd like to take it, too.", "Solution_1": "Here it is, in a word document.\r\n\r\n[i]Edited by admin -- once again, these documents are [b]copyrighted[/b]. Please don't post them.[/i]", "Solution_2": "darn i trashed my 10B\r\n\r\nI have a scanner which i use frequently HUZZAQHHH!!! OH YEAH SPECTACULAR!!!! \r\n\r\n\r\nbut i wasnt happy with my score :rotfl:", "Solution_3": "[quote=\"Klebian\"]Here it is, in a word document.\n\n[i]Edited by admin -- once again, these documents are [b]copyrighted[/b]. Please don't post them.[/i][/quote]\r\nMy apologies." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "geometry solved" ], "Problem": "in triangle ABC, we have $\\angle BAC=90^{\\circ}$. and D is a point on side BC, in wich $\\angle BDA=2\\angle BAD$.\r\nprove that: \r\n\r\n$\\frac{1}{AD}=\\frac12\\left(\\frac{1}{BD}+\\frac{1}{CD}\\right)$", "Solution_1": "Let's handle the problem with your favorite transformation. ;)\r\n\r\n[color=blue][b]Problem.[/b] Let ABC be a right-angled triangle with < CAB = 90\u00b0, and let D be a point on the side BC such that < BDA = 2 < BAD. Prove that $\\frac{1}{DA}=\\frac12\\cdot\\left(\\frac{1}{DB}+\\frac{1}{DC}\\right)$.[/color]\r\n\r\n[i]Solution.[/i] Let r be an arbitrary length, and let the inversion with respect to the circle with center D and radius r map the points A, B, C to the points A', B', C', respectively. Then, $DA\\cdot DA^{\\prime}=DB\\cdot DB^{\\prime}=DC\\cdot DC^{\\prime}=r^2$. Thus, $\\frac{1}{DA} : \\frac{1}{DB} : \\frac{1}{DC} = DA^{\\prime} : DB^{\\prime} : DC^{\\prime}$, and thus, the equation that we have to prove, $\\frac{1}{DA}=\\frac12\\cdot\\left(\\frac{1}{DB}+\\frac{1}{DC}\\right)$, is equivalent to $DA^{\\prime}=\\frac12\\cdot\\left(DB^{\\prime}+DC^{\\prime}\\right)$. Since obviously DB' + DC' = B'C', this becomes $DA^{\\prime}=\\frac12\\cdot B^{\\prime}C^{\\prime}$.\r\n\r\nSince $DA\\cdot DA^{\\prime}=DB\\cdot DB^{\\prime}$, we have DA : DB' = DB : DA'. Also, < ADB = < B'DA'. Thus, the triangles ADB and B'DA' are similar, so that < B'A'D = < ABD. Similarly, < C'A'D = < ACD. Hence, < C'A'B' = < C'A'D + < B'A'D = < ACD + < ABD = < ACB + < ABC. But since the triangle ABC is right-angled with < CAB = 90\u00b0, we have < ACB + < ABC = 90\u00b0, so that < C'A'B' = 90\u00b0, and thus the triangle A'B'C' is right-angled at A'. Hence, the midpoint M of its hypotenuse B'C' is simultaneously its circumcenter, and its circumradius MA' = MB' = MC' equals to the half of its hypotenuse B'C'. So we have $MA^{\\prime}=\\frac12\\cdot B^{\\prime}C^{\\prime}$.\r\n\r\nSince M is the circumcenter of triangle A'B'C', the central angle theorem yields < A'MC' = 2 < A'B'C'. But < A'B'C' = < A'B'D, and since the triangles ADB and B'DA' are similar, < A'B'D = < BAD. Hence, < A'MC' = 2 < A'B'C' = 2 < A'B'D = 2 < BAD = < BDA. In other words, < A'MD = < A'DM. Thus, the triangle MA'D is isosceles, so that DA' = MA'. Hence, the equation $MA^{\\prime}=\\frac12\\cdot B^{\\prime}C^{\\prime}$ becomes $DA^{\\prime}=\\frac12\\cdot B^{\\prime}C^{\\prime}$, and we are done. $\\blacksquare$\r\n\r\n darij", "Solution_2": "thanks alot dear darij :D", "Solution_3": "[b]I note [/b]$m(\\angle BAD)=x\\in (0^{\\circ},90^{\\circ})$. [b]Thus,[/b]\r\n\r\n$\\frac {2}{DA}=\\frac {1}{DB}+\\frac {1}{DC}\\Longleftrightarrow \\frac {DA}{DB}+\\frac {DA}{DC}=2\\Longleftrightarrow 2=\\frac {\\sin {(180^{\\circ}-3x)}}{\\sin x}+\\frac {\\sin ({3x-90^{\\circ})}}{\\sin ({90^{\\circ}-x)}}\\Longleftrightarrow$\r\n\r\n$\\sin {3x}\\cdot \\cos x -\\cos {3x}\\cdot \\sin x=\\sin {2x}\\Longleftrightarrow \\sin {(3x-x)}=\\sin {2x}.$", "Solution_4": "excellent dear levi ;)", "Solution_5": "[u]Remark.[/u] A general enunciation for the proposed problem:\r\n\r\n$\\triangle ABC,\\ w=C(O,R),\\ D\\in BC,\\ \\{A,A'\\}\\subset AD\\cap w,\\ m(\\angle ADB)=2(\\angle BAD)\\Longrightarrow$\r\n\r\n$\\underline {\\overline {\\left| \\frac{2}{DA}=\\frac{1}{DB}+\\frac{1}{DC} \\Longleftrightarrow DA'=\\frac 12 BC\\Longleftrightarrow A=90^{\\circ}\\right| }}$\r\n\r\n[b]It is very nice ![/b]", "Solution_6": "[quote=\"Ph-An\"][u]Remark.[/u] A general enunciation for the proposed problem:\n\n$\\triangle ABC,\\ w=C(O,R),\\ D\\in BC,\\ \\{A,A'\\}\\subset AD\\cap w,\\ m(\\angle ADB)=2(\\angle BAD)\\Longrightarrow$\n\n$\\underline {\\overline {\\left| \\frac{2}{DA}=\\frac{1}{DB}+\\frac{1}{DC} \\Longleftrightarrow DA'=\\frac 12 BC\\Longleftrightarrow A=90^{\\circ}\\right| }}$\n\n[b]It is very nice ![/b][/quote] \r\n\r\nThis is a problem \"IFF\"." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If a, b, c are three positive numbers with a + b + c = 3, then prove the inequality\r\n\r\n$\\frac{1+b+c^2}{1+c+b^2}+\\frac{1+c+a^2}{1+a+c^2}+\\frac{1+a+b^2}{1+b+a^2}\\geq 3$.\r\n\r\nUsing computer algebra, I have verified that it is true, but a nice proof is missing...\r\n\r\n Darij", "Solution_1": "So two and a half weeks past, anyone came up with a solution?\r\n\r\nI have been hesitating of multiplying and expanding everything--it's going to be like a 40-minute-boring-algebra practice. I am really not looking forward to it, does someone have a solution yet?", "Solution_2": "[quote=\"darij grinberg\"]If a, b, c are three positive numbers with a + b + c = 3, then prove the inequality\n\n\n\n$\\frac{1+b+c^2}{1+c+b^2}+\\frac{1+c+a^2}{1+a+c^2}+\\frac{1+a+b^2}{1+b+a^2}\\geq 3$.\n\n\n\nUsing computer algebra, I have verified that it is true, but a nice proof is missing...\n\n\n\n Darij[/quote]\n\nDear Darij, I have pursued some further investigations concerning this inequality, interestingly, the inequality is true for all positive $a,b,c$ such that $a + b + c \\geq k$, where $k \\approx 2.657$." } { "Tag": [ "geometry", "rectangle", "angle bisector", "geometry proposed" ], "Problem": "Let the triangle ABC Pythagoras (angle B = 90\u00ba) sides of A, B and C. It was built square on the sides AB, BC and AC. \r\nLets P and Q are centres of squares located on catetos AB and BC respectively. \r\n (i) Find length of PC in terms of a; b and c\r\n (ii) Find the length of AQ in terms of a; b and c\r\n(iii) Find the length of TR, with T is the intersection of PC and AQ, and R is the center of the square on the hypotenuse\r\n\r\nThanks\r\n :)", "Solution_1": "Note C' the point laying on AB and being a vertex of the square built on BC; similarly A' on BC.\r\nThen CP is median in triangle ACA', while AQ in CAC'.\r\nFor 3rd request, see BCRA is cyclic with AC diameter, hence BR is the angle bisector of 1$ one can select $2k$ distinct marked points, say $a_1,...,a_{2k}$, such that $a_{2i-1}$ and $a_{2i}$ are in the same row, $a_{2i}$ and $a_{2i+1}$ are in the same column, $\\forall i$, indices taken mod 2n.", "Solution_1": "We regard the $n$ lines and $n$ columns as the two classes of a bipartition in a bipartite graph, two vertices being connected iff the intersection of the corresponding row and column is marked. What we want to prove is that we can find a cycle in this graph, and this is obvious: the graph has $2n$ vertices and $2n$ edges.", "Solution_2": "had it in one of my exams:\n$\\color{yellow} \\rule{25cm}{3pt}$\nname each column $a_i$ and each row $b_i$\n,now create a bipartite graph such that on one side of it we have all $a_i$s and on the other side of it we have all $b_i$s. Then if the field $a_{i}b_{j}$ was selected draw an edge between $a_i$ and $b_j$ in the graph,now we should prove that we have a cycle which contains the same number of $a_i$s and the same number of $b_i$s. So we have exactly $2n$ edges which means that we have a cycle in our graph also each edge connects one $a_i$ to another $b_i$.as desired.\n\\[\\color{magenta} \\boxed{\\mathcal{Q.E.D}}\\]$\\color{yellow} \\rule{25cm}{3pt}$" } { "Tag": [ "algebra", "polynomial", "quadratics", "least common multiple", "number theory", "prime factorization" ], "Problem": "factor: 6x^4-7x 2 +2", "Solution_1": "Cindy: Is this a challenge or a request for help? I suggest everyone use the \"Message icons\" when you post a new topic.", "Solution_2": "[hide](3x^2 + __ )(2x^2 + __) (3x^2 + 2)(2x^2 + 1)[/hide]", "Solution_3": "[hide]$(3x^2 + 2)(2x^2 + 1)$[/hide]", "Solution_4": "[quote=\"rcv\"]Cindy: Is this a challenge or a request for help? I suggest everyone use the \"Message icons\" when you post a new topic.[/quote]\r\n\r\nwhy does it matter", "Solution_5": "Please avoid double posting like that. If I were a mod of this forum I would stick those two together.", "Solution_6": "how do you show work for factoring? We didn't learn factoring in school yet, but my dad taught me the way he learn it in China.", "Solution_7": "[quote=\"jli\"][quote=\"rcv\"]Cindy: Is this a challenge or a request for help? I suggest everyone use the \"Message icons\" when you post a new topic.[/quote]\n\nwhy does it matter[/quote]\r\n\r\nGenerally, if it's a challenge, you should hide your answers, and don't need to post extremely verbose solutions.\r\n\r\nIf it's a request for help, hiding your answers isn't needed, but showing how you got the solutions is.", "Solution_8": "http://artofproblemsolving.com/Forum/viewtopic.php?t=24224\r\n\r\nlooky here Lucy.... the way we talked about the least common multiple problem....\r\n\r\nit's the same....\r\n\r\nyou just use one integer instead of 3 like we did up there......\r\n\r\nyou try to divide them with the smallest prime number as possible......\r\n\r\nI'll draw a little picture for you...", "Solution_9": "so you find the GCM of the 3 numbers? anyways, I think i'll stick to the way my dad taught me until we learn it in school, think we learn it this year during summer. Anyways thanks! :D", "Solution_10": "hm..\r\n\r\nisn't that how your dad taught you?\r\n\r\nI just thought it was the easiest way ,,, :P :blush:", "Solution_11": "it's probably the same thing, but his way involves a square with 4 numbers and stuff...", "Solution_12": "can't think of it right now....\r\n\r\ndistracted by so many stuff... :D \r\n\r\nwell, it's the same as solving 2 + 2 as 2 * 2 or 2^2 or etc....\r\n\r\ndifferent ways, but the same solutions... ingenius! :D", "Solution_13": "[quote=\"cindy\"]factor: 6x 4 -7x 2 +2[/quote]\r\n\r\n[hide]\n(3x2-1)(2x2-2)\n[/hide]\r\n\r\nAnd shinwoo, I don't really understand what you were trying to say...so the prime factorization of a number helps to factor a polynomial? It looks really cool, so could you please explain it some more? Thanks! :)", "Solution_14": "Many attempts. But I don't see any correct answers to the original question.", "Solution_15": "This is my method of choice. Lucy, is this what you were talking about?", "Solution_16": "Now, we have a correct answer. Following is a much longer, more tedious solution. But the general method might help you crack some harder quadratic factorizations.\r\n\r\n[hide]Factor $6x^4-7x^2+2$\n\nIf we multiply by 6 (the leading coefficient), and make a substitution, we get a simpler factorization...\n\n\\begin{eqnarray*}\n6(6x^4-7x^2+2)&=&36x^4-42x^2+12 \\\\\n&&\\hbox{Let }y=6x^2 \\\\\n6(6x^4-7x^2+2)&=&y^2-7y+12 \\\\\n\\end{eqnarray*}\n\n\nNow, we try to factor $y^2-7y+12$\n\nAll possible factors of 12, along with the sums of those factors:\n\\begin{eqnarray*}\n12=12\\times 1 &\\Longrightarrow& 12+1=13 \\\\\n12=6\\times 2&\\Longrightarrow& 6+2=8\\\\\n12=4\\times 3 &\\Longrightarrow& 4+3=7\\\\\n12=-12\\times -1 &\\Longrightarrow& -12-1=-13\\\\\n12=-6\\times -2 &\\Longrightarrow& -6-2=-8\\\\\n12=-4\\times -3 &\\Longrightarrow& -4-3=-7 \\qquad \\Longleftarrow\n\\end{eqnarray*}\n\nThis gives us the factorization. We reverse the substitutions to get back to the original question...\n\n\\begin{eqnarray*}\ny^2-7y+12&=&(y-4)(y-3) \\\\\ny^2-7y+12&=&(6x^2-4)(6x^2-3) \\\\\n36x^4-42x^2+12&=&(6x^2-4)\\cdot(6x^2-3) \\\\\n6(6x^4-7x^2+2)&=&2(3x^2-2)\\cdot 3(2x^2-1) \\\\\n6x^4-7x^2+2&=&(3x^2-2)(2x^2-1)\n\\end{eqnarray*}[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Determine the positive integers $ m,n$ such that: $ 5^m\\plus{}6^n\\plus{}2$ is a perfect square.", "Solution_1": "Go to High School Basic.\r\n\r\nmod 4 we get $ n \\ge 2$ impossible. So $ n\\equal{}1$\r\nmod 8 $ 5^m\\plus{}8$ we have $ m\\equal{}2k$ is even.\r\n\r\nSo $ (x\\plus{}5^k)(x\\minus{}5^k)\\equal{}8$, gives no solution for positive integer $ k$, so no such $ m,n$ exist." } { "Tag": [ "geometry" ], "Problem": "If you tie a Cow at \"B\", a point of a circumference with \"R\" Radius, so that the cow can only eat half of the mown. What is C=R+d?\r\n\r\nhttp://img475.imageshack.us/img475/1682/vestibularc07ssdjkjfdsfic7.gif", "Solution_1": "ummm what are C and d? Also.. could you restate your question?", "Solution_2": "I can try. I got that from a riddle.\r\n\r\nI believe what it means is: They tied a cow with a rope with \"c\" length and ( c == R + d ) in the picture at \"B\".\r\nThe area of both colored and white sections of the circle are equal. What's the length of that rope( c == ? )?\r\n\r\nIs it clearer now?" } { "Tag": [ "function", "logarithms", "integration", "floor function", "calculus", "calculus computations" ], "Problem": "How to prove that $ \\ln n! \\equal{} \\int_{1}^{n}{\\frac {n \\minus{} \\left\\lfloor x \\right\\rfloor }{x}\\,dx}$ ?\r\n\r\n---------\r\n\r\nI proved that $ \\int_{1}^{n}{\\frac{\\left\\lfloor x \\right\\rfloor }{x}\\,dx}\\equal{}\\sum\\limits_{k\\equal{}1}^{n\\minus{}1}{k\\big(\\ln (k\\plus{}1)\\minus{}\\ln k\\big)},$ but I dunno how to solve the last sum.", "Solution_1": "$ \\sum\\limits_{k \\equal{} 1}^{n \\minus{} 1}{k\\big(\\ln (k \\plus{} 1) \\minus{} \\ln k\\big)}, \\\\\r\n\\equal{} 1(\\ln 2 \\minus{} \\ln 1) \\plus{} 2(\\ln 3 \\minus{} \\ln2) \\plus{} 3(\\ln 4 \\minus{} \\ln 3) \\plus{} \\ldots \\plus{} (n \\minus{} 1)(\\ln n \\minus{} \\ln (n \\minus{} 1)) \\vspace{10pt} \\\\\r\n\\equal{} (\\ln 2 \\minus{} \\ln 1) \\plus{} (\\ln 3 \\minus{} \\ln 2) \\plus{} (\\ln 4 \\minus{} \\ln 3) \\plus{} \\ldots \\plus{} (\\ln n \\minus{} \\ln (n \\minus{} 1)) \\\\\r\n\\phantom{\\equal{} (\\ln 2 \\minus{} \\ln 1)} \\plus{} (\\ln 3 \\minus{} \\ln 2) \\plus{} (\\ln 4 \\minus{} \\ln 3) \\plus{} \\ldots \\plus{} (\\ln n \\minus{} \\ln(n \\minus{} 1)) \\\\\r\n\\phantom{\\equal{} (\\ln 2 \\minus{} \\ln 1) \\plus{} (\\ln 3 \\minus{} \\ln 2) } \\plus{} (\\ln 4 \\minus{} \\ln 3) \\plus{} \\ldots \\plus{} (\\ln n \\minus{} \\ln(n \\minus{} 1))\\\\\r\n\\phantom{\\equal{} (\\ln 2 \\minus{} \\ln 1) \\plus{} (\\ln 3 \\minus{} \\ln 2) \\plus{} (\\ln 4 \\minus{} \\ln 3)} \\plus{} \\vdots \\\\\r\n\\phantom{\\equal{} (\\ln 2 \\minus{} \\ln 1) \\plus{} (\\ln 3 \\minus{} \\ln 2) \\plus{} (\\ln 4 \\minus{} \\ln 3) \\plus{} \\ldots} \\plus{} (\\ln n \\minus{} \\ln(n \\minus{} 1)) \\vspace{10pt} \\\\\r\n\\equal{} (\\ln n \\minus{} \\ln 1) \\plus{} (\\ln n \\minus{} \\ln 2) \\plus{} \\ldots \\plus{} (\\ln n \\minus{} \\ln(n \\minus{} 1)) \\\\\r\n\\equal{} (n \\minus{} 1)\\ln n \\minus{} \\ln (n \\minus{} 1)!$\r\n\r\n\r\nThen \r\n$ \\int_{1}^{n}{\\frac {n \\minus{} \\left\\lfloor x \\right\\rfloor }{x}\\,dx} \\equal{} n\\ln n \\minus{} (n \\minus{} 1)\\ln n \\plus{} \\ln(n \\minus{} 1)! \\equal{} \\ln n!$" } { "Tag": [ "adfghj" ], "Problem": "Let $a,b,c$ be positive integers such that $a$ divides $b^2$, $b$ divides $c^2$ and $c$ divides $a^2$. Prove that $abc$ divides $(a + b +c)^7$.", "Solution_1": "$a = \\prod p_i^{a_i}$, $b = \\prod p_i^{b_i}$, $c = \\prod p_i^{c_i}$\r\n\r\nWith $p_i$ prime.\r\n\r\nCondition is $a_i \\leq 2*b_i \\leq 4*c_i \\leq 8*a_i$\r\n\r\nWhen expandind $(a+b+c)^7$ we only have to consider monoms like $a^7$,$a^6*b$, ... If we take for instance $a^6*b$: $6a_i + b_i \\geq a_i+b_i+c_i$ so $abc | a^6*b$ and so on.", "Solution_2": "Please explain once again.. ", "Solution_3": "$a|b^2,b|c^2,c|a^2 \\Rightarrow abc|b^3 c,ac^3,a^3 b \\Rightarrow abc|c^7,a^7,b^7$\n\n$a|c^4,b|a^4,c|b^4\\Rightarrow abc|bc^5,a^5 c,ab^5$\n\n$abc|bc^6,b^2 c^5,b^3 c^4,b^4 c^3,b^5 c^2,b^6 c$\n\n$abc|ac^6,a^2 c^5,a^3 c^4,a^4 c^3,a^5 c^2,a^6 c$\n\n$abc|ab^6,a^2 b^5,a^3 b^4,a^4 b^3,a^5 b^2,a^6 b$", "Solution_4": "Easier solution is :\nb^2 = ka , c^2 = mb , a^2 = nc\nEliminate all to get a,b,c in terms of k ,n,m \nWe get a^7 + b^7 +c^7 = knm x (something else)\nand abc = knm\n", "Solution_5": "[quote=NitroZox]Easier solution is :\nb^2 = ka , c^2 = mb , a^2 = nc\nEliminate all to get a,b,c in terms of k ,n,m \nWe get a^7 + b^7 +c^7 = knm x (something else)\nand abc = knm[/quote]\n\nAre you trying to prove $abc|a^7+b^7+c^7$ ?\n\nI don't see how $a^7 + b^7 +c^7 = knm \\times (\\text{something else})$", "Solution_6": "[hide = Solution]We can expand the expression $(a+b+c)^7 $using binomial theorem twice:\nNote : I expanded the expression without multiplying the binomial coeffeciants, also terms of the form $a^sb^tc^q$ don't need to be included because $abc$ already divides those terms.\nExpansion : $a^7+ a^6b + a^6c + a^5b^2 + a^5c^2 + a^ 4b^3 + a^4c^3 + a^3b^4 +a^2b^5 + a^2c^5 + ab^6 + ac^6 + b^7 + b^6c+ b^5c^2 +b^4c^3 + b^3c^4 + b^2c^5 + bc^6 + c^7$\n$a|a ; b|a^4 ; c|a^2 \\Longrightarrow abc|a^7$\n$a|a ; b|b ; c|a^5 \\Longrightarrow abc|a^6b$\n$a|a^2 ; b|a^4 ; c|c \\Longrightarrow abc|a^6c$\n$a|a ; b|b^2 ; c|a^4 \\Longrightarrow abc| a^5b^2$\n$a|a^3 ; b|c^2 ; c|a^2 \\Longrightarrow abc|a^5c^2$\n$a|a ; b|b^3 ; c|a^3 \\Longrightarrow abc|a^4b^3$\n$a|a ; b|c^3 ; c|a^3 \\Longrightarrow abc|a^4c^3$\n$a|a ; b|b^4 ; c|a^2 \\Longrightarrow abc|a^3b^4$\n$a|b^2 ; b|c^4 ; c|a^3 \\Longrightarrow abc|a^3c^4$\n$a|b^2 ; b|b^3 ; c|a^2 \\Longrightarrow abc|a^2b^5$\n$a|b^2 ; b|c^5 ; c|a^2 \\Longrightarrow abc|a^2c^5$\n$a|a ; b|b^2 ; c|b^4 \\Longrightarrow abc|ab^6$\n$a|a ; b|c^2 ; c|c^4 \\Longrightarrow abc|ac^6$\n$a|b^2 ; b|b ; c|b^4 \\Longrightarrow abc|b^7$\n$a|b^2 ; b|b^4 ; c|c \\Longrightarrow abc|b^6c$\n$a|b^2 ; b|b^3 ; c|c^2 \\Longrightarrow abc|b^5c^2$\n$a|b^4 ; b|b^2 ; c|c^3 \\Longrightarrow abc|b^4c^3$\n$a|b^2 ; b|b ; c|c^4 \\Longrightarrow abc|b^3c^4$\n$a|b^2 ; b|c^4 ; c|c \\Longrightarrow abc|b^2c^5$\n$a|c^4 ; b|b ; c|c^2 \\Longrightarrow abc|bc^6$\n$a|b^4 ; b|c^2 ; c|c \\Longrightarrow abc|c^7$\nSince $abc$ divides each term of the multinomial expansion of $(a+b+c)^7$.\nThus, $abc|(a+b+c)^7$\n[/hide]", "Solution_7": "No binomial theorem solution: \n$v_p(a) \\leq 2v_p(b)$, $v_p(b) \\leq 2v_p(c)$, $v_p(c) \\leq 2v_p(a)$.\nCase: $v_p(a) \\leq v_p(b) \\leq v_p(c)$. \n$v_p(abc) = v_p(a) + v_p(b) + v_p(c) \\leq v_p(a)+2v_p(a)+2v_p(a) = 5v_p(a) < 7v_p(a) \\leq 7v_p(a+b+c)$\nCase: $v_p(a) \\leq v_p(c) \\leq v_p(b)$. \n$v_p(abc) = v_p(a) + v_p(b) + v_p(c) \\leq v_p(a) + 2v_p(a) + 4v_p(a) = 7v_p(a) \\leq 7v_p(a+b+c)$" } { "Tag": [ "algebra", "polynomial", "search", "geometry", "Ring Theory", "function", "linear algebra" ], "Problem": "Define a \"polynomial statement\" about matrices to be a statement of the form \"if square matrices $ A_1, A_2, ... A_n$ satisfy a polynomial identity $ p(A_1, ... A_n) \\equal{} 0$ with scalar coefficients, then they also satisfy the polynomial identity $ q(A_1, A_2, ... A_n) \\equal{} 0$.\" \r\n\r\nProve or give a counterexample: if a polynomial statement is true for arbitrary square matrices over a field of characteristic $ 0$, it is true in an arbitrary ring (edit: also of characteristic $ 0$) with identity.\r\n\r\n(Examples [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=956528026&t=242500]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=241701]here[/url]. I suppose this is really a question about representation theory.)", "Solution_1": "I don't really get your question: what about $ 2A_1\\equal{}0\\Longrightarrow A_1\\equal{}0$ ?\r\n\r\n darij", "Solution_2": "Edited. I suspect this statement is false because inverses are not unique in an arbitrary ring, but I haven't actually found a counterexample yet.", "Solution_3": "[quote=\"t0rajir0u\"]Define a \"polynomial statement\" about matrices to be a statement of the form \"if square matrices $ A_1, A_2, ... A_n$ satisfy a polynomial identity $ p(A_1, ... A_n) \\equal{} 0$ with scalar coefficients, then they also satisfy the polynomial identity $ q(A_1, A_2, ... A_n) \\equal{} 0$.\"[/quote]\nWhat are those \"scalar coefficients\"? \n\n[quote=\"t0rajir0u\"]Prove or give a counterexample: if a polynomial statement is true for arbitrary square matrices over a field of characteristic $ 0$, it is true in an arbitrary ring (edit: also of characteristic $ 0$) with identity.[/quote]\r\nI don't think I understand how to take a polynomial statement over, say, $ \\mathbb{Q}(\\sqrt{11},X)$ and re-interpret it over the ring $ \\mathbb{Z}[\\frac{1}{3}]$. Are the coefficients of the polynomials only allowed to be integers? That would make the concern about characteristic more clear to me, since then it's clear how to re-interpret an identity in various rings. \r\n\r\nSorry if I'm missing something here, but I think it's important to be annoyingly precise here :) more than anything else this strikes me as a problem in algebraic geometry, since I'm not convinced the matrices contribute much to the discussion - it's ultimately a family of polynomial identities. But it [i]is [/i]important to be careful about how to interpret a change of scalars, I think.", "Solution_4": "[quote=\"randomgraph\"]I think it's important to be annoyingly precise here :)[/quote]\r\nYou're right, of course :) My question is motivated by the observation that there exist non-commutative polynomial identities that can be shown to hold true for square matrices that also hold in an arbitrary ring with unity, and I'm wondering in what sense this observation generalizes.\r\n\r\nLet me be more specific, then: unless I'm terribly mistaken, any ring of characteristic zero contains a subring isomorphic to $ \\mathbb{Z}$. A polynomial identity in an arbitrary ring of characteristic zero is the statement $ p(a_1, a_2, ... a_n) \\equal{} 0$ where $ p \\in \\mathbb{Z} < x_1, x_2, ... x_n >$ (that's the correct notation for a non-commutative polynomial ring, right?). \r\n\r\nI agree that the question has an algebraic-geometric flavor to it, but I have no idea what concepts are necessary to develop algebraic geometry from non-commutative function fields.", "Solution_5": "What I had in mind when thinking of this problem as belonging to algebraic geometry is that there's nothing truly non-commutative here if you switch your point of view from the matrices to their coefficients. For a [i]fixed [/i]dimension $ d$, the expression $ p(A_1, \\ldots, A_n)\\equal{}0$ (where the $ A_i$ are $ d \\times d$ matrices) is equivalent to $ \\tilde{p}(x_k)\\equal{}0$ where the $ x_k$ are $ nd^2$ scalar variables in the ring or field over which we work, and $ \\tilde{p}$ is an ordinary, commutative polynomial. \r\n\r\nGiven polynomials $ p$ and $ q$ and a specific dimension, we can translate the implication question from a non-commutative one into a commutative one in $ \\tilde{p}$ and $ \\tilde{q}$. As such, I think the question is a special case of the following: suppose $ p, q \\in \\mathbb{Z}[X_1, \\ldots, X_m]$ are such that whenever $ x_i \\in \\mathbb{C}$ and $ p(x_i)\\equal{}0$, then $ q(x_i)\\equal{}0$. The question is then if the same implication holds when $ \\mathbb{C}$ is replaced by any ring containing $ \\mathbb{Z}$. \r\n\r\nSo this isn't the most basic type of algebraic geometry (since we're working over rings, rather than algebraically closed fields), but at least it is commutative. I don't remember much about alg geom over such rings, but this seems like a useful way to look at the problem. The condition should imply a simple condition about ideals in $ \\mathbb{Z}[X_1, \\ldots, X_m]$, and this in turn would imply the validity of the implication over rings containing $ \\mathbb{Z}$, perhaps with some slight additional restrictions.", "Solution_6": "So, how about this identity: in any matrix ring, $ BA\\minus{}1\\equal{}0$ if $ AB\\minus{}1\\equal{}0$. That's not true in general rings (with identity). Matrix rings are finite-dimensional, and that does imply some algebraic properties like this one." } { "Tag": [], "Problem": "\u0388\u03b3\u03b9\u03bd\u03b5 \u03c0\u03c1\u03b9\u03bd 2-3 \u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b5\u03c2. \u039f\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 [url=http://www.bmoc.maths.org/home/bmo2-2009.pdf]\u03b5\u03b4\u03ce[/url]", "Solution_1": "Vrika oti den yparxoun lyseis alla den eimai sigouros gia ayto pou ekana.\r\n\r\nAn ypsosoume kai ta dyo meli sto tetragono tote exoume:\r\n\r\n $ a \\plus{} b \\minus{} 2009 \\equal{} \\minus{} 2\\sqrt{ab}$ Epomenos to $ ab$ prepei na einai teleio tetragono pou simainei\r\n oti kathena ksexorista apo ta a,b prepei na einai teleio tetragono $ a \\equal{} m^2 and b \\equal{} n^2$\r\nalla tote exoume $ m \\plus{} n \\equal{} \\sqrt{2009}$ to opoio einai atopo mias kai to aristero melos einai integer eno\r\nto deksio mellos einai arritos", "Solution_2": "\u0394\u03b5\u03bd \u03b5\u03b9\u03c3\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03bf\u03c2.\u039f\u03c4\u03b1\u03bd \u03b1.\u03b2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf \u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c0\u03b1\u03c1\u03b1\u03b9\u03c4\u03b7\u03c4\u03bf \u03bf \u03ba\u03b1\u03b8\u03b5\u03bd\u03b1\u03c2 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b5\u03b9\u03bf \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf.\u0391\u03c5\u03c4\u03bf \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03bf\u03c4\u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03c9\u03c4\u03bf\u03b9 \u03bc\u03b5\u03c4\u03b1\u03be\u03c5 \u03c4\u03bf\u03c5\u03c2.\u0394\u03bf\u03ba\u03b9\u03bc\u03b1\u03c3\u03b5 \u03bd\u03b1 \u03c5\u03c8\u03c9\u03c3\u03b5\u03b9\u03c2 \u03b1\u03bb\u03bb\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5\u03bb\u03bf\u03c2 \u03bc\u03b9\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b9\u03c2 \u03b4\u03c5\u03bf \u03c1\u03b9\u03b6\u03b5\u03c2.\u03a4\u03bf\u03c4\u03b5 \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03c5\u03c0\u03bf\u03c1\u03c1\u03b9\u03b6\u03bf 2009\u03b1.\u039f\u03bc\u03c9\u03c2 2009=7.7.41 \u03b1\u03c1\u03b1$ a\\equal{}41k^2$.\u0397 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03b5\u03c7\u03b5\u03b9 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2.", "Solution_3": "\u0397 \u03c0\u03c1\u03c9\u03c4\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b7. \u039f\u03c0\u03c9\u03c2 \u03bb\u03b5\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03bf \u03c6\u03b9\u03bb\u03bf\u03c2 \u03bc\u03b1\u03c2 \u03b1\u03c0\u03bf \u03c0\u03b1\u03bd\u03c9 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 $ ab \\equal{} k^2$ \u03bf\u03c0\u03bf\u03c5 $ k$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c2 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03bf\u03c2. \u039c\u03b5\u03c4\u03b1 \u03bf\u03b9 $ a,b$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c1\u03b9\u03b6\u03b5\u03c2 \u03c4\u03c1\u03b9\u03c9\u03bd\u03c5\u03bc\u03bf\u03c5 2\u03bf\u03c5 \u03b2\u03b1\u03b8\u03bc\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03b5\u03c7\u03b5\u03b9 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03b9 \u03c4\u03bf\u03c5 $ k$, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03ba\u03c1\u03b9\u03bd\u03bf\u03c5\u03c3\u03b1 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03bf, \u03b5\u03c4\u03c3\u03b9 \u03b2\u03b3\u03b1\u03b6\u03b5\u03b9\u03c2 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b4\u03c5\u03bd\u03b1\u03c4\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf $ k$ \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 \u03b1\u03c0\u03bb\u03b1 \u03ba\u03bf\u03b9\u03c4\u03b1\u03c2 \u03c0\u03bf\u03b9\u03b5\u03c2 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03c5\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03b6\u03b5\u03b9\u03c2 \u03c4\u03b1 $ a,b$ \u03b1\u03c0\u03bf \u03c4\u03bf \u03c4\u03c1\u03b9\u03c9\u03bd\u03c5\u03bc\u03bf\r\n\r\n2\u03bf \u03ba\u03b1\u03b9 4\u03bf \u03b4\u03b5\u03bd \u03c4\u03b1 \u03b5\u03c7\u03c9 \u03ba\u03bf\u03b9\u03c4\u03b1\u03be\u03b5\u03b9 \u03b1\u03ba\u03bf\u03bc\u03b1\r\n\r\n3\u03bf \u03b5\u03c0\u03b9\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c7\u03b5\u03c4\u03b9\u03ba\u03b1 \u03b1\u03c0\u03bb\u03bf, \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03b7 \u03bf\u03c0\u03bf\u03c5 $ x$ \u03c4\u03b1 $ y, 1, \\minus{} y^2$ \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03b6\u03b5\u03b9\u03c2 3 \u03c3\u03c7\u03b5\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bf \u03c4\u03b9\u03c2 \u03bf\u03c0\u03bf\u03b9\u03b5\u03c2 \u03bc\u03b5 \u03bb\u03b9\u03b3\u03b5\u03c2 \u03c0\u03c1\u03b1\u03be\u03b5\u03b9\u03c2 \u03b2\u03b3\u03b1\u03b6\u03b5\u03b9\u03c2: $ f(x) \\equal{} cx$ \u03bf\u03c0\u03bf\u03c5 $ c \\in R$" } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "draw the line segments (0,0)-(1,1); (1,1)-(4,1); (4,1)-(4,10); (4,10)-(6,10); (6,10)-(6,6); and (6,6)-(7,7).\r\nwhat is the average slope of this graph?\r\n\r\nif you take it rise over run, the average slope is 1.\r\nbut there is a stretch of graph with undefined slope.\r\nif you take the average slope of the line segments and get the average slope of the whole graph you get (1+0+undefined+0+undefined+1)/5.", "Solution_1": "Umm what are you trying to say? Do you want help with the problem? Or is it for other people to solve? (as it says the source is your head) If it is for other people to solve, you shouldn't write the answer and solution unhidden in the first post.", "Solution_2": "By the way, if you're trying to find the average, you divide by the total number which is 6 in your case not 5", "Solution_3": "I think you might need to find a fitted line that is for that graph.\r\n\r\nThen you will get your answer.", "Solution_4": "i thought of it from my head, and i can't figure it out. \r\n\r\nif it's by six, it still comes out as undefined divided by 6.\r\n\r\nit's kind of like you have a 2-d mountain on a graph, and you are at the bottom. you have to find the slope of the mountain from where you are to halfway up the other side. you could take the two points, and get the slope. but that wouldn't be accurate, because you can't go through the mountain. the slope would first be steeper and then it would be negative. kind of like that. it'd be better with a picture.", "Solution_5": "You can't calculate slopes by the average because the length of the line segments are all different lengths, therefore not affecting individual slopes but the 'average' (and I use the term loosely) changes", "Solution_6": "[quote=\"mathgeniuse^ln(x)\"]I think you might need to find a fitted line that is for that graph.\n\nThen you will get your answer.[/quote]\r\n\r\n[hide]This method is wrong.\n\nHere is what I think it is.\n\nIf you graph these points notice how they are connected by line segments. So I think the average of the graph is just (7-0)/(7-0)=1.\n\n$1$[/hide]" } { "Tag": [ "inequalities", "logarithms", "calculus", "derivative", "inequalities unsolved" ], "Problem": "[quote=\"chuyue\"]If $a,b>0$ then $(a^a b^b)^{\\frac{2}{a+b}}\\geq \\frac{a^2+b^2}{2}$[/quote]\r\nOrigin:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=67464", "Solution_1": "ok,I have sloved this inequality months ago ;) ,but my proof have to use the largrange median theorem.\r\nhere is my solution:\r\nwe can assume $a>b$,(if $a=b$ it is easily proved)\r\n$(a^a b^b)^{\\frac{2}{a+b}}\\geq \\frac{a^2+b^2}{2} \\iff \\frac{a}{a+b} \\ln a^2+\\frac{b}{a+b} \\ln b^2 \\ge \\ln \\frac{a^2+b^2}{2}$ $(1)$\r\nuse largrange theorem: we have:\r\n$\\frac{a}{a+b} \\ln a^2-\\frac a {a+b}\\ln \\frac{a^2+b^2}{2}=\\frac a {a+b} (a^2-\\frac{a^2+b^2}{2})(\\frac 1 {\\xi_1})=\\frac a {a+b} (a^2-b^2)(\\frac 1 {\\xi_1})=(a-b)(\\frac a {\\xi_1})$,$\\xi_1 \\in (\\frac{a^2+b^2}2,a)$\r\nsimilar:\r\n$\\frac b {a+b}\\ln \\frac{a^2+b^2}{2}-\\frac{b}{a+b} \\ln b^2=(a-b)(\\frac b {\\xi_2})$,$\\xi_2 \\in (b,\\frac{a^2+b^2}2)$\r\nso we get \r\n$\\frac b {a+b}\\ln \\frac{a^2+b^2}{2}-\\frac{b}{a+b} \\ln b^2 \\le a-b \\le \\frac{a}{a+b} \\ln a^2-\\frac a {a+b}\\ln \\frac{a^2+b^2}{2}$\r\nwhich is $(1)$\r\nthe inequality has been proved :)", "Solution_2": "[quote=\"zhaobin\"]\n$\\frac{a}{a+b} \\ln a^2-\\frac a {a+b}\\ln \\frac{a^2+b^2}{2}=\\frac a {a+b} (a^2-\\frac{a^2+b^2}{2})(\\frac 1 {\\xi_1})=\\frac a {a+b} (a^2-b^2)(\\frac 1 {\\xi_1})=(a-b)(\\frac a {\\xi_1})$,$\\xi_1 \\in (\\frac{a^2+b^2}2,a)$\n[/quote]\r\nWhere $2$ ? ;)", "Solution_3": "Zhaobin, by your hint I proved it. Thank you.", "Solution_4": "hope my solution isn't wrong...\r\n\r\n$(a^a b^b)^{\\frac{2}{a+b}}\\geq \\frac{a^2+b^2}{2}\\Leftrightarrow \\frac{a\\ln a^2+b\\ln b^2}{a+b}\\geq \\ln\\frac{a^2+b^2}{2}$\r\n\r\nbut by weighted jensen we have that\r\n$\\frac{a\\ln a^2+b\\ln b^2}{a+b}\\geq \\ln\\frac{a^3+b^3}{a+b}$\r\n\r\nbut it can be easily verified that $\\ln\\frac{a^3+b^3}{a+b}\\geq \\ln\\frac{a^2+b^2}{2}$, so we're done!", "Solution_5": "[quote=\"campos\"]\nbut by weighted jensen we have that\n$\\frac{a\\ln a^2+b\\ln b^2}{a+b}\\geq \\ln\\frac{a^3+b^3}{a+b}$\n[/quote]\r\n$(lnx^2)''=-\\frac{2}{x^2}<0.$ ;)", "Solution_6": "Yes. It\u00b4s wrong because $\\frac{a\\ln a^2+b\\ln b^2}{a+b} \\leq \\ln\\frac{a^3+b^3}{a+b}$, because if $f(x)=ln(x^2)$ the second derivative o $f$ is $-\\frac{2}{x^2}$.", "Solution_7": "i thought this couldn't be so easy! :D \r\n\r\nbut there's a funny thing! zhaobin's solution contradicts the topic's name! :rotfl:", "Solution_8": "[quote=\"campos\"]\n... zhaobin's solution contradicts the topic's name![/quote]\r\nZhaobin's proof is wrong, but enable to end.", "Solution_9": "so nobody has posted a proof so far... arqady do you have a proof for it?", "Solution_10": "[quote=\"campos\"]... arqady do you have a proof for it?[/quote]\r\nYes. But I wait to Zhaobin. :P", "Solution_11": "Sorry there a little mitake in my proof,but I think it is not such big :P \r\n$\\frac{a}{a+b} \\ln a^2-\\frac a {a+b}\\ln \\frac{a^2+b^2}{2}=\\frac a {a+b} (a^2-\\frac{a^2+b^2}{2})(\\frac 1 {\\xi_1})=\\frac a {2(a+b)} (a^2-b^2)(\\frac 1 {\\xi_1})=(\\frac{a-b}2)(\\frac a {\\xi_1}),\\xi_1 \\in (\\frac{a^2+b^2}2,b)$\r\nand\r\n$\\frac b {a+b}\\ln \\frac{a^2+b^2}{2}-\\frac{b}{a+b} \\ln b^2=(\\frac{a-b}2)(\\frac b {\\xi_2}),\\xi_2 \\in (a,\\frac{a^2+b^2}2)$\r\nthen we get:\r\n$\\frac{a}{a+b} \\ln a^2-\\frac a {a+b}\\ln \\frac{a^2+b^2}{2} \\ge \\frac{a-b}2 \\ge \\frac b {a+b}\\ln \\frac{a^2+b^2}{2}-\\frac{b}{a+b} \\ln b^2$\r\nIs it right now :) (I think it is,but also not very sure)", "Solution_12": "[quote=\"arqady\"][quote=\"chuyue\"]If $a,b>0$ then $(a^a b^b)^{\\frac{2}{a+b}}\\geq \\frac{a^2+b^2}{2}$[/quote]\nOrigin:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=67464[/quote]\r\nSome times ago I have solved the similar problem on this forum...\r\nTake $f(a,b)=\\displaystyle\\frac{a^{\\frac{2a}{a+b}} b^{\\frac{2b}{a+b}}}{\\frac{a^2+b^2}{2}}$ then $f(a,b)=f(ka,kb)$ so take b=1,a>1 then the inequality is obvious :)", "Solution_13": "[quote=\"hardsoul\"]\nTake $f(a,b)=\\displaystyle\\frac{a^{\\frac{2a}{a+b}} b^{\\frac{2b}{a+b}}}{\\frac{a^2+b^2}{2}}$ then $f(a,b)=f(ka,kb)$ so take b=1,a>1 then the inequality is obvious :)[/quote]\r\nI tried it. This is very hard. Try! ;)\r\nTo Zhaobin. OK. Thank you.", "Solution_14": "I am sorry, but I think that your proof is wrong, zhaobin.\r\n[quote=\"zhaobin\"]Sorry there a little mitake in my proof,but I think it is not such big :P \n$\\frac{a}{a+b}\\ln a^{2}-\\frac a{a+b}\\ln \\frac{a^{2}+b^{2}}{2}=\\frac a{a+b}(a^{2}-\\frac{a^{2}+b^{2}}{2})(\\frac 1{\\xi_{1}})=\\frac a{2(a+b)}(a^{2}-b^{2})(\\frac 1{\\xi_{1}})=(\\frac{a-b}2)(\\frac a{\\xi_{1}}),\\xi_{1}\\in (\\frac{a^{2}+b^{2}}2,b)$[/quote]\n$\\xi_{1}\\in (\\frac{a^{2}+b^{2}}2,a^{2})$ :wink: \n[quote=\"zhaobin\"]\nand\n$\\frac b{a+b}\\ln \\frac{a^{2}+b^{2}}{2}-\\frac{b}{a+b}\\ln b^{2}=(\\frac{a-b}2)(\\frac b{\\xi_{2}}),\\xi_{2}\\in (a,\\frac{a^{2}+b^{2}}2)$[/quote]\n$\\xi_{2}\\in (b^{2},\\frac{a^{2}+b^{2}}2)$\n[quote=\"zhaobin\"]\nthen we get:\n$\\frac{a}{a+b}\\ln a^{2}-\\frac a{a+b}\\ln \\frac{a^{2}+b^{2}}{2}\\ge \\frac{a-b}2 \\ge \\frac b{a+b}\\ln \\frac{a^{2}+b^{2}}{2}-\\frac{b}{a+b}\\ln b^{2}$\nIs it right now :) (I think it is,but also not very sure)[/quote]\r\n$\\frac{a}{a+b}\\ln a^{2}-\\frac a{a+b}\\ln \\frac{a^{2}+b^{2}}{2}\\ge \\frac{a-b}{2a}$ and $\\frac{a-b}{2b}\\geq\\frac b{a+b}\\ln \\frac{a^{2}+b^{2}}{2}-\\frac{b}{a+b}\\ln b^{2}.$\r\nI think, there is no proof!\r\nTo hardsoul: you are right!" } { "Tag": [ "function", "algebra", "polynomial", "partial fractions", "system of equations" ], "Problem": "Express $\\frac{1}{x^{3}+(\\sqrt{2}+\\sqrt{3}+\\sqrt{5})x^{2}+(\\sqrt{10}+\\sqrt{15}+\\sqrt{6})x+\\sqrt{30}}$ as a sum of $3$ fractions whose numerators are $1$ and denominators are \r\n\r\nlinear functions of $x$.\r\nAlso, rationalize each fraction.", "Solution_1": "umm i'm guessing each is $\\frac{1}{x+\\sqrt{2,3,5}}$\r\nright? Now ratioanlizing is a mess", "Solution_2": "[quote=\"me@home\"]umm i'm guessing each is $\\frac{1}{x+\\sqrt{2,3,5}}$\nright? Now ratioanlizing is a mess[/quote]\r\nThe given fraction is the product of the three fractions you gave, not the sum.", "Solution_3": "$\\frac{1}{(\\sqrt{3}-\\sqrt{2})(\\sqrt{5}-\\sqrt{2})(x+\\sqrt{2})}+\\frac{1}{(\\sqrt{2}-\\sqrt{3})(\\sqrt{5}-\\sqrt{3})(x+\\sqrt{3})}+\\frac{1}{(\\sqrt{2}-\\sqrt{5})(\\sqrt{3}-\\sqrt{5})(x+\\sqrt{5})}$", "Solution_4": "how did you get[i] that[/i]?", "Solution_5": "[hide=\"Solution\"]\nPartial Fraction Decomposition:\n\nIf $n=\\frac{1}{ (x+\\sqrt5)(x+\\sqrt3)(x+\\sqrt2)}$, then $n= \\frac{A}{x+\\sqrt5}+\\frac{B}{x+\\sqrt3}+\\frac{C}{x+\\sqrt2}$, for constants $A,B,C$.\n\nMerge the three fractions in the second expression into one, and set the numerator equal to $1$. You'll get a system of equations for $A,B,C$ based on the powers of $x$ (ie. $A+B+C=0$ since $1$ has no second degree terms.)\n\nSolve the mess for $A,B,C$, then put these back into the partial fractions expression and rationalize.\n\nI dont have time to post the numbers, but i will when i return.[/hide]", "Solution_6": "That was the idea, indeed. But instead of solving the system of equations you get, there is a faster way that gives the values of $A,B$ AND $C$ immediately:\r\n$1=A(x+\\sqrt{3})(x+\\sqrt{5})+B(x+\\sqrt{2})(x+\\sqrt{5})+C(x+\\sqrt{2})(x+\\sqrt{3})$\r\nis the polynomial equation resulting from the considerations above. Now, since this is a equality between polynomials, lefthand and righthand side must be equal for any x. Now subtitute $x=-\\sqrt{2},-\\sqrt{3},-\\sqrt{5}$ respectively, and the values for $A,B$ and $C$ follow at once." } { "Tag": [], "Problem": "Hurray!!!!!!!!!!!!!!!!!!!! :gathering: :starwars:\r\n\r\nthis is officially the thread to talk about how cool SC is and NC isn't...lol\r\n\r\nwhat happened to NC anyway?", "Solution_1": "WOOHOO!!! WE FINALLY BROKE FREE!!!\r\n\r\nThank you Mr. Valentin :) :D", "Solution_2": "Is their no NC forum :rotfl: \r\n\r\nBTW, i would like to take credit for proposing such a brilliant idea....brilliant! :winner_first:", "Solution_3": "[quote=\"Iversonfan2005\"]Is their no NC forum :rotfl: :rotfl:[/quote]\r\n\r\nNot that I know of... I asked to have the Carolinas changed into South Carolina since 90% or so of it was SC anyway. If NC wants a forum they can ask Valentin for one...", "Solution_4": "[quote=\"joml88\"][quote=\"Iversonfan2005\"]Is their no NC forum :rotfl: :rotfl:[/quote]\n\nNot that I know of... I asked to have the Carolinas changed into South Carolina since 90% or so of it was SC anyway. If NC wants a forum they can ask Valentin for one...[/quote]\r\n\r\nHAHA...looks like we won the war! :rotfl:", "Solution_5": "We still have an NC flag :|", "Solution_6": "Ask them to change it to the flag of India...hehe", "Solution_7": "NO!!!! I actually started the secession idea in this post from last year:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=33955", "Solution_8": "i like how it isn't in alphabetical order anymore :D", "Solution_9": "[quote=\"furious\"]i like how it isn't in alphabetical order anymore :D[/quote]\r\n\r\nhmm, I hid the rest of the forums that I don't use as much as I could...so I didn't really notice that.", "Solution_10": "Is Valentin going to fix it?", "Solution_11": "[quote=\"joml88\"][quote=\"furious\"]i like how it isn't in alphabetical order anymore :D[/quote]\n\nhmm, I hid the rest of the forums that I don't use as much as I could...so I didn't really notice that.[/quote]\r\n\r\nI took care of it." } { "Tag": [ "geometry", "circumcircle", "incenter", "search", "geometry unsolved" ], "Problem": "Consider the triangle $ ABC$ and its circumcircle. A circle $ S$ is tangent to the sides $ CA,CB$ and internally tangent with the circumcircle of the triangle $ ABC$ at the points $ M$, $ N$ and $ P$, respectively.\r\nCall $ I$-the incenter of the triangle $ ABC$. Prove that $ I$ is midpoint of $ [MN]$.", "Solution_1": "$ PM\\cap$ circumcircle=$ B'$,$ PN\\cap$ circumcircle=$ A'$\r\nApply Pascal theorem for$ AA'PB'BC$", "Solution_2": "[quote=\"Inequalities Master\"]Consider the triangle $ ABC$ and its circumcircle. A circle $ S$ is tangent to the sides $ CA,CB$ and internally tangent with the circumcircle of the triangle $ ABC$ at the points $ M$, $ N$ and $ P$, respectively.\nCall $ I$-the incenter of the triangle $ ABC$. Prove that $ I$ is midpoint of $ [MN]$.[/quote]\r\nThe result you mentioned is well know, the cirlcle S is called \"C-mixtilinear incircle\". You can find its proofs in many documents.", "Solution_3": "[b]77ant[/b], could you detail your proof?\r\nThank you very much.", "Solution_4": "I have seen here ---> http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1164894600&t=31739" } { "Tag": [ "trigonometry", "calculus", "integration", "calculus computations" ], "Problem": "Show the following: :D \r\n\\[ \\int_{0}^{\\infty}\\sin(x^n)\\,dx \\equal{} \\Gamma\\left(1\\plus{}\\dfrac{1}{n}\\right)\\cdot \\sin\\left(\\dfrac{\\pi}{2n}\\right)\\]\r\n\r\nWhere $ n>1$.", "Solution_1": "put $ t\\equal{}x^n$ next\r\n$ \\textbf I \\equal{} \\frac{1}{n} \\int_0^\\infty t^{\\frac{1}{n} \\minus{} 1} \\sin t \\, \\textbf d t \\equal{} \\ldots$\r\n(see http://www.mathlinks.ro/viewtopic.php?p=1107526#1107526 )\r\n$ \\ldots \\equal{} \\frac{1}{n} \\sin \\left( \\frac{\\pi}{2n} \\right) \\Gamma \\left( \\frac{1}{n} \\right) \\equal{} \\boxed{\\Gamma\\left(1\\plus{}\\dfrac{1}{n}\\right)\\cdot \\sin\\left(\\dfrac{\\pi}{2n}\\right)}$" } { "Tag": [], "Problem": "Find all integer solutions to a(a + 1) = b(b + 1)(b + 2)(b + 3).", "Solution_1": "Do you mean a(a+2)=b(b+1)(b+2)(b+3)?", "Solution_2": "NO, a(a + 1) = b(b + 1)(b + 2)(b + 3).", "Solution_3": "That makes it harder then, doesn't it?", "Solution_4": "Bwah ... Indeed, but after all, it's still easy.", "Solution_5": "[hide] a(a+1)=b(b+1)(b+2)(b+3).....a(a+1)=(b :^2: +3b+1) :^2: -1, a(a+1)-1=(b :^2: +3b+1) :^2:.\n\n :sqrt: ([a+0.25] :^2: -1.25)....i dont know where to go from there... :cry: [/hide]\n\n\n\numm..", "Solution_6": "what i've got so far...hmm...[hide]Let a+.5=c and b+1.5 be d. Now let (2c):^2:=x and (2d):^2:=y. We get 16(a)(a+1)=16(c-.5)(c+.5)=4(x-1) and 16b(b+1)(b+2)(b+3)=16(d-1.5)(d-.5)(d+.5)(d+1.5)=(y-9)(y-1). So 4(x-1)=(y-9)(y-1). x-1=m, y-5=n => 4m=(n-4)(n+4)=n:^2:-16=>n:^2:=4(m+4) So m=k:^2:-4 and n=:pm:2k => x=k:^2:-3 and y=5:pm:2k => (k-2c)(k+2c)=3 and I'm getting tired. this could probably be continued easily as three is prime, but it's not very elegant...[/hide]", "Solution_7": "Well, s0mp was almost there !\r\n\r\nContinue from a + a + 1 = (b + 3b + 1).\r\n\r\nIt's really easy now !", "Solution_8": "I didn't finish it, but maybe we can get sth from here:\r\n\r\nWe put k=b^2+3b and we have (a-k)(a+k+1)=k. We take some cases (a<,>0 k<,>0 etc.). I think the only solutions are (a,b) in {0,-1}X{0,-1,-2,-3}, meaning that the 2 sides of the equality must be 0.\r\n\r\nThis is just a guess, I don't know if it's true...", "Solution_9": "Well, actually it's much simpler.\r\n\r\nThe condition\r\na + a + 1 = (b + 3b + 1)\r\nimplies that a + a + 1 is a perfect square. But for a > 0 we have\r\na < a + a + 1 < (a + 1)\r\nand for a < 0 we have\r\n(a + 1) < a + a + 1 < a.\r\nSo we have a contradiction (unless a = 0) since there are no squares between x and (x + 1) for any integer x.", "Solution_10": "Oh, didn't see that... :D", "Solution_11": "Oh, thats sweet...." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that for every $ n\\ge 6$ there are at least two primes in the set \r\n\\[ \\{n\\plus{}1, n\\plus{}2,...,2n\\minus{}1\\}\\]\r\n\r\n:)", "Solution_1": "See [url=http://en.wikipedia.org/wiki/Ramanujan_prime]Ramanujan prime[/url] for the generalization.", "Solution_2": "Thanks t0rajir0u it is a very interesting topic, I think. I haven't found the proof for that theorem until now. \r\n\r\nFor this particular case an Erd\u00f6s like proof works, the only difficulty (because of calculation) is to prove the theorem for \"the small numbers\". \r\n\r\n:)", "Solution_3": "Another similar result that can be proved by very elementary means is the following: \r\n\r\nFor all $ n\\ge 1$ there is at least a prime in the set $ \\{2n, 2n\\plus{}1,...,3n\\}$.\r\n\r\nWhich is a special case of the more general: for $ a>0$ between $ n$ and $ (1\\plus{}a)n$ for $ n$ greater than ..., \r\n:)", "Solution_4": "I'd very much love to see an elementary proof, will someone give one please?", "Solution_5": "This might help. \r\n\r\nhttp://www.nd.edu/~dgalvin1/pdf/bertrand.pdf\r\n\r\n:)", "Solution_6": "Thankyou Elchapin", "Solution_7": "is ramanujam's prime in some way connected to bertrands postulate" } { "Tag": [ "inequalities", "function" ], "Problem": "This is the last inequality I'll post here.\r\nI guess you've got enough inequalities now ... :D \r\nSo enjoy this one ! It's far from easy I must admit ... :wink: \r\n\r\nLet a, b, c be positive reals. Prove that\r\n\r\na/:sqrt:(a + 8bc) + b/:sqrt:(b + 8ca) + c/:sqrt:(c + 8ab) :ge: 1.\r\n\r\n:mrgreen::^3:", "Solution_1": "This is difficult. Here are three possible approaches (HINT) :\r\n\r\n(1) Determine a real number k such that for a, b, c > 0 :\r\n\r\na/:sqrt:(a + 8bc) :ge: (a^k)/(a^k + b^k + c^k).\r\n\r\nTricky !\r\nI wouldn't have thought of this. This is the official solution ...\r\n\r\n(2) Use CAUCHY ! (Be inventive, it's not obvious how to apply it.)\r\n \r\n(3) Use Jensen for the convex function f(x) = 1/:sqrt:x.\r\n\r\nHappy solving :)", "Solution_2": "Well, I for one have seen so many solns I'm not even sure I CAN come up with sth original. The idea of having every term n the left >=a^k/(a^k+b^k+c^k) (or b^k/(blah) or c^k/(blah) ie really cool. I tried applying it myself for some ineqs and it kind of worked, but I don't think you can apply it that often. Anyway, it's really nice and elegatn. In this case k is, I think, 4/3, but I'm not sure. It can't be actually fund like when you solve an eqn, you have to do a bit of guessing..", "Solution_3": "OK I'll post the solutions myself. \r\n\r\n[1st solution] Proving that \r\na/:sqrt:(a + 8bc) :ge: a^(4/3)/(a^(4/3) + b^(4/3) + c^(4/3)\r\nis quite straightforward. \r\n(The proof is straigthforward, the idea isn't straightforward at all :mrgreen:)\r\nNow the result follows from adding the three similar inequalities.\r\n\r\n[2nd solution] Cauchy gives :\r\n\r\n(1) LHS x (:Sigma:a:sqrt:(a + 8bc)) :ge: (a + b + c)\r\n\r\n(2) :Sigma:a:sqrt:(a + 8bc) = :Sigma::sqrt:a:sqrt:(a + 8abc)\r\n:le::sqrt:((a + b + c)(a + b + c + 24abc)\r\n:le::sqrt:((a + b + c)(a + b + c)) = (a + b + c)\r\nwhere a + b + c + 24abc :le: (a + b + c) follows from AM - GM.\r\n\r\nNow combining (1) and (2) gives the result. \r\n\r\n[3rd solution] WLOG assume a + b + c = 1.\r\nThen Jensen applied to the convex function 1/:sqrt:x gives :\r\n:Sigma:af(a + 8bc) :ge: f(:Sigma:a(a + 8bc)) = f(a + b + c + 24abc)\r\n:ge: f((a + b + c)) = f(1) = 1.\r\nsince a + b + c + 24abc :le: (a + b + c) and f is decreasing." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate $ \\int_0^{\\frac{\\pi ^2}{4}} (2\\sin \\sqrt{x}\\plus{}\\sqrt{x}\\cos \\sqrt{x})\\ dx$.", "Solution_1": "Let $ u \\equal{} \\sqrt{x}$. Then we have:\r\n$ \\int_0^{\\frac {\\pi ^2}{4}} (2\\sin \\sqrt {x} \\plus{} \\sqrt {x}\\cos \\sqrt {x})\\ dx$\r\n$ \\equal{} 2\\int_0^{\\frac {\\pi}{2}} (2u\\sin u \\plus{} u^{2}\\cos u)\\ du$\r\n$ \\equal{} 2\\int_0^{\\frac {\\pi}{2}} (u^{2}\\sin u)'\\ du$\r\n$ \\equal{} \\frac {\\pi ^2}{2}$.", "Solution_2": "That's right! The problem is very easy for you. :lol:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Consider 7 distinct positive integers not exceeding 1706. Prove that there are three of them say a,b,c, such that a < b+c < 4a.", "Solution_1": "Suppose there exist positive and pairwise distinct integers $x_1, x_2, \\ldots, x_7 \\le 1706$ not fulfilling the condition postulated by the problem. Wlog, you may suppose $1 \\le x_1 < x_2 < \\ldots < x_n \\le 1706$. Then necessarily $4x_i \\le x_{i+1} + x_1$, i.e. $x_{i+1} \\ge 4x_i - 1$, for any $i = 2, 3, \\ldots, 6$. This means $x_3 \\ge 8 \\cdot 2 - 1 = 7$; $x_4 \\ge 4 \\cdot 7 - 1 = 27$; $\\ldots$; $x_7 \\ge 4 \\cdot 427 - 1 = 1707$, absurd! Now conclude..." } { "Tag": [], "Problem": "original image from http://www.dlsonline.net\r\nhat image from http://www.neowin.net", "Solution_1": "Something must seriously wrong with my eyes , I see only one Image.", "Solution_2": "Please reply only to threads that have been active recently. You may start your topic if the subject of your discussion is not on one of those threads." } { "Tag": [], "Problem": "Zdravo svima....\r\nPostavljam ovaj topic, jer mislim da je predizvik za sve koji vole matematiku.\r\nNajcesci dokaz o ovu teoremu je onaj poznati dokaz koristeci nize .\r\nPa zeljela bi da zajedno dokazemo Teorema Picarda (za postojanje i jedinstvenost resenije diferencijalnih jednacine), us pomoc Teoremi Banach o fiksnoj tacki.\r\nMozda je ovo neko vec dokazao???\r\nMolim vas pisite....\r\nPozdrav....", "Solution_1": "Nadam se da je [url=http://www.mathlinks.ro/viewtopic.php?t=258706]odgovor profesora Merryfielda[/url] bio zadovoljavaju\u0107i. Ako ne, pogledaj jo\u0161 i ovo\r\n\r\nhttp://www.its.caltech.edu/~awrichar/HW/ACM100_Picard_ExistenceUniqueness.pdf\r\n\r\na mo\u017eda se jo\u0161 neko pojavi sa svojim dokazom :) \r\n\r\nDobrodo\u0161la na Mathlinks!" } { "Tag": [], "Problem": "Can anyone point me to a good link explaining what is the infimum of a set or can anyone explain it to me ?\r\n\r\nThank you! :)", "Solution_1": "http://mathworld.wolfram.com/Infimum.html\r\n\r\nmathworld is usually a good place to look for these things..", "Solution_2": "I'll give it a look! Thank you!", "Solution_3": "It's the greatest lower bound. \r\n\r\nFor example, $0$ is the infimum of the set $1,\\frac{1}{2},\\frac {1}{3},\\frac {1}{4}$, etc. This can be seen because if we pick any $x>0$, by the Archimedian property, there exists some $t$ such that $\\frac {1}{t}0$ is not a possibility and $x=0$ works, we conclude that $0$ is the inf of the set." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Prove that:\r\n$ (\\frac{\\pm\\sqrt{\\sqrt[3]{4}\\plus{}3\\sqrt{\\frac{3}{\\sqrt[3]{4}}}}\\minus{}\\sqrt{\\sqrt[3]{4}\\minus{}\\sqrt{\\frac{3}{\\sqrt[3]{4}}}}}{2})^3\\plus{}1\\equal{}\\pm\\sqrt{\\frac{3\\sqrt{3}}{2}\\plus{}1}$\r\nGood luck!. :lol:", "Solution_1": "[quote=\"Allnames\"]Prove that:\n$ (\\frac {\\pm\\sqrt {\\sqrt [3]{4} \\plus{} 3\\sqrt {\\frac {3}{\\sqrt [3]{4}}}} \\minus{} \\sqrt {\\sqrt [3]{4} \\minus{} \\sqrt {\\frac {3}{\\sqrt [3]{4}}}}}{2})^3 \\plus{} 1 \\equal{} \\pm\\sqrt {\\frac {3\\sqrt {3}}{2} \\plus{} 1}$\nGood luck!. :lol:[/quote]\r\n\r\nIn LHS, multiply upper and lower part of the fraction by $ \\sqrt[6]2$. When entering inside the square roots, this becomes $ \\sqrt[3]2$ and what we get is :\r\n\r\n\r\n$ (\\frac {\\pm\\sqrt {2 \\plus{} 3\\sqrt 3} \\minus{} \\sqrt {2 \\minus{} \\sqrt 3}}{2\\sqrt[6]2})^3 \\plus{} 1 \\equal{} \\pm\\sqrt {\\frac {3\\sqrt {3}}{2} \\plus{} 1}$ $ \\iff$ $ (\\pm\\sqrt {2 \\plus{} 3\\sqrt 3} \\minus{} \\sqrt {2 \\minus{} \\sqrt 3})^3 \\plus{} 8\\sqrt 2 \\equal{} \\pm 8\\sqrt {2\\plus{}3\\sqrt 3}$\r\n\r\nUsing the fact that this equality must be true for the two values of $ \\pm$, it would be enough to verify :\r\n\r\n$ (i)$ : $ (\\sqrt{2\\plus{}3\\sqrt 3})^3\\plus{}3\\sqrt{2\\plus{}3\\sqrt 3}(2\\minus{}\\sqrt 3)\\equal{}8\\sqrt{2\\plus{}3\\sqrt 3}$ $ \\iff$ $ (2\\plus{}3\\sqrt 3)\\plus{}3(2\\minus{}\\sqrt 3)\\equal{}8$, which is obviously true\r\n\r\n$ (ii)$ : $ \\minus{}3(2\\plus{}3\\sqrt 3)\\sqrt{2\\minus{}\\sqrt 3}\\minus{}(\\sqrt{2\\minus{}\\sqrt 3})^3\\plus{}8\\sqrt 2\\equal{}0$ $ \\iff$ $ \\sqrt{2\\minus{}\\sqrt 3}(1\\plus{}\\sqrt 3)\\equal{}\\sqrt 2$\r\n\r\nSquaring both sides (since positive), $ (ii)$ $ \\iff$ $ (2\\minus{}\\sqrt3)(4\\plus{}2\\sqrt 3)\\equal{}2$, which is obviously true.\r\n\r\nQ.E.D.\r\n\r\nBut ..... what could be the interest of such a problem ?", "Solution_2": "[quote=\"pco\"][quote=\"Allnames\"]Prove that:\n$ (\\frac {\\pm\\sqrt {\\sqrt [3]{4} \\plus{} 3\\sqrt {\\frac {3}{\\sqrt [3]{4}}}} \\minus{} \\sqrt {\\sqrt [3]{4} \\minus{} \\sqrt {\\frac {3}{\\sqrt [3]{4}}}}}{2})^3 \\plus{} 1 \\equal{} \\pm\\sqrt {\\frac {3\\sqrt {3}}{2} \\plus{} 1}$\nGood luck!. :lol:[/quote]\n\nIn LHS, multiply upper and lower part of the fraction by $ \\sqrt [6]2$. When entering inside the square roots, this becomes $ \\sqrt [3]2$ and what we get is :\n\n\n$ (\\frac {\\pm\\sqrt {2 \\plus{} 3\\sqrt 3} \\minus{} \\sqrt {2 \\minus{} \\sqrt 3}}{2\\sqrt [6]2})^3 \\plus{} 1 \\equal{} \\pm\\sqrt {\\frac {3\\sqrt {3}}{2} \\plus{} 1}$ $ \\iff$ $ (\\pm\\sqrt {2 \\plus{} 3\\sqrt 3} \\minus{} \\sqrt {2 \\minus{} \\sqrt 3})^3 \\plus{} 8\\sqrt 2 \\equal{} \\pm 8\\sqrt {2 \\plus{} 3\\sqrt 3}$\n\nUsing the fact that this equality must be true for the two values of $ \\pm$, it would be enough to verify :\n\n$ (i)$ : $ (\\sqrt {2 \\plus{} 3\\sqrt 3})^3 \\plus{} 3\\sqrt {2 \\plus{} 3\\sqrt 3}(2 \\minus{} \\sqrt 3) \\equal{} 8\\sqrt {2 \\plus{} 3\\sqrt 3}$ $ \\iff$ $ (2 \\plus{} 3\\sqrt 3) \\plus{} 3(2 \\minus{} \\sqrt 3) \\equal{} 8$, which is obviously true\n\n$ (ii)$ : $ \\minus{} 3(2 \\plus{} 3\\sqrt 3)\\sqrt {2 \\minus{} \\sqrt 3} \\minus{} (\\sqrt {2 \\minus{} \\sqrt 3})^3 \\plus{} 8\\sqrt 2 \\equal{} 0$ $ \\iff$ $ \\sqrt {2 \\minus{} \\sqrt 3}(1 \\plus{} \\sqrt 3) \\equal{} \\sqrt 2$\n\nSquaring both sides (since positive), $ (ii)$ $ \\iff$ $ (2 \\minus{} \\sqrt3)(4 \\plus{} 2\\sqrt 3) \\equal{} 2$, which is obviously true.\n\nQ.E.D.\n\nBut ..... what could be the interest of such a problem ?[/quote]\r\nThank for your fast reply. As I wrote. It is time for relax. This problem is posted for funny purpose (but not spam :lol: ). Did you surprise on it ?. And I think that is interested point!. :) \r\nI am waiting for another proof, my friends!", "Solution_3": "If you are interested in stange equalities, a classical way to build some is Cardan's solution for $ 3^{rd}$ degree equations :\r\n\r\nLet $ x^3 \\plus{} 3bx \\minus{} 8a^3 \\minus{} 6ab \\equal{} 0$; with $ b\\geq 0$. \r\n\r\nThis equation has a unique real root $ 2a$ and Cardan's formulas gives an unique real solution for this equation $ \\sqrt [3]{4a^3 \\plus{} 3ab \\plus{} \\sqrt {a^2(4a^2 \\plus{} 3b)^2 \\plus{} b^3}}$ +$ \\sqrt [3]{4a^3 \\plus{} 3ab \\minus{} \\sqrt {a^2(4a^2 \\plus{} 3b)^2 \\plus{} b^3}}$\r\n\r\nSo the identity : $ 2a \\equal{} \\sqrt [3]{4a^3 \\plus{} 3ab \\plus{} \\sqrt {a^2(4a^2 \\plus{} 3b)^2 \\plus{} b^3}}$ +$ \\sqrt [3]{4a^3 \\plus{} 3ab \\minus{} \\sqrt {a^2(4a^2 \\plus{} 3b)^2 \\plus{} b^3}}$ $ \\forall a,\\forall b\\geq 0$\r\n\r\nWith $ a \\equal{} b \\equal{} 1$, for example, you get $ 2 \\equal{} \\sqrt [3]{7 \\plus{} 5\\sqrt {2}}$ +$ \\sqrt [3]{7 \\minus{} 5\\sqrt {2}}$\r\n\r\nWith $ a \\equal{} 1$ and $ b \\equal{} 2$, you get $ 2 \\equal{} \\sqrt [3]{10 \\plus{} 6\\sqrt {3}}$ +$ \\sqrt [3]{10 \\minus{} 6\\sqrt {3}}$ which may also be written $ \\sqrt [3]4 \\equal{} \\sqrt [3]{5 \\plus{} 3\\sqrt {3}}$ +$ \\sqrt [3]{5 \\minus{} 3\\sqrt {3}}$", "Solution_4": "[quote=\"Allnames\"]I am waiting for another proof, my friends![/quote]\r\n\r\nIf you want another solution, here is one :\r\n\r\nIt's easy to check that $ (x\\minus{}a)^3\\minus{}8x\\plus{}8a(3\\minus{}a^2)\\equal{}(x^2\\plus{}3a^2\\minus{}8)(x\\minus{}3a)$ $ \\forall x,a$\r\n\r\nAnd so $ (\\pm\\sqrt{8\\minus{}3a^2}\\minus{}a)^3\\plus{}8a(3\\minus{}a^2)\\equal{}\\pm 8\\sqrt{8\\minus{}3a^2}$ $ \\forall a$ such that $ 3a^2\\leq 8$\r\n\r\nSetting $ a\\equal{}\\sqrt{2\\minus{}\\sqrt 3}$ is this identity gives the result $ (\\pm\\sqrt {2 \\plus{} 3\\sqrt 3} \\minus{} \\sqrt {2 \\minus{} \\sqrt 3})^3 \\plus{} 8\\sqrt 2 \\equal{} \\pm 8\\sqrt {2 \\plus{} 3\\sqrt 3}$ which is equivalent to the required result (see my previous post).\r\n\r\nUsing other values for $ a$ would lead to other \"identities\" ...", "Solution_5": "Thank you; dear pco. I got that equality when I solved the equation $ \\sqrt[3]{x\\minus{}1}\\plus{}\\sqrt[3]{x\\plus{}1}\\equal{}\\sqrt[3]{2}x$.\r\nI thought that everyone would work hard to get that equation above. But what you did makes my thinking be silly.\r\nThank you!" } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "let a,b,c>0 such that a+b+c=1 prove that\r\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3} \\leq 1$", "Solution_1": "[quote=\"tranquoc\"]let a,b,c>0 such that a+b+c=1 prove that\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3} \\leq 1$[/quote]\r\n\r\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3}\\le \\sqrt {2(a \\minus{} b \\plus{} b \\minus{} c \\plus{} c \\minus{} a) \\plus{} (\\sum a\\sqrt {a})^2} \\equal{} \\sqrt {\\sum (a\\sqrt {a})^2}$\r\n\r\nwe need prove that\r\n\r\n$ \\sum (a\\sqrt {a})\\le 1$\r\n\r\nwe have $ \\sum (a\\sqrt {a})\\le \\sqrt {(\\sum a^2)(\\sum a)} \\equal{} \\sqrt {\\sum a^2}\\le \\sqrt {\\sum a} \\equal{} 1$ \r\n\r\n(because $ a\\le 1,b\\le 1,c\\le 1$)\r\n\r\ndone~!", "Solution_2": "are you sure?", "Solution_3": "[quote=\"bunhiacovski\"][quote=\"tranquoc\"]let a,b,c>0 such that a+b+c=1 prove that\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3} \\leq 1$[/quote]\n\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3}\\le \\sqrt {2(a \\minus{} b \\plus{} b \\minus{} c \\plus{} c \\minus{} a) \\plus{} (\\sum a\\sqrt {a})^2} \\equal{} \\sqrt {\\sum (a\\sqrt {a})^2}$\n\nwe need prove that\n\n$ \\sum (a\\sqrt {a})\\le 1$\n\nwe have $ \\sum (a\\sqrt {a})\\le \\sqrt {(\\sum a^2)(\\sum a)} \\equal{} \\sqrt {\\sum a^2}\\le \\sqrt {\\sum a} \\equal{} 1$ \n\n(because $ a\\le 1,b\\le 1,c\\le 1$)\n\ndone~![/quote]\r\nWhen does equal occur ?", "Solution_4": "[quote=\"tranquoc\"]let a,b,c>0 such that a+b+c=1 prove that\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3} \\leq 1$[/quote]\r\n\r\n Try $ a\\rightarrow0, c\\rightarrow0, b\\rightarrow1$", "Solution_5": "[quote=\"Marius Mainea\"][quote=\"tranquoc\"]let a,b,c>0 such that a+b+c=1 prove that\n$ \\sum\\sqrt {a \\minus{} b \\plus{} c^3} \\leq 1$[/quote]\n\n Try $ a\\rightarrow0, c\\rightarrow0, b\\rightarrow1$[/quote]\r\nAny condition,tran quoc,\r\nThis ineq is quite nice and i thank if we have a suitable condition,this ineq will be true (i cant find it now,sorry)" } { "Tag": [ "geometry", "3D geometry", "number theory proposed", "number theory" ], "Problem": "Show that 1997^17 cannot be a sum of consecuative cubes.I think it's easy!!", "Solution_1": "[quote=\"a_vakilian\"]I think it's easy!![/quote]\r\nIt seems that it is not so esay because no one solved it,but I know it is realy simple :?", "Solution_2": "Let me tell you a common fact among mathlinks to you...problem remaining unsolved is always because of two reasons:\r\n-extremely extraordinary hard\r\n-too easy\r\n\r\nYou problem definitely belongs to the latter type.\r\n\r\nAnyway, the essential idea for the solution is just the formula $1^3+2^3+...n^3=(\\frac{n(n+1)}{2})^2$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "number theory proposed", "number theory" ], "Problem": "Find the lesat positive integer $n(n>1)$ such that $\\frac{1^{2}+2^{2}+...+n^{2}}{n}$ is a perfect square number.", "Solution_1": "$\\frac{1^{2}+2^{2}+...+n^{2}}{n}=\\frac{(n+1)(2n+1)}{6}$ \r\nIt is equavalent to Pell's equation $(4n+3)^{2}-48k^{2}=1$. \r\nTherefore $4n+3=\\frac{(7+\\sqrt{48})^{l}+(7-\\sqrt{48})^{l}}{2}, l\\in N$.", "Solution_2": "I don't understand.Can you post full solution?", "Solution_3": "its a USAMO problem and i think the answer is $n=13$. :)\r\n\r\nADD: oh Rust used the theory of pell equations. if you're not familiar, you can get it by trial. :wink:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Hello,\r\n\r\nLet $ a \\geq b \\geq c>0$\r\nProve that : $ a^2 \\times \\frac{b}{c}\\plus{}b^2 \\times \\frac{c}{a}\\plus{}c^2 \\times \\frac{a}{b} \\geq a^2\\plus{}b^2\\plus{}c^2$\r\n\r\nmhdi", "Solution_1": "ineq $ \\Leftrightarrow(b\\minus{}c)(\\frac{a^2}{c}\\minus{}\\frac{b^2}{a})\\plus{}(a\\minus{}b)(\\frac{c^2}{b}\\minus{}\\frac{b^2}{a})\\geq 0\\Leftrightarrow(b\\minus{}c)\\frac{a^3\\minus{}b^2c}{ca}\\plus{}(a\\minus{}b)\\frac{c^2a\\minus{}b^3}{ab}\\geq 0$\r\n$ \\Leftrightarrow(b\\minus{}c)\\frac{a^2\\minus{}b^2}{a}\\plus{}(a\\minus{}b)\\frac{c^2\\minus{}b^2}{a}\\geq 0\\Leftrightarrow\\frac{(a\\minus{}b)(b\\minus{}c)(a\\minus{}c)}{a}\\geq 0,Right$", "Solution_2": "Hello,\r\n\r\nI think there is a mistake here :[quote]$ (b \\minus{} c)(\\frac {a^2}{c} \\minus{} \\frac {b^2}{a})$[/quote]\r\nIf i expand i'll find this term : $ \\frac {b^3}{a}$ which doesn't exist!\r\n\r\nMhdi", "Solution_3": "You'll find a term $ \\minus{} \\frac {b^3}{a}$ and term $ \\frac {b^3}{a}$ ;)", "Solution_4": "Okay! Thank you!", "Solution_5": "$ \\Leftrightarrow$ $ a^3b^2\\plus{}b^3c^2\\plus{}c^3a^2\\geq a^3bc\\plus{}ab^3c\\plus{}abc^3\\geq 0$\r\n $ \\Leftrightarrow$\r\n $ a^3b(b\\minus{}c)\\plus{}b^2c^2(b\\minus{}c)\\plus{}z^3(a^2\\plus{}b^2\\minus{}2ab)\\minus{}abc(b^2\\minus{}c^2)\\geq0$\r\n $ \\Leftrightarrow$\r\n $ (b\\minus{}c)(a\\minus{}c)(a^2b\\plus{}bc(b\\minus{}c))\\plus{}a^3(b\\minus{}c)^2\\geq 0$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all function $ f: \\mathbb{R}\\to \\mathbb{R^\\plus{}}$ such that:\r\n$ \\sum \\sqrt{f(x)^2\\plus{}f(x)f(y)\\plus{}f(y)^2}\\sqrt{f(y)^2\\plus{}f(y)f(z)\\plus{}f(z)^2}\\equal{}(\\sum f(x))^2$\r\nfor all $ x,y,z \\in \\mathbb{R^\\plus{}}$.(Symbol $ \\mathbb{R^\\plus{}}$ denotes the set of all positive real number.)", "Solution_1": "I guess it's just an inequality.\r\n L.H.S.\u2265R.H.S. and equality holds only when f(x) is a constant." } { "Tag": [ "geometry", "advanced fields", "advanced fields unsolved" ], "Problem": "\\documentclass{article}\r\n\\usepackage{latexsym}\r\n\r\n\\begin{document}\r\n\r\nLet $\\omega$ be a symplectic differentiable 2-form which is globally \r\ndefined on ${\\rm R}^{4}$. Prove that there exist four 1-forms $\\theta_{1},\\;\\theta_{2},\\;\\theta_{3},\\;\\theta_{4}$ defined on ${\\rm R}^{4}$ so that the \r\nfollowing equality holds (globally) on ${\\rm R}^{4}$: $\\omega =\\theta_{1}\\wedge \\theta_{2}+\\theta_{3}\\wedge \\theta_{4}$. \r\n\r\nIf its true, where could I find the proof? If not, is there a \r\ncounterexample? \r\n\r\nThanks \r\n\r\n\\end{document}", "Solution_1": "If $\\omega$ had constant coefficients, you would find $\\theta_{i}$ using linear algebra. (Namely, the block-diagonal form of a skew-symmetric matrix.) Apply the same process to $\\omega(x)$ pointwise; the resulting 1-forms $\\theta_{i}(x)$ will be differentiable by virtue of nondegeneracy of $\\omega$." } { "Tag": [ "calculus", "integration", "function", "geometry", "inequalities", "calculus computations" ], "Problem": "Let $ f(x)$ be a continuous function such that $ f(x) \\equal{} \\frac {f(x \\plus{} y) \\plus{} f(x \\minus{} y)}{2}$ for all $ x,\\ y\\in\\mathbb{R}$. \r\n\r\nProve that $ \\int_a^b f(x)\\ dx \\equal{} \\frac {b \\minus{} a}{2}\\{f(a) \\plus{} f(b)\\}\\ (a < b)$.", "Solution_1": "It looks like the equation defines a straight line, the mid-value being the mid-point of the line segment joining $ f(x\\minus{}y)$ to $ f(x\\plus{}y)$ and so the integral is the area of a trapezium.", "Solution_2": "Yeah its Jensens functional equation with solution f(x)=ax+b", "Solution_3": "Could you show your work in detail, guys? :)", "Solution_4": "[quote=\"kunny\"]Could you show your work in detail, guys? :)[/quote]\r\n\r\nHello kunny! Here is a simple way :\r\n\r\nLet $ P(x,y)$ be the assertion $ f(x)\\equal{}\\frac{f(x\\plus{}y)\\plus{}f(x\\minus{}y)}2$\r\nLet $ g(x)\\equal{}f(x)\\minus{}f(0)$\r\n\r\nObviously, $ g(x)$ is too a solution and so matches $ P(x,y)$ and, since $ g(0)\\equal{}0$ :\r\n\r\n$ P(0,x)$ $ \\implies$ $ g(\\minus{}x)\\equal{}\\minus{}g(x)$\r\n$ P(x,x)$ $ \\implies$ $ g(2x)\\equal{}2g(x)$\r\n\r\n$ P(x,x\\plus{}2y)$ $ \\implies$ $ g(x)\\equal{}\\frac{g(2x\\plus{}2y)\\plus{}g(\\minus{}2y)}2$ $ \\equal{}\\frac{2g(x\\plus{}y)\\minus{}2g(y)}2$ and so $ g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y)$\r\n\r\nThis is a Cauchy equation whose only continuous solutions are $ g(x)\\equal{}ax$, which indeed are solutions of the original equation.\r\n\r\nHence the general solution $ f(x)\\equal{}ax\\plus{}b$\r\n\r\nHence the result.", "Solution_5": "Thank you, pco." } { "Tag": [], "Problem": "Hanna runs faster than Jenny but slower than Lim. They started at \r\nthe same time from the same place and ran around a circular path. After a while they stopped simultaneously at the same place from which they started. It turned out that Lim overtook Jenny ten times during this run. How many times did it occur that one of the girls overtook another? Assume that each of the girls ran at a constant speed. (The final stop together is not counted as an overtake.)", "Solution_1": "okay, if lim over took jenny ten times then he over took hanna 5 times hanna overtook jenny 5 times too. jenny never over took any one because she's the slowest so m guess is that all together they over took 20 times????? :?: \r\nis that right???? :thumbup: \r\nor not???? I'm not sure because i was assuming 5 :blush: \r\nit might have some thing to do with the circumference of the circle.and thats how u find the real answer, but i hope my answer is correct.", "Solution_2": "has nothing to do with dircumference or even the track shape\r\n\r\nas long as it is track \r\n\r\nso lim overtook Jenny 10 times which means lim did 11 more laps than Jenny *think about it yourself, it is commonsense after you think about it*\r\n\r\nhanna is somewhere between them, say she did x laps more than Jenny and 11-x laps less than lim\r\nshe overlapped jenny x-1 times and was lapped by lim 10-x times for a total of 9 times\r\nso 10+9=19\r\nim not sure about the number theaory but the point is as long as she is faster than Jenny and slower than lim (and ends on the finish line), it doesnt metter how much faster or slower.", "Solution_3": "Let's say that Jenny ran one lap. That means that Lim ran 12 laps. Let's say that Hanna ran 5 laps. Lim passed Jenny 10 times. Lim passed Hanna 6 times. Hanna passed Jenny 3 times. Therefore, a girl overtook another 19 times in all." } { "Tag": [ "function", "inequalities proposed", "inequalities" ], "Problem": "For $ n\\geq 3$, Let $ x_k\\ (k \\equal{} 1,2,\\ \\cdots n)$ be real numbers such that $ \\sum_{k \\equal{} 1}^n x_k \\equal{} 0,\\ \\sum_{k \\equal{} 1}^n x_k^2 \\equal{} 1$.\r\nFind the maximum value of $ \\sum_{k \\equal{} 1}^n x_k^3$.", "Solution_1": "Can anyone solve the problem? :lol:", "Solution_2": "[quote=\"kunny\"]For $ n\\geq 3$, Let $ x_k\\ (k \\equal{} 1,2,\\ \\cdots n)$ be real numbers such that $ \\sum_{k \\equal{} 1}^n x_k \\equal{} 0,\\ \\sum_{k \\equal{} 1}^n x_k^2 \\equal{} 1$.\nFind the maximum value of $ \\sum_{k \\equal{} 1}^n x_k^3$.[/quote]By Lagrange multipliers, :blush: it is easy to find $ max \\equal{} \\frac {n \\minus{} 2}{\\sqrt {n(n \\minus{} 1)}}$, \r\n\r\nwhich is attained for $ x_1 \\equal{} ... \\equal{} x_{n \\minus{} 1} \\equal{} \\minus{} \\frac {1}{\\sqrt {n(n \\minus{} 1)}}$ and $ x_n \\equal{} \\sqrt {\\frac {n \\minus{} 1}{n}}$ :lol:\r\n\r\nI wonder if there is a better method :maybe:", "Solution_3": "We can prove this problem using the following extension of EV-Theorem to real numbers.\r\n \r\n[b]EV-Theorem Extension.[/b]\r\n[i]Let $ p$ be an even positive integer, and let $ f(u)$ be a differentiable function on [b]R[/b] such that $ g(x)\\equal{}f'(x^\\frac{1}{p\\minus{}1})$ is strictly convex on [b]R[/b].\nLet $ a$ and $ b$ be real numbers such that $ |a|H(y)$.\r\n\r\nNow I will show that this satisfies the conditions.\r\n\r\nLemma. For any number $ a$, there are finitely many values $ b$ such that $ H(b)\\le a$.\r\n\r\nProof. First look only at positive $ b$. If $ b\\equal{}\\frac{m}{n}$ then $ m\\plus{}n\\le a$, so $ m\\le a$ and $ n\\le a$. Each of $ m$ and $ n$ will be of $ 0,1,2,...,a$ so there are $ a\\plus{}1$ possible values for each, so there are at most $ (a\\plus{}1)^2$ possible values of positive $ b$. The same applies for negative $ b$, so there are at most $ 2(a\\plus{}1)^2$, a finite amount, of values $ b$ satisfying $ H(b)\\le a$. This completes the lemma.\r\n\r\nNow, for a given value of $ y$, there are finitely many values of $ x$ such that $ H(x)\\le H(y)$, so on a given horizontal line, there are finitely many rational points in $ A$. Similarily, for a given value of $ x$, there are finitely many values of $ y$ such that $ H(x)>H(y)$, so on a given vertical line, there are finitely many rational points in $ B$, so this partition satisfies the conditions. QED." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "How is it that the following definitions are equivalent?:\r\n\r\n1. [i]A function $ f: \\left(X,\\mathcal{M}\\right)\\rightarrow\\left(\\mathbb{R},\\mathcal{B}_{\\mathbb{R}}\\right)$ is measurable if for each $ E\\in\\mathcal{B}_{\\mathbb{R}}$, $ f^{\\minus{}1}\\left(E\\right)\\in\\mathcal{M}$,[/i]\r\n\r\n2. [i]A function $ f: \\left(X,\\mathcal{M}\\right)\\rightarrow\\left(\\mathbb{R},\\mathcal{B}_{\\mathbb{R}}\\right)$ is measurable if the set $ \\left\\{x\\in X\\ : \\ f\\left(x\\right)>a\\right\\}$ is measurable for each $ a\\in\\mathbb{R}$.[/i]", "Solution_1": "$ \\mathcal{B}_{\\mathbb{R}}\\equal{}\\sigma (J_{oc} )\\equal{} \\sigma (J_{oo}) \\equal{}\\sigma (J_{co}) \\equal{}\\sigma (J_{cc})$\r\n\r\n$ J_{oc} \\equal{} \\{ (a,b]: a,b \\in \\mathbb{R} \\}$\r\n$ J_{oo} \\equal{} \\{ (a,b): a,b \\in \\mathbb{R} \\}$\r\n$ J_{co} \\equal{} \\{ [a,b): a,b \\in \\mathbb{R} \\}$\r\n$ J_{cc} \\equal{} \\{ [a,b]: a,b \\in \\mathbb{R} \\}$\r\nas well as the open and closed rays, which generate the same $ \\sigma$-algebra" } { "Tag": [], "Problem": "Use the following information: Q is in the interior of 0$), then for almost all points $ x\\in A$ the following holds:\r\n\\[ \\lim \\mu(A\\cap P)/\\mu(P)\\equal{}1,\\] where $ P$ is a coordinate parallelepiped, containing $ x$, and diameter of $ P$ tends to 0. Here coordiante parallelepiped is the parallelepiped with edges parallel to coordinate aces (i.e. $ P\\equal{}\\{(t_1,t_2\\dots,t_n): a_i\\leq t_i\\leq b_i\\}$). \r\n\r\nThe difference with the usual LDT is that we take not only regular guys like cubes or balls, but parallelepipeds with arbitrary different edges, the only requirement is that edges are small.\r\n\r\nIt is easy to prove, but I need a reference. Quite surprisingly, it is not well known and usually is absent in measure theory books and lecture curses.\r\n\r\nThanks!", "Solution_1": "Search Google. I did one search and came up with a lot of sites.", "Solution_2": "Thanks! Really, of course I tried with google, but without success. It suggests a book \"Geometry of Sets and Measures in Euclidean Spaces. Fractals and Rectifiability\" by Pertti Mattila, but it is not available neither online nor in my Institute. So I do not know, whether it contains such modification. \r\n\r\nWould you please give me a hint, what should I search for?", "Solution_3": "You're looking for the \"Strong Maximal Function\" and the theorem of Jessen, Marcinkiewicz, and Zygmund.\r\n\r\nFor a locally integrable function $ f,$ the strong maximal function is this:\r\n\\[ M_S(f)(x) \\equal{} \\sup_{P\\ni x}\\frac1{|P|}\\int_P|f|\\]\r\nwhere $ P$ ranges over the parallelepipeds that you mention.\r\n\r\nIf the dimension $ n$ equals $ 1,$ then this is the Hardy-Littlewood maximal function. The Hardy-Littlewood maximal operator maps $ L^p$ to $ L^p$ for $ p > 1$ and $ L^1$ to weak $ L^1.$ Because of the latter, we have the Lebesgue theorem on the differentiation of the integral, which works for every (locally) $ L^1$ function.\r\n\r\nThe key to the Jessen-Marcinkiewicz-Zygmund result is to somehow make the strong maximal operator appear as if you composed $ n$ sequential one-dimensional Hardy-Littlewood operators. As a result, we say that $ M_S$ maps $ L^p$ to $ L^p$ for $ p > 1.$ As a result, if $ f\\in L^p,\\ p > 1,$ we have\r\n\\[ \\lim_{P\\text{ shrinks to }x}\\frac1{|P|}\\int_Pf \\equal{} f(x)\\text{ for almost every }x.\\]\r\nSince you're applying this to the characteristic function of $ A,$ and that characteristic function is in $ L^{\\infty},$ hence locally $ L^p$ for any $ p,$ that's enough for your result.\r\n\r\nBut note that the strong maximal function does not send $ L^1$ to weak $ L^1,$ and as a result, strong differentiation (that is, this kind of differentiation, through rectangles), fails for some $ f\\in L^1.$ What we can say (and this is in the original J-M-Z paper) that if $ f\\in L\\log^{n \\minus{} 1}L,$ then $ M_S(f)$ is in weak $ L^1$ and the differentiation theorem applies.\r\n\r\nThis is all assuming that the edges of your paralellepiped are parallel to the axis. If you allow arbitrary orientations, then you're walking into the \"Kakeya problem\" and the results are far worse.", "Solution_4": "What exactly are you looking for? Just books that contain references to it? Here are two:\r\n\r\n\"Measure and Integral\" by Wheeden & Zygmund. They don't explicitly name it but it looks like they discuss it (see chapter 7).\r\n\r\n\"Real Analysis\" by Carothers. It mentions it by name and proves it (see chapter 20).", "Solution_5": "Dear Kent, thank you for the valuable comment. But I rather need the reference to this particular result, not the proof of it (I may prove it myself :blush: )...\r\n\r\nJRav, alas, I could not find it neither of these two books, only one-dimensional or \"uniform\" versions (with balls or cubes instead parallelepipeds).", "Solution_6": "Jessen, B., Marcinkiewicz, J. and Zygmund, A., Note on the differentiability of multiple integrals, [i]Fundamenta Mathematicae[/i] [b]25[/b] (1935), 217-34.\r\n\r\n(I have the paper itself as reprinted in the collected works of Marcinkiewicz.)\r\n\r\nYou might also see what searching for the words \"strong density\" gets you.", "Solution_7": "Thank you!" } { "Tag": [ "AMC", "AIME", "USA(J)MO", "USAMO", "AMC 12", "function", "inequalities" ], "Problem": "Post your tenative scores and discuss problems here. I think I got 8 at this point... Not sure about number 12.\r\n\r\n1)252\r\n2)025\r\n3)314\r\n4)080\r\n5)014\r\n6)017\r\n8)047\r\n12)375", "Solution_1": "Got 9. (problems 1-9)", "Solution_2": "8. Because I interpreted #5 weirdly just for the fun of it, off-by-one'd #6, and didn't attempt #12, #14, or #15 even though they were painfully easy.", "Solution_3": "7 because I didn't think much on 7 since I thought it was hard, and 12 because it looked beastly. Messed up number 9 because of addition error, was being dumb on number 11 :( I could've had a 9 at least...if I wasn't being dumb at all then 11. Congratulations to Albert for miserably failing.", "Solution_4": "1. 252\r\n2. 025\r\n3. 314\r\n4. 080\r\n5. 014\r\n8. 047\r\n9. 190\r\n\r\nfor me.", "Solution_5": "[quote=\"cognos599\"]1. 252\n2. 025\n3. 314\n4. 080\n5. 014\n8. 047\n9. 190\n\nfor me.[/quote]\r\nYo at least like have the courtesy of saying \"walks over to CA\" if you're gonna help spam up our forum. Don't you Indiana people have your own forum to spam up? I see you doing a pretty good job with that so I don't see why we need any help here.", "Solution_6": "i think i got a 6 woot woot!\r\n1-3, 5, 12\r\non 4 i thought the radius of the big circle was the height, not the slant height :wallbash: \r\nand on 7 i got $ \\frac {570}{3} \\equal{} 170$... :cursing: :cursing: \r\nbut im happy! i probably wont make USAMO but im only in 8th grade so its cool :D\r\n\r\n\r\nhehe i totally copied this post from my one in the other forum :P", "Solution_7": "here, in good spirit i edited my post and i am sorry for spamming the CA forum. :)", "Solution_8": "Wow, you two, just shut up. What does saying \"walking over to CA\" accomplish?\r\nSure, posting in a CA forum when you're in IN is a bit weird, but whatever, it's cool. It's not like it's taking up space on your computer, and it hardly qualifies as \"spamming\".", "Solution_9": "come on guys \r\nitsh all cool\r\n9 by the way \r\ngahhh missed 14\r\nthought that $ (9 \\plus{} x)^2 \\minus{} 81 \\equal{} 81 \\plus{} 18x$\r\noh dude \r\nAndrew, I think you mean \"miserably failing\"\r\nyeah quotes help\r\nholy bananas... daermon's in 8th \r\nI thought you were in 9th dude", "Solution_10": "Assuming the bubbling was okay, got a 7, and I doubt my error-ridden AMC will get me to USAMO.\r\nThe first five were okay. THen I got the right strategy for #6 but miscounted the number of entries in the vertical column. Then I missed #7 by one, because $ S_{999}$ obviously implied that there were 999 sets. 8 was very familiar, 9 was ok as well. I died on 10, messed up my casework for 11, and didn't really look at 12-15.", "Solution_11": "Good job everyone!?\r\nGot #1-7, 9, 12 correct\r\nGot the correct answer for 8, bubbled it in, thought it was wrong and changed it... (yes, i'm a failure) :( \r\nWas just being stupid on 10, 14, 15...and gave up on 13 after looking at it once :lol: \r\nHopefully I qualified for USAMO, keeping my fingers crossed :roll:", "Solution_12": "Fine fine fine sorry. Dude David the \"walks over to CA\" thing was a joke...\r\nWhat...archimedes you failed too? Dude this is really awesome for me, if I don't make USAMO, at least I have consolation that TWO beastly people didn't make USAMO (Brian Zhang has 126 AMC12, 7 AIME) :D But then its not as awesome as getting a good score and making USAMO so...", "Solution_13": "8 i think\r\nmaybe 9 or 10 if i have good luck\r\nmistakes\r\n23+24=57 on #8\r\nmisbubbled (maybe) on 12\r\ngah were 10,11 hard?\r\n9) 190\r\n12) 375?\r\n13) 012?????\r\n\r\nill hope for teh best", "Solution_14": "As mentioned before, I got #1 wrong. :noo: But to make up for that, I got 2-10, 14. So I got a score of [b]10 (assuming no bubbling errors)[/b]. On #10, it was weird since I assumed that AB=BC=CD without even thinking about it.\r\n\r\nThis is surprising. I did better than most people here, though I really didn't expect it.\r\n\r\nAnd... After the test, I got 1, 11, 13, and 15. Technically, if I sped up a bit, I could've pulled off a 14. But then again, there are many, many people who could've scored a 15 \"if they just had some more time\".\r\n\r\nDid anyone think that #12 was wierd? When working on it, I got something like a constant over a linear function of $ n$, and I thought, \"that can't be it. AIME never tests you on limits\", but that's exactly what they did on the solution. Maybe there's a nice AM-GM way or some inequality I'm not familiar with.", "Solution_15": "[quote=\"vishalarul\"]As mentioned before, I got #1 wrong. :noo: But to make up for that, I got 2-10, 14. So I got a score of [b]10 (assuming no bubbling errors)[/b]. On #10, it was weird since I assumed that AB=BC=CD without even thinking about it.\n\nThis is surprising. I did better than most people here, though I really didn't expect it.\n\nAnd... After the test, I got 1, 11, 13, and 15. Technically, if I sped up a bit, I could've pulled off a 14. But then again, there are many, many people who could've scored a 15 \"if they just had some more time\".\n\nDid anyone think that #12 was wierd? When working on it, I got something like a constant over a linear function of $ n$, and I thought, \"that can't be it. AIME never tests you on limits\", but that's exactly what they did on the solution. Maybe there's a nice AM-GM way or some inequality I'm not familiar with.[/quote]\r\nHmmmm...didn't it say 4 feet for each 15mph OR FRACTION THEREOF? So its like, \"stepped\" and the last step is at the asymptote.\r\nI could've gotten 11 but eh...I suck. 7, 12 literally took me 2 seconds each after the test...too bad it was after.\r\nYou're lucky you had calc and other bashing methods at your disposal for number 14...I attempted it but it quickly turned into stuff that was too advanced for my limited knowledge :( I really need to work on getting the simple solutions heh.", "Solution_16": "[quote=\"vishalarul\"]As mentioned before, I got #1 wrong. :noo: But to make up for that, I got 2-10, 14. So I got a score of [b]10 (assuming no bubbling errors)[/b]. On #10, it was weird since I assumed that AB=BC=CD without even thinking about it.\n\nThis is surprising. I did better than most people here, though I really didn't expect it.\n\nAnd... After the test, I got 1, 11, 13, and 15. Technically, if I sped up a bit, I could've pulled off a 14. But then again, there are many, many people who could've scored a 15 \"if they just had some more time\".\n\nDid anyone think that #12 was wierd? When working on it, I got something like a constant over a linear function of $ n$, and I thought, \"that can't be it. AIME never tests you on limits\", but that's exactly what they did on the solution. Maybe there's a nice AM-GM way or some inequality I'm not familiar with.[/quote]\r\n\r\nYeah, I wasted so much time on #12. Anyways, the way you did it isn't entirely correct, although a lot of people did that - there are no such things as limits on this problem. Cars have a finite speed (the phrase \"cars can go at any speed\" doesn't mean they can go at infinite speed), so clearly limits have nothing to do with the problem. The correct solution is what serial pointed out, that you need to have a car start off next to the eye, and I missed that :( .\r\n\r\nSo I didn't make USAMO :( :( :( .\r\n\r\nAlso, how did you do #13?", "Solution_17": "Dude, Vishal, seriously, nice job.\r\n\r\nHmm, perhaps I'll go into more detail about my AIME experience.\r\nSo at the 2-hour mark, I'd done #1-5, 7-9. (having skipped #6 temporarily because I was being dumb) So, as my goal was 8 anyway and I had done 8 problems, I was like, \"oh, I should check now.\"\r\n\r\n[b]Which was a good thing, because I had somehow managed to mess up #1 and #3.[/b]\r\n\r\nSo that left about 45 minutes left. Got #6 in about another 10 minutes, and then went to the bathroom, as I was feeling quite accomplished (after all, I had only got 5 on last year's AIME).\r\n\r\n30 minutes till the end of the test: Tried #10 without much luck because I couldn't figure out where point E was, and tried #12 half-heartedly, but made some random error and got a nonsensical answer. So, with 15 minutes left, I decided to check my answers again.\r\n\r\n[b]Which I'm glad I did, because it turns out that I had made an off-by-one on #6.[/b]\r\n\r\nFinally, the last five minutes were devoted to bubble-checking and guessing. :P\r\n\r\nSo in short, I got 9.\r\n\r\nP.S. Yeah, I was kinda pissed about other stuff when I made that other post above. I'm over it now, sorry about that. :wink:", "Solution_18": "[quote=\"davidyko\"]Dude, Vishal, seriously, nice job.\n\nHmm, perhaps I'll go into more detail about my AIME experience.\nSo at the 2-hour mark, I'd done #1-5, 7-9. (having skipped #6 temporarily because I was being dumb) So, as my goal was 8 anyway and I had done 8 problems, I was like, \"oh, I should check now.\"\n\n[b]Which was a good thing, because I had somehow managed to mess up #1 and #3.[/b]\n\nSo that left about 45 minutes left. Got #6 in about another 10 minutes, and then went to the bathroom, as I was feeling quite accomplished (after all, I had only got 5 on last year's AIME).\n\n30 minutes till the end of the test: Tried #10 without much luck because I couldn't figure out where point E was, and tried #12 half-heartedly, but made some random error and got a nonsensical answer. So, with 15 minutes left, I decided to check my answers again.\n\n[b]Which I'm glad I did, because it turns out that I had made an off-by-one on #6.[/b]\n\nFinally, the last five minutes were devoted to bubble-checking and guessing. :P\n\nSo in short, I got 9.\n\nP.S. Yeah, I was kinda pissed about other stuff when I made that other post above. I'm over it now, sorry about that. :wink:[/quote]\r\nNice, I got everything I actually did seriously and I was completely sure of correct on the first try.\r\nVishal that is pretty beast, forgot to mention that.\r\nOh and regarding a timeline of events, this is about how it went:\r\nStart: 8:14\r\nFinished 1-6 : 8:45ish, feeling pretty good about myself, because I thought that 7 was the usual cutoff, so I'm like YESH, I have plenty of time, gonna make USAMO.\r\nFinished 8: 8:50, feeling really great, I THINK I made USAMO already with my 7, even though a 7 in actuality is very dangerous, I go to bathroom.\r\nIn bathroom I start laughing, cuz I'm like, WAHOOO I MADE USAMO!!! ROFL THAT WAS EASY!!! and I keep that mentality for the rest of the test.\r\n8:55ish I come back, and I'm like, okay I'm ready to solve a few more and get an awesome score!!!\r\n[looks at 9, makes mistake, dies, but not aware]\r\n[looks at 10, draws diagram, decides too complicated for my small brain, gives up]\r\n[looks at 11, tries until about 10:00 with casework :O, thinks he's got the right answer, put its down, feeling really good]\r\n[looks at 12, thinks its beastly, gives up :mad:]\r\n[looks at 14, weeps at crap geo skill, moves on]\r\n[looks at 15, weeps at crap geo skill, goes to 13]\r\n[looks at 13, has no idea because of no experience on the type of problem, weeps, but still feeling good about qualifying for usamo (which was a false belief)]\r\n9:55 Hmmm...I should check some more. And of course start making up BS for what I'm gonna put for all the ones I have no idea how to do. I decide that all my answers are correct :roll: and move on to BSing 10, 12, 13, 14, 15. I decide that 159 looks like a cool number because it starts with a 1, and is an arithmetic progression! (the digits), put that for 15. Similar BS went to the other ones.\r\nHeh interesting eh. I think the issue was that I got way too excited over my false USAMO qualifying idea, and so I got every single problem after that wrong.", "Solution_19": "So, never mind. Official answers are up, and I actually got 9!\r\n1)252\r\n2)025\r\n3)314\r\n4)080\r\n5)014\r\n6)017\r\n8)047\r\n10)032\r\n12)375\r\n\r\nI was working on 7, 14, and 15... but I ran out of time on all 3 and ended up guessing. Oh well.", "Solution_20": "Short: I got 1-7, 9, 12, off-by-1 on 10, guessed on 8, 11, 13, and 14, and thought 15 was more simple than it was.\r\n\r\nLong: So first I did 1-3, skipped 4 because I thought it looked hard, did 5-7. Then I skipped 8, took far too long on 9, misread 10 (I thought 'diagonals' was 'legs' for some reason) so I didn't get anywhere.\r\n\r\nChecked my answers a buncha times, which took about 40 minutes.\r\n\r\nThen for the next hour or so I was just working randomly through the rest of the test. I realized 4 wasn't that hard, and re-read 10, and solved it... except I thought 25+7=31 :dry:\r\nI was thinking wrong on 12 and thought they had to go at an infinite speed, but luckily I got 375 anyways and left it even though I thought it was strange. On 15 I thought there was a tetrahedron so I got 2rt17 (019) as my answer, except there is no tetrahedron so....\r\n\r\nChecked my answers again, 30 minutes left. Somehow I didn't catch the 25+7.\r\n\r\nI worked randomly on 11 and 13 but didn't really get anywhere. I thought about 14 for a long time and figured out the answer had to be somewhere in the 400's, so I just guessed 450. And then with no more time left I just filled 8, 11, and 13 with 042 :P", "Solution_21": "Let's just say I'm lucky all the questions are weighted equally\r\n\r\nalso for 15 i got the answer, but i couldn't put it into the form they wanted.", "Solution_22": "I actually did decently on this AIME. It was good. :)", "Solution_23": "Meh. And then I tried the problems I didn't get after the test and I got 7, 9, 11,14,15...all the ones I missed on the actual thing. I just ran out of time :P Sheesh, I spend like an hour on 1,3, and 4... The only one I couldn't get in a reasonable amount of time was 13... :(", "Solution_24": "I did 1-5, 12, and 15 for a total of 7.\r\n\r\nI couldn't get 10 for quite a while, so I eventually just put down a random guess of 032. Right before the end of the 3 hours, I changed it to 023 for absolutely no reason. :|\r\n\r\nIf I don't make USAMO, I blame it on not trusting my instincts, even though that's quite stupid, considering that instincts won't tell you the correct answer out of 1000 possible choices.", "Solution_25": "Got a 9. Did 1-9, 11-12. My #5 was wrong because I added 1 instead of subtraacting. This year's #15 was actually pretty easy.", "Solution_26": "If the floor happens to be 7, then there'll definitely be a cutoff aplied to under-10. So paired with a rock-bottom official 120 for AMC 10, (five mistakes?? 5!?!), I have no USAMO chance.\r\n[quote=\"pythag011\"]Got a 9. Did 1-9, 11-12. My #5 was wrong because I added 1 instead of subtraacting. This year's #15 was actually pretty easy.[/quote]\r\nAll you need to do is improve by 2 points a year and you'll have a freshman 15.\r\nhahahahaha ..ouch", "Solution_27": "[quote=\"archimedes1\"]If the floor happens to be 7, then there'll definitely be a cutoff aplied to under-10. So paired with a rock-bottom official 120 for AMC 10, (five mistakes?? 5!?!), I have no USAMO chance.\n[quote=\"pythag011\"]Got a 9. Did 1-9, 11-12. My #5 was wrong because I added 1 instead of subtraacting. This year's #15 was actually pretty easy.[/quote]\nAll you need to do is improve by 2 points a year and you'll have a freshman 15.\nhahahahaha ..ouch[/quote]\r\n\r\nDude TWO OFF BY ONE ERRORS ON THE AIME.", "Solution_28": "Holy crap archimedes, you managed to fail worse than me at AMC10? I had 4 mistakes, but I decided to leave one blank because I caught it but had no idea how to get the right answer after.\r\nAlbert you shouldn't feel that bad. The Zuton Force made 8 computation errors but could've had perfect...now his USAMO chances are in jeopardy (and he's crazy beast)", "Solution_29": "[quote=\"serialk11r\"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]\r\nYes. I'm honored :P \r\n\r\nWait, what happened to ubemaya?", "Solution_30": "[quote=\"archimedes1\"][quote=\"serialk11r\"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]\nYes. I'm honored :P \n\nWait, what happened to ubemaya?[/quote]\r\nEDIT: By request of Ubeymaya :)\r\n\r\nMy birthday (24th, yesterday) was the Math Jam where I got to see how close I was to solving #15 (drew the correct lines, got the lengths, but forgot about one!!! AHH) & 14 :( that was pretty...pleasant.", "Solution_31": "[quote=\"archimedes1\"][quote=\"serialk11r\"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]\nYes. I'm honored :P \n\nWait, what happened to ubemaya?[/quote]\r\n\r\nFailed, two off by one errors :( .\r\n\r\nWait, who got the 12 in the poll?", "Solution_32": "[quote=\"Ubemaya\"][quote=\"archimedes1\"][quote=\"serialk11r\"]Holy crap archimedes, you managed to fail worse than me at AMC10? [/quote]\nYes. I'm honored :P \n\nWait, what happened to ubemaya?[/quote]\n\nFailed, two off by one errors :( .\n\nWait, who got the 12 in the poll?[/quote]\r\nFor all who don't know, that would be Albert...hence the smiley in my previous edited post.", "Solution_33": "I got an 8, but it's all good :) , since I'm only a 9th grader, and I got 135 on the 10B.\r\n\r\nI got #1-5 (which were way to easy), 7, 8, and 10. I skipped some easy problems, so that sucks. I also miscounted on #6. Oh well, I doubled my score from last year.", "Solution_34": "Hmmm I find it very wierd how tons of people have high AIME goals, and then they're like \"USAMO: 1\". Come on I didn't even qualify for AIME last year (111 phail :() and if I make USAMO my goal is like 12ish...at least aim higher even if you can't score that high.", "Solution_35": "Well, I don't really expect to do very well on the USAMO, since I'm really bad at proof problems. I would consider myself lucky if I could even solve one of those problems. Besides, I'm very used to those types of competitions. The only sort of olympiad-style contest I've done was the BAMO last year, and I think I got like 7 or 8.", "Solution_36": "Well, my original goal on USAMO was \"USAMO:0\"\r\n\r\nMeh. You never know. I might do so bad they might give me a negative score :P\r\n\r\nMy 12A score is 114 (fail) for a USAMO index of 204. Yay, its actually over 200! :)", "Solution_37": "[quote=\"serialk11r\"]Hmmm I find it very wierd how tons of people have high AIME goals, and then they're like \"USAMO: 1\". Come on I didn't even qualify for AIME last year (111 phail :() and if I make USAMO my goal is like 12ish...at least aim higher even if you can't score that high.[/quote]\r\n\r\nWho cares? my aime goal was low because I didn't expect much. My usamo score is around 7 because I don't expect I'd do well either. Maybe they'd jack up usamo like they did on the aime.", "Solution_38": "[quote=\"moogra\"][quote=\"serialk11r\"]Maybe they'd jack up usamo like they did on the aime.[/quote][/quote]\r\n\r\nI hope so.", "Solution_39": "number 15 on the AIME was f------ easy. Unfortunately, I wasted a lot of time by attempting problems that I got wrong, including number 11, where I was off by two. Unfortunately, for most of the problems, I was swamped by calculations (which means I got it wrong). I think I got a 4 or a 5, and I'm in 7th grade. Besides, I'm the only kid that took any AMC competitions.", "Solution_40": "agh... pythag is too pro... must beat him next year... lol\r\n\r\nanyways i got a 9 in 7th grade! (darnit pythag i can see you laughing) :mad:", "Solution_41": "Hmmm I see a second 12+...", "Solution_42": "I'm getting a 0 cause I'll be in Yosemite while I was SUPPOSED to be taking AIME.\r\n\r\nEpic phail." } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "Prove that there exists one and only one function $ f: \\mathbb{Q}^ \\plus{} \\to \\mathbb{Q}^ \\plus{}$ which satisfies the following conditions:\ni. For $ 0 < q < \\frac {1}{2}$, \\[ f(q) \\equal{} 1 \\plus{} f(\\frac {q}{1 \\minus{} 2q}).\\]\nii. For $ 1 < q < 2$, \\[ f(q) \\equal{} 1 \\plus{} f(q \\minus{} 1).\\]\niii. For every $ q\\in\\mathbb{Q}^ \\plus{}$, \n\\[ f(q)f\\left(\\frac {1}{q}\\right) \\equal{} 1\\].", "Solution_1": "[quote=\"Mashimaru\"]Let $ \\mathbb{Q}^ \\plus{}$ denoted the set of positive rational numbers. Prove that there exists one and only one function $ f: \\mathbb{Q}^ \\plus{} \\to \\mathbb{Q}^ \\plus{}$ satisfies the following condition:\ni. For $ 0 < q < \\frac {1}{2}$ then $ f(q) \\equal{} 1 \\plus{} f(\\frac {q}{1 \\minus{} 2q})$.\nii. For $ 1 < q < 2$ then $ f(q) \\equal{} 1 \\plus{} f(q \\minus{} 1)$.\niii For every $ q\\in\\mathbb{Q}^ \\plus{}$, $ f(q)f(\\frac {1}{q}) \\equal{} 1$.[/quote]\r\n\r\nI think the result is wrong.\r\n\r\nLet $ q\\equal{}\\frac ab$ with $ \\gcd(a,b)\\equal{}1$\r\n\r\nIf $ 00\\notin\\{\\frac 12,1,2\\}$, we can calculate in a unique manner from another rational $ \\frac cd$ such that $ c\\plus{}d2 + 1= 0, thus its roots are i and -i.\r\nIt follows that the set of solutions is the set of functions of the form f(x) = a*cos(x) + b*sin(x), where a,b are constants.\r\n\r\nPierre.", "Solution_2": "f+f''=0 is a standard ODE. we can either use the fact that x^2+1=0 has i and -i as roots, and therefore the general solution is that you mentioned.", "Solution_3": "Errr... What's an ODE?", "Solution_4": "Ordinal Differential Equation. Belive me. You don't want to know more. :D", "Solution_5": "he has too, now they can be subjects at the NMO in 12th grade :D:D", "Solution_6": "[quote=\"Valentin Vornicu\"]he has too, now they can be subjects at the NMO in 12th grade :D:D[/quote]\r\n\r\nvery incredible...\r\nand what's the diff. eqn. you are learning at university now? ODE or PDE?" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Can newton sums be generalized to calculate the sum of the powers of the roots higher than the degree of the polynomial?", "Solution_1": "Yes, I'm sure we can do that. So, if we have $f(x) = x^{2}-4x+3,$\r\n\r\n$s_{1}-4 = 0 \\Leftrightarrow s_{1}= 4$\r\n$s_{2}-4s_{1}+3(2) = 0 \\Leftrightarrow s_{2}= 10$\r\n$s_{3}-4s_{2}+3s_{1}= 0 \\Leftrightarrow s_{3}= 28 = 3^{3}+1^{3}$\r\n\r\nThat's not nearly a proof, but it's an example. I think other examples are shown in AoPS II.", "Solution_2": "When the powers of the roots are higher (or just as high) as the degree of the polynomial, you can reduce it to powers that are less than the degree.\r\n\r\nExample: $a,b,c$ are roots of $t^{3}-t^{2}-2007$, find $a^{4}+b^{4}+c^{4}$.\r\n\r\nWe have $a^{4}=a^{3}+2007a$ and $a^{3}=a^{2}+2007$, so the sum in question is actually\r\n\\[(a^{2}+2007a+2007)+(b^{2}+2007b+2007)+(c^{2}+2007c+2007)\\]\r\n\\[=(a^{2}+b^{2}+c^{2})+2007(a+b+c)+6021\\]\r\nNow it just reduces to computing $a+b+c$ and $a^{2}+b^{2}+c^{2}$. From $a+b+c=1$, $ab+bc+ca=0$, we get $a^{2}+b^{2}+c^{2}=1$, therefore $a^{4}+b^{4}+c^{4}=\\boxed{8029}$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c$ positive angles with $ a\\plus{}b\\plus{}c\\equal{}90$ prove that \r\n\r\n$ \\frac{cosa^{2}cosb^{2}cosc^{2}}{sinasinbsinc} \\geq \\frac{27}{8}$", "Solution_1": "Jensen's Inequality", "Solution_2": "Can you write full solution please?" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ ABC$ be an equilateral triangle and let $ D$ be a point on the minor arc $ \\widehat{BC}$ of the circumcircle. Lines $ BD$ and $ AC$ meet at point $ M$, and lines $ CD$ and $ AB$ meet at point $ N$. Show that $ (AD$ is a simedian in the triangle $ AMN$.", "Solution_1": "Let $ H,K$ be the projection of $ D$ onto $ AB, AC.$\r\n$ AD$ is the symmedian of triangle $ AMN$ iff $ \\frac{AN}{AM}\\equal{}\\frac{DH}{DK}$\r\nSo we will prove that $ \\Delta ADN\\sim \\Delta MDA$\r\n$ \\angle CDM\\equal{}60^o\\equal{}\\angle ADB, \\angle DCM\\equal{}\\angle ABD$ thus $ \\angle DMC\\equal{}\\angle BAD$\r\nSimilarly, $ \\angle BND\\equal{}\\angle DAC$\r\nTherefore $ (*)$ is true.\r\nWe end the proof." } { "Tag": [], "Problem": "Let $f(1)=1$ and for all natural numbers n, $f(1)+f(2)+...+f(n)=n^2f(n)$. What is $f(2006)$?", "Solution_1": "[hide=\"solution\"]$f(n+1)=(n+1)^2*f(n+1)-n^2*f(n)$\n$(n+2)*f(n+1)=n*f(n)$\n$f(n+1)=f(n)*\\frac{n}{n+2}$\n$f(2006)=\\prod_{i=1}^{2005}\\frac{i}{i+2}=\\frac{1*2}{2007*2006}=\\frac{1}{2013021}$[/hide]", "Solution_2": "[hide=\"Solution\"]All I can think of is induction.\n\nThe equation is equivalent to\n\n$f(1)+\\dots+f(n-1)=(n^2-1)f(n)\\iff f(n)=\\frac{1}{n^2-1}(f(1)+\\dots+f(n-1))$\n\nCalculating the first few terms, we get $f(2)={1\\over 3}, f(3)={1\\over 6}, f(4)={1\\over 10}, f(5)={1\\over 15}$. We note that the denominators are the triangular numbers, hence we assume $f(n)=\\frac{2}{n(n+1)}$\n\nFor $n=1$ we have $f(1)=\\frac{2}{1\\cdot 2}=1$\n\nInductive step:\n\n\\begin{eqnarray*}f(n+1) &=& \\frac{1}{(n+1)^2-1}\\left(\\frac{2}{1\\cdot 2}+\\frac{2}{2\\cdot 3}+\\dots+\\frac{2}{n(n+1)}\\right)\\\\ &=& \\frac{2}{(n+1)^2-1}\\left(1-{1\\over 2}+{1\\over 2}-{1\\over 3}+\\dots+\\frac{1}{n}-\\frac{1}{n+1}\\right)\\\\ &=& \\frac{2}{(n+1)^2-1}\\left(1-\\frac{1}{n+1}\\right)\\\\ &=& \\frac{2}{n(n+2)}\\cdot\\frac{n}{n+1}\\\\ &=& \\frac{2}{(n+1)(n+2)}\\end{eqnarray*}\n\nHence $f(2006)=\\frac{1}{1003\\cdot 2007}$[/hide]" } { "Tag": [ "USAMTS" ], "Problem": "Hello,\r\n\r\nThis might've been asked before, I don't know though. For the USAMTS entry forms and permission forms, I require a parent or guardian to sign for me. However, I can't get my parents to sign as they're a country away, and I don't really have a guardian here. What should I do?\r\n\r\nThanks.", "Solution_1": "Send a blank one now with your stuff filled out but without your parent's signature, and send one with your parent's signature with your Round 2 solutions." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "inequalities", "AIME I", "\\/closed" ], "Problem": "If you expect to qualify for the USAMO this year, but don't expect to do well (you got a 136 on the AMC 12 and a 10 on the AIME I), are you ready for intermediate courses or olympiad courses? I find from the Do You Know It? and Are You Ready? that I'm somewhat in the middle.", "Solution_1": "[quote=\"bubala\"]If you expect to qualify for the USAMO this year, but don't expect to do well (you got a 136 on the AMC 12 and a 10 on the AIME I), are you ready for intermediate courses or olympiad courses? I find from the Do You Know It? and Are You Ready? that I'm somewhat in the middle.[/quote]\r\n\r\nProbably either.\r\n\r\nReally it depends mostly on your weak and strong subjects. Any subject in which you feel weak, the Intermediate subject courses would be better for you. Any subject in which you feel strong (like you could answer easier Olympiad problems -- and maybe even a few harder ones), then the Olympiad class in that subject would benefit you more.\r\n\r\nThere are a couple of exceptions that I would currently mention. \r\n\r\nThe Olympiad Inequalities class is really 35% Intermedatie and 65% Olympiad. We'll eventually have 2 levels (one day...), but if you're very strong in Algebra, but don't have experience with Olympiad level Inequalities, that's still a good class.\r\n\r\nThe Intermediate Complex Numbers/Trig class is harder than our other Intermediate classes and gets into some Olympiad level material. That's another subject that might one day become two levels of classes." } { "Tag": [ "calculus", "derivative", "Vieta", "algebra", "polynomial", "function", "analytic geometry" ], "Problem": "If a and b are real, prove that $ x^4\\plus{}ax\\plus{}b \\equal{}0$ cannot have only real roots.", "Solution_1": "[b]Clarification of question:[/b]\r\n- $ a$ and $ b$ are not both $ 0$\r\n- Solution should not use differentiation\r\n\r\n[b]An approach:[/b]\r\n[hide]Equivalently, $ x^4 \\equal{} Ax \\plus{} B \\; : \\; A,B \\in \\mathbb{R}$.\n\nIt suffices to show that no line in the Cartesian plane can intersect the curve $ y \\equal{} x^4$ more than twice.\n[/hide]\r\n[b]Remark:[/b] A less obvious way to write the problem would have been to say, prove that there does not exist an ordered quadruple of non-zero real numbers $ (p,q,r,s)$ such that $ \\sum_{sym} p \\equal{} \\sum_{sym} pq \\equal{} 0$, or something along those lines.", "Solution_2": "Letting the roots be $ r_1,r_2,r_3,r_4$, we have using Vieta's, \r\n\r\n$ r_1^2\\plus{}r_2^2\\plus{}r_3^2\\plus{}r_4^2\\equal{}(r_1\\plus{}r_2\\plus{}r_3\\plus{}r_4)^2\\minus{}2(r_1r_2\\plus{}r_1r_3\\plus{}r_1r_4\\plus{}r_2r_3\\plus{}r_2r_4\\plus{}r_3r_4)\\equal{}0\\Rightarrow r_1\\equal{}r_2\\equal{}r_3\\equal{}r_4\\equal{}0$ or one of $ r_i\\not\\in\\mathbb{R}$. \r\n\r\nObviously, we can't have $ r_1\\equal{}r_2\\equal{}r_3\\equal{}r_4\\equal{}0$ since then the polynomial becomes $ x^4\\equal{}0$.", "Solution_3": "how do you show that no line in the Cartesian plane can intersect the curve more than twice?", "Solution_4": "Yeah! because i was thinking if you have a function y=ax, a being any positive number - it should intersect f(x)=x^4 at 0 and at a^(1/3) - and the addition of any constant to y show give you 2 non-zero intersections", "Solution_5": "The curve $ y \\equal{} x^4$ is [url=http://en.wikipedia.org/wiki/Convex_function]strictly convex[/url], so the portion of a curve over some x interval is below the secant line that spans that interval. So if your interval is the leftmost intersection to the rightmost intersection, we can't have any more intersections in between." } { "Tag": [], "Problem": "The walls of standard houses are constructed with two-by-four\nstuds placed 16 inches apart, center to center. How many studs are needed for a wall that is 48 feet wide?", "Solution_1": "$ 16$ inches is $ 1 \\frac{1}{3}$ feet. $ 48$ feet divided by $ 1 \\frac{1}{3}$ feet is 36. However, since the \r\n\r\nstuds are from center to center, only one of the ends should be calculated within 48. Thus, the answer is $ 36\\minus{}1\\equal{}\\fbox{35}$, unless I misinterpreted this hard-to-understand problem.", "Solution_2": "The answer says $ \\boxed{37}$ though..... I think that you divided 48 by $ 1\\frac{1}{3}$, and then add on one for on of the end posts.", "Solution_3": "Think about it like this, 16 inches is 1 1/3 feet. the width of the wall is 48 feet wide. 48 ft divided by 1 1/3 feet is 36 ft. but this is just the spaces between the studs (for example in a line of studs, if there are 4 spaces between the studs there are 5 studs). So the answer is 37 studs." } { "Tag": [ "AMC", "AIME", "AMC 10", "trigonometry", "AoPS Books" ], "Problem": "This year will be my first year taking the AMC 12. I own both the AOPS books but am not sure as to how I should spend my time studying throughout the year. At most, each day I will be able to spend 1.5 hours a day preparing/practicing for these competitions. I am a rising junior and would like advice as to what would probably be most beneficial for me to improve my math ability. My highest AMC 10 score was a 118.5 and my highest (practice) AIME score was a 5.", "Solution_1": "AMC 12 is beyond AoPS volume 1, but you should still read through it and master the basics. AoPS Volume 2 will help a lot, but you should also do lots of practice AMC12's and AIME's.", "Solution_2": "It really doesn't matter, as long as you're doing problems. If you find you have weaknesses, work on those. Do some practice tests to get comfortable with the format.", "Solution_3": "[quote=\"batteredbutnotdefeated\"]AMC 12 is beyond AoPS volume 1, but you should still read through it and master the basics. AoPS Volume 2 will help a lot, but you should also do lots of practice AMC12's and AIME's.[/quote]\r\n\r\nDo a lot of problems. Most of the AMC 12 is covered by Intro Series, except some algebra/trig", "Solution_4": "[quote=\"geomeister33\"]This year will be my first year taking the AMC 12. I own both the AOPS books but am not sure as to how I should spend my time studying throughout the year. At most, each day I will be able to spend 1.5 hours a day preparing/practicing for these competitions. I am a rising junior and would like advice as to what would probably be most beneficial for me to improve my math ability. My highest AMC 10 score was a 118.5 and my highest (practice) AIME score was a 5.[/quote]\r\n\r\nI think you should study more of the Introduction series and Volume 1. Also do a lot of problems. When I finished the Introduction series, my AMC10 score raised from 115.5 to a 144!!\r\n\r\nFor AMC 12, most is covered by the introduction series, but the trig and other topics you can get from some of the Intermediate books and Precalculus possibly. The Intermediate series is how I prepare for AIME.:)", "Solution_5": "Are the intermediate courses better or the books in your opinion? If you've taken the courses that is." } { "Tag": [ "geometry", "incenter", "trigonometry", "geometry solved" ], "Problem": "The incircle of triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $P$ be any point inside the incircle of triangle $ABC$, and let $X$, $Y$, $Z$ be the points where the segments $PA$, $PB$, $PC$, respectively, meet the incircle. Prove that the lines $DX$, $EY$, $FZ$ are concurrent.", "Solution_1": "I think I managed to prove the converse, that is: if DX and the others are concurrent, then AX and the others are cincurrent, but I think the converse is enough. I'll post more tonight.", "Solution_2": "I think the problem can be extended from the triangle ABC to a hexagon AECDBF circumscribed to circle W. A point P inside W. PA,PB,PC intersect W at X,Y,Z. Prove DX,... are concurent. \r\nI didn't prove this problem nor the first but I have no reason to belive otherwise.", "Solution_3": "hey\r\ni have only managed to prove it for point P being the incenter itself using pure trig.but it's getting difficult for the general case of P being any point inside the incircle.will post more later(if i make sum progress).ne1 else who has solved this one??\r\ncheers", "Solution_4": "Here's how I think the converse of the problem can be proved (that is: if X, Y, Z are points on the incircle such that DX, EY, FZ are concurrent, then AX, BY, CZ are also concurrent).\r\n\r\nFirst of all we use Ceva's theorem (trigonometric form) with sines on triangle DEF, and we get: (sinFDX/sinXDE)*(sinEFZ/sinZFD)*(sinDEY/sinYEF)=1 (*). But by applying sine law we get FX=2r*sinFDX, XE=2r*sinXDE etc., where r is the inradius. By replacing these in (*) we get FX/XE*EZ/ZD*DY/YF =1. So product of (AX/FX)/(AX/XE) and the other 2 analogous expressions is 1 (**). But AX/XF=sinAFX/sinFAX and because AF is tangent to the incircle AFX=FDX. Just use all these things in (**) and we get sinFAX/sinXAE*sinECZ/sinZCD*sinDBY/sinYBF=1, and fromthis we derive by the reciprocal of Ceva's thm with sines that AX, BY, CZ are concurrent at a point P. This, of course, is the reciprocal of the problem Alekk posted, but I think it can be used to prove the actual problem, or we could use a similar method.", "Solution_5": "hey, nice proof grobber.\r\n\r\nI think we have to use the fact that (AD),(CF),(BE) are \r\nconcurrent to finish the problem. I saw a similar problem (but much \r\nmore easy) wich is:\r\n**\r\nlet ABC be a triangle and D,E,F points on sides BC,CA,Ab such that \r\nthe cevians AD,BE,CF are concurrent . Show that if M,N,P are points \r\non sides EF,FD,DE then the lines Am,BN,CP concur iff the lines DM,EN,FP \r\nconcur.\r\n**\r\nbye.", "Solution_6": "hey, i used grobber method and i found a \"solution\"(i'm not sure):\r\n\r\nLEt G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)\r\nwe want to prove G=1\r\nby grobber method we have:\r\nsin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)\r\n =FX/XE*EZ/ZD*DY/YF=...\r\n =[sin(AFX)/sin(AEX)*sin(CEZ)/sin(CDZ)*sin(BDY)/sin(BFY)]*G\r\n or AFX=FDX etc..\r\n =[sin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)]*G\r\n\r\nthen\r\n G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)=1\r\nand we have what we want. (is it right? I'm not sure)\r\n\r\nregards.", "Solution_7": "sorry it was false.\r\nhere what i found (it's ugly then i'm not sure)\r\n\r\nLEt G=sin(XAE)/sin(FAX)*sin(DCZ)/sin(ECZ)*sin(YBF)/sin(DBY)\r\n H=sin(FDX)/sin(XDG)*sin(FZ)/sin(ZFD)*sin(DEY)/sin(YEX)\r\nwe want to prove H=1\r\nby grobber method we have:\r\nH = sin(FDX)/sin(XDE)*sin(EFZ)/sin(ZFD)*sin(DEY)/sin(YEX)\r\n =FX/XE*EZ/ZD*DY/YF\r\n =(FX/AX)/(XE/AX)*(EZ/CZ)/(ZD/CZ)*(DY/BY)/(YF/BY)\r\n =..\r\n =[sin(FAX)/sin(XAE)*sin(ECZ)/sin(DCZ)*sin(DBY)/sin(BFY)]/H\r\n\r\n1/G=H^2 or G=1\r\n\r\nthen\r\n H=1\r\nand we have what we want.(is it still false?)", "Solution_8": "what about the hexagon generalization??", "Solution_9": "[quote=\"amfulger\"]I think the problem can be extended from the triangle ABC to a hexagon AECDBF circumscribed to circle W. A point P inside W. PA,PB,PC intersect W at X,Y,Z. Prove DX,... are concurent. \nI didn't prove this problem nor the first but I have no reason to belive otherwise.[/quote]\r\n\r\nUnfortunately, my dynamic geometry sketch proves this theorem wrong. However, for the original (triangle) theorem, see also Theorem 2 of my note \"Variations of the Steinbart Theorem\" at http://www.cip.ifi.lmu.de/~grinberg/ .\r\n\r\n Darij Grinberg", "Solution_10": "Dear Mathlinkers,\r\nfor an overlook of this situation, you can see\r\n\r\nhttp://perso.orange.fr/jl.ayme vol. 3 Les points de Steinbart et de Rabinowitz\r\n\r\nSincerely\r\nJean-Louis" } { "Tag": [ "geometry", "circumcircle", "cyclic quadrilateral", "geometry unsolved" ], "Problem": "Let ABCD be a cyclic quadrilateral with AB=BC. If P is a point on CD such that AP=PB=BA, prove that AD=R, the circumradius of $ \\triangle ABC$.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=486680#486680\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1167502#1167502" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $f,g: \\Re\\to\\Re$ such that $g$ is strictly monotonic. If $\\lim_{t\\to x} f(t)=g(f(x))$ then $f$ is continous.", "Solution_1": "The main problem was: \r\n [quote=\"gm\"]Let $f: \\Re\\to\\Re$ a non-constant function such that there exists $g$ continous and strictly monotonic such that $\\lim_{t\\to x} f(t)=g(f(x)),\\forall x \\in \\Re$. Prove that there exists a non-degenerated interval formed by fix points of $g$.[/quote]\r\n\r\n\r\n Altough it may seem easier this is the original text the first post being given as a variation :roll:", "Solution_2": "I don't think it's easy...been working on it for a couple of days and NOTHING!!! there must be something i'm missing! :wallbash:", "Solution_3": "Here's some ideeas I have reached to. Maybe they will help you. \r\n\r\n1)I proved that $\\lim_{t\\to x}f(t)$ is continuous. Here http://www.mathlinks.ro/Forum/viewtopic.php?p=405226#p405226 i made a proof for this fact. Hence $gof$ is continuous. \r\n \r\n\r\n2) Assume that $g(a)> a+\\epsilon$ and prove that there exist $b>a$ such that $g(x) \\ge x+\\epsilon \\forall x\\in [a,b]$. Indeed, $\\forall x\\in [a,g(a)-\\epsilon]$ we have that $x>a \\Rightarrow g(x) > g(a) \\ge x-\\epsilon$. Similarly, If we have $g(a)\\lim_{t \\to x}f(t)+\\epsilon$(I took the \">\" sign without loss of generality). Hence, on this interval, for every $x$ and $\\delta$ there exists an $y$ such that $|x-y|\\le \\delta$ and $f(x)\\ge f(y)+\\epsilon$. \r\n\r\n4) For any $n$ we can define $x_1,x_2,..x_n$ by the method described in 3) such that $x_{i+1}-x_i \\le \\frac{max (|x_1-a|,|b-x_1|)}{n^2}$. This sequence is contained in interval $[a,b]$ hence it must have a limit point, name it $d$, and $f(x_k)$ is unbounded. Hence $g(f(d))=\\lim_{x\\to d}f(x)=\\lim_{n\\to infty} {x_{k_n}}=\\infty$. Contradiction. \r\nHence for any interior point of $Imf$ we have that $g(x)=x$.\r\n\r\n5) If $f(x_0)=a_0$ is boundary point of $Imf$ then there exist a sequence $x_k$ such that $f(x_k)\\to a_0$ and $f(x_k)> (<)a_0$ (the same sign holds for all $x_k$. I'll take it \">\" WLOG), or $a_0$ is a isolated point.\r\n\r\nMaybe the proof can be made even for these points (probably not so hard), but I'll give it a try another day.", "Solution_4": "I think it is something similar to 3). Suppose there is some $x \\in {\\mathbb R}$ such that $f(x) \\neq g(f(x))$. WLOG $f(x)>g(f(x))$ and $g$ is strictly increasing. Then there is $\\epsilon>0$ and $a>0$ such that $f(x)>f(t)+a$ for all $t \\in (x-\\epsilon,x+\\epsilon)$, $t \\neq x$. We have \\[ f(x)>f(t)+a \\Rightarrow f(x)-a>f(t) \\Rightarrow g(f(t)) g(f(t))+\\left(g(f(x))-g(f(x)-a)\\right), \\forall t \\in (x-\\epsilon,x+\\epsilon),t \\neq x, \\] with $g(f(x))-g(f(x)-a)>0$, which contradicts the fact that $\\displaystyle g \\circ f = \\lim_{t \\rightarrow x} {f(t)}$ is continuous in $x$, so $\\displaystyle f(x) = g(f(x)) = \\lim_{t \\rightarrow x} {f(t)}$ for all $x \\in {\\mathbb R}$, hence $f$ is continuous.", "Solution_5": "I see that the same method works even for boundary points which are not isolate. (because the interval [a,b] mntioned in 3),4) lies inside $Imf$). \r\n\r\nAssume that $Imf$ have an isolated point $a$. Hence, $f$ has a jump discontinuity in $f^{-1}(a)$ which implies that $gof$ has a jump discontinuity in $f^{-1}(a)$, Contradiction with the continuity of $gof$. So, the solution is finished. Pftiu!!\r\n\r\nAnyway, thank to dzeta for the beautifull solution.", "Solution_6": "rats...it was rather accessible .... anyway thanx dzeta and xirti :)" } { "Tag": [ "geometry", "parallelogram", "symmetry", "geometry proposed" ], "Problem": "Dear All My Friends,\r\nCutting one convex quadrilateral by two segments each connected midpoints of quadrilateral opposite sides we can divide the quadrilateral into four small convex quadrilaterals $ Q_1$, $ Q_2$, $ Q_3$, $ Q_4$. Suppose $ I_1$, $ I_2$, $ I_3$, $ I_4$ are intersections of diagonals of $ Q_1$, $ Q_2$, $ Q_3$, $ Q_4$ respectively and $ S_{1}$ is area of quadrilateral $ I_1I_2I_3I_4$\r\n\r\nBy rearranging four small convex quadrilaterals $ Q_1, Q_2, Q_3, Q_4$ (only moving, not upturn) we can bound one parallelogram (without gap, not overlapping). In this case their intersections of diagonals $ I_1$, $ I_2$, $ I_3$, $ I_4$ bound another convex quadrilateral with area $ S_{2}$\r\n\r\nPlease prove that $ S_{1}\\equal{}S_{2}$\r\nThank you and best regards,\r\nBui Quang Tuan", "Solution_1": "[quote=\"Quang Tuan Bui\"]\nBy rearranging four small convex quadrilaterals $ Q_1, Q_2, Q_3, Q_4$ (only moving, not upturn) we can bound one parallelogram (without gap, not overlapping). In this case their intersections of diagonals $ I_1$, $ I_2$, $ I_3$, $ I_4$ bound another convex quadrilateral with area $ S_{2}$\n[/quote]\r\n\r\nI don't understand this text but if I understand truely (as my figure) then two quarilateral are equal, here we use point symmetry only.", "Solution_2": "[quote=\"encyclopedia\"]\nI don't understand this text but if I understand truely (as my figure) then two quarilateral are equal, here we use point symmetry only.[/quote]\r\nDear encyclopedia,\r\nAs your figure you understand me truely! In any case two quarilaterals are not equal: they have equal area only!\r\nBest regards,\r\nBui Quang Tuan", "Solution_3": "oh you are right, thank you. I didn't see teh figure carefully, nice problem, I will try :) ." } { "Tag": [ "analytic geometry", "geometry unsolved", "geometry" ], "Problem": "Let $ABCD$ be a tetrahedron. Find the point $M$ in space satisfying the expession $P=MA^2+MB^2+MC^2+MD^2$ gets the least value.", "Solution_1": "[quote=\"Lovasz\"]Let $ABCD$ be a tetrahedron. Find the point $M$ in space satisfying the expession $P=MA^2+MB^2+MC^2+MD^2$ gets the least value.[/quote]\r\n\r\n\r\nThis point is the centre of gravity of the tetrahedron.\r\n\r\nLet's concider the coordinates:\r\n$P=\\sum (x_{A}-x_{M})^2 +\\sum (y_{A}-y_{M})^2 +\\sum (z_{A}-z_{M})^2 =$\r\n$4*( (x_{M}-\\frac{x_{A}+x_{B}+x_{C}+x_{D}}{4})^2 + (y_{M}-\\frac{y_{A}+y_{B}+y_{C}+y_{D}}{4})^2+$\r\n$(z_{M}-\\frac{z_{A}+z_{B}+z_{C}+z_{D}}{4})^2 + t)$ \r\nwhere $t$ is some constant.\r\nSo P is the least when equals to $4t \\Rightarrow$\r\n$x_{M}=\\frac{x_{A}+x_{B}+x_{C}+x_{D}}{4}$\r\n$y_{M}=\\frac{y_{A}+y_{B}+y_{C}+y_{D}}{4}$\r\n$z_{M}=\\frac{z_{A}+z_{B}+z_{C}+z_{D}}{4}$\r\n\r\nSo we can see that this point is the centre of gravity of ABCD.\r\nV:)", "Solution_2": "Yes :) \r\nBut there's not any proof without using coordinate?!?!?!?" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "find all functions wich satisfy $|f(x+y)-f(x)-f(y)| \\leq 1$.\r\nfor all reel nubers $x$ and $y$.", "Solution_1": "If $g: \\mathbb{R}\\to\\mathbb{R}$ is arbitrary solution of Cauchy equation $g(x+y)=g(x)+g(y)$ and $h: \\mathbb{R}\\to [-{1\\over 3},{1\\over 3}]$ arbitrary,\r\nthen $f(x)=g(x)+h(x)$ for $x\\in\\mathbb{R}$ defines a solution of above problem." } { "Tag": [ "trigonometry", "function", "algebra", "domain", "Gauss", "integration", "search" ], "Problem": "why can't a judge be disturbed at dinner?", "Solution_1": "This might be better off on the Puzzles and Brainteasers :wink:. Moved.", "Solution_2": "A mathematician, an engineer, and a chemist were walking down the road when they saw a pile of cans of beer. Unfortunately, they were the old-fashioned cans that do not have the tab at the top. One of them proposed that they split up and find can openers. The chemist went to his lab and concocted a magical chemical that dissolves the can top in an instant and evaporates the next instant so that the beer inside is not affected. The engineer went to his workshop and created a new HyperOpener that can open 25 cans per second.\r\n\r\nThey went back to the pile with their inventions and found the mathematician finishing the last can of beer. \"How did you manage that?\" they asked in astonishment. The mathematician answered, \"Oh, well, I assumed they were open and went from there.\"", "Solution_3": "Whenever I have a trig test, I always look for a $ sin$ from above.", "Solution_4": "[quote=\"7h3.D3m0n.117\"]This might be better off on the Puzzles and Brainteasers :wink:. Moved.[/quote]\r\n\r\nActually, shouldn't this go in the G&FF?\r\nThey should just make you a moderator so that you don't have to switch accounts all the time.", "Solution_5": "Well I assumed that we'd be swapping answers and jokes/brainteasers so I thought this might be better suited.\r\n\r\n[quote=\"Temperal\"]They should just make you a moderator so that you don't have to switch accounts all the time.[/quote]\r\n\r\n :?:", "Solution_6": "What is the volume of a pizza with z radius and a height?", "Solution_7": "I think this is very well known, but still I will write here since no one has yet :D :D \r\n\r\nTwo functions are walking in a domain, suddenly a robber function comes, and says:\r\n-I am the differentiator. you two there, throw me your ranges or else I will differentiate both of you.\r\nOne of the functions starts laughing really badly, so the robber function asks:\r\n-Hey why are you laughing, arent you scared?\r\nSo the function says:\r\n- hel no I aint scared, I am $ e^x$.\r\n\r\n :rotfl: :rotfl:", "Solution_8": "Here is a similar one:\r\n\r\nTwo functions are walking in a domain, suddenly another robber function comes and says:\r\n-I am the fourier transformer, so throw me your ranges or I will fourier transform you two.\r\nOne of the functions starts laughing and says:\r\n-I am the Gauss Curve hahaha.\r\n\r\n :rotfl: :rotfl:", "Solution_9": "[quote=\"Thaakisfox\"]I think this is very well known, but still I will write here since no one has yet :D :D \n\nTwo functions are walking in a domain, suddenly a robber function comes, and says:\n-I am the differentiator. you two there, throw me your ranges or else I will differentiate both of you.\nOne of the functions starts laughing really badly, so the robber function asks:\n-Hey why are you laughing, arent you scared?\nSo the function says:\n- hel no I aint scared, I am $ e^x$.\n\n :rotfl: :rotfl:[/quote]\r\n\r\nWhat's wrong with being reduced to a constant?", "Solution_10": "The differentiator says, \"I'm $ \\frac{d}{dy}$ !\" :rotfl:", "Solution_11": "What is \r\n$ \\displaystyle\\int\\frac {1}{cabin}\\,d(cabin)$?\r\n\r\n[hide=\"click here after you found the answer\"]\nIf you forgot the $ \\plus{} C$, shame on you![/hide]", "Solution_12": "[quote=\"Thaakisfox\"]I think this is very well known, but still I will write here since no one has yet :D :D \n\nTwo functions are walking in a domain, suddenly a robber function comes, and says:\n-I am the differentiator. you two there, throw me your ranges or else I will differentiate both of you.\nOne of the functions starts laughing really badly, so the robber function asks:\n-Hey why are you laughing, arent you scared?\nSo the function says:\n- hel no I aint scared, I am $ e^x$.\n\n :rotfl: :rotfl:[/quote]\r\n\r\nSlight variation\r\nRobber: Who says I differentiate on X", "Solution_13": "Well, we wouldn't have a problem if the function was\r\n\r\n\\[ e^{(\\prod\\limits_{cyc}a_i})\\]\r\n\r\nwhich covers pretty much every variable possibly dnoted by one letter and a subscript.", "Solution_14": "[quote=\"Temperal\"]Well, we wouldn't have a problem if the function was\n\\[ e^{(\\prod\\limits_{cyc}a_i})\n\\]\nwhich covers pretty much every variable possibly dnoted by one letter and a subscript.[/quote]\r\nWell that doesn't differentiate to zero, no... but it doesn't differentiate to itself either, which is still a problem.\r\n\r\n(unless that product symbol was supposed to be summation ;))\r\n\r\n---\r\n\r\nRobber: I'll differentiate you until you're zero!\r\n\r\nFunction: Too bad, I'm f(x)=1 if x is rational, 0 otherwise", "Solution_15": "[quote=\"Hamster1800\"][quote=\"lingomaniac88\"]If and only if $ |\\text{for Dummies}| < 1$. Otherwise it doesn't converge.[/quote]\n\nWhat?\n\n$ 1 \\plus{} 2 \\plus{} 4 \\plus{} 8 \\plus{} 16 \\plus{} \\cdots \\equal{} \\minus{} 1$. How can you say that doesn't converge?[/quote]\r\n\r\n :rotfl: \r\nbut can be demostrated:\r\n\r\nlet C=1+2+4+...\r\n\r\nC=1+2+4+8+...=1+(2+4+8+...)=1+2(1+2+4+...)=1+2C\r\nC=1+2C\r\n0=1+C\r\n-1=C\r\nQED :D \r\n\r\nWhy don't we transform it in the \"rightful demostration of wrong things\"?", "Solution_16": "Compute: $ \\dfrac{1}{0}\\times{0}$.\r\nAny number multiplied by 0 is 0. Therefore it is 0.\r\nThe 0 cancels the denominator of 1/0. Therefore it is also 1.\r\n0=1", "Solution_17": "Why do you say that $ 1\\plus{}2\\plus{}4\\plus{}8\\plus{}\\cdots\\equal{}\\minus{}1$ is a \"wrong thing\"? It is very much a right thing.\r\n\r\nNow let's compute $ \\sum_{n\\equal{}0}^{\\infty} n2^n$. Well, we simply divide by $ 2$ and then integrate with respect to $ 2$ to get the following equalities:\r\n\r\n$ \\sum_{n\\equal{}0}^{\\infty} n2^n \\equal{} 2*\\frac{d}{d2}\\left(\\sum_{n\\equal{}0}^{\\infty} 2^n\\right) \\equal{} 2*\\frac{d}{d2}\\frac{1}{1\\minus{}2} \\equal{} \\frac{2}{(1\\minus{}2)^2} \\equal{} 2$", "Solution_18": "[quote=\"Hamster1800\"]$ 1 \\plus{} 2 \\plus{} 4 \\plus{} 8 \\plus{} 16 \\plus{} \\cdots \\equal{} \\minus{} 1$. How can you say that doesn't converge?[/quote]\r\nPlease tell me how a bunch of positive numbers, when added together, can result in a negative number without bending the rules of mathematics for the purpose of humor.\r\n\r\nThen again, it is well-known that $ 2\\equal{}1$.", "Solution_19": "It's the same way a bunch of rational numbers can be added together to get an irrational number.", "Solution_20": "[quote=\"Lawrence Wu\"]Compute: $ \\dfrac{1}{0}\\times{0}$.\nAny number multiplied by 0 is 0. Therefore it is 0.\nThe 0 cancels the denominator of 1/0. Therefore it is also 1.\n0=1[/quote]\r\n\r\nalso, you can never divide by 0. Thus, it is also undefined\r\n\r\n0=1=undefined", "Solution_21": "x^2=x+x+x+x..... (x times). Taking the derivative of both sides gives 2x=1+1+1...(x times)=x, so 1=2.", "Solution_22": "easier\r\n\r\nlet A=B=1\r\n\r\n$ A\\equal{}B$\r\n$ A^2\\equal{}AB$\r\n$ A^2\\minus{}B^2\\equal{}AB\\minus{}B^2$\r\n$ (A\\minus{}B)(A\\plus{}B)\\equal{}B(A\\minus{}B)$\r\n$ A\\plus{}B\\equal{}B$\r\n\r\nso replacing A=B=1,\r\n\r\n2=1", "Solution_23": "wowwww... (a-b)=(1-1)=0...\r\n\r\nno dividing by zero. i've seen that bogus proof like fifty times.\r\nso has everyone else.. xD", "Solution_24": "here's one of the cooler ones:\r\n\r\nconsider a wheel with circumference k, and a wheel with circumference k/2 that's fixed to the first wheel's center so that it turns with it. when the first wheel makes a complete revolution, the second wheel does the same, so the circumferences are equal.\r\nk=2k, so 2=1\r\n\r\nthere's another one involving trig and integrals, but i forget it.", "Solution_25": "$ \\sin{\\pi}\\equal{}\\sin{2\\pi}$\r\nlol, yeah that doesn't work but thats as close as I can get.", "Solution_26": "[quote=\"Temperal\"]here's one of the cooler ones:\n\nconsider a wheel with circumference k, and a wheel with circumference k/2 that's fixed to the first wheel's center so that it turns with it. when the first wheel makes a complete revolution, the second wheel does the same, so the circumferences are equal.\nk=2k, so 2=1\n\nthere's another one involving trig and integrals, but i forget it.[/quote]\r\n\r\nHere are the calculus ones: http://en.wikipedia.org/wiki/Invalid_proof#Proof_that_2_.3D_1_2", "Solution_27": "[hide=\"All numbers are equal\"]To prove any two real (or complex, I haven't tried) numbers equal...\nLet the two numbers be $ a$ and $ b$. We have...\n\\begin{align*}a^2-ab+b^2&=a^2-ab+b^2\\\\\na^2-a^2-ab&=b^2-b^2-ab\\\\\na^2-2a\\left(\\frac{a+b}{2}\\right)&=b^2-2b\\left(\\frac{a+b}{2}\\right)\\\\\na^2-2a\\left(\\frac{a+b}{2}\\right)+\\left(\\frac{a+b}{2}\\right)^2&=b^2-2b\\left(\\frac{a+b}{2}\\right)+\\left(\\frac{a+b}{2}\\right)^2\\\\\n\\left(a-\\frac{a+b}{2}\\right)^2&=\\left(b-\\frac{a+b}{2}\\right)^2\\\\\na-\\frac{a+b}{2}&=b-\\frac{a+b}{2}\\\\\na&=b\n\\end{align*}[/hide]", "Solution_28": "the trouble with the proof is that in the third to last line the term s inside the square are additive inverses to each other: when square rooting there has to be a plus-minus sign(in this case minus sign)", "Solution_29": "http://www.kent.k12.wa.us/staff/DavidWright/calculus/book/55/jokes.html\r\n\r\nhttp://www.workjoke.com/projoke22.htm\r\n\r\nSome of these are good." } { "Tag": [ "induction" ], "Problem": "Find the value of $ c$ such that $ \\sum_{n \\equal{} 1}^{11} (n \\minus{} c)(2n \\minus{} c)$ is minimal and find the minimum value $ m$.", "Solution_1": "note that $ \\sum_{n\\equal{}1}^x n\\equal{}x(x\\plus{}1)/2$ and $ \\sum_{n\\equal{}1}^x n^2 \\equal{} x(x\\plus{}1)(2x\\plus{}1)/6$ (both well known facts which are easily proven by induction)\r\n\r\n$ \\sum_{n\\equal{}1}^{11} (n\\minus{}c)(2n\\minus{}c) \\equal{} \\sum_{n\\equal{}1}^{11} 2n^2\\minus{}3nc\\plus{}c^2 \\equal{} 2*11*12*23/6 \\minus{} 3*11*12/2*c \\plus{} 11c^2$\r\n$ \\equal{} 1012 \\minus{}198c\\plus{}11c^2$.\r\n$ \\equal{} 11(92\\minus{}18c\\plus{}c^2)$\r\n$ \\equal{} 11((c\\minus{}9)^2 \\plus{}11)$\r\n\r\nThus the mininum occurs at $ c\\equal{}9$ and is $ m\\equal{}121$.", "Solution_2": "That's correct answer.\r\nLet me say one thing, $ \\sum_{n \\equal{} 1}^{11} 2n^2 \\minus{} 3nc \\plus{} c^2$ should be written as $ \\sum_{n \\equal{} 1}^{11} (2n^2 \\minus{} 3nc \\plus{} c^2)$." } { "Tag": [ "LaTeX", "USAMTS", "geometry", "rectangle" ], "Problem": "hmmmm, I got marked down for not showing the diagram in the first round of the USAMTS. could someone please show me how to position latex diagrams properly. so the text fits nicely around them? thanks. :D basically, all i've done is copy and paste from the original latex file of the problem set, obviously, the diagram doesn't go where i want it to.", "Solution_1": "[quote=\"nameisong\"]hmmmm, I got marked down for not showing the diagram in the first round of the USAMTS. could someone please show me how to position latex diagrams properly. so the text fits nicely around them? thanks. :D basically, all i've done is copy and paste from the original latex file of the problem set, obviously, the diagram doesn't go where i want it to.[/quote]\n\nNone of the Round 1 solutions required a diagram. For example, in Problem 1/1/16, a solution could have said that the numbers 1, 10, 3, 6, 5, 7, 2, 8, 4, 9 wrap around the pentagon in that order with the 1, 3, 5, 2, and 4 at the vertices, without drawing the pentagon. However, many confusing descriptions can be made clearer by adding a diagram. Perhaps that is what your USAMTS grader meant. An alternative to mastering diagrams would be to instead write a clear, well-organized solution.\n\nWhen I typeset last year's USAMTS (Year 15) in LaTeX, I put the diagrams to the right of the paragraphs mainly because I could not afford to waste room on the problem sets. A problem set is easier to look over when it fits onto a single page. Richard Rusczyk (or someone else at AoPS) did likewise for this year's USAMTS. However, when you submit a solution, you may take several pages. So feel free to put your diagrams between paragraphs and not worry about making the text flow around a diagram.\n\nLet's get back to your LaTeX question. The LaTeX code for the pentagon in Problem 1/1/16 is:\n[quote]\\USprob{1/1/16.}{ \n\\rightskip 2in\nThe numbers 1 through 10 can be arranged along the vertices and sides of a pentagon so that the sum of the three numbers along each side is the same. The diagram on the right shows an arrangement with sum~16. Find, with proof, the smallest possible value for a sum and give an example of an arrangement with that sum.\n\\vskip-1.2in\\hskip4.5in\\raisebox{-70pt}[0pt]{\\setlength{\\unitlength}{1pt}\n\\begin{picture}(90,80)\n\\put(15,46){\\line(1,-3){12}}\n\\put(27,10){\\line(1,0){38}}\n\\put(65,10){\\line(1,3){12}}\n\\put(15,46){\\line(4,3){31}}\n\\put(77,46){\\line(-4,3){31}}\n\\put(20,0){4}\n\\put(43,0){9}\n\\put(65,0){3}\n\\put(73,22){6}\n\\put(80,44){7}\n\\put(63,59){8}\n\\put(43,71){1}\n\\put(22,59){5}\n\\put(1,44){10}\n\\put(12,22){2}\n\\end{picture}}}[/quote]\n\nThe portion that begins with \\begin{picture} and ends with \\end{picture} is the diagram itself. LaTeX treats it as a solid rectangle (a \"box\" in TeX nomenclature) measuring 90 units wide and 80 units high, due to the (90,80) after the \\begin{picture}. The \\setlength command before the diagram specifies that 1 unit equals 1 point (0.013837 inch), so the box is 90 points wide and 80 points high. That is all you need to put a copy of the diagram into your LaTeX solution:\n[quote]\\setlength{\\unitlength}{1pt}\n\\begin{picture}(90,80)\n\\put(15,46){\\line(1,-3){12}}\n\\put(27,10){\\line(1,0){38}}\n\\put(65,10){\\line(1,3){12}}\n\\put(15,46){\\line(4,3){31}}\n\\put(77,46){\\line(-4,3){31}}\n\\put(20,0){4}\n\\put(43,0){9}\n\\put(65,0){3}\n\\put(73,22){6}\n\\put(80,44){7}\n\\put(63,59){8}\n\\put(43,71){1}\n\\put(22,59){5}\n\\put(1,44){10}\n\\put(12,22){2}\n\\end{picture}[/quote]\r\n\r\nAll the other LaTeX code is to make the text and diagram fit together. The \\rightskip 2in at the beginning of the paragraph tells LaTeX to leave 2 inches of white space to the right of the paragraph before the right margin. The left bracket \"{\" before the \\rightskip command matches to the last right bracket \"}\" after the \\end{picture} command. The purpose of that pair of brackets it to make sure the \\rightskip 2in command does not apply to any other paragraph. The \\vskip-1.2in\\hskip4.5in\\raisebox{-70pt}[0pt]{...} serves to move the diagram into that white space to the right of the paragraph. The exact size, 1.2 inches, of the \\vskip depends on the length of the paragraph and the spacing between the lines of text, but it is negative because we are moving the diagram upward. The \\raisebox command does the same thing as the \\vskip command, but without leaving horizontal mode (LaTeX calls this LR mode). It is sloppy to use both: that problem had been edited a few times and accumulated redundant commands. The \\hskip command moves the diagram rightward 4.5 inches because the paragraph is 4.5 inches wide.\r\n\r\nDo you have any other questions? The picture environment is complicated. An alternative is to create a diagram with other software and save it as a JPEG or PDF image file. The \\includegraphics command can insert the image into the PDF file created from the LaTeX, but you still have to mess with \\vskip and \\hskip to postion the image. Furthermore, when you e-mail your LaTeX solution to the USAMTS, you have to e-mail the image file too.\r\n\r\nErin Schram\r\nFormer typesetter of the USAMTS", "Solution_2": "Thanks a lot for this clear, informative explanation! props :P" } { "Tag": [ "geometry", "circumcircle", "trigonometry" ], "Problem": "Point P lies on minor arc AB of the circumcircle of square ABCD. AB=5 and AP=4. Find PD.\r\n\r\nthe picture is easy to draw.", "Solution_1": "[hide=\"hint\"]\nLaw of Cosines on APB to find PB, then use the fact that DPB is a right triangle\n[/hide]", "Solution_2": "i would hate to use trig in a geometry problem :P", "Solution_3": "Non-trig approach\r\n[hide]\nConnect $ AC, PC$. Find $ PC$ in the right triangle $ APC$, then use Ptolemy's Theorem to find $ DP$ in the cyclid quad $ ACDP$.[/hide]", "Solution_4": "[quote=\"timwu\"]Non-trig approach\n[hide]\nConnect $ AC, PC$. Find $ PC$ in the right triangle $ APC$, then use Ptolemy's Theorem to find $ DP$ in the cyclid quad $ ACDP$.[/hide][/quote]\r\n\r\nwow so easy... :) :o" } { "Tag": [ "calculus" ], "Problem": "Any one able to help me with this.\r\nUse newtons method to find as many rootd as you can of the nonlinear system:\r\nx+y+z=6\r\nx^2 + y^2 + Z^2=14\r\nx^3 + y^3 + z^3=37\r\nproduce 3m-fikes\r\n@ each inteation step solve the equation\r\nJd=-f\r\nwhere d=x^(k+1)-x^k in matlab by:\r\nd=-J/f", "Solution_1": "This is an interesting question, but this probably isn't the best forum to post it on. Maybe I should move this thread to another forum: which forum would fit better?", "Solution_2": "hey for matlab questions go to http://www.mathworks.com and go to the user community. they are really good at answering matlab questions as they are professional programmers :)", "Solution_3": "Other problem solving topics or the calculus computations and tutorials, I say." } { "Tag": [ "geometry", "romania", "incenter", "trigonometry", "conics", "hyperbola" ], "Problem": "The circle of center $I$ is inscribed in the convex quadrilateral $ABCD$. Let $M$ and $N$ be points on the segments $AI$ and $CI$, respectively, such that $\\angle MBN = \\frac 12 \\angle ABC$. Prove that $\\angle MDN = \\frac 12 \\angle ADC$.", "Solution_1": "This one is easy.\r\n\r\nMove $M$ on $AI$. Then map $BM \\mapsto BN$ is projective from pencil of lines through $B$ onto it self, since it is rotation for angle ${1\\over 2}\\angle ABC$ around $B$. There for map $M \\mapsto N$ is projective form line $AI$ to line $CI$. Now this map induces projective map $\\rho: DM \\mapsto DN$ from pencil of lines through $D$ onto itself. \r\n\r\nNow, when $M= A$ then $N= I$, so $\\angle MDN = {1\\over 2} \\angle ADC$ \r\nand when $M= I$ then $N= C$ so again $\\angle MDN = {1\\over 2} \\angle ADC$,\r\nand when $M$ is incenter of triangle $ABD$, $N$ is incenter of triangle $BCD$, from where we get again $\\angle MDN = {1\\over 2} \\angle ADC$.\r\n\r\nThis implys that map $\\rho$ is actualy rotation around $D$ for fixed angle ${1\\over 2}\\angle ADC$ since every projective map is exactly determined with 3 points.", "Solution_2": "my solution :\r\n\r\nlet $\\alpha$ be $\\angle ABM$ and let $\\beta$ be angle $CBN$. Then $\\angle IBM=\\beta$ and $\\angle IBN=\\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get :\r\n$\\displaystyle\\frac{AI}{MI}\\cdot\\frac{CI}{NI}=\\frac{\\sin BIA\\sin BIC}{\\sin A/2\\sin C/2}$. \r\nAnd :\r\n\r\n$\\displaystyle\\frac{AI}{MI}\\cdot\\frac{CI}{NI}=\\frac{\\sin p\\sin s}{\\sin r\\sin q}\\cdot\\frac{\\sin DIA\\sin DIC}{\\sin A/2\\sin C/2}$.\r\n\r\nBut $\\sin BIA\\sin BIC=\\sin DIA\\sin DIC$, because it will be the same with $\\cos\\frac{A-C}{2} +\\cos(D+\\frac{A+C}{2})=\\cos\\frac{C-A}{2} +\\cos(B+\\frac{A+C}{2})$, but this is true because $\\cos x=\\cos (-x)=\\cos (2\\pi-x)$.\r\n\r\nwe will obtain that $\\sin p\\sin s=\\sin r\\sin q$ and because $p+q=r+s=D/2$, we transform into sum of cosines and we will obtain easy that $p=r$ and this solves the problem.", "Solution_3": "[quote=\"maky\"]my solution :\n\nlet $\\alpha$ be $\\angle ABM$ and let $\\beta$ be angle $CBN$. Then $\\angle IBM=\\beta$ and $\\angle IBN=\\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get :\n$\\displaystyle\\frac{AI}{MI}\\cdot\\frac{CI}{NI}=\\frac{\\sin BIA\\sin BIC}{\\sin A/2\\sin C/2}$. [/quote]\r\n\r\n\r\nCan you show this step by step how to get\r\nAbdurashid", "Solution_4": "well, first :\r\n\r\n$\\alpha=\\angle ABM$ and $\\beta=\\angle CBN$ (notations).\r\nthen, because $\\angle BMN$ is half of $\\angle B$, we get that $\\angle MBI=\\beta$ and $\\angle NBI=\\alpha$.\r\n\r\nalso,\r\n\r\n$p=\\angle CDN$, $q=\\angle NDI$, $r=\\angle IDM$ and $s=\\angle MDA$.\r\n\r\nThen, sine theorem in $AMB$ and we get :\r\n\r\n$\\frac{AM}{MB}=\\frac{\\sin\\alpha}{\\sin A/2}$.\r\n\r\nSine theorem in $BMI$ :\r\n\r\n$\\frac{MB}{MI}=\\frac{\\sin BIA}{\\sin\\beta}$.\r\n\r\nSine theorem in $BNI$ :\r\n\r\n$\\frac{BN}{IC}=\\frac{\\sin BIC}{\\sin\\alpha}$.\r\n\r\nSine theorem in $BNC$ :\r\n\r\n$\\frac{CN}{BN}=\\frac{\\sin\\beta}{\\sin C/2}$.\r\n\r\nWe multiply this four relations and there goes the desired result :\r\n\r\n$\\frac{AI}{MI}\\cdot\\frac{CI}{NI}=\\frac{\\sin BIA\\sin BIC}{\\sin A/2\\sin C/2}$\r\n\r\nIn a similar way, we get the other part :\r\n\r\n$\\frac{AI}{MI}\\cdot\\frac{CI}{NI}=\\frac{\\sin p\\sin s}{\\sin r\\sin q}\\cdot\\frac{\\sin DIA\\sin DIC}{\\sin A/2\\sin C/2}$.\r\n\r\nNow we show that $\\sin BIA\\sin BIC=\\sin DIA\\sin DIC$. (*)\r\nLet $A,B,C,D$ be the angles of $ABCD$. \r\n\r\nWe get $\\sin BIA=\\sin\\frac{A+B}{2}$, and similar relations.\r\n\r\nThen (*) becomes equivalent with :\r\n\r\n$\\cos(\\frac{A+B}{2}-\\frac{C+B}{2})-\\cos(\\frac{B+A}{2}+\\frac{B+C}{2})$ $=$ $\\cos(\\frac{A+D}{2}-\\frac{C+D}{2})-\\cos(\\frac{D+A}{2}+\\frac{D+C}{2})$.\r\n\r\nBut we have $\\cos\\frac{A-C}{2}=\\cos\\frac{C-A}{2}$ and $\\cos(B+\\frac{A+C}{2})$ $=$ $\\cos(2\\pi-B-\\frac{A+C}{2})=\\cos(A+B+C+D-B-\\frac{A+C}{2})=\\cos(D+\\frac{A+C}{2})$.\r\n\r\nBut using the above facts, this becomes $\\sin p\\sin s=\\sin q\\sin r$. (**)\r\nNow let`s look at the angles $p,q,r,s$. We have $p+q=r+s=D/2$, and because $ABCD$ is convex, $D/2<\\pi/2$, so $p+q<\\pi/2$ and $r+s<\\pi/2$. Also, $p,q,r,s$ are all smaller than $D/2$. \r\nLet $t=D/2$.\r\nThen $q=t-p$ and $s=t-r$. \r\n(**) means $\\sin p\\sin(t-r)=\\sin r\\sin(t-p)$, equivalent with $\\cos(p-t+r) - \\cos(p+t-r)=\\cos(r-t+p)-\\cos(r+t-p)$, which means $\\cos(t+r-p)=\\cos(t+p-r)$. We also have $0 2(n/2)$ (so the best choice is take its half part) ; if $ n$ is odd, it's better to split it into $ (n \\minus{} 1)/2$ and $ (n \\plus{} 1)/2$: $ n < (n^2 \\minus{} 1)/4$.\r\n\r\nSo, there are $ a$ 2 and $ b$ 3 in the product. Besides, see that 3+3 has a greater product than 2+2+2, so we can substitute 3 \"2\" in 2 \"3\".\r\n\r\nIf the sum is 40,\r\nthe best choice is to write 40 as $ 3 \\cdot 12 \\plus{} 2\\cdot2$ and the product becomes $ 4\\cdot 3^{12}$ :)\r\n\r\n\r\nIs it right?\r\n\r\n\r\nEDIT: harwi, why are you taking into consideration just two integers? Have I misread the question?", "Solution_3": "very nice work Gabuntu94.", "Solution_4": "sorry , I didn't read the question well." } { "Tag": [], "Problem": "Ean $n_{1},n_{2},\\ldots,n_{k}$ opou $k\\in \\mathbb{N}^{*}, k\\geq 2$, einai diaforetikoi metaksi tous thetikoi akeraioi oi opoioi den diairountai apo prwtous arithmous megaliterous tou $3$, tote na deixthei oti \r\n\r\n$S=\\frac{1}{n_{1}}+\\frac{1}{n_{2}}+\\cdots+\\frac{1}{n_{k}}<3$.\r\n\r\nAlexandros", "Solution_1": "\u0398\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $n_{i}=2^{k_{j}}\\cdot 3^{l_{m}}$ \r\n\u0388\u03c3\u03c4\u03c9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 $k_{1}=max{k_{j}}$ \u03ba\u03b1\u03b9 $l_{1}=max{l_{m}}$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \r\n$S=\\frac{1}{n_{1}}+...+\\frac{1}{n_{k}}$ \r\n$S_{1}=1+\\frac{1}{2}+\\frac{1}{2^{2}}+...+\\frac{1}{2^{k_{1}}}$\r\n$S_{2}=1+\\frac{1}{3}+\\frac{1}{3^{2}}+...+\\frac{1}{3^{l_{1}}}$\r\n\r\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \r\n$S< S_{1}\\cdot S_{2}=\\frac{1}{1-\\frac{1}{2}}\\cdot\\frac{1}{1-\\frac{1}{3}}=3$\r\n\r\n\u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf", "Solution_2": "[quote=\"silouan\"]\n\u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \n$S< S_{1}\\cdot S_{2}=\\frac{1}{1-\\frac{1}{2}}\\cdot\\frac{1}{1-\\frac{1}{3}}=3$\n\n\u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf[/quote]\r\n\r\nApolitws swsto me mia mikri paratirisi. Einai: $S_{1}<\\sum_{i=0}^{\\infty}\\frac{1}{2^{i}}=\\frac{1}{1-\\frac{1}{2}}$\r\nkai $S_{2}<\\sum_{i=0}^{\\infty}\\frac{1}{3^{i}}=\\frac{1}{1-\\frac{1}{3}}$\r\n\r\nAra $S\\leq S_{1}S_{2}<\\frac{1}{1-\\frac{1}{2}}\\cdot\\frac{1}{1-\\frac{1}{3}}=3$\r\n\r\nPoli wraia Silouan,\r\n\r\nAlexandros" } { "Tag": [], "Problem": ":surf: let triangle ABC: $C\\leq B \\leq A \\leq 90$\r\nand $M=cos(\\frac{A-B}{2})sin(\\frac{A}{2})sin(\\frac{B}{2})$\r\nFind min of M :rotfl:", "Solution_1": "nobody find this problem????\r\nWe prove $cos(A-B)\\leq cosA+cosB$", "Solution_2": ":( more hints, please?", "Solution_3": "first i guess min of M=$\\frac{1}{4}$\r\nwe need to prove $M\\geq\\frac{1}4{}$\r\n$\\Leftrightarrow cos(A-B)\\geq cosA+cosB$\r\n(Sorry!!!I made mistake \"$\\leq$\" in second reply)\r\nwe have $2sin(A+B)cos(A-B)=sin2A+sin2B$\r\n$\\Leftrightarrow sinCcos(A-B)=sinAcosA+sinBcosB$\r\n$\\Leftrightarrow cos(A-B)=\\frac{sinA}{sinC}cosA+\\frac{sinB}{sinC}cosB$\r\n$\\Leftrightarrow cos(A-B)\\geq cosA+cosB$\r\nBecause $\\frac{\\pi}{2}\\geq A\\geq B\\geq C$\r\nBesides I have another solution but long use function." } { "Tag": [ "geometry", "rectangle", "analytic geometry" ], "Problem": "I seen this problem awhile ago, I find it hard. (I'm gonna write the problem so that even color blind can understand).\r\n\r\n[i]Suppose every point in $ \\mathbb{R}^2$ is colored either black or white. Prove there exists a rectangle with all four vertices the same color[/i].\r\n\r\nThis has been bugging me, what if [u]rectangle[/u] is replaced by [u]square[/u]. Is it still true?", "Solution_1": "we set ourselves coordinate axes. of the two colors, at least one, say red, must occur infinitely many times on the $ x$ axis. denote by $ S$ the set of these red points. now, let $ X$ be the set of points that are directly above the points in $ S$ on the line $ y\\equal{}1$. if more than one point in $ X$ is red, we're done; otherwise, all points in $ X$ except possibly one of them are black. let $ Y$ be the set of points directly below the points in $ S$ on the line $ y\\equal{}\\minus{}1$. if more than one point in $ Y$ is red, we're done; otherwise, all points in $ Y$ except possibly one of them are black. but with all points but one in $ X$ and all points in $ Y$ but one black, there must exist a black rectangle, and again we're done.", "Solution_2": "I always find great pleasure in solving coloring problems in $ \\mathbb{R}^n$. I will generalize the problem using $ n$ colors instead of two.\r\n\r\n[u]Solution.[/u] Consider a set of $ n^{n\\plus{}1}\\plus{}1$ parallel lines with $ n\\plus{}1$ points that lie atop one another. Then there exist at least two monochromatic points on each line. Since there are $ n^{n\\plus{}1}$ sequences of colors, and we have $ n^{n\\plus{}1}\\plus{}1$ parallel lines, two of them must have the same color sequence. And the conlcusion follows. $ \\Box$", "Solution_3": "right. though... there's really no need for it to be $ \\mathbb{R}^n$ for a specific $ n$.", "Solution_4": "The $ \\mathbb{R}^n$ in the first sentence is not related to the solution. $ n$ in the solution is simply the number of colors used in $ \\mathbb{R}^2$.", "Solution_5": "[quote=\"sylow_theory\"]I seen this problem awhile ago, I find it hard. (I'm gonna write the problem so that even color blind can understand).\n\n[i]Suppose every point in $ \\mathbb{R}^2$ is colored either black or white. Prove there exists a rectangle with all four vertices the same color[/i].\n\nThis has been bugging me, what if [u]rectangle[/u] is replaced by [u]square[/u]. Is it still true?[/quote]\r\nThe original problem is still true if you replace $ \\mathbb{R}^2$ with a 4x7 lattice grid. (A 4x6 lattice grid will not suffice.)\r\n\r\nAs for the part that's bugging you, it's bugging me too..." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let a, b, c are the besides in a triangle. Prove that\r\n$2<\\frac{a+b}{c}+\\frac{b+c}{a}+\\frac{c+a}{b}-\\frac{a^{3}+b^{3}+c^{3}}{abc}\\leq 3$", "Solution_1": "2nd inequality is equivalent to $\\sum a(a-b)(a-c)\\geq 0$, it is true by Schur's theorem.", "Solution_2": "[quote=\"N.T.TUAN\"]2nd inequality is equivalent to $\\sum a(a-b)(a-c)\\geq 0$, it is true by Schur's theorem.[/quote]\r\nThanks, but i need a solution for 1st ineq.", "Solution_3": "[quote=\"kcbu\"][quote=\"N.T.TUAN\"]2nd inequality is equivalent to $\\sum a(a-b)(a-c)\\geq 0$, it is true by Schur's theorem.[/quote]\nThanks, but i need a solution for 1st ineq.[/quote]\r\n\r\nIt is well known (not a known inequality but it is famous and in fact the only one I know of :) )\r\n\r\nThat,\r\n\r\n$\\frac{a}{b+c}+\\frac{b}{a+c}+\\frac{c}{a+b}\\geq \\frac{3}{2}$\r\n\r\nNow the Harmonic mean (HM-AM),\r\n\r\n$\\frac{3}{\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c}}\\geq \\frac{3}{2}$\r\n\r\nThus,\r\n\r\n$\\frac{1}{\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c}}\\leq \\frac{1}{2}$\r\n\r\nTake reciprocals,\r\n\r\n$\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c}\\geq 2$\r\n\r\n(I hope I did this right, I am brand new to these inequalities stuff).", "Solution_4": "sylow_theory,\r\n\r\nWhat do you do? I don't understand :huh: \r\n\r\nFirst maybe equivalent to $(a+b-c)(b+c-a)(c+a-b)>0$, I think so.", "Solution_5": "How are a,b,c the angles? Meaning radians in terms of pi? This seems really unnatural. Do you mean sides?\r\n\r\nAlso, why did you write \"easy\" if you can't solve it...", "Solution_6": "[quote=\"me@home\"]How are a,b,c the angles? Meaning radians in terms of pi? This seems really unnatural. Do you mean sides?\n\nAlso, why did you write \"easy\" if you can't solve it...[/quote]\r\nOMG, sorry, a, b, c, are besides not angles.\r\nI solved but i need another solution :oops:", "Solution_7": "I am true :D\r\n[quote=\"N.T.TUAN\"]\n\nFirst maybe equivalent to $(a+b-c)(b+c-a)(c+a-b)>0$, I think so.[/quote]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ a_i>0$\r\n$ \\sum_{i\\equal{}1}^{n}{a_i}\\equal{}1$\r\nProve:\r\n$ \\sum_{i\\equal{}1}^{n}{\\frac{a_i}{\\sqrt{1\\minus{}a_i}}} \\ge \\sqrt{\\frac{n}{n\\minus{}1}}$", "Solution_1": "[quote=\"alpha-beta\"]$ a_i > 0$\n$ \\sum_{i \\equal{} 1}^{n}{a_i} \\equal{} 1$\nProve:\n$ \\sum_{i \\equal{} 1}^{n}{\\frac {a_i}{\\sqrt {1 \\minus{} a_i}}} \\ge \\sqrt {\\frac {n}{n \\minus{} 1}}$[/quote]\r\nIf $ f(x)\\equal{}\\frac{x}{\\sqrt{1\\minus{}x}}$ then, $ f''(x)\\equal{}\\frac{4\\minus{}x}{4\\sqrt{(1\\minus{}x)^5}},$ which is positive for $ 0k>n\\(n-1) real hunghtn \r\nand this generalisation just follows form suranji itself \r\n(put $a_i ^k = b_i ^n$ ;) )\r\n :)", "Solution_5": "[quote=\"Albanian Eagle\"]we should have n>k>n\\(n-1) real hunghtn \nand this generalisation just follows form suranji itself \n(put $a_i ^k = b_i ^n$ ;) )\n :)[/quote]\r\n\r\nOh, surprisingly, I don't realize this relation. Very good, Albanian. :) \r\n\r\nSo this problem was solved....", "Solution_6": "But I'm still bemused.\r\n[b]circumstantiate[/b],please.Thanks.", "Solution_7": "How to prove Suranji's Inequality? I mean a solution different from Harazi's solution in Old and New Inequality? I'm have a very nice solution (in my book :).", "Solution_8": "well, some days ago Hawk Tiger showed me a nice way to prove Suranyi (again with induction) and that's the shortest proof I've ever seen for it. I hope he will post it :)", "Solution_9": "I solved it by Karamara Inequality, no induction. It is in Hungkhtn's English Version book but I will wait by two month before post my solution.", "Solution_10": "Can anyone post the proof of Suranyi's inequality now, please? :maybe: \r\nThank you.", "Solution_11": "search because it is already in the forum. :wink:", "Solution_12": "Can you please post a link, because I cannot find it..." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "hi find all f:R--->R that for any real x,y \r\n(x^y)-f(f(y))=y*f(x)", "Solution_1": "[quote=\"irantst\"]hi find all f:R--->R that for any real x,y \n(x^y)-f(f(y))=y*f(x)[/quote]\r\n\r\nWhere is this problem coming from ?\r\n\r\n$ x^y$ is not defined for any $ x,y\\in\\mathbb R$, so this is meaningless. In what kind of contest did you encounter it ?", "Solution_2": "excuse me f:N--->N", "Solution_3": "[quote=\"irantst\"]hi find all f:N--->N that for any real x,y \n(x^y)-f(f(y))=y*f(x)[/quote]\r\n\r\nDo you just try to solve the problems you found in your strange class / contest / book ?\r\n\r\nLet $ P(x,y)$ be the assertion $ x^y\\minus{}f(f(y))\\equal{}yf(x)$\r\n\r\n$ P(1,1)$ $ \\implies$ $ f(1)\\equal{}1\\minus{}f(f(1))<1\\notin\\mathbb N$ and so no solution ." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a+b>c\r\nb+c>a\r\na+c>b\r\na>0 b>0 c>0\r\n\r\nPROVE $ (a\\plus{}b\\plus{}c)(a^2 \\plus{}b^2 \\plus{}c^2 )\\ge a^3 \\plus{}b^3 \\plus{}c^3$", "Solution_1": "hello, after expanding your inequality \r\n$ (a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2)\\minus{}(a^3\\plus{}b^3\\plus{}c^3) \\geq0$ we get $ ab^2\\plus{}ac^2\\plus{}a^2b\\plus{}bc^2\\plus{}a^2c\\plus{}b^2c\\geq0$ which is true because the conditions $ a>0,b>0,c>0$.\r\nSonnhard.", "Solution_2": "Sorry! i made a mistake. $ (a\\plus{}b\\plus{}c)(a^2 \\plus{}b^2 \\plus{}c^2)\\ge2(a^3 \\plus{}b^3 \\plus{}c^3 )$", "Solution_3": "hello, afteryour correction is your inequality equivalent to\r\n$ \\minus{} a^3 \\plus{} ab^2 \\plus{} ac^2 \\plus{} a^2b \\minus{} b^3 \\plus{} bc^2 \\plus{} ca^2 \\plus{} cb^2 \\minus{} b^3\\geq0$ This is true because we can rewrite in the following form\r\n$ a^2( \\minus{} a \\plus{} b \\plus{} c) \\plus{} b^2(a \\plus{} c \\minus{} b) \\plus{} c^2(a \\plus{} b \\minus{} c)\\geq0$ and this is true because of the triangle inequality.\r\nSonnhard." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Prove:\r\n1 - $ \\frac{1}{2}$ + $ \\frac{1}{3}$ - $ \\frac{1}{4}$ + $ \\frac{1}{5}$ -.......... + $ \\frac{1}{2007}$ - $ \\frac{1}{2008}$ = $ \\frac{1}{2005}$ + $ \\frac{1}{1006}$ +........+ $ \\frac {1}{2008}$\r\nThank you for reading. :lol:", "Solution_1": "Not $ \\equal{} \\frac {1}{2005} \\plus{} \\frac {1}{1006}\\cdots$ but $ \\equal{} \\frac {1}{1005} \\plus{} \\frac {1}{1006}\\cdots$ \r\n$ 1 \\minus{} \\frac {1}{2} \\plus{} \\frac {1}{3} \\minus{} \\frac {1}{4} \\plus{} \\cdots \\plus{} \\frac {1}{2n \\minus{} 1} \\minus{} \\frac {1}{2n} \\equal{}$\r\n$ \\equal{}1 \\plus{} \\frac {1}{2} \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{4} \\plus{} \\cdots \\plus{} \\frac {1}{2n \\minus{} 1} \\plus{} \\frac {1}{2n} \\minus{}$\r\n$ \\minus{} \\frac {2}{2} \\minus{} \\frac {2}{4} \\minus{} \\frac {2}{6}\\cdots \\minus{} \\frac {2}{2n} \\equal{}$\r\n$ \\equal{} \\frac {1}{n \\plus{} 1} \\plus{} \\frac {1}{n \\plus{} 2} \\plus{} \\cdots \\plus{} \\frac {1}{2n}$.\r\nSet $ n \\equal{} 1004.$\r\n :)" } { "Tag": [ "trigonometry", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "study the convergence of \r\n\r\n$u(n)=\\sin(1+\\sin(2+\\sin(3+...+\\sin(n)...)))$", "Solution_1": "This is a nice problem. It is a pity that nobody replied for a month. Here is a \r\n[hide=\"Hint\"] We always have $|\\sin x-\\sin y|\\le |x-y|$ and for most $x,y$, this inequality can be improved quite a lot. Use it to estimate $|u(n)-u(n+1)|$[/hide]", "Solution_2": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=121224[/url]\r\nIn fact it's easy to prove that $|u_{n+1}-u_{n}|\\leqslant C\\cdot q^{n}$ with some $q\\in(0;1)$", "Solution_3": "Here's proof of my statement. In [b]Rust[/b]'s notation \\[|x_{n+1}-x_{n}|=|x_{n+1}(y_{n,1})-x_{n+1}(0)|=|\\frac{\\partial x_{n+1}}{\\partial x}(\\theta y_{n,1})|\\cdot|y_{n,1}|\\le\\sup_{[-1;1]}|\\frac{\\partial x_{n+m}}{\\partial x}|\\]\r\n\\[\\frac{\\partial x_{n+m}}{\\partial x}=\\cos(1+\\sin(2+\\ldots)\\cdot\\cos(2+\\sin(3+\\ldots)\\cdot\\ldots\\]\r\nFor $0\\le k\\le\\frac{n-\\frac{\\pi}2-1/2}{\\pi}$ let $m$ be the nearest integer to $\\frac{\\pi}2+\\pi k$. For such $m$\r\n\\[|\\cos(m+\\ldots)|\\le\\max_{t\\in[\\frac{\\pi}2+\\pi k-1,5;\\frac{\\pi}2+\\pi k+1,5]}|\\cos t|=0,9974949866\\ldots.\\]" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Thanks you for helping me for a long time.\r\nGiven two circles $(O_{1})$, $(O_{2})$ which meet together at $A$ and $B$. Let $PQ$ and $MN$ are two common tangent ($P,M \\in (O_{1})$, $Q,N \\in (O_{1})$). $MM'$ is diameter of $O_{1}$. \r\nLet $O_{3}$, $O_{4}$ are the circumcircle of $APQ$ and $AMN$ respectively. Prove that \r\n$O_{1}O_{2}$, $O_{3}O_{4}$, $M'N$ are concurrent .", "Solution_1": "$\\angle AMN = \\angle MBA, \\angle ANM = \\angle NBA \\Rightarrow \\angle MAN+\\angle PAQ = 180^{o}\\\\ \\Rightarrow \\angle MAN+\\angle MBN = 180^{o}\\Rightarrow R_{(AMN)}= R_{(APQ)}\\text{(Of course }MN = PQ ) \\\\ \\angle O_{3}AP = 90^{o}-\\angle PQA = 90^{o}-\\angle QNA \\\\ \\angle O_{4}AM = 90^{o}-\\angle MNA = 90^{o}-\\angle NQA \\\\ \\angle PAM = 360^{o}-\\angle MAP-\\angle PAQ-\\angle QAN = 180^{o}-\\angle QAN \\\\ \\Rightarrow O_{3}AO_{4}= 180^{o}\\Rightarrow (O_{3}) \\text{ touches }(O_{4}) \\text{ at }A \\\\ \\text{We have }\\angle PAO_{3}-\\angle QAO_{3}= (90^{o}-\\angle PQA)-(90^{o}-\\angle QPA )\\\\ = \\angle QPA-\\angle PQA = \\frac{\\angle AO_{1}P-\\angle AO_{2}Q}{2}= \\angle O_{2}AQ-\\angle O_{1}AP \\\\ \\text{So }\\angle O_{1}AO_{3}= \\angle O_{2}AO_{3}\\Rightarrow \\frac{\\bar{SO_{1}}}{\\bar{SO_{2}}}=-\\frac{R_{1}}{R_{2}}\\text{ (\\{S\\}=}O_{3}O_{4}\\cap O_{1}O_{2}) \\\\ \\text{Easy to prove that }S \\in M'N$", "Solution_2": "Yes, good solution. Congratulation and thanks for solved my problems. :) \r\n\r\nP/s: I wonder that why I can't illustrate this problem by Skectchpad or Cabri. It seem wrong if I draw in computer.\r\nCan anybody help me?", "Solution_3": "I've drawn it by Sketch Pad and it's true.\r\nIt's a RAR file because MathLinks doesn't allowed to attach GSP file :D", "Solution_4": "Yes, thanks a lot. I'm to bad." } { "Tag": [ "percent" ], "Problem": "If a is 50% larger than c and b is 25% larger than c, then a is what percent larger than b?", "Solution_1": "$ a\\equal{}\\frac{3c}{2}, b\\equal{}\\frac{5c}{4}$\r\n\r\n$ c\\equal{}\\frac{4b}{5}$\r\n\r\n$ a\\equal{}\\frac{3}{2}\\times \\frac{4b}{5} \\equal{} \\frac{6b}{5}$\r\n\r\nTherefore $ 20\\%$." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "we considere a function $ f: I\\minus{}>R$ ($ I$ is an intervalle)\r\n\r\n$ 1$) let $ E$ be the differential equation :$ x'\\equal{}f(x)$\r\nProof that a solution $ \\phi$ of $ E$ has to be monotonic.\r\n\r\n$ 2$) let $ F$ be the differential equation : $ x\\equal{}f(x')$\r\nproof that a solution $ \\psi$ of $ F$ has to be convexe or concave.", "Solution_1": "no one ? :rotfl:", "Solution_2": "No need to bump it up :P. This is fairly standard, so I'm also surprised nobody posted a solution yet. Perhaps, it just doesn't look attractive enough... :?", "Solution_3": "I don't like this \"BAD HABIT\" of saying that everything is STANDARD or easy,just because you have read a lot of problems(maybe without even trying to solve them) :)", "Solution_4": "[quote=\"fedja\"]No need to bump it up :P. This is fairly standard, so I'm also surprised nobody posted a solution yet. Perhaps, it just doesn't look attractive enough... :?[/quote]\r\n\r\nOr maybe it is incomplete. I think it will be possible to find degenerate f, for which there is no solution to either E or F.", "Solution_5": "[quote=\"othman\"]I don't like this \"BAD HABIT\" of saying that everything is STANDARD or easy,just because you have read a lot of problems(maybe without even trying to solve them) :)[/quote]\r\n\r\nNot everything, far from it. But look for yourself. Take part 1. What does the condition really say, if you drop all antourage? It says that $ x(s)\\equal{}x(t)\\implies x'(s)\\equal{}x'(t)$. This forces us to try to find two equal values of $ x$ with different values of $ x'$ for any differentiable non-monotone function. What is our intuition here? If $ x$ is not monotone, it should cross some level going both up and down. Essentially we are done. The rest is just routine \r\n[hide=\"technicalities\"] \nIf $ x$ is not monotone, we can find $ ax(b)$ and $ x(b)0 , b>1 and c>=1 .", "Solution_1": "[quote=\"trbst\"]Solve in R : $ c^{\\log_b x} \\plus{} a\\cdot c^{\\frac {a\\cdot \\log_b x \\plus{} 1}{a}} \\plus{} a\\cdot\\log_b x\\cdot c^{\\log_b x} \\equal{} a$ , a>0 , b>1 and c>=1 .[/quote]\r\n\r\nSo $ c^{\\log_b x}(1\\plus{} a\\cdot c^{\\frac 1a} \\plus{} a\\cdot\\log_b x) \\equal{} a$\r\n\r\nSo $ 1\\plus{} a\\cdot c^{\\frac 1a} \\plus{} ay \\equal{} ac^{\\minus{}y}$ where $ y\\equal{}\\log_b x$\r\n\r\nLHS is an increasing function while RHS is a decreasing one. So we have at most one solution to this equation.\r\n\r\nAnd since we have the trivial one $ y\\equal{}\\minus{}\\frac 1a$, we got the unique solution $ \\boxed{x\\equal{}b^{\\minus{}\\frac 1a}}$" } { "Tag": [ "number theory", "least common multiple" ], "Problem": "Joe and Millie each have some baseball cards, and $ \\frac34$ of the number of cards Joe has is equal to $ \\frac23$ of the number of cards Millie has. What is the least number of cards they could have altogether?", "Solution_1": "First, we must find $ \\text{lcm}(3,4)$. This is clearly $ 12$.\r\n\r\nNow, you have $ 12 \\div \\frac34 \\plus{} 12 \\div \\frac23$ as our answer. This is $ 16 \\plus{} 18$, or $ \\boxed{34}$.", "Solution_2": "$ \\frac {3}{4}\\cdot J \\equal{} \\frac {2}{3}\\cdot M$\r\n\r\n$ 9J \\equal{} 8M$\r\n\r\nTherefore J=8 and M=9 when we minimize. 8+9=17\r\n\r\n\r\nEDIT: Thanks math!", "Solution_3": "[quote=\"mewto55555\"]$ \\frac {3}{4}\\cdot J = \\frac {2}{3}\\cdot M$[/quote]\r\n\r\nWhat you really want is:\r\n\r\n[code]$\\frac34J=\\frac23M$ or $\\frac34\\cdot J=\\frac23\\cdot M$[/code]\r\n\r\n$ \\frac34J=\\frac23M$ or $ \\frac34\\cdot J=\\frac23\\cdot M$\r\n\r\nI know, I'm fanatical about optimization...\r\n\r\nBut in general, you must separate all commands from variables by a space or a bracket around the variable, etc.", "Solution_4": "You could just cross multiply to get $\\frac{3}{4}x={\\frac{2}{3}y}$. You'd get $\\frac{9+8}{12}=\\frac{\\boxed{17}}{12}$" } { "Tag": [ "geometry", "3D geometry", "geometric transformation", "rotation" ], "Problem": "Carl has no life and glues together four unit cubes, lined up face to face. How many non-congruent solids can he make? Rotations do not form new solids.\r\n\r\n[hide=\"comments\"]The answer key sayd 8, but I only count 7...[/hide]", "Solution_1": "[hide]I get 8. You probably forgot that the weird 3D one (an isosceles L with an extra cube stuck on top (3rd dimension) of the leg) AND ITS MIRROR IMAGE are two different ones. You cannot get the mirror image with any number of rotations. (i thin k)\n\nThe list:\n2D figures:\n1. line\n2. long L\n3. T\n4: Z - ish\n5. 2x2 square\nnow for the 3D ones:\n6.isosc. L with a cube on top at the vertex\n7. '' on top at the leg-end\n8. MIRROR IMAGE of 7.[/hide]", "Solution_2": "Oh, I was suspecting that. Gah I can't imagine these things, and I didn't have any blocks handy...", "Solution_3": "Yeah... these spatial things are hard to imagine sometimes. Spatial problems are not like any other MC problems; you should practice them separately." } { "Tag": [ "inequalities", "inequalities unsolved", "133" ], "Problem": "Let a,b,c>0 prove that:\r\n $ \\frac{a^3}{b^2\\minus{}bc\\plus{}c^2}\\plus{}\\frac{b^3}{c^2\\minus{}ca\\plus{}a^2}\\plus{}\\frac{c^3}{a^2\\minus{}ab\\plus{}b^2}\\geq{a\\plus{}b\\plus{}c}$", "Solution_1": "it is true by chebysev if a,b,c is in triangle :wink:", "Solution_2": "I have deleted because mistake", "Solution_3": "great! :oops:", "Solution_4": "[quote=\"xiaohui\"]Let a,b,c>0 prove that:\n $ \\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\minus{} ca \\plus{} a^2} \\plus{} \\frac {c^3}{a^2 \\minus{} ab \\plus{} b^2}\\geq{a \\plus{} b \\plus{} c}$[/quote]\r\n$ \\sum_{cyc}\\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2}\\equal{}\\sum_{cyc}\\frac {a^4}{ab^2 \\minus{} abc \\plus{} ac^2}\\geq\\frac{(a^2\\plus{}b^2\\plus{}c^2)^2}{\\sum(a^2b\\plus{}a^2c\\minus{}abc)}\\geq a\\plus{}b\\plus{}c,$\r\nwhere the last inequality is true by Schur.", "Solution_5": "[quote=\"arqady\"][quote=\"xiaohui\"]Let a,b,c>0 prove that:\n $ \\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\minus{} ca \\plus{} a^2} \\plus{} \\frac {c^3}{a^2 \\minus{} ab \\plus{} b^2}\\geq{a \\plus{} b \\plus{} c}$[/quote]\n$ \\sum_{cyc}\\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2} \\equal{} \\sum_{cyc}\\frac {a^4}{ab^2 \\minus{} abc \\plus{} ac^2}\\geq\\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^2b \\plus{} a^2c \\minus{} abc)}\\geq a \\plus{} b \\plus{} c,$\nwhere the last inequality is true by Schur.[/quote]\r\nCan U detail ? If use Schur ; follow me ; we have to prove that \r\n$ (a^\\plus{}b^2\\plus{}c^2)^2 \\ge\\ (a^3\\plus{}b^3\\plus{}c^3)(a\\plus{}b\\plus{}c)$\r\nBut This Ineq is wrong :maybe:", "Solution_6": "[quote=\"onlylove_math\"][quote=\"arqady\"][quote=\"xiaohui\"]Let a,b,c>0 prove that:\n $ \\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2} \\plus{} \\frac {b^3}{c^2 \\minus{} ca \\plus{} a^2} \\plus{} \\frac {c^3}{a^2 \\minus{} ab \\plus{} b^2}\\geq{a \\plus{} b \\plus{} c}$[/quote]\n$ \\sum_{cyc}\\frac {a^3}{b^2 \\minus{} bc \\plus{} c^2} \\equal{} \\sum_{cyc}\\frac {a^4}{ab^2 \\minus{} abc \\plus{} ac^2}\\geq\\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{\\sum(a^2b \\plus{} a^2c \\minus{} abc)}\\geq a \\plus{} b \\plus{} c,$\nwhere the last inequality is true by Schur.[/quote]\nCan U detail ? If use Schur ; follow me ; we have to prove that \n$ (a^ \\plus{} b^2 \\plus{} c^2)^2 \\ge\\ (a^3 \\plus{} b^3 \\plus{} c^3)(a \\plus{} b \\plus{} c)$\nBut This Ineq is wrong :maybe:[/quote]\r\n$ (a^2\\plus{}b^2\\plus{}c^2)^2\\geq(a\\plus{}b\\plus{}c)\\sum_{cyc}(a^2b\\plus{}a^2c\\minus{}abc)\\Leftrightarrow\\sum_{cyc}(x^4\\minus{}x^3y\\minus{}x^3z\\plus{}x^2yz)\\geq0,$ which Schur.", "Solution_7": "I think the problem is wrong . Because With $ a\\equal{}1 , b\\equal{}10 , c \\equal{} 100$ the problem is wrong", "Solution_8": "[quote=\"thanhak5\"]I think the problem is wrong . Because With $ a \\equal{} 1 , b \\equal{} 10 , c \\equal{} 100$ the problem is wrong[/quote]\r\nAre you sure?", "Solution_9": "[quote=\"xiaohui\"]Let $ x,y,z>0$, prove that\n\n$ \\frac {x^3}{y^2 \\minus{} yz \\plus{} z^2} \\plus{} \\frac {y^3}{z^2 \\minus{} zx \\plus{} x^2} \\plus{} \\frac {z^3}{x^2 \\minus{} xy \\plus{} y^2}\\geq{x \\plus{} y \\plus{} z}$.[/quote]\r\nWithout lost the general, we may assume $ x\\equal{}\\min\\{x,y,z\\}$. Using the arithmetic mean - geometric mean inequality,\r\n\r\n$ \\sum{\\frac{x^3}{y^2\\minus{}yz\\plus{}z^2}}\\plus{}\\frac{4}{(y\\plus{}z)^2}\\sum{x\\left(y^2\\minus{}yz\\plus{}z^2\\right)}\\geq\\frac{4}{y\\plus{}z}\\sum{x^2}$,\r\n\r\nwe only need to prove :\r\n\r\n$ \\frac{4}{y\\plus{}z}\\sum{x^2}\\geq\\sum{x}\\plus{}\\frac{4}{(y\\plus{}z)^2}\\sum{x\\left(y^2\\minus{}yz\\plus{}z^2\\right)}$.\r\n\r\n[hide=\"Proof.\"]$ 4(y\\plus{}z)\\sum{x^2}\\minus{}(y\\plus{}z)^2\\sum{x}\\minus{}4\\sum{x\\left(y^2\\minus{}yz\\plus{}z^2\\right)}\\equal{}(y\\minus{}z)^2(3y\\plus{}3z\\minus{}5x)\\geq 0$.[/hide]" } { "Tag": [ "symmetry", "LaTeX" ], "Problem": "Find the value of \\[ \\frac{1}{5!} \\plus{} \\frac{1}{4!} \\plus{} \\frac{1}{2!3!} \\plus{} \\frac{1}{3!2!} \\plus{} \\frac{1}{4!} \\plus{} \\frac{1}{5!}\\]", "Solution_1": "We realize that this expression is symmetrical at $ \\frac {1}{2!3!}$. That means we only need to evaluate $ 2\\left(\\frac {1}{5!} \\plus{} \\frac {1}{4!} \\plus{} \\frac {1}{2!3!}\\right) \\equal{} 2\\left(\\frac {1}{120} \\plus{} \\frac {1}{24} \\plus{} \\frac {1}{12}\\right) \\equal{} 2\\left(\\frac {1}{120} \\plus{} \\frac {5}{120} \\plus{} \\frac {10}{120}\\right) \\equal{} 2\\left(\\frac {16}{120}\\right) \\equal{} 2\\left(\\frac {2}{15}\\right) \\equal{} \\boxed{\\frac {4}{15}}$", "Solution_2": "[quote=\"RoFlLoLcOpT\"]We realize that this expression is symmetrical at $ 2\\left(\\frac {1}{120} \\plus{} \\frac {1}{24} \\plus{} \\frac {1}{12}\\right) \\equal{} 2\\left(\\frac {1}{120} \\plus{} \\frac {4}{120} \\plus{} \\frac {10}{120}\\right) \\equal{} 2\\left(\\frac {15}{120}\\right) \\equal{} 2\\left(\\frac {1}{8}\\right) \\equal{} \\boxed{\\frac {1}{4}}$[/quote]\r\n\r\nI never thought of the symmetry, so that's why I had no idea how to do this :blush: But the problem here is that $ \\frac{1}{24}\\equal{}\\frac{5}{120}$, and not $ \\frac{4}{120}$. Therefore, the answer is actually $ \\boxed{\\frac{4}{15}}$", "Solution_3": "ah thanks for catching that :D", "Solution_4": "Well, there's another thing you could have seen.\nIf you look at the terms again, you notice something related to values of 5C1 5C2 5C3..etc, only 5! is missing, so multiply by it and divide by it and it becomes 1/5! * ( 2^5) --> Using summation of nCr = 2^n\nwhich is 32/120 = 4/15 :)", "Solution_5": "LaTeX edited:\n\n$\\frac{5!}{5!}\\cdot\\left(\\frac{1}{5!}+\\frac{1}{4!}+\\frac{1}{3!2!}+\\frac{1}{2!3!}+\\frac{1}{4!}+\\frac{1}{5!}\\right)$\n\n$=\\frac{\\dbinom{5}{0}+\\dbinom{5}{1}+\\dbinom{5}{2}+\\dbinom{5}{3}+\\dbinom{5}{4}+\\dbinom{5}{5}}{5!}$\n\n$=\\frac{2^5}{5!}$\n\n$=\\frac{32}{120}$\n\n$=\\boxed{\\frac{4}{15}}$\n\nThis method tends to be easier when there are a lot of terms in the expression." } { "Tag": [ "geometry", "rectangle", "trigonometry" ], "Problem": "a circle passes through the vertex C of a rectangle ABCD and touches AB and AD at M and N respectively.If the distance between C and MN is 5 cm. find the area of the rectangle.\r\n\r\nthank you.....", "Solution_1": "someone plz reply\r\nthank you", "Solution_2": "are you sure there is enough information?", "Solution_3": "i dnt know.......it was given in a school test", "Solution_4": "[hide=\"diagram\"][asy]pair A=(0,0), B=(0,45+45*sin(65/180*pi)), C=(45+45*cos(65/180*pi),45+45*sin(65/180*pi)), D=(45+45*cos(65/180*pi),0), E=(45-45*cos(65/180*pi),45+45*sin(65/180*pi)), F=(45+45*cos(65/180*pi),45-45*sin(65/180*pi)), M=(0,45), N=(45,0), O, P; draw(circumcircle(M, N, C)); O=circumcenter(M,N,C); draw(A--B--C--D--cycle); draw(O--M--N--O--C--M--F--C--N--D--F--N--E--F); draw(M--E); P = intersectionpoint(M--N,C--(C+3*(M-N)*dir(90))); draw(P--C); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NW); label(\"F\",F,NE); label(\"M\",M,NW); label(\"N\",N,SW); label(\"P\",P,SW); label(\"O\",O,SW); dot(A^^B^^C^^D^^E^^F^^M^^N^^P^^O);[/asy][/hide]Let a perpendicular down from $ C$ onto $ MN$, call this point $ P$. \r\nLet the center of the circle be $ O$. \r\nLet $ E$ and $ F$ be the intersection of $ BC$ and $ CD$ with circle $ O$, respectively.\r\nNow we will do some angle chasing. Let $ \\angle BME = \\beta$ and $ \\angle DNF = \\alpha$. Since $ ON\\perp ND$, $ \\angle DNF = \\angle NMF = \\angle NEF = \\angle NCF = \\alpha$ and similarly $ \\angle BME = \\angle MNE = \\angle MFE = \\angle MCE = \\beta$.\r\nSince $ \\triangle OEN$ is isosceles, $ \\angle ONE=\\angle OEN=\\alpha$. $ ON\\perp ND, CD \\perp ND \\implies ON \\parallel CD \\implies \\angle ONC = \\angle NCD=\\alpha$. So $ \\angle MNC = \\angle MNE+\\angle ENO+\\angle ONC = \\beta+\\alpha+\\alpha$.\r\nWe already know that $ \\angle ECF = 90$ and $ \\angle MCN = \\frac{1}{2}\\angle MON = 45$, so $ \\alpha+\\beta = 45$. This means $ \\angle PCN = \\beta$. $ PC = 5$ by definition so $ NC = \\frac{5}{\\cos\\beta}\\implies DC = \\frac{5\\cos\\alpha}{\\cos\\beta}$.\r\nSimilarly we obtain $ \\angle PCM = \\alpha \\implies MC = \\frac{5}{\\cos\\alpha}\\implies BC = \\frac{5\\cos\\beta}{\\cos\\alpha}$.\r\nSo $ [ABCD]=BC\\cdot DC = 25$", "Solution_5": "thank you for the solution....but i have a question\r\nis it necessary that EF will pass through the centre O", "Solution_6": "yes, because $ \\angle ECF = 90$, so $ \\angle EOF=2\\cdot \\angle ECF = 180$, which means $ EF$ passes through $ O$." } { "Tag": [], "Problem": "Determine the least natural number $n$ for which the following result holds:\r\n\r\nNo matter how the elements of the set $\\{1,2, \\dots ,n\\}$ are coloured red or blue, we can find integers $w,x,y,z$ in the set (not necessarily distinct) of the same colour such that\r\n$w+x+y=z$", "Solution_1": "[hide]\nWe claim that $n = 11$. We will attempt to construct a failing case for $n = 11$ and show that it is impossible. First, we note that $1+1+1 = 3$, $2+2+2 = 6$, and $3+3+3 = 9$ so none of the pairs $(1,3)$, $(2,6)$, or $(3,9)$ can be the same color. Also, $3+3+2 = 6+1+1$ so $(3,2)$ and $(1,6)$ cannot be red-red and blue-blue or vice versa. Consequently, we can obtain (assuming WLOG that $1$ is blue):\n\n[code]1 2 3 4 5 6 7 8 9 10 11\nB B R X X R X X B X X\n[/code]\n\nsince if $2$ was red and $6$ was blue we would have the red-red and blue-blue case. Now $2+1+1 = 4$, so $4$ is red. And $3+4+4 = 11$, so $11$ is blue.\n\n[code]1 2 3 4 5 6 7 8 9 10 11\nB B R R X R X X B X B\n[/code]\n\nBut this gives us $9+1+1 = 11$ and they are all blue, so contradiction. Hence we cannot construct a failing coloring with $n = 11$.\n\nFor $n = 10$, we have\n\n[code]1 2 3 4 5 6 7 8 9 10\nB B R R R R R R B B\n[/code]\n\nSo $n = 11$ is the smallest.\n\n[/hide]", "Solution_2": "Paladin: It's a good proof, but how did you find the smallest n? Did you just keep trying cases?", "Solution_3": "Well, I constructed a case for like $n = 5$ or something... then I realized it was too small so I added numbers one-by-one until it didn't work, tried to recolor and if it worked I kept going. So pretty much just experimenting to find the answer; a good way to solve olympiad problems." } { "Tag": [ "function", "number theory open", "number theory" ], "Problem": "In the following $x_0$ and $x_1$ are integers and $A,B$ are given rational numbers. By [ X ] is denoted the integer part of $X$ . Suppose that \r\n\\[ x_{k+1}= \\left[ Ax_k + Bx_{k-1} \\right] \\; ,\\; \\; k=1,2,...,n,... . \\]\r\n[color=green] [b]It's possible to find $x_n$ as a function of variables $n,x_0 ,x_1,A,B$ ? [/b] [/color]", "Solution_1": "Is this an open problem?", "Solution_2": "I think flip meant Unsolved instead of Open?\r\n\r\nAnyway I know there must be one for every value of A,B although I guess the general way of writing will be extremely complex...\r\n\r\nGuess it'll start with:\r\n\r\n...in the fibonacci sequence the n-th term is given by: $\\displaystyle F(n)=\\frac{(\\frac{1+\\sqrt{5}}2)^n-(\\frac{1-\\sqrt{5}}2)^n}{\\sqrt{5}}$\r\n\r\nand writing it as a lineary combination of fibonacci sequences.\r\nAlthough I have to admit I didn't get it fixed myself yet.", "Solution_3": "I have'nt a solution of this problem.Regards,proposer.", "Solution_4": "[quote=\"flip2004\"]I have'nt a solution of this problem.Regards,proposer.[/quote]Then please next time post the problem in the open problems section. :)", "Solution_5": "Valentin, this [b]IS[/b] the open problems section :D:D", "Solution_6": "[quote=\"Peter VDD\"]Valentin, this [b]IS[/b] the open problems section :D:D[/quote]yes, because I have moved the topic. ;)", "Solution_7": "lol\r\n\r\nwell, he originally posted it in the open problems section, and you replied: \"Is this an open question??\"\r\n\r\nAnd the times I replied were in open section as well :P (never even noticed it's been gone since then) :)" } { "Tag": [ "Euler", "geometry proposed", "geometry" ], "Problem": "Given an acute triangle $ ABC$. Let $ P$ be the interior point of $ ABC$ such that $ \\hat{APB}\\equal{}\\hat{BPC}\\equal{}\\hat{CPA}\\equal{}120^o$. \r\nProve that the Euler Line of triangles $ ABP,BCP,CAP$ are concurrent at the centriod of Napoleon Triangle.", "Solution_1": "see here:\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6188[/url]", "Solution_2": "[quote=\"\u00ac[\u0192(Gabriel)\u00b3\u00b2\u00b9\u00ba]\u00bc\"]see here:\n\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6188[/url][/quote]\r\n\r\nThank's Very much :)" } { "Tag": [ "geometry", "circumcircle", "perpendicular bisector", "geometry proposed" ], "Problem": "[b]let ABC be a triangle. and let P be a point inside. let M=max{PA,PB,PC}. Find P that M has its min value! :maybe: [/b]", "Solution_1": "[b] Any Idea?[/b]", "Solution_2": "for an acute triangle , i believe it is the cicuncenter of the triangle :|", "Solution_3": "i think we need to think about the perpendiculars bisectors\r\nbut i\u00b4m not sure of it :maybe:\r\nand , by inside of the triangle , you mean not at it\u00b4s sides?", "Solution_4": "I don\u00b4t whether this is right or not...my guess is that it is not totally right,or maybe it is all wrong :blush: .can anyone tell me whether i\u00b4m wrong or not?\r\n\r\n for an acute triangle ABC, P is the cicuncenter of ABC .suppose for absurd that it is not the circumcenter O and call this point X.\r\n Then , WLOG X lies in the triangle OAB . But then CX is bigger than CO , absurd.\r\n Now , if the triangle is obtuse, let AB be it\u00b4s longest side .\r\nConsider the perpendicular bisector of AB . Let D be the point such that |AD-CD| is as least as possible.If CD>AD=BD,choose one point E at CD such that |EC-EB| is as least as possible.\r\nSince ABC is obtuse, we\u00b4ll have that EA B'C' = AP sin(A)\r\n\r\nAnd we find it the same for others. So the problem is changed to what I said At first. :blush: :blush: :roll: :maybe: \r\n\r\nI hope some one solve it for me! I am really interested in it! :huuh: :help:\r\n\r\nSorry I don't have pro painting programs!:D", "Solution_6": "I don't have a good program to draw a good one! sorry!\r\nAnd the names have problem in the pic! But you can understand what I tried to say!:D" } { "Tag": [ "calculus", "derivative", "limit", "function", "calculus computations" ], "Problem": "Suppose f(x) =\r\nx^3 if x < 1\r\nax^2 + bx + c if x \u2265 1\r\n\r\nFind the values of a, b and c such that f''(1) exists.", "Solution_1": "$f(x)=\\left\\{\\begin{array}{cc}x^{3},&\\mbox{ if }x<1\\\\ax^{2}+bx+c, & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n$f^{\\prime}(x) = \\left\\{\\begin{array}{cc}3x^{2},&\\mbox{ if }x<1\\\\2ax+b , & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n$f^{\\prime\\prime}(x) = \\left\\{\\begin{array}{cc}6x,&\\mbox{ if }x<1\\\\2a, & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n\r\nFirst derivatives first! In order that $f^{\\prime}(1)$ exist, we require that $f_{-}^{\\prime}(x)=f_{+}^{\\prime}(x)$ where\r\n\r\n$f_{-}^{\\prime}(x)=\\lim_{x\\to1^{-}}\\frac{f(x)-f(1)}{x-1}= \\lim_{x\\to1^{-}}\\frac{x^{3}-(a+b+c)}{x-1}$\r\n\r\nand \r\n\r\n$f_{+}^{\\prime}(x)=\\lim_{x\\to1^{+}}\\frac{f(x)-f(1)}{x-1}= \\lim_{x\\to1^{+}}\\frac{(ax^{2}+bx+c)-(a+b+c)}{x-1}$\r\n$=\\lim_{x\\to1^{+}}\\frac{a(x^{2}-1)+b(x-1)}{x-1}=\\lim_{x\\to1^{+}}\\frac{a(x+1)(x-1)+b(x-1)}{x-1}$\r\n$=\\lim_{x\\to1^{+}}\\left(a(x+1)+b\\right)=2a+b$\r\n\r\nFor $f^{\\prime\\prime}(1)$ to exist, we require that $f_{-}^{\\prime\\prime}(x)=f_{+}^{\\prime\\prime}(x)$ where\r\n\r\n$f_{-}^{\\prime\\prime}(x)=\\lim_{x\\to1^{-}}\\frac{f^{\\prime}(x)-f^{\\prime}(1)}{x-1}= \\lim_{x\\to1^{-}}\\frac{3x^{2}-(2a+b)}{x-1}$\r\n\r\nand \r\n\r\n$f_{+}^{\\prime\\prime}(x)=\\lim_{x\\to1^{+}}\\frac{f^{\\prime}(x)-f^{\\prime}(1)}{x-1}= \\lim_{x\\to1^{+}}\\frac{(2ax+b)-(2a+b)}{x-1}$\r\n$= \\lim_{x\\to1^{+}}\\frac{2a(x-1)}{x-1}=2a$\r\n\r\nI left some blanks, see if you can fill them in...", "Solution_2": "You don't have to calculate the limits explicitly, by mean value theorem you can just pass limits in your other expressions for the derivatives to get the left and right derivatives at the problematic point.", "Solution_3": "Not sure about this one, but perhaps we require that f(x) be continuous at x=1, i.e. require that $\\lim_{x\\to1^{-}}f(x)=\\lim_{x\\to1^{+}}f(x)=f(1)$ that is require that\r\n$\\lim_{x\\to1^{-}}f(x)=\\lim_{x\\to1^{-}}x^{3}=1=\\lim_{x\\to1^{+}}f(x)=\\lim_{x\\to 1^{+}}(ax^{2}+bx+c)=a+b+c=f(1)$ \r\nand hence that\r\n\r\nEqn. #1) $a+b+c=1$.\r\n\r\nThis fills in some blanks:\r\n\r\n$f(x)=\\left\\{\\begin{array}{cc}x^{3},&\\mbox{ if }x<1\\\\ax^{2}+bx+c, & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n$f^{\\prime}(x) = \\left\\{\\begin{array}{cc}3x^{2},&\\mbox{ if }x<1\\\\2ax+b , & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n$f^{\\prime\\prime}(x) = \\left\\{\\begin{array}{cc}6x,&\\mbox{ if }x<1\\\\2a, & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n\r\nFirst derivatives first! In order that $f^{\\prime}(1)$ exist, we require that $f_{-}^{\\prime}(x)=f_{+}^{\\prime}(x)$ where\r\n\r\n$f_{-}^{\\prime}(x)=\\lim_{x\\to1^{-}}\\frac{f(x)-f(1)}{x-1}= \\lim_{x\\to1^{-}}\\frac{x^{3}-(a+b+c)}{x-1}$\r\n$= \\lim_{x\\to1^{-}}\\frac{x^{3}-1}{x-1}=\\lim_{x\\to1^{-}}\\frac{(x-1)(x^{2}+x+1)}{x-1}$\r\n$=\\lim_{x\\to1^{-}}(x^{2}+x+1)=3$\r\n\r\nand \r\n\r\n$f_{+}^{\\prime}(x)=\\lim_{x\\to1^{+}}\\frac{f(x)-f(1)}{x-1}= \\lim_{x\\to1^{+}}\\frac{(ax^{2}+bx+c)-(a+b+c)}{x-1}$\r\n$=\\lim_{x\\to1^{+}}\\frac{a(x^{2}-1)+b(x-1)}{x-1}=\\lim_{x\\to1^{+}}\\frac{a(x+1)(x-1)+b(x-1)}{x-1}$\r\n$=\\lim_{x\\to1^{+}}\\left(a(x+1)+b\\right)=2a+b$\r\n\r\nSo Eqn. #2) $2a+b=3$\r\n\r\nFor $f^{\\prime\\prime}(1)$ to exist, we require that $f_{-}^{\\prime\\prime}(x)=f_{+}^{\\prime\\prime}(x)$ where\r\n\r\n$f_{-}^{\\prime\\prime}(x)=\\lim_{x\\to1^{-}}\\frac{f^{\\prime}(x)-f^{\\prime}(1)}{x-1}= \\lim_{x\\to1^{-}}\\frac{3x^{2}-(2a+b)}{x-1}$\r\n$=\\lim_{x\\to1^{-}}\\frac{3x^{2}-3}{x-1}=\\lim_{x\\to1^{-}}\\frac{3(x-1)(x+1)}{x-1}=\\lim_{x\\to1^{-}}3(x+1)=6$\r\n\r\nand \r\n\r\n$f_{+}^{\\prime\\prime}(x)=\\lim_{x\\to1^{+}}\\frac{f^{\\prime}(x)-f^{\\prime}(1)}{x-1}= \\lim_{x\\to1^{+}}\\frac{(2ax+b)-(2a+b)}{x-1}$\r\n$= \\lim_{x\\to1^{+}}\\frac{2a(x-1)}{x-1}=2a$\r\n\r\nSo Eqn. #3) $2a=6$\r\n\r\nSolve the system $a+b+c=1$, $2a+b=3$, $2a=6$ to get the solution: $a = 3$, $b =-3$, $c = 1$.\r\n\r\nSo the function is $f(x)=\\left\\{\\begin{array}{cc}x^{3},&\\mbox{ if }x<1\\\\3x^{2}-3x+1, & \\mbox{ if }x\\geq 1\\end{array}\\right.$\r\n\r\nA plot of it:\r\n\r\n[url=http://ezimagecenter.com/out.php/i214_SecondDerivativeProblem.jpg][img]http://ezimagecenter.com/out.php/i214_SecondDerivativeProblem.jpg[/img][/url]" } { "Tag": [ "inequalities" ], "Problem": "X,Y and Z are positive real numbers such that XYZ=1.\r\n\r\nFind the least values of \r\n\r\n1. I/(X+1) + 1/(Y+1) +1/(Z+1)\r\n\r\n2. 1(/X+1)(Y+1) + 1/(Y+1)(Z+1) +1/(Z+1)(X+1)", "Solution_1": "you mean:\r\n[quote]Given $x,y,z \\ | \\ x,y,z \\in \\mathbb{R^+} \\ | \\ xyz=1$ minimize:\n$\\mathfrak{1) } \\sum_{cyc}\\frac{1}{x+1}$\n$\\mathfrak{2) } \\sum_{cyc}\\frac{1}{(x+1)(y+1)}$[/quote]\r\n :P\r\n\r\nWell here's how I see it:\r\nWe're most certainly gonna use: $\\prod_{cyc}{(x+1)}=xyz+\\sum_{cyc}{xy}+\\sum_{cyc}x+1=2+\\sum_{cyc}{xy}+\\sum_{cyc}x$\r\n\r\nI'll finish up tomorrow, but I'll give [hide=\"my unproved answers:\"]Just set $x=y=z \\implies x^3=1 \\implies x=y=z=1$\nSo answer to #1: $\\frac32$\n#2: $\\frac34$[/hide]\r\nThose are just my first guesses...", "Solution_2": "for the first one , if $x,y\\rightarrow \\infty$ then $z\\rightarrow 0$ so the whole expression will $\\rightarrow 1$\r\n\r\nfor the second one , we do the same way and the expression $\\rightarrow 0$", "Solution_3": "[hide=\"1\"]\n3-term AM-GM yields\n$\\frac{S}{3} \\ge \\sqrt[3]{\\frac{1}{xyz+xy+yz+zx+x+y+z+1}}$, $S = LHS$\n$\\frac{S}{3} \\ge \\sqrt[3]{\\frac{1}{xy+yz+zx+x+y+z+2}}$\n$S \\ge 3\\sqrt[3]{\\frac{1}{6+2}}$, since $xy+yz+zx+x+y+z \\ge \\sqrt[6]{[xyz]^3} = 6$\n$S \\ge 3 \\cdot \\frac{1}{2} = \\frac{3}{2}$\n[/hide]", "Solution_4": "[quote=\"xxreddevilzxx\"][hide=\"1\"]\n3-term AM-GM yields\n$\\frac{S}{3} \\ge \\sqrt[3]{\\frac{1}{xyz+xy+yz+zx+x+y+z+1}}$, $S = LHS$\n$\\frac{S}{3} \\ge \\sqrt[3]{\\frac{1}{xy+yz+zx+x+y+z+2}}$\n$S \\ge 3\\sqrt[3]{\\frac{1}{6+2}}$, since $xy+yz+zx+x+y+z \\ge \\sqrt[6]{[xyz]^3} = 6$\n$S \\ge 3 \\cdot \\frac{1}{2} = \\frac{3}{2}$\n[/hide][/quote]\r\n\r\nif $xy+yz+xz+x+y+z\\geq 6$ then $\\frac{1}{xy+yz+xz+x+y+z}\\leq \\frac{1}{6}$\r\nso your solution is not correct . Moreover , see my above post which implies minimum at $1$\r\n\r\n ;)", "Solution_5": "According to shyong's post, there is [b]no[/b] minimum because $x,y,z$ are positive real numbers. 1 is a limit point but it's not actually achieved by the inequality (least upper bound)", "Solution_6": "[quote=\"10000th User\"]1 is a limit point but it's not actually achieved by the inequality (least upper bound)[/quote]\r\n\r\nAre you sure you understand what limit point means? There are better phrases you could use there." } { "Tag": [], "Problem": "The numerator of a fraction is 1 less than the denominator. if the numerator and the denominator are both increased by 4 ,the new fraction will be 1/8 more than the original fraction . find the original fraction", "Solution_1": "[hide] Let $ x$ be the denominator of the first fraction. Then, \n\n$ \\frac{x\\minus{}1}{x}\\plus{}\\frac{1}{8}\\equal{}\\frac{x\\plus{}3}{x\\plus{}4}$.\n\nSolving, we have\n\n$ \\frac{x^2\\plus{}3x\\minus{}4\\minus{}x^2\\minus{}3x}{x^2\\plus{}4x}\\equal{}\\minus{}\\frac{1}{8}$\n\n$ \\implies x^2\\plus{}4x\\equal{}32 \\implies (x\\plus{}8)(x\\minus{}4)\\equal{}0$.\n\nUpon inspection, $ x\\equal{}\\minus{}8$ does not work as a solution, so the original fraction is $ \\boxed{\\frac34}$.[/hide]" } { "Tag": [], "Problem": "Two distinct positive integers $ x$ and $ y$ are factors of 36. What is the least product $ xy$ that is not a factor of 36?", "Solution_1": "both 2 and 4 are factors of 36, but 8 is not.\r\n\r\nanswer : 8" } { "Tag": [ "calculus", "derivative", "integration", "calculus computations" ], "Problem": "$ f \\in C^0 $ , differentiate\r\n\\[\r\n\\displaystyle \\int^{\\int^x_0 f(t) \\, dt}_0 f(u) \\, du\r\n\\]", "Solution_1": "The answer is \r\n\r\n\\[f\\left(\\int^x_0 f(t) dt\\right) f(x)\\]." } { "Tag": [], "Problem": "Hello, I have a question that I have help with. \r\n\r\nIt states: If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?", "Solution_1": "[hide]\n$3n, 30 = 5 \\cdot 6$\n$2n, 28 = 4 \\cdot 7$\n$6n, 35 = 5 \\cdot 7$\n[/hide]", "Solution_2": "[hide=\"Full explanation\"]If we can factor $n=2^{a}\\cdot 3^{b}\\cdot 5^{c}\\cdots$, then, by combinatorics, $n$ has $d(n)=(a+1)\\cdot (b+1)\\cdot (c+1)\\cdots$ divisors.\nSince only $2$ and $3$ are important in this problem, let's say\n$d(n)=(a+1)\\cdot (b+1)\\cdot k$\n$d(2n)=(a+2)\\cdot (b+1)\\cdot k$\n$d(3n)=(a+1)\\cdot (b+2)\\cdot k$\n$d(6n)=(a+2)\\cdot (b+2)\\cdot k$\n\nLet's factor $d(2n)=28$ and $d(3n)=30$ in three different factors with all the different possibilities:\n$28=1\\cdot 1\\cdot 28=1\\cdot 2\\cdot 14=\\boxed{1\\cdot 4\\cdot 7}=2\\cdot 2\\cdot 7$\n$30=1\\cdot 1\\cdot 30=1\\cdot 2\\cdot 15=1\\cdot 3\\cdot 10=\\boxed{1\\cdot 5\\cdot 6}=2\\cdot 3\\cdot 5$\n\nI boxed the only way (just a little casework) that we can get\n$d(2n)=(a+2)\\cdot (b+1)\\cdot k=28$\n$d(3n)=(a+1)\\cdot (b+2)\\cdot k=30$\nwith $a=5$, $b=3$, $k=1$ ($n=2^{5}\\cdot 3^{3}=864$).\n\nTherefore, $d(6n)=7\\cdot 5\\cdot 1=35$.[/hide]", "Solution_3": "[quote=\"lordWings\"]\nNotice that there are two possible values of $n$:\n$n=2^{3}\\cdot 3^{5}=1944$\n$n=2^{5}\\cdot 3^{3}=864$[/quote]\r\n\r\nNot quite. For the first value, the numbers of divisors of $2n$ and $3n$ are reversed.", "Solution_4": "You're right. I just edited the mistake so that the reasoning will be clear." } { "Tag": [ "inequalities", "6th edition" ], "Problem": "I think we can just prove:\r\n$(a+b)(b+c)(c+a) \\ge (a+1)(b+1)(c+1)$\r\nam I right?", "Solution_1": "I think so, and this is easy, it's simply\r\n$(a+1)(b+1)(c+1) \\ge 8$ and $(a+a)(a+b)(a+c) \\ge (a+1)^3$", "Solution_2": "[quote=\"zhaobin\"]I think we can just prove:\n$(a+b)(b+c)(c+a) \\ge (a+1)(b+1)(c+1)$\nam I right?[/quote]\r\n\r\nYes, you're right. The problem was easy, although it took me about a day to think about AM-GM :blush: \r\n\r\nValentin, can you tell me now whose problem is this? :?", "Solution_3": "[quote=\"Soarer\"]I think so, and this is easy, it's simply\n$(a+1)(b+1)(c+1) \\ge 8$ and $(a+a)(a+b)(a+c) \\ge (a+1)^3$[/quote]\r\nyour solution are nicer than mine :P", "Solution_4": "I didn't understand what Soarer said in his post... Not thinking clearly now. can you develop?", "Solution_5": "ok.my pleasure\r\n$(a+a)(a+b)(a+c) \\ge (a+\\sqrt[3]{abc})^3=(a+1)^3$ the first one is holder.\r\nwrite any two similar.then we have \r\n$2a2b2c(a+b)^2(a+c)^2(b+c)^2 \\ge (a+1)^3(b+1)^3(c+1)^3 \\ge 8 (a+1)^2(b+1)^2(c+1)^2$\r\nwhich is equal to $(a+b)(b+c)(c+a) \\ge (a+1)(b+1)(c+1)$", "Solution_6": "Actually, the inequality holds for more general constraint\r\n$ab+bc+ca \\geq 3$.", "Solution_7": "I didn't think, that after using AM-GM inequality our ineq would true, and I didn't use it.", "Solution_8": "[quote=\"Vasc\"]Actually, the inequality holds for more general constraint\n$ab+bc+ca \\geq 3$.[/quote]\r\nyes.with it we can get\r\n$\\frac 1 3(a+b+c)(ab+bc+ca) \\ge ab+bc+ca$\r\nand $\\frac 1 3(a+b+c)(ab+bc+ca) \\ge a+b+c$\r\nso \r\n$\\sum_{syms}a^2b \\ge \\frac 2 3(a+b+c)(ab+bc+ca) \\ge ab+bc+ca+a+b+c$\r\nwhich is equal to what I wrote :)", "Solution_9": "[quote=\"zhaobin\"] so \n$\\sum_{syms}a^2b \\ge \\frac 2 3(a+b+c)(ab+bc+ca) \\ge ab+bc+ca+a+b+c$\nwhich is equal to what I wrote :)[/quote]But this result doesn't imply the inequality \r\n$(a+b)(b+c)(c+a) \\geq (a+1)(b+1)(c+1)$.", "Solution_10": "OK\uff0cit is wrong. :) \r\nbut another one:\r\nI think we can wlog $ab+bc+ca=3$ if $ab+bc+ca=3k^2(k>1)$,then we can multiply $\\frac 1 {k^3}$ in both side,let $x=\\frac a {k}$,then we need to prove $(x+y)(y+z)(z+x) \\ge (x+\\frac 1 k)(y+\\frac 1 k)(z+\\frac 1 k)$ $xy+yz+zx=3$ which is weaker than the case $ab+bc+ca=3$.\r\n$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$\r\nso we need to prove:\r\n$(a+b+c)(ab+bc+ca) \\ge a+b+c+ab+bc+ca+1+2abc$\r\nequal to \r\n$(a+b+c-1)(ab+bc+ca-1) \\ge 2+2abc$\r\nwith $ab+bc+ca-1 = 2$so we need to prove \r\n$2(a+b+c) \\ge 4+2abc$\r\nwhich is true(because $a+b+c \\ge 3$,$abc \\le 1$)\r\nhope it is right this time :)", "Solution_11": "OK. It is right now. :lol:" } { "Tag": [ "trigonometry" ], "Problem": "$ \\frac{1}{sin^{10}x}\\plus{}\\frac{1}{cos^{10}x}\\equal{}64$", "Solution_1": "[quote=\"thanhnam2902\"]$ \\frac {1}{sin^{10}x} + \\frac {1}{cos^{10}x} = 64$[/quote]\r\n\r\nLet $ sin^2xcos^2x = t \\in\\ [0;\\frac{1}{4}]$\r\n\r\n$ \\frac {1}{sin^{10}x} + \\frac {1}{cos^{10}x} = 64$\r\n\r\n$ \\leftrightarrow (sin^2x+cos^2x)\n(sin^8x+cos^8x)-sin^2xcos^2(cos^6x+sin^6x)=64sin^{10}xcos^{10}x$\r\n\r\n$ \\leftrightarrow 64t^5-5t^2+5t-1=0$ :wink:", "Solution_2": "$ \\frac{sin^{10}x\\plus{}cos^{10}x}{4}\\equal{}\\frac{sin^6x\\plus{}cos^6x}{4cos^22x\\plus{}sin^22x}$", "Solution_3": "hello, your equation is equivalent with\r\n$ {\\displaystyle (1915+76\\cos(4x)+68\\cos(8x)-12\\cos(12x)+\\cos(16x))\\csc(x)^{10}\\sec(x)^{10}\\sin\\left(\\frac{\\pi}{4}-x^\\right)^2\\sin\\left(\\frac{\\pi}{4}+x\\right)^2=0}$\r\nSonnhard.", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, your equation is equivalent with\n$ {\\displaystyle (1915 + 76\\cos(4x) + 68\\cos(8x) - 12\\cos(12x) + \\cos(16x))\\csc(x)^{10}\\sec(x)^{10}\\sin\\left(\\frac {\\pi}{4} - x^\\right)^2\\sin\\left(\\frac {\\pi}{4} + x\\right)^2 = 0}$\nSonnhard.[/quote]\r\nYes! I see, thank you. :P" } { "Tag": [ "probability", "blogs" ], "Problem": "Natural languages have a huge amount of redundancy and words are often unnecessarily long. These both limit the efficiency in which people may communicate, but help to prevent ambiguity and confusion.\r\n\r\nMany have tried to construct languages or coding schemes in which it is faster to communicate. For example\r\n[url]http://en.wikipedia.org/wiki/Speedtalk[/url]\r\n[url]http://en.wikipedia.org/wiki/Arahau[/url]\r\n[url]http://en.wikipedia.org/wiki/Ithkuil[/url]\r\n[url]http://www.iiap.res.in/personnel/srik/Lin.html[/url]\r\n\r\nI have read somewhere, though I cannot remember, that these attempts of efficiency are doomed to fail. Is it possible to communicate at a significantly higher rate than natural languages but still retain the same accuracy and low probability of error?\r\n\r\nFor that matter, do there exist natural languages that differ significantly (i.e. two times faster) in the speed of communication? It seems to me that languages with fewer phonemes (major ones being Japanese, Spanish) have fewer options for word construction, and thus must have longer words, but compensate through a high speed of speech. This is possible because the speaker doesn't have enunciate so much and distinguish each phoneme clearly from the next, since there are fewer of them.", "Solution_1": "Here are some Language Log posts that are relevant to your question. BTW, Language Log is a collaborative linguistics blog run by a number of prominent linguists. It's both wonderfully entertaining and a good jumping off point for research on wide-ranging linguistics questions like yours.\r\n\r\nThis one, \"Speech rate and per-syllable information across languages,\" looks right on target for what you were asking:\r\n[url]http://languagelog.ldc.upenn.edu/nll/?p=22[/url]\r\n\r\nThis set of posts, about comparative efficiency of different languages measured in terms of text size, might also be interesting: [url]http://languagelog.ldc.upenn.edu/nll/?p=93[/url]" } { "Tag": [ "logarithms", "inequalities", "function", "calculus", "calculus computations" ], "Problem": "Prove the following inequality for $00$, then $ \\prod(\\sqrt{a}\\plus{}\\sqrt[3]{a}\\plus{}\\sqrt[6]{a})\\leq\\prod(b\\plus{}2)$", "Solution_1": "[quote=\"lipako\"]If $ a,b,c > 0$, then $ \\prod(\\sqrt {a} \\plus{} \\sqrt [3]{a} \\plus{} \\sqrt [6]{a})\\leq\\prod(b \\plus{} 2)$[/quote]\r\n\r\n\r\n$ a\\equal{}x^6,b\\equal{}y^6,c\\equal{}z^6$($ x,y,z>0$)\r\n\r\n\r\n$ x^6\\plus{}2\\minus{}(x^3\\plus{}x^2\\plus{}x)\\equal{}(x^4\\plus{}2x^3\\plus{}3x^2\\plus{}3x\\plus{}2)(x\\minus{}1)^2\\geq 0$\r\n\r\n\r\nso..............", "Solution_2": "We also have $ \\prod_{cyc}(\\sqrt{a}\\plus{}\\sqrt[3]{b}\\plus{}\\sqrt[6]{c})\\leq\\prod(a\\plus{}2)$:)", "Solution_3": "[quote=\"bakerbakura\"]We also have $ \\prod_{cyc}(\\sqrt {a} \\plus{} \\sqrt [3]{b} \\plus{} \\sqrt [6]{c})\\leq\\prod(a \\plus{} 2)$:)[/quote]\r\n\r\nBy Holder inequality we have \r\n\r\n$ (a^6\\plus{}1\\plus{}1)(a^6\\plus{}1\\plus{}1)(a^6\\plus{}1\\plus{}1)(1\\plus{}b^6\\plus{}1)(1\\plus{}b^6\\plus{}1)(1\\plus{}1\\plus{}c^6) \\ge (a^3\\plus{}b^2\\plus{}c)^6$\r\n\r\n$ (b^6\\plus{}1\\plus{}1)(b^6\\plus{}1\\plus{}1)(b^6\\plus{}1\\plus{}1)(1\\plus{}c^6\\plus{}1)(1\\plus{}c^6\\plus{}1)(1\\plus{}1\\plus{}a^6) \\ge (b^3\\plus{}c^2\\plus{}a)^6$\r\n\r\n$ (c^6\\plus{}1\\plus{}1)(c^6\\plus{}1\\plus{}1)(c^6\\plus{}1\\plus{}1)(1\\plus{}a^6\\plus{}1)(1\\plus{}a^6\\plus{}1)(1\\plus{}1\\plus{}b^6) \\ge (c^3\\plus{}a^2\\plus{}b)^6$\r\n\r\nMultiplynig the three inequalities we obtain the desired result" } { "Tag": [ "ratio", "geometry", "invariant", "analytic geometry", "trigonometry", "geometry theorems" ], "Problem": "Could someone post good proofs for these theorems?", "Solution_1": "[url]http://www.google.com[/url]", "Solution_2": "Ok... this might help more. Hem Hem. :lol: \r\n\r\n[url=http://planetmath.org/encyclopedia/ProofOfCevasTheorem.html]Proof of Ceva's[/url]", "Solution_3": "the simplest proof for Ceva's theorem is to\r\nexpress the product in terms of the \r\nratios of the areas of the 6 pieces\r\ninto which the lines divide the triangle.\r\nThis is in a sense the \"correct\" proof, as the\r\nstatement of the theorem is affine (i.e. \r\ninvariant under linear transformations,\r\nand so dependent only on ratios of areas).\r\n\r\nYou can prove menelaus theorem in the same\r\nway. \r\n\r\nbarycentric coordinates have a lot to do with\r\nthis, as well.", "Solution_4": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14581]www.mathlinks.ro/Forum/viewtopic.php?t=14581[/url]", "Solution_5": "i have a very nice solution by using inversive geometry.but it may take time to write it.i will post it as soon as \r\n\r\npossible ;)", "Solution_6": "there is another version of ceva using trigonometry. Every cevian dives each angle of the $\\triangle{ABC}$ in two Angles. \r\n$A$ into $\\angle {a_1}$ and $\\angle {a_2}$\r\n$B$ into $\\angle {b_1}$ and $\\angle {b_2}$ \r\n$C$ into $\\angle {c_1}$ and $\\angle {c_2}$ \r\n\r\nThen we have that the 3 cevians are concurrent if anf only if:\r\n$\\frac{sin a_1}{sin a_2} \\frac{sin b_1}{sin b_2} \\frac{sin c_1}{sin c_2}=1$\r\n you can prove using the sine law.", "Solution_7": "Hello\r\nPaul A. Clement, found in The American\r\nMathematical Monthly, v. 65, num.8, October, 1958 the following Theorem.\r\n\r\n\r\n Given D on BC, E on CA, and F on AB (possibly extended sides) of\r\ntriangle ABC the perpendiculars to the sides at D, E, and F are concurrent\r\niff\r\n\r\n (AF)^2 + (BD)^2 + (CE)^2 = (FB)^2 + (DC)^2 + (EA)^2.\r\n\r\nAnd the proof???", "Solution_8": "[quote] Given D on BC, E on CA, and F on AB (possibly extended sides) of\ntriangle ABC the perpendiculars to the sides at D, E, and F are concurrent\niff $(AF)^2 + (BD)^2 + (CE)^2 = (FB)^2 + (DC)^2 + (EA)^2$.[/quote]\r\nThis was known long before 1958. It follows from the Pythagoras theorem applied to the 6 right triangles with vertex at the point of concurrency." } { "Tag": [], "Problem": "There are exactly $4n+2$ paths in space, and there are $2n^{2}+6n+1$ intersections of exactly two of them. What is the maximum number of intersections of three paths?\r\n\r\nI failed the SMT :( .", "Solution_1": "hard.... :o", "Solution_2": "Well it was super hard for a stupid guy like me, but you can probably solve it in under a minute :( .", "Solution_3": "is the answer 6n^2 by any chance?", "Solution_4": "'Paths' seems sort of vague but maybe...\r\n\r\n${{{4n+2}\\choose 2}= 8n^{2}+6n+1}$ Ways that two paths intersect. Since exactly two intersect at $2n^{2}+6n+1$ points, there's still $6n^{2}$ pairs left. However, each intersection of three, say A,B, and C, includes 3C2 = 3 pairs (AB, BC, AC). So that would be $2n^{2}$ intersections to cover all the pairs." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "let us choose arbitrarily n vertices of a regular 2n-gon \r\nand colour them red. the remaining vertices colour blue.\r\nnext arrange all red-red distances into a non-decreasing sequence and do the same with blue-blue distances.\r\nprove that the sequences are equal.", "Solution_1": "Let $ A$ be the set of red vertices, $ B$ be the set of blue vertices. Label the vertices $ v_1 , v_2 ... v{}_2{}_n$ going clockwise around the polygon.\r\n\r\nLet $ E_k = \\{ (v_i, v_i_ + _k) | v_i, v_i_ + _k \\in A \\}$\r\nLet $ F_k = \\{ (v_i, v_i_ + _k) | v_i, v_i_ + _k \\in B \\}$\r\n\r\nSo $ E_k$ is the set of red-red edges where the vertices are k steps around the polygon from each other and $ F_k$ is the set of blue-blue edges where the vertices are k steps around the polygon from one another.\r\nSo the problem is equivalent to showing that:\r\n\r\n$ | E_k | = | F_k |$ for any k\r\n\r\nDefine $ S_k = \\{ (v_i, v_i_ + _k) | v_i \\in A , v_i_ + _k \\in B \\}$\r\n\r\nClaim that $ | E_k | = n - | S_k |$ \r\n\r\nThis is because for each red vertex, the vertex k steps clockwise round the polygon must be either red or blue, so $ | E_k | + | S_k | = |R| = n$\r\n\r\nBut then similarly $ | F_k | = n - | S_k |$ \r\n(this time we look at the vertex k steps [b]anticlockwise[/b] round the polygon from each blue vertex)\r\n\r\nSo we are done. I made a bit of a meal out of this, it's pretty simple really." } { "Tag": [], "Problem": "in the city of medlers there is $100$ medlers.one day we told each medlers one different news.(none of news are same).\r\n$A$ can tell anything that he know to $B$ with calling him.in each call only one of the medlers can tell the other everything that he knows.\r\nwith how many phone calls at least all of them will know all of the news?", "Solution_1": "i thinks this problem is nice and i think the answer is $198$ or $197$.if anybody has any idea plz send. :alien:" } { "Tag": [], "Problem": "A pulley that has a diameter of 8 inches is belted to a pulley that has a diameter of 12 inches. The 8-inch-diameter pulley is running at1,548 revolutions per minute. If the speeds of the pulleys vary inverselyto their diameters, how many revolutions per minute does the larger pulley make?", "Solution_1": "since they are inversely proportional, \r\n8*1548 = k = 12 * r\r\nr = 1032", "Solution_2": "Great equation." } { "Tag": [ "function", "algebra", "floor function", "complex analysis", "complex analysis unsolved" ], "Problem": "Does exist a function $ f$ which is analytic on $ \\mathbb{D}$- the unit disc such that : $ f(\\frac{1}{n})\\equal{}\\frac{1}{[\\sqrt{n}]}$ ? where [x] is the floor function of $ x$", "Solution_1": "Just think of what $ f(0)$ and $ f'(0)$ should be in this case." } { "Tag": [], "Problem": "I'm trying to solve this nice equality :\r\n\r\nC(n, 0)^2 + C(n, 1)^2 + C(n, 2)^2 + ... + C(n, n)^2 = C(2n, n)\r\n\r\nwhere :\r\n\r\n n!\r\nC(n, k) = ___________\r\n k! (n - k)!\r\n\r\n\r\nCan anybody please give me a hint...?", "Solution_1": "This is actually a very famous formula, and it probably should be in Intermediate, not Advanced. Anyhow, I don't want to give you too big a hint, but you're going to need to use a counting argument somewhere -- you know, show that the two sides each count (in different ways) the number of ways to do something, and since the number of ways of doing that thing is fixed, you will show that the two sides are equal.", "Solution_2": "[hide]Hints :\n\n\n\ndon't forget that C(n,k) = C(n,n-k) \n\n\n\na) combinatorics way :\n\n\n\ncount (in two different ways) the number of ways to choose n persons from a group of n men and n women.\n\n\n\n\n\nb) algebraic way :\n\n\n\nlook from coefficient of x^n in (1+x)^2n = (1+x)^n*(1+x)^n[/hide]", "Solution_3": "It's easier to prove (try a conbinatorial way) a more general formula\r\nC(m+n,k)= C(m,0)*C(n,k) + C(m,1)*C(n,k-1)+...+ C(m,k-1)*C(n,1)+C(m,k)*C(n,0).\r\nIt's Valdermode formula if i'm not mistaken.\r\nThen for m=n=k and C(n,k)=C(n,n-k). You get result.", "Solution_4": "The name of the formula is Vandermonde's Equation. It's not particularly easier or harder to prove the generalization.", "Solution_5": "[hide]\n\n\n\nConsider 2n objects, of which we want to choose n.\n\n\n\nThen we can choose them in C(2n,n) ways:\n\n\n\nWe also can choose them in:\n\n\n\nC(n,0)*C(n,n)+C(n,1)*C(n,n-1)+.......+C(n,n)*C(n,0) ways\n\n\n\nBut since C(n,k) = C(n,n-k), we have \n\n\n\nC(n,0)*C(n,0)+C(n,1)*C(n,1)+.......+C(n,n)*C(n,n)\n\n= C(n,0)^2+C(n,1)^2+.....+C(n,n)^2 ways\n\n\n\nAnd we are done.\n\n[/hide]", "Solution_6": "Thanks guys, but I think that i mislead something here.\r\n\r\nWell the idea is that by observing :\r\n\r\n(1 + x)^a * (1 + x)^b = (1 + x)^(a+b) \r\n\r\n\r\nto prove the equality......\r\n\r\nbut thanks anyway, now i got a lot of information that i didn't know before.\r\n\r\nOn the other hand. Why are you telling me that that problem belong to Interm., not to Advance, if that kind of problems belong to Discret Mathematic?", "Solution_7": "[quote=\"Andy\"]Well the idea is that by observing :\n\n(1 + x)^a * (1 + x)^b = (1 + x)^(a+b) \n\nto prove the equality......[/quote]\r\n\r\nTry expanding out each of the three terms and see.\r\n\r\n\r\n\r\n\r\nJust because something has a fancy name like \"Discrete Mathematics\" doesn't make it difficult -- this question is a fairly standard combinatorics result, and the Intermediate forum has lots of combinatorics on it. I mean, it could go either place -- I just learned it a long time ago, so I think it's Intermediate :)", "Solution_8": "I trust Joel's judgment, so off it goes to Intermediate.", "Solution_9": "Isn't this one of the first things stated in AoPS Vol. 2 Combinatorics Chapter" } { "Tag": [ "calculus", "analytic geometry", "graphing lines", "slope", "ratio" ], "Problem": "Could someone help me with this problem:\r\n\r\nFind an equation for the family of lines tangent to the circle with center at the origin and radius 3.", "Solution_1": "I cheated and used calculus, so I'll just post my answer. Hopefully someone can find a nice geometrical or algebraic solution.\r\n\r\n[hide=\"general equation\"]On the $xy\\text{-plane}$ the general equation I think would be \\[\\sqrt{r^2-x_0^2}\\left(y\\pm\\sqrt{r^2-x_0^2}\\right)=\\pm{x}_0(x-x_0).\\]\n\n$r$ is a positive real number, $3$ in this case. Also, $0\\le{x}_0\\le{r}$.[/hide]", "Solution_2": "I have a solution that is in the form [tex]y = mx + b[/tex] which is the slope-intercept form of the equation of a straight line.\r\n\r\n[hide]\n\nFirst, draw the circle given in the problem and then draw a line tangent to the circle to represent the general case of a line tangent to a circle of radius 3. Draw the tangent so it will intercept the y-axis at [tex]b[/tex].\n\nNext draw a line from the origin or center of the circle to the point where the tangent touches the circle. This line will be normal or perpendicular to the tangent. We will call this point [tex](x,y)[/tex].\n\nNow draw a line from [tex](x,y)[/tex] perpendicularly down to the x-axis.\n\nThis will form two similar right triangles.\n\nthe larger triangle will have its hypotenuse along the y-axis with a length of [tex]b[/tex] and the line perpendicular to the tangent from the origin will form one of the legs of that right triangle. The length of that leg is [tex]3[/tex].\n\nThe smaller triangle's hypotenuse is the same line that forms the given leg on the larger triangle, which is [tex]3[/tex]. It's legs are lengths [tex]x[/tex] and [tex]y[/tex].\n\nWe can now find the other leg of the larger triangle in terms of [tex]b[/tex].\n\nUsing Pythagorean's Theorem, the other leg is [tex]\\sqrt {b^2 - 9}[/tex]\n\nWe can now set up ratios between the larger triangle and the smaller triangle.\n\n[tex] \\frac {b}{3} = \\frac {3}{y}[/tex] so [tex]y = \\frac {9}{b}[/tex]; also [tex]\\frac {b}{\\sqrt {b^2 - 9}} = \\frac {3}{x}[/tex] so [tex] x = \\frac {3 \\sqrt {b^2 - 9}}{b}[/tex]\n\nThe next step is to find [tex]m[/tex] or the slope in the formula in terms of [tex]b[/tex]\n\nSubstituting into the formula:\n\n[tex]\\frac {9}{b} = m(\\frac {3 \\sqrt {b^2 - 9}}{b}) + b[/tex]\n\n[tex]m = \\frac {9 - b^2}{3 \\sqrt {b^2 - 9}}[/tex]\n\nThe general equation would then be [tex]y = \\frac {9 - b^2}{3 \\sqrt {b^2 - 9}}x + b[/tex]\n\nThis is the solution you would have if you were to make it in terms of the y-intercept or [tex]b[/tex].\n\nThere may be other ways to represent a general equation.\n\nThis is for values [tex] 3 \\leq \\ b \\ < \\ \\infty[/tex] and [tex] -3 \\geq \\ b \\ > \\ - \\ \\infty[/tex]\n\n[/hide]", "Solution_3": "[hide]If you could find the x- and y-intercepts, you could say $b|x|+a|y|=ab$, where $a,b$ are the x- and y-intercepts, respectively. But I'm too lazy. :) [/hide]" } { "Tag": [ "AMC", "AIME", "\\/closed" ], "Problem": "Please do not discuss the AIME solutions or any \"cheating\" on this post.\r\n\r\n=========================================\r\n\r\nMr. Dunbar, are you saying that we cannot discuss it at all, including \"I think I did well\" and \"That problem was hard\"?", "Solution_1": "I am taking him to mean \"Do not discuss it at all.\"" } { "Tag": [], "Problem": "1) Prove $ \\displaystyle{\\frac{a^3\\plus{}a}{a^4\\plus{}3a^2\\plus{}1}, a\\in \\mathbb{N}}$ is fraction in its lowerst terms\r\n\r\n2)Find $ a,b$ coprime such that $ \\displaystyle{\\frac{a\\plus{}b}{a^2\\minus{}ab\\plus{}b^2}\\equal{}\\frac{7}{13}}$", "Solution_1": "[u][b]Solution for number 1[/b][/u]:\r\n\r\nIt suffices to show that $ gcd(a^3\\plus{}a,a^4\\plus{}3a^2\\plus{}1)\\equal{}1$.\r\n\r\nIn the contraty,let's suppose that there is some $ d>1$ such as:\r\n\r\n $ d|a^3\\plus{}a$ and $ d|a^4\\plus{}3a^2\\plus{}1$.\r\n\r\nThen $ d$ also divided the difference: $ d|a^4\\plus{}3a^2\\plus{}1\\minus{}a(a^3\\plus{}a)\\equal{}2a^2\\plus{}1$.\r\n\r\nMoreover $ d$ also divides: $ d|2(a^3\\plus{}a)\\minus{}a(2a^2\\plus{}1)\\equal{}a$.\r\n\r\nFinally $ d|(2a^2\\plus{}1)\\minus{}2a^2\\equal{}1$,therefore d=1 and our proof is complete.", "Solution_2": "Let me try the Number 2 question\r\nI assume that they are positive(is it??).\r\n\r\nExpression equivalent to $ 13(a \\plus{} b) \\plus{} 21ab \\equal{} 7(a \\plus{} b)^2$. In $ (\\bmod (a \\plus{} b))$ we get that $ 21ab \\equiv 0(\\bmod (a \\plus{} b))$ or $ 21ab \\equal{} k(a \\plus{} b)$ or $ {{21ab} \\over {(a \\plus{} b)}} \\equal{} k$ for positive $ k$. Since $ ab$ is coprime with $ (a \\plus{} b)$ then we got only $ 21$ divisible by $ (a\\plus{}b)$, which means $ a\\plus{}b\\equal{}3$ which has not solutions and $ a\\plus{}b\\equal{}7$ for which we get 2 solutions set $ (a,b) \\equal{} \\left\\{ {(3,4),(4,3)} \\right\\}$.\r\n\r\nSorry if I mistaken.\r\nI hope this solution is right... :blush:" } { "Tag": [ "geometry", "parallelogram", "rhombus", "trigonometry", "rectangle" ], "Problem": "The area of a parallelogram is 28 and the length of the largest diagonal is 14. Find the length of the smallest diagonal.\r\n\r\nI think this problem has solution only if we consider that parallelogram is a rhombus. I'm wrong?", "Solution_1": "Um, there are numerous possible side lengths for the parallelogram. The smallest occurs when the small diagonal is length 4 and we have a rhombus. \r\n\r\nIs that what your question was?", "Solution_2": "[quote=\"4everwise\"]Um, there are numerous possible side lengths for the parallelogram. The smallest occurs when the small diagonal is length 4 and we have a rhombus. \n\nIs that what your question was?[/quote]\r\n\r\nI think i used the wrong word. \r\n\r\nTrying again: If $D$ and $d$ are the diagonals of that parallelogram and $D=14$ find $d$ if the area of the parallelogram is 28.\r\n\r\nSo, i guess the author of this question supposed that parallelogram is a rhombus, but that is not necessarily one, right?", "Solution_3": "[hide]The diagonals of the parallelogram bisect each other so it can be split into 4 triangles with side lengths 7, d/2, and some other length. It $\\theta$ denotes one of the angles of the intersection of the diagonals, then the other three angles have measures $180^\\circ -\\theta$, $\\theta$, and $180^\\circ -\\theta$. The total area will be $7d\\sin \\theta=28\\Rightarrow d=4\\csc \\theta$.\n\nMaybe the problem wanted the minimum value of d (which will occur at $\\theta=90^\\circ$, which would make the parallelogram a rhombus), which is 4.[/hide]", "Solution_4": "Thank you scorpius. Thats what i thought. And no, the problem don't want the minimum value of d.", "Solution_5": "Also, can't you write the area of a parellagram as the product of the diagonals over 2\r\ntherefore\r\n$\\frac{14*d}{2}=28$\r\nso\r\n$d=\\boxed{4}$", "Solution_6": "That formula is only for a rhombus, I think.", "Solution_7": "Well, wouldn't the formula always work for any parallelagram?\r\nIf anyone could clarify, that would be great. :)", "Solution_8": "[quote=\"AMPIStiffler\"]Well, wouldn't the formula always work for any parallelagram?\nIf anyone could clarify, that would be great. :)[/quote]\r\n\r\nNo that's not true, it's only for a rhombus (that's how the formula is derived). And if you want to see proof, think about a rectangle with sides $3$ and $4$ ;)", "Solution_9": "Generally, for any quadrilateral, we have $(\\text{Area})=\\frac{1}{2}d_1d_2\\sin{\\theta}$, where $d_1$ and $d_2$ are the diagonals and $\\theta$ is the angle formed by the two diagonals.\r\n\r\nIn a rhombus, $\\theta=90^\\circ$, so $\\sin{\\theta}$ becomes $1$ and thus the formula becomes $\\frac{1}{2}d_1d_2$.\r\n\r\nEdit: scorpius119 has already posted the same idea :P", "Solution_10": "[quote=\"frt\"]Generally, for any quadrilateral, we have $(\\text{Area})=\\frac{1}{2}d_1d_2\\sin{\\theta}$, where $d_1$ and $d_2$ are the diagonals and $\\theta$ is the angle formed by the two diagonals.\n\nIn a rhombus, $\\theta=90^\\circ$, so $\\sin{\\theta}$ becomes $1$ and thus the formula becomes $\\frac{1}{2}d_1d_2$.[/quote]I was just thinking about using that in a problem! Too bad it didn't turn out as well as I wanted, hehe. :blush:" } { "Tag": [], "Problem": "We are given a circle $ \\Omega $, a chord $ AB $, the midpoint $ M $ of $ AB $ and a chord $ CMD $ through $ M $. If the tangents to the circle at $ C $and $ D $ interest line $ AB $ at $ E $ and $ F $ , prove that $AE = BF$.\r\n\r\n[i]This is a not a new problem .It is very old and very known, but it is a good introductory problem for Olympiads and especially for pupils 13-15 years old. I post it for our new members and visitors. Enjoy it. [/i]\r\n\r\nBabis", "Solution_1": "Pretty cute. Note that F, M, O, D are concyclic as are E, M, O, C. Then $\\angle MEO = \\angle MCO = \\angle MDO = \\angle MFO$ so OEF is isosceles. Then the result follows.", "Solution_2": "Let me tell you : Congratulations!\r\n\r\nI think that this is the best (probably the unique) solution.\r\n\r\n Babis", "Solution_3": "I didnt understand,please explain,a figure would help.", "Solution_4": "Please tell me : Do you know the theory of inscriptible quadrilaterals ?If no , do not try any more. If yes , then O is the center of the circle , so , as in the hint(post 2) , two quadrilaterals are inscriptible. Play with the angles ...\r\n\r\n If you have problem , write again. I' ll do something for you.\r\n\r\nBabis", "Solution_5": "[quote=\"stergiu\"]Please tell me : Do you know the theory of inscriptible quadrilaterals?[/quote]\r\nNo. :? \r\nCan you please explain?", "Solution_6": "Stergiu is talking about inscribed quadrilaterals, more commonly known as cyclic quadrilaterals. A cyclic quadirlateral is a quadrilateral for which there exists a circle that goes through all four vertices. Cyclic quads have some special properties that other quads do not have. For instance, ABCD is cyclic iff angle CBD equals angle ACB. Cyclic quads can often be very useful is solving vertain types of problems, as probability1.01 demonstrates. Try to draw the diagram yourself and follow the proof.", "Solution_7": "There is somewhere a topic with the same problem , but in a little different formulation. I think it was << North China - 2005 >>. You can see there 3 nice proofs, not all new.\r\n\r\n Babis", "Solution_8": "I suppose $A\\in (ME),\\ B\\in (MF)$ and I note $L\\in CE\\cap DF$. From the Menelaus's theorem applied\r\n\r\nto the transversal $\\overline {DMC}$ in the triangle $LEF$ we obtain the relation $\\frac {DF}{DL}\\cdot \\frac {CL}{CE}\\cdot \\frac {ME}{MF}=1$. But $LC=LD$.\r\n\r\nThus, $\\frac {ME}{MF}=\\frac {CE}{DF}$. But $EC^2=EA\\cdot EB,\\ FD^2=FA\\cdot FB$. Therefore, $\\left(\\frac {ME}{MF}\\right)^2=\\frac {EA\\cdot EB}{FB\\cdot FA}\\ \\ (1)$.\r\n\r\nI note $EA=y,\\ MA=MB=x,\\ BF=z$. The relation (1) becomes\r\n\r\n$(x+y)^2z(z+2x)=(x+z)^2y(y+2x)\\Longrightarrow$\r\n\r\n$[2(x^2+yz)+x(y+z)](y-z)=0\\Longrightarrow y=z\\Longrightarrow AE=BF.$", "Solution_9": "Levi thanks!\r\n You taught me that in this forum I must never say << ... probably the unique solution ! ) >> ;)", "Solution_10": "Here's a diagram (made in KSEG)\r\n\r\n[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=274[/img]" } { "Tag": [], "Problem": "Hi, i'm sorry if there is a previous thread similar to this one, but i couldn't find one, so i thought I'd ask. So: taking the SAT, I did very well on math (800) and a 630 on verbal, which placed me in the 80-something percentile. I'm thinking of taking the SAT again because there is such a huge gap, between my scores. However, English is my second language (i only live in the U.S. since 8th grade), so I was wondering if 630 was a good enough score and what you think. \r\n\r\nursina", "Solution_1": "630 is more than fine. It also matters what grade you're in and how competetive the schools you're looking at are. But an 800 and 630 (as an immigrant) will get you in virtually anywhere.", "Solution_2": "I didn't catch how old you are in your message, and of course that makes a difference, but it is clear that you are above eighth grade. I'm wondering from how you signed your message if you are a girl. If you intend to major in a field such as math or engineering that has traditionally had more male applicants than female applicants, being a girl should be an admissions advantage in itself. \r\n\r\nAn 800 math with a 630 verbal is indeed already quite a good score for a first-generation immigrant. I think if you indicate what kind of reading you are doing and what else you are doing to make an effort to improve your English, admissions officers will take that into account when evaluating your scores. Certainly you are writing better English than many participants (who possibly are younger than you) who post on this forum, so you are making a good start in your learning of English. I think if you keep up what you are doing, and especially if you work hard at something that increases your English ability in your school course work or your extracurricular activities (for example, debate or speech), you should have little to worry about when applying for college. \r\n\r\nGood luck.", "Solution_3": "I might recommend that you take the TOEFL or equivalent exam. That's the \"out\" for people for whom English is a second language.", "Solution_4": "Ok, to clarify, yes i am a girl and i'm a junior in high school. I'm in Honors English and was even offered to take the AP English language this year (I reclined, deciding to focus on my other three APs). I'm taking the SAT II writing in june. I am also on the model UN team, which involves a lot of speaking. \r\nI'm thinking of studying business or economics.", "Solution_5": "You would have a particularly good-looking application for a business program with your previous international experience, and there are certainly many strong economics programs that would be glad to get a young woman with the scores you have mentioned. It sounds like you are making good preparations already." } { "Tag": [ "probability", "number theory", "\\/closed" ], "Problem": "To DPatrick: \r\n\r\nIn the \"are you ready\" test for Intermediate Counting and Probability, it mentions that you have a book recommendation for sigma and sets. I wanted to ask you what that is. \r\n\r\nAlso, I really liked having the textbook to prepare for class in Introduction to Counting and Probability. I'm wondering how I'll be able to prepare for class without a textbook for Intermediate. \r\n\r\nI have really enjoyed Intro. Thanks!", "Solution_1": "The whole \"having a book to supplement the class\" is actually very new. AoPS classes have gone on for a long time, and they are extremely enjoyable and worthwhile without a book. I think it would be a very bad idea not to take a class for that reason.", "Solution_2": "The bit about \"sigma and sets\" is just a mild warning that the level of algebra in the Intermediate Course is a bit more sophisticated than that of the Introduction course.\r\n\r\nIn AoPS Volume 1, you can read Chapter 24 for sigma notation and Chapter 27 for the basics of set theory. This will also be covered in our forthcoming Introduction to Algebra book, which will be available later this year.\r\n\r\nThe Intermediate Counting textbook should be available in late 2006 or early 2007, and the online class will likely be offered Spring 2007 using the new book. I agree that the classes are enhanced with the book. If you prefer that style of class, you might consider taking Intro Number Theory this summer instead (which should have its book available later this spring) and deferring Intermediate Counting until next spring. Or you could take the Intermediate Counting class now and follow it up by reading the book in the spring.", "Solution_3": "I took intermediate c&p last fall, and it's great without a book. I don't think you should wait to take it just because the book isn't out yet.", "Solution_4": "Though I might suggest trying to familiarize yourself with the topics a little beforehand... it really helps you out alot. I personally wish I had haha. \r\n\r\nSame applies for pretty much any class. If you know what the topic is beforehand.. its the best thing you can do to help yourself... learn some of the material before class starts and get a head start. \r\n\r\nI personally didn't do that well in Intermediate Counting, but I did wind up learning alot more than I thought I had, just by looking at the transcripts any time I was sorta fuzzy in my understanding on a topic." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "MATHCOUNTS", "AIME", "ARML", "USAMTS" ], "Problem": "[size=150]Hey everybody! [/size]Okay, so here's my problem. I enjoy doing math competitions, but I dont really have time to do all of them. So I was wondering what your guys's favorite math competitions were and why... because if I can only do one, it would be nice to do the best :) thanks!\r\n~[size=117]becky[/size]", "Solution_1": "What type of competition do you mean? In-school, mathmeets, AMC sort of stuff, etc.\r\n\r\nPersonally, I (duh) like the Mandelbrot for in-school, I like Charleston for mathmeets (that doesn't mean much, though, because I only go to two), and I like the USAMO best of all national competitions, even though I only got like a 2 this year.", "Solution_2": "Well, my favorite competition was MATHCOUNTS, that is before I got to 8th grade and got killed at nationals. Now, I look forward to next year and I am looking to do AMCs, Mandelbrot, and ARML. 8-)", "Solution_3": "I'd have to say MathCounts, since I won it, but my second favorite is the AIME (Didn't make USAMO this year)", "Solution_4": "good job winning nationals. great comeback against andrew chien.", "Solution_5": "Thanks. Were you there? I still can't believe it...", "Solution_6": "nope. the top 4 CA people were too good... thats y. AND i missed the first sprint question... sad", "Solution_7": "My favorite's Mathcounts(excluding the countdown round - totally stink at it, but I'm working on it). Then, when you're from Oklahoma, your school doesn't really suggest any competitive math contests to you, but I've been trying to get the school to let me do AMC 10 next year instead of AMC 8. And I was a Nats. You made a great comeback! Congrats!", "Solution_8": "china boi............ you most likely will be able to. and if you AND I can we dont have anything to lose cause amc 8 is kinda pointless all you get is a certificate thing. and who knows maybe you AND I can take the aime or amc 12 or somat. well nothing to lose eh. lau. lool. nuff said", "Solution_9": "Neal (or Synatx), would you please make more sense by using right grammar. What can I do if they don't let me take the AMC 10 or 12? Complain (I don't think they'll listen to a kid) or just pass up the whole chance by taking AMC 8? Or is there also an alternative? :?: :?: :?: :?:", "Solution_10": "blurg! just bringing it back up", "Solution_11": "[quote=\"Chinaboy\"]Neal (or Synatx), would you please make more sense by using right grammar. What can I do if they don't let me take the AMC 10 or 12? Complain (I don't think they'll listen to a kid) or just pass up the whole chance by taking AMC 8? Or is there also an alternative? :?: :?: :?: :?:[/quote]\r\nagreed :) \r\n\r\nI definitely suggest complaining, not to some random teacher but directly to the principal. It takes some guts but you don't have anything to lose!", "Solution_12": "my favorite contest is imo, but obviously ive never taken it. :D", "Solution_13": "AIME is the most important one for me right now.", "Solution_14": "same for me. fiery are you practicing from aime test in the test format or just do individual problems?", "Solution_15": "[quote=\"tetrahedr0n\"]my favorite contest is imo, but obviously ive never taken it. :D[/quote]\r\nHow do you know you'll like it if you've never taken it?\r\n\r\nI have no favorites currently, but if I must take one at this point I'd choose the middle school Math Team since it's fun, entertaining, and gives out tons of food/drink/soda/candy/etc. and it's just so much fun, especially with the hair spray and all :)", "Solution_16": "[quote]fiery are you practicing from aime test in the test format or just do individual problems?[/quote]\r\n\r\ntest format of course. If i wanted to do individual problems, I can find tons of those from other sources, but there are only a limited number of past AIME's. Plus, I find it very important to have experience in actually taking the test under test conditions.", "Solution_17": "The Mandelbrot Competition (http://www.mandelbrot.org) is a really wonderful contest, and unlike many of these that people are naming (particularly IMO and AIME), you don't need to have already done something in order to do it. It's taken 4 times a year, and you've unfortunately missed the first date already. It really is a great contest, though.", "Solution_18": "[quote=\"Tare\"][quote=\"tetrahedr0n\"]my favorite contest is imo, but obviously ive never taken it. :D[/quote]\nHow do you know you'll like it if you've never taken it?[/quote]\r\n\r\nProbably just the novelty of being good enough to get to it. That is, everyone at IMO is either 1) really good at math or 2) in a country with approximately six people that are less than 18 years of age. :wink: Since the latter case almost never happens, making it to IMO is a sign of math superiority, which is obviously a good thing. If that made sense at all.\r\nWhy are we revitalizing all these old threads?", "Solution_19": "What a long time since someone posted here!\r\n\r\nI personally like ARML and USAMTS, but the AMCs are OK too. Those are the only contests that I know of.", "Solution_20": "USAMO, cuz its a big confidence booster and last competition of year b4 ARML", "Solution_21": "i like amc8 its really easy ^.^ mathcounts is okay too. and kramier or wtv, our class (from miller middle) watched you countdown on tv last year. mucho congRats" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "please send me [i][b]direct[/b][/i] prove of steiner-lehmus theorem(if two angle bisectors of a triangle are equal, then the triangle\r\n is isosceles.)", "Solution_1": "Let [AD] and [BE] the angle bisectors of triangle ABC, then: $ |AD|^2 \\equal{} \\frac {bc.(a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)}{(b \\plus{} c)^2}$ and $ |BE|^2 \\equal{} \\frac {ac.(a \\plus{} b \\plus{} c)(a \\plus{} c \\minus{} b)}{(a \\plus{} c)^2}.$\r\nIf |AD|=|BE|, then: \r\n$ \\frac {bc.(a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)}{(b \\plus{} c)^2} \\equal{} \\frac {ac.(a \\plus{} b \\plus{} c)(a \\plus{} c \\minus{} b)}{(a \\plus{} c)^2}\\Rightarrow \\\\\r\n\\Rightarrow b.(a \\plus{} c)^2.(b \\plus{} c \\minus{} a) \\equal{} a.(b \\plus{} c)^2.(a \\plus{} c \\minus{} b)\\Rightarrow \\\\\r\n\\Rightarrow b.(a \\plus{} c)^2.(b \\plus{} c) \\minus{} ab.(a \\plus{} c)^2 \\minus{} a.(a \\plus{} c)(b \\plus{} c)^2 \\plus{} ab(b \\plus{} c)^2 \\equal{} 0\\Rightarrow \\\\\r\n\\Rightarrow (a \\plus{} c)(b \\plus{} c).[b(a \\plus{} c) \\minus{} a(b \\plus{} c)]\\plus{}ab.[(b\\plus{}c)^2\\minus{}(a\\plus{}c)^2] \\equal{} 0\\Rightarrow\\\\\r\n\\Rightarrow c(a \\plus{} c)(b \\plus{} c)(b \\minus{} a)\\plus{}ab(b\\minus{}a)(a\\plus{}b\\plus{}2c) \\equal{} 0 \\Rightarrow a \\equal{} b.$", "Solution_2": "thank you!but I dont want Algebraic solution.I want Geometric solution! :(", "Solution_3": "Dear Mathlinkers\r\nafter Coexter's book \"Geometry revisited\" p. 14-16 no direct proof exist or is known...\r\nWhat do you think?\r\nSincerely\r\nJean-Louis", "Solution_4": "http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus\r\n\r\nLots of proofs. Find one for yourself :D", "Solution_5": "See and [url=http://forumgeom.fau.edu/FG2005volume5/FG200525.pdf][color=red][b]here[/b][/color][/url] ...." } { "Tag": [ "geometry", "3D geometry", "probability" ], "Problem": "Two number cubes, each with the digits 1-6 on the six faces, are rolled. What is the probability that the product of the numbers on the top faces will be greater than 12? Express your answer as a common fraction.", "Solution_1": "The total number of products are $ 6\\cdot6\\equal{}36$. Instead of finding the direct case, let us use complimentary probability. The ways to get a product less than or equal to $ 12$ are as follows: \r\n[hide=\"List\"]\\[ (1,1)\\]\\[ (1,2)\\]\\[ (1,3)\\]\\[ (1,4)\\]\\[ (1,5)\\]\\[ (1,6)\\]\\[ (2,1)\\]\\[ (2,2)\\]\\[ (2,3)\\]\\[ (2,4)\\]\\[ (2,5)\\]\\[ (2,6)\\]\\[ (3,1)\\]\\[ (3,2)\\]\\[ (3,3)\\]\\[ (3,4)\\]\\[ 4,1)\\]\\[ (4,2)\\]\\[ (4,3)\\]\\[ (5,1)\\]\\[ (5,2)\\]\\[ (6,1)\\]\\[ (6,2)\\] Whew! We find there to be $ 36\\minus{}23\\equal{}13$ desired cases.[/hide]\r\nHence, our answer is $ \\boxed{\\frac{13}{36}}$.", "Solution_2": "Since you have to count out 23 sets, wouldn't it be easier to just count the pairs that have a product $ >12?$", "Solution_3": "it might be harder because you have to check twice as many numbers for greater than 12.\r\nwhere as if you have a number less than 12, you can find factor pairs easily.", "Solution_4": "First, find all the possible ways to get 2 numbers, so 6*6=36.\r\nNow, find all the ways 2 numbers on a die can be multiplied to get a number >12.\r\n3 can be multiplied by 2 numbers.\r\n4 can be multiplied by 3 numbers.\r\n5 can be multiplied by 4 numbers.\r\n6 can be multiplied by 4 numbers.\r\n2+3+4+4=13\r\n\r\nSo the probability is 13/36." } { "Tag": [ "calculus", "integration", "trigonometry", "trig identities", "calculus computations" ], "Problem": "Hi all, I've recently encountered an integral which I eventually solved. However, I'm searching for any alternative novel solutions, especially one's that utilize a unique combination of power substitutions.\r\n\r\n$I= \\int [ \\frac{15x-20}{x^2 } ]^{ \\frac{1}{6}}$\r\n\r\nit turns out that the power 1/6 turns out to be the major inconvenience. I took out the 15 factor and substituted x=1.5sec^2@ ,this way one is left with soley sin2@ as the fractional power term, not to mention the 1/6 being reduced to 1/3. Then subsequently u^3=sin2@ (to be general), this get's rid of the annoying 1/6 exponent to merely terms consisting of square root exponents through the usage of trig identities and algebraic simplifications....trig substitutions and so on. \r\n\r\nI was quite fortunate to solve it since I'm not too familiar with power substitutions.", "Solution_1": "hello, do you mean $ \\int\\sqrt[6]{\\frac{15x\\minus{}20}{x^6}}\\,dx$?\r\nSonnhard." } { "Tag": [ "modular arithmetic" ], "Problem": "There are $n$ cards lying down on a table. Persons $x$ and $y$ switch turns ($x$ goes first). When it is either $x$ or $y$'s turn, they must flip either 1, 3, or 4 cards (Flipped cards cannot be flipped again; only laid down cards can). The winner is one who flips the last card. How many cards $n$ are recquired for $y$ to win the game no matter what $x$ does?\r\n\r\nMasoud Zargar", "Solution_1": "For $y$ to win, it must be $n\\equiv 2\\pmod{4}.$\r\n\r\nProof:\r\n\r\nLet's look one move by $x$ and the subsequent move by $y.$ If $x$ flips $1$ card, $y$ will flip $3$. If $x$ flips $3$ cards, $y$ will flip $1.$ If $x$ flips $4$ cards, $y$ will flip $4$ if possible (or otherwise win the game - see later). In any case, after every $y$'s move the number of unflipped cards remains congruent modulo $4$ to the initial number of cards. After some $y$'s move the number of unflipped cards will necessarily reach $2$, as $n\\equiv 2\\pmod{4}$, and then $x$ will be forced to flip $1$ card. In the next move $y$ wins.", "Solution_2": "Would you algeabraically please explain more?", "Solution_3": "Tell me what part isn't clear, since there's not much algebra there.\r\n\r\nAfter every $y$'s move the number of unflipped cards is smaller by $4$ or $8$ ($1+3$ or $3+1$ or $4+4$), because this is his strategy. Therefore, since the initial number of cards is congruent to $2$ modulo $4$, eventually there will remain only $2$ unflipped cards, in which situation $y$ wins as described.\r\n\r\n[b]ADDITIONAL NOTE[/b]: Actually, after some more analyzing, I see that $y$ can win ALSO if $n\\equiv 0\\pmod{8}$, in the following manner: If $x$ flips $1$ card in his first move, $y$ will also flip $1$ and face him with the number of cards of the form $4k+2$, which is the winning situation for $y$, as we've seen. If $x$ flips $3$ cards in his first move, $y$ will also flip $3$ and again reach $4k+2$ unflipped cards. If $x$ flips $4$, then $y$ should also flip $4$ and force $x$ either to flip $1$ or $3$ and thus to lose, or if $x$ keeps flipping $4$, $y$ will keep doing that too, and the last four cards will go to $y$ since $n\\equiv 0\\pmod{8}$.\r\n\r\nIf $n=4k+1$ or $n=4k+3$, then $x$ can assure his win by flipping $3$ cards in the first case or $1$ card in the second case and facing $y$ with $4k+2$ cards, which we know is a losing situation for him.\r\n\r\nIf $n=8k+4$, then the game is lost by whoever flips $1$ or $3$ cards first, as already seen in the case $n\\equiv 0\\pmod{8}$, and if both keep flipping $4$, the last four will go to $x$.\r\n\r\n[u]The final conclusion[/u]: $y$ wins if $n\\equiv 0,2,6\\pmod{8}$ and $x$ wins if $n\\equiv 1,3,4,5,7\\pmod{8}$", "Solution_4": "Oh, sorry, it is my stupidity. :blush: I got it, thanks.", "Solution_5": "Please see the additional note in my previous post.", "Solution_6": "Only one more thing, this is similar to a contest question that had multiple choice answers that included both 32 and 34, which one would be correct, since both 32 and 34 satisfy $n\\equiv 0,2,6\\pmod 8$?", "Solution_7": "Then the author of the problem must be held responsible :)\r\n\r\nFirst move if $n=32=8\\cdot 4$:\r\na) If $x$ takes $1$, $y$ takes $1$ and reaches $30=4\\cdot 7+2$ and hence he wins.\r\nb) If $x$ takes $3$, $y$ takes $3$ and reaches $26=4\\cdot 6+2$ and hence he wins.\r\nc) If $x$ takes $4$, $y$ takes $4$ and reaches $24=8\\cdot 3$. If in the second move $x$ takes one or three, see cases a) and b). If $x$ keeps taking four, $y$ will be a copycat and finally take the last $4$.\r\n\r\nFirst move if $n=34$:\r\nThis is $4\\cdot 8+2$ and is already described in the first post.\r\n\r\nTherefore BOTH answers are correct.", "Solution_8": "But actually, I've just discovered I was wrong. For example, if $n=6$, then $x$ wins by taking four.\r\n\r\nLet me think some more. I'll patch it up." } { "Tag": [ "algebra", "polynomial", "quadratics", "Rational Root Theorem" ], "Problem": "Factor completely: $x^3-7x^2-85x+91.$\r\n\r\nI did it with synthetic division and got lucky, but is there a better way?", "Solution_1": "I would do it with synthetic division too. I don't know if there's an easier way.", "Solution_2": "Okay, try this: I just discovered it.\r\n\r\nExpanding $(x-a)(x-b)(x-c)$ and collecting like terms, we have $x^3-x^2(a+b+c)+x(ab+ac+bc)-abc$!\r\n\r\nSo we can set $abc=-91=7 \\cdot 13 \\cdot -1$ and $a+b+c=7$, so by a little experimentation, we have $(x-1)(x+7)(x-13)$\r\n\r\nAwesome! Now I can factor monic trinomials with degree 3! \\/\\/o0+!", "Solution_3": "NIce way, that's just like quadratics! Except a little longer.", "Solution_4": "$(a, b, c) = (-7, 13, 1)$ because the last term is positive 91.", "Solution_5": "[quote=\"dakyru\"]$(a, b, c) = (-7, 13, 1)$ because the last term is positive 91.[/quote]\r\n\r\nNo, $(a,b,c)=(1,-7,13)$. Then $abc=-91$, $ab+bc+ac=-7-91+13=-85$, and $a+b+c=7.$", "Solution_6": "The order is insignificant because the values are interchangeable. I just thought your previous post said 7, 13, and -1 as factors.", "Solution_7": "He stated that $-91=7\\cdot 13\\cdot -1$, but his factored equation uses $(13, -7, 1)$.", "Solution_8": "A third degree polynomial factors (over the rationals) if and only if it has a linear factor, and that linear factor gives you a root. This makes the Rational Root theorem your best tool, since it immediately reduces the problem to testing a finite collection of possible linear factors. If none of them work, the cubic has no rational roots and is guaranteed to be ugly. If one works, you divide it out and get a quadratic which is easy to handle.", "Solution_9": "in this case particularly, just check whether $-1$ or $1$ are roots of the polynomial (it's easy to add up al co\u00ebfficients)...\r\nthen you're left with a second degree polynomial, which is easy to solve :)", "Solution_10": "Chess's idea is really cool! I useed to just guess for the real root because there has to be at least one and use synthetic division, but that's nice!\r\n\r\n :10:\r\n\r\nAnd of course, we can all factor out the leading co-efficient to allow us to find the roots of any trinomials of degree 3. ;)", "Solution_11": "[quote=\"4everwise\"]Chess's idea is really cool! I useed to just guess for the real root because there has to be at least one and use synthetic division, but that's nice!\n\n :10:\n\nAnd of course, we can all factor out the leading co-efficient to allow us to find the roots of any trinomials of degree 3. ;)[/quote]\r\nTry to guess as little as possible. \r\nFor the polynomial $P(x)=a_1x^n+a_2x^{n-1}+...+a_nx+a_{n+1}$ a could strategy is to try the factors of $\\frac{a_{n+1}}{a_1}$ as zeros", "Solution_12": "Could you explain this further? I haven't factored in a while, but this seems really familiar.", "Solution_13": "If you look at chess's post you can see that the last term of the polynomial is the product of the roots, so you try the factors of the x^0 term as roots" } { "Tag": [ "LaTeX" ], "Problem": "-1(x-1)=((3/x)-1)(x-3)\r\nsolve for x.\r\n\r\n\r\nim having some trouble with this after i multiply it out.", "Solution_1": "Thats a bit hard to decipher w/o $\\LaTeX$, but I'll give it a shot:\r\n\r\n[hide]$-1(x-1)=(\\frac{3}{x}-1)(x-3)$\nUse distributive property on left and FOIL on right\n$-x+1=\\frac{3x}{x}-x-\\frac{9}{x}+3$\nNow combine like terms and solve\n$1=6-\\frac{9}{x}$\n$-5=-\\frac{9}{x}$\n$\\boxed{x=\\frac{9}{5}}$[/hide]", "Solution_2": "yup thats it...thanks a bunch =]" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Nine lines, parallel to the base of a triangle, divide the other sides into 10 equal segments and the area into 10 distinct parts. Find the area of the original triangle, if the area of the largest of these parts is 76.", "Solution_1": "$S(\\triangle ABC) - \\left(\\frac{9}{10}\\right)^2S(\\triangle ABC) = \\frac{19}{100}\\ S(\\triangle ABC) = 76$\r\n\r\n$S(\\triangle ABC) = 400$" } { "Tag": [ "geometric series" ], "Problem": "Factor \r\n\r\n $ b^8 \\plus{}b^6 \\plus{} b^4 \\plus{}b^2\\plus{} 1$ .\r\n\r\n\r\n[hide=\"Here is the solution\"] $ (b^4\\plus{}b^2\\plus{}1)^2\\minus{}b^2(b^4\\minus{}1)^2$\n\nIt is trivial to factor it from here...[/hide]\r\n\r\nBut i found it by guessing . i'm interested in how one could come up with something like that.", "Solution_1": "hello, i have $ (1\\minus{}b\\plus{}b^3\\minus{}b^3\\plus{}b^4)(1\\plus{}b\\plus{}b^2\\plus{}b^3\\plus{}b^4)$.\r\nSonnhard.", "Solution_2": "Let $ P(b) \\equal{} b^8 \\plus{} b^6 \\plus{} b^4 \\plus{} b^2 \\plus{} 1$ and let $ \\omega \\equal{} e^{2\\pi i/5}$ be the fifth root of unity; then you could note that $ P(\\omega) \\equal{} P(\\omega^2) \\equal{} P(\\omega^3) \\equal{} P(\\omega^4) \\equal{} 0$, so $ (b\\minus{}\\omega)(b\\minus{}\\omega^2)(b\\minus{}\\omega^3)(b\\minus{}\\omega^4) \\equal{} b^4\\plus{}b^3\\plus{}b^2\\plus{}b\\plus{}1 | b^8 \\plus{} b^6 \\plus{} b^4 \\plus{} b^2 \\plus{} 1$. Then long divide.", "Solution_3": "See (33) and (34) [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]here[/url].", "Solution_4": "You could just treat it as a geometric series, sum it, and then divide out the denominator factors in the resulting fraction. :)", "Solution_5": "Consider the equation $ b^8 \\plus{} b^6 \\plus{} b^4 \\plus{} b^2 \\plus{} 1 \\equal{} 0$. As b = 0 is not a solution of this equation, divide both terms by $ b^4$. Then we get\r\n\r\n$ b^4 \\plus{} b^2 \\plus{} 1 \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{b^4} \\equal{} 0$\r\n\r\n$ \\equal{} > \\left(b^2 \\plus{} \\frac {1}{b^2}\\right)^2 \\minus{} 2 \\plus{} \\left(b^2 \\plus{} \\frac {1}{b^2}\\right) \\plus{} 1 \\equal{} 0$\r\n\r\n$ \\equal{}> \\left(b^2 \\plus{} \\frac {1}{b^2}\\right)^2 \\plus{} \\left(b^2 \\plus{} \\frac {1}{b^2}\\right) \\minus{} 1 \\equal{} 0$\r\n\r\nand now make $ b^2 \\plus{} \\frac {1}{b^2} \\equal{} x$ to get $ x^2 \\plus{} x \\minus{} 1 \\equal{} 0$ whose solutions are $ x \\equal{} \\minus{} \\phi$ and $ x \\equal{} \\phi \\minus{} 1$. Now for each of these solutions $ x_i$ solve the equation $ b^4 \\minus{} x_i b^2 \\plus{} 1 \\equal{} 0$ by making $ b^2 \\equal{} u$.", "Solution_6": "treat as a geometric series with $ r \\equal{} x^2, a_1 \\equal{} 1$. Then the sum of the first 5 terms is equal to the desired expression $ \\equal{} \\frac {a(1 \\minus{} r^5)}{1 \\minus{} r} \\implies \\frac {(1 \\minus{} x^{10})}{1 \\minus{} x^2},$ dividing out gives the desired result.", "Solution_7": "[quote=\"Wichking\"]Factor \n\n $ b^8 \\plus{} b^6 \\plus{} b^4 \\plus{} b^2 \\plus{} 1$ . [/quote]\r\n\r\n$ b^6\\plus{}b^4\\plus{}b^2\\equal{}b^2(b^4\\plus{}b^2\\plus{}1)$\r\n$ b^8\\plus{}1\\equal{}(b^4\\plus{}1)^2\\minus{}b^4\\equal{}(b^4\\plus{}b^2\\plus{}1)(b^4\\minus{}b^2\\plus{}1)$\r\n\r\n$ \\therefore b^8\\plus{}b^6\\plus{}b^4\\plus{}b^2\\plus{}1\\equal{}(b^4\\plus{}b^2\\plus{}1)(b^4\\plus{}1)$\r\n$ \\equal{}(b^2\\plus{}b\\plus{}1)(b^2\\minus{}b\\plus{}1)(b^2\\plus{}\\sqrt{2}b\\plus{}1)(b^2\\minus{}\\sqrt{2}b\\plus{}1)$." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "find all matrix A of order n $ (n \\geq 2)$ such that det(A+M) =detA +det M for all matrix M", "Solution_1": "You didn't specify what type of matrices, so I'll assume $ \\mathcal{M}_n(\\mathbb{C})$, but you can change $ \\mathbb{C}$ with a field of characteristic zero easily below.\r\n\r\nFirst, $ A \\equal{} 0$ is solution. We then show this is the only solution. \r\nAssume $ A\\neq 0$ is solution.\r\nIf $ A\\in \\mathrm{GL}_n(\\mathbb{C})$, then taking $ M \\equal{} A$ you get $ 2^n \\det A \\equal{} 2 \\det A$ which is impossible.\r\nIf $ A\\in \\mathcal{M}_n(\\mathbb{C}) \\setminus \\mathrm{GL}_n(\\mathbb{C})$. Take $ M \\equal{} \\lambda I_n$, $ A$ is nilpotent because $ 0$ is its only eigenvalue. To conclude look at a Jordan block with $ 1$ is position $ (i,i \\plus{} 1)$ and $ 0$ everywhere else, take $ M \\equal{} \\mathrm{diag}(K, 1,1,\\ldots, 1)$ with $ K \\equal{} \\begin{pmatrix} 1 & 0 \\\\\r\n1 & 1 \\end{pmatrix}$.", "Solution_2": "Any field of characteristic zero? Try any field.\r\n\r\nConsider cases in which $ M$ has exactly $ k$ ($ 0=4{(OA+OB+OC)^2}", "Solution_1": "[quote=\"Michael Niland\"]Triangle $ABC$ is acute-angled and $O$ is a point in its plane.\n\n$U,V,W$ are the distances from $O$ to the sides of the triangle.\n\nProve that $4(U+V+W)^2 +(AB+BC+CA)^2 \\geq 4(OA+OB+OC)^2$[/quote]\r\n\r\n :)" } { "Tag": [ "ratio", "algebra unsolved", "algebra" ], "Problem": "Starting with any three digit number $ n$ (such as $ n\\equal{}625$) we obtain a new number $ f(n)$ which is equal to the sum of the three digits of $ n$, their three products in pairs and the product of all three digits.\r\n\r\n(i) Find the value of $ \\frac{n}{f(n)}$ when $ n\\equal{}625$. (The answer is an integer!)\r\n\r\n(ii) Find all three digit numbers such that the ratio $ \\frac{n}{f(n)} \\equal{}1$.", "Solution_1": "[quote=\"AndrewTom\"]Starting with any three digit number $ n$ (such as $ n \\equal{} 625$) we obtain a new number $ f(n)$ which is equal to the sum of the three digits of $ n$, their three products in pairs and the product of all three digits.\n\n(i) Find the value of $ \\frac {n}{f(n)}$ when $ n \\equal{} 625$. (The answer is an integer!)\n\n(ii) Find all three digit numbers such that the ratio $ \\frac {n}{f(n)} \\equal{} 1$.[/quote]\r\n\r\nSo $ f(\\overline{abc})\\equal{}(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)\\minus{}1$\r\n\r\n(i) $ f(625)\\equal{}7\\cdot 3\\cdot 6\\minus{}1\\equal{}125$ and $ \\frac{625}{f(625)}\\equal{}5$\r\n\r\n(ii) We have to solve $ 100a\\plus{}10b\\plus{}c\\equal{}(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)\\minus{}1$\r\nSo $ 100a\\plus{}10b\\equal{}a(b\\plus{}1)(c\\plus{}1)\\plus{}b(c\\plus{}1)$\r\n\r\nIf $ (b\\plus{}1)(c\\plus{}1)<100$, we need $ 10b9$, impossible.\r\nSo $ (b\\plus{}1)(c\\plus{}1)\\ge 100$ and so $ b\\equal{}c\\equal{}9$\r\n\r\nAnd it is easy to see that this is indeed a solution. Hence the answer $ \\boxed{f(\\overline{a99})\\equal{}\\overline{a99}}$ for any $ a\\in\\{1,2,3,4,5,6,7,8,9\\}$" } { "Tag": [ "geometry", "3D geometry", "sphere", "puzzles" ], "Problem": "i sorta derived this from the \"Turtles\" topic. anyways, there are 3 ants, ant A says there are 2 ants behind me, and B says there is one ant in front and behind me, and ant C says there are 2 ants ahead of me and 1 behind me.How is this possible? no, ant C is not lying. this riddle might be kinda stupid, but oh well.", "Solution_1": "[hide=\"maybe\"]ants can't count?[/hide]", "Solution_2": "there is a fourth ant. It simply wasn't mentioned by the first two or the problem. I posted that in the turtles topic too.", "Solution_3": "That doesn't work. All I can think of is that they are going sideways with some unmentioned ants. Directions they're facing doesn't help either.", "Solution_4": "i says there are three ants", "Solution_5": "So there are three ants. Does that mean that there cannot be a fourth?", "Solution_6": "i think so", "Solution_7": "should i tell you guys the answer?", "Solution_8": "yeah.", "Solution_9": "maybe ant C is staying on top of ant B?", "Solution_10": "Obviously Ants A and B are lying.", "Solution_11": "We could be on the surface of a sphere!", "Solution_12": "Or even better, there is a big number 1 behind them all...", "Solution_13": "[quote=\"seamusoboyle\"]We could be on the surface of a sphere![/quote]\r\nyou're warm!!", "Solution_14": "we could be in hyperspherical space? :P \r\n\r\nor we could be on a circular line....but I prefer hyperspherical space", "Solution_15": "is it of any importance that now we are talking of ants and not turtles anymore?\r\n :)", "Solution_16": "On a circular line?", "Solution_17": "yea, that's it!!... i know, it's kinda stupid.", "Solution_18": "darn....I wanted it to be something like hyperspace curved in the fifth dimension lol.\r\n\r\nActually that's what a circular line is...1 dimensional space curved in the second dimension. Any n dimensional space curved in the n+1 dimension would work here.", "Solution_19": "I guess the hyperdimential concept could make this problem a possiblity,without one ant over the other", "Solution_20": "But my answer makes more sense and its a better answer!!!", "Solution_21": "[quote=\"GoBraves\"]But my answer makes more sense and its a better answer!!![/quote]\r\nI would agree, if only the riddle hadn't said that there were exactly three ants (well actually it just said 'three ants' but gr8tsk8r clarified that there were exactly three afterwards, I believe).", "Solution_22": "Hi guys, kinda new here...\r\n\r\nI can't help myself but to react on this topic... How lucky the ants are, they were the center of discussion here.\r\n\r\nI think there's somethin wrong with the problem. It should be clear you know to avoid opinions :)", "Solution_23": "what's wrong with da problemo?" } { "Tag": [ "algorithm", "videos", "function", "Duke", "college", "factorial", "Mafia" ], "Problem": "[b] FOR THOSE WHO ARE READING THIS THREAD FOR THE FIRST TIME \nAOPSTI is usually held Friday and Saturday. 8PM EST over AOL Instant Messenger. We ask participants to show up at around 7:45PM so they can receive invites. Expect the games to last till about 9:30.\nWanna join? Just leave your AIM in this thread and come online at the appropriate time. Or talk to Lucky (AIM: hezgotzbeenz)[/b]\r\n\r\nBased on the success of the AOPS Trivia Challenge, I have compiled a list of 30 or so questions for use for the first ever \"AOPS Trivia Invitational\".\r\n\r\nBasically, you may remember whitehorseking88 planning something like this many many months ago but never getting around to actually doing it. Well... I have the questions and am ready to administer them.\r\n\r\nThis would last about an hour and would occur over AIM (Since that seems to be what most everyone has). I'm still not sure whether I would do it in teams (probably) or individual.\r\n\r\nI have not set a date yet because I need to see if there is interest for this first.\r\n\r\nYes, it says the AOPS Trivia [i]Invitational[/i] but truth be told, [b]you're all invited [/b].\r\n\r\nSo is anyone interested? Reply here or contact me on AIM (hezgotzbeenz) if you are. I'd need at least 8 people or so for this to run smoothly.\r\n\r\n(And if you can't make it this time and there's enough interest, I'd be perfectly fine with holding more of these events)\r\n\r\n[b] Quick Stats After 30 Trivia Invitationals [/b]\r\n\r\nMost Games Played (by Games) \r\nGoBraves - 22\r\nyif man 12 - 19\r\nhello - 17\r\nDarkKnight - 15\r\nmcalderbank - 15\r\nmetsfan001 - 14\r\nhspotter2002 - 13\r\npaladin314159 - 12\r\nchesspro - 12\r\nwhitehorseking88 - 12\r\n\r\nMost Games Played (by Questions) \r\nGoBraves - 646\r\nyif man 12 - 533\r\nDarkKnight - 431\r\nhello - 398\r\nmcalderbank - 384\r\nmetsfan001 - 380\r\npaladin314159 - 345\r\nhspotter2002 - 342\r\nchesspro - 339\r\nLucky707 - 270\r\n\r\nTotal Points \r\nGoBraves - 1950\r\nyif man 12 - 1949\r\nmetsfan001 - 1926\r\nLucky707 - 1475\r\nDarkKnight - 1447\r\npaladin314159 - 1417\r\nAznphatso - 1086\r\nhello - 913\r\nmcalderbank - 867\r\nchesspro - 860\r\n\r\nPoints Per Game \r\nLucky707 - 163.89\r\nmetsfan001 - 137.57\r\npaladin314159 - 118.08\r\nAznphatso - 108.60\r\nyif man 12 - 102.58\r\nAnkur87 - 97.25\r\nDarkKnight - 96.47\r\nmiraculouspostmaster - 96.00\r\nChigr - 94.60\r\nJSRosen3 - 94.00\r\n\r\nPoints Per Question \r\nLucky707 - 5.463\r\nmetsfan001 - 5.068\r\nAznphatso - 4.180\r\npaladin314159 - 4.107\r\nmiraculouspostmaster - 3.934\r\nComplexZeta - 3.807\r\nyif man 12 - 3.657\r\nAnkur87 - 3.458\r\nDarkKnight - 3.357\r\nscrambled - 3.255\r\nwhitehorseking88 - 3.224\r\nDutchfreak81 - 3.167\r\nChigr - 3.153\r\nJSRosen3 - 3.133\r\nGoBraves - 3.019\r\nbakwardz41 - 2.867\r\nPoignantPianist1 - 2.733\r\nYashaBK - 2.694\r\nhenryyangrui - 2.548\r\nchesspro - 2.537\r\n\r\nBest Record (by Wins)\r\nchesspro - 10-1\r\nDarkKnight - 8-3\r\nyif man 12 - 8-9\r\nsheepwarrior - 7-2\r\nGoBraves - 7-12\r\nLucky707 - 6-3\r\nmetsfan001 - 6-4\r\nmcalderbank - 6-7\r\nAznphatso - 5-2\r\npaladin314159 - 5-3\r\nhello - 5-9\r\n\r\nWinning %\r\nbakwardz41 - 100.0% (1-0)\r\nMiChFrEaK15 - 100.0% (1-0)\r\nweiyangmiami - 100.0% (1-0)\r\nchesspro - 90.9% (10-1)\r\nsheepwarrior - 77.8% (7-2)\r\nDarkKnight - 72.7% (8-3)\r\naznphatso - 71.4% (5-2)\r\nLucky707 - 66.7% (6-3)\r\npaladin314159 - 62.5% (5-3)\r\nmetsfan001 - 60.0% (6-4)\r\n\r\nThe 100 Club -\r\n(People whose highest scores exceed 100, highest score is listed beside name)\r\nyif man 12 - 201\r\nmetsfan001 - 201\r\nLucky707 - 189\r\nAznphatso - 183\r\nGoBraves - 173\r\npaladin314159 - 154\r\nmiraculouspostmaster - 138\r\nDarkKnight - 134\r\nAnkur87 - 133\r\nComplexZeta - 131\r\nmcalderbank - 128\r\nJSRosen3 - 120\r\nwhitehorseking88 - 118\r\nchesspro - 113\r\nhspotter2002 - 112\r\nChigr - 110\r\nhello - 106\r\nsheepwarrior - 104\r\n\r\nMVP's \r\nmetsfan001 - 7\r\nhezgotzbeenz - 6\r\nyif man 12, aznphatso, paladin314159 - 3\r\nGoBraves, miraculouspostmaster, Ankur87 - 2\r\nComplexZeta, JSRosen3 - 1\r\n\r\nOverall Rating (using Lucky's algorithm ... based on capped experience and 50% of rating is based on points per question)\r\n(all ratings above 70 are shown)\r\nLucky707 - 96.15\r\nmetsfan001 - 92.62\r\nAznphatso - 86.15\r\npaladin314159 - 84.40\r\nyif man 12 - 83.64\r\nDarkKnight - 79.70\r\nGoBraves - 77.86\r\nAnkur87 - 76.91\r\nmirauclouspostmaster - 73.20\r\nchesspro - 71.95\r\nChigr - 71.85\r\nwhitehorseking88 - 70.14", "Solution_1": "Awesome. I'm in. I'm looking forward to being the first to lose or getting the lowest or whatever.", "Solution_2": "I'm in too.", "Solution_3": "count me in", "Solution_4": "I am in if the time is fine.", "Solution_5": "Yay! Dream come true!\r\n\r\nI'm in.", "Solution_6": "Sounds fun! :) But I'll lose of course :P", "Solution_7": "[hide]Im in, adentro, po, binnen,dentro,\u03bc\u03ad\u03c3\u03b1,\u0432\u043d\u0443\u0442\u0440\u0438[/hide]\r\n\r\nIM= AmeemShk", "Solution_8": "im in if the date and time are good", "Solution_9": "Hmm... lol, I count 8 people already. [b] This does not mean that sign-ups are closed [/b]\r\n\r\nThe time is this Monday at 8 PM EST.\r\n\r\nIf you want in, I'll be on at ~7:45 PM. Send me a message and I'll invite you in.\r\n\r\nOnce in, I will decide on the teams and we will go at it for an hour.\r\n\r\nLeave your AIM here if you want an invitation at the time of the AIT. This goes for all interested people, not just for the ones that replied above.", "Solution_10": "My AIM is metzphan001.", "Solution_11": "my aim is tennisguru51", "Solution_12": "If it's Monday night, I won't be able to make it. Sorry.", "Solution_13": "Don't worry about it. I'll hold more of them.\r\n\r\nSO FAR: (TOTAL: 8 with 1 maybe)\r\nplokoon\r\nmetsfan\r\nGoBraves\r\nyif man 12\r\nhello\r\nqwertyeq\r\nmccalderbank\r\nwhitehorseking88\r\nsheepwarrior ???\r\n\r\n\r\nIf you'd like to join, just leave your AIM here so I can invite you on Monday at 8PM EST.", "Solution_14": "caocoprez", "Solution_15": "I'll be your partner lucky. :yup:", "Solution_16": "I'll sign up, whoever wants to be my partner, im fine with it", "Solution_17": "I'll play.\r\n\r\nEDIT: actually, i'll be gone friday, so i cant play.", "Solution_18": "I'm in.", "Solution_19": "10 more ppl..or we can play singles", "Solution_20": "AOPSTI is scheduled for this Friday at 8 EST\r\n\r\nIt won't be a typical AOPSTI, rather it will be more like the old format, except with buzzing. ie. most questions are categories or regular questions.\r\n\r\nCome if you'd like.", "Solution_21": "Anyone up for another AOPS trivia invitational? I will host tomorrow and same time as always. 7:30 - 9 EST\r\nOf course, I'd prefer it earlier. That's up to you.", "Solution_22": "Er... which format are you hosting under?", "Solution_23": "[b] Final Results of Tournament 2 If Anyone Cares [/b]\r\n\r\nROUND ONE:\r\n*Phatso/Penguin 49, Hello/Potter 15\r\nMets/Dark 37, Montoya/Leo 32\r\nAmeem/Ankur 29, Yif/Rosen 44\r\nSponge/Perry 22, Chigr/Paladin 58\r\n\r\nROUND TWO:\r\n**Mets/Dark 41, Paladin/Chigr 42\r\n***Phatso/Penguin 38, Yif/Rosen 34\r\n\r\nCHAMPIONSHIP:\r\nPhatso/Penguin 51, Chigr/Paladin 26\r\n\r\n* - Denotes match was split into halves. Phatso faced Hello 1v1 in first half and Penguin faced Potter 1v1 in second half.\r\n** - Denotes match went to a 3 question overtime.\r\n*** - Denotes match was not only split into halves but went to a 5 question overtime with all players participating.", "Solution_24": "I've been gone for a while.\r\nI don't understand the new format, so I'm just hosting under the two teams format. Come and separate into two teams. You know, the old way. :roll:", "Solution_25": "Yay, haven't had trivia for a long time. Still at 8est?\r\n\r\nNew way's just each person buzzes if they know and each team can only buzz once on a question.", "Solution_26": "[quote=\"white_horse_king88\"]Anyone up for another AOPS trivia invitational? I will host tomorrow and same time as always. 7:30 - 9 EST\nOf course, I'd prefer it earlier. That's up to you.[/quote]\r\n\r\nI wouldn't like it earlier, since the mock AMC B is ending at 7:15.", "Solution_27": "I maybe won't do it because I'm not seeing much interest.", "Solution_28": "i would play, but ill probably have a unexpected event when you do the game", "Solution_29": "this is also old 16 year bump" } { "Tag": [ "abstract algebra", "function", "Ring Theory", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "How do I prove that the rings Z2 x Z2 AND Z4 are not isomorphic?\r\nFrom this, is there a ring homomorphism Z4 to (Z2 x Z2) and is there a ring homomorphism (Z2 x Z2) to Z4?\r\nNote, for me, a ring homomorphism takes 1 to 1.\r\nThanks in advance for help.", "Solution_1": "Use the classification theorem of finite abelian groups to prove that the two aren't even isomorphic as groups.\r\nEdit: if you don't know the theorem, show one is cyclic and the other is not. Also, you can show that in $ \\mathbb{Z}_4$ there is a nilpotent element where as there are no nilpotent elements in $ \\mathbb{Z}_2 \\times \\mathbb{Z}_2$ The second part shouldn't be so hard once you understand the structure of the rings better.", "Solution_2": "I figured out the isomorphism part. I'm guessing that there does exist a ring homomorphism for both in the second part. But how do I define the functions? f(x) = x? I don't get how you set up the function for the product ring. Obviously, I'm just learning abstract algebra.", "Solution_3": "Well in this specific case, you really only have four elements so you could in principle write out the function element by element. So you said that for this problem, homomorphism means that it takes 1 to 1 (and 0 to 0 also).\r\nThis takes care of 2 of the four elements. That only leaves us with two more to mess with.\r\nSo lets list the elements of $ \\mathbb{Z}_2 \\times \\mathbb{Z}_2$\r\nWe have: (0,0) (1,1) (1,0) (0,1)\r\nWhat elements do we have in $ \\mathbb{Z}_4$? Well we have 0,1,2,3\r\n\r\nWe know that f(0) = (0,0) and f(1) = (1,1) so where can we send 2 and 3?\r\nWell, 2 = 1+1 so the only place we can send f(2) is f(1+1) = f(1)+f(1) = (1,1)+(1,1) = 0\r\n\r\nNow what about 3 = 1+1+1. by a similar argument f(3) = (1,1)+(1,1)+(1,1) which is (1,1)\r\n\r\nthis is automatically a group homomorphism because we defined everything based on where the generator goes. does this make a ring homomorphism though?\r\n\r\nLets check f(2*3) = f(2). This should equal f(2)*f(3). This is in fact true.\r\nHow about f(1*x)? Well this is f(1)*f(x) but f(1) is the unit so this of course is f(x)\r\nHow about f(0*x)? Well f(0) is the 0 in the image so this is also equal f(0).\r\n\r\nTry the reverse process yourself and see if theres a map going the other way.\r\n\r\nEdit: if you're curious the argument i just posted is expressed in a more general phenomenon known as a \"universal property\". Heuristically, it says that if you can find two projections of a ring onto two other rings, there is a unique map from that ring onto the product of the two rings (satisfying some \"niceness\" conditions). In my opinion, this would be the quickest way to show there is a map from $ \\mathbb{Z}_4$ to $ \\mathbb{Z}_2 \\times \\mathbb{Z}_2$" } { "Tag": [ "function", "algebra", "domain", "real analysis", "real analysis unsolved" ], "Problem": "Does there exists a function $f: \\mathbb{R}\\setminus \\{ 0 \\}\\to \\mathbb{R}\\setminus \\{ 0 \\}$ such that\r\n$f(f(x))=-\\frac1x$", "Solution_1": "$f(x)=\\frac{x-1}{x+1}$ is interesting, but it's got the wrong domain and the wrong range.", "Solution_2": "There are such functions, but they can't be continuous, or even continuous except at a finite set of points. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113408]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=52068]here[/url] for a pair of closely related problems." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find the condition of $a_{0},a_{1},a_{2}$ such that the sequence ${a_n}$ satisfies:\r\n $a_{n+2}=\\frac {a_{n+1}a_{n}a_{n-1}}{6a_{n}a_{n-1}-11a_{n-1}a_{n+1}+6a_{n}a_{n+1}}$ for all $n \\in N \\geq 1$ has the infinite integer terms. :)", "Solution_1": "$a_n=\\frac{2a_0a_1a_2}{(a_0a_1-3a_0a_2+2a_1a_2)3^n-(a_0a_1-4a_0a_2+3a_1a_2)2^{n+1}+a_0a_1-5a_0a_2+6a_1a_2}$" } { "Tag": [ "inequalities", "function", "logarithms", "number theory proposed", "number theory" ], "Problem": "Let $ H$ denote the set of those natural numbers for which $ \\tau(n)$ divides $ n$, where $ \\tau(n)$ is the number of divisors of $ n$. Show that \r\n\r\na) $ n! \\in H$ for all sufficiently large $ n$, \r\n\r\nb)$ H$ has density $ 0$. \r\n\r\n[i]P. Erdos[/i]", "Solution_1": "(a) has already been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=586&t=290291]here[/url].\n\n(b) Let $S_p$ denote the set of positive integers with maximal prime divisor $p$ and $T_p$ denote the set of positive integers with all prime divisors $\\le p$ (so obviously $S_p,T_p$ both have zero density within $\\mathbb{N}$). If $n\\in H_p=S_p\\cap H$, then $\\tau(n)\\mid n\\implies v_p(n)+1\\mid n$, so $v_p(n)+1>1\\in T_p$. Hence $H_p$ has zero density within $S_p$ (because $T_p$ has zero density within $\\mathbb{N}$), so averaging over all primes, $H=H_{p_1}\\cup H_{p_2}\\cup\\cdots$ (should have?) zero density within $\\mathbb{N}=S_{p_1}\\cup S_{p_2}\\cup\\cdots$.\n\nI'm clueless as to how to rigorize (b), and I think the whole argument might be wrong in the first place, so can someone check it?\n\n[b]Edit:[/b] Actually, this is trivial, because for all $\\epsilon>0$ there exists $N$ such that for all $n>N$ and $k\\ge1$, $|H_{p_k}(\\le n)|\\le \\epsilon|S_{p_k}(\\le n)|$, so because both $|H_{p_k}(\\le n)|,|S_{p_k}(\\le n)|$ eventually become zero we can just sum these inequalities up.\n\nBut we do need to prove $|H_{p_k}(\\le n)|/|S_{p_k}(\\le n)| = o(1)$ more rigorously (for fixed $p$). Let $\\Psi(x,y)$ denote the $y$-[url=http://en.wikipedia.org/wiki/Smooth_number#Distribution]smooth number counting function[/url]. Because $\\Psi(x,p_k)$ is the number of nonnegative integer solutions to $a_1\\log{p_1}+\\cdots+a_k\\log{p_k}\\le \\log{x}$, simplex volume comparisons (both lower and upper lattice point bounds are easy to estimate) yield\n\\[\\Psi(x,p_k)=\\frac{\\log^k{x}}{k!\\log{p_1}\\cdots\\log{p_k}}+O(\\log^{k-1}{x}).\\]Because all numbers in $S_{p_k}$ and $H_{p_k}\\subset S_{p_k}$ can be written as a positive power of $p_k$ times a $p_{k-1}$-smooth number, it's easy to see that because $T_{p_k}$ has zero density within $\\mathbb{N}$ and $v_p(H_{p_k})+1\\subseteq T_{p_k}$,\n\\begin{align*}\n\\frac{|H_{p_k}(\\le n)|}{|S_{p_k}(\\le n)|}\n&\\le \\frac{\\sum_{r=1}^{o(1)\\log_{p_k}{n}}\\Psi(n/p_k^r,p_{k-1})}{\\sum_{r=1}^{\\log_{p_k}{n}}\\Psi(n/p_k^r,p_{k-1})} \\\\\n&= \\frac{O(\\log{n}\\log^{k-2}{n})+\\sum_{r=1}^{o(1)\\log_{p_k}{n}}\\log^{k-1}{n/p_k^r}}{O(\\log{n}\\log^{k-2}{n})+\\sum_{r=1}^{\\log_{p_k}{n}}\\log^{k-1}{n/p_k^r}} \\\\\n&= \\frac{O(\\log{n}\\log^{k-2}{n})+\\sum_{r=1}^{o(1)\\log_{p_k}{n}}\\log^{k-1}{n}}{O(\\log{n}\\log^{k-2}{n})+\\sum_{r=1}^{\\log_{p_k}{n}}\\log^{k-1}{n}} \\\\\n&= \\frac{O(\\log^{k-1}{n})+o(1)\\log_{p_k}{n}\\log^{k-1}{n}}{O(\\log^{k-1}{n})+\\log_{p_k}{n}\\log^{k-1}{n}} \\\\\n&= o(1),\n\\end{align*}where we have used the fact that $\\log^{k-1}{n/p_k^r}=\\log^{k-1}{n}+O(\\log^{k-2}{n})$." } { "Tag": [ "percent", "calculus", "calculus computations" ], "Problem": "Can anyone help me with this series problem? We've never done word problems and I guess I'm setting it up incorrectly becuase I'm having it diverge to infinity. \r\n\r\nA ball drops from a height of 15 feet. Each time it hits the ground, it bounces up 65 percents of the height it fall. Assume it goes on forever, find the total distance it travels. \r\n\r\nThanks =)\r\n\r\n-----\r\nSorry, I always forget that part!\r\nI know it's the summation of a series from 1 (first drop) to infinity\r\nEach time the height is 65% or 0.65 of the height before so the series would be:\r\n15 + 9.75 + 6.3375 + ...\r\nFor the series I was doing this is 15(.65n)\r\nBut I'm not sure this really makes sense. Or if I'm even going about this right! Sorry, we just started this stuff Friday I'm still pretty new at it =o)", "Solution_1": "Why don't you write down what you've done? Then we can see where you've gone wrong.", "Solution_2": "15+2[(15)(.65)+(15)(.65)^2+...]\r\nthat's just 15 plus twice a geometric series- how did you get that it diverges?", "Solution_3": "Factors of 2 are needed because in each bounce the ball bounces up and THEN falls down.\r\n\r\nThe answer is 15 + 2*15*[0.65 + 0.65^2 + 0.65^3 + 0.65^4 + ...]\r\n= 15 + 2 * 15 * 0.65 * [1 + 0.65 + 0.65^2 + 0.65^3 + ...]\r\n= 15 + 2 * 15 * 0.65 / (1 - 0.65) = 70.7... feet." } { "Tag": [], "Problem": "Each bottle of perfume at The Perfume Parlor contains $ \\frac3{16}$ ounces of perfume. How many of these bottles of perfume could be made from 96 ounces of perfume?", "Solution_1": "[quote=\"GameBot\"]Each bottle of perfume at The Perfume Parlor contains $ \\frac3{16}$ ounces of perfume. How many of these bottles of perfume could be made from 96 ounces of perfume?[/quote]\r\n\r\nOur answer is $ 96 \\div \\frac{3}{16} \\equal{} \\boxed{512}$." } { "Tag": [ "geometry", "rectangle", "geometry solved" ], "Problem": "show that for every integer $n\\ge 6$ , there exists a convex hexagon wich can be dissected into exactly $n$ congruent triangles.\r\n\r\n[color=red][Moderator edit: See also http://www.kalva.demon.co.uk/apmo/asoln/asol905.html for this problem.][/color]", "Solution_1": "trivial. and an old problem. (I remember doing it 4-5 years ago)\r\nThe trick is that you can use right angled triangles with sides a, b and (na) with $\\left(na\\right)^2=a^2+b^2$\r\n(you only require n to be integer $\\geq 2$) to make a rectangle of sides b and na. you need 2n triangles for this. \r\nthen you add two of these triangles to the sides to get a convex hexagon.\r\nso we can do it for any $\\left(2n+2\\right)\\geq 6$ triangles.\r\n-Ali" } { "Tag": [ "floor function", "inequalities", "Euler", "integration", "calculus", "function", "arithmetic sequence" ], "Problem": "Let $ A \\equal{} \\sum_{k \\equal{} 1}^{10^{289}}{\\frac {1}{\\sqrt [17]{k}}}$ \r\n\r\nDetermine the integer part of $ A$\r\n\r\n\r\n :)", "Solution_1": "i believe $ \\lfloor A \\rfloor \\equal{} \\frac {17}{16}\\cdot 10^{17\\cdot 16} \\minus{} 1$.", "Solution_2": ":lol: Yes that's the correct answer.\r\n\r\nI wonder if your method is the same as mine???\r\n\r\nWhat did you d0?\r\n\r\nI was working with this kind of problem, but with square and cubic roots so I generalized an inequality wich works now for every real greater than 1, and of the form $ \\frac {1}{p}$ where $ p$ is an integer greater than one.\r\n\r\n\r\n :)", "Solution_3": "A straightforward solution would be just to bash the problem with Euler-Maclaurin (Note that the term is almost the same as $ \\int_1^{10^{289}} \\frac{1}{\\sqrt[17]{x}}\\,\\mathrm dx$) and estimate the remainder, which is most likely less than one.", "Solution_4": "Here's another interesting problem:\r\n\r\nFind all $ (a,n,m)\\in(\\mathbb{Z}^{+})^3$ for wich the integer part of $ B=\\sum_{k=1}^{n^{m(a+1)}}{\\frac{1}{k^{\\frac{a}{a+1}}}}$ is a multiple of $ 2007$.\r\n\r\n\r\n :)", "Solution_5": "I don't know much about calculus so I haven't read with much attention bertram's post, but suddenly I realized that my inequality give the same result as that integral, in fact my inequality is like an integration, so this nice result came to my mind:\r\n\r\nLet $ f$ be a continuous, decreasing function in the interval $ [a,b]$ wich contains $ x$, $ x \\plus{} \\alpha$ and $ x \\minus{} \\alpha$.\r\n\r\n$ \\frac {f(x \\plus{} \\alpha) \\plus{} f(x)}{2} \\le f(x) \\le \\frac {f(x) \\plus{} f(x \\minus{} \\alpha)}{2} \\Leftrightarrow \\frac {1}{\\alpha}(\\alpha\\frac {f(x \\plus{} \\alpha) \\plus{} f(x)}{2}) \\le f(x) \\le \\frac {1}{\\alpha}(\\frac {f(x) \\plus{} f(x \\minus{} \\alpha)}{2})$\r\n\r\nIf the function is convex:\r\n$ \\frac {1}{\\alpha}\\int_x^{x \\plus{} \\alpha}{f(x) \\mathrm dx}\\le\\frac {1}{\\alpha}(\\alpha\\frac {f(x \\plus{} \\alpha) \\plus{} f(x)}{2})\\le f(x)$\r\n\r\nSimilar if $ f$ is concave, increasing..., when you have an arithmetic progression, and you sum this inequalitys for consecutive values of the progression you are like telescoping so it may be very useful for certain functions and values, as the problems posted before.\r\n\r\nFrom this point the only advantage that my inequality has is that it gives and upper and lower bound for the same function, and it can be easily proved by AM GM.\r\n\r\nHere it is:\r\n\r\n$ \\frac {k}{k \\minus{} 1}((n \\plus{} 1)^{\\frac {k \\minus{} 1}{k}} \\minus{} n^{\\frac {k \\minus{} 1}{k}}) < \\frac {1}{n^{\\frac {1}{k}}} < \\frac {k}{k \\minus{} 1}(n^{\\frac {k \\minus{} 1}{k}} \\minus{} (n \\minus{} 1)^{\\frac {k \\minus{} 1}{k}})$\r\n\r\nI hope someone replies to the problems, with a complete solution.\r\n\r\n :)" } { "Tag": [], "Problem": "\u03c3\u03b7\u03bc\u03b5\u03c1\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03b1 \u03b3\u03b5\u03bd\u03b5\u03b8\u03bb\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03bf\u03c5 \u03c6\u03bf\u03c1\u03bf\u03c5\u03bc.\u0395\u039d\u0391\u03a3 \u03c7\u03c1\u03bf\u03bd\u03bf\u03c2 \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03bf....\u03ba\u03b1\u03bb\u03b7 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b5\u03b9\u03b1 \u03c3\u03b5 \u03bf\u03bb\u03bf\u03c5\u03c2 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd.... :clap:", "Solution_1": "\u03a7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac ! :lol:", "Solution_2": "\u03c0\u03cc\u03c4\u03b5 \u03b8\u03b1 :gathering: \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf \u03b3\u03b9\u03bf\u03c1\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5?\r\n\u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac...", "Solution_3": "\u0398\u03b1\u03bd\u03bf ,\u03c1\u03b5 \u03c6\u03b9\u03bb\u03b5 \u03bc\u03b1\u03ba\u03b1\u03c1\u03b9 \u03bd\u03b1 \u03b5\u03b9\u03c7\u03b1\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03b7\u03bb\u03b5\u03ba\u03c4\u03c1\u03bf\u03bd\u03b9\u03ba\u03b7 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1...\u03ae \u03bc\u03b7\u03c0\u03c9\u03c2 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03b1\u03ba\u03c1\u03b1\u03b9\u03bf\u03c2? :P", "Solution_4": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c6\u03af\u03bb\u03b5\u03c2 \u03ba\u03b9 \u03c6\u03af\u03bb\u03bf\u03b9 \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03ae\u03bc\u03b5\u03c1\u03b1 4/9/06 \u03ad\u03bd\u03b1\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u0395\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03cc \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c4\u03bf\u03c5 \u03c0\u03bf\u03bb\u03bb\u03ac \u03ba\u03b9 \u03c0\u03c1\u03b9\u03bd \u03ba\u03bb\u03b5\u03af\u03c3\u03c9 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1 \u03b8\u03b1 \u03ae\u03b8\u03b5\u03bb\u03b1 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03ce \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc\u03bd \u03c4\u03bf\u03bd 1 \u03c7\u03c1\u03cc\u03bd\u03bf \u03bf \u03bf\u03c0\u03bf\u03af\u03bf\u03c2 \u03c6\u03b1\u03bd\u03c4\u03ac\u03b6\u03bf\u03bc\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03b1\u03c2 \u03ae\u03c4\u03b1\u03bd \u03b5\u03c0\u03bf\u03b9\u03ba\u03bf\u03b4\u03bf\u03bc\u03b7\u03c4\u03b9\u03ba\u03cc\u03c2 \u03a0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03ae\u03c4\u03b1\u03bd \u03ad\u03bd\u03b1\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03bc\u03b5 \u03ba\u03b1\u03bb\u03ac \u03c3\u03c4\u03bf\u03b9\u03c7\u03b5\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03b2\u03ac\u03b8\u03bc\u03b9\u03c3\u03b1\u03bd \u03c4\u03bf \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03cc \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03cc\u03c0\u03c9\u03c2 1 \u0397 \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 \u03ba\u03b9 \u03b7 \u03c0\u03c1\u03bf\u03c3\u03ae\u03bb\u03c9\u03c3\u03b7 \u03c4\u03c9\u03bd \u03bc\u03b5\u03bb\u03ce\u03bd \u03c4\u03bf\u03c5 \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b1\u03be\u03b9\u03bf\u03b8\u03b1\u03cd\u03bc\u03b1\u03c3\u03c4\u03b7 \u03ba\u03b9 2 \u03b7 \u03b4\u03b7\u03bc\u03b9\u03bf\u03c5\u03c1\u03b3\u03af\u03b1 \u03c4\u03bf\u03c5 \u03c3\u03b1\u03b9\u03c4 \u03c4\u03bf\u03c5 \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c5 \u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03c3\u03b7\u03bc\u03b1\u03c4\u03bf\u03b4\u03bf\u03c4\u03b5\u03af \u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03cd\u03c6\u03c9\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03c5\u03c0\u03b5\u03c1\u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1\u03c2 \u03b1\u03c0\u03cc \u03bc\u03ad\u03c1\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5 \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b4\u03b9\u03ac\u03b4\u03bf\u03c3\u03b7 \u03c9\u03c6\u03b5\u03bb\u03af\u03bc\u03c9\u03bd \u03b2\u03b9\u03b2\u03bb\u03af\u03c9\u03bd \u03b3\u03b9\u03b1 \u039f\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 \u03a4\u03ad\u03bb\u03bf\u03c2 \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03b5\u03c5\u03c7\u03b7\u03b8\u03ce \u03ba\u03b1\u03bb\u03ae \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03b5 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03b1 \u03ba\u03b9 \u03c0\u03c1\u03bf\u03c3\u03ae\u03bb\u03c9\u03c3\u03b7 \u03ba\u03b9 \u03b8\u03ad\u03bb\u03c9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03bd\u03b1 \u03c0\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b5\u03b9\u03c3\u03c6\u03bf\u03c1\u03ac \u03ae\u03c4\u03b1\u03bd \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b5\u03c1\u03b9\u03c3\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd \u03ba\u03b9 \u03c5\u03c0\u03cc\u03c3\u03c7\u03bf\u03bc\u03b1\u03b9 \u03bd\u03b1 \u03b2\u03b5\u03bb\u03c4\u03b9\u03c9\u03b8\u03ce \r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03ba\u03b9 \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac \u03be\u03b1\u03bd\u03ac\r\n\u039c\u03b5\u03c4\u03ac \u03c4\u03b9\u03bc\u03ae\u03c2\r\n\u0391\u03bc\u03b2\u03c1\u03cc\u03c3\u03b9\u03bf\u03c2 \u03a3\u03b1\u03b2\u03b2\u03af\u03b4\u03b7\u03c2 (ambros)\r\nps \u03c5\u03c0\u03b5\u03c1\u03b2\u03ac\u03bb\u03bb\u03b5\u03b9\u03c2 \u03ba\u03ce\u03c3\u03c4\u03b1", "Solution_5": "\u03a0\u03c9\u03c2 \u03c3\u03b1\u03c2 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 :P :D", "Solution_6": "|\r\n \u0399 \u0399\r\n \u0399 \u0399\r\n_______________\u0399_\u0399______________\r\n| |\r\n| |\r\n| aops greek forum |\r\n| |\r\n| |\r\n| |\r\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\r\n\r\n\u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03b1\u03bd\u03ac\u03c8\u03b5\u03b9 \u03c4\u03bf \u03ba\u03b5\u03c1\u03ac\u03ba\u03b9...\r\n\u03c4\u03b9\u03c0\u03bf\u03c4\u03b1 \u03ac\u03bb\u03bb\u03bf \u03b8\u03ad\u03bb\u03b5\u03c4\u03b5?", "Solution_7": "\u03a7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf \u03b5\u03bc\u03b5\u03bd\u03b1,\u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03b5\u03bd\u03b1\u03bd \u03c7\u03c1\u03bf\u03bd\u03bf \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03bf forum!!!!\r\nA,\u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03b1\u03c4\u03b9 \u03b1\u03bb\u03bb\u03bf..... 7 \u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 alter!!!! :rotfl: \r\n\u0391,\u03c9\u03c1\u03b1\u03b9\u03b1 \u03b7 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03b5\u03b9\u03c3\u03c4\u03b5 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03bf\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b3\u03b5\u03bd\u03b5\u03b8\u03bb\u03b9\u03c9\u03bd,\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b5\u03bc\u03b5\u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b1\u03c0\u03bf\u03ba\u03c1\u03b9\u03b1\u03c2...... :rotfl: \r\n\r\n\u03a1\u0391\u039b\u039b\u0397\u03a3 :D", "Solution_8": "[quote=\"RAKAR\"]\u03a7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03bb\u03bf\u03b9\u03c0\u03bf\u03bd \u03ba\u03b1\u03b9 \u03b1\u03c0\u03bf \u03b5\u03bc\u03b5\u03bd\u03b1,\u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03b5\u03bd\u03b1\u03bd \u03c7\u03c1\u03bf\u03bd\u03bf \u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03bf forum!!!!\nA,\u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03b1\u03c4\u03b9 \u03b1\u03bb\u03bb\u03bf..... 7 \u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 alter!!!! :rotfl: \n\u0391,\u03c9\u03c1\u03b1\u03b9\u03b1 \u03b7 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b1\u03bb\u03bb\u03b1 \u03b5\u03b9\u03c3\u03c4\u03b5 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03bf\u03c4\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b3\u03b5\u03bd\u03b5\u03b8\u03bb\u03b9\u03c9\u03bd,\u03b3\u03b9\u03b1\u03c4\u03b9 \u03b5\u03bc\u03b5\u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c1\u03c4\u03b1 \u03b1\u03c0\u03bf\u03ba\u03c1\u03b9\u03b1\u03c2...... :rotfl: \n\n\u03a1\u0391\u039b\u039b\u0397\u03a3 :D[/quote]\r\n\r\n :D :lol:", "Solution_9": "\u03b3\u03b9\u03b1 \u03c0\u03bf\u03b9\u03b1 \u03c4\u03bf\u03cd\u03c1\u03c4\u03b1 \u03bc\u03b9\u03bb\u03ac\u03c2???", "Solution_10": "\u03c1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9\u03b1 \u03b5\u03b9\u03c3\u03c4\u03b5 \u03c5\u03c0\u03b5\u03c1\u03bf\u03c7\u03bf\u03b9. \u03c1\u03b5 \u03b1\u03c0\u03bf\u03c3\u03c4\u03bf\u03bb\u03b7 \u03bc\u03c0\u03c1\u03b1\u03b2\u03bf...\u03b1\u03c5\u03c4\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 7 \u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03b1\u03bb\u03c4\u03b5\u03c1 \u03bd\u03b1\u03b9\u03b9\u03b9\u03b9\u03b9\u03b9\u03b9...\u03c7\u03b1\u03c7\u03c7\u03b1\u03c7\u03b1\u03c7\u03b1\u03c7\u03b1 \u03ba\u03b1\u03b9, \u03bb\u03b9\u03b3\u03b7 \u03bc\u03bf\u03c5\u03c3\u03b9\u03ba\u03b7 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 :lol: :rotfl:", "Solution_11": "na zisis aopsgf kai y polla\r\noo na gineis me topic polla\r\nktl", "Solution_12": "\u03ba\u03c1\u03b9\u03bc\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b5\u03b9\u03bc\u03b1\u03b9 \u03b5\u03ba\u03b5\u03b9\u03bd\u03b7 \u03c4\u03b7\u03bd \u03b7\u03bc\u03b5\u03c1\u03b1 \u03b5\u03ba\u03b5\u03b9.\u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03b1\u03b6\u03b5\u03b9 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03c4\u03c9\u03c1\u03b1 \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd.\u03c6\u03b1\u03bd\u03c4\u03b1\u03c3\u03c4\u03b5\u03b9\u03c4\u03b5 \u03bf\u03c4\u03b9 \u03c4\u03bf \u03b2\u03c1\u03b9\u03ba\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b8\u03bf\u03c2 \u03c4\u03bf \u03bc\u03b1\u03b8\u03bb\u03b9\u03bd\u03ba\u03c2.\u03bc\u03b5\u03c3\u03c9 \u03b1\u03bd\u03b1\u03b6\u03b7\u03c4\u03b7\u03c3\u03b7\u03c2 :o \u03c6\u03b1\u03bd\u03c4\u03b1\u03b6\u03bf\u03bc\u03b1\u03b9 \u03bf\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03bf\u03c4\u03b5\u03c1\u03bf\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03b8\u03b1 \u03c4\u03bf \u03b2\u03c1\u03b7\u03ba\u03b1\u03bd.\u03bf \u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03bf\u03c2 \u03b5\u03ba\u03b1\u03bd\u03b5 \u03ba\u03b1\u03c4\u03b9 \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03bf:\u03b5\u03bd\u03b1 \u03b1\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03bc\u03bf....\u03c3\u03c9\u03c3\u03c4\u03bf\u03c2 \u03bf\u03bc\u03c9\u03c2 \u03b5\u03ba\u03b5\u03b9 \u03c0\u03bf\u03c5 \u03bc\u03b9\u03bb\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03b5\u03b1\u03c5\u03c4\u03bf \u03c4\u03bf\u03c5 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03b4\u03b9\u03ba\u03bf\u03c2.\u03c0\u03b9\u03c3\u03c4\u03b5\u03c8\u03b5 \u03bc\u03b5 \u03c1\u03b5 \u03bc\u03b5\u03b3\u03b1\u03bb\u03b5 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03b4\u03c9\u03c3\u03b5\u03b9 \u03c0\u03bf\u03bb\u03bb\u03b1 .\u0391\u03bc\u03b2\u03bf\u03c3\u03b9\u03bf\u03c2 for ever !", "Solution_13": "\u03a7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03b1 \u03c7\u03b9\u03bb\u03b9\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5!!!!!!!!!!!!\r\n :thumbup: \r\n\r\n\u03a0\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03ba\u03cc\u03c8\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03bf\u03cd\u03c1\u03c4\u03b1; :lol:\r\n\r\nP.S.\u03a1\u03ac\u03bb\u03bb\u03b7 \u03b5\u03af\u03b4\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c3\u03bf\u03c5 \u03ad\u03bb\u03b5\u03b3\u03b1 \"\u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac\".\u039a\u03ac\u03c4\u03b9 \u03ae\u03be\u03b5\u03c1\u03b1...! :wink:\r\n\r\nEdit:\u039f\u03c5\u03c6,\u03c3\u03c4\u03bf \u03c4\u03c3\u03b1\u03c6 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b1! 12:00 am \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03ce\u03c1\u03b1! :P", "Solution_14": "Kathws i giorti krataei (opws lene) 40 meres, kathws den exw alli dikaiologia gia tin kathisterisi mou, kai telos epeidi stous fisikous vazw kai to 0 (ara simera kleinei 1 xrono to forum) thelw na euxithw sto elliniko forum xronia polla kai sinexws na anavathmizetai apo idees, atoma kai askiseis. Auto pou euxomai omws olopsixa einai na voithisei tous ellines mathites gia na anavathmistei i olimpiaki paideia sti xwra mas... as min perimenoume tous \"megalous\" na kanoun kiniseis gi'auto... as prospathisei o kathenas mathitis monos tou na parei ta efodia gia na sinexisei!! Se auti tin prospatheia tou, oloi emeis tha stathoume sinodiporoi!!\r\n\r\nNa eiste kala,\r\nAlexandros", "Solution_15": "\u0391\u03b3\u03b1\u03c0\u03b7\u03c4\u03bf\u03af \u03c6\u03af\u03bb\u03bf\u03b9 \u03ad\u03c7\u03b5\u03c4\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03b1\u03bd\u03b5\u03be\u03b1\u03b9\u03c1\u03ad\u03c4\u03c9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03b5\u03ba\u03c0\u03bb\u03b7\u03ba\u03c4\u03b9\u03ba\u03ae \u03b7 \u03c4\u03bf\u03cd\u03c1\u03c4\u03b1 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03ba\u03ce\u03c3\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03bb\u03ac \u03c3\u03bf\u03c5 \u03bb\u03cc\u03b3\u03b9\u03b1 \u03ba\u03b9 \u03b8\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03c9 \u03bc\u03b5 \u03c0\u03b9\u03bf \u03cc\u03c1\u03b5\u03be\u03b7 \r\n\u03c7\u03b1\u03b9\u03c1\u03b5\u03c4\u03ce", "Solution_16": "\u039d\u03b1 \u03c0\u03c9 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03c4\u03b1 \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac \u03c3\u03c4\u03bf <<\u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03ac>> .\u0389\u03c4\u03b1\u03bd \u03ad\u03bd\u03b1\u03c2 \u03c0\u03bf\u03bb\u03c5\u03c4\u03ac\u03c1\u03b1\u03c7\u03bf\u03c2 \u03c7\u03c1\u03cc\u03bd\u03bf\u03c2 \u03b8\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b1 \u03bd\u03b1 \u03c0\u03c9 .\u039e\u03b5\u03ba\u03b9\u03bd\u03ae\u03c3\u03b1\u03bc\u03b5 \u03bc\u03b5 \u03b5\u03bd\u03b8\u03bf\u03c5\u03c3\u03b9\u03b1\u03c3\u03bc\u03cc \u03bc\u03b5\u03c4\u03ac \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03ce\u03c1\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 6 \u03bc\u03ae\u03bd\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c0\u03ac\u03bb\u03b9 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03cc \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf .\u0388\u03c7\u03bf\u03c5\u03bd \u03c3\u03c5\u03b6\u03b7\u03c4\u03b7\u03b8\u03b5\u03af \u03c6\u03bf\u03b2\u03b5\u03c1\u03ad\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 , \u03ad\u03c7\u03bf\u03c5\u03bd \u03bb\u03c5\u03b8\u03b5\u03af \u03b1\u03c0\u03cc \u03c3\u03c0\u03bf\u03c5\u03b4\u03b1\u03af\u03b1 \u03bc\u03ad\u03bb\u03b7 (\u03bc\u03b7\u03bd \u03be\u03b5\u03c7\u03bd\u03ac\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03bf\u03cd\u03c2 medalist \u0399\u039c\u039f \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03b8\u03ad\u03bd\u03c4\u03b5\u03c2 \u03c3\u03b5 \u0399\u039c\u039f) \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03bf\u03c5\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c5\u03c2 \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03bc\u03ad\u03c4\u03c1\u03b9\u03bf\u03c5\u03c2 (\u03c0.\u03c7 \u03b5\u03b3\u03ce) \u03c0\u03bf\u03c5 \u03cc\u03bc\u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03bf\u03cd\u03bc\u03b5 \u03bc\u03c0\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c4\u03b9 .\r\n\u0391\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ad\u03c4\u03c3\u03b9 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae \u03cc\u03bc\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b5\u03b8\u03cc\u03b4\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03cc\u03bb\u03bf\u03b9 \u03c4\u03b9\u03c2 \u03b5\u03c0\u03b9\u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5 .\r\n\u0398\u03b1 \u03ae\u03c4\u03b1\u03bd \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03ba\u03b1\u03bb\u03cc \u03bd\u03b1 \u03b3\u03af\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bb\u03af\u03b3\u03bf \u03c0\u03b9\u03bf \u03bf\u03c1\u03b3\u03b1\u03bd\u03c9\u03bc\u03ad\u03bd\u03bf\u03b9 (\u03c7\u03c9\u03c1\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03c3\u03b5 \u03c5\u03c0\u03bf-\u03c6\u03cc\u03c1\u03bf\u03c5\u03bc )(\u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b6\u03b7\u03c4\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd Vornicu ).\r\nNa \u03b5\u03af\u03c3\u03c4\u03b5 \u03ba\u03b1\u03bb\u03ac \u03cc\u03bb\u03bf\u03b9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03b5\u03c0\u03b9\u03c4\u03c5\u03c7\u03af\u03b1 \r\n\r\n\u03a5.\u0393. \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bf \u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03bf \u039d\u03af\u03ba\u03bf\u03c2 (\u03a1\u03ac\u03c0\u03b1\u03bd\u03bf\u03c2 ) \u03bd\u03b1 \u03b3\u03c1\u03b1\u03c6\u03bf\u03c5\u03bd \u03ba\u03ac\u03c4\u03b9 \u03c4\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 .", "Solution_17": "\u03ba\u03b9 \u03ad\u03bd\u03b1\u03c2 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf\u03c2\r\n\r\n>\u03c7\u03c1\u03ae\u03c3\u03c4\u03bf\u03c2\r\n>\u03b8\u03b5\u03c3\u03c3\u03b1\u03bb\u03bf\u03bd\u03af\u03ba\u03b7", "Solution_18": "sorry \u03be\u03ad\u03c7\u03b1\u03c3\u03b1 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 upload... enjoy!\r\nchristinho \u03b1\u03bd \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 topic \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03c5\u03c3\u03c4\u03b7\u03b8\u03b5\u03af\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf\u03bd \u03ba\u03b1\u03c4\u03ac\u03bb\u03bf\u03b3\u03bf \u03c4\u03c9\u03bd \u03bc\u03b5\u03bb\u03ce\u03bd. \u0391\u03c0\u03bb\u03ce\u03c2 \u03c3\u03c4\u03b5\u03af\u03bb\u03b5 \u03ad\u03bd\u03b1 post. \u0394\u03b5 \u03c3\u03b5 \u03b2\u03ac\u03b6\u03c9 \u03bc\u03b5 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03c0\u03c1\u03c9\u03c4\u03bf\u03b2\u03bf\u03c5\u03bb\u03af\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03b7\u03bd \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c6\u03bf\u03c1\u03ac \u03c0\u03bf\u03c5 \u03c4\u03bf \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ad\u03bc\u03c0\u03bb\u03b5\u03be\u03b1 (\u03bc\u03b5 M stedes)", "Solution_19": "\u03a9\u03c1\u03b1\u03af\u03b1 \u03c4\u03bf\u03cd\u03c1\u03c4\u03b1! :rotfl: \r\n\u0395\u03c3\u03cd \u03c4\u03b7\u03bd \u03ad\u03c6\u03c4\u03b9\u03b1\u03be\u03b5\u03c2; \u039a\u03b9 \u03b5\u03ba\u03b5\u03af\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9; \u0395\u03bb\u03b9\u03ad\u03c2; :rotfl:", "Solution_20": "\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9! \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03ac\u03c4\u03b9 \u03c3\u03b1\u03bd \u03bc\u03c0\u03b9\u03c3\u03ba\u03bf\u03c4\u03ac\u03ba\u03b9\u03b1", "Solution_21": "\u039c\u03c0\u03b9\u03c3\u03ba\u03bf\u03c4\u03ac\u03ba\u03b9\u03b1 \u03bc\u03bc\u03bc, \u03bc\u03bf\u03c5 \u03ac\u03bd\u03bf\u03b9\u03be\u03b5\u03c2 \u03c4\u03b7\u03bd \u03cc\u03c1\u03b5\u03be\u03b7! \r\n\u0391\u03bb\u03ae\u03b8\u03b5\u03b9\u03b1,\u03c4\u03bf \u039c \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5\u03c2 \u03c0\u03b9\u03bf \u03c0\u03ac\u03bd\u03c9 \u03c4\u03b9 \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9;;;; :maybe:", "Solution_22": "\u03c4\u03bf F \u03c4\u03b9 \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9?\r\n[color=#f8f8f8]http://home.comcast.net/~wolfand/[/color]", "Solution_23": "\u0394\u03b5\u03bd \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bd\u03b1 \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03c3\u03b5\u03c2 \u03b1\u03c5\u03c4\u03cc \u03cc\u03c4\u03b1\u03bd \u03c4\u03bf \u03ad\u03b3\u03c1\u03b1\u03c6\u03b5\u03c2.\r\n[color=#f0f0f0]http://www.humor.gr/modules.php?name=News&file=article&sid=45[/color] :mad:", "Solution_24": "giati??????", "Solution_25": "\u0393\u03b9\u03b1\u03c4\u03af \u03b1\u03c0\u03bb\u03ac \u03b4\u03b5\u03bd \u03ba\u03bf\u03bb\u03bb\u03ac\u03b5\u03b9 \u03c3'\u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2.", "Solution_26": "\u03ad\u03bc\u03c0\u03bb\u03b5\u03be\u03b1 \u03bc\u03b5 \u03c3\u03ad\u03bd\u03b1... \u0394\u0395\u039c?", "Solution_27": "\u03a6\u039a\u0395!\r\n\u0391\u03bb\u03bb\u03ac \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03bf \u03ad\u03b3\u03c1\u03b1\u03c8\u03b5\u03c2;", "Solution_28": "\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03bc\u03c0\u03bb\u03ad\u03be\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03ac\u03bb\u03bb\u03bf\u03b9... :rotfl: \r\n\u039a\u039d&SD \u03c4\u03ce\u03c1\u03b1, \u03bf\u03ba? bye!", "Solution_29": "Ok, KN!\r\n\r\n&sd" } { "Tag": [ "LaTeX" ], "Problem": "$\\triangle ABC$ is a right triangle with $\\angle A=90^\\circ$. $K$, $L$ are the points outside of $\\triangle ABC$ such that $\\triangle KAB$ and $\\triangle LAC$ are equilateral triangles.\r\n\r\nLet $M$ be the intersection of $KC$ and $LB$. Find $\\angle KML$.\r\n\r\nEDIT- Since my previous figure was a little bit confusing, I attached a new picture. ($AB$ is not necessarily equal to $AC$.)", "Solution_1": "Since KA=BA, LA=CA, and angle KAC=LAB (Both are equal to 60+90), triangles KAC and LAB are congruent. \r\nSince angle AKC=ABL, quadrilateral AKBM is concylic.\r\nThen we get angle BMK=BAK=60. So, angle KML is 180-60=120", "Solution_2": "Is it $120$ degrees? The way I did it was that since triangle $ALC$ is equilateral, all the angles are $60$, therefore, angle $LAB$ was $150$ and angle $ALB$ and angle $ABL$ are $15$, and then since angle $ABC$ and $ACB$ are equal, they both are $45$, so then angle $MBC$ and $MCB$ are each $30$ and so angle $BMC$ is $120$ and using verticle angles, angle $KML=\\boxed{120}$.\r\n\r\n\r\nP.S. What's the latex code for angles?", "Solution_3": "You can't assume that the two angles of the right triangle are equal, cuz it might be a 30,60,90 triangle, for example. But 120 is right.", "Solution_4": "[quote=\"krustyteklown\"]You can't assume that the two angles of the right triangle are equal, cuz it might be a 30,60,90 triangle, for example. But 120 is right.[/quote]\r\n\r\nI assumed it was a $45,45,90$ triangle because triangles $KAC$ and $BAL$ were congruent using the diagram.", "Solution_5": "[quote=\"Iversonfan2005\"]\nP.S. What's the latex code for angles?[/quote]\r\n\r\nit's \\angle $\\angle$\r\nfor degrees you can use ^{\\circ}, EX: $120^{\\circ}$\r\n\r\nalso i do not understand how you can make such an assumption that the right triangle is 45-90-45. or anything else....\r\n\r\nthere is nothing that confirms that, and you are not allowed to do such things in math, only because the diagram looks like it.. :huuh:", "Solution_6": "The educated guessing strategy is much faster than all of your slow solutions :P It only takes $\\le 3$ seconds.\r\n\r\nFirst notice that it's talking about equilateral triangles. Then notice that it's an obtuse angle. No other angle measurements are given. $120^{\\circ}$ comes to mind. :D \r\n\r\nActually, that's a good strategy before you start, so that you'll have an $\\approx$ for the answer.", "Solution_7": "Sorry that my picture caused some confusions :oops: \r\nBut anyway, $120^\\circ$ is correct.\r\n\r\nEDIT- I changed my picutre now. :)", "Solution_8": "When I did the problem, I also thought \"Well, let's just say the right triangle is 45-90-45...\"", "Solution_9": "[quote=\"krustyteklown\"]Then we get angle BMK=BAK=60.[/quote]\r\n\r\nWhere did that come from? :? :blush:", "Solution_10": "It's from the cyclic-ness of AKMB", "Solution_11": "[quote=\"nr1337\"]When I did the problem, I also thought \"Well, let's just say the right triangle is 45-90-45...\"[/quote]\r\nThat's actually a good strategy especially if the problem only asks the answer, not complete solution :laugh:", "Solution_12": "[quote=\"krustyteklown\"]It's from the cyclic-ness of AKMB[/quote]\r\n\r\nOkay I see...I'm inexperienced with cyclicness. Can you explain how you proved that it was cyclic? :blush:", "Solution_13": "Since angle AKC=ABL, quadrilateral AKBM is concylic.", "Solution_14": "We don't need the right angle. The conclution is true for any triangles.", "Solution_15": "[quote=\"krustyteklown\"]Since angle AKC=ABL, quadrilateral AKBM is concylic.[/quote]\r\n\r\nWell yeah, I still don't understand how that proves anything. ;)", "Solution_16": "[quote=\"shobber\"]We don't need the right angle. The conclution is true for any triangles.[/quote]\r\n\r\nNice, I didn't see that, although I didn't use the right angle in my proof, hehe.", "Solution_17": "There are more solution paths given that $ABC$ is a right triangle. My original method, before I read the replies to the question, did utilize the right angle. Since I didn't see that triangles $KAC$ and $BAL$ were equilateral, I said that $K$, $A$, and the midpoint of $CL$ were collinear (beacause $60^\\circ + 90^\\circ + 30^\\circ = 180^\\circ$). $KA$ is perpendicular to $CL$ because triangle $ACL$ is equilateral, so the altitude and median to side $CL$ will be concurrent. Hence, triangle $KCL$ is isoceles, making $\\angle KLA = \\angle ACK$. Since triangle $KBL$ is isoceles for similar reasons, $\\angle KLA = \\angle ALB = \\angle ACK$. The latter two of these angles being equal tells us that quadrilateral $AMCL$ is cyclic, so $\\angle LAC = \\angle LMC = 60^\\circ$, which tells us that $\\angle KML = 180^\\circ - 60^\\circ = 120^\\circ$." } { "Tag": [], "Problem": "is there a formula for a question where there r 6 people standing in a line and 3 wont stand next to each other. how many ways are there for them to stand", "Solution_1": "yeah there is\r\n\r\nso, first you do the total amount of ways that they could go in order which is\r\n\r\n6! or 6*5*4...*1 or 720\r\n\r\nThen, you find the amount of ways that they could stand together for this, pretend that the three are one block, so pretend there are 4 people, so 4! is 24. But, the 3 could be in different orders within their group, so it is \r\n\r\n4!3! or 144\r\n\r\nThen you do 720-144, or 576.\r\n\r\nEdit.", "Solution_2": "I'm quite sure 4!3! doesn't equal 140. :lol:\r\n\r\nSorry if this is spam.", "Solution_3": "haha, oops a typo. :blush:", "Solution_4": "how i would do it is\r\nthere are 3 people that can stand anywhere in 6 ways. the other 3 people can go inbetween them in 4 ways. so 6*4=24\r\n\r\ndid i misread it or somethhing? can two be together but just not all 3?", "Solution_5": "u got the first part right, with 24, but remember, they can be ordered in different ways inside of their 3-person group. so it is 24*3! or 144.\r\n\r\nOne other thing that you missed is the 3 are NOT supposed to be together, not that they are.", "Solution_6": "if the \"o\"s wont go together then you could have:\r\n\r\nononon\r\nononno\r\nonnono\r\nnonono\r\n\r\nand both the n's and the o's could be arranged 6 ways so 4*6*6=144" } { "Tag": [ "inequalities", "function", "inequalities unsolved" ], "Problem": "let aa,b,c be positeve real number prove that \r\n \\sum a 2 (b+c) \\geq 2 (a 2 b 2 c 2 )^1/3*(a+b+c) :? :(", "Solution_1": "By Cauchy\r\na 2 b+a 2 c+abc \\geq 3(a^5*b^2*c^2)^(1/3)\r\nThe similar \\sum (sym)a 2 b+3abc \\geq 3( \\sum a^5*b^2*c^2)^(1/3) =2( \\sum a^5*b^2*c^2)^(1/3)+( \\sum a^5*b^2*c^2)^(1/3) \\geq 2( \\sum a^5*b^2*c^2)^(1/3)+3abc\r\n<=>Thanhliem's inequality.\r\nIs it right?????", "Solution_2": "Can you tell me the source of the problem ?\r\nI think you should tell more clearly about it .\r\nIf you don't know the source , write Unknown.", "Solution_3": "[quote=\"Minh Thang\"]By Cauchy\na 2 b+a 2 c+abc \\geq 3(a^5*b^2*c^2)^(1/3)\nThe similar \\sum (sym)a 2 b+3abc \\geq 3( \\sum a^5*b^2*c^2)^(1/3) =2( \\sum a^5*b^2*c^2)^(1/3)+( \\sum a^5*b^2*c^2)^(1/3) \\geq 2( \\sum a^5*b^2*c^2)^(1/3)+3abc[/quote]\r\n\r\nI think this should be\r\n\r\nThe similar \\sum (sym)a 2 b+3abc \\geq 3 \\sum (a^5*b^2*c^2)^(1/3) =2 \\sum (a^5*b^2*c^2)^(1/3)+ \\sum (a^5*b^2*c^2)^(1/3) \\geq 2 \\sum (a^5*b^2*c^2)^(1/3)+3abc.\r\n\r\nNice solution indeed (and yes, it's true). Somehow I failed to note your reply two months ago and kept searching for a solution on my own. Here is what I came up with:\r\n\r\nWe have to show the inequality\r\n\r\n$a^{2}\\left( b+c\\right) +b^{2}\\left( c+a\\right) +c^{2}\\left( a+b\\right) \\geq 2\\root{3}\\of{a^{2}b^{2}c^{2}}\\cdot \\left( a+b+c\\right) $\r\n\r\nfor all positive numbers a, b, c.\r\n\r\nThis is obviously equivalent to $f\\left( a,b,c\\right) \\geq 0$, where the function f(a, b, c) is defined as follows:\r\n\r\n[tex]f\\left( a,b,c\\right) =\\frac{a^{2}\\left( b+c\\right) +b^{2}\\left( c+a\\right) +c^{2}\\left( a+b\\right) }{a+b+c}-2\\root{3}\\of{a^{2}b^{2}c^{2}}[/tex].\r\n\r\nNow, we want to reduce the problem of proving $f\\left( a,b,c\\right) \\geq 0$ to a two-variable inequality by \"mixing variables\". The best way to do so is to replace both of the variables b and c by their geometric mean $\\sqrt{bc}$; in fact, if we compute the difference $f\\left( a,b,c\\right) -f\\left( a,\\sqrt{bc},\\sqrt{bc}\\right) $, we see that the nasty $\\root{3}\\of{...}$ radicals cancel each other. Also, it's wise to consider $a^2$, $b^2$, $c^2$ instead of a, b, c here, so that the $\\sqrt{...}$ radicals disappear, too. Here is the solution in detail:\r\n\r\nInstead of proving $f\\left( a,b,c\\right) \\geq 0$, we will show $f\\left( a^{2},b^{2},c^{2}\\right) \\geq 0$. This is, of course, the same, since a, b, c are assumed positive.\r\n\r\nWe have\r\n\r\n[tex]f\\left( a^{2},b^{2},c^{2}\\right) =\\frac{a^{4}\\left( b^{2}+c^{2}\\right) +b^{4}\\left( c^{2}+a^{2}\\right) +c^{4}\\left( a^{2}+b^{2}\\right) }{a^{2}+b^{2}+c^{2}}-2\\root{3}\\of{a^{4}b^{4}c^{4}}[/tex].\r\n\r\nWe will calculate the difference $f\\left( a^{2},b^{2},c^{2}\\right) -f\\left( a^{2},bc,bc\\right) $. Since $a^{2}b^{2}c^{2}=a^{2}bcbc$, the radicals cancel each other, and we get\r\n\r\n[tex]f\\left( a^{2},b^{2},c^{2}\\right) -f\\left( a^{2},bc,bc\\right) =\\frac{a^{4}\\left( b^{2}+c^{2}\\right) +b^{4}\\left( c^{2}+a^{2}\\right) +c^{4}\\left( a^{2}+b^{2}\\right) }{a^{2}+b^{2}+c^{2}}-\\frac{a^{4}\\left( bc+bc\\right) +b^{2}c^{2}\\left( bc+a^{2}\\right) +b^{2}c^{2}\\left( a^{2}+bc\\right) }{a^{2}+bc+bc}[/tex].\r\n\r\nThis simplifies to\r\n\r\n[tex]f\\left( a^{2},b^{2},c^{2}\\right) -f\\left( a^{2},bc,bc\\right) =a^{2}\\left( b-c\\right) ^{2}\\frac{2b^{3}c+a^{2}b^{2}+3b^{2}c^{2}+2bc^{3}+2bca^{2}+a^{4}+a^{2}c^{2}}{\\left( a^{2}+b^{2}+c^{2}\\right) \\left( a^{2}+2bc\\right) }[/tex],\r\n\r\nwhat is obviously non-negative. Hence, $f\\left( a^{2},b^{2},c^{2}\\right) \\geq f\\left( a^{2},bc,bc\\right) $. Now, it remains only to prove that $f\\left( a^{2},bc,bc\\right) \\geq 0$. Denoting $a^{2}=x^{3}$ and $bc=y^{3}$, we get\r\n\r\n[tex]f\\left( a^{2},bc,bc\\right) =f\\left( x^{3},y^{3},y^{3}\\right) =2y^{3}\\left( x+y\\right) \\left( x-y\\right) ^{2}\\frac{x^{3}+y^{2}x+y^{3}}{x^{3}+2y^{3}}[/tex],\r\n\r\nwhat is $\\geq 0$ again, so we are done.\r\n\r\nOf course, it's by far not so nice as your proof, but when I have seen yours, I had already written up the mine, and I didn't want it to get lost, hence this posting...\r\n\r\n Darij", "Solution_4": "Or: [tex]a^2(b+c)+b^2(c+a)+c^2(a+b)=(a+b+c)(ab+bc+ca)-3abc[/tex], and so we have to prove that [tex](a+b+c)(ab+bc+ca-2\\sqrt[3]{a^2b^2c^2})\\geq 3abc[/tex], which is true because [tex]a+b+c\\geq 3\\sqrt[3]{abc}[/tex] and [tex]ab+bc+ca\\geq3\\sqrt[3]{a^2b^2c^2}[/tex]. ;)", "Solution_5": "Hmm ! Chebychev inequality should help:\r\n We have\r\n $ \\sum a^2(b+c) = \\sum bc(b+c) $\r\n WLOG, we assume $a \\geq b \\geq c$. Thus\r\n $ab \\geq ac \\geq bc $ and $a+b \\geq a+c \\geq b+c $\r\n Hence\r\n $ RHS \\geq \\frac {1}{3}(bc+ca+ab)(a+b+c) \\geq LHS $" } { "Tag": [ "geometry", "perimeter", "Why the bump" ], "Problem": "A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are $ 6.2$ cm, $ 8.3$ cm, and $ 9.5$ cm. The area of the square is\r\n\r\n\\[ \\textbf{(A)}\\ 24 \\text{ cm}^2 \\qquad\r\n\\textbf{(B)}\\ 36 \\text{ cm}^2 \\qquad\r\n\\textbf{(C)}\\ 48 \\text{ cm}^2 \\qquad\r\n\\textbf{(D)}\\ 64 \\text{ cm}^2 \\qquad\r\n\\textbf{(E)}\\ 144 \\text{ cm}^2\r\n\\]", "Solution_1": "Then the square has perimeter $ 24$. Hence each side has length $ 6$, meaning the area of the square is $ 36$.", "Solution_2": "Just to clarify, to get that the perimeter is 24, you need to add 6.2+8.3+9.5=24." } { "Tag": [ "limit", "function", "integration", "inequalities", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "Show that\r\n\r\nif $ f \\in L_{\\infty}$ and $ f \\in L_q$ for some q,\r\n\r\nthen $ f \\in L_p$ for all $ p > q$ and $ \\| f \\|_{\\infty} \\equal{} \\lim_{p \\to \\infty} \\|f\\|_p$", "Solution_1": "suppose that $ f$ is a bounded measurable function on $ [0,1]$\r\nwe have $ \\parallel{}f\\parallel{}_{p} \\equal{} (\\int_{[0,1]}|f|^{p})^{\\frac {1}{p}}\\leq (\\int_{[0,1]}\\parallel{}f\\parallel{}_{\\infty}^{p})^{\\frac {1}{p}} \\equal{} \\parallel{}f\\parallel{}_{\\infty}$\r\nHence $ \\overline{lim}_{p\\rightarrow\\infty}\\parallel{}f\\parallel{}_{p}\\leq \\parallel{}f\\parallel{}_{\\infty}$\r\n\r\nfor $ \\epsilon > 0$ let $ E \\equal{} \\{x: |f(x)| > \\parallel{}f\\parallel{}_{\\infty} \\minus{} \\epsilon\\}$\r\n\r\n$ \\parallel{}f\\parallel{}_{p}\\equal{}(\\int_{[0,1]}|f|^{p})^{\\frac {1}{p}}\\geq (\\int_{E}|f|^{p})^{\\frac {1}{p}}\\geq (\\parallel{}f\\parallel{}_{\\infty} \\minus{} \\epsilon\\})m(E)^{1/p}$ , from here you have $ \\underline{lim}_{p\\rightarrow\\infty}\\parallel{}f\\parallel{}_{p}\\geq \\parallel{}f\\parallel{}_{\\infty}\\minus{}\\epsilon$", "Solution_2": "No, you do not have to assume that $ m(E)<\\infty$ or that $ f$ is a function on $ [0,1]$ or whatever. Use Holder's inequality to control the $ p\\minus{}$norm $ (p>q)$ of $ f$ in terms of $ \\|f\\|_\\infty$ and $ \\|f\\|_q$ instead.", "Solution_3": "Huh? You are implicitly assuming that $ f$ is a function from $ [0,1]$ to $ \\mathbb{R}$ (or $ \\mathbb{C}$), and then using the fact that the $ L^p$ norm can be controlled strictly by the $ L^\\infty$ norm in this setting. This is not true if $ f$ is a function on $ \\mathbb{R}$ or on some other measure space $ \\Omega$ in which the underlying space has infinite measure. There you need to use the assumption that there exists $ q < \\infty$ with $ f\\in L^q(\\Omega)$. In this setting (as I said, you can use Holder's inequality to get the bound $ \\|f\\|_p \\leq \\|f\\|_q^\\alpha\\|f\\|_\\infty^\\beta$ for $ p > q$ (figure out what $ \\alpha$ and $ \\beta$ have to be).\r\n\r\ne: the post I replied to was deleted while I was composing this message.", "Solution_4": "[quote=\"blahblahblah\"]Huh? You are implicitly assuming that $ f$ is a function from $ [0,1]$ to $ \\mathbb{R}$ (or $ \\mathbb{C}$), and then using the fact that the $ L^p$ norm can be controlled strictly by the $ L^\\infty$ norm in this setting. This is not true if $ f$ is a function on $ \\mathbb{R}$ or on some other measure space $ \\Omega$ in which the underlying space has infinite measure. There you need to use the assumption that there exists $ q < \\infty$ with $ f\\in L^q(\\Omega)$. In this setting (as I said, you can use Holder's inequality to get the bound $ \\|f\\|_p \\leq \\|f\\|_q^\\alpha\\|f\\|_\\infty^\\beta$ for $ p > q$ (figure out what $ \\alpha$ and $ \\beta$ have to be).\n\ne: the post I replied to was deleted while I was composing this message.[/quote]\r\n\r\nI can't figure it out..\r\n\r\nSorry but.. could you explain that explicitly?", "Solution_5": "I think we don't have to use Holder's inequality:\r\n\r\nSince $ f\\in L_{\\infty}, |f|\\le \\|f\\|_{\\infty}$ a.e.\r\n\r\nSo $ |f|^p\\le \\|f\\|_{\\infty}^{p\\minus{}q}.|f|^q$\r\n\r\nHence $ \\|f\\|_p \\leq \\|f\\|_q^{q/p}\\|f\\|_\\infty^{1\\minus{}q/p}$ Done!", "Solution_6": "Sure. I think of that as Holder's inequality with exponents $ 1$ and $ \\infty$.", "Solution_7": "blahblahblah:\r\n\r\n Ok from that it is clear that $ \\lim_{p \\to \\infty} \\|f\\|_p\\le \\| f \\|_{\\infty}$, but what about the reversed inequality?", "Solution_8": "Notice that 1234567a didn't use the fact that $ \\Omega$ has finite measure in his proof of the reversed inequality :wink: .", "Solution_9": "[quote=\"limsup\"]blahblahblah:\n\n Ok from that it is clear that $ \\lim_{p \\to \\infty} \\|f\\|_p\\le \\| f \\|_{\\infty}$, but what about the reversed inequality?[/quote]\r\n\r\nWell, pick $ t\\in [0,\\|f\\|_\\infty)$. By Chebyshev's inequality, the integral of $ |f|^p$ over $ A_t: \\equal{} \\{x: f(x)\\geq t\\}$ is at least $ t^pm(A_t)$, and (by the definition of $ \\|f\\|_\\infty$), $ m(A_t) > 0$. So the $ L^p$ norm of $ f$ is at least $ tm(A_t)^{1/p}$. For fixed $ t$, letting $ p\\to\\infty$ shows that $ \\liminf_{p\\to\\infty} \\|f\\|_p \\geq t$ (since $ m(A_t)^{1/p}\\to 1$). Since $ t\\in [0,\\|f\\|_\\infty)$ is arbitrary, $ \\liminf_{p\\to\\infty} \\|f\\|_p \\geq \\|f\\|_\\infty$." } { "Tag": [ "Gamebot" ], "Problem": "Suppose I have 6 shirts, 4 ties, and 3 pairs of pants. If an outfit requires a shirt and pants, and can either have a tie or not have a tie, how many outfits can I make?", "Solution_1": "[hide]This gets a little confusing because of the tie/no tie part. It seems like you could have 4 ties and 4 no ties, but there is only 1 true no tie![/hide]", "Solution_2": "The answer is easily seen to be $ 6\\cdot(4\\plus{}1)\\cdot3\\equal{}\\boxed{90}$.", "Solution_3": "I get confused by this also.\r\nThe questions says: \"Suppose I have $ 6$ shirts, $ 4$ ties, and $ 3$ pairs of pants.\" then \"how many outfits can I make?\" I think this last statement should be replaced by \"how many way?\" because if you have $ 3$ pairs of pants that's mean that you can't make more than $ 3$ outfits because you have only $ 3$ pants???", "Solution_4": "[hide=My personal solution :)]The choices for having a tie are $6 \\cdot 3 \\cdot 4 = 72$. The choices without the ties are $6 \\cdot 3 = 18$. $72+18=\\boxed{90}$.[/hide]", "Solution_5": "I did the same as g1zq. \n[hide= Here\u2019s How!]With ties: $6 \\cdot 3 \\cdot 4 = 72$\nWithout ties: $6 \\cdot 3 = 18$\nAltogether: $72 + 18 = \\boxed{90}$[/hide]", "Solution_6": "[hide][hide] :wacko: Do not understand [/hide][/hide]", "Solution_7": "[quote=limath1014][hide][hide] :wacko: Do not understand [/hide][/hide][/quote]\n\n[hide=Hint]\nUse the Multiplication Principle in Counting. That makes things much easier. :) \n[/hide]", "Solution_8": "[hide=my sol..?]I thought of considering the ties as Tie1, Tie2, Tie3, Tie4, and NoTie (Tie5). So, there are essentially 5 ties. 6 x 3 x 5 = 90[/hide]", "Solution_9": "uh\n\ndon't post solutions that are the same thing others have posted, even abstractly\n\nits sort of a postfarm :/", "Solution_10": "[hide=My sol]I figured since [b]not[/b] wearing a tie counts as a \"clothing\" I added that to the tie count of $4$ and got,\n\n[center]$6 \\cdot 5 \\cdot 3 = \\boxed{90}$.[/center]\n\nAnd we're done. $\\smiley$[/hide]", "Solution_11": "[quote=son7][hide=My sol]I figured since [b]not[/b] wearing a tie counts as a \"clothing\" I added that to the tie count of $4$ and got,\n\n[center]$6 \\cdot 5 \\cdot 3 = \\boxed{90}.$[/center]\n\nAnd we're done. $\\smiley$[/hide][/quote]\n\nfixed the slashes\n\nedit: oh oops" } { "Tag": [ "real analysis", "geometry", "rectangle", "induction", "analytic geometry", "real analysis solved" ], "Problem": "Let $X$ is a measurable set and $f : X\\times \\mathbb{R} \\to \\{0,1\\}$. It is known that for each $a \\in X$ we have $f(a,y)=0$ except countably many $y\\in \\mathbb{R}$. Also for each $b\\in R$, $f(x,b)=0$ for almost all $x\\in X$. Is it true that $f(x,y)=0$ a.e. in $X\\times \\mathbb{R}$?", "Solution_1": "There exists a non-measurable set $E$ on the plane such that every straight line intersects $E$ at no more than 2 points. So the answer is \"No\".", "Solution_2": "I deleted my reply the other night because I thought there was something wrong with it, but I don't think there was :).\r\n\r\nHere's a construction that works here, I think. It's set with Lebesgue external measure $\\ge 1$ (actually, it's not measurable, but we don't need that for the problem) which every horizontal or vertical line cuts in at most one point.\r\n\r\nLet $\\alpha$ be the smallest ordinal of continuum cardinal ($c$), and let $(C_u)_{u<\\alpha}$ be an indexing of all countable unions of open rectangles with sides parallel to the axes, rational vertices, and total area $\\le 1$ (there are clearly $c$ such unions). We can now perform transfinite induction as follows: for every ordinal $u<\\alpha$ choose one of the $c$ vertical lines which have not been chosen at any previous step (\"previous step\" here means having chosen the line for some $u' 0$.\r\n\r\nAfter a bit of searching I found:\r\n\r\n$ 2 \\ce{NO} \\plus{} \\ce {O2}$ $ \\rightarrow 2 \\ce{NO2}$ \r\n\r\nis a reaction whose rate decreases with temperature.\r\n\r\nIn fact I believe, from a bit more searching, that the explanation lies in the fact that the $ E_a$ value is only \r\n\r\n[i]apparently[/i] negative. That is to say we have to look at the reactio mechanism.\r\n\r\n$ 2 \\ce{NO}$ $ \\rightleftharpoons \\ce{N2O2}$\r\n\r\n$ \\ce{N2O2} \\plus{} \\ce{O2}$ $ \\rightarrow 2 \\ce{NO2}$\r\n\r\nThe second process is the rate limiting step and we can thus derive the rate equation (I will leave that up to you and denote $ K_c$ as the equilibrium constant and $ k'$ as the rate constant for the second process).\r\n\r\n$ R \\equal{} k[NO]^{2}[O_2]$ where $ k \\equal{} k'K_c$. This is the rate equation. Now $ K_c$ decreases with increasing temperature (dimerization is bond formation which is exothermic) and hence explains the anomalous behaviour of the process.\r\n\r\nMy conclusion would be that an actual negative $ E_a$ does not exist. After all the notion of the activation energy is just a model to explain the thermodynamics of a chemical reaction. No model is infallible.", "Solution_3": "Any more opinions/facts? :oops:", "Solution_4": "[quote=\"BanishedTraitor\"]\nAfter a bit of searching I found:\n\n$ 2 \\ce{NO} \\plus{} \\ce {O2}$ $ \\rightarrow 2 \\ce{NO2}$ \n\nis a reaction whose rate decreases with temperature.\n\nIn fact I believe, from a bit more searching, that the explanation lies in the fact that the $ E_a$ value is only \n\n[i]apparently[/i] negative. That is to say we have to look at the reaction mechanism.\n\n$ 2 \\ce{NO}$ $ \\rightleftharpoons \\ce{N2O2}$\n\n$ \\ce{N2O2} \\plus{} \\ce{O2}$ $ \\rightarrow 2 \\ce{NO2}$\n[/quote]\r\nYes I was expecting someone to post that reaction ( My Prof also told me that )\r\nBut one has to accept that Eact is negative bcos we define it as such an expr.\r\nCan someone answer this question.\r\nWhat can you say abt the energy of activation of an exothermic reaction?\r\na. negative b. positive c. zero d. cant say", "Solution_5": "If the energy of activation of a reaction is negative, that would appear to imply that the reactants take an intermediate form more stable than the final product. It would seem that the reactions to and from the stable form need to be studied independently (since one is exothermic and the other endothermic); I believe this is related to BanishedTraitor's point.", "Solution_6": "[quote=\"t0rajir0u\"]If the energy of activation of a reaction is negative, that would appear to imply that the reactants take an intermediate form more stable than the final product. [/quote]\r\n\r\nMaybe we should empahasize it is [i]thermodynamically[/i] more stable. \r\n\r\nAnyhow, we might have to wait for Carcul or someone else to make further comments.", "Solution_7": "I do agree that your reaction has a negative Ea.\r\n\r\nBut one clarification is the reaction exothermic???", "Solution_8": "I did not say that my reaction has a negative $ E_a$. I basically said that I don't believe that Arrehnius' model involving an energy profile (also the equation $ k\\equal{}Ae^{\\frac{\\minus{}E_a}{RT}}$) is applicable in this case and we should treat it separately using rate equations.\r\n\r\nI am not sure about the thermodynamic properties of the overall process. All I know is that the dimerization equilibrium should be exothermic (forward) because it is a bond forming process.", "Solution_9": "My answer: no, unless you have some really weird substance. Think about what the activation energy of a reaction actually is. It's the amount of energy difference between a [i]reactant[/i] (or intermediate) and the [i]transition state[/i]. So a negative activation energy would mean that the reactant has a higher energy than the transition state. But this can only happen if our reactant is at a relative maximum on the potential energy curve, meaning that it would be extremely reactive... kind of like balancing a ball on the point of a needle, any slight disturbance would be enough to cause reaction.", "Solution_10": "I searched the net this is what [b]wikipedia[/b] said :-\r\n\r\n[b][i]In some cases rates of reaction decrease with increasing temperature. When following an approximately exponential relationship so the rate constant can still be fit to an Arrhenius expression, this results in a negative value of Ea. Reactions exhibiting these negative activation energies are typically barrierless reactions, in which the reaction proceeding relies on the capture of the molecules in a potential well. Increasing the temperature leads to a reduced probability of the colliding molecules capturing one another (with more glancing collisions not leading to reaction as the higher momentum carries the colliding particles out of the potential well), expressed as a reaction cross section that decreases with increasing temperature. Such a situation no longer leads itself to direct interpretations as the height of a potential barrier.\n\n[/i][/b]\r\n\r\nThough I totally agree with greg. I think that would contradict one more theory relating to kinetics\r\nwhich states that all the reactant molecules in that T.S. have a choice or it is like a see-saw.\r\nIt can either go to Product side or come back to reactants. If Ea were to be negative then this\r\nwhole theory of toppling fails as the T.S. becomes very stable and the Product will be the intermediate T.S. which is very absurd.", "Solution_11": "Hence the next question might be whether there is an alternative model which can add to/replace Arrehnius' Model of a Reaction (involving an activation energy barrier and a transition state/activated complex).", "Solution_12": "I agree to your views everyone but the fact is that Eact has been defined as so , so one must accept that Eact is negative in those cases or find another alternate model for this. :P", "Solution_13": "[quote=\"JRav\"]Well technically, energy is a vector.[/quote]\r\n\r\nAs far as I know, energy is a scalar quantity, not a vector. Regarding the issue of this thread, here's my opinion: the activation energy $ E_{act}$ is usually defined as the minimum amount of energy required for a chemical reaction to occur, where that energy is to be provided by collision between species. With this definition at hand, then only two things can happen: $ E_{act}$ is positive, and so there is in fact an energy barrier to overcome in order for reaction to occur, or $ E_{act} \\equal{} 0$, in which case there is no barrier for the reaction to occur. Most of the processes with $ E_{act} \\equal{} 0$ are reactions which involve the \"capturing\" of two species A and B, where the \"complex\" AB formed corresponds in the usual potential energy diagram to a \"potential well\", and not to a \"potential hill\". Obviously, for these reactions increasing the temperature correspond to decrease the probability of the two species A and B to capture one another in a colision, because now they have increasing energy. Therefore, for these reactions increasing the temperature leads to a decreasing reaction rate. However, if we make a plot using Arrhenius equation (ln k versus 1/T) then we will obviously get a positive slope, which means that $ E_{act} < 0$, but we already know that in this case the activation energy (as we defined above) is zero, because species A and B do not have to overcome any energy barrier. Geting a negative $ E_{act}$ is just a result of modeling the reaction with Arrhenius equation.", "Solution_14": "Yes all that is ok but what answer will you write in an exam which asks the foll. question:\r\nCan Eact be negative?", "Solution_15": "The post I wrote above is a possible answer to that question.", "Solution_16": "Okay I understand your post but in an objective question if there are only 2 options a. can be negative\r\nb. cannot be negative c. Cannot be determined and you have to write an ans and there is no question of reasoning with the examiner then what will be your answer? :maybe:", "Solution_17": "Still my answer is that it will be positive unless the equation like the \r\none you referred to comes up.We only need to go with the majority.\r\nSuppose in an organic reaction the product is asked we always take the \r\nmajor product as the answer even though other options do occur in the reaction.", "Solution_18": "theiritically speaking i would say yes but practically know\r\nwell it clearly depends what u want to choos reactant species as \r\nsay if my reactant species were a carbocation then if the final product say is a halide which obviously has a lower potential energy than the internediate hence this transition has a negative Activation Energy .\r\ni remember reading Olah's super Acid experiment where proabbly he was able to 'capture' a Carbocation hence if i were to take this as a reactant then the my activation energy would be negative", "Solution_19": "No.1 first understand what we are arguing.\r\nWhen you a cabocation is chosen as the reactant then this reaction is a barrierless reaction and trying to define a hypothetical Ea for it is wrong.The carbocation is very reactive and so don't talk of an\r\nEa for it.", "Solution_20": "@Chemrock\r\ni was answering madness's question to whether activation energy can be negativein a question paper\r\nwell tah depends on what u want to choose as ur reactant and as far as a carbocation goes it can equally revert back to the thing fromn which it originated from so there is a choice factor here also but both lead top a path of negative activation energy\r\nfor ex let's consider dehydration were the removal of $ \\ce{H2O}^{\\plus{}}$ is reversible \r\nHere the Carbocation has a choice of reverting back to the original thing from where it is formed and that can in some sense have a negative activation energy \r\nthough this thing itself is meaning less(i agree with Carcul on this)\r\nBut if i were asked this 'indian' Question Paper i would write yes and state this", "Solution_21": "No its an objective paper with \" NO MARKS MADE ANYWHERE ELSE IN THE ANSWER SHEET \" :P :D", "Solution_22": "I would go for $ \\text{NO}$. \r\n\r\nAfter reading the Chemistry dictionary I think the Activation Energy is [i]defined[/i] to be positive.", "Solution_23": "I apologize for reviving this thread. But something I don't quite understand struck me. It concerns the potential well. So are we saying consider the following reaction:\r\n\r\n$ A\\plus{}B \\rightarrow [AB] \\rightarrow P$ \r\n\r\nThe \"transition\" state lies in the potential well?", "Solution_24": "the conversion of ethyl free radical on reaction with hydrogen bromide to give ethane and bromine free radical has negative Eact", "Solution_25": "for a reaction to happen , it should have some activation energy...\r\n\r\nI don think that Eact is negatiuve....", "Solution_26": "but it comes out to be in one of those examples Banished Traitor has given :huh:", "Solution_27": "that is just a hypothetical concept....\r\n\r\nand @ madness- where is the example...", "Solution_28": "see the third post of this thread da :) \r\nAnd if the model is not infalliable, then we mustn't use the model but as we are using it, we need to use it here also n say that the activation energy comes to be negative here :)", "Solution_29": "actually the result i stated was due toa research work by P.W.Seakings,M.J.Pilling and co" } { "Tag": [ "inequalities", "inequalities theorems" ], "Problem": "What's AM-GM???\r\nit's too new for me.\r\nplease explain!!!", "Solution_1": "AM-GM is very famous inequality in maths, and you should learn it if you want to do competition problems. It states the following: Let $x_1,\\ldots,x_n$ be non-negative real numbers. Then $\\frac{x_1+\\ldots +x_n}{n}\\geq \\sqrt[n]{x_1\\cdot \\ldots \\cdot x_n}$, i.e. the arithmetic mean of non-negative numbers is larger or equal than the geometric mean of those numbers. Equality holds if and only if $x_1=\\cdots =x_n$." } { "Tag": [ "trigonometry", "ratio", "quadratics", "geometry", "angle bisector", "similar triangles", "algebra" ], "Problem": "Source: A Mathematical Mosiac--by Ravi Vakil\r\n\r\nA regular decagon of side 1 is inscribed in a circle. What is the radius of the circle? \r\n\r\nThis is pretty easy once you get the right idea.", "Solution_1": "[hide=\"My answer\"]\n\\[\\frac{\\frac12}{\\sin 18 \\text{ Degrees}}=1.618033989. . . . . \\]\n[/hide]", "Solution_2": "Yes, but I forgot to mention no calculators. \r\n\r\nHint: [hide]Why is your answer the golden ratio???? Think about that[/hide]", "Solution_3": "Let $O$ be the radius, and $\\overline{AB}$ be a side of the dodecagon, so we have a isosceles triangle $\\triangle OAB$ with $AB = 1$, $\\overline{OA} \\cong \\overline{OB}$, $\\angle O = \\frac{\\pi}{5}$, and $\\angle A \\cong \\angle B$. Now draw the angle bisector from $A$ intersecting $\\overline{OB}$ at $C$, henceforth $\\triangle ABC \\sim \\triangle OAB$ and $BC = \\frac{1}{OA}$. Using the angle bisector theorem, $OA - \\frac{1}{OA} = 1$. Solving, we find that $OA = \\frac{1 + \\sqrt{5}}{2}$.", "Solution_4": "That is a very, very clever way to do it. My way (or Ravi Vakil's way) was different\r\n\r\n\r\n[hide] You have a triangle with angles 72, 72, and 36, where the side opposite the 36 has lenght 1. Now draw a pentagon and draw all the diagonals in. Let the sides of the inner pentagon have length 1, and fill in the measures of all the angles. Locate one of the 72, 72, 36 triangles. Use similar triangles and the quadratic formula, you find the length of the sides opposite the 72 degree angles are ${1 + \\sqrt(5)}/2$ [/hide]", "Solution_5": "Can you explain how to \"Use similar triangles and the quadratic formula\" to find out the answer??", "Solution_6": "Yeah, well call the outer pentagon ABCDE and the inner pentagon formed by the intersection of the diagonals FGHIJ. \r\n\r\nOk, now, we want to find the lenghts of the sides of triangle AHI given that $AB=1$. Sorry, in my last post I accidentally wrote that the lengths of the inner pentagon have length 1--I meant the outer pentagon--sorry. Now, let $x$ denote the length of $AC$. Okay, now triangle AIB is similar to traingle ABC, so $AI/AB = AB/AC$. Now, we substitute $AB=1$ and $AC=x$ to get that $AI=1/x$. Now tirangle IBC is isosceles, so $IC=BC$. Since $BC=1$, $IC=1$. Now, $x=AC=AI+IC=1/x + 1$, so $x^2 - x - 1 = 0$. The roots of this quadratic equation are x = $(1+\\sqrt{5})/2$ and $(1-\\sqrt{5})/2$. Since distance is positive, the only solution is $x=(1+\\sqrt{5})/2$." } { "Tag": [ "percent" ], "Problem": "What percent of the natural numbers from 1 to 100, inclusive, have at least one digit which is a 5?", "Solution_1": "You have 10 numbers with a 5 in the units digit, and 10 with a 5 in the tens digit. The only number you have double counted is 55, giving you $ \\boxed{19}$" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "1. Find all $ f: \\mathbb {R}\\to\\mathbb {R}$ such that\n\\[ f(x^2 \\plus{} y) \\plus{} f(f(x) \\minus{} y) \\equal{} 2f(f(x)) \\plus{} 2y^2,\\]\nhappens for all $ x,y\\in \\mathbb {R}$.\n\n2. Find all $ f: \\mathbb {R}\\to\\mathbb {R}$ such that for all $ x,y\\in \\mathbb {R}$, we have\n\\[f(f(x) \\plus{} y) \\equal{} 2x \\plus{} f(f(y) \\minus{} x).\\]", "Solution_1": "[quote=\"mathVNpro\"][u]Problem 1:[/u] Let $ f: \\mathbb {R}\\longrightarrow \\mathbb {R}$. Find all $ f$ such that:\n$ f(x^2 \\plus{} y) \\plus{} f(f(x) \\minus{} y) \\equal{} 2f(f(x)) \\plus{} 2y^2$, $ \\forall$ $ x,y\\in \\mathbb {R}$.[/quote]\r\n\r\nLet $ P(x,y)$ be the assertion $ f(x^2 \\plus{} y) \\plus{} f(f(x) \\minus{} y) \\equal{} 2f(f(x)) \\plus{} 2y^2$\r\n\r\n$ P(x,0)$ $ \\implies$ $ f(f(x))\\equal{}f(x^2)$\r\n\r\n$ P(x,f(x)\\minus{}x^2)$ $ \\implies$ $ f(f(x))\\plus{}f(x^2)\\equal{}2f(f(x))\\plus{}2(f(x)\\minus{}x^2)^2$ and so, since $ f(f(x))\\equal{}f(x^2)$ : $ 2(f(x)\\minus{}x^2)^2\\equal{}0$ and so $ \\boxed{f(x)\\equal{}x^2}$\r\n\r\nAnd it's easy to check that this function indeed is a solution.", "Solution_2": "Probleme 1:\r\n\r\nlet $ P(x,y)$ be the assertioon : $ f(x^2\\plus{}y)\\plus{}f(f(x)\\minus{}y)\\equal{}2f(f(x))\\plus{}2y^2$\r\n\r\ndenote $ a\\equal{}f(0)$\r\n\r\n$ P(0,0)\\implies f(a)\\equal{}a$\r\n\r\n$ P(0,a) \\implies a\\plus{}a\\equal{}2a\\plus{}2a^2 \\implies a\\equal{}0$\r\n\r\nso $ f(0)\\equal{}0$\r\n\r\n$ P(x,f(x))\\implies f(x^2\\plus{}f(x))\\equal{}2f(f(x))\\plus{}2f^2(x)$ (1)\r\n\r\n$ P(x,\\minus{}x^2)\\implies f(f(x)\\plus{}x^2)\\equal{}2f(f(x))\\plus{}2x^4$ (2)\r\n\r\n$ P(0,x)\\implies f(x)\\plus{}f(\\minus{}x)\\equal{}2x^2$ (3)\r\n\r\nfrom (1) ,(2) and (3) we get $ f(x)\\equal{}x^2 ,\\forall x$\r\n\r\nMehdi", "Solution_3": "[quote=\"mehdi cherif\"]Probleme 1:\n\nlet $ P(x,y)$ be the assertioon : $ f(x^2 \\plus{} y) \\plus{} f(f(x) \\minus{} y) \\equal{} 2f(f(x)) \\plus{} 2y^2$\n\ndenote $ a \\equal{} f(0)$\n\n$ P(0,0)\\implies f(a) \\equal{} a$\n\n$ P(0,a) \\implies a \\plus{} a \\equal{} 2a \\plus{} 2a^2 \\implies a \\equal{} 0$\n\nso $ f(0) \\equal{} 0$\n\n$ P(x,f(x))\\implies f(x^2 \\plus{} f(x)) \\equal{} 2f(f(x)) \\plus{} 2f^2(x)$ (1)\n\n$ P(x, \\minus{} x^2)\\implies f(f(x) \\plus{} x^2) \\equal{} 2f(f(x)) \\plus{} 2x^4$ (2)\n\n$ P(0,x)\\implies f(x) \\plus{} f( \\minus{} x) \\equal{} 2x^2$ (3)\n\nfrom (1) ,(2) and (3) we get $ f(x) \\equal{} x^2 ,\\forall x$\n\nMehdi[/quote]\r\n\r\nHello Mehdi !\r\n\r\nQuite nice :) (I appreciated the usage of assertion 3 for solving $ f(x)^2 \\equal{} x^4$)\r\n\r\nCongrats.\r\n(and sorry to have posted some seconds before you) :)", "Solution_4": "[quote=\"pco\"][quote=\"mehdi cherif\"]Probleme 1:\n\nlet $ P(x,y)$ be the assertioon : $ f(x^2 \\plus{} y) \\plus{} f(f(x) \\minus{} y) \\equal{} 2f(f(x)) \\plus{} 2y^2$\n\ndenote $ a \\equal{} f(0)$\n\n$ P(0,0)\\implies f(a) \\equal{} a$\n\n$ P(0,a) \\implies a \\plus{} a \\equal{} 2a \\plus{} 2a^2 \\implies a \\equal{} 0$\n\nso $ f(0) \\equal{} 0$\n\n$ P(x,f(x))\\implies f(x^2 \\plus{} f(x)) \\equal{} 2f(f(x)) \\plus{} 2f^2(x)$ (1)\n\n$ P(x, \\minus{} x^2)\\implies f(f(x) \\plus{} x^2) \\equal{} 2f(f(x)) \\plus{} 2x^4$ (2)\n\n$ P(0,x)\\implies f(x) \\plus{} f( \\minus{} x) \\equal{} 2x^2$ (3)\n\nfrom (1) ,(2) and (3) we get $ f(x) \\equal{} x^2 ,\\forall x$\n\nMehdi[/quote]\n\nHello Mehdi !\n\nQuite nice :) (I appreciated the usage of assertion 3 for solving $ f(x)^2 \\equal{} x^4$)\n\nCongrats.\n(and sorry to have posted some seconds before you) :)[/quote]\r\n\r\nthank's mr.Patrick :)", "Solution_5": "[quote=\"mathVNpro\"][u][u]Problem 2:[/u] Let $ f: \\mathbb {R}\\longrightarrow \\mathbb {R}$. Find all $ f$ such that:\n$ f(f(x) \\plus{} y) \\equal{} 2x \\plus{} f(f(y) \\minus{} x)$, $ \\forall$ $ x,y\\in \\mathbb {R}$[/quote]\r\n\r\nLet $ P(x,y)$ be the assertion $ f(f(x)\\plus{}y)\\equal{}2x\\plus{}f(f(y)\\minus{}x)$\r\n\r\n$ P(\\frac{f(v)}2,u\\minus{}f(\\frac{f(v)}2))$ $ \\implies$ $ f(u)\\equal{}f(v)\\plus{}f(\\text{something})$ and so $ f(u)\\minus{}f(v)\\in f(\\mathbb R)$ $ \\forall u,v$\r\n\r\nSo $ f(f(x)\\plus{}y)\\minus{}f(f(y)\\minus{}x)\\equal{}2x\\in f(\\mathbb R)$ and so $ f(x)$ is surjective.\r\n\r\nIf $ f(y_1)\\equal{}f(y_2)$, then comparing $ P(x,y_1)$ and $ P(x,y_2)$, we get $ f(y_1\\plus{}f(x))\\equal{}f(y_2\\plus{}f(x))$ and, since $ f(x)$ is surjective, $ f(x\\plus{}T)\\equal{}f(x)$ $ \\forall x$ with $ T\\equal{}y_1\\minus{}y_2$\r\n\r\nComparing then $ P(x,y)$ and $ P(x\\plus{}T,y)$, we get $ T\\equal{}0$ and so $ f(x)$ is injective.\r\n\r\nThen $ P(0,x)$ $ \\implies$ $ f(x\\plus{}f(0))\\equal{}f(f(x))$ and, since $ f(x)$ is injective, $ f(x)\\equal{}x\\plus{}f(0)$\r\n\r\nAnd it's immediate to check back that this indeed is a solution\r\n\r\nHence the result : $ \\boxed{f(x)\\equal{}x\\plus{}a}$ $ \\forall x$", "Solution_6": "[quote=\"mathVNpro\"]\n[u]Problem 2:[/u] Let $ f: \\mathbb {R}\\longrightarrow \\mathbb {R}$. Find all $ f$ such that:\n$ f(f(x) \\plus{} y) \\equal{} 2x \\plus{} f(f(y) \\minus{} x)$, $ \\forall$ $ x,y\\in \\mathbb {R}$.\n\nBest regard,\n[b]mathVNpro[/b][/quote]\r\n$ P(x,\\minus{}f(x))\\rightarrow f(0)\\equal{}2x\\plus{}f(f(\\minus{}f(x))\\minus{}x)$, so $ f$ is surjective.\r\nLet $ a\\in \\mathbb R$ such that $ f(a)\\equal{}0$.\r\n$ P(a,y)\\rightarrow f(y)\\equal{}2a\\plus{}f(f(y)\\minus{}a)\\Rightarrow f(y)\\minus{}a\\equal{}a\\plus{}f(f(y)\\minus{}a)$\r\nSince $ f$ is surjective, for each real $ x$, there exist a real $ y$ for $ x\\equal{}f(y)\\minus{}a$. Then, $ x\\equal{}a\\plus{}f(x)$ for every $ x\\in \\mathbb R$.\r\nThe result $ f(x)\\equal{}x\\plus{}c$, $ c\\in \\mathbb R$", "Solution_7": "[quote=\"mr.danh\"][quote=\"mathVNpro\"]\n[u]Problem 2:[/u] Let $ f: \\mathbb {R}\\longrightarrow \\mathbb {R}$. Find all $ f$ such that:\n$ f(f(x) \\plus{} y) \\equal{} 2x \\plus{} f(f(y) \\minus{} x)$, $ \\forall$ $ x,y\\in \\mathbb {R}$.\n\nBest regard,\n[b]mathVNpro[/b][/quote]\n$ P(x, \\minus{} f(x))\\rightarrow f(0) \\equal{} 2x \\plus{} f(f( \\minus{} f(x)) \\minus{} x)$, so $ f$ is surjective.\nLet $ a\\in \\mathbb R$ such that $ f(a) \\equal{} 0$.\n$ P(a,y)\\rightarrow f(y) \\equal{} 2a \\plus{} f(f(y) \\minus{} a)\\Rightarrow f(y) \\minus{} a \\equal{} a \\plus{} f(f(y) \\minus{} a)$\nSince $ f$ is surjective, for each real $ x$, there exist a real $ y$ for $ x \\equal{} f(y) \\minus{} a$. Then, $ x \\equal{} a \\plus{} f(x)$ for every $ x\\in \\mathbb R$.\nThe result $ f(x) \\equal{} x \\plus{} c$, $ c\\in \\mathbb R$[/quote]\r\n\r\nQuite OK. Nice and quick , and better than mine !\r\n\r\nCongrats! :)", "Solution_8": "How can I edit my text thus he saty like yours( R -> R) hehe\r\nThanks" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $x,y,z \\ge 0,x+y+z = 1,n \\in N,n \\ge 2$\r\nProve that \r\n$x^{n}y+y^{n}z+z^{n}x \\le \\frac{{n^{n}}}{{(n+1)^{n+1}}}$", "Solution_1": "[quote=\"scpajmb\"]Let $x,y,z \\ge 0,x+y+z = 1,n \\in N,n \\ge 2$\nProve that \n$n^{n}y+y^{n}z+z^{n}x \\le \\frac{{n^{n}}}{{(n+1)^{n+1}}}$[/quote]\r\nIt must is $x^{n}y+...$ :wink:" } { "Tag": [ "geometry" ], "Problem": "An equilateral triangular of side $ 1$ is cuted into $ 3$ pieces, which are arrangened such that they form a square. What is the maximum lenght of the square?", "Solution_1": "[hide]\ncalculate the area of the triangle.\nIt is the area of the corresponding square.\nTake the square root and u have the length of a side of the square.\n[/hide]", "Solution_2": "[quote=\"Fraenkel\"][hide]\ncalculate the area of the triangle.\nIt is the area of the corresponding square.\nTake the square root and u have the length of a side of the square.\n[/hide][/quote]\r\n\r\nFollowing...\r\n\r\nThe formula for area is $ \\frac{s^2\\sqrt{3}}{4}$\r\n\r\nwe have $ \\frac{\\sqrt{3}}{4}$ as the area.\r\n\r\nThus the side of the square is $ \\sqrt{\\frac{\\sqrt{3}}{4}}\\equal{}\\boxed{\\frac{3^\\frac{1}{4}}{2}}$", "Solution_3": "And that can be simplified $ \\implies \\boxed {\\frac{13}{8}}$ or $ \\boxed {1\\frac{5}{8}}$", "Solution_4": "[quote=\"isabella2296\"]And that can be simplified $ \\implies \\boxed {\\frac {13}{8}}$ or $ \\boxed {1\\frac {5}{8}}$[/quote]\r\n\r\nHe was saying 3 to the 1/4th power, not 3 and 1/4. 3 to the 1/4th power is equal to the 4th root of 3.\r\n\r\n$ 3^{\\frac 14} \\equal{} \\sqrt[4]{3}$", "Solution_5": "Ah, I see. My bad. Sorry. :blush:", "Solution_6": "well it can be simplified to 3/16 by quarting the top and the bottom.", "Solution_7": "Umm... I think that would be equivalent of raising the whole fraction to the 4th power, which is definitely NOT the same as the original fraction!" } { "Tag": [ "HMMT", "AMC" ], "Problem": "So it seems my school is probably going to only offer the B test. Is there any way for me to take the A test at another school, or should I just beg the head of the math department to change it? It's really discouraging, especially since when I talked to them about HMMT they said \"We'll think about it\" as the first registration deadline passed silently. The same thing will probably happen here.", "Solution_1": "Either ask politely, and if they reject, say that you are going to take it at a different school, and that's final.\r\n\r\n[quote=\"worthawholebean\"]It's really discouraging, especially since when I talked to them about HMMT they said \"We'll think about it\" as the first registration deadline passed silently.[/quote]\r\n\r\nAsk the head of the math department every other day until he/she gives you an answer.", "Solution_2": "The problem is that there really isn't anywhere good to take the A nearby. All the other schools are taking it during the school day, and I can't leave then, and the colleges also only offer the B.", "Solution_3": "The advice in AMC Frequently Asked Question #4 generally applies here.\r\nSee http://www.unl.edu/amc/f-miscellaneous/faq.shtml\r\n\r\nAsk a local college or university to give the A date. Offer to pay for the registration.\r\n\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "probability", "geometry", "rectangle" ], "Problem": "A point $ (x,y)$ is randomly selected such that $ 0 \\le x \\le 8$ and $ 0 \\le y \\le 4$. What is the probability that $ x\\plus{}y \\le 4$? Express your answer as a common fraction.", "Solution_1": "For $ x\\plus{}y\\le 4$, the region within the rectangle with vertices $ (0,0)$, $ (4,0)$, $ (4,8)$, and $ (0,8)$ is the isosceles right triangle with vertices $ (0,0)$, $ (4,0)$, and $ (0,4)$.\r\n\r\nGeometric probability of this occurring is just $ \\frac{\\frac{1}{2}(4)^2}{(8)(4)}\\equal{}\\frac{8}{32}\\equal{}\\boxed{\\frac{1}{4}}$.", "Solution_2": "The [hide=answer] is 1/4[/hide]\n" } { "Tag": [ "function", "geometry proposed", "geometry" ], "Problem": "$ABC$ is a triangle with $AB=min\\{AB,AC,BC\\}$ and $P$ is a point in the interior of the triangle .\r\nProve that $PA+PB+PC0 ... nope\r\nb: 2-1 = 1> 0 ... nope\r\nc. 2-2 = 0 which isnt greater than 0... nope\r\nd. 2-3 = -1<0... and there you have it", "Solution_2": "[quote=\"feynman_wannabe\"]\n\nWhere $ W \\equal{} P \\Delta V$\n\n[/quote]\r\n\r\nHmmmm, not sure but I though it's supposed to be $ W \\equal{} \\minus{} P \\Delta V$. :maybe: \r\n\r\nBy the way the $ E$ in this case is internal energy right?", "Solution_3": "It doesnt really matter in this case whether its PdV or -PdV in this case since we just need E0$\r\n\r\nThus we have \r\n\r\n$ \\Delta E \\plus{}k\\equal{} \\Delta H \\rightarrow k\\equal{} \\Delta H \\minus{} \\Delta E\\equal{}\\minus{}P\\Delta V$ Which must be greater than 0 (E H-E>0). Therefore, we need $ k>0 \\rightarrow P\\Delta V <0$ \r\n\r\netc..\r\n\r\nAnd yes... $ \\Delta E$ would refer to the Internal energy of the system. This whole premise of this question is the first law of thermodynamics", "Solution_4": "EDIT...\r\nOK my first response is flawed twice to get a correct answer...\r\n\r\n\r\n[quote] Thus, in order for the first inequality to hold, [/quote]\r\n\r\nyea.. that should be $ W>0 \\rightarrow P\\Delta V<0$ etc...\r\n\r\nThanks for the catch banished traitor..", "Solution_5": "ahhhh i get it. thanks!", "Solution_6": "Is the change in internal energy related at all to the change in free energy?\r\nAlso the change in free energy for the first reaction (A), would be zero right?", "Solution_7": "$ \\Delta G^o \\equal{} \\Delta H^o \\minus{} T\\Delta S^o$\r\n\r\nI guess you could substitute variables in and find a relation\r\n\r\n\r\nand if $ \\Delta G \\equal{} 0$, then the reaction is in equilibrium.", "Solution_8": "Okay, I'm really sorry for reviving an old thread, but I was really confused.\n\n[quote=\"feynman_wannabe\"]It doesnt really matter in this case whether its PdV or -PdV in this case since we just need E0$\n\nThus we have \n\n$ \\Delta E \\plus{}k\\equal{} \\Delta H \\rightarrow k\\equal{} \\Delta H \\minus{} \\Delta E\\equal{}\\minus{}P\\Delta V$ Which must be greater than 0 (E H-E>0). Therefore, we need $ k>0 \\rightarrow P\\Delta V <0$ \n\netc..\n\nAnd yes... $ \\Delta E$ would refer to the Internal energy of the system. This whole premise of this question is the first law of thermodynamics[/quote]\n\nI thought that the equation was $\\Delta H = \\Delta E + P \\Delta V$, so that $\\Delta H - \\Delta E = P \\Delta V$. So why does it say $ k=\\Delta H-\\Delta E=-P\\Delta V $?" } { "Tag": [ "geometry", "perimeter" ], "Problem": "The perimeter of a square is 44 meters. How many square\nmeters are in the area of the square?", "Solution_1": "hello, let $ a$ the side length of the given square then we have $ 4a\\equal{}44 \\Leftrightarrow a\\equal{}11$ and the area is $ a^2\\equal{}121m^2$.\r\nSonnhard." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "First of all, I apologise to Harazi and Arne. From now on, I'll write full solutions. :oops: \r\n Here is an inequality. From what I know so far this problem has a brute force solution. I have found a nice solution for this one, but no more comments, here it is:\r\n\r\n Let $a,b,c>0$ with $abc=1$.Prove that: $2(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a})\\geq a+b+c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$ ;)", "Solution_1": "$\\frac{a}{b}+\\frac{b}{c}+\\frac{b}{c} \\ge 3\\sqrt[3]{\\frac{ab}{c^2}} = \\frac{3}{c}$\r\n$\\frac{a}{b}+\\frac{a}{b}+\\frac{b}{c} \\ge 3\\sqrt[3]{\\frac{a^2}{bc}} = 3a$\r\nSum up, $\\frac{a}{b}+\\frac{b}{c} \\ge a+\\frac{1}{c}$\r\nAdd up similar inequalities and done.", "Solution_2": "Very nice. This is shorter than my solution. ;)" } { "Tag": [ "inequalities", "logarithms", "function", "inequalities proposed" ], "Problem": "$a,b,c$ are positive numbers.prove that :\r\n $(\\frac{a+b+c}{3})^{a+b+c}\\geq a^{b}b^{c}c^{a}$", "Solution_1": "Beautiful!!!\r\nTaking $\\ln$ of both sides we get: \\[(a+b+c)\\ln\\frac{a+b+c}{3}\\geq\\sum b\\ln a\\iff\\] \\[\\iff \\sum b\\ln a-b\\ln\\frac{a+b+c}{3}\\leq 0\\iff\\] \\[\\iff \\sum\\frac{b}{a+b+c}\\ln\\frac{a}{\\frac{a+b+c}{3}}\\leq 0.\\] Applying Jensen for concave function $\\ln t$ we get \\[\\sum\\frac{b}{a+b+c}\\ln\\frac{a}{\\frac{a+b+c}{3}}\\leq \\ln \\frac{\\sum\\frac{ab}{a+b+c}}{\\frac{a+b+c}{3}}.\\] We have to prove \\[\\ln \\frac{\\sum\\frac{ab}{a+b+c}}{\\frac{a+b+c}{3}}\\leq 0\\iff\\] \\[\\iff \\frac{\\sum\\frac{ab}{a+b+c}}{\\frac{a+b+c}{3}}\\leq 1 \\iff\\] \\[\\iff \\sum\\frac{ab}{a+b+c}\\leq \\frac{a+b+c}{3}\\iff\\] \\[\\iff 3\\sum ab\\leq (a+b+c)^{2}\\iff 3\\sum ab\\leq a^{2}+b^{2}+c^{2}+2\\sum ab\\iff\\] \\[\\iff a^{2}+b^{2}+c^{2}\\geq ab+bc+ca,\\] which is true by rearrangement and we are done.", "Solution_2": "We may suppose that $a+b+c=3.$ Then by weighted AM-GM, \\[a^{b}b^{c}c^{a}\\leq \\left(\\frac{ab+bc+ca}{3}\\right)^{a+b+c}\\leq \\left(\\frac{(a+b+c)^{2}}{9}\\right)^{a+b+c}=\\left(\\frac{a+b+c}{3}\\right)^{a+b+c}.\\]", "Solution_3": "both of your solution are very nice.and i think this problem is so nice too.thank you.if anyone have another solution plz send it.thanks.", "Solution_4": "here is my solution for this problem:\r\n we have:\r\n $\\frac{a}{a+b+c}+\\frac{b}{a+b+c}+\\frac{c}{a+b+c}=1$\r\nand we have:\r\n $a+b+c=\\frac{(a+b+c)^{2}}{a+b+c}=\\frac{a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}{a+b+c}\\geq 3(\\frac{ab}{a+b+c}+\\frac{bc}{a+b+c}+\\frac{ca}{a+b+c}) \\geq 3(a^{\\frac{b}{a+b+c}}\\times b^{\\frac{c}{a+b+c}}\\times c^{\\frac{a}{a+b+c}})$\r\nso we get :\r\n $(\\frac{a+b+c}{3})^{a+b+c}\\geq a^{b}b^{c}c^{a}$\r\nand we R done.", "Solution_5": "More general, see http://www.mathlinks.ro/Forum/viewtopic.php?t=32451", "Solution_6": "[quote=\"Lovasz\"]More general, see http://www.mathlinks.ro/Forum/viewtopic.php?t=32451[/quote]thanks alot. :D" } { "Tag": [ "function", "floor function", "ceiling function" ], "Problem": "From AoPS I:\r\n\r\nIf $ f(x) \\equal{} \\lfloor{x}\\rfloor$ and $ g(x) \\equal{} \\lceil{x}\\rceil$, express $ g(x)$ in terms of $ f(x)$ and $ x$.\r\n\r\nBasically, I got stuck on the second step: (I got the rest of the proof)\r\n\r\n\r\n1. If $ x$ is an integer: then $ \\lfloor{x}\\rfloor \\equal{} \\lceil{x}\\rceil \\equal{} x$\r\n\r\n2. If $ x$ isn't an integer than $ \\lfloor{x}\\rfloor \\plus{} 1 \\equal{} \\lceil{x}\\rceil$ (it's \"common sense\", but how do we prove this?)\r\n\r\nI'm probably missing something really basic :)", "Solution_1": "For 2, let $ \\lfloor x\\rfloor\\equal{}x\\minus{}\\{x\\},$ where $ \\{x\\}$ is the fractional part of $ x.$ Then let $ \\lceil x\\rceil\\equal{}x\\plus{}(1\\minus{}\\{x\\}).$ It becomes obvious that \\[ (x\\minus{}\\{x\\})\\plus{}1\\equal{}x\\plus{}(1\\minus{}\\{x\\}).\\]\r\n\r\nNot really sure if this is rigorous or not.", "Solution_2": "since x is not an integer.\r\nx= y+p/q(pm$.", "Solution_3": "https://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=384021\nhttps://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=304536&p=1648889#p1648889", "Solution_4": "Posted verbatim at JBL's first link - the poster did not even take the trouble to change notations. The problem is indeed interesting, but this does not mean we have to repeat it forever ...\nThe mentioned link above sends, in turn, to JBL's second link, where I presented a proof.\n\nTopic locked; there is no interest to reopen the discussion in a new thread - all contributions must go to former postings." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find the minimum value of $ a_{1}a_{2}a_{3}+b_{1}b_{2}b_{3}+c_{1}c_{2}c_{3}$, where the nine numbers are permutations of $ 1,2,3,...,9$.", "Solution_1": "[hide=\"Suprisingly simple solution\"]We have by AM-GM that the expression is greater than $ 3\\cdot \\sqrt[3]{9!}\\approx 213.98$ for every permutation. Thus, the theoretical minimum is 214. But we can actually achieve this, with all the permutations which give us a final expression like $ 9\\cdot8\\cdot1+7\\cdot5\\cdot2+6\\cdot4\\cdot3$.[/hide]", "Solution_2": "If this problem were given to you on a contest, how would you go about calculating the value you got with no calculating instruments? Since it's so close to the lower bound, approximating might be difficult because you might go over.\r\n\r\nLike, suppose someone taking the contest was really bad at computation (there are many at the olympiad level). Would you be able to recommend a way for them to use this method to write up a solution? It seems like there might be a better way to solve this.", "Solution_3": "The solution process without being able to compute cube roots would look like this:\r\n\r\nDiscover the arrangement $ 9\\cdot8\\cdot1+7\\cdot5\\cdot2+6\\cdot4\\cdot3$. Remark to oneself that 72, 70 and 72 are very close together, suggesting that we might be able to argue by AM-GM. Observe that in order to show that 214 is the minimum, it suffices to show that there is some lower bound on the possible values which is larger than 213. From AM-GM, note that every possible value must be at least $ 3\\sqrt[3]{9!}$. Suppose that there is a possible value $ x \\leq 213$. Then $ x^{3}= 27\\cdot 9! \\leq 213^{3}$, so $ 9! \\leq 71^{3}$. Now compute $ 9! = 326880$ and $ 71^{3}= 357911$, a contradiction. Thus there are no solutions less than or equal to 213, so 214 is indeed the minimum." } { "Tag": [ "calculus", "integration", "probability and stats" ], "Problem": "Let $ W$ be standard $ 1$-d Brownian Motion, $ T$ a stopping time with $ E(T) < \\infty$. Prove that $ E(W_T) \\equal{} 0$ and $ E(W_T^2) \\equal{} E(T)$. Thanks", "Solution_1": "use that $ W_{\\min{t,T}}^2 \\minus{} \\min(t,T)$ is a martingale to show that $ W_{\\min{t,T}}$ is a uniformly integral martingale, hence the conclusion." } { "Tag": [ "calculus", "integration", "function", "derivative", "calculus computations" ], "Problem": "Hello Everyone :) \r\n\r\nI'm kind of stuck on this exercise because I'm not sure whether or not I can integrate. The exercise:\r\nLet $ f$ be a differentiable function on $ \\mathbb{R}$ whose derivative satisfies: $ |f'(x)|<\\frac{5}{6}$ for all $ x \\in \\mathbb{R}$.\r\nDefine the sequence: $ a_0 \\equal{} 0$ and $ a_n \\equal{} f(a_{n\\minus{}1})$ for any natural $ n$. Prove that the sequence $ (a_n)$ converges. \r\n\r\nIf I can integrate $ f'(x)$ then I think I can prove that $ |{a_{n\\plus{}1} \\minus{} a_n}| < \\frac{5}{6}|a_n \\minus{} a_{n\\minus{}1}|$. However, not every function that has an antiderivative is Riemann Integrable. So, is it possible to integrate here?\r\n\r\nThanks in advance!\r\nPerrin", "Solution_1": "The right way around this one is to avoid integrals entirely: use the Mean Value theorem to get $ |f(x)\\minus{}f(y)|\\equal{}|f'(c)|\\cdot |x\\minus{}y|$ for some $ c$, and $ |f'(c)|<\\frac56$.", "Solution_2": "Of course! I was thinking about integrals so much that I completely missed it. Thank you very much, you've really helped me! :lol:" } { "Tag": [ "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Major problems with this proof:\r\n\r\nLet G be a finite cyclic group of order n. If d is a positive divisor of n, prove that the equation x^d=e has exactly d distinct solutions in G, where e is the identity in G.", "Solution_1": "Assume you have an x = g^m such that x^d = 0. Then g^md = 0, so n | md, so n/d | m, so f | m, so all solutions are of the form g^(in/d), of which there are d such elements (as g^n = e, and nothing less than n does that)." } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "Consider the function $ f(\\theta) \\equal{} \\int_0^1 |\\sqrt {1 \\minus{} x^2} \\minus{} \\sin \\theta|dx$ in the interval of $ 0\\leq \\theta \\leq \\frac {\\pi}{2}$.\r\n\r\n(1) Find the maximum and minimum values of $ f(\\theta)$.\r\n\r\n(2) Evaluate $ \\int_0^{\\frac {\\pi}{2}} f(\\theta)\\ d\\theta$.", "Solution_1": "hello, we get $ f(\\theta)=%Error. \"signum\" is a bad command.\n(\\sin(\\theta))\\sin(\\theta)-\\frac{1}{4}%Error. \"signum\" is a bad command.\n(\\sin(\\theta))\\pi$ and\r\n$ -\\frac{1}{4}\\pi\\le f(\\theta)\\le 1-\\frac{1}{4}\\pi$, further we have $ \\int_{0}^{\\frac{\\pi}{2}}f(\\theta)\\,d\\theta=1-\\frac{1}{8}\\pi^2$.\r\nSonnhard.", "Solution_2": "[quote=\"1605272\"]Let $ x \\equal{} \\cos t$\n$ f(\\theta) \\equal{} \\int_0^{\\frac {\\pi}{2}}|\\sin t \\minus{} \\sin\\theta|dt$\n$ \\equal{} \\int_0^{\\theta}(\\sin\\theta \\minus{} \\sin t)dt \\plus{} \\int_{\\theta}^{\\frac {\\pi}{2}}(\\sin t \\minus{} \\sin\\theta)dt$\n$ \\equal{} (2\\theta \\minus{} \\frac {\\pi}{2})\\sin\\theta \\plus{} 2\\cos\\theta \\minus{} 1$\n\n(1)$ f'(\\theta) \\equal{} (2\\theta \\minus{} \\frac {\\pi}{2})\\cos\\theta$\n\nlet$ f'(\\theta) \\equal{} (2\\theta \\minus{} \\frac {\\pi}{2})\\cos\\theta \\equal{} 0$,We get $ \\theta \\equal{} \\frac {\\pi}{4}$,Caculate\n\n$ f(\\frac {\\pi}{4}) \\equal{} \\sqrt {2} \\minus{} 1, f(0) \\equal{} 1, f(\\frac {\\pi}{2}) \\equal{} \\frac {\\pi}{2} \\minus{} 1$\n\nso\n\n$ \\max_{0 < \\theta < \\frac {\\pi}{2}}f(x) \\equal{} 1,\\min_{0 < \\theta < \\frac {\\pi}{2}}f(x) \\equal{} \\sqrt {2} \\minus{} 1$\n\n(2)$ \\int_0^{\\frac {\\pi}{2}}f(\\theta)d\\theta$\n$ \\equal{} \\int_0^{\\frac {\\pi}{2}}[(2\\theta \\minus{} \\frac {\\pi}{2})\\sin\\theta \\plus{} 2\\cos\\theta \\minus{} 1]d\\theta$\n$ \\equal{} 4 \\minus{} \\pi$[/quote]" } { "Tag": [], "Problem": "I have a spreadsheet that has three columns: \r\n\r\n1. Loan Amount, 2. Fees ($\\$$ amt), and 3. Fees % (fee amt/Loan amount)\r\n\r\nand at the last row, there's an average of each column.\r\n\r\nHere's my question: Why doesn't the (average Loan Amt/average Fee Amt) = average fee % ?\r\n\r\nAll of the averages work out (ie. The bottom rows are \"correct\"), but I would think that the Avg Loan amount / Avg Fee Amt = Avg Fee %. What's up? Thanks in advance!", "Solution_1": ":oops: I'm not sue I think that\r\nThe average fee%=average fee amt/average loan amt\r\n\r\n\r\n\r\n\r\n\r\nMath is my life :gathering:", "Solution_2": "[quote=\"TRAN THAI HUNG\"]:oops: I'm not sue I think that\nThe average fee%=average fee amt/average loan amt\n\n\n\n\n\nMath is my life :gathering:[/quote]\r\n\r\nI believe you're correct, but I'm having trouble firguring out why. I've ran through a few examples and most of them worked out. \r\n\r\nIs there an exaplaination?", "Solution_3": "Okay, I've figured it out. My brain is very wonky today... :blush:", "Solution_4": "The total fee over the total loan amount will give you the average percentage. This number will (probably) be [i]different[/i] than the number you get by averaging the percentages for each loan. This is an example of [url=http://mathworld.wolfram.com/SimpsonsParadox.html]Simpson's Paradox[/url]" } { "Tag": [ "geometry", "geometric transformation", "rotation", "LaTeX" ], "Problem": "I'm not very sure about how to find the angle for this one.\r\n\r\nExpress the angle that the second hand on a watch turns through 1s. a) in degree measure b) in radian measure in exact form.", "Solution_1": "[hide] There are 60s in a minute (or in 1 rotation) and also 360 degrees in one rotation. 1 sec. is 1/60 of a rotation and 1/60 of 360 degrees (one rotation) is 6 degrees.\nTo solve for radians, remember that 180 deg. = pi radians. So (6/180)*pi or (1/30)*pi is the answer (I really need to learn LaTeX!))[/hide]", "Solution_2": "[quote=\"Mathematicus\"][hide] There are 60s in a minute (or in 1 rotation) and also 360 degrees in one rotation. 1 sec. is 1/60 of a rotation and 1/60 of 360 degrees (one rotation) is 6 degrees.\nTo solve for radians, remember that 180 deg. = pi radians. So (6/180)*pi or (1/30)*pi is the answer (I really need to learn LaTeX!))[/hide][/quote]\n\n[hide=\"helping out mathematicus :D\"][b]Degrees[/b]:$360^\\circ(\\frac{1}{60})=\\boxed{6^\\circ}$\n\n[b]Radians[/b]: $(6^\\circ)(\\frac{\\pi}{180^\\circ})=\\boxed{ \\frac{\\pi^{rad}}{30}}$.[/hide]\r\n(sorry if the text is huge, theres somethin wrong...)" } { "Tag": [ "articles" ], "Problem": "Do you think that you can determine which is morally better --- letting someone die or making someone die --- in terms of medical field?", "Solution_1": "Why do people always think there is a \"one-line-exact-rule\" to follow\u00bf :wink: \r\n\r\nThis strongly depends on the circumstances. In general, decissions should ba made by wagering the different possible outcomes.", "Solution_2": "[quote=\"ZetaX\"]Why do people always think there is a \"one-line-exact-rule\" to follow\u00bf :wink: \n\nThis strongly depends on the circumstances. In general, decissions should ba made by wagering the different possible outcomes.[/quote]\r\n\r\nThe origin of this question stems from an article I read in [i]Journal of Applied Philosophy[/i]. Like, this was the example they gave, which wasn't exactly medical, but related:\r\n\r\nLet's say we have two people -- Smith and Jones. Smith intentionally drowned his nephew (I think?) for money. Jones, with same intention, tried to do this, but the nephew hit his head, fell in the water, face down and unconscious, and Jones watched the child die. So, the former case was killing someone while the latter was watching someone die. The article then went to see if we can justify which one was more moral or not.", "Solution_3": "[quote=\"Silverfalcon\"]But, to me, life is something that shouldn't be taken away unless by God.[/quote] Because reducing questions to determining the will of a god has worked so well to generate consensus all throughout human history?\r\n\r\nEthics is very hard. It seems to me that toy examples such as the one you mentioned typically just illustrate the unlikelyhood that there are objective moral principles that apply in all circumstances.", "Solution_4": "[quote=\"Silverfalcon\"] We can't create life, so we shall not take one away.[/quote]\r\n\r\nBut what if someone has a terminal illness and begs a doctor to humanely kill him or her, in order to avoid more pain that would serve no cause?", "Solution_5": "Thanks JBL. I agree... This is a field that I just walked on recently by reading one article.\r\n\r\nisabella2296, do you know that what you are just asking is one of the most often debated question in bioethics? I don't think you should expect an answer, and thus, I won't answer it especially when I don't have an adequate amount of information.\r\n\r\nP.S. Can JBL lock this? I think that I'm not going to see any too much intellectual debate in this thread just because the concept is too difficult to answer." } { "Tag": [ "limit", "inequalities", "inequalities proposed" ], "Problem": "Prove that for every $ n \\in N$ :\r\n\r\n$ \\sqrt{1\\plus{}\\sqrt{2\\plus{}\\sqrt{3\\plus{}...\\plus{}\\sqrt{n}}}}<2$", "Solution_1": "Let $ (a_n)$ be a sequence ,and $ a_n >a_{n\\minus{}1}$ ,such that \r\n\r\n$ a_n\\equal{}\\sqrt {1 \\plus{} \\sqrt {2 \\plus{} \\sqrt {3 \\plus{} ... \\plus{} \\sqrt {n}}}}$\r\n\r\n\r\nI can not prove ,but I think it is true that $ \\lim_{n \\rightarrow \\infty}{a_n}\\leq2$", "Solution_2": "observe that your inequality is equivalent to\r\n\r\n$ \\sqrt{2\\plus{}\\sqrt3{....\\plus{}\\sqrt{n}}}<3$\r\n\r\n$ \\Leftrightarrow\\sqrt{3\\plus{}\\sqrt{4..\\plus{}\\sqrt{n}}}<7$\r\n....\r\n$ \\sqrt{n}<(\\text{something})$\r\n\r\nLet be the sequence $ (a_n)$ such that: $ a_{n\\plus{}1}\\equal{}a_n^2\\minus{}n$ and $ a_0\\equal{}2$\r\n\r\nwe need only prove that $ n1 \\Leftrightarrow a_n>1$ which is true since $ a_n>a_0$", "Solution_3": "Let be the sequence $ (a_n)$ such that: $ a_{n \\plus{} 1} \\equal{} a_n^2 \\minus{} n$ and $ a_0 \\equal{} 2$\r\n\r\nwhy is that ?", "Solution_4": "[quote=\"Hamddi\"]Let be the sequence $ (a_n)$ such that: $ a_{n \\plus{} 1} \\equal{} a_n^2 \\minus{} n$ and $ a_0 \\equal{} 2$\n\nwhy is that ?[/quote]\r\n\r\nobserve that : $ \\sqrt{1\\plus{}\\sqrt{2\\plus{}\\sqrt{3\\plus{}...\\plus{}\\sqrt{n}}}}<2\\equal{}a_0 \\Leftrightarrow \\sqrt{2\\plus{}\\sqrt{3\\plus{}...\\plus{}\\sqrt{n}}}<3\\equal{}a_1$\r\n\r\n$ \\Leftrightarrow \\sqrt{3\\plus{}...\\plus{}\\sqrt{n}}<7\\equal{}a_2 \\rightarrow......\\sqrt{n}1009.$\r\nChoose any prime $r$ not dividing $b.$ Then the congruence $1009a+bx \\equiv 0\\pmod{r}$ has a solution $x\\equiv c\\pmod{r}.$\r\nAccording to Dirichlet's theorem, there are infinitely many primes in the sequence $c+r,\\ c+2r,\\dots.$ Let $q>1009$ be a prime from this sequence. Then $1009a+bq>r$ and it is divisible by $r$ (thus, not prime).\r\nContradiction proves that there are no such $a,b.$", "Solution_2": "You don't need Dirichlet. There are infinitely many primes greater than 1000. Thus there are 2 distinct primes $ p=3. By trial and error I was able to get the solutions, and noticed that if there was another one, the sum would be greater than 1000. So that means there aren't any more, since AIME questions are in the range from 0 to 999. And this was in the last five problems. \r\n\r\nNow thats using a test to your advantage, but I felt very cheap afterward. (But I didnt know the number theory then I know now, so not that cheap...)", "Solution_8": "It's rather funny to hear about how people got that problem (#10, AIME I, 2003) correct by using a good diagram, because I missed it due to a bad one.\r\nThis might sound kind of odd, but what happened was this: I had drawn a quick-and-dirty diagram (I'm not much good with constructions and such) in order to visualize things, and (eventually) I worked out a solution, using (IIRC) the law of sines. I then looked back at my diagram, and thought that the answer that I had gotten was impossible, not suspecting that in fact my diagram was VERY far off - so far off as to make the correct answer look impossible. (Needless to say, I was not at my sanest that morning, and in fact missed two other problems on that test due to misreading them. Go figure.) I have a feeling that there is a lesson there, but I'm not quite certain what it is.\r\n\r\nOn the other hand, I got the fence-painting problem correct (#9, AIME I, 2002) correct in the same manner as Nukular, although I managed to figure out why those two solutions were the only ones after I knew the fact. I don't really consider that sort of tactic (taking into account the contest format) \"cheap\" on tests like the AIME -- it's part of the game -- but managing to answer questions by drawing scale figures bothers me a bit. Not that I wouldn't do such a thing (if I could...), but it seems to be a different sort of tactic.\r\n\r\nOh well. That's just my two cents' worth.", "Solution_9": "yes it was only AFTER the fence painting problem that I learned the Chinese Remainder Theorem. VERY USEFUL !!! :)", "Solution_10": "Its also helpful in any other type of competition if you have to prove the answer.. eg in a geometry problem, you could draw a diagram and see what equals what. Then knowing that, you may find it easier to prove it. Very useful.", "Solution_11": "[hide]\n\nrotate M 60 degrees about C towards B. CMM' is equilateral. CM'B is congruent to CMA (SAS) and hence CM'B=150, and MM'B=360-60-150=150, hence MM'B is congruent to CMB and CMB=M'MB+60=23+60=83\n\n[/hide]\n\nI dont see how trig Ceva gets you anywhere:\n\nsin87*sinx*sin7=sin25*siny*sinz\n\nand x+y=7+z\n\nwhere goes from here?", "Solution_12": "Is the answer to that problem above with the fenceposts \n\n\n\n\n\n\n\n[hide]780? I found 001, 333, 424, and 022 as solutions[/hide]", "Solution_13": "[quote=\"jelyman\"]did anyone get that problem from last year's aime with the triangle and it asks for the angle made by some intersection (sorry bout the bad description but im in a hurry and i dont think it is needed to be too exact because there wasnt another like this)? I'm talking about getting it with a proper method, not a construction.[/quote]\r\nIf you're talking about the problem I think you are - yes, I got it. I drew a very nice diagram and measured the angle needed.", "Solution_14": "You can't have 0s. What does every 0th fencepost mean? The ones you have without the 0s are the only ones though.", "Solution_15": "I figured the way it is written, u start with that 1, 2, or 3, and then after that, is the ath, bth or cth, so if one of them is zero, then u just have the original 1, 2, or 3,", "Solution_16": "You can look at a,b,c with 1/a+1/b+1/c=1. That limits the possibilities greatly.", "Solution_17": "BTW: My math teacher at the time was the guy who wrote the problem\r\n\r\nThe intended solution is an application of the trigonometric form of Ceva's theorem, followed by some cancellations and a quick trigonometric equation to solve. It really shouldn't have taken more than a few minutes.", "Solution_18": "Actually since there was a nice construction, I would think Penev woulda liked that way more.\r\n\r\n\r\n^ you never miss a chance to say a problem is easy do you?", "Solution_19": "[quote=\"God\"]BTW: My math teacher at the time was the guy who wrote the problem\n\nThe intended solution is an application of the trigonometric form of Ceva's theorem, followed by some cancellations and a quick trigonometric equation to solve. It really shouldn't have taken more than a few minutes.[/quote]\r\n\r\nI have yet to see a solution using trig Ceva. plz show", "Solution_20": "Fencepost solution:\n\n[hide]\n\nIf we take 1/a+1/b+1/c =1, we see that min(a,b,c)<=3. Also, we need gcd(a,b)>1, gcd(a,c)>1, and gcd(b,c)>1 (otherwise some fencepost will get painted twice). Then we just check the cases: a=3, b=2, b=3, c=2, and c=3. The only solutions are (a,b,c)=(4,2,4) and (a,b,c)=(3,3,3). So the answer is 757.\n\n[/hide]" } { "Tag": [], "Problem": "Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that\r\n\r\n(A) No member of $S$ is divisible by 2.\r\n(B) No member of $S$ is divisible by 3 but some member is divisible by 11.\r\n(C) no member of $S$ is divisible by 3 or by 5.\r\n(D) No member of $S$ is divisible by 3 or by 7.\r\n(E) None of these.", "Solution_1": "Should I post an answer if someone else did the problem?\r\n\r\nIs there a \"do it yourself\" policy?", "Solution_2": "[hide]\n(B)\nI just guessed and checked\n$1^2 + 2^2 + 3^2 = 14$ so it can't be A\n$4^2 + 5^2 + 6^2 = 77$ b looks good\n$3^2 + 4^2 + 5^2 = 50$ cant be c\n$4^2 + 5^2 + 6^2 = 77$ can't be d\n\nSo it must be B.\n\nI guess maybe none are divisble by 3 because\n$(m+1)^2 + (m+2)^2 + (m+3)^2 = N$\nSo dividing by 3 would give their average. None of them are an equal distance away (ie 1 (3) 4 (5) 9); they'll always be consecutive odds apart. Because of this, their average will never be a whole number so none of the sums will be divisible by 3.\n[/hide]", "Solution_3": "SnowStorm: \r\n[hide][quote=\"SnowStorm\"]\n\nI guess maybe none are divisble by 3 because\n$(m+1)^2 + (m+2)^2 + (m+3)^2 = N$\n \n [/quote]\n\nYou could have selected $(n-1)^2+n^2+(n+1)^2$ to make it a little easier as $3n^2+2$ to see that you will always get \na remainder of 2 when divided by 3.\n\nThe expression you selected is little lengthy but still equal to $3m^2+12m+ 14 = 3*(m^2+4m+4) + 2 $ which again can not be a multiple of 3.\nHope this helps.\n\nGreat work![/hide]", "Solution_4": "[quote=\"hiitsme\"]Should I post an answer if someone else did the problem?\n\nIs there a \"do it yourself\" policy?[/quote]\r\n\r\nThanks for asking. Students post solutions, but use the \"spoiler text\" to wrap their solutions up so that nobody sees them unless they choose.", "Solution_5": "I meant if someone else did the problem for you, and you did'nt solve the problem yourself, can you still post a solution?\r\n\r\nEg. Someone at your house, school, etc.", "Solution_6": "Sure, though it's probably fair to give proper credit.", "Solution_7": "[quote=\"SnowStorm\"][hide]\n(B)\nI just guessed and checked\n$1^2 + 2^2 + 3^2 = 14$ so it can't be A\n$4^2 + 5^2 + 6^2 = 77$ b looks good\n$3^2 + 4^2 + 5^2 = 50$ cant be c\n$4^2 + 5^2 + 6^2 = 77$ can't be d\n\nSo it must be B.\n\nI guess maybe none are divisble by 3 because\n$(m+1)^2 + (m+2)^2 + (m+3)^2 = N$\nSo dividing by 3 would give their average. None of them are an equal distance away (ie 1 (3) 4 (5) 9); they'll always be consecutive odds apart. Because of this, their average will never be a whole number so none of the sums will be divisible by 3.\n[/hide][/quote]\r\n\r\nhow do you know it cant be (E)?(not saying it is but....)" } { "Tag": [ "AMC", "AIME" ], "Problem": "Is it possible to take both the AIME and the Alternate AIME?", "Solution_1": "I highly doubt it. There are two AMCs probably just so people can take the 10 and 12, or for people who mess up really bad (unwanted side effect, probably).\r\nTaking 2 AIMEs doesn't make much sense.", "Solution_2": "No, it is not allowed.", "Solution_3": "[quote=\"not_trig\"]I highly doubt it. There are two AMCs probably just so people can take the 10 and 12, or for people who mess up really bad (unwanted side effect, probably).\nTaking 2 AIMEs doesn't make much sense.[/quote]\r\n\r\nBy this same logic I don't see why it wouldn't make sense to be able to take both AIMEs. The only problem is that people who could only take one would be very disadvantaged.", "Solution_4": "It is not possible to take both officially, as in, have both count for official scores from the AMC", "Solution_5": "You could, but that'd be cheating... and NO CHEATING ALLOWED.\r\n\r\nSo, no.", "Solution_6": "[quote=\"Phelpedo\"][quote=\"not_trig\"]I highly doubt it. There are two AMCs probably just so people can take the 10 and 12, or for people who mess up really bad (unwanted side effect, probably).\nTaking 2 AIMEs doesn't make much sense.[/quote]\n\nBy this same logic I don't see why it wouldn't make sense to be able to take both AIMEs. The only problem is that people who could only take one would be very disadvantaged.[/quote]\r\n\r\n\r\nNote the parenthetical statement.", "Solution_7": "You don't think that people can \"mess up\" on the AIME?" } { "Tag": [ "geometry", "3D geometry", "sphere", "pyramid", "ratio", "Pythagorean Theorem" ], "Problem": "A sphere is inscribed inside a pyramid with a square as a base whose height is sqrt(15)/2 times the length of one edge of the base. A cube is inscribed inside the sphere. What is the ratio of the volume of the pyramid to the volume of the cube?\r\n\r\n\r\nI have the answer, but would someone like to show me a DETAILED solution?\r\n\r\nthx :)", "Solution_1": "This isn't exactly a detailed solution...\r\n\r\n[hide] Basically it boils down to finding the inradius. Do this by breaking the big pyramid into four tetrahedra and one square pyramid, using the incenter. Then, find the volume in two ways.[/hide]", "Solution_2": "[hide=\"Solution\"]Let the side length of the base of the pyramid be $ s$. Then, the height is $ \\frac {s\\sqrt {15}}{2}$, and by the Pythagorean Theorem, the slant height is $ 2s$.\n\nSetting up a ratio to solve for $ r$, the radius of the inscribed sphere, we have:\n\\[ \\frac {\\frac {s\\sqrt {15}}{2} \\minus{} r}{r} \\equal{} \\frac {2s}{\\frac {s}{2}}\n\\]\nThus,\n\\[ r \\equal{} \\frac {s\\sqrt {15}}{10}\n\\]\nTherefore, the volume of the inscribed cube is:\n\\[ {\\left(\\frac {\\frac {2s\\sqrt {15}}{10}}{\\sqrt {3}}\\right)}^{3} \\equal{} \\frac {s^{3}}{5^{\\frac {3}{2}}}\n\\]\nThe desired ratio of the volume of the pyramid to the volume of the sphere is then:\n\\[ \\frac {\\frac {s^3\\sqrt {15}}{6}}{\\frac {s^{3}}{5^{\\frac {3}{2}}}} \\equal{} \\boxed{\\frac {25\\sqrt {3}}{6}}\n\\]\n[/hide]", "Solution_3": "I believe that, in general, the radius of a sphere inscribed inside a square pyramid with height $ h$ and base side length $ s$ is given by:\r\n\\[ r \\equal{} \\frac {hs}{\\sqrt {4h^2 \\plus{} s^2} \\plus{} s}\r\n\\]\r\nAm I right?" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "find all (a,b,f,g) such that hold condition 1,2,3\r\n1) $a,b\\in\\mathbb{N}$and$1.>", "Solution_19": "i guess it's sorta a majority\r\n\r\n\r\npoke...\r\n\r\nno post for 4 days", "Solution_20": "I am working on this behind the scenes, don't worry.\r\n\r\nLet's just say that a comod is being instated...", "Solution_21": "ok\r\n\r\nPlease lengthen this message...", "Solution_22": "[quote=\"perfect628\"]\nI am working on this behind the scenes, don't worry.\n\nLet's just say that a comod is being instated...\n[/quote]\r\n\r\nfor one month...\r\n\r\ncan we at least get a definitive answer on whether this is abandoned or not?", "Solution_23": "i think it's safe to say it has been mod-abandoned.", "Solution_24": "I did try and get this started again but it's getting a bit ridiculous now isn't it.\r\n\r\n[hide=\"Abandon\"]The big twist was: there were [b]two[/b] Moles. They were...zapi2007 and kimmystar94. (rep Catalyst). At F6, the one who had been answered the least on Q11 was to move on, the other one revealed and executed. You were scored as follows: I scored your quizzes as if each were the Mole, and added. Kimmy as winning the Mole total thing 1-6. [/hide]", "Solution_25": "Perfect, please respond to my pm!", "Solution_26": "o, but then i would still remain the mole as we kept going down to the winners.", "Solution_27": "dang so this game was supposed to have ended on june 30", "Solution_28": "well, to keep it short, i am not a very good liar\r\nbut that does explain why i kept pushing for this game to continue", "Solution_29": "I was really suspicious that there were 2 moles 'cause of the double eliminations. But I never found the mole :(" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>0$ prove that\r\n\r\n\\[ \\frac{a}{(b\\plus{}c)^2}\\plus{}\\frac{b}{(a\\plus{}c)^2}\\plus{}\\frac{c}{(b\\plus{}a)^2}\\ge \\frac{3(a^2\\plus{}b^2\\plus{}c^2)(a\\plus{}b\\plus{}c)}{4(ab\\plus{}bc\\plus{}ca)^2}\\]", "Solution_1": "[quote=\"M.A\"]Let $ a,b,c > 0$ prove that\n\\[ \\frac {a}{(b \\plus{} c)^2} \\plus{} \\frac {b}{(a \\plus{} c)^2} \\plus{} \\frac {c}{(b \\plus{} a)^2}\\ge \\frac {(a^2 \\plus{} b^2 \\plus{} c^2)(a \\plus{} b \\plus{} c)}{2(ab \\plus{} bc \\plus{} ca)^2}\n\\]\n[/quote]\r\n\r\nI'm sorry i edit the problem.", "Solution_2": "Which is the \"final version\" of the problem? The one with the $ \\frac{3}{4}$ or the one with the $ \\frac{1}{2}$?" } { "Tag": [ "algebra", "system of equations", "algebra proposed" ], "Problem": "$a,b,c,d>0$ and satisfy following system of equations :\r\n\r\n\\[ a^{3}+b^{3}+c^{3}=3d^{3} \\]\r\n\\[ b^{4}+c^{4}+d^{4}=3a^{4} \\]\r\n\\[ c^{5}+d^{5}+a^{5}=3b^{5} \\]\r\n\r\nProve that $a=b=c=d$\r\n\r\nEDIT : :blush:", "Solution_1": "Change all the variables to a's because of what you want to show\r\n(a = b = c = d):\r\n\r\na^3 + a^3 + a^3 = a^3\r\n\r\na^4 + a^4 + a^4 = a^4\r\n\r\na^5 + a^5 + a^5 = a^5\r\n\r\n& they become\r\n\r\n3a^3 = a^3\r\n\r\n3a^4 = a^4\r\n\r\n3a^5 = a^5\r\n\r\n& they become\r\n\r\n2a^3 = 0\r\n\r\n2a^4 = 0\r\n\r\n2a^5 = 0\r\n\r\n& they become\r\n\r\na^3 = 0.......which begets a = 0, but a > 0 in the set-up\r\n\r\na^4 = 0.......which begets a = 0, but a > 0 in the set-up\r\n\r\na^5 = 0.......which begets a = 0, but a > 0 in the set-up\r\n\r\n\r\n**** NOTE: The above work was presented here by me BEFORE the \r\n author made an edit to the problem. ****" } { "Tag": [ "geometry", "geometric transformation", "reflection", "analytic geometry", "exterior angle", "number theory" ], "Problem": "the older one is getting lengthy so i think u can discuss further here.......", "Solution_1": "the older one is getting really lengthy...\r\ni just wanted to clarify a few things... \r\n\r\n\"From 2008, INMO awardees applying for B. Stat or B. Math courses of the Indian Statistical Institute will be directly called for the interview without going through the written test.\" (and also CMI)\r\nlooks like this was written b4 it was actually implemented.. ny1 has ny idea how one might be able to confirm this?\r\n\r\nAlso, just keep posting the updates on the training camp as and when u come to know of it... \r\nCurrently i only know the dates... thru the RC\r\n29th Apr - 27th May", "Solution_2": "Please check my answer to question no. 5\r\n\r\nLet the greatest altitude be AH \r\nTherefore $ BC \\le AB$ as well as $ AC$\r\n\r\nNow produce $ HD$ to $ H$[b]'[/b] such that $ HD \\equal{} H$[b]'[/b]$ D$\r\nTherefore $ H$[b]'[/b] is the reflection of $ H$ \r\nSo $ CH \\equal{} CH$[b]'[/b] and $ BH \\equal{} BH$[b]'[/b] \r\nThis implies that $ \\angle 1 \\equal{} \\angle 2$\r\n$ \\equal{} \\angle 3$ (since $ \\angle 2$ is the exterior angle of cyclic quad $ BDHF$)\r\nSo $ ACH$[b]'[/b]$ B$ are cyclic\r\n\r\nApplying Ptolemy\u2019s theorem we get\r\n$ AH$[b]'[/b] x $ BC$$ \\equal{} AB$ x $ CH$[b]'[/b]$ \\plus{} AC$ x $ BH$[b]'[/b]\r\nThis implies that $ AD \\plus{} DH \\equal{} (AB/BC)$ x $ CH \\plus{} (AC/BC)$ x $ BH$\r\nSince each of the fractions is less than or equal to 1\r\n$ AD \\plus{} DH \\ge CH \\plus{} BH$\r\nAdding $ AH$ to both sides the result follows.\r\n$ Q.E.D.$\r\n\r\nEDIT: the figure will appear black coloured--click on the figure to see it", "Solution_3": "i guess u marked angle '3' incorrrectly in the diagram.. \r\nbut the solution is really nice.. elegant", "Solution_4": "Woot! I got 4 in my INMO performance card! Looks like the examiners couldn't make out even the 1 1/2 solns. I did write :rotfl:", "Solution_5": "hey what's the INMO performance card.......please explain......\r\n\r\n@keshav-thanx..", "Solution_6": "[hide=\"@keshav\"] -would you please be so kind to post your coordinate geometry solution in the question linked to the resources section......... :maybe: \nalso can you post some of your other solutions here...thanx :lol: \nfinally i was thinking that it would be great if we had a chat online so it would be great(for me)if you could come online on msn,google talk or yahoo messenger....thanx again :blush: \n[/hide]\n[hide=\"to others\"]this thread is turning out to be dead one.......do something :o[/hide]", "Solution_7": "[quote=\"Rijul saini\"]\nwould you please be so kind to post your coordinate geometry solution in the question linked to the resources section......... :maybe: \nalso can you post some of your other solutions here...thanx :lol: \nfinally i was thinking that it would be great if we had a chat online so it would be great(for me)if you could come online on msn,google talk or yahoo messenger....thanx again :blush: \n[/quote]\r\n\r\nresources section??\r\n\r\ni have already posted 2 of the three sums i solved...\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1380362#1380362]Geometry: Problem 5[/url]\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251942&start=48]Co-ordinate Geometry: Problem 1[/url]\r\n\r\ni have sent u a private message with my id..", "Solution_8": "hey can anyone clear my doubt about the first question.\r\n\r\nif we take $ P$ as $ (0,0)$\r\n$ B \\equiv (0,2b)$\r\n$ C \\equiv (2c,0)$\r\n\r\nthen $ \\overline{BP} \\equal{} 2b$ \r\n\r\nnow taking $ P$ as centre and Half $ \\overline{BP}$ as radius draw the arc in only the fourth quadrant\r\n\r\nFrom $ C$ we make the tangent to the circle and choose the tangent cutting the circle in the fourth quadrant , name the point of intersection as $ M$ and extend $ CM$ to $ A$ such that $ CM \\equal{} AM$\r\nthus we get the required points $ A,B,C,P$ without making $ \\angle PCB \\equal{} \\angle PAB$ .......\r\n\r\nP.S.- i'm choosing only the fourth quadrant since if we get $ A$ from the other tangent $ P$ will not be inside the triangle", "Solution_9": "@Rijul...\r\nsorry for the incorrect explaining on gtalk... \r\ndidnt notice something...\r\n\r\nu cant take M to be on the tangent from C to the circle... \r\nM can be any point on the circle... as long as u extend it to A such that AM = CM\r\nthe given construction will always have PM perpendicular to AC... \r\nwhich is NOT NECESSARY acc. to the given problem... \r\n\r\nout of a many M's possible for given b and c\r\nu need to choose the one which gives A such that the given angles are equal.\r\n\r\ni hope i am clear enough...", "Solution_10": "[quote=\"nuclear_alchemist\"]Woot! I got 4 in my INMO performance card! Looks like the examiners couldn't make out even the 1 1/2 solns. I did write :rotfl:[/quote]\r\n\r\nIs there anything like INMO performance card??....coz i havent received it yet??\r\nplus any idea from where can we get our marks....they use to tell the awardees their marks a couple pf years back..", "Solution_11": "Ummm I got my marks so I guess the performance report exists. Allso is anyone applying for re-evaluation? 2 of my friends are considering it coz they got 4 ques correct...", "Solution_12": "[quote=\"nuclear_alchemist\"]Ummm I got my marks so I guess the performance report exists. Allso is anyone applying for re-evaluation? 2 of my friends are considering it coz they got 4 ques correct...[/quote]\r\n\r\nDid you get the performance card from HBCSE or did u get it from ur local coordinator??...plus how long has it been since u received the card...", "Solution_13": "CAN WE APPLY FOR RE EVALUATION?\r\nIS IT POSSIBLE?", "Solution_14": "well, arvind, how much did u get in INMO acc to performance card? i think this idea is catching up like in other INOs :wink: !!!\r\nBut like in other INOs the results shud be announced after re-evaluation!! imagine the uproar one wud cause if he has initially got selected but after re-checking of others, he didn't qualify!!! :lol:", "Solution_15": "congrats to everyone who got through to the camp.\r\nI've been planning to drop in at the camp each year, but not getting around to it. :( \r\nI hear now they have no soccer and stuff like we used to in the good old days.\r\nanyway, as for the direct admission to B.Stat. or B.Math, I can personally confirm this to be true. But while the written test is waived, appearing for the interview is still compulsory (though that'll hardly be a problem around here).", "Solution_16": "Hey when is the imotc starting?", "Solution_17": "must have already started (on april 27th) :P", "Solution_18": "Can anybody tell that what is the INMO performance card?\r\nWhen did you receive it?\r\nDid you get it directly from HBCSE through post or from the regional coordinator.\r\nAlso,for those who have received it- is anyone from the delhi centre?(bcos I am and I havn't received mine)", "Solution_19": "Please reply :(", "Solution_20": "RC\r\nchar lim", "Solution_21": "Hey!!!!!!!! :mad: \r\nPlease reply to my post.\r\nIs it a fashion to not reply to the posts of new members.\r\n\r\nMy question was(and is there still)\r\nCan anybody tell me what is the INMO performance card? \r\nWhen did you receive it? \r\nDid you get it directly from HBCSE through post or from the regional coordinator?\r\nAlso,for those who have received it- is anyone from the delhi centre?(bcos I am and I havn't received mine) :(", "Solution_22": "[quote=\"nuclear_alchemist\"]RC\nchar lim[/quote]\r\nRC=Regional Coordinator(Obviously by post, he won't come to my doorstep)\r\nchar lim= character limit(my post was too short for aops to accept originally)\r\nwhen did I receive it?\r\naround the same time as the big guys got their INO cards...\r\nwhat is it?\r\njust a piece of paper with your name and score I guess....no certificate", "Solution_23": "i have not got my NSEP and NSEB scorecards yet :rotfl:", "Solution_24": "Hey anybody who's currently attending the camp,\r\nCan anybody tell the possible IMO team now that all the Selection Tests are over\r\n\r\nand bw after how many posts would this tag of 'New Member' be removed from underneath my name.", "Solution_25": "Gaurav Patil\r\nShubhashish\r\nAkshay Degvekar\r\nAnand Degvekar\r\nSameer wagh\r\nGanesh Ajjangadde\r\n[/hide]\r\nI am not in the camp :( \r\nbut this is the general opinion of the campers!", "Solution_26": "I heard that Ananth shankar from Tamil Nadu is in with a very good chance :?: \r\n\r\n@abhitayal: 20 posts and you're no longer a 'New Member'.", "Solution_27": "How many camps are Akshay Degvekar and Anand Degvekar (DES school Pune)attending??\r\nAnd they are in 11th \r\nThey were selected for Informatics camp in 10th :roll: and now 11th.\r\n\r\nAny idea how Keshav has done??", "Solution_28": "i want Akshay Degvekar , Anand Degvekar and Keshav Dhandhania to go to IMO .\r\nThere'll be less competition in IOITC . :D :D :D", "Solution_29": "I heard something about Ankush Das,Aakansh Gupta,Akshay Mittal that they are strong contenders for being in the team\r\nAccording to my source, Ankush and Akshay have done 6 questions each out of 12 in the selection tests\r\nand only 4 have done better that them highest being 9 questions", "Solution_30": "BTW, is Ganesh Ajjangadde some ninth(or eight ) grader from karnataka ?", "Solution_31": "well usually they prefer 12thies for IMO (i m not sure)\r\n\r\ncoz others hav more chances.... :)", "Solution_32": "NO.The selection is as per the merit.\r\n\r\nGanesh is in 9th and 9th graders have been in tne IMO teams in the past and got medals!", "Solution_33": "Idont think 6 is enough for getting inot the team. All the geometry were easy in the TSTs and there were some easy number theory and algebra problems too.\r\nSo I think 8 or 9 out of 12 would get one through to the team.\r\nI hope there is some one from TN....Maybe Ashwath or Ananth Shankar.", "Solution_34": "i just hope Ashwath and Nikhil/Akshay mittal are in the team :D" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If a,b,c are sides of a triangle then prove that,\r\n[b] a/(b+c)+b/(c+a)+c/(a+b) < 2[/b]", "Solution_1": "We have : $ \\frac{a}{b\\plus{}c} \\le \\frac{2a}{a\\plus{}b\\plus{}c}$\r\n Thus, $ LHS \\le 2$", "Solution_2": "Oh! It was easy! So silly of me to post such an easy problem! :blush:", "Solution_3": "[b]1). [/b]If m,n,x>0 and $ \\frac {m}{n} < 1$, then: $ \\frac {m}{n} < \\frac {m \\plus{} x}{n \\plus{} x} < 1$.\r\n[b]2). [/b]If m,n,x>0 and $ \\frac {m}{n} > 1$, then: $ 1 < \\frac {m \\plus{} x}{n \\plus{} x} < \\frac {m}{n}$.\r\n[b]3).[/b] If a,b and c side of the triangle, then: $ a < b \\plus{} c \\Longrightarrow \\frac {a}{b \\plus{} c} < 1 \\Longrightarrow \\frac {a}{b \\plus{} c} < \\frac {2a}{a \\plus{} b \\plus{} c}.$" } { "Tag": [ "geometry", "3D geometry", "calculus", "integration" ], "Problem": "Hi, I'm preparing my final when coming across this problem. I have no idea to solve it. The problem goes like this: \r\n\r\nA cone with base R and heigh h is located on a horizontal table. A horizontal uniform field E penetrates the cone. Determine the electric flux that enters the left-hand side of the cone. \r\n\r\nI've tried to use Guass's law but I can't find the area and the angle is changing.", "Solution_1": "Hmm ... isn't it just $\\Phi = ERh$?\r\n\r\n$\\Phi = E.S$ and since the flux entering the cone is the same as the flux going through its midsection (vertical section prependicular to the field, just a triangle with base $2R$ and altidude $h$), it's just $E.(2R.h)/2 = E.R.h$.", "Solution_2": "How can you prove them though? My teacher requires a formal proof which includes intergration", "Solution_3": "Well. If your teacher requires integration, I'm sorry...\r\n\r\nIf a formal proof is enough, then take a half-cone = cone sliced by a vertical cut through the vertex, perpendiculer to the E-field. There is no charge inside this half-cone, therefore the total flux through it is zero. There is no flux through the base, because it is perpendicular. The outward flux through the triangular cut is $\\Phi = ERh$ as above. Hence the total flux through the rest of the half-cone (which is exactly the area we want to consider) is $ERh$ inwards." } { "Tag": [ "geometry", "trigonometry", "ratio", "perimeter", "angle bisector" ], "Problem": "I'm not really good at math, cuz everybody in art of problem solving is ahead of me. Does anybody know how to measure areas of a pentagon, hexagon, hectagon, octagon, nonogan, or decagon? :|", "Solution_1": "There is a general formula for an $n$-gon (which can be found on Wikipedia), but it involves trig, so I won't post it here.\r\n\r\nYou might want to ask this in HSB or Intermediate, or ask a mod to move it.", "Solution_2": "(1/2)ap, where a is the apothem and p is the perimeter. So that's the formula if you're actually given a side length and apothem. But you would have to use trig to find the apothem in many cases.", "Solution_3": "[img]http://intermath.coe.uga.edu/topics/geometry/polygons/resrces/apothem.gif[/img]\r\n\r\nTake for example, the regular hexagon above. The quintessential formula would be $ AP$/$ 2$ in which A stands for the apothem, and P for the perimeter. [b](Note: This formula only works for regular convex polygons)[/b]\r\n\r\nNow generally in the problems I receive, I'm given only the apothem, so this is how you would use trignometry to solve for the perimeter.\r\n\r\nLet's assume that there is a regular polygon such that, the apothem is 6. Find the area?\r\n\r\nFirst what's the measure of each angle of a hexagon? $ (4*180)$/$ 6$= $ 120$\r\n\r\nTake the radius of the hexagon. This serves as an angle bisector, and with the apothem becomes a 30-60-90 triangle.\r\n\r\nThe properties of the sides of 30-60-90 triangle are in the ratio:\r\n[img]http://mathcentral.uregina.ca/QQ/database/QQ.09.05/gary1.1.gif[/img]\r\n\r\nBecause we know the apothem is 6, we can make out from the above image, that half the side of the hexagon $ 6$/$ \\sqrt{3}$ or $ 2$ $ \\sqrt{3}$.\r\n\r\nWe thus know, the perimeter is $ 24$$ \\sqrt{3}$. \r\n\r\nThuse the area would be by $ AP$/$ 2$ : $ 72$ $ \\sqrt{3}$", "Solution_4": "Are you talking about regular (equiangular and equilateral) polygons? Then yes, there is a general formula. \r\n\r\nMore useful than a general formula is a general [i]technique.[/i] What you want to do is split up a complicated area into simple areas (usually triangles if you're talking about polygons). There are lots of ways to find the area of triangles (depending on what you're given), some of which involve trigonometry. One of those ways ($ [ABC] \\equal{} \\frac {1}{2} ab \\sin C$) is used to prove the general regular $ n$-gon formula.", "Solution_5": "yeah...for example take a hexagon with side length \"s.\" divide the hexagon's area into 6 equilateral triangles. The formula for the area of an equilateral triangle is (\"s\" squared times the square root of 3)/4... now that you have the area of the single equilateral triangle, multiply it by 6 and voila now you have the area of the whole hexagon.", "Solution_6": "man, how did u guys manage to dig up a one-year old post? LOL. i don't remember posting this...o well. :lol:" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "The incircle of a triangle $ ABC$ is tangent to its sides $ AB,BC,CA$ at $ M,N,K,$ respectively. A line $ l$ through the midpoint $ D$ of $ AC$ is parallel to $ MN$ and intersects the lines $ BC$ and $ BD$ at $ T$ and $ S$, respectively. Prove that $ TC\\equal{}KD\\equal{}AS.$", "Solution_1": "I think S in the intersection of BA and the parallal line\r\n\r\n1.it is obvious KD=1/2(a-c)\r\n2.using sin law we can calculate CK=AS \r\nthen CK+AS=BS-c+a-BT=c-a\r\nCK=AS=KD=(a-c)/2", "Solution_2": "i think this problem is not difficult, . we can compute the length of $ MN$ and use the Thales 's theorem.\r\nit's Ok :wink:", "Solution_3": "Draw a line through A parallel to MN, and let it meet BC at P. Note that BMN is isosceles. We get AS=PT. Since M is midpoint of AC we have PT=TC, so AS=TC as desired. Now note that CN=CK, and we have CN=2TC+NP and CK=2CD-NP (to see the last assertion observe that AK=AM=NP). This yields AD=CD=TC+NP=NT=MS. Subtract AM=AK and the result follows." } { "Tag": [ "search", "superior algebra", "superior algebra theorems" ], "Problem": "Hello! I was wondering if anybody could tell me what Algebric closure is , what are the algebric closed bodies (it is for Romanian NMO)", "Solution_1": "Hi! :)\r\nhttp://mathworld.wolfram.com/AlgebraicClosure.html\r\nhttp://planetmath.org/encyclopedia/AlgebraicClosure.html\r\nhttp://en.wikipedia.org/wiki/Algebraic_closure\r\n\r\nAlso, searching for \"Algebraic closure\" here: http://www.mathlinks.ro/Forum/search.php, will give you results that may interest you, such as : http://www.mathlinks.ro/Forum/viewtopic.php?t=66543 ;)\r\n\r\nTo put it in one word : [i]search[/i]." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all the $ n\\in \\mathbb{N}$ such that $ 1205\\ |\\ 8^{n }\\minus{} 1$ .", "Solution_1": "You probably made a typo by writing $ 1205|8^{n\\minus{}1}$, so I assume you meant $ 1205|(8^n\\minus{}1)$. Since $ 8^2\\equiv \\minus{}1\\bmod{5}$ the order of $ 8$ modulo $ 5$ is four. Since $ 8^4\\equal{}4097\\equiv \\minus{}1\\bmod{241}$ holds, the order of $ 8$ modulo $ 241$ is eight. Due to $ 1205\\equal{}5\\cdot 241$ we have $ 1205|8^{n}\\minus{}1\\Longleftrightarrow 8|n$.", "Solution_2": "You're right . I was wrong :blush:" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "The following facts are known on a mathematical contest:\r\n(i) There were $ n \\geq 4$ problems;\r\n(ii) Each problem was solved by exactly four contestants;\r\n(iii) For any two problems there is exactly one contestant who solved both\r\nproblems.\r\nAssuming that there were at least $ 4n$ contestants, find the minimum value\r\nof $ n$ for which there always exists a contestant who solved all the problems.", "Solution_1": "Anybody? :)" } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometry solved" ], "Problem": "Let be given a triangle ABC with sides a,b,c,R is circumradius. M is a point inside ABC. Prove that:\r\n[b]a)[/b] [tex] \\frac{MA.MB}{MC.c}+\\frac{MB.MC}{MA.a}+\\frac{MC.MA}{MB.b}\\geq\\sqrt{3}[/tex]\r\n[b]b)[/b] [tex] \\frac{MA}{a^2}+\\frac{MB}{b^2}+\\frac{MC}{c^2}\\geq\\frac{1}{R}[/tex]", "Solution_1": "First one is on the forum.", "Solution_2": "Yes(a) is not hard,but what about (b)?", "Solution_3": "Is $\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}\\geq \\frac{1}{R^2}$ a known inequality?", "Solution_4": "yes\r\n $\\sum{\\frac{1}{a^2}} \\geq \\frac{9}{a^2+b^2+c^2}$\r\n\r\n$9R^2\\geq a^2+b^2+c^2$", "Solution_5": "the last is true since\r\n$OH^2=9R^2-(a^2+b^2+c^2)$\r\nwhere $O$ and $H$ mean the usual", "Solution_6": "Also, by Rearrangement: \\[\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2} \\geq \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}\\]So, Myth's inequality follows from\r\n\\[\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}\\geq \\frac{1}{R^2}\\]which has been posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15798[/url]", "Solution_7": "It is not my inequality :P I just substituted $M=O$ into (b).", "Solution_8": "So let's solve the initial problems [b](a)[/b] and [b](b)[/b]:\r\n\r\nThe point M can be an arbitrary point in the plane of triangle ABC, not necessarily inside $\\triangle ABC$. So we get the following extended problem:\r\n\r\n[color=blue][b]Extended problem.[/b] Let ABC be a triangle with sidelengths a, b, c and circumradius R. Prove the inequalities\n\n$\\frac{MB\\cdot MC}{MA\\cdot a}+\\frac{MC\\cdot MA}{MB\\cdot b}+\\frac{MA\\cdot MB}{MC\\cdot c}\\geq\\sqrt3$\n\nand\n\n$\\frac{MA}{a^2}+\\frac{MB}{b^2}+\\frac{MC}{c^2}\\geq\\frac{1}{R}$,\n\nwhere M is an arbitrary point in the plane of triangle ABC.[/color]\r\n\r\n[i]Solution.[/i] The second inequality was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=6073 posts #2 and #6, so we will only prove the first inequality here.\r\n\r\nBy http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 post #1, for any three reals x, y, z, we have the inequality\r\n\r\n$\\left(x+y+z\\right)\\left(x\\cdot MA^2+y\\cdot MB^2+z\\cdot MC^2\\right)\\geq yza^2+zxb^2+xyc^2$.\r\n\r\nParticularly, if we take x = a, y = b, z = c, this becomes\r\n\r\n$\\left(a+b+c\\right)\\left(a\\cdot MA^2+b\\cdot MB^2+c\\cdot MC^2\\right)\\geq bca^2+cab^2+abc^2$.\r\n\r\nThis is equivalent to\r\n\r\n$\\left(a+b+c\\right)\\left(a\\cdot MA^2+b\\cdot MB^2+c\\cdot MC^2\\right)\\geq abc\\left(a+b+c\\right)$,\r\n\r\nand upon division by abc (a + b + c), this becomes\r\n\r\n$\\frac{a\\cdot MA^2+b\\cdot MB^2+c\\cdot MC^2}{abc}\\geq 1$.\r\n\r\nBut now, for any three reals u, v, w, we have $\\left(u+v+w\\right)^2\\geq 3\\left(vw+wu+uv\\right)$ (since $\\left(u+v+w\\right)^2-3\\left(vw+wu+uv\\right)=\\frac12\\left(\\left(v-w\\right)^2+\\left(w-u\\right)^2+\\left(u-v\\right)^2\\right)\\geq 0$); applying this to the reals $u=\\frac{MB\\cdot MC}{MA\\cdot a}$, $v=\\frac{MC\\cdot MA}{MB\\cdot b}$ and $w=\\frac{MA\\cdot MB}{MC\\cdot c}$, we get\r\n\r\n$\\left(\\frac{MB\\cdot MC}{MA\\cdot a}+\\frac{MC\\cdot MA}{MB\\cdot b}+\\frac{MA\\cdot MB}{MC\\cdot c}\\right)^2$\r\n$\\geq3\\left(\\frac{MC\\cdot MA}{MB\\cdot b}\\cdot\\frac{MA\\cdot MB}{MC\\cdot c}+\\frac{MA\\cdot MB}{MC\\cdot c}\\cdot\\frac{MB\\cdot MC}{MA\\cdot a}+\\frac{MB\\cdot MC}{MA\\cdot a}\\cdot\\frac{MC\\cdot MA}{MB\\cdot b}\\right)$\r\n$=3\\left(\\frac{MA^2}{bc}+\\frac{MB^2}{ca}+\\frac{MC^2}{ab}\\right)=3\\cdot\\frac{a\\cdot MA^2+b\\cdot MB^2+c\\cdot MC^2}{abc}\\geq 3$,\r\n\r\nsince $\\frac{a\\cdot MA^2+b\\cdot MB^2+c\\cdot MC^2}{abc}\\geq 1$. Thus,\r\n\r\n$\\frac{MB\\cdot MC}{MA\\cdot a}+\\frac{MC\\cdot MA}{MB\\cdot b}+\\frac{MA\\cdot MB}{MC\\cdot c}\\geq\\sqrt3$,\r\n\r\nand the first inequality is proven. This solves the problem.\r\n\r\n Darij" } { "Tag": [ "inequalities", "linear algebra", "linear algebra unsolved" ], "Problem": "Let $ A$, $ B$, $ C$ be three $ n\\times n$ matrices such that $ C$ is commutative with $ A$ and $ B$, $ C^2 \\equal{} I$ and $ AB \\equal{} 2(A \\plus{} B)C$. \r\n1. Prove that: $ AB \\equal{} BA$.\r\n2. If we have another condition: $ A \\plus{} B \\plus{} C \\equal{} 0$, then prove that\r\n\\[ \\text{rank} (A \\minus{} C) \\plus{} \\text{rank} (B \\minus{} C) \\equal{} n\r\n\\]", "Solution_1": "1)We have $ 4I_n \\equal{} AB \\minus{} 2(A \\plus{} B)C \\plus{} 4C^2 \\equal{} (A \\minus{} 2C)(B \\minus{} 2C)$ therefore $ (A \\minus{} 2C),(B \\minus{} 2C)$ is commutative\r\nThen $ (A \\minus{} 2C)(B \\minus{} 2C) \\equal{} (B \\minus{} 2C)(A \\minus{} 2C)$ implies $ AB \\minus{} 2(A \\plus{} B)C \\plus{} 4C^2 \\equal{} BA \\minus{} 2(A \\plus{} B)C \\plus{} 4C^2$ or $ AB \\equal{} BA$\r\n2)We have $ (A \\minus{} C)(B \\minus{} C) \\equal{} AB \\minus{} (A \\plus{} B)C \\plus{} C^2 \\equal{} (A \\plus{} B)C \\plus{} C^2 \\equal{} ( \\minus{} C)C \\plus{} C^2 \\equal{} O$ and $ (A \\minus{} C) \\plus{} (B \\minus{} C) \\equal{} \\minus{} 3C$\r\n Use inequality $ rank(X \\plus{} Y)\\leq rank(X) \\plus{} rank(Y) \\leq rank(XY) \\plus{} n$ ,$ \\forall X,Y \\in M_n{(\\mathbb{C})}$\r\nWe have :\r\n $ rank( \\minus{} 3C) \\leq rank(A \\minus{} C) \\plus{} rank(B \\minus{} C) \\leq rank((A \\minus{} C)(B \\minus{} C)) \\plus{} n$ or $ n \\leq rank(A \\minus{} C) \\plus{} rank(B \\minus{} C) \\leq n$ or $ rank(A \\minus{} C) \\plus{} rank(B \\minus{} C) \\equal{} n$\r\n :P", "Solution_2": "More precisely:\r\n$ C^2\\equal{}I,CA\\equal{}AC,BC\\equal{}CB$ then we may assume $ C\\equal{}diag(I_x,\\minus{}I_y),A\\equal{}diag(U_1,U_2),B\\equal{}diag(V_1,V_2)$. $ AB\\equal{}2(A\\plus{}B)C$ implies $ (U_1\\minus{}2I_x)(V_1\\minus{}2I_x)\\equal{}4I_x,(U_2\\plus{}2I_y)(V_2\\plus{}2I_y)\\equal{}4I_y$ or $ V_1\\equal{}4(U_1\\minus{}2I_x)^{\\minus{}1}\\plus{}2I_x,V_2\\equal{}4(U_2\\plus{}2I_y)^{\\minus{}1}\\minus{}2I_y$.\r\nMoreover $ A\\plus{}B\\plus{}C\\equal{}0$ implies $ U_1^2\\plus{}U_1\\minus{}2I_x\\equal{}0,U_2^2\\minus{}U_2\\minus{}2I_y\\equal{}0$.\r\nThus we may assume \r\n$ C\\equal{}diag(I_p,I_q,\\minus{}I_r,\\minus{}I_s)$,\r\n$ A\\equal{}diag(I_p,\\minus{}2I_q,\\minus{}I_r,2I_s)$,\r\n$ B\\equal{}diag(\\minus{}2I_p,I_q,2I_r,\\minus{}I_s)$ \r\nwith $ p\\plus{}q\\equal{}x,r\\plus{}s\\equal{}y$." } { "Tag": [ "linear algebra", "matrix", "search" ], "Problem": "let A be a $ n\\times n$ matrix, n>1, rank(A)=n-1. prove that $ rank(A^*)\\equal{}1$", "Solution_1": "Which $ A^*$ is this? Is it perhaps the classical adjoint - the matrix of deteminants of minors?", "Solution_2": "Let's assume that sunlight means classical adjoint of a matrix. I think the solution is best illustrated in the second post of Prof. Merryfield [url=http://www.mathlinks.ro/viewtopic.php?search_id=994667900&t=153199]here[/url].\r\n\r\n@sunlight You'll find the solution to your problem at the end of the page whose link I've given above, but I recommend you to read to whole solution there.", "Solution_3": "$ A \\,\\, adj(A) \\equal{} det(A) \\, I_n$\r\n\r\nTherefore, as rank(A) < n we have det(A) = 0, and this implies $ A \\,\\, adj(A) \\equal{} 0$. Now, if we have a matrix A m x n with rank r and if there is a matrix B n x p such that AB = 0, what can we say about the rank of B? Use the answer to this question, together with the obvious fact that rank[adj(A)] > 0, to complete this proof." } { "Tag": [ "trigonometry" ], "Problem": "What is the power factor of a 20 millihenry coil at 950 hertz if the resistance of the coil is 40 ohms?", "Solution_1": "basically, we want the cosine of the phase angle, and cos71= 0.32" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $x,y,z$ positive reals prove that\r\n\r\n$\\sqrt{\\frac{x(y+z)}{3}}+\\sqrt{\\frac{y(x+z)}{3}}+\\sqrt{\\frac{z(y+x)}{3}} \\leq \\frac{5}{6}(x+y+z) $", "Solution_1": "From Cauchy,\r\n\r\n$LHS \\leq \\sqrt{(1+1+1) \\sum \\frac{x(y+z)}{3}} = \\sqrt{2(xy+yz+zx)}$\r\n\r\n$\\leq \\sqrt{\\frac{2}{3}(x+y+z)^2} = \\frac{\\sqrt{6}}{3}(x+y+z) \\leq \\frac{5}{6}(x+y+z)$" } { "Tag": [ "vector" ], "Problem": "The non-collinear points $ A$, $ B$ and $ C$ have position vectors $ a$, $ b$ and $ c$, respectively. The points $ P$ and $ Q$ have position vectors $ p$ and $ q$ respectively, given by\r\n\r\n$ p\\equal{}\\lambda a \\plus{} ( 1\\minus{} \\lambda) b$ and $ q \\equal{} \\mu a \\plus{} (1\\minus{} \\mu) c$\r\n\r\nwhere $ 0 < \\lambda < 1$ and $ \\mu > 1$. Draw a diagram showing $ A$, $ B$, $ C$, $ P$ and $ Q$.\r\n\r\nGiven that $ CQ$ X $ BP \\equal{} AB$ X $ AC$, find $ \\mu$ in terms of $ \\lambda$, and show that, for all values of $ \\lambda$, the line $ PQ$ passes through the fixed point $ D$, with position vector $ d$ given by $ d\\equal{}\\minus{}a \\plus{} b \\plus{} c$. What can be said about the quadrilateral $ ABDC$?", "Solution_1": "Assuming the equality of cross-products refers to their magnitudes (as there are no arrows over the letters), I get $ \\mu \\equal{} \\frac{1}{\\lambda}$ (and not $ \\mu \\equal{} \\frac{\\minus{}1}{\\lambda}$). Substituting this into the vector equation of the line $ PQ$ and equating cofficients with $ \\minus{}a\\plus{}b\\plus{}c$, this works, and finally, $ d\\minus{}c \\equal{} b\\minus{}a$ makes $ ABCD$ a parallelogram." } { "Tag": [ "MATHCOUNTS", "logarithms", "function", "algebra", "domain", "AoPS Books", "number theory" ], "Problem": "Hi people!\r\nHere's a test to your ingeniosity: Prove that not all numbers are rational.\r\n\r\nPlease do not look in the AoPS books unless necessary.\r\nTry to be original.", "Solution_1": "I don't think this belongs in the MathCounts forum.\r\n\r\nAnyway, for my solution:\r\n\r\n[hide=\"Proof\"]Assume $\\left(\\frac{a}{b}\\right)^2=2$, where $a$ and $b$ are relatively prime integers. For this to be the case, either $a$ or $b$ must be odd. Multiplying out the equation, we have $a^2=2b^2$. However, if we divide both sides by $2$, we have $\\frac{a^2}{2}=b^2$. For $b$ to be an integer, we must have that $a$ is even, so it is evenly divided by $2$. However, $a^2$ must contain at least two factors of $2$, so $\\frac{a^2}{2}$ contains at least one factor of $2$. Therefore, $b^2$ is even, but the square of an odd number is never even. Therefore, we have both $a$ and $b$ even, which is a contradiction with our original assumption, and therefore, $\\sqrt{2}$ is irrational, so not all numbers are rational.\n\nI could go into a more detailed proof of the existence of other rational numbers, but this suffices for the given problem.[/hide]", "Solution_2": "No proofs in the middle school forums ;) so to Getting Started this goes.", "Solution_3": "Another number easily proven irrational:\r\n\r\n[hide]$\\log_2 3$. Suppose $\\log_2 3=\\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers.\n\nIn exponential form this is $2^\\frac{p}{q}=3$. Raise both sides to the $q$ power to get $2^p=3^q$.\n\nThe left side is even, the right side is odd. Contradiction.[/hide]", "Solution_4": "I don't mind keeping this is Getting Started, but sticking to moderating rules, this belongs in the Intermediate forum. :P \r\n\r\nSo off it goes. :arrow:", "Solution_5": "What about taking any rational number, and changing each of its decimal places as needed so that there's no pattern, and going off to infinity with this procedure? (just keep going after the decmial place)", "Solution_6": "[hide]Proof: consider the number i, it cannot be expressed by p/q where p and q are positibe integers. :D [/hide]", "Solution_7": "Taking \"number\" to mean \"any number expressible as an infinite sequence of digits,\" the proof is straightforward:\r\n\r\n[hide=\"Proof that irrationals exist\"]\nFirst, we examine the property of rationals. We note that for any number such that its decimal expansion terminates (i.e. it can be written $0.d_1d_2d_3...d_n$ for finite $n$), it can be written $\\frac{d_1d_2d_3...d_n}{10^n}$, so it is rational. For numbers whose decimal expansions do not terminate:\n\nLemma: $1 = .99999...$\n\nProof: $x = .999999... \\implies 10x = 9.99999... \\implies 10x - x = 9 \\implies x = 1$\n\nHence, we can see that for any rational number $\\frac{p}{q}$, we have $\\frac{p}{q} = \\frac{p \\times .9999999...}{q}$. By Euler's Theorem, we have $10^{\\phi(q)} \\equiv 1 \\bmod q$ (if $(10, q) = 1$; if not, take the largest factor of $q$ such that $(10, f) = 1$ and repeat the analysis), so we know that $9999...$ repeated $\\phi(q)$ times is divisible by $q$. \n\nThis means that $\\frac{.9999...}{q}$ repeated an infinite number of times will yield a sequence of repeating digits (since the $9$s can be taken to be in $\\phi(q)$-sized blocks). This establishes that $\\frac{p \\times .99999...}{q}$ consists of a set of repeating decimal digits that repeat, at maximum, after $\\phi(q)$ digits.\n\nProving that irrationals exist is simply a matter of proving that there exist numbers whose decimal digits do not repeat, which is trivial. Consider the Champernowne constant, $c = .123456789101112131415...$.[/hide]\n\nSecondly (a stronger proof):\n\n[hide=\"Proof that there are more irrationals than rationals\"]A proof I learned while reading about Cantor.\n\nLet us suppose that the number of rationals is equivalent to the number of irrationals. Then there is a bijective function $f(q), q \\in \\mathbb{Q}$, such that $f(q) \\in \\mathbb{R}$. \n\nNow let us construct an irrational number $n$ that cannot be generated by $f(q)$. List the input-output pairings created by $f(q)$ in order of the size of the denominator first (least to greatest), and then in order of the size of the numerator (least to greatest; sign is irrelevant). For the first pair (which would map $\\frac{1}{1}$ to some irrational number $R_{1, 1}$), let us set the first digit of $n$ such that $d(n, 1) \\neq d(R_{1, 1}, 1)$, where $d(n, k)$ denotes the $k^{th}$ decimal digit of $n$. \n\nNow, look at the next pair (which would map $\\frac{2}{1}$ to some irrational number $R_{2, 1}$), let us set the second digit of $n$ such that $d(n, 2) \\neq d(R_{2, 1}, 2)$. \n\nRepeat this process for every rational number in the domain of $f(q)$. Then you have constructed $n$ such that $n$ is not in the range of $f(q)$ (because at least one digit of $n$ is not equal to that same digit in every other number in the range), even though $f(q)$ was supposed to map to all $\\mathbb{R}$, thereby proving that the original assumption was wrong (no such bijective function exists), and so there must be more irrationals than rationals. Q.E.D.[/hide]" } { "Tag": [], "Problem": "If Esmerelda becomes positively charged when petting her pet chicken, then the chicken...\r\n\r\nA. lost electrons\r\nB. gained electrons\r\nC. lost protons\r\nD. gained protons", "Solution_1": "Gained electrons. Protons don't move, and to gain positive charge, Esmerelda had to lose electrons and so the chicken had to gain them.", "Solution_2": "[quote=\"JRav\"]Protons don't move[/quote]\r\n\r\nHow's that?", "Solution_3": "I think he's trying to say that usually charge is induced due to motion of electrons than protons.", "Solution_4": "That's it. Protons can obviously move but the phenomena of charge is due to the movement of electrons.", "Solution_5": "so is the answer still B. (gaining electrons)" } { "Tag": [], "Problem": "Find all positive integers $x , y$ which are roots of the equation \r\n\r\n $2 x^y-y= 2005$\r\n[u] Babis[/u]", "Solution_1": "Obviously $x\\geqslant 2$ and $y$ is odd. Put $y=2a-1, a\\geqslant 1$. Then we get\r\n\r\n$x^{2a-1}=1002+a$\r\n\r\nAssume that for some $a=a_0$ we have $2^{2a_0-1}>1002+a_0$. Then for $a=a_0+1$ we have $2^{2a_0+1}=4\\cdot 2^{2a_0-1}>4(1002+a_0)>1003+a_0=1002+(a_0+1)$\r\n\r\nSince $x^{2a-1}\\geqslant 2^{2a-1}$, inductively we conclude that for all $a\\geqslant a_0$ the equality can't be satisfied.\r\n\r\nWe can take $a_0=6$ since $2^{11}>1002+6$, so it's enough to check the cases $a=1,2,3,4,5$. We find that we have solution only for $a=1$, and that is $x=1003, y=1$", "Solution_2": "I worked modulo 2 and I found that y it should be 1 modulo 2.\r\nAlso notice that it should $y<11$ and take the cases.", "Solution_3": "[quote=\"silouan\"]\nAlso notice that it should $y<11$ and take the cases.[/quote]\r\n\r\nWhy so?", "Solution_4": "$x = 1$ makes the LHS $\\le 2$ because $x, y$ positive. $x \\ge 2$ means $y < 11$ or else $2x^y$ is too large.", "Solution_5": "[quote=\"t0rajir0u\"]$x = 1$ makes the LHS $\\le 2$ because $x, y$ positive. $x \\ge 2$ means $y < 11$ or else $2x^y$ is too large.[/quote]\r\n\r\nExactly, thanks for the explanation t0rajir0u" } { "Tag": [ "AMC", "AIME" ], "Problem": "I just took the AMC for the first time with the amc 12a ( I can't believe my school never had it before!) when are we supposed to find out our official results and if we get to take the aime?", "Solution_1": "This should be moved to the AMC forums", "Solution_2": "[quote=\"matt276eagles\"]This should be moved to the AMC forums[/quote]\r\n\r\nDone.", "Solution_3": "[quote=\"quantum leap\"]I just took the AMC for the first time with the amc 12a ( I can't believe my school never had it before!) when are we supposed to find out our official results and if we get to take the aime?[/quote]\r\n\r\nProbably in 2-3 weeks. The AIME qualifiers should come with the AMC official results, though most people know already from checking with the solutions manual sent to your school with the AMC test." } { "Tag": [ "puzzles" ], "Problem": "Prove that there can be no knight's tour of a 4*4 chess board starting and ending at the same corner.", "Solution_1": "What's a knight's tour? :blush:", "Solution_2": "Knight's tour = A sequence of moves in which a knight moves to every single square of the board.\r\nAs for the question, I know you can't do it, but I'm not sure how to prove it.", "Solution_3": "If such tour does exist, we are forced to include the two possible path from a corner in the tour(wolog from 1A to 3B and 1A to 2C). This goes same for the opposite corner, so contradiction.", "Solution_4": "There are $16$ squares on the board. Knights jump from black squares to white squares. Let's say the orignal corner was white. Thus the 16th jump would need to land on black, however, the square we started on was white.", "Solution_5": "Similarly, one can prove that a knight's tour is impossible on a $2n\\times2n$ chessboard for any natural number $n$.", "Solution_6": "[quote=\"miyomiyo\"]There are $16$ squares on the board. Knights jump from black squares to white squares. Let's say the orignal corner was white. Thus the 16th jump would need to land on black, however, the square we started on was white.[/quote]I think you counted it wrongly. That's not a contradiction.", "Solution_7": "We need to end on the square we started on, which is white.", "Solution_8": "[quote=\"lingomaniac88\"]Similarly, one can prove that a knight's tour is impossible on a $2n\\times2n$ chessboard for any natural number $n$.[/quote]\r\nCouldn't the tour be done on a standard 8x8 chessboard?", "Solution_9": "Not ending on the starting square as the question asks.", "Solution_10": "[url]http://mathworld.wolfram.com/KnightsTour.html[/url] gives 5 examples that work.\r\nThe fault in your proof is your counting, as OHO said. The 1st jump lands on the 2nd square, so the 16th jump lands on the 17th square, which is the same as the 1st. There is no contradiction.", "Solution_11": "[quote=\"Bictor717\"]16th jump lands on the 17th square, which is the same as the 1st.[/quote]The 16th jump lands on a black square; the square on which we started was white.", "Solution_12": "Starting on a white square, the first jump lands on a black square, the second lands on a white, the third lands on black... Odd numbered jumps land on black squares and even numbered jumps land on white squares.", "Solution_13": "Oops :blush: you're right. Did 1=2 make a mistake or was the question supposedto be stated like this?", "Solution_14": "Lightrhee gave the proof for the original proposition above.\r\n\r\nConsider opposite corner squares. You can only jump into and out of them from two of the center squares.\r\nIf you don't begin on a corner square, then after jumping to and leaving from it, there is no way to get to the other corner square, since the two appropriate center squares were visited.\r\nIf you begin from a corner square and you do not land on the opposite corner square on the third jump, then by the time you get to it, there would be no way to leave, since one center square was visited immediately after leaving the first corner and the second was visited on the way in.\r\nIf the third jump does land on the opposite corner, the only way to leave is by the other center square, so you can't end on the first square." } { "Tag": [ "USAMTS", "geometry", "\\/closed" ], "Problem": "Many students request that we repost class materials (transcripts, problem set, solutions) from past classes. This happens for a variety of reasons from \"I didn't get around to printing them\" to \"the files got corrupted on my computer\".\r\n\r\nAltering our database takes valuable work time away from building courses, working with students, working with the website, working on the USAMTS, and all the other many projects that we work on.\r\n\r\nWe have made the decision to repost all materials from past classes for one day only. Take advantage of this opportunity because we have no intention of doing this again. The date has not yet been scheduled but I will post the time and date that the files will be available as soon as we make a decision.\r\n\r\nIn the future students should know that it is absolutely their own responsibility to take care of all class materials themselves. Ample time is given during and after classes to save all class materials.", "Solution_1": "I have saved all the materials for Olympiad Geometry, but are the pictures that go along with the problems still on the interntet? If not, then I should probably download all the pictures and diagrams also.\r\n\r\nOut of curiosity, why aren't the materials and transcripts just kept online for a few years? Unless it slows down the site, having past transcripts and problem sets online makes it easy for students to go back and review, for example, right before a competition.", "Solution_2": "Materials are not kept indefinitely for a number of reasons:\r\n\r\nMaintaining all the materials and the site takes valuable time. We already have a number of classes to deal with at all times and having a huge database full of classes and materials would halt any progress we can make as a small company. Materials evolve as we continue to improve our courses. We're not even sure if we have over-written some of the old materials with new files.\r\n\r\nWhen we enter transcript names in the database for instance, it is natural to continue giving the transcripts standard names so that we don't have to re-enter a completely new set of transcript names each time we add a new class to the database. When a new class begins the old files could be written over. There is no reason to take lengthy pains to avoid a possibility like this when students can simply take it upon themselves to organize their own matierals.\r\n\r\nIf students can simply take the time to print or save the materials on their own then it greatly simplifies our tasks and helps us move forward without having to redesign and implement a new system. This means better classes and materials for everyone.\r\n\r\nThere is also the issue of material duplication. While we aren't trying to make copyrights an issue or a headache for anyone, but we certainly can't have our materials floating around the world for free. We place a lot of trust in our students, but permanent access seems unnecessary. If we cannot find ways to fund the classes and this site, the entire community will cease to exist.", "Solution_3": "Should we also save the pictures or diagrams that are linked to in the transcripts, i.e. will they still be online or not?", "Solution_4": "Most of the images should stick around - I'd save any you just couldn't live without, but most of them should be available persistently." } { "Tag": [ "geometry", "circumcircle", "angle bisector", "power of a point", "radical axis" ], "Problem": "subet konid dar mosalase ABC:\r\n1)dayerehaye apollonius hammehvarand\r\n\r\n2)O markaze dayere mohiti ABC roye mehvar asli 3 dayere appolonius gharar darad\r\n\r\n3) sabet konid $X,Y,S,O$ Hamsaz hastand.($X,Y$ noghate barkhorde 3 dayere va $S$ mahale hamresiye 3 miane motagharen ast)", "Solution_1": "rahe hal:\r\n\r\nlet circumcircle of $ABC$ be $\\mathcal C$\r\nlet angle bisector of $A$ intersect $BC$ at $D,E$ therefore circumcircle of $\\triangle AEF$ is one of the apollonius circles.we know that applonius circles are orthogonal to $\\mathcal C$ therfore $O$ and two isodynamic points are collinear.\r\n\r\nlet tangents at $B,C$ wrt $\\mathcal C$ intersect at $G$ cause $(GM,A'A'')=-1$ and $\\angle A'AA''=\\frac \\pi 2$ hence , $AA'$ is angle bisector of $\\angle MAG$ so $AG$ is symmedian of triangle $ABC$ we know $O'T\\perp OT$ and polar of $G$ pass through $O'$ therefore polar of $O'$ pass thtough $G$ it meanes that $G,T,A$ are collinear.\r\n\r\nso we understood that symmedian wrt $A$ is radical axis of $\\mathcal C$ and apollonius circle wrt $A$.\r\n\r\nnow consider three circle [color=blue]1.[/color]$\\mathcal C$ [color=blue]2.[/color] apollonius circle wrt $A$ [color=blue]3.[/color]apllonius circle wrt $B$ we understood that $S$(intersection point of symmedian) is radical center of these circles and because apollonian circles have the same radical-axis hence $S$ is radical center of these circles.\r\n\r\nlet $X,Y$ be isodynamic points we proved up to now that $X,Y,O,S$ are collinear , now notice that $AT$ is polar of $O$ wrt apllonius circle ,\r\n\r\ntherefore points $\\text{X,Y,S,O}$ are harmonic.\r\n\r\n\r\n\r\n[color=red]roye ax click konid.[/color]", "Solution_2": "albatteh in ha bar migardeh be ye masale dar theorems and formulas.\r\n\r\nkeh huch az ma khasteh keh ba enekas hall konim , na chize digeh. :wink:" } { "Tag": [ "function" ], "Problem": "Let $P$ be a point inside or on the sides of a square $ABCD$. Determine the minimum and maximum possible values of \\[ f(P)=\\angle ABP+\\angle BCP+\\angle CDP+\\angle DAP. \\]", "Solution_1": "[hide]I got a minimum of $135^\\circ$ and a maximum of $180^\\circ$.\n\nI'm not positive if that's right though...\nI put $P$ on any of the vertices for the minimum, and in the center for the maximum...[/hide]" } { "Tag": [ "inequalities", "integration", "LaTeX", "vector", "real analysis", "real analysis unsolved" ], "Problem": "how to show: if $ 1<\\equal{}p<\\infty$,\r\n$ \\sum_{i\\equal{}1}^{n} (\\int_0^1 |f_i|^2)^{p/2} <\\equal{}C \\sum_{i\\equal{}1}^{n} \\int_0^1 |f_i|^p$ for some constant C (independent of $ f_i$)\r\n\r\ncase $ n\\equal{}1$ is trivial: $ \\parallel{}f\\parallel{}_{L_1}<\\equal{}C\\parallel{}f\\parallel{}_{L_p}$. how to show for general $ n$?", "Solution_1": "The inequality is false even for $ n = 1$, in which case it claims $ \\|f\\|_2\\le \\|f\\|_p$.\r\n\r\n$ \\text{\\LaTeX}$ tip: use \\| instead of || for norms.", "Solution_2": "[quote=\"mlok\"]The inequality is false even for $ n = 2$, in which case it claims $ \\|f\\|_2\\le \\|f\\|_p$.\n\n$ \\text{\\LaTeX}$ tip: use \\| instead of || for norms.[/quote]\r\n\r\nmade a mistake in my post:\r\n\r\nit should be:\r\n\r\n$ \\Big (\\sum_{i = 1}^{n} \\Big( \\int_0^1 |f_i| \\Big )^2 \\Big )^{p/2} <= C \\sum_{i = 1}^{n} \\int_0^1 |f_i|^p$\r\n\r\nhow to prove this?", "Solution_3": "Do you see the Euclidean norm of a vector on the left? Since all norms in finite-dimensional spaces are equivalent, it may as well be the maximum norm, for which the inequality is \"trivial\" (follows from Holder's). Formally, \r\n$ \\Big (\\sum_{i \\equal{} 1}^{n} \\Big( \\int_0^1 |f_i| \\Big )^2 \\Big )^{p/2}\\le n^{p/2} \\max_{1\\le i\\le n}\\Big(\\int_0^1 |f_i|\\Big)^{p} \\le n^{p/2}\\max_{1\\le i\\le n}\\int_0^1 |f_i|^p \\le n^{p/2}\\sum_{i\\equal{}1}^{n}\\int_0^1 |f_i|^p$\r\nSometimes an argument like this is undesirable because we'd rather have a dimension-independent constant. But here $ C\\equal{}n^{p/2}$ is best possible." } { "Tag": [], "Problem": "There's a bit of tyranny over in Fun and Games. Seems like the mods are using a desperate burst of totalitarianism in a last ditch attempt to preserve their power. So here it is, out of their reach.\r\n\r\nI believe this thread has been deleted FIVE TIMES already. This illustrates why we need new moderators soon. So far, the nominations are:\r\nTripleM\r\nConfuted\r\nMysticTerminator\r\nJBL\r\nMathfanatic\r\nTare\r\n\r\nThe voting topic will be opened later today.", "Solution_1": "Are you sure this will work though?", "Solution_2": "What do you mean will it work? As soon as an admin finds out what they have been doing, they're long gone :)", "Solution_3": "What have i done wrong???", "Solution_4": "take me off. I go by what the admins say and they say not to do this I want out!", "Solution_5": "It isn't me, not MathFiend who deleted your threads. I understand you're upset over this, but so am I. Now it was (is) either Mith or Mr. Crawford, and I'm not blaming either one of them. Just know that it wasn't me nor MathFiend, OK? So please, stop the false accusations.", "Solution_6": "Revoke that: It could've also been MathFiend (the last one (15th thread) was deleted after Mr. Crawford and Mith left)", "Solution_7": "You deleted all the threads at first, Tare. Don't lie.", "Solution_8": "um...tare...what makes you so sure it isn't me??? Because it is. I have been deleting the nominations threads ont eh grounds that they can't change anything. Mr. Crawford said that there are better things to do than change moderators. There's nothing really wrong with what we're doing and if you have a problem, talk to an administrator, don't complain to people who can't do anything about it.", "Solution_9": "Are you accusing me of lying? I swear on my family's honor that I haven't laid a [b]finger[/b] on any of the nomination threads. You take that back. :-x", "Solution_10": "Let's not get embroiled in some sort of flame-and-dagger war here. Tare says he didn't do it, MathFiend said she did, we'll leave it at this.", "Solution_11": "Well put Hann :)", "Solution_12": "[quote=\"MathFiend\"]um...tare...what makes you so sure it isn't me??? Because it is. I have been deleting the nominations threads ont eh grounds that they can't change anything. Mr. Crawford said that there are better things to do than change moderators. There's nothing really wrong with what we're doing and if you have a problem, talk to an administrator, don't complain to people who can't do anything about it.[/quote]\r\n\r\nOk. So Mr Crawford said that he doesn't think we should change moderators. So why does that mean we're not allowed to talk about it? It doesn't give you the right to go and delete our posts..for goodness sakes, you even deleted my posts asking you for an explanation!" } { "Tag": [ "group theory", "abstract algebra", "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Given a group G and a set J. J={phi: G-->G : phi is an isomorphism}. Prove J is a group (NOT a subgroup!}\r\n\r\nSo out of four properties of a Group, I've been able to get 2.\r\n\r\n1.) Identity:\r\n Here we have the identity function. phi_e(g)=g such that every g put in comes out. Here we know the function is a bijection (defined in class). Since the elements are not changed, we know that it preservers the operation. Hence, phi_e is in J.\r\n \r\n2.) Inverse:\r\n Given any phi, since it is a bijection it lends itself to having an inverse. In addition to that, we know from class that an inverse isomorphism is an isomorphism. So, for every phi in J there is a phi^{-1}.\r\n\r\nI'm stuck on proving closure and associativity.", "Solution_1": "What are you using as your operation? You should be using function composition. Then just check that the composition of \r\n2 isomorphisms is an isomorphisms. One gets associativity from the associativity of function composition.", "Solution_2": "My worry here is that I have a sort of \"come on, it must be true\" kind of proof. Basically, given two isomorphisms in J we know that each of them sends elements from G to G. So , composing them would take elements from G to G. Since G is a group and closed under its operation, that sort of lends to closure in A.", "Solution_3": "Well, go the full way through it then. Show the composition is a homomorphism, one-to one, and onto. This should take all of a couple of minutes.", "Solution_4": "Haha, I guess I'm being dense here. Doesn't it sort of lend itself to being a homomorphism?", "Solution_5": "It all depends on what you want to accept. I believe the goal of this type of question is for you to check all the details!", "Solution_6": "Do you mind lending a hand on how to demonstrate that the homomorphism property holds. I sort of know what to do if we were talking about two different groups (a G and H). I'm really not sure except just by saying, it lends it to itself.", "Solution_7": "Let $ \\phi, \\psi \\in J$ and $ x, y \\in G$. You have $ \\psi(xy) \\equal{} \\psi(x)\\psi(y)$ where $ \\psi(x), \\psi(y) \\in G$\r\n\r\n$ \\phi \\circ \\psi (xy) \\equal{} \\phi(\\psi(xy)) \\equal{} \\phi(\\psi(x)\\psi(y))$. Now, use that $ \\phi$ is a homomorphism.", "Solution_8": "$ \\phi\\psi(x)\\phi\\psi(y)$ which is in G\r\n\r\nTherefore $ \\phi$ and $ \\psi$ are in J.", "Solution_9": "Does that finish off the proof?" } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "Let a and b be positive reals with a < 1 and b < 1. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\\displaystyle \\frac{1}{1-ax-by}\\leq \\frac{a}{1-x}+\\frac{b}{1-y}$ holds.\r\n\r\n Darij", "Solution_1": "If $a+b=1$ the inequality holds from Jensen's inequality and the convex function $f(x) = \\frac 1 {1-x}$ on $]0,1[$.\r\n\r\nNow, if the inequality holds for all $x,y$ such that blabla...then for $x=y=0$ we deduce that $a+b \\geq 1$.\r\nNow, for $x=y$, and clearing the denominators, we have $(a+b-1)(1-x(1+a+b)) \\geq 0$.\r\nThus we only have to choose $x$ such that $\\frac 1 {a+b+1} < x < \\frac 1 {a+b}$ to deduce that $a+b-1 \\leq 0$ and we are done.\r\n\r\nPierre.", "Solution_2": "If $a+b=1$, it's clear that $\\displaystyle \\frac{1}{1-ax-by}\\leq \\frac{a}{1-x}+\\frac{b}{1-y}$ holds by applying Jensen to the convex function $\\displaystyle f(x)=\\frac{1}{1-x}$ on the interval $(0,\\,1)$.\r\n\r\nFor the other direction, taking the limit $(x,\\,y) \\rightarrow (0,\\,0)$ immediately gives us $a+b \\geq 1$. \r\n\r\nNow, we assume that $a+b=r >1$. If we let $x=y$ then the inequality implies that $\\displaystyle \\frac{1}{1-rx} \\leq \\frac{r}{1-x}$ whenever $ rx<1$. Solve for $x$, we get $\\displaystyle x\\le \\frac{1}{1+r}$. Therefore if we let $\\displaystyle \\frac{1}{1+r} < x=y < \\frac{1}{r}$, then the inequality fails. So we must have $a+b=r=1$.\r\n\r\nOops, somebody beats me while I am typing the solution. :( :(", "Solution_3": "Huh, I never thought it would be so trivial using Jensen!\r\n\r\nMy proof used Cauchy-Schwarz in the Engel form:\r\n\r\n$\\displaystyle \\frac{a}{1-x}+\\frac{b}{1-y}=\\frac{a^{2}}{a-ax}+\\frac{b^{2}}{b-by}\\geq \\frac{\\left( a+b\\right) ^{2}}{\\left( a-ax\\right) +\\left( b-by\\right) }=\\frac{\\left( a+b\\right) ^{2}}{\\left( a+b\\right) -ax-by}=\\frac{1}{1-ax-by}$\r\n\r\nin the case when a + b = 1.\r\n\r\nWhen $a+b\\neq 1$, then I set $x=y=\\frac{1}{2}$ and get $\\displaystyle \\frac{1}{1-\\frac{1}{2}\\left( a+b\\right) }\\leq 2a+2b$, what quickly leads to $\\left( a+b-1\\right) ^{2}\\leq 0$. This entails a + b - 1 = 0 and a + b = 1, contradiction.\r\n\r\n Darij", "Solution_4": "[quote=\"darij grinberg\"]Let a and b be positive reals with a < 1 and b < 1. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\\displaystyle \\frac{1}{1-ax-by}\\leq \\frac{a}{1-x}+\\frac{b}{1-y}$ holds.\n[/quote]\r\n\r\nThis problem can be changed to: \r\n\r\nLet a and b be positive reals. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\\displaystyle \\frac{1}{1-ax-by}\\leq \\frac{a}{1-x}+\\frac{b}{1-y}$ holds.\r\n\r\nDo you notice the subtle difference?", "Solution_5": "[quote=\"fuzzylogic\"]This problem can be changed to: \n\nLet a and b be positive reals. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\\displaystyle \\frac{1}{1-ax-by}\\leq \\frac{a}{1-x}+\\frac{b}{1-y}$ holds.\n\nDo you notice the subtle difference?[/quote]\r\n\r\nIndeed, the conditions a < 1 and b < 1 are unnecessary! Your and Pierre's proofs work without the conditions a < 1 and b < 1 as well, while mine does not :( \r\n\r\n Darij", "Solution_6": "when you take x=y=0 you obtain a+b>=0\r\nassume that a+b>1\r\n\r\n\r\nby x=y \r\nyou have for all x<1/(a+b)\r\n1/(1-(a+b)x)<=(a+b)/1-x\r\n\r\nso \r\n\r\n\r\nx((a+b)-1)<=(a+b)-1\r\n\r\nx<=1/(a+b+1)\r\nwhen x approche 1/(a+b) it's kindov false \r\nso a+b=1\r\n\r\nfor the other application it's jensen who can solve that \r\nbye" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Could it be possible that the sum $ 2^{2^{2}}\\plus{}3^{3^{3}}\\plus{}...\\plus{}2008^{2008^{2008}}$ is a perfect square?", "Solution_1": "I think it's easy to see that $ 2^{2^{2}}\\plus{}3^{3^{3}}\\plus{}...\\plus{}2008^{2008^{2008}} \\equiv \\minus{}1 \\ (mod4)$ which is not possible if it has to be a square :wink:", "Solution_2": "[quote=\"polskimisiek\"]I think it's easy to see that $ 2^{2^{2}} \\plus{} 3^{3^{3}} \\plus{} ... \\plus{} 2008^{2008^{2008}} \\equiv \\minus{} 1 \\ (mod4)$ which is not possible if it has to be a square :wink:[/quote]\r\nWhy $ ?$", "Solution_3": "Well, it's obvious that for every integer n:\r\na) if $ n\\equal{}2k$ than $ n^{n^{n}}\\equiv 0 \\ (mod4)$\r\nb)if $ n\\equal{}2k\\plus{}1$ than we have two cases:\r\nI)$ n\\equiv 1 \\ (mod4)$ than $ n^{n^{n}}\\equiv 1 \\ (mod4)$\r\nII)$ n\\equiv (\\minus{}1) \\ (mod4)$ than $ n^{n^{n}}\\equiv (\\minus{}1) \\ (mod4)$ since $ n^{n}$ is odd\r\nSo we acquire:\r\n$ 2^{2^{2}}\\plus{}3^{3^{3}}\\plus{}...\\plus{}2008^{2008^{2008}} \\equiv 0\\minus{}1\\plus{}0\\plus{}1\\plus{}0\\minus{}1\\plus{}0\\plus{}1...\\minus{}1\\plus{}0\\equiv \\minus{}1 \\ (mod4)$ :wink:", "Solution_4": "ok,thank you.And what about problems:\r\nCould it be possible that the sum $ 2^2 \\plus{} 3^{3^{3}} \\plus{} 4^{4^{4^{4}}} \\plus{} ... \\plus{} 2008^{^{2008^{...^{2008}}}}$ is a perfect square?", "Solution_5": "Well, it's easy to see that it's pretty much the same method, because \r\n$ \\forall_{n\\equal{}2k\\plus{}1}n^{n^{...^{n}}}\\equal{}2l\\plus{}1$ :wink:" } { "Tag": [ "trigonometry", "quadratics" ], "Problem": "[b]Find all $ m \\in \\mathbb{R}$ such the following equation has at least one real root[/b]\r\n$ z^{3}+\\left(3+i \\right) z^{2}-3z-\\left(m+i \\right) =0$.\r\n\r\n[b]Find all $ z \\in \\mathbb{C}$ such that $ \\left( z-2 \\right) \\left( \\bar{z}+i \\right) \\in \\mathbb{R}$.[/b]\r\n[hide=\"Possibly?\"]\nFor $ z=a+bi$\n$ \\left( z-2 \\right) \\left( \\bar{z}+i \\right) = |z|^{2}-2a+2bi+ai-b-2i$\nWhere the imaginary parts are zero so $ \\boxed{ \\forall \\{a,b | a,b \\in \\mathbb{R}\\ , \\ 2b+a=2 \\}}$.\n[/hide]\n\n[b]Find all $ z \\in \\mathbb{C}$ such that $ |z| = \\left| \\frac{1}{z}\\right|$.[/b]\n[hide]\n$ z=a+bi$\n$ \\left| z \\right|^{2}-1 =0$\n$ \\left( \\left| z \\right|-1 \\right) \\left( \\left| z \\right|+1 \\right)= 0$\nSince the magnitude can not be negative\n$ \\boxed{ \\forall \\{ a,b | a,b \\in \\mathbb{R}\\ , \\ a^{2}+b^{2}= 1 \\}}$.\n[/hide]\n\n[b]Find all $ n$ such that[/b]\n$ \\left( \\frac{-1+i \\sqrt{3}}{2}\\right)^{n}+\\left( \\frac{-1-i \\sqrt{3}}{2}\\right)^{n}= 2$.\n[hide=\"Is it...?\"]\nAll even $ n$? There appears to be a pattern making the statement true for every such $ n$.\n[/hide]\r\n\r\n[b]For $ n \\in \\mathbb{Z}\\ , \\ n>2$ how many solutions are there to $ z^{n-1}=i \\bar{z}$.[/b]", "Solution_1": "[quote=\"AstroPhys\"]\n\n[b]Find all $ n$ such that[/b]\n$ \\left( \\frac{-1+i \\sqrt{3}}{2}\\right)^{n}+\\left( \\frac{-1-i \\sqrt{3}}{2}\\right)^{n}= 2$.\n[hide=\"Is it...?\"]\nAll even $ n$? There appears to be a pattern making the statement true for every such $ n$.\n[/hide]\n\n[/quote]\n[hide=\"maybe...\"]Note that these are the roots of $ x^{2}+x+1$. Now try Newton's Sums.[/hide]", "Solution_2": "Problem 3 is just like $ \\text{cis}\\, 120 n+\\text{cis}\\,-120n=2\\cos 120n=2$.", "Solution_3": "Instead of starting a new topic\r\n\r\nSolve in $ \\mathbb{C}$\r\n$ z^{2}-8 \\left( 1-i \\right) z+63-16i=0$\r\n[hide=\"Questions\"]\nUsing the method where $ \\left( 2az+b \\right)^{2}= y^{2}\\equiv\\Delta = b^{2}-4ac \\equiv u+vi$ where $ u,v \\in \\mathbb{R}$\nThe roots for$ y$ are $ y_{1,2}= \\pm \\left( \\sqrt{\\frac{\\left| \\Delta \\right|+u}{2}}+i \\left(\\text{sgn}\\ v \\right) \\sqrt{\\frac{\\left| \\Delta \\right|-u}{2}}\\right)$ and $ \\text{sgn}\\ v$ is the sign of $ v$\nThen it follows the roots for $ z$ are $ z_{1,2}= \\frac{y_{1,2}-b}{2a}$\n\n[b]In the solution they define $ \\Delta \\equiv \\Delta ' = \\frac{\\Delta}{4}= \\left( \\frac{b}{2}\\right)^{2}-ac$, how they known to be equivalent before finishing the solution? Since it is obvious it works in the last step of the solution $ z_{1,2}=\\frac{8-8i \\pm 2 \\left( 1-8i \\right)}{2}$ the two in the numerator is where the four appears again.[/b]\n\n(More of the solution is in Answers) [b]About $ \\text{sgn}\\ v$, should not it be negative because of $ -63i$ not $ \\pm \\left( \\sqrt{\\frac{65-63}{2}}\\boxed{+i }\\sqrt{\\frac{65+63}{2}}\\right)$?[/b]\n\n[b]Also how is $ y_{1,2}= \\pm \\left( \\sqrt{\\frac{\\left| \\Delta \\right|+u}{2}}+i \\left( \\text{sgn}\\ v \\right) \\sqrt{\\frac{\\left| \\Delta \\right|-u}{2}}\\right)$ derived?[/b]\n[/hide]\n[hide=\"Answers\"]\n$ \\Delta ' =-63-16i$\n$ \\left| \\Delta ' \\right| = \\sqrt{ 63^{2}+16^{2}}= 65$\n$ y_{1,2}=\\pm \\left( \\sqrt{\\frac{65-63}{2}}+i \\sqrt{\\frac{65+63}{2}}\\right)= \\pm \\left( 1-8i \\right)$\n$ z_{1,2}=\\frac{4-4i \\pm \\left( 1-8i \\right)}$\n$ z_{1}=5-12i$ and $ z_{2}=3+4i$.\n[/hide]", "Solution_4": "[quote=\"AstroPhys\"]\n[b]Find all $ n$ such that[/b]\n$ \\left( \\frac{-1+i \\sqrt{3}}{2}\\right)^{n}+\\left( \\frac{-1-i \\sqrt{3}}{2}\\right)^{n}= 2$.\n[hide=\"Is it...?\"]\nAll even $ n$? There appears to be a pattern making the statement true for every such $ n$.\n[/hide]\n[/quote]\r\nNotice that $ \\zeta$ being a 3-rd root of unity have,\r\n$ \\zeta = \\frac{-1+i\\sqrt{3}}{2}$ and $ \\zeta^{2}= \\frac{-1-i\\sqrt{3}}{2}$\r\nNow,\r\n$ \\zeta = \\cos \\frac{2\\pi}{3}+i\\sin \\frac{2\\pi}{3}$ and $ \\zeta^{2}= \\cos \\frac{2\\pi}{3}-i\\sin \\frac{2\\pi}{3}$ by de Moivre.\r\n\r\nSo,\r\n$ \\zeta^{n}+(\\zeta^{2})^{n}= \\cos \\frac{2\\pi n}{3}+i\\sin \\frac{2\\pi n}{3}+\\cos \\frac{2\\pi n}{3}-i \\sin \\frac{2\\pi n}{3}= 2\\cos \\frac{2\\pi n}{3}$\r\n\r\nI guess it is all $ n$ so that $ n\\equiv 0 (\\bmod 3)$", "Solution_5": "[quote=\"AstroPhys\"]\n\n[b]For $ n \\in \\mathbb{Z}\\ , \\ n>2$ how many solutions are there to $ z^{n-1}=i \\bar{z}$.[/b][/quote]\r\n\r\nLet $ |z|=r$, then $ r^{n-1}=r$, so $ r=0,1$. $ r=0$ gives one solution, $ r=1$ gives $ \\cis (n-1)x=\\cis-x+90$. Just solve from there.\r\n\r\n--\r\n\r\nI think you are confusing yourself in that next problem. You just apply the quadratic formula. The other part is just a way of getting rid of radicals. You might as well put the stuff under the squareroot into polar form and finish that way. That accomplishes the same thing with less memorization.", "Solution_6": "[quote=\"archimedes1\"][quote=\"AstroPhys\"]\n\n[b]Find all $ n$ such that[/b]\n$ \\left( \\frac{-1+i \\sqrt{3}}{2}\\right)^{n}+\\left( \\frac{-1-i \\sqrt{3}}{2}\\right)^{n}= 2$.\n[hide=\"Is it...?\"]\nAll even $ n$? There appears to be a pattern making the statement true for every such $ n$.\n[/hide]\n\n[/quote]\n[hide=\"maybe...\"]Note that these are the roots of $ x^{2}+x+1$. Now try Newton's Sums.[/hide][/quote]\r\n\r\nWrite it in polar form!\r\n\r\n$ LHS= (cis 120^{\\circ})^{n}+(cis(-120^{\\circ}))^{n}= 2 \\cos{120^{\\circ}n}= 2$\r\n\r\nThus $ \\cos{120^{\\circ}n}= 1$ giving us $ n=0,3,6,......$", "Solution_7": "[hide= ][/hide] Bump." } { "Tag": [], "Problem": "There exists 4 circles, $ a,b,c,d$, such that $ a$ is tangent to both $ b$ and $ d$, $ b$ is tangent to both $ a$ and $ c$, $ c$ is both tangent to $ b$ and $ d$, and $ d$ is both tangent to $ a$ and $ c$. Show that all these tangent points are located on a circle.", "Solution_1": "let the radii of the circles be a,b,c,d\n\nThe quadrilateral whose vertices are the centres of the circles, has side lengths a+b, b+c, c+d, d+a. Opposite pairs of sides sum to the same length [since (a+b)+(c+d) = (b+c)+(a+d) ], which is the condition required to enable a circle to be inscribed in the quadrilateral, tangent to each side.\n\nNow it's obvious. The four points at which the circles are tangent to each other are precisely the four points at which the inscribed circle is tangent to the quadrilateral's sides.\n\nMerlin" } { "Tag": [], "Problem": "I accidently changed my TI-84 Plus Silver Edition's language to Dutch. And now I can't figure out how to change it back to English. Please help! \r\nFertig.", "Solution_1": "[quote=\"mysmartmouth\"]I accidently changed my TI-84 Plus Silver Edition's language to Dutch. And now I can't figure out how to change it back to English. Please help! \nFertig.[/quote]\r\n\r\nclick apps, scroll to any language app you recognize, press enter, and then click 1", "Solution_2": "Thank you so much! Wow, I thought I was dead for a second when I saw Espanol, but no English. Then I reread your instructions, clicked Espanol, and went from there. (tip to fututre readers).", "Solution_3": "if you want to prevent it from happening again, go to to memory, then memory management, scroll to apps, under apps select dutch, then press delete", "Solution_4": "Errr, thanks, but I'm not THAT stupid. :rotfl:\r\n\r\nEDIT: kstan, what i meant was that i won't mess it up again. hopefully. :wink: :rotfl:", "Solution_5": "sorry then :blush:", "Solution_6": "or you can just delete all the programs that have no purpose at all (science in spanish, french, etc and periodic tables in multiple languages)", "Solution_7": "Or you can just delete all Apps, period.\r\n\r\nThat's what I did.", "Solution_8": "why would you want to delete all of the apps? :huh: some of them are very useful(e.g., any games) :D", "Solution_9": "Or you can just learn Dutch.", "Solution_10": "that should be on a TV show. Someone accidentally turns his calculator to Dutch, then spends the next 78 weeks learning Dutch. After he finally learns it perfectly, his calculator runs out of battery power. Then he thinks he needs a new calculator. He takes the batteries out of the new calculator, throws it away, and puts the batteries in his old calculator. The screen is powering up... there it is.... but its still in Dutch... still :rotfl: :rotfl:", "Solution_11": "[quote=\"kstan013\"]that should be on a TV show. Someone accidentally turns his calculator to Dutch, then spends the next 78 weeks learning Dutch. After he finally learns it perfectly, his calculator runs out of battery power. Then he thinks he needs a new calculator. He takes the batteries out of the new calculator, throws it away, and puts the batteries in his old calculator. The screen is powering up... there it is.... but its still in Dutch... still :rotfl: :rotfl:[/quote]\r\n\r\nCan we not spam? Kthx.\r\n\r\nNow that the problem has been solved, I suggest this topic be locked.", "Solution_12": "I don't really think it's spam, it's not the funniest remark ever, but it's still better than that what a lot of users do in other threads (without giving names :roll: ). And it especially doesn't annoy anyone who wants an answer to his problem.\r\n\r\nWe already had that discussion twice, so just a short remark: there is no point in locking topics that have done it's task.", "Solution_13": "[quote=\"joml88\"]Or you can just learn Dutch.[/quote]\r\n\r\nlol and win.", "Solution_14": "Ga eenvoudig aan de Nederlandse toepassing terug en selecteer het Engels. :D (Courtesy of Babelfish--probably very poor grammar)\r\n\r\n[quote=\"kstan013\"]if you want to prevent it from happening again, go to to memory, then memory management, scroll to apps, under apps select dutch, then press delete[/quote]\r\n\r\nThat's basically what I did: deleted all the applications I probably won't use. Wonder why TI doesn't put all the languages in one application...", "Solution_15": "LOL that happened to my ipod...now it is in french and i don't want to change it..and my cell phone is in i don't know...." } { "Tag": [ "geometry", "rotation", "vector", "trigonometry", "trapezoid", "analytic geometry" ], "Problem": "on a wall there are 2 clocks, they have the same shape(radius), with the same speed. they may not show the same hour. we know that the minimum distance between the edges of their hands is $m$, and the maximum is $M$. what is the distance between their centers?", "Solution_1": "here I give you a picture. the blue line is the distance between the edges of their hands. you know that some time it in its minimum and some time in its maximum, and you are given that numbers. what is the distance between their centers?", "Solution_2": "I think I found a condition that satisfies you require:\r\nwhen the left clock's pointing 9 and right clock's pointing 3, we have the maximum distance, M. And we have minimum distance,m, after 6 hours, when left clock's pointing 3 and right clock's pointing 9. In both conditions they are in same line, so calculating is easy.Just putting M and m where they belong to and have the distance between their centers,d, as $\\frac{M+m}{2}$", "Solution_3": "you find it! but you didn't prove it :(", "Solution_4": "[hide]If we take the two clocks and put one directly above the other, then as they rotate around, the vector from one hand's end to that of the other remains the same in magnitude, but changes in direction. So the vector in the original picture going from the end of one hand to the other is something of the form \\[(a+b\\cos{t}, b\\sin{t})\\] where $b$ is the length of that distance vector when the two clocks are placed over one another, and $a$ is the distance between the centers. Then it's clear $a=\\frac{M+m}{2}$.[/hide]", "Solution_5": "[quote=\"srulikbd\"]you find it! but you didn't prove it :([/quote]\r\nBut, the lengths I found are the maximum and the minimum ones which two circles can have, if we are sure that, there will be no more need for proving whether there is bigger than maximum and smaller than minimum.", "Solution_6": "[quote=\"K81o7\"][hide]If we take the two clocks and put one directly above the other, then as they rotate around, the vector from one hand's end to that of the other remains the same in magnitude, but changes in direction. So the vector in the original picture going from the end of one hand to the other is something of the form\n\\[(a+b\\cos{t}, b\\sin{t}) \\]\nwhere $b$ is the length of that distance vector when the two clocks are placed over one another, and $a$ is the distance between the centers. Then it's clear $a=\\frac{M+m}{2}$.[/hide][/quote]\r\nmm maybe. anyway there is 2 nice geometry solutions! also there is an ugly technical solution with Analytic geometry", "Solution_7": "Sorry for misunderstanding :wink: \r\n\r\nThe first image shows an arbitrarily chosen state for clocks. \r\nWe have to determine how the distance between the points of arrows changes. In the second figure, the arrow in counterclockwise direction moves more vertical path than the other dashed arrow as seen in figure 3.\r\nSo, we have to imagine the condition when they are getting closer each other and getting away from each other. As in figure 2, the arrows do not move counterclockwise but when one moves at the left side, the other sometimes move at right side. This situation is similar to figure 2. So, one travels less horizantal path and they change duties when small angles between the arrow and x axis are equal. Then, we obtain the 4th figure. Here, we see the maximum length M. The minimum state can be seen in symetric situation easily. So, we get 5th figure and the trapezoid with bases with M and m. Then the distance between the centers is $\\frac{M+m}{2}$ using Thales Theorem.", "Solution_8": "good! this is the solution I found. you only need to prove that this is the minimum and the maximum, which isn't very difficult", "Solution_9": "[quote=\"srulikbd\"]good! this is the solution I found. you only need to prove that this is the minimum and the maximum, which isn't very difficult[/quote]\r\nI thought I explained why it's max ro min but anyway\r\n :) I leave it here.", "Solution_10": "it doesn't seem to me as a proof, anyway, here is mine. we need to prove that the diagnoal in a Trapezoid is larger than its central median." } { "Tag": [ "function", "superior algebra", "superior algebra theorems" ], "Problem": "Let $A$ be a unitary ring in which the relation $ab=1$ implies $ba=1$. Prove that the same property has $ A[X]$.", "Solution_1": "I'll use a theorem, also due to Kaplansky (I'm sure he discovered these two results in connection to one another :)), saying that if in an infinite ring $a$ has a right inverse but no left inverse, then it has infinitely many right inverses.\r\n\r\nWe apply the above-mentioned theorem to $A[X]$: \r\n\r\nAssume we have $(a_0+a_1x+\\ldots+a_nx^n)(b_0+b_1x+\\ldots+b_mx^m)=1$. Then $a_0b_0=1\\Rightarrow b_0a_0=1$ (from the hypothesis that $ab=1\\Rightarrow ba=1$), and $b_0$ is uniquely determined (being the inverse of $a_0$). After that, we get $a_0b_1+a_1b_0=0$. If we multiply this to the left with $b_0$, we see that $b_1$ is uniquely determined. In the exact same manner we go to the next power of $x$, and see that $b_2$ is also uniquely determined, and so on. In the end, we find that $a_0+\\ldots+a_nx^n$, if it has a right inverse, it has a unique right inverse, so, by the cited theorem, it also has a left inverse, which can only be equal to its right inverse.\r\n\r\nI don't know how hard it would be to prove the theorem (I've never tried :)), but I don't think it should be that hard.", "Solution_2": "Very nice. I saw that theorem given in oral examination at X, but I could not find a solution. I find this type of problem very tricky, since it always uses either a function taken from the hat or I don't know what identity.", "Solution_3": "I remeber in the algebra book written by Batacharia (i dont know right spelling) solved with easy and nice idea which i dont remember correctly right now :) but this one is straight forward approach. suppose we have already constructed n distinct righ inverse of $a$ ,$b_1,b_2,b_3,\\dots,b_n$.we want to increase them.consider $c_i=1-b_i.a(1\\leq i\\leq n)$ .then $ac_i=0$ if we have $c_i=c_j$.then $b_i.a=b_j.a$.and right multiplication by any $b_k$ implies that $b_i=b_j$.this shows that $c_1,\\dots,c_n$are distinct.we know that $a$ has no left inverse.so $b_k,b_k+c_1,\\dots,b_k+c_n$ are $n+1$ distinct right inverse." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "The result obtained from Mathematica is as follows\r\nhttp://integrals.wolfram.com/index.jsp?expr=sqrt%28sqrt%28x%5E4+%2B+1%29+-+x%5E2%29&random=false\r\n\r\ncould anyone show me how to integrate \u221a[\u221a(x^4 + 1) - x\u00b2]?", "Solution_1": "Use the substitution $ u\\equal{}\\sqrt{x^4\\plus{}1}\\minus{}x^2$" } { "Tag": [ "trigonometry", "logarithms", "function", "calculus", "derivative", "calculus computations" ], "Problem": "Answer the following question.\r\n\r\n[1] $\\frac{d}{dx} (x\\cos x+\\ln \\sqrt{1+x^2})$\r\n\r\n[2] $\\frac{d}{dx} (x^2+x+1)^x$\r\n\r\n[3] Let $f(x)$ be the inverse function of $\\sin x\\left(-\\frac{\\pi}{2}\\leq x\\leq \\frac{\\pi}{2}\\right)$. Find $f'(x)$.\r\n\r\n[4] $\\frac{d}{dx} \\ln \\frac{\\sqrt{1+e^x}-1}{\\sqrt{1+e^x}+1}$\r\n\r\n[5] Let $f(x)=x^{n-1}\\ln x$. Find the value of natural number $n$ such that $f^{(n)}(6)=120$", "Solution_1": "[1] $\\displaystyle \\cos x - x\\sin x + \\frac{x}{1+x^2}$", "Solution_2": "[quote=\"ambierona\"][1] $\\displaystyle \\cos x - x\\sin x + \\frac{x}{1+x^2}$[/quote] :)", "Solution_3": "[2] $\\displaystyle (x^2+x+1)^x \\ln (x^2+x+1) (2x+1)$? i'm not sure...", "Solution_4": "[quote=\"ambierona\"][2] $\\displaystyle (x^2+x+1)^x \\ln (x^2+x+1) (2x+1)$? i'm not sure...[/quote] :( \r\n\r\nTaking natural logarithm, $\\ln f(x)=x\\ln (x^2+x+1)$, then $\\frac{d}{dx}\\ln f(x)=.....$\r\n\r\nkunny", "Solution_5": "ohhh\r\n[2] $\\displaystyle \\frac{d}{dx}\\ln f(x)=\\ln (x^2+x+1) + \\frac{x}{x^2+x+1}(2x+1)$\r\n\r\n$\\displaystyle \\frac{1}{f(x)}(f'(x))=\\ln (x^2+x+1) + \\frac{2x^2+x}{x^2+x+1}$\r\n\r\n$\\displaystyle f'(x)=f(x)[\\ln (x^2+x+1) + \\frac{2x^2+x}{x^2+x+1}]$\r\n\r\n$\\displaystyle =(x^2+x+1)^x[\\ln (x^2+x+1) + \\frac{2x^2+x}{x^2+x+1}]$\r\n\r\ni think", "Solution_6": "Yes, that's right, ambierona. :)", "Solution_7": "[5]\r\n[hide]\nYou will quickly see that all monmial terms that branch off will all go to zero when the $n-1$th derivative is taken, thus after $n-1$ derivatives we are left with $(n-1)!\\ln{x}$. So of course $\\displaystyle f^{(n)}(x) = \\frac{(n-1)!}{x}$.\nThen,\n\n$(n-1)! = 6!$, $n=7$.[/hide]", "Solution_8": "Yes,you are right, Spoon :)" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "inequalities", "geometry proposed" ], "Problem": "Everybody knows that if positive real numbers $x$, $y$, $z$, where $x \\geq y \\geq z$, satisfy the inequality $x \\leq y+z$ then there exist a tringle with sides of lengths $x$, $y$, $z$.\r\n\r\nHere is a natural question. Let $a$, $b$, $c$, $x$, $y$, $z$ be positive real numbers. What is the sufficient conditions to existance of a tetrahedron with edges of lengths $a$, $b$, $c$, $x$, $y$, $z$?\r\n\r\nThe question is really natural and really not so simple. :)", "Solution_1": "I know there is a condition that $ax, by, cz$ - products of opposite sides\r\nin tetrahedron are sides of a triangle.", "Solution_2": "[quote=\"prowler\"]I know there is a condition that $ax, by, cz$ - products of opposite sides\nin tetrahedron are sides of a triangle.[/quote]\r\nIs this condition sufficient? I have an enormous doubt!\r\n\r\nI know another condition, that IS sufficient! And the inequalities in that condition have a degree $>2$." } { "Tag": [ "geometry", "angle bisector" ], "Problem": "In $\\triangle ABC$, the point $I$ is the incentre. Select an arbitrary point $O$. Prove that:\r\n\r\n$\\vec{OI}= \\frac{a}{a+b+c}\\vec{OA}+\\frac{b}{a+b+c}\\vec{OB}+\\frac{c}{a+b+c}\\vec{OC}$ where $a,b,c$ are the length of $BC, AC, AB$ respectively.", "Solution_1": "Let $D$ the intersection point of the line $AI$ and $BC.$ By angle bisector theorem we have $BD: DC=c: b,$ yielding $BD=\\frac{c}{b+c}a.$\r\n\r\nThus use the theorem again $DI: IA=\\frac{c}{b+c}a: c=a: b+c.$\r\n\r\n$\\therefore \\overrightarrow{OI}=\\frac{(b+c)\\overrightarrow{OD}+a\\overrightarrow{OA}}{a+(b+c)}=\\frac{1}{a+b+c}\\left[(b+c)\\cdot \\frac{b\\overrightarrow{OB}+c\\overrightarrow{OC}}{c+b}+a\\overrightarrow{OA}\\right]$\r\n\r\n$=\\frac{a\\overrightarrow{OA}+b\\overrightarrow{OB}+c\\overrightarrow{OC}}{a+b+c}.\\ \\ Q.E.D.$", "Solution_2": "Also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109152]here[/url] (in less detail)", "Solution_3": "This problem is also posed at Doshisya University entrance exam/Economics 1973 in Japan." } { "Tag": [], "Problem": "Solve in IR the equation:\r\n\r\n$\\sqrt{\\frac{x-1977}{23}}+\\sqrt{\\frac{x-1978}{22}}+\\sqrt{\\frac{x-1979}{21}}=\\sqrt{\\frac{x-23}{1977}}+\\sqrt{\\frac{x-22}{1978}}+\\sqrt{\\frac{x-21}{1979}}$", "Solution_1": "In IR? What is that supposed to be? Is that just supposed to be the fancy R for the reals?", "Solution_2": "IR means $\\mathbb{R}$, the real numbers." } { "Tag": [ "inequalities", "conics", "parabola", "LaTeX" ], "Problem": "Answer the following questions.\r\n\r\n(1) Find the residue when $x^{2007}$ is devided by $x^{2}-1.$\r\n(2) Solve the inequality $|x|+|x+1|\\geq 2.$\r\n(3) Find the value of $k$ such that the line $y=3x+1$ is touched to the palabola $y=x^{2}+k.$", "Solution_1": "[quote=\"kunny\"]Answer the following questions.\n\n(1) Find the residue when $x^{2007}$ is devided by $x^{2}-1.$\n(2) Solve the inequality $|x|+|x+1|\\geq 2.$\n(3) Find the value of $k$ such that the line $y=3x+1$ is touched to the palabola $y=x^{2}+k.$[/quote]\r\nWhats the residue? Is that like the remainder?", "Solution_2": "[hide=\"3\"]\n$3x+1=x^{2}+k\\implies x^{2}-3x+(k-1)=0$\n\n$(-3)^{2}-4\\cdot 1\\cdot (k-1)=0\\implies\\boxed{k=\\frac{13}{4}}$.[/hide]", "Solution_3": "[quote=\"ckck\"][quote=\"kunny\"]Answer the following questions.\n\n(1) Find the residue when $x^{2007}$ is devided by $x^{2}-1.$\n(2) Solve the inequality $|x|+|x+1|\\geq 2.$\n(3) Find the value of $k$ such that the line $y=3x+1$ is touched to the palabola $y=x^{2}+k.$[/quote]\nWhats the residue? Is that like the remainder?[/quote]\r\n\r\nYes, it is. :)", "Solution_4": "[quote=\"rnwang2\"][hide=\"3\"]\n$3x+1=x^{2}+k\\implies x^{2}-3x+(k-1)=0$\n\n$(-3)^{2}-4\\cdot 1\\cdot (k-1)=0\\implies\\boxed{k=\\frac{13}{4}}$.[/hide][/quote]\r\n\r\nYour anwer is correct,rnwang2 :)", "Solution_5": "[hide=\"2\"] \n[b]Case 1.[/b] $x<0$\n\n$\\implies-x-x-1\\geq 2\\implies x\\leq-\\frac{3}{2}$\n\n[b]Case 2.[/b] $0\\leq x<1$\n\n$\\implies-1\\geq 2$ :o \n\n[b]Case 3.[/b] $1\\leq x$\n\n$\\implies x+x+1\\geq 2\\implies x\\geq \\frac{1}{2}$\n\nThe solution set is $\\boxed{x\\leq-\\frac{3}{2}\\,\\,\\,\\,\\text{or}\\,\\, \\,\\, x\\geq \\frac{1}{2}}$.[/hide]", "Solution_6": "Your answer is correct, rnwang2 :) \r\nLet me say one thing, you should state that for case 1, since $x<0,$ we have that your result.\r\n\r\nkunny", "Solution_7": "[hide=\"1\"]This isn't rigorous, but I believe this brute force method is accurate.\n\nUsing long division (which I don't know how to represent in $LaTeX$), we see a pattern. The quotient is $x^{2005}+x^{2003}+x^{2001}+...+x^{3}$. The remainder after each partial step of division is $x^{2005}$, then $x^{2003}$, then $x^{2001}$... etc. until we reach $x$, which $x^{2}-1$ cannot divide evenly. Thus, our remainder is $x$.[/hide]", "Solution_8": "Your answer is correct, I Am Me :) \r\n\r\nYou can set $x^{2007}=(x^{2}-1)Q(x)+ax+b,$ where $Q(x)$ is quotient." } { "Tag": [ "inequalities", "triangle inequality", "inequalities proposed" ], "Problem": "Let $a,b,c,d$ be positive numbers.Prove the following inequality. When the equality holds?\r\n\r\n\\[ (a+b+c+d)^3-(a^3+b^3+c^3+d^3)\\geqq 15abcd\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\right). \\]", "Solution_1": "Murhead. :)", "Solution_2": "Thank you for your reply,arqady.\r\n\r\nSorry for that I myself can't use Murhead.\r\n\r\nThis problem is for Japanese high school students those for university entrance examination , so could you please solve this ellementaly.\r\n\r\nIn Japan Tokyo and kyoto University's entrance examination is high level, though the problem in relation to inequality is limited to A.M-G.M. triangle inequality, Cauchy-Schwalz, Jensen at most Chevichev, unless students who are preparing for I.M.O.\r\n\r\nJudging from this site, I think oversea's level with respect to inequality is very high. \r\n\r\nkunny", "Solution_3": "[quote=\"kunny\"]Let $a,b,c,d$ be positive numbers.Prove the following inequality. When the equality holds?\n\n\\[ (a+b+c+d)^3-(a^3+b^3+c^3+d^3)\\geqq 15abcd\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\right). \\][/quote]\r\n\r\nOf course, It can be proven by Muihard ! :)", "Solution_4": "Well, yes, but kunny wants a non-muirhead solution ;)", "Solution_5": "[quote=\"mathmanman\"]Well, yes, but kunny wants a non-muirhead solution ;)[/quote]\r\nAll Muirhead equal $\\sum x^2\\geq0.$ :)", "Solution_6": "[quote=\"mathmanman\"]Well, yes, but kunny wants a non-muirhead solution ;)[/quote]\r\n\r\nYes, can anyone solve this problem without using Muirhead? :D", "Solution_7": "[quote=\"kunny\"]\nYes, can anyone solve this problem without using Muirhead? :D[/quote]\r\n$(a+b+c+d)^3-(a^3+b^3+c^3+d^3)\\geqq 15abcd\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow a^3+b^3+c^3+d^3+3(a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c)+6(abc+abd+acd+bcd)-(a^3+b^3+c^3+d^3)\\geq15(abc+abd+acd+bcd)\\Leftrightarrow$\r\n$2(a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c)-6(abc+abd+acd+bcd)\\geq0\\Leftrightarrow$\r\n$a(b-c)^2+a(b-d)^2+a(c-d)^2+b(a-c)^2+b(a-d)^2+b(c-d)^2+$\r\n$+c(a-b)^2+c(a-d)^2+c(b-d)^2+d(a-b)^2+d(a-c)^2+d(b-c)^2\\geq0$ :(\r\nOK?", "Solution_8": "Indeed Muirhead is just an application of AM-GM.", "Solution_9": "Awesome! Thank you, arqady.\n\nkunny", "Solution_10": "If we open parantheses, we get\n$(a^{2}b+a^{2}c+a^{2}d+b^{2}a+b^{2}c+b^{2}d+c^{2}a+c^{2}b+c^{2}d+d^{2}a+d^{2}b+d^{2}c)\\geq$\n$3(abc+abd+acd+bcd)$\n\nit is easy by am-gm\n\n$ a^{2}b+b^{2}c+c^{2}a\\geq 3abc$\n$a^{2}d+d^{2}b+b^{2}a\\geq3abd$\n$a^{2}c+c^{2}d+d^{2}a\\geq 3acd$\n$b^{2}d+d^{2}c+c^{2}b\\geq 3bcd$\n\nIt seemed very hard when i saw it first time. It got easy when i opened parantheses." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve for x:\r\n$ \\frac{{x^2 \\plus{} 6x}}{{x \\plus{} 1}}(x \\minus{} \\frac{{x \\plus{} 6}}{{x \\plus{} 1}}) \\equal{} 27$", "Solution_1": "[quote=\"Lanyes\"]Solve for x:\n$ \\frac {{x^2 \\plus{} 6x}}{{x \\plus{} 1}}(x \\minus{} \\frac {{x \\plus{} 6}}{{x \\plus{} 1}}) \\equal{} 27$[/quote]\r\n\r\nLet $ y\\equal{}x \\minus{} \\frac {{x \\plus{} 6}}{{x \\plus{} 1}}\\equal{}\\frac{x^2\\minus{}6}{x\\plus{}1}$.\r\n\r\nThen the equation $ \\Leftrightarrow (y\\plus{}6)y\\equal{}27 \\Leftrightarrow y\\equal{}3$ or $ \\minus{}9$\r\n\r\n$ y\\equal{}3 \\Leftrightarrow x^2\\minus{}6\\equal{}3(x\\plus{}1) \\Leftrightarrow x\\equal{}\\frac{3 \\pm 3 \\sqrt{5}}{2}$\r\n$ y\\equal{}\\minus{}9 \\Leftrightarrow x^2\\minus{}6\\equal{}\\minus{}9(x\\plus{}1) \\Leftrightarrow x\\equal{}\\frac{\\minus{}9 \\pm \\sqrt{69}}{2}$\r\n\r\nSo $ x\\equal{}\\frac{3 \\pm 3 \\sqrt{5}}{2}, \\frac{\\minus{}9 \\pm \\sqrt{69}}{2}$ are the solutions.", "Solution_2": "Checking by substituting into the original equation and also by sketching LHS and seeing that it meets $ y\\equal{}27$ in 4 real points, we see that all 4 solutions work." } { "Tag": [ "calculus", "integration", "algebra", "function", "domain", "Ring Theory", "superior algebra" ], "Problem": "suppose R is a boolean ring and with unity.M is a maximal idael of R and M$ %Error. \"display\" is a bad command.\n\\not=$R.show that for every a$ %Error. \"display\" is a bad command.\n\\in$R we have:a$ %Error. \"display\" is a bad command.\n\\in$ M or 1_a$ %Error. \"display\" is a bad command.\n\\in$M .", "Solution_1": "$ R/M\\cong\\mathbb{F}_2$, since it is an integral domain and a boolean ring. If $ 1\\minus{}a,a\\not\\in M$, then the identity $ 1\\minus{}a\\equiv a\\bmod{M}$ in $ R/M$ yields $ 1\\equiv 1\\minus{}a\\minus{}a\\equiv 0\\bmod{M}$, a contradiction, since $ 1\\not\\in M$. To be more precise, we have the stronger result: $ a\\in M\\Longleftrightarrow 1\\minus{}a\\not\\in M$" } { "Tag": [], "Problem": "In six years, Sam will be older than he is now by half his current\nage. Expressed in years, what is Sam's current age?", "Solution_1": "Hello\r\n$ \\frac{x}{2} \\equal{} 6$\r\nhis current age=12 years.", "Solution_2": "You should explain more.\r\n\r\nLet his age now be $ x$. Then we have $ x\\plus{}6\\equal{}x\\plus{}\\frac{x}{2} \\implies 6\\equal{}\\frac{x}{2}$, so $ x\\equal{}\\boxed{12}$ as you described." } { "Tag": [], "Problem": "One train is 100 meters, another train is 200 meters long. \r\n\r\nIf each of them travels with a constant speed towards each other, \r\n\r\nthey can pass each other completely in 5 seconds. \r\n\r\nWith the same speeds, traveling in the same direction, \r\n\r\none can pass the other in 15 seconds. \r\n\r\n(In this case passing means that it takes the front of the faster train \r\n\r\n15 seconds to get from the back of the slower train to the front of the \r\n\r\nslower train.) What are the speeds of the two trains?", "Solution_1": "[hide]$15(y-x)=200$ and $5(y+x)=300$\n\n$y=x+\\frac{40}{3}$ and $y=60-x$\n\n$x=\\frac{70}{3}\\,\\text{m/s}$\n\n$y=\\frac{110}{3}\\,\\text{m/s}$[/hide]", "Solution_2": "[quote=\"i_like_pie\"][hide]$15(y-x)=200$ and $5(y+x)=300$\n\n$y=x+\\frac{40}{3}$ and $y=60-x$\n\n$x=\\frac{70}{3}\\,\\text{m/s}$\n\n$y=\\frac{110}{3}\\,\\text{m/s}$[/hide][/quote]\n\nYou're wrong. The first equation\n\n[hide=\"should be\"]$15(y-x)=300$, because from when the front ends of the trains meet to when the rear ends split away, the front end of either train has traveled $100+200=300$ meters. Answer: $x=20, y=40$.[/hide]", "Solution_3": "[hide]The two equations are\n$300=5(x+y)$ and $300=15(x-y)$\n$\\implies 60=x+y,20=x-y$\n$\\implies x=40, y=20$[/hide]" } { "Tag": [], "Problem": "Solve equation in $ \\mathbb{Z}$:\r\n\\[ x.y \\equal{} z^2\\]", "Solution_1": "but it's basic!", "Solution_2": "$ x\\equal{}y\\equal{}z$...?\r\n\r\nReally, $ x$ and $ y$ just have to be factors of $ z^2$. Maybe I'm missing something.", "Solution_3": "Let gcd(x,y)=d, we have that d|z, so\r\nx=dn,y=dm,z=dk. \r\nNow $ mn \\equal{} k^2$ with m,n coprime. They cannot have common factor => they ara both squares, $ m \\equal{} t^2$, $ n \\equal{} s^2$, $ k\\equal{}st$.\r\nSolutions are $ (ds^2,dt^2,dst)$ where d,s,t are integers.", "Solution_4": "[quote=\"Kondr\"]\nSolutions are $ (ds^2,dt^2,dk)$ where d,s,t,k are integers.[/quote]\r\n\r\nYea you are right [b]Kondr[/b]. This gives infinitely many solutions.", "Solution_5": "[quote=\"Kondr\"]Let gcd(x,y)=d, we have that d|z, so\nx=dn,y=dm,z=dk. \nNow $ mn \\equal{} k^2$ with m,n coprime. They cannot have common factor => they ara both squares, $ m \\equal{} t^2$, $ n \\equal{} s^2$.\n\n>> Solutions are $ (ds^2,dt^2,dk)$ where d,s,t,k are integers. << [/quote]\r\n\r\nThen $ z$ would have a factor of $ d^2$ in it, instead of just a $ d$.\r\n\r\nWhat about eliminating $ k$ in the solution and expressing it as:\r\n\r\n$ (ds^2, dt^2, d^2s^2t^2)?$", "Solution_6": "$ x \\equal{} (a \\plus{} b)^2,\\ y \\equal{} (a \\minus{} b)^2,\\ z \\equal{} a^2 \\minus{} b^2$ for $ a,\\ b\\in\\mathbb{Z}.$", "Solution_7": "Um Kunny, that misses all solutions when $ x$ and $ y$ are of different parity, such as $ (x, y, z)\\equal{}(4, 1, 2)$. The best way to put it is $ (ds^2, dt^2, dst)$ like what Kondr has.", "Solution_8": "You are right." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c \\geq\\ 0$ satisfy $ ab \\plus{} bc \\plus{} ca \\equal{} 3.$\r\nProve that: $ (a \\plus{} b^2)(b \\plus{} c^2)(c \\plus{} a^2) \\geq\\ 8$", "Solution_1": "And I want to ask that: the stronger following is trues, or not?\r\n$ (a\\plus{}2b\\minus{}1)(b\\plus{}2c\\minus{}1)(c\\plus{}2a\\minus{}1) \\geq\\ 8$\r\n(with the same condition)\r\n :wink:", "Solution_2": "[quote=\"nguoivn\"]And I want to ask that: the stronger following is trues, or not?\n$ (a \\plus{} 2b \\minus{} 1)(b \\plus{} 2c \\minus{} 1)(c \\plus{} 2a \\minus{} 1) \\geq\\ 8$\n(with the same condition)\n :wink:[/quote]\r\nOf course not. Try $ b\\equal{}c\\equal{}\\frac{1}{3},a \\equal{}\\frac{13}{3}$ ;)", "Solution_3": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0$ satisfy $ ab \\plus{} bc \\plus{} ca \\equal{} 3.$\nProve that: $ (a \\plus{} b^2)(b \\plus{} c^2)(c \\plus{} a^2) \\geq\\ 8$[/quote]\r\nToday, I checked your inequality and I found this inequality is valid, nguoivn. And actually, it is not hard to prove, we just need AM-GM to prove it. ;) :)", "Solution_4": "Yes, I also found an easy proof by Am-Gm. But indeed, it's so long becausee I estimated many steps :wink:", "Solution_5": "Dear can_hang, I think we should write both of proofs in our file as one of the gifts for the friends.\r\n(I'll send my proof for you).\r\nI want to make some surprises for the next time. So, please don't post it now :oops:", "Solution_6": "And here is my proof (it's not long but not nice with me). Hope that we'll find the nicer solutions for this ineq :)" } { "Tag": [ "algorithm" ], "Problem": "Could someone explain to me why the Newton method and extracting methods for getting square roots? Our teacher said she'd tell us in 8th grade, but I was curious, so yeah... :D Thanks!", "Solution_1": "[url=http://en.wikipedia.org/wiki/Newton's_method#Square_root_of_a_number]Newton's method[/url] actually has far more applications than just finding square roots.", "Solution_2": "http://www.mathlinks.ro/Forum/posting.php?f=300&mode=smilies", "Solution_3": "Usually, the methods taught in middle school are either the Babylonian method or the simple digit-by-digit square root extraction.\r\n\r\nHere is a link to a site (in advanced middle school level) with a wonderful explanation of both algorithms:\r\n\r\n[url]http://www.mathpath.org/Algor/squareroot/algor.square.root.htm[/url]" } { "Tag": [ "trigonometry", "inequalities proposed", "inequalities" ], "Problem": "Prove that $2cos(\\frac{\\pi-C}{2})+sinC\\le\\frac{3\\sqrt{3}}{2}$, where\r\n$C \\in(0,\\pi)$.", "Solution_1": "[hide]We'll prove $2\\sin{\\frac{C}{2}}(1+\\cos{\\frac{C}{2}})\\leq\\frac{3\\sqrt{3}}{2}$ \n\nBy AM-GM with four positive numbers $3-3\\cos{\\frac{C}{2}},1+\\cos{\\frac{C}{2}},1+\\cos{\\frac{C}{2}},1+\\cos{\\frac{C}{2}}$,...[/hide]" } { "Tag": [ "analytic geometry", "function", "geometry", "3D geometry", "sphere", "trigonometry", "6th edition" ], "Problem": "For this problem I think inversion would be useful ,But I didn't do it ;) \r\nDoes any body have solution?", "Solution_1": "This is the only problem that I managed to solve. :( \r\n\r\nBelow is my super-messy solution. :D \r\nWe are allowed to post solutions now, right?", "Solution_2": "For this problem I just used calculations:\r\n\r\nFirst, calculate $\\overrightarrow{MN}$ as a linear combination of $\\overrightarrow{OA},\\ldots,\\overrightarrow{OD}$ and $\\overrightarrow{OV}$. Try it, you will obtain something interesting. $(*)$\r\n\r\nNow, set up a coordinate system (with $V(0,0,1)$ etc.) and calculate the coordinates of $A^{\\prime}, \\ldots$ as a function of $VA^{\\prime},\\ldots$. Use the $\\det$ formula for spheres and you'll soon obtain the same condition $(*)$.\r\n\r\n\r\nIf you want, I will post the full solution.\r\n\r\nAlso, lightrhee's solution seems nice, but I don't have time to look over it.", "Solution_3": "I think I did it a bit like lightrhee... but with more trig :P", "Solution_4": "This is the only problem I maneged to solve :P \r\n\r\nIt wasn't so difficult using analytic geometry!\r\n\r\nPut $V(0,0,0)$ and $A,B,C,D$ in $(\\pm r , \\pm s, -h)$\r\n \r\nNow put $A'=h_A *A$ and so on. Obiviously $MN//ABCD$ iff $z_M=z_N$ where $z_X$ is the third coordinate of the point $X$.\r\n\r\nNow we have that this is true iff $h_A+h_C=h_B+h_D$.\r\n\r\nNow take the sphere that pass trough $A',B',C',V'$ that has equation $x^2+y^2+z^2+ax+by+cz=0$\r\nNow we have that:\r\n$ra+sb-ch = -h_A (r^2+s^2+h^2)$\r\n$-ra+sb-ch = -h_B (r^2+s^2+h^2)$\r\n$-ra-sb-ch = -h_C (r^2+s^2+h^2)$\r\nand now summing up (1) and (3) and subtracting (2) we get:\r\n$ra-sb-ch=(h_B-h_C-h_A) (r^2+s^2+h^2)$\r\n\r\nBut $h_B-h_C-h_A=-h_D$ and so $D'$ lie on the sphere.\r\n\r\nThe converse is the same" } { "Tag": [], "Problem": "A greeting card is six inches wide and eight inches tall. Point $ A$\nis three inches from the fold, as shown. As the card is opened to\nan angle of 45 degrees, through how many more inches than point\n$ A$ does point $ B$ travel? Express your answer as a common\nfraction in terms of $ \\pi$.\n\n[asy]draw((0,0)--(21,-4)--(21,28)--(0,32)--cycle);\ndraw((21,0)--(24,0)--(24,32)--(0,32));\ndot((21,28));\ndot((10.5,30));\nlabel(\"$A$\",(10.5,30),S);\nlabel(\"$B$\",(21,28),S);\nlabel(\"6''\",(10.5,-2),SW);\nlabel(\"8''\",(0,16),W);[/asy]", "Solution_1": "point B travels $ 6*2*\\pi*\\frac{1}{8}$ inches\r\n\r\npoint A travels $ 3*2*\\pi*\\frac{1}{8}$ inches\r\n\r\nB - A = $ \\frac{3\\pi}{4}$\r\n\r\nanswer : $ \\frac{3\\pi}{4}$", "Solution_2": "tricky question because u use two circles" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Find all integers, satisfying the system of equations:\r\n$xt-2yz=3;$\r\n$xz+yt=1$.", "Solution_1": "We have $x=\\frac{3t+2z}{t^2+2z^2},y=\\frac{t-3z}{t^2+2z^2},3x+2y=\\frac{11t}{t^2+2z^2},x-3y=\\frac{11z}{t^2+2z^2}$.\r\n1. z=0 give $z=0,t=\\pm 1=y,x=3t.$\r\n2. y=0 have not solution.\r\n3. $t=\\pm 1$ have not solution.\r\n4. $z=\\pm 1$ give $y=0,t=3z,x=z=\\pm 1$\r\n5.$|z|\\ge 2, |t| \\ge 2$ have not solution.", "Solution_2": "I have another, more siple solution. Let's notice, that \r\n$(x^2+2y^2)(t^2+2z^2)=(xt-2yz)^2+2(xz+yt)^2=11$\r\nWe have two variants:\r\n1) $x^2+2y^2=1$ :arrow: $y=0, x=\\pm 1$ :arrow: $z=x=\\pm 1, t=3z, y=0$\r\n2) $t^2+2z^2=1$ :arrow: $z=0, t=\\pm 1$ :arrow: $y=t=\\pm 1, x=3y, z=0$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Is there a good way to count the number of graphs with n vertices? I was trying to solve this the other day when I misunderstood a homework problem and it seems more difficult than I thought.", "Solution_1": "That depends on what you mean by counting the number of graphs. \r\n\r\nIf, your vertices are labeled somehow, and you treat the graph with only a single edge that goes from $ a$ to $ b$ different from that with only a single edge going from $ x$ to $ y$, then it's not too hard (hint: Every edge is either in the graph or not in the graph).\r\n\r\nIf your vertices are unlabeled (meaning you count all graphs containing only one edge as the same graph), it's a very hard problem. You may want to look at entry A000088 in the online encyclopedia of integer sequences for small values and some other details", "Solution_2": "Yes, I meant unlabeled vertices. Thanks for the info, I'll check it out." } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions $ f : \\mathbb{R} \\ \\to \\ \\mathbb{R}$ such that\n\\[f((x\\plus{}1)f(y)) \\ \\equal{} \\ y (f(x) \\plus{}1) , \\quad\\forall x,y \\in \\mathbb{R}.\\]", "Solution_1": "[quote=\"nguyenvuthanhha\"][i]Find all function $ f : \\mathbb{R} \\ \\to \\ \\mathbb{R}$ such that :\n\n $ f((x \\plus{} 1)f(y)) \\ \\equal{} \\ y (f(x) \\plus{} 1) \\ \\forall \\ \\ x ; y \\ \\in \\ \\mathbb{R}$\n\n [hide]I think the problem will become very easy if we can prove $ f(1) \\equal{} 1$ but i can't reach that arguement [/hide][/i][/quote]\nThanks for the gift :)\n\n[hide=\"My solution\"]\nLet $ P(x,y)$ be the assertion $ f((x \\plus{} 1)f(y)) \\equal{} y(f(x) \\plus{} 1)$\nLet $ a \\equal{} f(1)$\n$ f(x) \\equal{} 0$ $ \\forall x$ is not a solution. So $ \\exists c$ such that $ f(c)\\neq 0$\n\nIf $ f(0)\\neq 0$, $ P(\\frac {c}{f(0)} \\minus{} 1,0)$ $ \\implies$ $ f(c) \\equal{} 0$, which is wrong. So $ f(0) \\equal{} 0$\n$ P(0,x)$ $ \\implies$ $ f(f(x)) \\equal{} x$ and $ f(x)$ is an involution, and so is bijective.\n$ P( \\minus{} 1,1)$ $ \\implies$ $ f( \\minus{} 1) \\equal{} \\minus{} 1$\nSince $ f( \\minus{} 1) \\equal{} \\minus{} 1$ and $ f(0) \\equal{} 0$ and $ f(x)$ is bijective, $ a \\equal{} f(1)\\notin\\{ \\minus{} 1,0\\}$\n\nSince $ f(f(x)) \\equal{} x$, we get $ f(a) \\equal{} 1$\n$ P(x \\minus{} 1,a)$ $ \\implies$ $ f(x) \\equal{} a(f(x \\minus{} 1) \\plus{} 1)$ and so $ f(x \\minus{} 1) \\plus{} 1 \\equal{} \\frac {f(x)}a$\n$ P(x \\minus{} 1,f(y))$ $ \\implies$ $ f(xy) \\equal{} f(y)(f(x \\minus{} 1) \\plus{} 1)$ $ \\equal{} \\frac {f(x)f(y)}{a}$\n\nWe can compute then $ f(4)$ in two ways :\n$ P(1,a)$ $ \\implies$ $ f(2) \\equal{} a^2 \\plus{} a$\n$ P(2,a)$ $ \\implies$ $ f(3) \\equal{} a^3 \\plus{} a^2 \\plus{} a$\n$ P(3,a)$ $ \\implies$ $ f(4) \\equal{} a^4 \\plus{} a^3 \\plus{} a^2 \\plus{} a$\n\nBut, using $ x \\equal{} y \\equal{} 2$ in $ f(xy) \\equal{} \\frac {f(x)f(y)}{a}$ implies $ f(4) \\equal{} \\frac {(a^2 \\plus{} a)^2}a \\equal{} a^3 \\plus{} 2a^2 \\plus{} a$\n\nSo $ a^4 \\plus{} a^3 \\plus{} a^2 \\plus{} a \\equal{} a^3 \\plus{} 2a^2 \\plus{} a$ and so $ a^4 \\equal{} a^2$ and so $ a\\in\\{ \\minus{} 1,0,1\\}$ but we already know that $ a\\notin\\{ \\minus{} 1,0\\}$ and so $ a \\equal{} 1$\n\nSo $ f(xy) \\equal{} f(x)f(y)$\nThen $ P(x,f(y))$ $ \\implies$ $ f((x \\plus{} 1)y) \\equal{} f(y)(f(x) \\plus{} 1) \\equal{} f(x)f(y) \\plus{} f(y)$ $ \\equal{} f(xy) \\plus{} f(y)$ and $ f(xy \\plus{} y) \\equal{} f(xy) \\plus{} f(y)$ and so $ f(u \\plus{} v) \\equal{} f(u) \\plus{} f(v)$\n\nSince we also have $ f(xy) \\equal{} f(x)f(y)$ $ \\implies$ $ f(x^2) \\equal{} f(x)^2$, we get that $ f(x)$ and $ x$ have same signs and so $ f(x)$ is increasing.\n\nHence $ f(x) \\equal{} x$ (increasing solution of Cauchy equation with $ f(1)\\equal{}1$) and it's immediate to verify that this mandatory result indeed is a solution.\n[/hide]", "Solution_2": "[quote=\"nguyenvuthanhha\"][i]Find all function $ f : \\mathbb{R} \\ \\to \\ \\mathbb{R}$ such that :\n\n $ f((x \\plus{} 1)f(y)) \\ \\equal{} \\ y (f(x) \\plus{} 1) \\ \\forall \\ \\ x ; y \\ \\in \\ \\mathbb{R}$\n\n I think the problem will become very easy if we can prove $ f(1) \\equal{} 1$ but i can't reach that arguement [/i][/quote]\r\n\r\nlet $ P(x,y)$ be the assertion $ f((x \\plus{} 1)f(y)) \\equal{} y(f(x) \\plus{} 1) \\forall x,y$.\r\n\r\n$ P( \\minus{} 1,0)\\implies f(0) \\equal{} 0$\r\n\r\n$ P( \\minus{} 1, 1)\\implies f( \\minus{} 1) \\equal{} \\minus{} 1$\r\n\r\n$ P(0,y)\\implies f(f(y)) \\equal{} y$\r\n\r\n$ P( \\minus{} 2, \\minus{} 1)\\implies f( \\minus{} f( \\minus{} 1)) \\equal{} f(1) \\equal{} \\minus{} (f( \\minus{} 2) \\plus{} 1)$ so $ f( \\minus{} 2) \\equal{} \\minus{} 1 \\minus{} f(1)$ $ (1)$\r\n\r\n$ P( \\minus{} 2,f(1))\\implies f(( \\minus{} 2 \\plus{} 1)f(f(1))) \\equal{} f( \\minus{} 1) \\equal{} \\minus{} 1 \\equal{} f(1)f( \\minus{} 2) \\plus{} f(1)$ $ (2)$\r\n\r\nfrom $ (1)$ and $ (2)$ we have $ f(1)^2 \\equal{} 1$\r\n\r\nand since $ f(f(1)) \\equal{} 1$ , $ f(1) \\equal{} 1$\r\n\r\n.....\r\n\r\n :wink:", "Solution_3": "[quote=\"mehdi cherif\"][quote=\"nguyenvuthanhha\"][i]Find all function $ f : \\mathbb{R} \\ \\to \\ \\mathbb{R}$ such that :\n\n $ f((x \\plus{} 1)f(y)) \\ \\equal{} \\ y (f(x) \\plus{} 1) \\ \\forall \\ \\ x ; y \\ \\in \\ \\mathbb{R}$\n\n I think the problem will become very easy if we can prove $ f(1) \\equal{} 1$ but i can't reach that arguement [/i][/quote]\n\nlet $ P(x,y)$ be the assertion $ f((x \\plus{} 1)f(y)) \\equal{} y(f(x) \\plus{} 1) \\forall x,y$.\n\n$ P( \\minus{} 1,0)\\implies f(0) \\equal{} 0$\n\n$ P( \\minus{} 1, 1)\\implies f( \\minus{} 1) \\equal{} \\minus{} 1$\n\n$ P(0,y)\\implies f(f(y)) \\equal{} y$\n\n$ P( \\minus{} 2, \\minus{} 1)\\implies f( \\minus{} f( \\minus{} 1)) \\equal{} f(1) \\equal{} \\minus{} (f( \\minus{} 2) \\plus{} 1)$ so $ f( \\minus{} 2) \\equal{} \\minus{} 1 \\minus{} f(1)$ $ (1)$\n\n$ P( \\minus{} 2,f(1))\\implies f(( \\minus{} 2 \\plus{} 1)f(f(1))) \\equal{} f( \\minus{} 1) \\equal{} \\minus{} 1 \\equal{} f(1)f( \\minus{} 2) \\plus{} f(1)$ $ (2)$\n\nfrom $ (1)$ and $ (2)$ we have $ f(1)^2 \\equal{} 1$\n\nand since $ f(f(1)) \\equal{} 1$ , $ f(1) \\equal{} 1$\n\n.....\n\n :wink:[/quote]\r\n\r\nYess, nicer and quicker than my own path to $ f(1)\\equal{}1$\r\nCongrats :)", "Solution_4": "thank's :blush:", "Solution_5": "[i]Yes , after prove $ f(1) \\equal{} 1$\n\n We have $ f(x \\plus{} 1) \\equal{} f(x) \\plus{} 1$ ( put $ y \\equal{} 1$ into the initial equation )\n\n Consider $ P( x \\minus{} 1; f(y) ) \\rightarrow f(xy) \\equal{} f(x) f(y)$\n\n $ f(x \\plus{} y) \\equal{} f \\left(y \\left( 1 \\plus{} \\frac {x}{y} \\right) \\right) \\equal{} f \\left( \\left( 1 \\plus{} \\frac {x}{y} \\right) f(f(y))\\right) \\equal{} f(y)\\left( f \\left( \\frac {x}{y} \\right) \\plus{} 1 \\right) \\equal{} f(x) \\plus{} f(y) \\forall x ; y \\ \\in \\ \\mathbb{R} ; y \\ \\neq \\ 0$\n \nBut $ f(0) \\equal{} 0$\n So we have $ f(xy) \\equal{} f(x) f(y) ; f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y) \\ \\forall x ; y \\ \\in \\ \\mathbb{R}$\n\n By a well - known lemma , we have known that $ f(x) \\equal{} x \\forall x \\ \\in \\ \\mathbb{R}$\nor $ f(x) \\equal{} 0 \\forall x \\ \\in \\ \\mathbb{R}$\n\n After an easy check , we can eliminate the case $ f(x) \\equal{} 0 \\forall x \\ \\in \\ \\mathbb{R}$\n\n and $ f(x) \\equal{} x \\forall x \\ \\in \\ \\mathbb{R}$ is the unique root of the functional equation :lol: [/i]" } { "Tag": [ "USAMTS", "geometry", "ratio" ], "Problem": "Assume that [tex]\\displaystyle{{CN\\over NA}={CM\\over MB}=r}[/tex] instead of assuming N and M are the midpoints of AC and CB respectively. Prove that [tex]a^2+b^2=kc^2[/tex] for some constant [tex]k[/tex]. Also find the value of [tex]k[/tex] in term of [tex]r[/tex].\n\n\n\nNote: [hide]it's not necessary to find the ratio of NP/PB and MP/PA as in the solutions.[/hide]", "Solution_1": "Let AP be x, BP be y. [tex]c^2=x^2+y^2[/tex]. \r\nAN=[tex]\\frac{a}{r+1} [/tex] BM=[tex]\\frac{b}{r+1}[/tex]\r\nNC=[tex]\\frac{ar}{r+1}, [/tex]CM = [tex]\\frac{br}{r+1}[/tex]\r\nSince, [tex]\\Delta CMN \\sim \\Delta CBA[/tex]\r\nNP=[tex]\\frac{yr}{r+1}[/tex] PM=[tex]\\frac{xr}{r+1}[/tex]\r\n[tex]b^2=(r+1)^2(({\\frac{yr}{r+1})}^2+x^2)[/tex]\r\n[tex]a^2=(r+1)^2(({\\frac{xr}{r+1})}^2+y^2)[/tex]\r\n\r\nadd them together we have\r\n\r\n[tex]x^2(2r^2+2r+1))+y^2(2r^2+2r+1)=(2r^2+r+1)(x^2+y^2)[/tex]=[tex](2r^2+r+1)(c^2)[/tex]. Therefore[tex] k=2r^2+2r+1)[/tex]\r\n\r\nedit--I fixed the mistake", "Solution_2": "beta, I think there is a calculation error somewhere since k=5 when r=1 according to the solution on the AoPS site. My answer is:\n\n[hide]k = r^2+(r+1)^2 = 2r^2+2r+1[/hide]\n\nI will give a hint here without finding NP and MP in terms of AP and BP resp.\n\nHint: [hide]It's well known that the diagonals of a convex quadrilateral are perpendicular to each other if and only if the sums of squares of the correponding opposite sides are equal. Consider the (convex) quadrilateral ANMB, we have AN^2+MB^2 = NM^2+AB^2. The rest is obvious.[/hide]", "Solution_3": "I don't know about how \"well known\" that fact is, but:\n\n\n\n[hide]It helps a lot to note that MN is parallel to AB, and that MN and AB are both part of right triangles...[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "1) Let $ a$ be an element of $ Z_n$. State and prove something of interest concerning the cyclic groups and <-a>.\r\n\r\nMy conjecture: $ \\equal{} <\\minus{}a>$\r\n\r\nIm not really good at this, so please comment on my proof. I don't have any idea if what im doing is correct. \r\n\r\n[hide=\"My Proof\"]\nLet {$ a, a^2,a^3, ..., a^n$} be elements of $ Z_n$. The group generated by $ $ will be {$ a^m, a^{2m},a^{3m}, ..., a^{km}, e, a^{\\minus{}m}, a^{\\minus{}2m},a^{\\minus{}3m}, ...$}, $ m,k$ belongs to $ Z$ and where the terms negative exponents are inverses of those with postive exponents, which are elements of $ Z$.\n\nNow, the group generated by <-a> which is the inverse of $ a$ will be {$ ...,a^{\\minus{}m}, a^{\\minus{}2m},a^{\\minus{}3m}$} its inverse elements,{$ a^m, a^{2m},a^{3m}, ..., a^{km},...$} and the identity $ e$.\n\nTherefore, $ \\equal{} <\\minus{}a>$ [/hide]\n\n2.) Let $ $ be an abelian group. Let $ H$ and $ K$ be subgroups of $ G$. Define $ H\\plus{}K \\equal{} \\{h\\plus{}k|h \\in H, k \\in K \\}$. Prove that $ H\\plus{}K$ is a subgroup of $ G$.\n\n[hide=\"My Proof\"]Since $ H$ and $ K$ are subgroups of $ G$, they contain both the identity element $ e$. But $ e \\plus{} e \\equal{} e$, therefore, $ e \\in H\\plus{}K$ \n\n(This is the part of the proof which might have a flaw.)\n$ h$ is a unique element in $ H$ and $ k$ is a unique element in $ k$. Hence, $ h \\plus{} k$ is unique in $ H \\plus{} K$ which still belongs to the group $ G$ since both $ H$ and $ K$ are subgroups of $ G$. Therefore, $ H \\plus{} K$ is closed under $ \\plus{}$.\n\nFor each $ h \\in H$ and $ k \\in K$, there is a corresponding inverses $ \\minus{}h \\in H$ and $ \\minus{}k \\in K$. Therefore, each element $ h\\plus{}k \\in H \\plus{} K$ has a corresponding inverse $ \\minus{}h \\minus{} k$. \n\nHence, $ H \\plus{} K$ is a subgroup of $ $[/hide]", "Solution_1": "For the second one, you didn't show closure. It means showing two elements in $ H\\plus{}K$ have a sum that is also in $ H\\plus{}K$.\r\n\r\nTo do this, say two elements are in $ H\\plus{}K$, the first element can be written (not necessarily uniquely) as $ h_1\\plus{}k_1$, and the second element can be written as $ h_2\\plus{}k_2$.\r\n\r\nThen find $ h,k$ such that the sum of $ h_1\\plus{}k_1$ and $ h_2\\plus{}k_2$ is $ h\\plus{}k$.\r\n\r\n(Here, $ h_1,h_2\\in H, k_1,k_2\\in K$, and you need $ h\\in H,k\\in K$, of course.)\r\n\r\nThe first one looks right but there's a simpler way, using: $ $ is a subgroup of a group $ G$ if and only if $ x\\in G$.", "Solution_2": "[quote=\"scorpius119\"]For the second one, you didn't show closure. It means showing two elements in $ H \\plus{} K$ have a sum that is also in $ H \\plus{} K$.\n\nTo do this, say two elements are in $ H \\plus{} K$, the first element can be written (not necessarily uniquely) as $ h_1 \\plus{} k_1$, and the second element can be written as $ h_2 \\plus{} k_2$.\n\nThen find $ h,k$ such that the sum of $ h_1 \\plus{} k_1$ and $ h_2 \\plus{} k_2$ is $ h \\plus{} k$.\n\n(Here, $ h_1,h_2\\in H, k_1,k_2\\in K$, and you need $ h\\in H,k\\in K$, of course.)\n\nThe first one looks right but there's a simpler way, using: $ < x >$ is a subgroup of a group $ G$ if and only if $ x\\in G$.[/quote]\r\n\r\nThanks. You know, the problem, is that the books here in our school is very limited, and the examples in most of the books are more or less the same, so you don't really know, if your proof is correct. That is the reason why you can't practice." } { "Tag": [ "symmetry", "geometry", "geometric transformation", "reflection" ], "Problem": "Prove that a line tangent to a circle is perpendicular to the radius at the point of tangency.", "Solution_1": "Suppose that line $ \\ell$ is tangent to circle $ \\Gamma$ with center $ O$ at the point of tangency $ T$.\r\n\r\nChoose an arbitrary point $ X\\in\\ell$. Observe that point $ X$ is outside of the circle since $ \\ell$ is tangent to $ \\Gamma$.\r\n\r\nDraw $ XO$ and let $ C$ be the point of intersection of $ XO$ and $ \\Gamma$. This means that point $ C$ is between $ X$ and $ O$ and so $ XO\\equal{}XC\\plus{}CO$.\r\n\r\n$ T\\in\\Gamma$ and $ C\\in\\Gamma$, and so, by definition of a radius, $ TO$ and $ CO$ are radii so that $ TO\\equal{}CO$.\r\n\r\n$ XC>0$ (because it is outside of the circle and not on $ \\Gamma$), then $ XO>CO\\equal{}TO$.\r\n\r\nThus, $ TO$ is the shortest segment from line $ \\ell$ to the center $ O$.\r\n\r\nBy definition of distance from a point to a line, $ TO\\perp\\ell$ at $ T$.", "Solution_2": "[b]Symmetry:[/b] You can use the fact that the entire circle O is reflectivly symmetric across the line containing the radius. From here you can argue that the normal at any point on the circle, say C, must be its own reflection over OC, and must thus be in the direction of OC. Then, the tangent line is clearly perpendicular to OC." } { "Tag": [], "Problem": "Suppose $ P$ is the point $ (5,3)$ and $ Q$ is the point $ (-3,6)$. Find point $ T$ such that $ Q$ is the midpoint of $ \\overline{PT}$.", "Solution_1": "There are two equations you must solve:\r\n\\[ \\frac{5\\plus{}x}{2}\\equal{}\\minus{}3,\\frac{3\\plus{}y}{2}\\equal{}6\r\n\\]\r\n\r\nThis implies $ (\\minus{}11,9)$" } { "Tag": [ "function", "combinatorics proposed", "combinatorics" ], "Problem": "I'll post some nice combinatorics problems here, taken from the wonderful training book \"Les olympiades de mathmatiques\" (in French) written by Tarik Belhaj Soulami.\r\nHere goes the first one:\r\n\r\nLet $\\mathbb{I}$ be a non-empty subset of $\\mathbb{Z}$ and let $f$ and $g$ be two functions defined on $\\mathbb{I}$. Let $m$ be the number of pairs $(x,\\;y)$ for which $f(x) = g(y)$, let $n$ be the number of pairs $(x,\\;y)$ for which $f(x) = f(y)$ and let $k$ be the number of pairs $(x,\\;y)$ for which $g(x) = g(y)$. Show that \\[2m \\leq n + k.\\]", "Solution_1": "I've the book.... No interest for me to answer :( \r\n\r\nPierre.", "Solution_2": "The book does not give any specific source for this problem.", "Solution_3": "This is problem 10.6 from the Russian olympiad in 1994.\r\n\r\nFor each $t$, let $n_f(t)$ be the number of values of $x$ for which $f(x)=t$, and define $n_g(t)$ similarly, except that this time we use $g$ instead of $f$. We have $m=\\sum_t n_f(t)n_g(t),\\ n=\\sum_t n_f^2(t),\\ k=\\sum_t n_g^2(t)$, and the result follows (the sums are taken after all those $t$ for which at $n_f(t)n_g(t)>0$)." } { "Tag": [ "number theory", "number theory unsolved" ], "Problem": "Let $ f(x)\\equal{}5x^{13}\\plus{}13x^5\\plus{}9ax$. Find the least positive integer $ a$ such that $ 65$ divides $ f(x)$ for every integer $ x$.", "Solution_1": "$ 65\\equal{}5\\cdot13$ so we just need it to be divisible by both.\r\nFor divisiblity by $ 5$ let's assume $ 5$ doesn't divide $ x$, then by Fermat's little theorem we need $ 5|13x\\plus{}9ax \\Longrightarrow 5|1\\plus{}3a$\r\nSimilarly we need $ 13|5\\plus{}9a$. Now we just use Chinese Remainder theorem and find the least positive $ a$.", "Solution_2": " We note that by the Chinese Remainder Theorem, it suffices to find $a$ such that $f(x) \\equiv 0$ modulo 5 and modulo 13. \n \n First, modulo 5 we have\n \\begin{align*}\n f(x) = 5x^{13} + 13x^5 + 9ax \\equiv 13x^5 - ax \\pmod{5}.\n \\end{align*}\n We note that if $5 \\mid x$, then $f(x) \\equiv 0$, so we can assume that $5 \\nmid x$ and apply Fermat's little theorem to obtain\n \\begin{align*}\n f(x) \\equiv 13x - ax \\equiv (3 - a)x \\pmod{5}\n \\end{align*}\n so if $f(x) \\equiv 0$ for all $x$, we need $3 - a \\equiv 0 \\pmod{5}$, that is, $a = 3 \\pmod{5}$.\n\n Now modulo 13, we again ignore the case of $13 \\mid x$ as $f(x) \\equiv 0$ if that is true, then we apply Fermat's little theorem as before.\n $$\n f(x) = 5x^13 + 13x^5 + 9ax \\equiv 5x + 9ax \\equiv (5 + 9a)x \\equiv 0 \\pmod{13},\n $$\n so we need $9a \\equiv 8 \\pmod{13}$.\n\n Thus we solve the system of congruences\n \\begin{align*}\n a &\\equiv 3 \\pmod{5}\\\\\n 9a &\\equiv 8 \\pmod{13},\n \\end{align*}\n and applying the Chinese remainder theorem, we have $a$ is unique modulo 65, thus $a \\equiv 63 \\pmod{65}$, and $a = 63$ is the smallest solution.", "Solution_3": "Another solution:\n\n$f(x)=5x^{13}-5x+5x+13x^{5}-13x+13x+9ax=5(x^{13}-x)+13(x^{5}-x)+9(a+2)x$\n\nBy Fermat's little theorem, we have $x^{13}-x\\equiv 0 \\pmod {13}$, so $5(x^{13}-x)\\equiv 0 \\pmod{65}$ and $x^{5}-x\\equiv 0 \\pmod{5}$, so $13(x^{5}-x)\\equiv 0 \\pmod{65}$. $65 \\nmid 9$, then $65 \\mid a+2$ and its easy to see that $a=63$ is the solution." } { "Tag": [ "number theory", "combinatorics proposed", "combinatorics" ], "Problem": "Given 2 piles of stones. 2 players alternately make moves. In a move player can take any number of stones from 1 pile or equally from 2 piles. Player that takes the last stone wins. Find, in which initial positions first player wins. Find his winning strategy.", "Solution_1": "This has been discussed before. Look for the topic \"A golden game\" (started by Mindflyer, I think) in the Number Theory section. It's called [url=http://mathworld.wolfram.com/WythoffsGame.html]Wythoff's game[/url].", "Solution_2": "OK. What about if we have n piles and player can take equally from k piles (1<=k<=n)?", "Solution_3": "Is my generalization too crazy?", "Solution_4": "[quote=\"Sasha\"]Is my generalization too crazy?[/quote]\r\n\r\nNo. This game has been analyzed completely by Fraenkel and Zusman:\r\n\r\n Aviezri Fraenkel and Dmitri Zusman\r\n A new heap game.\r\n Theoretical Computer Science 252 (2001), pp 5-12.", "Solution_5": "Can you tell more about it?" } { "Tag": [], "Problem": "I was looking, and I discovered basically the answer to the first problem on Wikipedia. It says that there are 288 ways to fill a $2 \\times 2$ Sudoku thing, and well it's pretty easy to see that one divides that by $4!$ to get the answer to problem 1.\r\n\r\n[url=http://en.wikipedia.org/wiki/Mathematics_of_Sudoku#Sudoku_with_rectangular_regions]Wikipedia Hint at #1[/url]\r\n\r\nI guess it's like how last year one of the problems was actually on mathworld...\r\n\r\n\r\nNOTE that I found this after I had submitted my solutions", "Solution_1": "After I finished my solution, I googled 2x2 sudoku and found the same thing." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "sphere", "geometric transformation", "reflection", "geometry unsolved" ], "Problem": "Could anyone help me with some English terms in solid geometry?\r\n\r\n1.) In the plane geometry, we have the term median to refer to the line segm\r\nent joining a vertex to the midpoint of the opposite side. In solid geometry, can we have the term median plane of a tetrahedron: through an edge and the midpoint of the opposite edge?\r\n\r\n\r\n2.) Similarly, can we have the term bisector plane of a tetrahedra? or other term?\r\n\r\n3.) In a sphere, the center of the sphere is the intersection of what planes?\r\n\r\n4.) A plane that is the reflection of median plane about the bisector plane ... how do we call this plane?\r\n\r\n5.) We call $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ a sphere, but how we call the set of points such that $(x-a)^2+(y-b)^2+(z-c)^2\\leq r^2$", "Solution_1": "In my math (our country) the [u]bimedian[/u] is a segment which unites the midpoints of the opposite edges (there are three medians) and the [u]median[/u] is the segment which unites a vertex with the centroid of the opposite face (there are four medians). The common point of all seven above segments is the [u]centroid[/u] of the tetraedron. Such is here, in Romania.", "Solution_2": "Let me help where I can:\r\n\r\n[quote=\"pvthuan\"]3.) In a sphere, the center of the sphere is the intersection of what planes?[/quote]\n\nFor any two points A and B, the locus of all points P in space such that PA = PB is called the [i]perpendicular bisecting plane[/i] of the segment AB.\n\nThe center of a sphere lies on the perpendicular bisecting plane of every chord of the sphere.\n\n[quote=\"pvthuan\"]5.) We call $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ a sphere, but how we call the set of points such that $(x-a)^2+(y-b)^2+(z-c)^2\\leq r^2$[/quote]\r\n\r\nThis set is called a [i]ball[/i] (more exactly, a [i]closed ball[/i]; an [i]open ball[/i] would be the set of all points such that $\\left(x-a\\right)^2+\\left(y-b\\right)^2+\\left(z-c\\right)^20$\r\n\r\nProve $\\frac{a}{\\sqrt{b+c}}+\\frac{b}{\\sqrt{c+a}}+\\frac{c}{\\sqrt{a+b}}\\geq \\sqrt{\\frac{3}{2}}$", "Solution_1": "http://www.mathlinks.ro/Forum/topic-87487.html :)", "Solution_2": "[hide]\n\nsince $a+b+c=1$, the inequality becomes\n\n$\\frac{a}{\\sqrt{1-a}}+\\frac{b}{\\sqrt{1-b}}+\\frac{c}{\\sqrt{1-c}}\\geq \\sqrt{\\frac{3}{2}}$\n\nlet $f(a)=\\frac{a}{\\sqrt{1-a}}$. This function is convex, cause $f''(a)=\\frac{3}{4}*\\frac{1}{(1-a)^{\\frac{5}{2}}}$.\n\nSo, by Jensen, $f\\left (\\frac{a+b+c}{3} \\right)\\leq \\frac{1}{3}(f(a)+f(b)+f(c))$, and we have the proof\n\n[/hide]" } { "Tag": [ "MATHCOUNTS", "probability", "blogs", "number theory", "relatively prime", "prime numbers" ], "Problem": "Hmm I got this question: Of the set {20,21,22,23,24,25,26,27,28,29,30}, exactly 2 members are selected. What is the probability that the two numbers are relatively prime?\r\n\r\nIts confusing to me! \r\n\r\nCan someone help?", "Solution_1": "Well, in total there's $ \\binom{11}{2} \\equal{} 55$\r\n\r\nNow I would proceed with casework....blah. I don't have a better method. You do know what releatively prime means...correct?\r\n\r\n[hide=\"Here goes....\"] Ok, so with 20, we have 21, 23, 27, 29 who are relatively prime...\n\n21: 22, 23, 25, 26, 29 who are relatively prime.\n\n22: 23, 25, 27, 29.\n\n23: 24, 25, 26, 27, 28, 29, 30\n\n24: 25, 29.\n\n25: 26, 27, 28, 29\n\n26: 27, 29,\n\n27: 28, 29\n\n28: 29\n\n29: 30\n\nAdd them up, perhaps? $ \\boxed{32}$ in total I believe....[/hide]", "Solution_2": "well....i do know what relatively prime is...but is there an easier way to do this w/o casework (i hate casework :P )", "Solution_3": "First of all, the number of ways to choose two numbers, as has been noted, is $ \\binom{11}{2}\\equal{}55$. We next proceed by complementary counting. We thus consider all combinations of multiples of $ 2$, multiples of $ 3$, multiples of $ 5$, etc. (we can limit our case work to prime numbers since all composite numbers can be expressed as the product of prime numbers). So first, we consider even numbers. There are $ 6$ even numbers, and hence the number of ways to choose $ 2$ of them is $ \\binom{6}{2}\\equal{}15$. Next we consider multiples of $ 3$. There are $ 4$ multiples of $ 3$, and hence in order to choose two of them we have $ \\binom{4}{2}\\equal{}6$. However, we must neglect the case of $ (24,30)$, since we have already counted it. In the same way, there are $ 3$ multiples of $ 5$, and thus $ \\binom{3}{2}\\equal{}3$. This time we neglect the case of $ (20,30)$, since it was already counted. We only have one case for $ 7$: namely, $ (21,28)$. We don;t have any more cases to consider, so we have\r\n\r\n$ \\frac{55\\minus{}15\\minus{}5\\minus{}2\\minus{}1}{55}\\equal{}\\boxed{\\frac{32}{55}}$.", "Solution_4": "Thanks!\r\n\r\nI like complementary counting - but for this one I think casework is better (I think i wouldve overcounted)", "Solution_5": "[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=324971#comments]http://www.artofproblemsolving.com/Wiki/index.php/Main_Page/Relatively_Prime#Problems[/url]\r\nmight help.", "Solution_6": "Erm that link .... sends me to pinkys blog", "Solution_7": "I sent but I do not know what happens beyond", "Solution_8": "[quote=\"CRICKET229\"]Erm that link .... sends me to pinkys blog[/quote]\r\n\r\nYou are probably clicking the wrong link.", "Solution_9": "It's true, the link doesn't work" } { "Tag": [ "symmetry", "algebra", "function", "domain", "conics", "ellipse" ], "Problem": "I have come across a very nice problem, that can be solved using high school physics but is quite challenging, so I thought you'd like it:\r\n\r\nWe have an infinite superconductive plate in empty space and a poing charge Q in distance R from this plate. There is a very small hole in this plate exactly in the closest point to the charge. If we let the charge move freely, it is attracted to the plate and starts to oscillate back and forth (it can pass thorugh the hole). Calculate the frequency (or period, whatever) of this movement.", "Solution_1": "... still nothing?\r\n\r\n[hide=\"hint\"]The infinite conductive plate guarantees a zero potential everywhere on it. That means all the electric field must be of the kind that there is zero potential in that plane, and also the field is perpendicular to the plane ... There is another simple situation (with no such ugly thing as an infinite plane) where this condition si satisfied. The field around the charge is the same, so the force must also be the same ...[/hide]", "Solution_2": "this is the most basic image problem (though I do not consider it a high school level), you place a charge -Q a distance R on the other side of the plane and then pretend the plane is gone. This satisfies the boundary conditions. The existance and uniqueness theorem guarantees this is a unique solution. The force on the charge can be calculated to be $\\frac{-q^{2}}{16\\pi \\epsilon R^{2}}$. This does not however apply to moving charges, so I don't see where you are going with this problem.", "Solution_3": "I have encountered this problem in a high-school competition, it does involve some tricks, especially if you're a high-school student, but I think it's a nice one.\r\n\r\nAnd I admit, that the non-stationary effects may affect the result, but let's assume that they won't. For example, let's set the mass of the charged thing to be sufficiently large so that it doesn't move too fast. Can you somehow quantify those effects? I don't see through electromagnetism that much, (I am still in high school;-) but on a first glance, it seems like there should be no magnetic force, from the symmetry of the whole situation .. or is there?", "Solution_4": "Umm, well if it's superconducting it will repel the induced magnetic charge. Just making a conducting plane. Anyway, if you want to ignore the physics (mass has nothing to do with it in E&M), then your motion equation is:\r\n\r\n$F=m\\frac{d^2x}{dt^2}=\\frac{-q^2}{16\\pi \\epsilon x^2}$\r\n\r\nI do not know of any elementary solution to this equation solved in the time domain. I suppose if you only want the period, then you change it to a radial domain and perhaps the bounds could be obtained.", "Solution_5": "What do you mean by \"it will repel the induced magnetic charge.\"? If you can explain, I would like to know more about that - what happens when the charge starts moving.\r\n\r\nI said the mass is large enough so that the force on a static charge is small and it starts to move slowly and the magnetic effects are not too big. Do you think that the other effects are there regardless of this?!", "Solution_6": "what are you unclear about, the superconducting plate replelling? or the induced charge in the first place. Unfortunately, a moving point charge is much more complex than can be explained at the high school level. A whole world of things happen when the charge starts moving, but all of the physics is very mathematically rigorous, as it deals with relativistic effects and electromagnetic waves. \r\n\r\nI am curious about the original problem though, let's assume for a second that no magnetic charge is induced, and the moving charge is treated like the static case, the equation I gave above represents the necessary equation for motion. As you should note, it is analogous to a mass traveling in a gravitational field with a different proportional constant. In general, this differential equation cannot be solved. I am very curious how such a problem could be put on a high school exam.", "Solution_7": "... well, i admit, that the problem was one of the \"stoppers\" - the last problems that ensure that noone solves everything in the time limit ...\r\n\r\nThis was my solution - the situation is analogous to two masses that start 2R apart and attract each other. To find out the movement of one of them, instead of the second you can put a fixed mass in between them, with proper mass. Now solve the system. It is like if it was was orbiting the central fixed mass, but with a very eccentric orbit. According to the Kepler's ?th law, its period is the same as if the orbit was circular with radius R (because the main axis of the quasi-ellipse is 2R). This is an ordinary satellite-motion problem. I know my explanation is a bit fuzzy, but I can give you figures if it's too bad.\r\n\r\nAnd, if you can explain, I'm really curious about what happens when the charge starts to move, at least qualitatively (yes, the repelling and effects of the motion of the charge) . You don't have to keep at high school level, the worst thing that can happen is that I don't get it;-)", "Solution_8": "Your solution seems like it'd be the only way to solve the problem, using kepler's laws and making the analogy to a planet in orbit.\r\n\r\nAs far as what happens, I would suggest reading Griffiths, Introduction to Electrodynamics for the answers you seek.\r\n\r\nI do see what you're saying with a very large mass now though, it would cause the object to move slower and thus better approximate a point charge. I suppose that in the limit where R is extremely small compared to M, the problem might be ok.\r\n\r\nCheck this out, it shows the field of a moving point charge. Trust me, the math behind them is intense.\r\n\r\nhttp://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html", "Solution_9": "Will that magnetic field not influence the motion of charge? I'm not sure about that.", "Solution_10": "i think that pretty much any problem done with high school physics is flawed.. but since thats all i know im going to try it with that: (i am going to assume that it was meant to be an infinite plane of charge)\r\n[hide=\"is this right?\"]say it is a charged plane with surface charge density $-\\sigma$ instead of a superconductor.. \n\nso there is a point charge of mass $m$ and charge $q$.\n\n$E=\\frac{\\sigma}{2\\epsilon_o}$ (twoards the plane)\n\nfrom that, $F=\\frac{q\\sigma}{2\\epsilon_o}=ma$\n\ntherefore, $a = \\frac{q\\sigma}{2m\\epsilon_o}$\n\nnow, since the electric field is constant wrt $x$ (the distance from the plate), $U=\\frac{qx\\sigma}{2\\epsilon_o}$\n\nnow to find the velocity at the center, $v_o$\n\n$\\frac{qx\\sigma}{2\\epsilon_o}=\\frac{1}{2}mv_o^2$\n\nfrom that, we have $v_o = \\sqrt{\\frac{qx\\sigma}{m\\epsilon_o}}$\n\nnow we know the acceleration is constant, so we can use $\\Delta x = v_ot+\\frac{1}{2}at^2$\n\nafter the charge goes out and comes back, $\\Delta x = 0$\n\ntherefore, $0 = \\frac{1}{2}at^2+v_ot=t(\\frac{1}{2}at+v_o)$\n\nignoring the trivial $t=0$, we have $t = \\frac{-2v_o}{a}$\n\nbecause $a$ is twoards the plane, and $v_o$ is away from the plane, the negative sign goes away, and the magnatudes of both can be used...\n\ntherefore..\n\n$t=2\\sqrt{\\frac{qx\\sigma}{m\\epsilon_o}}\\frac{2m\\epsilon_o}{q\\sigma} = 4\\sqrt{\\frac{qx\\sigma m^2 \\epsilon_o^2}{m \\epsilon_o q^2 \\sigma^2}} = 4\\sqrt{\\frac{x m \\epsilon_o}{q \\sigma}}$\n\nbecause this is only half, then period, $T$ is given by $T=2t$\n\ntherefore, $T=8\\sqrt{\\frac{x m \\epsilon_o}{q \\sigma}}$, or $f=\\frac{1}{8}\\sqrt{\\frac{q \\sigma}{x m \\epsilon_o}}$\n\n[/hide]", "Solution_11": "Sky, unfortunately that is not how it works. You are neglecting the induced surface charge density from the point charge onto the plane. This induced surface charge is not constant, it's a function of the radial distance away from the point as well as the distance from the point to the plane. \r\n\r\nYou typically solve this problem using what's called \"The method of images\". Basically, you seek to find a configuration such that the potential on the plane is zero everywhere. There is a methamtical theorem known as the Existance and Uniqueness Theorem that guarantees that if you find such a configuration such that the boundary conditions are satisfied, then you have a unique solution. This means that once you find a configuration such that the potential on the plane is zero, and the potential at infinity (in the direction perpendicular to the plane) is 0, you are done.\r\n\r\nConsider the having a point charge +Q a distance d away from a plane in the +z direction, and you then have have a point charge -Q a distance d away from the plan in the -z direction, and pretend the plane is gone. This artificial -Q is known as the \"image charge\".\r\n\r\nCalculate the potential when z=0. Calculate the potential at z = $\\pm \\infty$. Are the conditions satisfied? If so, then this situation is equivalent to the original. You can then calculate the force and you will find that the force does indeed depend on the distance of the point from the plane, thus making stuff very complicated.", "Solution_12": "ohh. lol. :( now i understand what people were talking about before..", "Solution_13": "Sorry to interrupt... what category of Arts and Sciencesdoes engineering fit into?" } { "Tag": [ "algorithm", "function" ], "Problem": "I'm not sure if this is really the right place to ask but I'm stuck on a problem, and I'm not that good with math.. I'm working on a mod for a game called Tribes, and I need help figuring out the time to impact for a projectile. The things I have to work with are Initial Velocity, Acceleration (per second), Maximum Velocity and the distance to the target.\r\n\r\nThe projectile starts off at Initial Velocity, and increases speed by Acceleration every second untill it reaches Maximum Velocity. I hope that's enough info, any help would be greatly appreciated.", "Solution_1": "Hm, I program myself, but am not too well with making algorithms as fast as possible for computing. And I don't know what Tribes's scripting is like, so I hope you understand basic C++/Java notation. If not, I can tell you what it's doing.\r\n\r\n[code]\n//function returns the amount of frames it will take for projectile modeled by inital velocity, maximum velocity,\n//and acceleratoin will take to reach a certain distance\npublic int calculateTime(double initialVelocity, double maximumVelocity, double acceleration, double distance)\n{\n double framesUntilMax, tempDistance = 0;\n int totalFrames = 0;\n if( (framesUntilMax = (maximumVelocity - initialVelocity) / acceleration) < 0)\n framesUntilMax = 0;\n \n while(tempDistance < distance)\n {\n dist += (totalFrames >= framesUntilMax) ? maximumVelocity :\n totalFrames * acceleration + initialVelocity;\n\n totalFrames += 1;\n }\n\n return totalFrames;\n}\n[/code]\r\n\r\nNow I know that isn't very elegant, and maybe not very fast, but I think it should work, returning the number of frames that must pass before the projectile reaches the distance that you want. I didn't test it, though, so I'm not completely sure.", "Solution_2": "There's actually a whole forum for Computer Science... it's towards the bottom of the main page. It's times like these I wish I were a mod :(", "Solution_3": "No kidding? I never traveled down that far on the page :blush: I'll have to go there some time... I know this algorithm can be done faster, and I'm sure a lot of mine can too.", "Solution_4": "That's plenty good enough, thanks [b]alot[/b] for the help. :)" } { "Tag": [ "LaTeX", "articles" ], "Problem": "Hi Guyz,\r\n I got a new project to create some latex documents for publications for my professor, as I am new to this language, I tried to gather all the information from the net and thru tutorials and almost am done with the project. I am now stuck with inserting pictures into the latex documents. I see many figures in the source soft copy of the file( which is to be converted into a latex document) which are graphs and mathematical related diagrams. All the source document is in word document. I am using the Winedit software to edit the latex documents. I saw different ways to include the pictures in the latex documents. The diagrams should be in eps or pdf forms. Please let me know how i can produce the figures which are in the word document into the eps or pdf forms. I tried to paste the document in the photoshop and later changed it to the .eps form then I used the following code:\r\n\r\n--------------EPS form-----------\r\n\r\n\\documentclass[8pt]{article}\r\n\\usepackage{natbib}\r\n\\usepackage{graphicx}\r\n\\usepackage{graphics}\r\n\\begin{figure}[htp]\r\n\\includegraphics[height=60mm, width=45mm]{fig1}\r\n\\end{figure}\r\n \r\n Then I tried to copy the figure into another word document and convert it directly into pdf by clicking the icon on the word document; \"convert to pdf\". well I got the fig1.pdf file now with me. Later I tried using the following code\r\n\r\n------------PDF form ---------------\r\n\r\n\\documentclass[8pt]{article}\r\n\\usepackage{natbib}\r\n\\usepackage[pdftex]{graphicx}\r\n\\begin{figure}[htp]\r\n\\includegraphics*[viewport=100 200 500 650]{fg1}\r\n\\end{figure}\r\n\r\n In either case I am getting the follwing error:\r\n\r\nCannot determine the size of the grapic in fig1 \r\n\r\n Please help me with this....please guyz", "Solution_1": "See the solutions in [url=http://groups.google.com/group/comp.text.tex/browse_thread/thread/6e048135606fb4a2/116bb8364e02312d#116bb8364e02312d]comp.text.tex newsgroup[/url] on using EPS generated by Photoshop\r\n\r\nPersonally I use the option to save as EPS in OpenOffice's drawing program and that seems to work well with LaTeX." } { "Tag": [ "algebra", "polynomial", "search", "number theory proposed", "number theory" ], "Problem": "find all polinomials P with real coefficientes such that iff P(r) is an integer, then r is an integer.", "Solution_1": "It was already discussedand anyway it is allmost trivial.", "Solution_2": "It is Iran 2002", "Solution_3": "Where can i get the Iran Math Olympiad???", "Solution_4": "Here is the other topic where you will find a solution.\r\nhttp://www.mathlinks.ro/viewtopic.php?t=6122&highlight=iran\r\n\r\nPierre.", "Solution_5": "[quote=\"Leonardo\"]Where can i get the Iran Math Olympiad???[/quote] search after Iran in the problem source field. You will find tons of Iran problems." } { "Tag": [ "calculus", "advanced fields", "advanced fields unsolved" ], "Problem": "I am in a lower math class and am being tutored by a friend of mine who is in higher math (calculus). She has done an amazing job helping me out, but I feel it has caused her problems with her own class. She was given a take home quiz and she started crying because she is not so clear on this section (derivatives). I would love to help her as a way of returning the favor. Is there anyone that will work out a few problems for me? It would mean the world to her and I would feel better about her falling behind. She told me the answers she has gotten just do not make sense. Thank you for the consideration!", "Solution_1": "Welcome to AoPS! \r\n\r\n1) There is a separate forum for calculus. The sort of problems it sounds like you want to post will go in the Calculus Computations and Tutorials forum. \r\n\r\n2) We on this forum prefer not to do other people's homework for them. We're happy to help people who are having trouble with math (at any level), but most of us are more inclined to explain and stuff rather than do someone's homework. If you post some problems, you will receive (from me at least, if I answer) an explanation of how to do that sort of problem, not just free answers." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Two players play a game on a $2000 * 2001$ board. Each has one piece and the players move their pieces alternately.\r\n A short move is one square in any direction (including diagonally) or no move at all.\r\n On his first turn each player makes a short move. \r\nOn subsequent turns a player must make the same move as on his previous turn followed by a short move. \r\nThis is treated as a single move. \r\nThe board is assumed to wrap in both directions so a player on the edge of the board can move to the opposite edge. \r\nThe first player wins if he can move his piece onto the same square as his opponent's piece. \r\nFor example, suppose we label the squares from $(0, 0)$ to $(1999, 2000)$, and the first player's piece is initially at $(0, 0)$ and the second player's at $(1996, 3)$. \r\nThe first player could move to $(1999, 2000)$, then the second player to $(1996, 2)$. \r\nThen the first player could move to $(1998, 1998)$, then the second player to $(1995, 1)$.\r\n Can the first player always win irrespective of the initial positions of the two pieces?", "Solution_1": "no one? :?:", "Solution_2": ":( ;) :mad:" } { "Tag": [ "LaTeX", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "In my Hilbert space course, i came across the question \"write down an explicit isomorhism from $ l^2(\\mathbb{Z})$ to $ l^2$ \". I cannot find such an isomorphism. Is it something very obvious?\r\nThanks.", "Solution_1": "Can you find a bijection between $ \\mathbb Z$ and $ \\mathbb N$?", "Solution_2": "I think that a bijection from $ \\mathbb{N}\\to \\mathbb{Z}$ is $ n\\to k$ if $ n\\equal{}2k$, and $ n\\to \\minus{}k$ if $ n\\equal{}2k\\plus{}1$. Is this correct?", "Solution_3": "That's only correct if your version of $ \\mathbb{N}$ includes $ 0.$ I would normally define $ \\mathbb{N}$ so as not to include zero.\r\n\r\nBut anyway - use the same idea, tweak it slightly, you'll make it work. Do you see how do get the desired isomorphism of Hilbert spaces from this?\r\n\r\n$ \\text{\\LaTeX}$ tip: for these Hilbert spaces, I like \\ell better than l. That is, $ \\ell^2$ rather than $ l^2.$ It's just a preference about appearances." } { "Tag": [ "geometry", "trapezoid", "circumcircle", "geometry proposed" ], "Problem": "In an hexagon $ ABCDEF$ the oposite sides are parallel and the diagonals $ AD,BE,CZ$ are equal.Prove that the hexagon is inscribed in a circle.\r\n\r\n Babis.", "Solution_1": "Consider ABDE: it is a trapezium, as AB//DE(they are opposite sides). But its diagonals AD and BE are equal; that means, ABDE is an isosceles trapezium. Similarly, one can obtain that AFDC and FBCE are both isosceles trapeziums. But every isosceles trapezium can be inscribed in a circle. We notice that the circumcircles of the three trapeziums share, one by one, a string: AD,BE,CF. Thus, these strings are the root axis of the three circles, and so are concurrent. But that common point is a point belonging also to three perpendicular bisectors: the common of AB and DE, the common of AF and CD, and the common of FE and BC(common because of the isosceles trapeziums). That is, the hexagon is inscribed in a circle whose center is the root center of the three circles described above(that now appear to be the same circle)." } { "Tag": [ "geometry", "\\/closed" ], "Problem": "How much would it help, if I took it this year, and the SAT two times this year?\r\n\r\n\r\n\r\n\r\n\r\nPretty much, is this enough to get 600-700s on the SAT verbal and math:\r\nSome algebra 1\r\nIntro to Geomatry\r\nIntro to problem solving", "Solution_1": "I don't see how it would help on the verbal...\r\n\r\nOtherwise, yes, though I'd say most people who really understand them should get 800, or very close.", "Solution_2": "Heh. Meant to say math and that I was also doing verbal :X" } { "Tag": [ "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO", "trigonometry", "ARML" ], "Problem": "How many of last year's top 8 finishers were eighth graders?", "Solution_1": "I don't think this really matters, does it? But probably 5-6 of them. Just because you're in 6th or 7th doesn't mean you can't be better than 8th graders... and just because you're an 8th grader doesn't mean you can't beat seniors on USAMO (Evan O'Dorney... who also won the national spelling bee...)", "Solution_2": "dude that was a pointless comment\r\n\r\neighth graders obviously beat seniors on the USAMO every year\r\n\r\nby virtue of the fact that eighth graders often make USAMO\r\n\r\nWHILE MOST SENIORS DO NOT IN FACT TAKE USAMO GASP\r\n\r\nalso it was pointless since you referenced that other competition which we shall not deign to mention here", "Solution_3": "Anyway, to answer the question,\r\n\r\nPlaces 1-5 were 8th grade for certain. There were probably some more 8th graders 6-8, though not_trig and MysticTerminator are correct (I did my best at state mathcounts in 6th grade).", "Solution_4": "me and calvin (7th and 6th graders last year respectively) got 7th and 8th. I think bryce got 5th and jenny C. got 2nd and Grey Symon got 1st and this homeschool kid and 'insert name that I know but don't remember' got 3rd and 4th. Arjun got 6th. \r\n\r\nI think the homeschool dude wasn't in 8th, (he might of been), so probably 5, but maybe 6...\r\n\r\nI still am mad of how i didn't get a trophy thing (7th ind. and 6+ got trophies, 5th team and 4+ got trophies! :furious: )\r\n\r\n\r\n[hide]\nbtw, WHY AM I NOT ON THE CENTRAL NC MATHCOUNTS PAGE!! >_> i should be... I got first...\n\nheres the website (unfortunately without my picture) http://www.penc.org/Student-Youth/MATHCOUNTS/MCCentralCarolina.aspx\n[/hide]", "Solution_5": "I suspect this \"homeschool dude\" was Brendan Fletcher? Or David Lucia (who went to nats but may or may not be homeschooled?) Also Gray Symon is not that good... well maybe at MC... w/e.\r\n\r\nWhen you get out of middle school, you see how silly MathCounts was.", "Solution_6": "1. Gray Symon - 41???\r\n2. Jenny Chen - 39?\r\n3. Brendan Fletcher - 38?\r\n4. David Lucia - 37\r\n5. Bryce Taylor - 37\r\n\r\nAlso, david goes to Providence Day, I think.", "Solution_7": "why do you know this? and why does it matter? it was last year's state mathcounts... which has no bearing whatsoever on this year because they were all 8th graders. Oh, well I guess that's the relationship? No one in the top 5 last year is gonna be there this year? ... ok", "Solution_8": "[quote=\"not_trig\"]why do you know this? and why does it matter? it was last year's state mathcounts... which has no bearing whatsoever on this year because they were all 8th graders. Oh, well I guess that's the relationship? No one in the top 5 last year is gonna be there this year? ... ok[/quote]\r\n\r\nMemory. Yes, comparing the results of old 8th graders is rather pointless, but someone asked for it, so I gave them the information.", "Solution_9": "I think david goes to hanes middle?\r\n\r\nso brendan must be the homeschool dude", "Solution_10": "[quote=\"PI-Dimension\"]I think david goes to hanes middle?\n\nso brendan must be the homeschool dude[/quote]\r\n\r\nUmm, no. David goes to Providence Day High (in Charlotte) and went to some middle school in Charlotte when he was in middle school. Hanes Middle School is in Winston-Salem (I went there for middle school).\r\n\r\nHowever, Brendan is indeed home schooled.", "Solution_11": "So, in all, the only people who were in top 8 last year in NC who was not in 8th grade was a 6th grader (me) and a seventh grader (PI-Dimension)?", "Solution_12": "[quote=\"dnkywin\"]So, in all, the only people who were in top 8 last year in NC who was not in 8th grade was a 6th grader (me) and a seventh grader (PI-Dimension)?[/quote]\r\n\r\n\r\nPretty much, yeah.", "Solution_13": "Wait...\r\nWhat grade was Christine Hong in?", "Solution_14": "[quote=\"dnkywin\"]Wait...\nWhat grade was Christine Hong in?[/quote]\r\n\r\nShe is a freshman this year (i.e. 8th last year).", "Solution_15": "Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .", "Solution_16": "[quote=\"dnkywin\"]Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .[/quote]\r\n\r\nThe talent was not lost (with the exception of Jenny); it's just been reallocated to use for different purposes (i.e. arml, AMCs, etc) :)", "Solution_17": "EDIT: [i]Middle School[/i] talent", "Solution_18": "[quote=\"b-flat\"][quote=\"dnkywin\"]Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .[/quote]\n\nThe talent was not lost (with the exception of Jenny); it's just been reallocated to use for different purposes (i.e. arml, AMCs, etc) :)[/quote]\r\n\r\nha, ha\r\n\r\nso 8 seniors graduating from ARML A team don't count? (5 or 6 of which were USAMO qualifiers? and arnav/pardon/yasha/misha included?)", "Solution_19": "No. Didn't you get the memo? This is now the NC middle school forum.", "Solution_20": "Well, if it's the NC [i]Middle School[/i] forum, then NC [i]did[/i] lose a lot of talent. :D :lol:", "Solution_21": "It's nice to see the next generation emerging... I feel old\r\nI'd better do well my final year :/", "Solution_22": "[quote=\"dnkywin\"]Wow. NC lost a [i][b]lot[/b][/i] talent this year :!: Increases my chances of getting to nats :lol: .[/quote]\r\n\r\nyou $ \\ne$ grammar", "Solution_23": "Well, math $ \\neq$ gramm[i]e[/i]r :lol:", "Solution_24": "[quote=\"dnkywin\"]Well, math $ \\neq$ gramm[i]e[/i]r :lol:[/quote]\r\n\r\nI assume also that $ \\text{math} \\neq \\text{spelling}$?", "Solution_25": "Yep, that[b][u]''[/u][/b]s right :lol:", "Solution_26": "How very hilarious. I assume this thread has served its purpose. Locked." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c > 0 $ and $ ab+ac +bc =3 $ .Prove that : \r\n $ a^2+b^2+c^2+abc \\geq a+b+c+1 $", "Solution_1": "Multiply both sides by two and add $2(ab+bc+ca)$ gives\r\n$(a+b+c)^2+a^2+b^2+c^2+2abc-6 \\ge 2(a+b+c)+2$\r\nRecall that $a^2+b^2+c^2+2abc + 1 \\ge 2(ab+bc+ca)=6$, \r\n$L.H.S. \\ge (a+b+c)^2 -1$\r\nIt remains to prove $(a+b+c)^2-2(a+b+c)-3 \\ge 0$ or $(a+b+c-3)(a+b+c+1) \\ge 0$\r\nBut this simple, since $a+b+c \\ge \\sqrt{3(ab+bc+ca)} = 3$", "Solution_2": "Nice solution ! but you used the ineq $ a^2+b^2+c^2+2abc + 1 \\geq 2(ab+bc+ca) $ which is more difficult to prove (of course mixing variables ;) ) than the original ineq \r\n\r\n This problem seemed to be posted a long time ago .\r\n Of course , this problem has a short solution but try thinking about my solution :\r\n $ a^2+b^2+c^2+abc \\geq a+b+c+1 \\iff (a+b+c-3)^2 +5(a+b+c) +abc -16 \\geq 0 $ \r\n It's enough to prove : $ 5(a+b+c) +abc -16 \\geq 0 $ \r\n \r\n Let's post solutions for the last ineq :)", "Solution_3": "[quote=\"nttu\"]Nice solution ! but you used the ineq $ a^2+b^2+c^2+2abc + 1 \\geq 2(ab+bc+ca) $ which is more difficult to prove (of course mixing variables ;) ) than the original ineq [/quote]\r\n\r\nNono, we only need to use $2abc+1 \\ge 3\\sqrt[3]{a^2b^2c^2}$ and then schur! It has become famous after apmo 04 no.5 :lol:", "Solution_4": "[quote=\"siuhochung\"][quote=\"nttu\"]Nice solution ! but you used the ineq $ a^2+b^2+c^2+2abc + 1 \\geq 2(ab+bc+ca) $ which is more difficult to prove (of course mixing variables ;) ) than the original ineq [/quote]\n\nNono, we only need to use $2abc+1 \\ge 3\\sqrt[3]{a^2b^2c^2}$ and then schur! It has become famous after apmo 04 no.5 :lol:[/quote] \r\n \r\n Really nice , Siuhochung ! I saw this ineq many times on ML , but i forgot its nice solution :( till you remind me ;)" } { "Tag": [ "probability", "geometry", "3D geometry", "modular arithmetic" ], "Problem": "1. The sum of three binomials each of the form ma+nb, where m and n are natural numberds, is 4a +8b. how many distinct sets of such three binomials exist?\r\n\r\n2. Emily randomly selected a diagonal of a regular octagon. Tim randomly selected one of the remaining diagonals. Find the probability that the diagonals they selected have the same length. Expres answer as comon fraction.\r\n\r\n3. Exactly one ordered pair of positive integers (x,y) satisfies the equation 37x + 73y = 2016. What is the sum of x+y ? (Is there any other way than brute force)\r\n\r\n4. What is the least two-digit integer whose cube has the same tens and units digits as the integer?\r\n\r\n5. What is the smallest positive integer that cannot be expressed as the sum of eight or fewer cubes of positive integers?\r\n\r\n6. A rectangular floor is covered with square tiles. The floor is 81 tiles long and 63 tiles wide. If a diagonal is drawn across the floor, how many tiles will it cross? \r\n\r\nAnswers:\r\n\r\n[hide]1. 12\n\n2. 31/95\n\n3. 36\n\n4. -99\n\n5. 23\n\n6. 135[/hide]", "Solution_1": "[hide=\"1\"]\n\nso find total ways to get 4, total ways to get 8, and multiply.\n\nso for 4, there is 2,1,1\n\nso for 8, there is 6,1,1 5,2,1 4,2,2 4,3,1 3,3,2 \n\n1*5=5\n(I bet I missed something\n[/hide]\n\n\n[hide=\"2\"]\nso there are 20 diagonals, by n(n-3)/2\n\nso, use casework, there are 4 long diagonals, so 1/5*3/19 chance of that happening\nthere are 8 semi long diagonals, so 2/5*7/19 chance of that\nThere are 8 short diagonals, so 2/5*7/19 chance of that\nso 3/95+14/95+14/95=31/95\n\n[/hide]\n\n[hide=\"3\"]\n\nso you have 37x+73y=2016\n(The only way I can find out how is brute force)\n\n[/hide]\n\n\n[hide=\"4\"]\n\nremember negatives. So basically, first, the last digit cubed has to equal itself.\nso the choices are 0,1,4,5,6,9. In addition, 9times the units digit times the first digit squared has the units digit as the tens digit. (sorry if this is unclear) So if this is a target, try them out, starting at the nineties. so, 9*x*9^2=9 (or something with a units digit of 9. so taking only units digits,9x=9. So take the highest possible one out of the units digits I stated above (since they are negative) to get the lowest answer. It turns out to be -99.\n\n[/hide]\n\n\n[hide=\"5\"]\nso list out the cubes, 1,8,27,64.....\nso first, you know that it can't be under 8. So try over it. Since 24, wouldn't work, try numbers under that. You can quickly find that 23 is the answer.\n[/hide]\n\n[hide=\"6\"]\nThe equation for these is m+n-GCF(m,n) where m and n are the dimensions of the floor.\n81+63-9=135.[/hide]", "Solution_2": "1. The sum of three binomials each of the form ma+nb, where m and n are natural numberds, is 4a +8b. how many distinct sets of such three binomials exist? \r\n\r\n#1 is 12, not 5. It even says on the answers.\r\n\r\n4 can be expressed as a)$ 1 \\plus{} 1 \\plus{} 2$\r\n8 can be experssed as b)$ 1 \\plus{} 1 \\plus{} 6$,___c)$ 1 \\plus{} 2 \\plus{} 5$,___d)$ ,1 \\plus{} 3 \\plus{} 4$,___ e)$ ,2 \\plus{} 2 \\plus{} 4$,___ f)$ 2 \\plus{} 3 \\plus{} 3$. the 2 in a) can match up with either $ 1$ or $ 6$ in b). The other 1's are forced to pair up with the two left in b). That's $ 2$ ways. For c), 2 in a) can pair with 1, 2, or 5, and again, 1's are forced to pair with the rest. That's $ 3$ possibilities. For d), there are 3, 2 for e, 2 for f. $ 2 \\plus{} 3 \\plus{} 3 \\plus{} 2 \\plus{} 2 \\equal{} [12]$", "Solution_3": "oops, sorry, #1 is 5", "Solution_4": "[quote=\"kl2836\"]oops, sorry, #1 is 5[/quote]\r\n\r\n#1 is 12. Keep the answer key as it is. It's Ihatepie who messed up. :D", "Solution_5": "argh this is getting me dizzy. oh well, its either 5 or 12 but mostly likely 12 :roll:", "Solution_6": "[quote=\"l Bob l\"]1. The sum of three binomials each of the form ma+nb, where m and n are natural numberds, is 4a +8b. how many distinct sets of such three binomials exist? \n\n#1 is 12, not 5. It even says on the answers.\n\n4 can be expressed as a)$ 1 \\plus{} 1 \\plus{} 2$\n8 can be experssed as b)$ 1 \\plus{} 1 \\plus{} 6$,___c)$ 1 \\plus{} 2 \\plus{} 5$,___d)$ ,1 \\plus{} 3 \\plus{} 4$,___ e)$ ,2 \\plus{} 2 \\plus{} 4$,___ f)$ 2 \\plus{} 3 \\plus{} 3$. the 2 in a) can match up with either $ 1$ or $ 6$ in b). The other 1's are forced to pair up with the two left in b). That's $ 2$ ways. For c), 2 in a) can pair with 1, 2, or 5, and again, 1's are forced to pair with the rest. That's $ 3$ possibilities. For d), there are 3, 2 for e, 2 for f. $ 2 \\plus{} 3 \\plus{} 3 \\plus{} 2 \\plus{} 2 \\equal{} [12]$[/quote]\r\n\r\no, I see.", "Solution_7": "[hide=\"3\"]#3 - diophantine equations\n\n$ x\\equal{}\\frac{2016\\minus{}73y}{37}$\n\nThus, the top must be divisible by 37, so take that mod 37\n\n$ 2016\\minus{}73y \\equiv 18\\minus{}y \\mod 37$\n\nSo the smallest, and only $ y$ that will yield a positive pair is $ 18$.\n\n$ x\\equal{}\\frac{2016\\minus{}18*73}{37}$.\n\nWhich should give the desired answer.[/hide]", "Solution_8": "HA HA, I'm going to make you all look stupid (actually, Mr. Frost is :P)\r\n\r\nNotice that the coefficients are all multiples of 36 plus 1 and that 2016 is divisible by 36. So take both sides mod 36...\r\n\r\n$ x \\plus{} y \\equiv 0 \\pmod{36}$\r\n\r\nand that makes the problem much simpler! :D", "Solution_9": "Slight Brute force method but easier for #3 (doesn't require mods and other stuff):\r\n\r\n[hide]The least multiple of 73 less than 2016 is 73*27=1971.\nThe two numbers have a difference of 45.\nNow, what happens if you add 37 3 times and then subtract 1 73? \nYou get +38.\nAdd 2 more 37's and subtract 1 more 73's.\nYou get +39.\nContinue this on and on until 45 find that there are \n17 37's and 27-8=19 73's\n\n17*37+19*73=2016\n17+19=36[/hide]\r\n\r\nnice yongyi... I should've seen that", "Solution_10": "[hide=\"number 3\"]$ 37x \\plus{} 73y \\equal{} 2016$\n$ 37x \\plus{} 73y \\equal{} 36(56)$\n$ 36x \\plus{} x \\plus{} 72y \\plus{} y \\equal{} 36(56)$\n$ 36(x \\plus{} 2y) \\plus{} x \\plus{} y \\equal{} 36(56)$\n\nThus x+y is a multiple of 36.\nIf $ x \\plus{} y$ is greater than or equal to 72, $ x \\plus{} 2y > 72$.\nBut $ x \\plus{} 2y < 56$ since $ 36(x \\plus{} 2y) \\plus{} x \\plus{} y \\equal{} 36(56)$ and x,y are positive integers.\nSince $ x \\plus{} y$ cannot be 0, it is 36.\nPlease correct me if i'm wrong :) [/hide]" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "We consider a prism with 6 faces, 5 of which are circumscriptible quadrilaterals. Prove that all the faces of the prism are circumscriptible quadrilaterals.", "Solution_1": "[hide=\"hint\"]just use Pithot's lemma (did i spelled corectly?)[/hide]", "Solution_2": "What's that lemma? I googled it, but no results...", "Solution_3": "$ABCD$ is a circumscriptibile quadrilateral $\\Longleftrightarrow AB+CD = AD+BC$.", "Solution_4": "using it you will get the result faster then using the official solution.\r\nis this problem a geometry or combinatorics one?", "Solution_5": "Geometry (solid geometry)." } { "Tag": [ "videos" ], "Problem": "Counting with Multiplication Part 1\n\nWe visualize how we can use multiplication in basic counting problems with a tree.", "Solution_1": "I thought that the material in the lecture was taught clearly.", "Solution_2": "i like the way he used a very simple topic to make his point. I have seen people teach this with like 15 of each thing making it very complicated.", "Solution_3": "I agree with Gamebot. Using trees to solve combinatorics problems are quite useful.", "Solution_4": "hey guys ... i know i'm a bigginer ...\n\nbut i really need to watch the vedio but i have a stupid question :D\n\nwhere is the link ...???\n\nbest wishes", "Solution_5": "On the top, where there's all the links, click Resources. Then, among the subtitles (like where you see Profile and Members and all that) click Videos. Then, click the first link, \"A collection of over 70 videos\". Then, scroll down into Chapter 1 Section and find Counting with Multiplication Part 1. Now, just watch! :D", "Solution_6": "if ur lazy here is link [url]http://www.artofproblemsolving.com/Resources/video.php?video_id=18[/url]" } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "modular arithmetic", "quadratics", "superior algebra" ], "Problem": "Let $ \\mathbb{Q}(\\sqrt{n})$ be an extension of field $ \\mathbb{Q}$. Prove that if $ 4$ does not divide $ n\\minus{}1$, then $ \\mathbb{Z}(\\sqrt{n})$ is the integral closure of $ \\mathbb{Z}$ in $ \\mathbb{Q}(\\sqrt{n})$. In other words, if a polynomial over $ \\mathbb{Z}$ admits a root the form $ a\\plus{}b\\sqrt{n}, a,b \\in \\mathbb{Q}$, then actually $ a,b \\in \\mathbb{Z}$, What if $ 4$ divides $ n\\minus{}1$?", "Solution_1": "If $ 4$ divides $ n\\minus{}1$, the integral closure (if I understand it correctly) is $ \\mathbb{Z}\\left[ \\frac{1 \\plus{} \\sqrt{n}}{2} \\right]$.", "Solution_2": "if $ D\\equiv 1\\pmod{4}$, $ D$ square-free, the algebraic integers in the quadratic field $ \\mathbb{Q}(\\sqrt {D})$ are those numbers of the form $ \\frac {m \\plus{} n\\sqrt {D}}{2}$, where $ m$ and $ n$ are of the same parity. recall that an algebraic integer is a number whose minimal polynomial has integer coefficients. so yes." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be the sidelengths of a triangle. Prove that\r\n\\[ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\plus{}\\frac{3(ab\\plus{}bc\\plus{}ca)^2}{a^3b\\plus{}b^3c\\plus{}c^3a\\minus{}abc(a\\plus{}b\\plus{}c)} \\ge 5.\\]\r\nWhen do we have equality?", "Solution_1": "[quote=\"can_hang2007\"]Let $ a,b,c$ be the sidelengths of a triangle. Prove that\n\\[ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\plus{} \\frac {3(ab \\plus{} bc \\plus{} ca)^2}{a^3b \\plus{} b^3c \\plus{} c^3a \\minus{} abc(a \\plus{} b \\plus{} c)} \\ge 5.\n\\]\nWhen do we have equality?[/quote]\r\n\r\nWe assume that $ a \\ge b \\ge c$ then \r\n\\[ a^3 b \\plus{} b^3 c \\plus{} c^3 a \\minus{} abc\\left( {a \\plus{} b \\plus{} c} \\right) \\equal{} a^2 b\\left( {a \\minus{} c} \\right) \\plus{} b^2 c\\left( {b \\minus{} a} \\right) \\plus{} c^2 a\\left( {c \\minus{} b} \\right) \\le a^2 b\\left( {a \\minus{} c} \\right) \\le a^2 b^2 \r\n\\]\r\n\r\nIt suffices to prove \r\n\\[ \\frac{{a^2 \\plus{} b^2 \\plus{} c^2 }}{{ab \\plus{} bc \\plus{} ca}} \\plus{} \\frac{{3\\left( {ab \\plus{} bc \\plus{} ca} \\right)^2 }}{{a^2 b^2 }} \\ge 5\r\n\\]\r\nwith $ a \\plus{} b \\equal{} 1$ and $ x \\equal{} ab$ then the inequality can rewrite as \r\n\\[ \\frac{{c^2 \\plus{} 1 \\minus{} 2x}}{{c \\plus{} x}} \\plus{} \\frac{{3\\left( {c \\plus{} x} \\right)^2 }}{{x^2 }} \\ge 5\r\n\\]\r\n\\[ \\frac{{1 \\minus{} \\left( {c \\plus{} 2} \\right)x}}{{c \\plus{} x}} \\plus{} \\frac{{3c^2 }}{{x^2 }} \\plus{} \\frac{{6c}}{x} \\ge 2 \\minus{} c\r\n\\]\r\nHence it's enough to prove in the case $ x \\equal{} \\frac{1}{4}$ which is equivalent with \r\n\r\n\\[ \\frac{{2 \\minus{} c}}{{4c \\plus{} 1}} \\plus{} 48c^2 \\plus{} 26c \\minus{} 2 \\ge 0\r\n\\]\r\n\r\n\\[ \\Leftrightarrow c\\left( {48c \\plus{} 26 \\minus{} \\frac{9}{{c \\plus{} 1}}} \\right) \\ge 0\r\n\\]\r\n\r\nObvious true. The inequality is proved. \r\nEquality holds if and only if $ a\\equal{}b,c\\equal{}0$." } { "Tag": [ "quadratics", "algebra" ], "Problem": "What is the pH of the solution when\r\n\r\n(a) $0.2$ mol/l $H_{2}SO_{4}$ mix with $0.1$ mol/l $Na_{2}CO_{3}$, in $1$l solution.\r\n\r\n(b) $0.2$ mol/l $H_{2}SO_{4}$ mix with $0.3$ mol/l $Na_{2}CO_{3}$, in $1$l solution.", "Solution_1": "(a) The concentration of hydronium ion from sulphuric acid is given by \r\n\r\n$[H_{3}O^{+}] = \\frac{\\sqrt{C^{2}+6KC+K^{2}}+C-K}{2}$\r\n\r\nwhere C is the concentration of $H_{2}SO_{4}$ and K is the ionization constant of hydrogensulphate ion (K = 0.012). In this case, C = 0.2 and substituting gives $[H_{3}O^{+}] = 0.21 mol/L$. This concentration of acid is enough to convert all the carbonate ion in carbonic acid, so in the \"end\" we will have a solution of $H_{2}CO_{3}$ with concentration 0.20 mol/L (this is not correct, since a reasonable amount of carbon dioxide will bubble out of the solution, but for the present case that doesn't matter) in which $[H_{3}O^{+}] = 0.01 mol/L$. Then we have the following equilibrium in solution\r\n\r\n$H_{2}CO_{3}+H_{2}O \\rightleftharpoons HCO_{3}^{-}+H_{3}O^{+}$\r\n\r\nand the equilibrium concentrations will be\r\n\r\n$[H_{2}CO_{3}] = 0.2-x \\approx 0.2$ [since $K_{a}(H_{2}CO_{3})$ is very small, $4.3 \\times 10^{-7}$\r\n$[HCO_{3}^{-}] = x$, $[H_{3}O^{+}] = 0.01+x$.\r\n\r\nSolving the resultant quadratic equation we get $x = 8.6 \\times 10^{-6}$ and so $[H_{3}O^{+}] \\approx 0.01 \\Rightarrow pH = 2$." } { "Tag": [ "geometry", "ratio", "calculus", "derivative", "probability", "expected value" ], "Problem": "The spectroscopic methods of structure determination \u2013 infrared, NMR, and UV spectroscopies and mass spectrometry \u2013 are very useful tools available to the organic chemist for the complete determination of the structure of an unknown compound. The study of the basic aspects of these techniques is an important part of any standard Organic Chemistry course, and some spectroscopy problems also sometimes appear in Chemistry Olympiads. In this thread, several problems regarding those four spectroscopic methods are presented in order to provide suitable training for those interested. The problems are roughly organised by increasing difficulty: however, a more challenging exercise may appear in the first problems, and an easy one can come up in the last problems. Most problems include a molecular formula, but for some this information is not provided: instead, data from combustion/elemental analysis and mass spectral information is provided. The first 19 problems address basic knowledge of NMR and IR spectroscopy. The next 37 problems are mainly structure determination problems, where the task is to deduce the structure of one or more compounds from the spectral data provided (sometimes, some chemical reactivity information is also given). Most of these problems include NMR information. Problems 57 to 66 provide specific training in IR spectroscopy, Mass spectrometry, and UV spectroscopy. From problem 67 onwards, additional problems and more advanced problems are provided. Whenever not specifically stated, NMR data is given as ppm units, IR data in cm-1, and MS data in m/z values. For the description of the intensity and shape of the IR absorption bands, the following abbreviations were adopted: w = weak; m = medium; s = strong; vs = very strong; vbr = very broadened band (like an \u2013OH band from alcohols, phenols, and carboxylic acids); br = medium broadened band (like the C=O absorption bands), sbr = small broadened band (like the N-H, C-O, and some aromatic bands); sh = sharp. Above all, don\u2019t forget: try to have some fun with these problems!\r\n\r\n------------------------------------------------------------------------------------------------------\r\n\r\n[size=150][b]Problem 1[/b][/size]\r\n\r\nHow many types of non-equivalent protons are there in each of the following molecules?\r\n\r\n(a) $ CH_{3}CH_{2}Br$.\r\n(b) $ CH_{3}CH_{2}OH$.\r\n(c) $ CH_{3}\\minus{}O\\minus{}CH_{3}$.\r\n(d) The isomers of formula $ C_{2}H_{4}Cl_{2}$.\r\n(e) $ CH_{3}CH_{2}\\minus{}O\\minus{}CH_{2}CH_{3}$.\r\n(f) $ CH_{3}\\minus{}O\\minus{}CH_{2}CH_{2}CH_{3}$.\r\n(g) $ CH_{3}\\minus{}O\\minus{}CH(CH_{3})_{2}$.\r\n(h) $ CH_{3}CH_{2}CH_{2}CH_{2}OH$.\r\n(i) The four isomers of formula $ C_{3}H_{6}Br_{2}$.\r\n(j) $ CH_{3}CH_{2}CH_{2}NO_{2}$.\r\n(k) $ CH_{3}\\minus{}O\\minus{}CH_{2}CH(CH_{3})_{2}$.\r\n(l) 2-Methylbut-1-ene.\r\n(m) Acetone.\r\n(n) $ CH_{3}CH_{2}CHO$.\r\n(o) Toluene.\r\n(p) $ CH_{2}\\equal{}CHCH_{2}OH$.\r\n(q) 2-Chlorobutan-2-ol.\r\n(r) $ C_{6}H_{5}CH_{2}CH_{3}$.\r\n(s) 1,2-Dichloropropane.\r\n(t) Methylcyclopropane.\r\n(u) cis-3-Hexene.\r\n(v) p-Methoxyaniline.\r\n(w) Styrene ($ C_{6}H_{5}CH\\equal{}CH_{2}$).\r\n(x) N,N-Dimethylacetamide ($ CH_{3}CON(CH_{3})_{2}$).\r\n(y) 3-Methylpen-1-yne.\r\n(z) $ CH_{2}\\equal{}CH\\minus{}CO_{2}\\minus{}CH_{2}CH_{3}$.\r\n\r\n\r\n[size=150][b]Problem 2[/b][/size]\r\n\r\nHow many 1H- and 13C-NMR signals are expected in the spectrum of each of the following compounds?\r\n\r\n(a) Ethane.\r\n(b) Propane.\r\n(c) $ CH_{3}\\minus{}O\\minus{}C(CH_{3})_{3}$.\r\n(d) 2,3-Dimethylbut-2-ene.\r\n(e) Z-But-2-ene.\r\n(f) E-But-2-ene.\r\n(g) n-Butane.\r\n(h) Ethanol.\r\n(i) Propene.\r\n(j) 1,2-Dibromopropane.\r\n(k) 1,1-Dimethylcyclopropane.\r\n(l) trans-1,2-Dimethylcyclopropane.\r\n(m) cis-Dimethylcyclopropane.\r\n(n) Pen-1-ene.\r\n(o) 1-Chloropropan-2-ol.\r\n(p) n-Propylbenzene.\r\n(q) 1,1-Dimethylcyclohexane.\r\n(r) Naphthalene.\r\n(s) 1-Methylcyclohexene.\r\n(t) p-Diethylbenzene.\r\n(u) $ (CH_{3})_{3}C\\minus{}CO_{2}\\minus{}CH_{3}$.\r\n(v) 3-Mehylpentanal.\r\n(w) 1,1,4-Tribromobutane.\r\n(x) 2-Chloropropene.\r\n(y) $ p\\minus{}CH_{3}O\\minus{}C_{6}H_{4}\\minus{}CH\\equal{}CHCH_{2}CH_{3}$.\r\n(z) 1,1-Dichlorocyclopropane.", "Solution_1": "[size=150][b]Problem 3[/b][/size]\r\n\r\nIdentify the compound(s) that meet each of the following descriptions:\r\n\r\n(a) $ C_{2}H_{6}O$; only one singlet in its 1H-NMR spectrum.\r\n(b) $ C_{5}H_{8}$; IR absorptions at 3300 and 2150m cm-1.\r\n(c) An hydrocarbon with 7 signals in its 13C-NMR spectrum.\r\n(d) $ C_{5}H_{12}$; only one signal in its 1H-NMR spectrum.\r\n(e) $ C_{4}H_{8}O$; strong IR absorption at 3400 cm-1.\r\n(f) $ C_{3}H_{7}Cl$; one doublet and one septet (1H-NMR).\r\n(g) A compound with six carbons and only 5 signals in its 13C-NMR spectrum.\r\n(h) $ C_{4}H_{8}O$; strong IR absorption at 1715 cm-1.\r\n(i) $ C_{4}H_{8}Cl_{2}$; two triplets (1H-NMR).\r\n(j) $ C_{5}H_{10}$; only one signal in its 1H-NMR spectrum.\r\n(k) A compound with 4 carbons that shows 3 signals in its 13C-NMR spectrum.\r\n(l) $ C_{8}H_{10}$; IR absorptions at 1600 and 1500 cm-1.\r\n(m) $ C_{4}H_{8}O_{2}$; one singlet, one triplet, and one quartet (1H-NMR). \r\n(n) $ C_{4}H_{8}O_{2}$; only one signal in its 1H-NMR spectrum.\r\n\r\n[size=150][b]Problem 4[/b][/size]\r\n\r\nHow would you use infrared spectroscopy to distinguish between these pairs of isomers?\r\n\r\n(a) $ HC\\equiv CCH_{2}NH_{2}$ and $ CH_{3}CH_{2}C\\equiv N$\r\n(b) $ CH_{3}COCH_{3}$ and $ CH_{3}CH_{2}CHO$\r\n(c) $ CH_{3}C\\equiv CCH_{3}$ and $ CH_{3}CH_{2}C\\equiv CH$\r\n(d) $ CH_{3}COCH\\equal{}CHCH_{3}$ and $ CH_{3}COCH_{2}CH\\equal{}CH_{2}$\r\n(e) $ H_{2}C\\equal{}CH\\minus{}O\\minus{}CH_{3}$ and $ CH_{3}CH_{2}CHO$\r\n\r\n[size=150][b]Problem 5[/b][/size]\r\n\r\nThree isomeric dimethylcyclopropanes give, respectively, 2, 3, and 4 signals in their 1H-NMR spectra. Identify the isomer that gives each of those numbers of signals.\r\n\r\n[size=150][b]Problem 6[/b][/size]\r\n\r\nEach of the following compounds have an 1H-NMR spectrum characterized by having only one signal, which appear at the indicated chemical shift. Identify each compound. \r\n\r\n(a) $ C_{8}H_{18}$; 0.9 ppm. \r\n(b) $ C_{5}H_{10}$; 1.5 ppm. \r\n(c) $ C_{8}H_{8}$; 5.8 ppm. \r\n(d) $ C_{4}H_{9}Br$; 1.8 ppm. \r\n(e) $ C_{2}H_{4}Cl_{2}$; 3.7 ppm. \r\n(f) $ C_{2}H_{3}Cl_{3}$; 2.7 ppm. \r\n(g) $ C_{5}H_{8}Cl_{4}$; 3.7 ppm. \r\n(h) $ C_{12}H_{18}$; 2.2 ppm. \r\n(i) $ C_{3}H_{6}Br_{2}$; 2.6 ppm.", "Solution_2": "[size=150][b]Problem 7[/b][/size]\r\n\r\nFor each of the following compounds, predict the approximate positions, shapes, and intensities of as many bands as you can that are expected to be observed in their infrared spectra:\r\n\r\n(a) Butanone.\r\n(b) 4-Methylpen-1-ene.\r\n(c) 4-Methylpen-1-yne.\r\n(d) Methyl butanoate.\r\n(e) 2-Methyloctan-2-ol.\r\n(f) m-Diethylbenzene.\r\n(g) 4-Methylbenzaldehyde.\r\n(h) Acetophenone.\r\n(i) cis-Pen-2-ene.\r\n(j) Styrene.\r\n(k) 4-Methylphenol.\r\n(l) Anisole (methoxibenzene).\r\n(m) Isobutyric acid.\r\n(n) Vinyl acetate.\r\n(o) Propionamide.\r\n(p) N-methylaniline\r\n(q) Methyl benzoate.\r\n(r) Phenyl acetonitrile.\r\n(s) 4-Nitrotoluene.\r\n\r\n\r\n[size=150][b]Problem 8[/b][/size]\r\n\r\nDistinguish the stereoisomers of 1,3-dibromo-1,3-dimethylcyclobutane on basis on their 1H-NMR spectra:\r\n\r\n- isomer X: 2.13 ppm (singlet, 6H), 3.21 ppm (singlet, 4H).\r\n\r\n- isomer Y: 1.88 ppm (singlet, 6H), 2.84 ppm (doublet, 2H), 3.54 ppm (doublet, 2H) (both doublets have equal spacings).\r\n\r\n\r\n[size=150][b]Problem 9[/b][/size]\r\n\r\nIn the 1H-NMR spectrum of each of the following compounds there are only two signals (both singlets), having the indicated chemical shifts (in ppm). Identify each compound.\r\n\r\n(a) $ C_{6}H_{8}$; 2.7 (4H) and 5.6 (4H). \r\n(b) $ C_{5}H_{11}Br$; 1.1 (9H) and 3.3 (2H). \r\n(c) $ C_{6}H_{12}O$; 1.1 (9H) and 2.1 (3H). \r\n(d) $ C_{6}H_{10}O_{2}$; 2.2 (6H) and 2.7 (4H).\r\n\r\n\r\n[size=150][b]Problem 10[/b][/size]\r\n\r\nHow many 1H-NMR signals are expected in the spectrum of each of the following compounds?\r\n\r\n(a) 1-Chloro-2-metilpropane.\r\n(b) 1-Bromobutane.\r\n(c) 1-Butanol.\r\n(d) Butane.\r\n(e) 1,4-Dibromobutane.\r\n(f) 2,2-Dibromobutane.\r\n(g) 2,2,3,3-Tetrabromobutane.\r\n(h) 1,1,4-Tribromobutane.\r\n(i) 1,1,1-Tribromobutane.\r\n(j) Vinyl bromide ($ CH_{2}\\equal{} CH\\minus{}Br$).\r\n(k) 1,1-Dibromoethene.\r\n(l) cis-1,2-Dibromoethene.\r\n(m) trans-1,2-Dibromoethene.\r\n(n) Allyl bromide ($ CH_{2}\\equal{} CHCH_{2}Br$).\r\n(o) 2-Methylbut-2-ene.\r\n(p) 2-Bromopropene.\r\n(q) 1-Butyne.\r\n\r\n\r\n[size=150][b]Problem 11[/b][/size]\r\n\r\nOne of the following compounds has only one peak in its 1H-NMR spectrum and only two peaks in its 13C-NMR spectrum. Which one?\r\n\r\n(a) Cyclohexane.\r\n(b) Cyclopropane.\r\n(c) Ethane.\r\n(d) 2,3-Dichlorobutane.\r\n(e) 2,2-Difluoro-1,3-Dioxacyclopentane.\r\n\r\n\r\n[size=150][b]Problem 12[/b][/size]\r\n\r\nEach of the following molecules shows only one peak in their 1H-NMR spectra. Order them according to increasing chemical shift: methane, benzene, acetylene, acetone, dichloromethane, cyclohexane, and ethylene.", "Solution_3": "[size=150][b]Problem 13[/b][/size]\r\n\r\nThe infrared spectrum of cis-cyclopentan-1,2-diol shows an \u2013OH band at a lower frequency relative to the position of a free \u2013OH group band. It is also observed that the band doesn\u2019t disappear even at high dilution. On the other hand, trans-cyclopentan-1,2-diol doesn\u2019t show that band. Can you suggest an explanation for this observation?\r\n\r\n\r\n[size=150][b]Problem 14[/b][/size]\r\n\r\nWhat kind of 1H-NMR spectrum is expected for the compound $ (Cl_{2}CH)_{3}CH$? Make a simple drawing of the spectrum, showing the relative position of each signal and its splitting pattern.\r\n\r\n\r\n[size=150][b]Problem 15[/b][/size]\r\n\r\nA compound of molecular formula $ C_{3}H_{6}O$ contains a carbonyl group. How could we used 1H-NMR spectroscopy to ascertain if the compound is an aldehyde or a ketone?\r\n\r\n\r\n[size=150][b]Problem 16[/b][/size]\r\n\r\nConsider the isomers with molecular formula $ C_{4}H_{9}Cl$, and select the one that has in its 1H-NMR spectrum:\r\n\r\n(a) just one peak.\r\n(b) several peaks, including one doublet at 3.4 ppm.\r\n(c) several peaks, including one triplet at 3.5 ppm.\r\n(d) several peaks, including two distinct peaks (each one integrating for 3H): a triplet at 1.0 ppm and a doublet at 1.5 ppm.\r\n\r\n\r\n[size=150][b]Problem 17[/b][/size]\r\n\r\nThe 1H-NMR spectra of formic acid ($ HCO_{2}H$), maleic acid ($ cis\\minus{}HO_{2}C\\minus{}CH\\equal{}CH\\minus{}CO_{2}H$), and malonic acid ($ HO_{2}C\\minus{}CH_{2}\\minus{}CO_{2}H$) are very similar because they all are made up of just two singlets with equal intensity. Make the correspondence between compounds A, B, C and the appropriate acid:\r\n\r\nCompound A: peaks at 3.2 and 12.1 ppm.\r\nCompound B: peaks at 6.3 and 12.4 ppm.\r\nCompound C: peaks at 8.0 and 11.4 ppm.\r\n\r\n\r\n[size=150][b]Problem 18[/b][/size]\r\n\r\nShow how one could use 1) 1H-NMR spectroscopy, and 2) IR spectroscopy to distinguish the following pairs of isomers:\r\n\r\n(a) Pent-2-ene and ethylcyclopropane.\r\n(b) Diethyl ether and methyl n-propyl ether.\r\n(c) Ethyl acetate and methyl propanoate.\r\n(d) 3-Methylbut-3-en-2-one and pent-3-en-2-one.\r\n(e) 3-Methylcyclohex-2-enone and 4-acetylcyclopentene.\r\n(f) p-Dimethylbenzene and ethylbenzene.\r\n\r\n\r\n[size=150][b]Problem 19[/b][/size]\r\n\r\nHow many peaks are expected in the 1H- and 13C-NMR spectra of the following brominated cyclopropanes?\r\n\r\n(a) Bromocyclopropane.\r\n(b) 1,1-Dibromocyclopropane.\r\n(c) cis-1,2-Dibromocyclopropane.\r\n(d) trans-1,2-Dibromocyclopropane.\r\n(e) 1,1,2-Tribromocyclopropane.", "Solution_4": "[size=150][b]Problem 20[/b][/size]\r\n\r\nPropose a structure for each of the following compounds, consistent with the spectroscopic data provided.\r\n\r\n(1) $ C_{4}H_{10}O$; 1H-NMR: 1.28 (singlet, 9H), 1.35 (singlet, 1H).\r\n\r\n(2) $ C_{4}H_{10}O_{2}$; 1H-NMR: 3.25 (singlet, 6H), 3.45 (singlet, 4H).\r\n\r\n(3) $ C_{4}H_{8}O_{3}$; 1H-NMR: 1.27 (triplet, 3H), 3.66 (quartet, 2H), 4.13 (singlet, 2H), 10.95 (singlet, 1H); IR: 2500-3000 (vs, vbr), 1715 (vs).\r\n\r\n(4) $ C_{2}H_{2}O_{2}Cl_{2}$; 1H-NMR: 6.0 (singlet), 11.70 (singlet); IR: 2500-2700 (vbr), 1705 (vs).\r\n\r\n(5) $ C_{3}H_{7}Br$; 1H-NMR: 1.71 (doublet, 6H), 4.32 (septet, 1H).\r\n\r\n(6) $ C_{15}H_{14}O$; 1H-NMR: 2.20 (singlet, 3H), 5.08 (singlet, 1H), 7.25 (multiplet, 10H); IR: 1720 (vs).\r\n\r\n(7) $ C_{4}H_{8}O$; 1H-NMR: 1.05 (triplet, 3H), 2.13 (singlet, 3H), 2.47 (quartet, 2H); IR: 1720 (vs).\r\n\r\n(8) $ C_{8}H_{14}O_{4}$; 1H-NMR: 1.2 (triplet, 6H), 2.5 (singlet, 4H), 4.1 (quartet, 4H); IR: 1740 (vs).\r\n\r\n(9) $ C_{7}H_{8}O$; 1H-NMR: 2.43 (singlet, 1H), 4.58 (singlet, 2H), 7.28 (multiplet, 5H); IR: 3200-3550 (s, vbr).\r\n\r\n(10) $ C_{4}H_{9}Cl$; 1H-NMR: 1.04 (doublet, 6H), 1.95 (multiplet, 1H), 3.35 (doublet, 2H). \r\n\r\n(11) $ C_{4}H_{7}O_{2}Cl$; 1H-NMR: 1.3 (triplet), 4.0 (singlet), 4.2 (quartet); IR: 1745 (vs).\r\n\r\n(12) $ C_{4}H_{7}O_{2}Br$; 1H-NMR: 1.08 (triplet, 3H), 2.07 (quintet, 2H), 4.23 (triplet, 1H), 10.97 (singlet, 1H); IR: 2500-3000 (vbr, s), 1715 (vs).\r\n\r\n(13) $ C_{4}H_{7}O_{2}Br$; 1H-NMR: 1.30 (triplet, 3H), 3.77 (singlet, 2H), 4.23 (quartet, 2H).\r\n\r\n(14) $ C_{8}H_{10}$; 1H-NMR: 1.25 (triplet, 3H), 2.68 (quartet, 2H), 7.23 (multiplet, 5H).\r\n\r\n(15) $ C_{3}H_{5}O_{2}Cl$; 1H-NMR: 1.73 (doublet, 3H), 4.47 (quartet, 1H), 11.22 (singlet, 1H).\r\n\r\n(16) $ C_{3}H_{5}O_{2}Cl$; 1H-NMR: 3.81 (singlet, 3H), 4.08 (singlet, 2H).\r\n\r\n(17) $ C_{11}H_{14}O_{2}$; 1H-NMR: 1.0 (doublet, 6H), 2.1 (multiplet, 1H), 4.1 (doublet, 2H), 7.8 (multiplet, 5H); IR: 1720 (vs). \r\n\r\n(18) $ C_{3}H_{7}O_{2}N$; 1H-NMR: 1.55 (doublet, 6H), 4.67 (septet, 1H). \r\n\r\n(19) $ C_{5}H_{10}O$; 1H-NMR: 1.10 (doublet, 6H), 2.10 (singlet, 3H), 2.50 (septet, 1H); IR: 1720 (vs). \r\n\r\n(20) $ C_{10}H_{12}O_{2}$; 1H-NMR: 1.2 (triplet, 3H), 3.5 (singlet, 2H), 4.1 (quartet, 2H), 7.3 (multiplet, 5H); IR: 1740 (vs). \r\n\r\n(21) $ C_{3}H_{3}Cl_{5}$; 1H-NMR: 4.52 (triplet, 1H), 6.07 (doublet, 2H). \r\n\r\n(22) $ C_{10}H_{14}$; 1H-NMR: 1.20 (triplet, 6H, J = 7 Hz), 2.70 (quartet, 4H, J = 7 Hz), 7.18 (broad singlet, 4H); IR: 745 (vs). \r\n\r\n(23) $ C_{10}H_{14}$; 1H-NMR: 0.88 (doublet, 6H), 1.86 (multiplet, 1H), 2.45 (doublet, 2H), 7.12 (singlet, 5H).\r\n\r\n(24) $ C_{10}H_{14}$; 1H-NMR: 1.30 (singlet, 9H), 7.28 (singlet, 5H).\r\n\r\n(25) $ C_{10}H_{14}$; 1H-NMR: 1.20 (doublet, 6H, J = 8 Hz), 2.28 (singlet, 3H), 2.85 (septet, 1H, J = 8 Hz), 7.0 (broad singlet, 4H); IR: 825 (s). \r\n\r\n(26) $ C_{10}H_{12}$; 1H-NMR: 0.65 (multiplet, 2H), 0.81 (multiplet, 2H), 1.37 (singlet, 3H), 7.17 (singlet, 5H).\r\n\r\n(27) $ C_{9}H_{11}Br$; 1H-NMR: 2.15 (quintet, 2H), 2.75 (triplet, 2H), 3.38 (triplet, 2H), 7.22 (singlet, 5H).\r\n\r\n(28) $ C_{3}H_{5}Br$; 1H-NMR: 2.32 (singlet, 3H), 5.35 (broad singlet, 1H), 5.54 (broad singlet, 1H).\r\n\r\n(29) $ C_{4}H_{9}Br$; 1H-NMR: 1.7 (singlet).\r\n\r\n(30) $ C_{4}H_{6}Cl_{2}$; 1H-NMR: 2.18 (singlet, 3H), 4.16 (doublet, 2H, J = 7 Hz), 5.71 (triplet, 1H, J = 7 Hz).\r\n\r\n(31) $ C_{4}H_{7}OBr$; 1H-NMR: 2.11 (singlet, 3H), 3.52 (triplet, 2H, J = 6 Hz), 4.40 (triplet, 2H, J = 6 Hz).\r\n\r\n(32) $ C_{4}H_{7}Br_{3}$; 1H-NMR: 1.95 (singlet, 3H), 3.9 (singlet, 4H).\r\n\r\n(33) $ C_{6}H_{14}$; 1H-NMR: 0.8 (doublet, 12H), 1.4 (septet, 2H). \r\n\r\n(34) $ C_{6}H_{12}$; 1H-NMR: 0.9 (triplet, 3H), 1.6 (singlet, 3H), 1.7 (singlet, 3H), 2.0 (quintet, 2H), 5.1 (triplet, 1H).\r\n\r\n(35) $ C_{4}H_{6}Cl_{4}$; 1H-NMR: 3.9 (doublet, 4H), 4.6 (triplet, 2H).\r\n\r\n(36) $ C_{5}H_{7}O_{2}N$; 1H-NMR: 1.2 (triplet, 3H), 3.5 (singlet, 2H), 4.2 (quartet, 2H); IR: 2980 (s), 2260 (s), 1750 (s).\r\n\r\n(37) $ C_{3}H_{7}OCl$; 1H-NMR: 2.0 (quintet, 2H), 2.8 (singlet, 1H), 3.7 (triplet, 2H), 3.8 (triplet, 2H).\r\n\r\n(38) $ C_{8}H_{9}Br$; 1H-NMR: 2.0 (doublet, 3H), 5.15 (quartet, 1H), 7.35 (multiplet, 5H). \r\n\r\n(39) $ C_{8}H_{16}$; 1H-NMR: 1.0 (singlet, 9H), 1.75 (singlet, 3H), 1.9 (singlet, 2H), 4.6 (singlet, 1H), 4.8 (singlet, 1H); IR: 3040 (m), 2950 (s), 1640 (s).\r\n\r\n(40) $ C_{9}H_{10}$; 1H-NMR: 2.04 (quintet, 2H), 2.91 (triplet, 4H), 7.17 (singlet, 4H).\r\n\r\n(41) $ C_{9}H_{10}O$; 1H-NMR: 2.0 (singlet, 3H), 3.75 (singlet, 2H), 7.2 (singlet, 5H); IR: 3100 (m), 3000 (s), 1720 (vs), 740 (s), 700 (s).\r\n\r\n(42) $ C_{8}H_{18}O_{2}$; 1H-NMR: 1.24 (singlet, 12H), 1.56 (singlet, 4H), 1.95 (singlet, 2H); IR: 3350. \r\n\r\n(43) $ C_{3}H_{5}Cl_{3}$; 1H-NMR: 2.20 (singlet, 3H), 4.02 (singlet, 2H).\r\n\r\n(44) $ C_{14}H_{14}$; 1H-NMR: 2.9 (singlet, 4H), 7.1 (broad singlet, 10H). \r\n\r\n(45) $ C_{10}H_{13}Cl$; 1H-NMR: 1.57 (singlet, 6H), 3.07 (singlet, 2H), 7.27 (singlet, 5H).\r\n\r\n(46) $ C_{3}H_{5}ClF_{2}$; 1H-NMR: 1.75 (triplet, 3H), 3.63 (triplet, 2H). \r\n\r\n(47) $ C_{9}H_{12}$; 1H-NMR: 0.90, 1.50, 2.20, 7.10.\r\n\r\n(48) $ C_{7}H_{14}O$; 1H-NMR: 1.0 (singlet, 9H), 2.1 (singlet, 3H), 2.3 (singlet, 2H). IR: 1710 (vs).\r\n\r\n(49) Hydrocarbon; MS: molecular ion at m/z = 120; 1H-NMR: 1.22 (doublet, 6H, J = 7 Hz), 2.90 (septet, 1H, J = 7 Hz), 7.25 (broad singlet, 5H).\r\n\r\n(50) MS: molecular ion at m/z = 88; 1H-NMR: 1.11 (triplet, 3H, J = 7 Hz), 2.32 (quartet, 2H, J = 7 Hz), 3.65 (singlet, 1H); 13C-NMR: 9.3, 27.6, 51.4, 174.6; IR: 1735 (vs).", "Solution_5": "[size=150][b]Problem 21[/b][/size]\r\n\r\nCompound A, $ C_{8}H_{10}$, yields three monobrominated substitution products, $ C_{8}H_{9}Br$, by reaction with bromine. The 1H-NMR spectrum of A shows a complex multiplet at 7.0 ppm (4H) and a singlet at 2.30 ppm (6H). What is the structure of A?\r\n\r\n\r\n[size=150][b]Problem 22[/b][/size]\r\n\r\nThe red fox ([i]Vulpes vulpes[/i]) uses a chemical communication system based on scent markers in the urine. A recent investigation showed that one of the components of the fox\u2019s urine, compound B, is an organic sulphide. Spectral analysis of pure compound B yielded the following results: MS: molecular ion at m/z = 116; IR: 890 cm-1 (strong band); 1H-NMR: 1.74 (singlet, 3H), 2.11 (singlet, 3H), 2.27 (triplet, 2H, J = 4.2 Hz), 2.57 (triplet, 2H, J = 4.2 Hz), 4.73 (broad peak, 2H). Propose a structure for compound B that is consistent with all these spectral properties.\r\n\r\n\r\n[size=150][b]Problem 23[/b][/size]\r\n\r\nTreating butanone (1 eq.) with molecular bromine (2 eqs.) in aqueous HBr gives a compound with molecular formula $ C_{4}H_{6}OBr_{2}$. The 1H-NMR spectrum of this compound shows the following peaks: 1.9 (doublet, 3H), 4.6 (singlet, 2H), e 5.2 (quartet, 1H). Identify the product.\r\n\r\n\r\n[size=150][b]Problem 24[/b][/size]\r\n\r\nCompounds C and D are isomeric diketones of molecular formula $ C_{6}H_{10}O_{2}$. The 1H-NMR spectrum of C shows two peaks, both singlets, at 2.2 ppm (6H) and 2.8 ppm (4H). The 1H-NMR spectrum of D also shows two peaks: 1.3 ppm (triplet, 6H) and 2.8 ppm (quartet, 4H). What are the structures of C and D?\r\n\r\n\r\n[size=150][b]Problem 25[/b][/size]\r\n\r\nThe 1H-NMR spectrum of compound E ($ C_{8}H_{8}O$) consists in two singlets of equal area at 5.1 ppm (sharp) and 7.2 ppm (broad). By treatment with excess HBr, compound E yields a single dibromide ($ C_{8}H_{8}Br_{2}$). The 1H-NMR spectrum of this dibromide is similar to the spectrum of E, as it also shows two singlets of equal area at 4.7 ppm (sharp) and 7.3 ppm (broad). Suggest reasonable structures for E and its dibromide.", "Solution_6": "[size=150][b]Problem 26[/b][/size]\r\n\r\nCompound F has molecular formula $ C_{10}H_{14}O_{3}$. Its 1H-NMR spectrum shows the following peaks: 3.25 (singlet, 6H), 4.38 (singlet, 1H), two doublets at ~ 4.6 (1H each), 7.29 (multiplet, 5H). By hydrolysis in diluted HCl, F yields compound G (m. p. 164-165\u00baC), which has molecular formula $ C_{16}H_{16}O_{4}$. In the infra-red spectrum of G there is a band at 3550 cm-1, but there is no band in the C=O region. What are the structures of F and G?\r\n\r\n\r\n[size=150][b]Problem 27[/b][/size]\r\n\r\nThe acid catalysed dehydration of 1-methylcyclohexanol yields a mixture of two alkenes as products. Who are them? After you separated them by chromatography, how would you use 1) 1H-NMR and 2) infrared spectroscopy to distinguish them?\r\n\r\n\r\n[size=150][b]Problem 28[/b][/size]\r\n\r\nBy bromination of 3-methylbutanone two isomeric compounds of molecular formula $ C_{5}H_{9}OBr$ are obtained in a ratio 95:5. The 1H-NMR spectrum of the major isomer H is the following: 1.2 (doublet, 6H), 3.0 (septet, 1H), 4.1 (singlet, 2H). The 1H-NMR spectrum of the minor isomer I shows the following peaks: 1.9 (singlet), 2.5 (singlet), with the peak at lower field having half the area of the peak at higher field. Suggest reasonable structures for isomers H and I.\r\n\r\n\r\n[size=150][b]Problem 29[/b][/size]\r\n\r\nCompounds J and K are isomers of molecular formula $ C_{4}H_{8}O_{3}$. Identify them through their 1H-NMR spectra:\r\n\r\nCompound J: 1.3 (triplet, 3H), 3.6 (quartet, 2H), 4.1 (singlet, 2H), 11.1 (broad singlet, 1H).\r\nCompound K: 2.6 (triplet, 2H), 3.4 (singlet, 3H), 3.7 (triplet, 2H), 11.3 (broad singlet, 1H).\r\n\r\n\r\n[size=150][b]Problem 30[/b][/size]\r\n\r\nExplain why 3-methylbutan-2-ol shows [i]five[/i] peaks in its 13C-NMR spectrum, at 17.90, 18.15, 20.00, 35.05, and 72.75 ppm, and make the correspondence between each signal and the appropriate carbon.\r\n\r\n\r\n[size=150][b]Problem 31[/b][/size]\r\n\r\nThe 13C-NMR spectrum of cyclooctatetraene shows only one peak, but different 13C-NMR spectra are observed for 1,2-dimethylcyclooctatetraene and 2,3-dimethylcyclooctatetraene, which indicates that these molecules are isomers and not resonance structures. Examine the structures of these three compounds (models may help) and explain 1) why cyclooctatetraene shows only one peak, and 2) the expected number of 13C peaks of each dimethyl cyclooctatetraene.", "Solution_7": "[size=150][b]Problem 32[/b][/size]\r\n\r\nCompound L ($ C_{10}H_{14}O$) is soluble in aqueous NaHO but insoluble in aqueous sodium bicarbonate. This compound reacts with bromine in water to yield a dibrominated derivative, $ C_{10}H_{12}OBr_{2}$. The infrared spectrum of L shows a broad band centred at 3250 cm-1 and a strong band at 830 cm-1. The 1H-NMR spectrum of L is the following: 1.3 (singlet, 9H), 4.9 (singlet, 1H), 7.0 (multiplet, 4H). What is the structure of L?\r\n\r\n\r\n[size=150][b]Problem 33[/b][/size]\r\n\r\nThe 13C-NMR spectrum of a commercial sample of 2,4-pentanediol shows not three but [i]five[/i] peaks, at 23.3, 23.9, 46.5, 64.8, and 68.1 ppm. Explain why.\r\n\r\n\r\n[size=150][b]Problem 34[/b][/size]\r\n\r\nCompound M ($ C_{9}H_{10}$) reacts with bromine in carbon tetrachloride. Some spectral data of M follows: \r\n\r\nIR: 3035 (m), 3020 (m), 2925 (m), 2853 (s), 1640 (m), 990 (s), 915 (s), 740 (s), 695 (s). \r\n\r\n1H-NMR: 3.1 (doublet, 2H), 4.8 (multiplet), 5.1 (multiplet), 5.8 (multiplet), 7.1 (multiplet, 5H). \r\n\r\nThe UV spectrum of M shows a maximum at 255 nm. Propose a structure for M.\r\n\r\n\r\n[size=150][b]Problem 35[/b][/size]\r\n\r\nCompound N ($ C_{9}H_{10}O$) gives a negative iodoform test, and have the following spectral properties: IR: 1690 (s); 1H-NMR: 1.2 (triplet, 3H), 3.0 (quartet, 2H), 7.7 (multiplet, 5H). What is the structure of N?\r\n\r\n\r\n[size=150][b]Problem 36[/b][/size]\r\n\r\nCompound O is an isomer of compound N from the previous problem, but gives a positive iodoform test. Spectral data for O: IR: 1705 (s); 1H-NMR: 2.0 (singlet, 3H), 3.5 (singlet, 2H), 7.1 (multiplet, 5H). What is the structure of O?\r\n\r\n\r\n[size=150][b]Problem 37[/b][/size]\r\n\r\nIdentify each of the following two isomers of molecular formula $ C_{20}H_{18}O$ from the 1H-NMR data given:\r\n\r\nIsomer P: 2.23 (singlet, 1H), 3.92 (doublet, 1H, J = 7 Hz), 4.98 (doublet, 1H, J = 7 Hz), 6.81 (singlet, 10H), 6.99 (singlet, 5H).\r\n\r\nIsomer Q: 2.14 (singlet, 1H), 3.55 (singlet, 2H), 7.25 (broad peak, 15H).\r\n\r\nThese two isomers could be distinguished by a single chemical test. What is it?", "Solution_8": "[size=150][b]Problem 38[/b][/size]\r\n\r\nBy treatment of methallyl chloride, $ H_{2}C\\equal{}C(CH_{3})CH_{2}Cl$, with a solution of sodium amide in THF, a hydrocarbon with molecular formula $ C_{4}H_{6}$ is obtained. The 1H-NMR spectrum of this hydrocarbon shows the following peaks: 0.83 (doublet, 2H, J = 2 Hz), 2.13 (doublet, 3H, J = 1 Hz), 6.40 (multiplet, 1H). What is the most likely structure for this hydrocarbon, and what is the mechanism of its formation?\r\n\r\n\r\n[size=150][b]Problem 39[/b][/size]\r\n\r\nCompound R have molecular formula $ C_{8}H_{18}O_{2}$ and does not react with periodic acid. Its 1H-NMR spectrum shows three singlets at 1.2 (12H), 1.6 (4H), and 2.0 (2H). What is the structure of R?\r\n\r\n\r\n[size=150][b]Problem 40[/b][/size]\r\n\r\nBy treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid a compound with molecular formula $ C_{18}H_{29}OBr$ is obtained quantitatively. The infrared spectrum of this compound shows bands at 1630 and 1655 cm-1. Its 1H-NMR spectrum shows only three singlets at 1.2, 1.3, and 6.9 ppm, with peak areas of 9:18:2, respectively. What is the structure of this compound?\r\n\r\n\r\n[size=150][b]Problem 41[/b][/size]\r\n\r\nCompound S, with molecular formula $ C_{9}H_{8}O_{2}$, absorbs two moles of hydrogen (with a deactivated palladium catalyst) to form compound T. By hydrogenation over an active platinum catalyst, compound T absorbs another two moles of hydrogen to give compound U, which also forms directly from S by hydrogenation with the platinum catalyst. All three compounds react with hydroxylamine to form oximes. Compound T, by ozonolysis, yields two moles of a substance and one mol of another substance. The 1H-NMR spectrum of S shows only two singlets at 2.1 and 2.6 ppm, with peak areas of 3:1, respectively. Identify compounds S, T, and U.\r\n\r\n\r\n[size=150][b]Problem 42[/b][/size]\r\n\r\nOne of the many products that can be obtained from the chlorination of propane is a pentachloropropane with the following 1H-NMR spectrum: 4.5 (triplet), 6.1 (doublet), with peak areas 1:2, respectively. Identify this compound.\r\n\r\n\r\n[size=150][b]Problem 43[/b][/size]\r\n\r\nCompound V has molecular formula $ C_{9}H_{11}NO$ and the following spectral properties: IR: 1680 (vs); 1H-NMR: 2.8 (singlet, 3H), 2.9 (singlet, 3H), 6.7-7.7 (multiplet, 4H), 9.7 (singlet, 1H). When $ D_{2}O$ is added to the sample, the peak at 9.7 ppm disappears. Identify compound V.\r\n\r\n\r\n[size=150][b]Problem 44[/b][/size]\r\n\r\nCompound W is an isomer of compound V from the previous problem. In its IR spectrum the most prominent bands appear at 1652 (s), 1610 (vs), 1511 (vs), 1246 (vs), 1029 (s), and 829 (s). A weak band appears at 2760 cm-1, but this is due to a contamination from a precursor of compound W. When $ D_{2}O$ is added to a sample of W, there is no change in its 1H-NMR spectrum. Identify compound W.", "Solution_9": "[size=150][b]Problem 45[/b][/size]\r\n\r\nBy treatment of neopentyl chloride with sodium amide, an hydrocarbon with molecular formula $ C_{5}H_{10}$ was obtained. This hydrocarbon is easily dissolved in concentrated sulphuric acid, but is not oxidized by cold, neutral, diluted solutions of potassium permanganate. The 1H-NMR spectrum of this hydrocarbon shows peaks at 0.20 and 1.05 ppm, with relative areas of 2:3, respectively. When the isotopically labelled alkyl halide $ (CH_{3})_{3}CCD_{2}Cl$ was allowed to react under the same conditions, the product obtained had its molecular ion at m/z = 71. Identify the hydrocarbon and propose a mechanism for its formation.\r\n\r\n\r\n[size=150][b]Problem 46[/b][/size]\r\n\r\nA compound with molecular formula $ C_{4}H_{6}$ gives a 13C-NMR spectrum with two peaks of nearly equal intensity: one, at 30.2 ppm, corresponds to a CH2 carbon; the other, at 136 ppm, is of a CH carbon. Identify this compound.", "Solution_10": "[size=150][b]Problem 47[/b][/size]\r\n\r\nA compound with molecular formula $ C_{10}H_{16}O_{2}$ absorbs one mol of hydrogen by catalytic hydrogenation. In the UV region it shows an absorption maximum at 236 nm ($ \\epsilon \\equal{} 2\\times 10^{4}$), and its IR spectrum shows characteristic absorptions at 1667, 1620, and 1100 cm-1 (the last band is strong and broad). Its 1H-NMR spectrum shows the following peaks: 1.0 (triplet, 3H, J = 7 Hz), 1.0-2.0 (multiplet, 6H), 2.4 (singlet, 3H), 3.2 (quartet, 2H, J = 7 Hz), 4.8 (triplet, 1H), 6.2 (triplet, 1H). Propose a structure for this compound.\r\n\r\n\r\n[size=150][b]Problem 48[/b][/size]\r\n\r\nCompound X reacts with magnesium metal to form a product which, when added to acetone, produces an alcohol. Elemental analysis shows it to be 15.40% C and 3.23% H. IR spectrum: 2973 (s), 2910 (m), 1440 (m), 1378 (m), 1202 (vs, sbr), 951 (s). Mass spectrum: m/z = 27, 29 (base peak), 127, 156 (molecular ion). Identify compound X.", "Solution_11": "[size=150][b]Problem 49[/b][/size]\r\n\r\nCompound Y has the following elemental composition: 71.4% C, 9.5% H, 19.0% O. Spectral data for Y: \r\n\r\nMass spectrum: m/z = 43, 69 (base peak), 83, 84 (molecular ion, vanishingly small peak, almost undetectable). \r\n\r\nIR spectrum: 3400 (s, vbr), 3300 (sh, vs), 2987 (sh, s), 2200 (w), 1460 (m), near 1378 (two sharp bands \u2013 close enough to each other that they almost give a single broadened band -, the low frequency one stronger than the other), 1170 (s, sbr), 640 (s, sbr). \r\n\r\nIdentify compound Y.\r\n\r\n\r\n[size=150][b]Problem 50[/b][/size]\r\n\r\nCompound Z has the following elemental composition: 62.1% C, 10.3% H, 27.6% O. Spectral data for Z: \r\n\r\nMass spectrum: m/z = 26, 27, 28, 29 (base peak), 57 (weak), 58 (molecular ion). \r\n\r\nIR spectrum: 2989 (m), 2813 (m), 2716 (m), 1739 (vs), 1461 (m), 1415 (m), 1391 (m). \r\n\r\nWhat is the structure of Z?", "Solution_12": "[size=150][b]Problem 51[/b][/size]\r\n\r\nIdentify the following isomers of molecular formula $ C_{3}H_{5}Br$ from the spectral information provided:\r\n\r\n(a) Isomer A\u2019 (1H-NMR): 2.31 (singlet, 3H), 5.38 (singlet, 1H), 5.59 (singlet, 1H).\r\n\r\n(b) Isomer B\u2019 (13C-NMR): 32.6 (CH2), 118.8 (CH2), 134.2 (CH).\r\n\r\n(c) Isomer C\u2019 (13C-NMR): 12.0 (CH2), 16.8 (CH). The low field peak has half the intensity of the other peak.\r\n\r\n\r\n[size=150][b]Problem 52[/b][/size]\r\n\r\nTwo isomeric lactones D\u2019 and E\u2019, with molecular formula $ C_{5}H_{8}O_{2}$, show a characteristic IR absorption at 1810 cm-1. In their 1H-NMR spectra, isomer D\u2019 shows two singlets at 1.1 and 2.2 ppm (peak areas 3:1, respectively), while isomer E\u2019 also shows two singlets but at 1.2 and 4.0 ppm (peak areas 3:1, respectively). Suggest reasonable structures for D\u2019 and E\u2019.", "Solution_13": "[size=150][b]Problem 53[/b][/size]\r\n\r\n(a) By catalytic hydrogenation, compound F\u2019 ($ C_{5}H_{8}$) gave cis-1,2-dimethylcyclopropane. On the basis of this result, three possible isomeric structures were proposed for F\u2019. Which were they?\r\n\r\n(b) The lack of an absorption band near 890 cm-1 in the IR spectrum of F\u2019 turned one of the proposed structures unlikely. Which one?\r\n\r\n(c) The 1H-NMR spectrum of F\u2019 showed peaks at 1.04 and 2.22 ppm with relative areas 1:3, respectively. Which of the three structures from (a) is compatible with this data?\r\n\r\n(d) The base peak in the mass spectrum of F\u2019 appears at m/z = 67. Which ion is very probably responsible for this peak? How do you explain its abundance?\r\n\r\n(e) Compound F\u2019 can be prepared from open chain compound(s) in a single step. How?", "Solution_14": "[size=150][b]Problem 54[/b][/size]\r\n\r\nTwo isomers G\u2019 and H\u2019, with molecular formula $ C_{8}H_{12}O_{4}$, are easily converted in compound I\u2019 by catalytic hydrogenation. Spectral data (1H-NMR) is given below for all three compounds. Identify them.\r\n\r\nG\u2019: 1.30 (triplet, 3H, J = 7 Hz), 4.28 (quartet, 2H, J = 7 Hz), 6.28 (singlet, 1H).\r\nH\u2019: 1.32 (triplet, 3H, J = 7 Hz), 4.27 (quartet, 2H, J = 7 Hz), 6.83 (singlet, 1H).\r\nI\u2019: 1.25 (triplet, 3H, J = 7 Hz), 2.62 (singlet, 2H), 4.15 (quartet, 2H, J = 7 Hz).\r\n\r\n\r\n[size=150][b]Problem 55[/b][/size]\r\n\r\nElemental analysis of compound J\u2019 showed only the presence of C, H, and O. By combustion analysis, the following composition was obtained: 79.37% C, 8.88% H. The spectral properties of J\u2019 are the following: \r\n\r\nIR: 3360 (s, vbr); three sharp, weak-medium bands between 3100 and 3000 cm-1; three strong and sharp bands between 3000 and 2850 cm-1; 1600 (m); 1500 (s); 760 (m); 700 (vs). \r\n\r\n1H-NMR: 0.91 (triplet, 3H), 1.75 (quartet, 2H), 2.0 (singlet, 1H, broadened peak), 4.5 (triplet, 1H), 7.3 (singlet, 5H).\r\n\r\n13C-NMR: 10, 31, 70, 125, 127, 128, 146 (short peak).\r\n\r\nIdentify compound J\u2019. \r\n\r\n\r\n[size=150][b]Problem 56[/b][/size]\r\n\r\nHydrocarbon K\u2019, with molecular formula $ C_{6}H_{6}$, has an 1H-NMR spectrum with only two signals, at 3.84 ppm and 6.55 ppm, with peak areas 1:2, respectively. By heating in pyridine for 3 hours, K\u2019 is quantitatively converted in benzene. By moderate hydrogenation K\u2019 gives L\u2019, whose spectra give the following information: MS \u2013 molecular ion at m/z = 82; IR \u2013 there are no double bonds; 1H-NMR \u2013 broad peak at 2.34 ppm.\r\n\r\n(a) Propose structures for K\u2019 and L\u2019.\r\n(b) In the 1H-NMR spectrum of K\u2019, the high field peak is a quintet, while the low field peak is a triplet. How can these splittings be explained?" } { "Tag": [ "geometry", "arithmetic sequence" ], "Problem": "Here is #18 from 1998 MMPC.\r\n\r\nA long strip of paper 0.01 cm thick and 10 cm wide is wound tightly about a wooden cylinder 5 cm in diameter. The resulting reel of paper is 10 cm in diameter. Which of the following numbers is closest to the length of the strip of paper (in cm)?\r\n\r\n(A)4700 (B)5100 (C)5500 (D)5900 (E)6300\r\n\r\nCould you please give a detailed solution and answer?\r\n\r\nThanks", "Solution_1": "If I give detailed solution, you won't be able to learn to develop your skills. Instead, I'll give you steps on how to solve it.\r\n\r\n[hide=\"Steps\"]Always draw pictures. It gives you an idea what it looks like.\n1.- Get the collective thickness of the paper rolled around the cylinder.\n2.- Now determine the number of times the paper is rolled around the cylinder.\n3.- Use the formula for the length of circumference, find the length of the paper. Remember each circumference of paper around the cylinder increases a little for every roll around the cylinder due to the thickness of the paper.[/hide]", "Solution_2": "[hide]\nFind area of combined paper and cylinder - the cylinder which is $\\frac{75\\pi}{4}$. Then, since thickness if 0.01cm, multiply by 100 to get $1875\\pi$. Finally, multiply by $\\pi$ and round to the nearest hundred to get $\\boxed{D}$.[/hide]", "Solution_3": "[hide=\"Slighty longer solution\"] The radius must increase from 2.5 cm to 5 cm at 0.01 cm per layer. Therefore, there must be 250 layers. The layers have a length, or circumference, of $5\\pi$, $5.02\\pi$, $5.04\\pi$,..., $10\\pi$. Using the formula for arithmetic sequence, we get $\\frac{250}{2}\\cdot15\\pi=1875\\pi\\approx5890.5$. So the answer is $\\boxed{(D) 5900}$. [/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ a > 0$ , $ b > 0$ , $ a\\ne b\\ \\implies\\ \\boxed {\\ a^b\\ \\plus{} \\ b^a\\ \\ge\\ 1\\ \\plus{} \\ ab\\ \\plus{} \\ (1 \\minus{} a)\\ (1 \\minus{} b)\\ \\min\\ \\{\\ 1\\ ,\\ ab\\ \\}\\ }\\ .$", "Solution_1": "Use Bernoulli Inequality" } { "Tag": [ "trigonometry" ], "Problem": "I'm looking for information and uses for something called Mollweide's Formula, $\\frac{a+b}{c}=\\frac{\\cos\\frac{1}{2}(\\alpha-\\beta)}{\\sin\\frac{1}{2}\\gamma}$.\r\n\r\nDerivation, general uses, and contest problems would be appreciated. Thank you. :)", "Solution_1": "this is a random formula, this is not useful, this is much like the law of tangents", "Solution_2": "The expression on the right was used with the same expression in some other cyclic permutation of the variables on some old mandelbrot problem...it's derivation is probably just LOC and LOS." } { "Tag": [ "geometry", "ratio" ], "Problem": "The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle. Many triangles are formed when this is done. What is the ratio, expressed as a common fraction, of the area of the original triangle? Express your answer as a common fraction.", "Solution_1": "Ratio of what?\r\n\r\nMy guess:\r\n[hide]1/9[/hide]", "Solution_2": "I must have mistyped the problem. I don't know if I'll get back to this." } { "Tag": [], "Problem": "I have suggested Mr. Goebel set up a math forum under his current web page. An example is as follows:\r\n\r\nhttp://forum.texasmath.org/index.php\r\n\r\nMr. Goebel thinks it was a good idea but he does not know how to do it.\r\n\r\nDo you know what is involved to set on of these up? If you know something about this and like to help, please contact Mr. Goebel at goebel@ncssm.edu \r\n\r\nThanks.", "Solution_1": "Just run a quick google search. [url]http://www.createforum.net/[/url] There are plenty of other options too.\r\n\r\nOr you could ask the folks at the Texas forum how they made theirs. I'm sure they would be glad to share.", "Solution_2": "Here is one that I just created for test purpose only. It is free. Mr. Goebel may need to pay to buy better one ( I really do not know).\r\n\r\nhttp://ncmathcontest.informe.com/" } { "Tag": [ "geometry", "trigonometry" ], "Problem": "If one side of a regualr hexagon has a side of length $r$, then what is the area of the hexagon?\r\n\r\nThere should be two ways to do this problem.", "Solution_1": "[hide=\"HINT\"] Notice something special about a regular hexagon. Divide it into triangles. Now, for those special triangles, what is the area for each one? Times it by 6. This is one way. \n\nYou could also use trigonometry, or you could put the hexagon inside a rectangle. I think that works also. [/hide]", "Solution_2": "[hide]3 ( r ^ 2 )[/hide]", "Solution_3": "[quote=\"shinwoo\"][hide]3 ( r ^ 2 )[/hide][/quote]\r\n\r\nThats not right. Remember you should always post how you got the solution as well, not just the final answer - then other people can tell you where you went wrong.", "Solution_4": "Oh I'm sorry, I made a mistake :blush: ...\r\n\r\nI will do it again..\r\n\r\n[hide] you first have to find out the regular triangle's height...\n\nthe height is :sqrt: -r/2+r^2 \n\n3(r * :sqrt: -r/2+r^2)\n\ndid I get it right?\n\nCould I simplyfy it little farther?\n[/hide]", "Solution_5": "I'm pretty sure that you have the height wrong shinwoo.", "Solution_6": "[hide]split it into 6 equilateral triangles. Area of an equilateral triangle is s^2*sqrt(3)/4 (use the 30-60-90 triangle created by dropping the altitude) then just do times 6 or 3s^2*sqrt(3)/2.[/hide]\r\n\r\nelvenchamp777", "Solution_7": "Please post in spoiler until everyone who wants to try the problem has already tried it. Thanks.", "Solution_8": "sry i forgot. :blush: i'll remember next time.\r\n\r\nelvenchamp", "Solution_9": "Ahhhhh!!!\r\n\r\nI hate me!!!", "Solution_10": "There's still another way if you want to try it.\r\n\r\n[hide=\"HINT\"]Try putting the hexagon in a triangle[/hide]", "Solution_11": "[quote]Try putting the hexagon in a triangle [/quote]\r\n\r\nThat's what I did\r\n\r\nand I had to find the height of the triangle in order to get the area, but I screwed up that part" } { "Tag": [ "Support", "MIT", "college", "Harvard", "Yale", "geometry", "geometric transformation" ], "Problem": "Hi folks, I am currently an undergraduate student. \r\n\r\nI have some queries pertaining to Spring admission as well as Financial Assistance in Spring Semester.\r\n\r\nQuery 1)\r\n\r\nHow many schools in the U.S. admit International students in the Spring Semester (for the Computer Science program, can anyone suggest some good/above average universities that admit international students in Spring semester)?\r\n\r\nQuery 2)\r\n\r\nI would also like to know how many schools ( notably those who admit students in the Spring semester) proffer financial assistance to International Students in Spring Semester? Can anyone suggest any such schools that offer financial assistance as well as an admit to an International student in Spring?\r\n\r\nQuery 3)\r\n\r\nHow much \"weightage\" do the standardized tests (GRE,TOEFL) have while applying for financial assistance?\r\n\r\nI currently don't have an extraordinary GPA, assuming If I manage to secure an outstanding GRE/TOEFL score, can I be guaranteed financial aid while applying for graduate program in Spring semester?\r\n\r\nI had already perused sites of some universities, but I would like to seek someone's advice (people in this forum).\r\n\r\n\r\nI ASK PEOPLE IN THIS FORUM TO SUGGEST SOME UNIVERSITIES THAT MEET THE ABOVE CRITERION.\r\n\r\nI thank you all in advance!", "Solution_1": "[quote=\"Gen8\"]\nQuery 1)\n\nHow many schools in the U.S. admit International students in the Spring Semester (for the Computer Science program, can anyone suggest some good/above average universities that admit international students in Spring semester)?\n[/quote]\n\n[url=http://grad-schools.usnews.rankingsandreviews.com/grad/com/search]Here[/url] is a (not necessarily definitive) list of the best graduate programs in CS. [url=http://www.google.com/]Here[/url] is a tool which will allow you to find the webpages of the graduate programs on that list and see whether they offer spring admission.\n\n[quote=\"Gen8\"]\nQuery 2)\n\nI would also like to know how many schools ( notably those who admit students in the Spring semester) proffer financial assistance to International Students in Spring Semester? Can anyone suggest any such schools that offer financial assistance as well as an admit to an International student in Spring?\n[/quote]\n\nUse the above tool again! The answer may be different depending on whether you are pursuing a Master's or a Ph.D.\n\n[quote=\"Gen8\"]\nQuery 3)\n\nHow much \"weightage\" do the standardized tests (GRE,TOEFL) have while applying for financial assistance?\n\nI currently don't have an extraordinary GPA, assuming If I manage to secure an outstanding GRE/TOEFL score, can I be guaranteed financial aid while applying for graduate program in Spring semester?\n[/quote]\r\n\r\nPretty much none. At a reputable program, if you are getting in, there should be some sort of financial package attached (some combination of TAship + RAship + stipend + tuition waivers). Some M.Sc programs won't offer financial support though. These will be given to everyone admitted and are not dependent on GRE or TOEFL, although the admissions decision itself may be. It's pretty absurd to think that you would get any sort of benefit from having a good TOEFL score though -- a bad one might disqualify you, but even a perfect one basically says that your english level may almost be as good as a native speaker, and I don't think that's a qualification that graduate programs get particularly excited over!", "Solution_2": "Hmm, Then I would like to know the importance of these standardized tests. I had earlier perused through the site of a university ( a pretty nice one at that!) which proffered a fellowship for a good GRE score/ a good GPA.\r\n\r\nI think securing a good GRE/TOEFL score does have some \"weightage\".\r\n\r\nWell, I was talking about some less putative schools (in relative to MIT, Harvard, Yale etc), which I feel proffer some financial assistance based on the GRE/TOEFL score.\r\n\r\nI think a TA is awarded based solely on the TOEFL score, ain't it?\r\n\r\n\r\nHow do you convert % into GPA? Is there any generalized formula?", "Solution_3": "As blahblahblah said, GRE (the general one) and TOEFL scores only matter in the sense that very poor scores might hurt your chances. Getting a score above some minimum to show that you are capable of communicating in English is all that's really needed, and anything more than that is nice but ultimately irrelevant to the admission decision. I'm not sure what you mean by TAships being awarded based 'solely on TOEFL' scores, but I don't believe that's the case (except that, again, if you score very poorly it might preclude you from getting a TAship because the committee might conclude that you will not be able to communicate with students). \r\n\r\nThe standardized test that [i]might[/i] be of somewhat more importance is the GRE subject test in Computer Science or Mathematics (some programs don't require you to take subject tests, but 'strongly encourage it'). Admission and fellowship committees do seem to put weight on these, though not as much as on the GPA. If your GPA is poor then scoring very well on the GRE subject test might help in making the case that for whatever reason your grades don't accurately reflect your ability to learn. Similarly, if your GRE subject test score is poor, it may nullify a solid GPA by creating the impression that grades notwithstanding you simply don't know the basics.\r\n\r\nPerhaps even more important than GPA, and almost certainly more important than any standardized test scores, are the recommendations that you get, and the interest you show in research. Ultimately admission committees are looking for self-motivated people who are going to succeed at research - not at getting grades, and not at scoring high on tests. If you can demonstrate a genuine interest in the field by engaging yourself in undergraduate research, you'll be doing far more to improve your chances of getting into the school you want than by cramming for GREs.", "Solution_4": "Okay, Then what are the criterion that are evaluated before granting a Teaching Assistantship?", "Solution_5": "[quote=\"Gen8\"]Okay, Then what are the criterion that are evaluated before granting a Teaching Assistantship?[/quote]\r\n\r\num, basically none at a lot of (most?) schools. some will require some sort of course to be taken before one can actually teach a class, but TAships are given to everyone at the schools I have gone to.", "Solution_6": "I wonder what schools you had attended! Then, on what basis are R.A.'s awarded? Can a person be awarded with a T.A. as well as an R.A. simultaneously? \r\n\r\nI had earlier perused through the websites of some schools in which they were some minimum thresholds regarding GRE and TOEFL score, GPA , so provided one has surpassed these thresholds is he/she likely to get admitted at that particular school or do they also look for something more?", "Solution_7": "[quote=\"Gen8\"]I wonder what schools you had attended! Then, on what basis are R.A.'s awarded? Can a person be awarded with a T.A. as well as an R.A. simultaneously? \n[/quote]\n\nRAs are generally paid out of individual faculty member's (usually your supervisor!) research grants, whereas TAs are paid out of the department's budget. At my school people who do not have external funding usually get a full TAship during September to April (12 hours/week) and then some sort of RAship from their supervisor during the summer.\n\n[quote=\"Gen8\"]I had earlier perused through the websites of some schools in which they were some minimum thresholds regarding GRE and TOEFL score, GPA , so provided one has surpassed these thresholds is he/she likely to get admitted at that particular school or do they also look for something more?[/quote]\r\n\r\nNo, in general surpassing the thresholds is not enough This is why they are called 'minimum thresholds' and not 'necessary and sufficient requirements' for admission. References are important, as are research experience and various similar things. This has already been said by others in this thread!", "Solution_8": "Hiii blahblah..., I attempted the GRE some time ago, since I was busy, I couldn't access this forum. I had perused some random posts relating to GRE, and also some posts pertaining to IIT-JEE, I observed some posts made by you in which you claimed (or could have) solved some IIT-JEE problems in minutes ( You said a particular intricate (that was someone else's opinion) JEE problem took you 2 minutes to solve!).\r\n\r\nHere's the link regarding the problem and your response. \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=42989&search_id=889566661&start=20\r\n\r\n\r\nSecondly, I also read a post of yours on GRE-Verbal in which you mention that-- the GRE Verbal ability section is extremely tough (I think you also mentioned that you managed to secure 730 (mostly through some guessing as you have said)).\r\n\r\nI think you are very adept at problem solving, then, I find it very weird that you found GRE Verbal as extremely tough, the GRE Verbal and JEE problems aren't comparable at all! \r\n\r\nSo, may I know, How can a person like you who had managed to solve JEE problems in minutes find the GRE-Verbal ability section as being extremely hard (The GRE Verbal is all about memorizing some words)?\r\n\r\n[b]I am not trying to offend you in any sense, I just wanted to know how come an adept problem solver like you find GRE Verbal extremely hard? [/b]\r\n\r\nBetween,I didn't manage a stellar score on the Verbal, I just got only a 700.\r\n\r\nThanks!", "Solution_9": "Well, if you are doing well on the verbal section of the GRE, the problems get harder very quickly, to the point where I think I ended up having to compare two words that I had never seen before. I am saying that the verbal section of the GRE is hard in the sense that it must be very difficult to score 800.\r\n\r\nIt's also a test in which problem-solving skills aren't that useful -- certainly you can use logic to increase your chances of getting the correct answer, but it mostly boils down to how many words you know and how many definitions you know for the words you know. Any mathematical problem-solving skills I may or may not have don't translate particularly well to that setting.\r\n\r\nIn any event any score in the 700s is quite good though -- I think that 690 and above are 99th percentile, or something similar. It's certainly not worth retaking.", "Solution_10": "Haha, You are smart! You have viewed my previous post (might be the case!) in which I mentioned my GRE score, and have been able to discern that I might be contemplating of re-taking the GRE (mostly in order to bolster my Quants score), that's pretty good!\r\n\r\nBut I would like to ask the graduate program(also the school) you are pursuing (That is if you like to tell, I observed that you are very proficient in Math) and I also realized a possible mistake of mine,in a certain PM I might have offended you through the usage of certain words that might not actually depict your personality in reality. But one should accept that every person makes (or is prone to making mistakes), it might have been the case that at that point of time I might have reacted in a strong manner (or might have been touchy)." } { "Tag": [], "Problem": "This is a simple game based on the prisoners' dilemma. \r\n\r\nObjective: You are a prisoner, among others, incarcerated by an evil master. To observe the spirit of democracy, the master holds a voting for the prisoners: each prisoner can either (a) vote to save himself and kill others or (b) vote to kill himself and save others. Then each prisoner will be killed or saved according to the voting (that is, a person is killed if more people vote to kill him than to save him and saved otherwise). Communication between prisoners are allowed. Your goal is simply survival. \r\n\r\nScoring (Cumulative): # survival / # total rounds played. \r\n\r\nRound 1: PM me your choice (a) or (b) by Friday, Aug. 7. The results will be posted afterwards.\r\n\r\nThe rules are subject to future adjustment.", "Solution_1": "Round 1 Results:\r\nSuperNerd123 (a) killed\r\nmaybach (a) killed\r\nzmesdr (a) killed\r\n\r\nCumulative Ranking\r\n1. \r\nSuperNerd123 0/1\r\nmaybach 0/1\r\nzmesdr 0/1\r\n\r\nRound 2: PM me your choice (a) or (b) by Wed, Aug. 12. The results will be posted afterwards.", "Solution_2": "NOTE: you are allowed to communicate with each other. Feel free to post in this thread for others to know your name.", "Solution_3": "i'm not in the game, but if everyone votes b, then won't everyone survive?", "Solution_4": "I don't get the point of voting b. You just forfeit points.", "Solution_5": "[quote=\"funtwo\"]i'm not in the game, but if everyone votes b, then won't everyone survive?[/quote]\r\n\r\nThe point is that not everybody will do that.\r\n\r\nIn the case of 2 players, this game is akin to the Prisoner's Dilemma.", "Solution_6": "[quote=\"maybach\"]I don't get the point of voting b. You just forfeit points.[/quote]\r\n\r\nUmm look at what happened the first round when no one voted b." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all composite numbers $n$ satisfying the following property:\r\nif $1=d_1 d_k$ and has to be one of the $d_i$'s) which means that $n$ is a perfect number. It is well-known that an even perfect number has form $n=2^{p-1}(2^p-1)$ where $2^p - 1$ is prime.\r\nSince $3$ divides $n$ it follows that $2^p-1 = 3$ so that $p=2$ and $n=6.$\r\n\r\nIt is easy to verify that $n=6$ is indeed a solution.\r\n\r\nPierre.", "Solution_2": "You can solve it without using that theorem.", "Solution_3": "Too late. I used it :P \r\n\r\nPierre." } { "Tag": [ "group theory", "abstract algebra", "search", "superior algebra", "superior algebra solved" ], "Problem": "G is a group of order 3825, Prove that if H is a normal subgroup of order 17 in G then H$\\leq Z(G)$", "Solution_1": "I think it has already been discussed (in the general case).. try to search .." } { "Tag": [ "geometry", "circumcircle", "inequalities" ], "Problem": "Show that in any triangle we have:\r\n\r\n(b^2+c^2)\\m_a+(c^2+a^2)\\m_b+(a^2+b^2)\\m_c <=12R.", "Solution_1": "what is m_a, m_b, m_c, and R?", "Solution_2": "I'd guess that m_a, m_b, and m_c are the medians to sides a, b, and c respectively, and R is the circumradius.", "Solution_3": "I'll just [tex]\\TeX[/tex] this thing real quick:\r\n\r\n[tex]\\displaystyle \\mbox{Prove that}[/tex]\r\n\r\n[tex]\\displaystyle \\frac{a^2+b^2}{m_c}+\\frac{b^2+c^2}{m_a}+\\frac{c^2+a^2}{m_b} \\le 12R[/tex]\r\n\r\nwhere [tex]m_a[/tex], [tex]m_b[/tex], and [tex]m_c[/tex] are the medians to sides a, b, and c, respectively, and R is the circumradius.", "Solution_4": "[hide]\n\nLet F be the midpoint of AB and X be the point on the circumcircle such that the points C, F and X are colinear. Then [tex]CF\\cdot FX = AF\\cdot FB[/tex], hence [tex]FX=\\frac{c^2}{4m_c}[/tex]. We have [tex]a^2(\\frac{1}{2}c) + b^2(\\frac{1}{2}c) = c(m_c^2 + (\\frac{1}{2}c)^2)[/tex] by Stewart's Theorem. So [tex]a^2+b^2 = 2m_c^2 + \\frac{1}{2}c^2[/tex]. Then [tex]\\frac{a^2+b^2}{m_c} = 2m_c + \\frac{c^2}{2m_c}= 2(m_c + \\frac{c^2}{4m_c}) = 2(CF + FX) = 2 AX \\leq 4R[/tex]. Similarly, [tex]\\frac{b^2+c^2}{m_a} \\leq 4R[/tex] and [tex]\\frac{c^2+a^2}{m_b} \\leq 4R[/tex]. Adding the three inequalities together to get the required result.\n\nQ.E.D\n\n[/hide]" } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "The space can not be represented as a finite reunion of planes. (search didn't give anything)", "Solution_1": "Actually, grobber has proved [url=http://www.mathlinks.ro/viewtopic.php?p=624250#624250]here[/url] a generalization.\r\nIf one's not comfortable with the language, grobber suggests picking a plane $\\pi$ that is different from all the planes that presumably cover the whole space. Now, each plane intersects $\\pi$ along a line (either that, or it doesn't intersect it all). Hence, we have finitely many lines that cover the plane and we can use the same procedure to prove that this isn't possible." } { "Tag": [], "Problem": "An object falls freely from rest near the surface of Earth. What is the speed of the object after having fallen a distance at 4.90 meters?\r\n\r\n(1) 4.90 m/s (3) 24.0 m/s\r\n(2) 9.80 m/s (4) 96.1 m/s", "Solution_1": "[hide]\nFrom the formula $v_{f}^{2}=v_{0}^{2}+2ad$, we find that $v_{f}=9.8\\,\\,\\text{m/s}$.[/hide]", "Solution_2": "I do not understand the formula provided.\r\n\r\nCan you break it down piece by piece?\r\n\r\nThanks!", "Solution_3": "If you consider distance as the under a velocity graph, we have $d=v_{0}t+gt^{2}/2$, where $g=9.8\\ m/s^{2},\\ v_{0}=0$. So $t=1$. After 1 second, the velocity is $1(9.8)\\ m/s =9.8\\ m/s$", "Solution_4": "I can see this to be a physics question, right?", "Solution_5": "[quote=\"Interval\"]I do not understand the formula provided.\n\nCan you break it down piece by piece?\n\nThanks![/quote]\r\nIt is easily derived using some basic physics.", "Solution_6": "[quote=\"rnwang2\"][quote=\"Interval\"]I do not understand the formula provided.\n\nCan you break it down piece by piece?\n\nThanks![/quote]\nIt is easily derived using some basic physics.[/quote]\r\nBasic math. :P", "Solution_7": "Starting with the equation\r\n\\[d = vt, \\]\r\nwhere $v$ is the average velocity of the object. If the object undergoes constant acceleration, the average velocity is then equal to the average of $v_{i}$ and $v_{f}$, the initial and final velocities. So the above equation can be rewritten as \r\n\\begin{eqnarray*}d &=& \\left (\\frac{v_{i}+v_{f}}{2}\\right) t\\\\ &=& \\left (\\frac{v_{i}+(v_{i}+at)}{2}\\right) t\\\\ &=& v_{i}t+\\frac{at^{2}}{2}\\end{eqnarray*}\r\nSquaring both sides of the equation $v_{f}= v_{i}+at$,\r\n\\begin{eqnarray*}v_{f}^{2}&=& v_{i}^{2}+2v_{i}at+a^{2}t^{2}\\\\ &=& v_{i}^{2}+2a\\left(v_{i}t+\\frac{at^{2}}{2}\\right)\\\\ &=& \\boxed{v_{i}^{2}+2ad}\\end{eqnarray*}", "Solution_8": "Guys, you complicate things too much...\r\nStart with the law of energy conservation.\r\n$E_{gp}=E_{k}$ (gravitational energy transforms into kinetical)\r\n$mg\\Delta h=\\frac{mv^{2}}{2}$\r\n$v=\\sqrt{2g\\Delta h}$\r\nPlug in $\\Delta h=4.9\\ m \\Longrightarrow v=9.8\\ m/s$" } { "Tag": [ "linear algebra", "matrix", "inequalities" ], "Problem": "I was just wondering:\r\n\r\nif we denot R(A) be the reduced echelon form of the matrix A is it tru that:\r\n\r\nR(A*B) = R(A)*R(B)\r\n\r\nLets Add one more:\r\n\r\nlets say that A = B*C where A = m by n, B = m by w, and C = w by n (w is the least possible) than this holds true:\r\n \r\nRank(A) = Rank(B) * Rank(C)\r\n\r\nThanx!", "Solution_1": "Could anyone please help me with the second problem!", "Solution_2": "Isn't it obvious? Multiplying ranks can't possibly be right. We have the inequalities\r\n$\\text{rank }B+\\text{rank }C-w\\le\\text{rank }A\\le\\min\\{\\text{rank }B,\\text{rank }C\\}$,\r\nso we can only have $\\text{rank }A=\\text{rank }B\\cdot\\text{rank }C$ when one of $B,C$ is zero or when $\\text{rank }B=\\text{rank }C=\\text{rank }A=1$.\r\n\r\nOn the echelon form:\r\n\r\nTry $A=\\begin{bmatrix}1&1\\\\1&1\\end{bmatrix}$ and $B=\\begin{bmatrix}1&-1\\\\-1&1\\end{bmatrix}$.", "Solution_3": "the matrix A that u gave will be decomposed into [1]\r\n [1] * [1 1] = [1 1]\r\n [1 1] and thus the problem holds. I think there is a mistake in what u wrote. Thanx for the help though!", "Solution_4": "I am sorry i think that u are right!\r\nI found that:\r\nRank(A) + Rank(B) >= Rank(A*B) (I think it is Sylvester Inequality)\r\nwell than we have :L\r\nRank(A) + Rank(B) >= Rank(A*B) = Rank(A)*Rank(B) or\r\nRank A =Rank B = 1,2. \r\nAnd each case can be solved separately!" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ n\\in \\mathbb{N}$ and $ n\\geq 3$.Prove or disprove that between numbers of following secuences \r\n\r\n$ 2n \\minus{} 1,2n,2n \\plus{} 1,...,3n \\minus{} 2$\r\n\r\n$ 2n,2n \\plus{} 1,2n \\plus{} 2,...,3n \\minus{} 1$\r\n\r\n$ 2n \\plus{} 1,2n \\plus{} 2,2n \\plus{} 3 ,...,3n$ \r\n\r\n $ .$\r\n $ .$\r\n $ .$\r\n\r\nThere exist a perfect square in each secuence.", "Solution_1": "[quote=\"enndb0x\"]Let $ n\\in \\mathbb{N}$ and $ n\\geq 3$.Prove or disprove that between numbers of following secuences \n\n$ 2n \\minus{} 1,2n,2n \\plus{} 1,...,3n \\minus{} 2$\n\n$ 2n,2n \\plus{} 1,2n \\plus{} 2,...,3n \\minus{} 1$\n\n$ 2n \\plus{} 1,2n \\plus{} 2,2n \\plus{} 3 ,...,3n$ \n\n $ .$\n $ .$\n $ .$\n\nThere exist a perfect square in each secuence.[/quote]\r\n\r\nYou want to prove or disprove if, given $ n\\in \\mathbb{N}$ and $ n\\geq 3$, exists a perfect square in $ [2n \\plus{} k \\minus{} 1,3n \\plus{} k \\minus{} 2]$ $ \\forall k\\geq 0$\r\n\r\nTake for instance $ k \\equal{} n^2 \\minus{} 2n \\plus{} 2$. We have $ n^2 < n^2 \\plus{} 1 \\equal{} 2n \\plus{} k \\minus{} 1 < 3n \\plus{} k \\minus{} 2 \\equal{} n^2 \\plus{} n < (n \\plus{} 1)^2$ and so no perfect square in $ [2n \\plus{} k \\minus{} 1,3n \\plus{} k \\minus{} 2]$\r\n\r\n(and, btw, quicker : this is wrong for $ n\\equal{}3$ : the first sequence $ (5,6,7)$ contains no perfect square)" } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "I know this may seem annoying, but can any one direct me to the thread which discussed the evaluation of this integral (or something similar, I may not be remembering exactly):\r\n\r\n$ \\displaystyle\\int_0^1 \\ln{(1\\minus{}x)}\\ln{x} dx$\r\n\r\nI recall that the solution involved converting one of the ln's into a power series, so maybe my rough memory of this integral is flawed. Any direction will be appreciated, thank you!", "Solution_1": "Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106770]here[/url]. It's always a good practice to keep track of your own posts.", "Solution_2": "Or look [url=http://www.mathlinks.ro/viewtopic.php?t=191806]here[/url]." } { "Tag": [ "AMC", "AIME", "\\/closed" ], "Problem": "We will be doing a Math Jam as soon as we can after each AMC test. Unfortunately, the day after the AMC B coincides with one of our courses. Which of the following times would you like to see us hold the AMC B review Math Jam?", "Solution_1": "The day after! I will die if I don't know the definitive answers =D.", "Solution_2": "are teachers not allowed to let you see the solutions manual or tell you the answers the day of the test? I thought they were.", "Solution_3": "I am not positive, but I believe that a set of solutions is available to your teacher after the AMC. I am not sure this is true for the AIME.\r\n\r\nWe hope we can add to the process by explaining how to think along the lines that will produce the solution.", "Solution_4": "I think you have to pay for solutions for the AIME, whereas you get them automatically with the AMCs.", "Solution_5": "We've opted for the Saturday slot. The last one went 3 hours, and we can't ask folks on the east coast to stay up til midnight on Thursday...", "Solution_6": "Is it possible that someone could post the answers thursday like last time on the board?" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Let G = (V,E) be a graph. Then the line graph L(G) of G has the edges of G as the vertex, and two edges are joined by a line if they are incident with a common vertex of G.\r\n\r\nProve that a connected graph is isomorphic to its line graph iff it is a cycle.\r\nThen determine the triangles (3-cycles) in L(G) in terms of the number of triangles of G and the degrees of the vertices.", "Solution_1": "[quote=\"cooley\"]Prove that a connected graph is isomorphic to its line graph iff it is a cycle.[/quote]\n\nIf $G=L(G)$,\n\tIf $G$ is not a cycle,\n\t\tConsider the largest cycle $C$ of length $x$ in $G$\n\t\tSince $G$ is connected and is not a cycle, there exists a vertex $V$ on $C$ that has degree $\\geq 3$\n\t\tThe edges of $C$ form the vertices of a cycle $C'$ on length $x$ in $L(G)$\n\t\tThe $2$ edges of $V$ on $C$ and $1$ other edge form the vertices of a triangle $T'$ in $L(G)$\n\t\tBut $C'$ and $T'$ share an edge\n\t\tSo $XOR(C',T')$ is a cycle in $L(G)$ with length $x+1$ $\\Rightarrow\\Leftarrow$\n\tThus $G$ is a cycle\nIf $G$ is a cycle,\n\t$L(G)$ is obviously a cycle\nTherefore $G=L(G)$\n\n\n[quote]Then determine the triangles (3-cycles) in L(G) in terms of the number of triangles of G and the degrees of the vertices.[/quote]\r\n\r\nFor any triangle $X'Y'Z'$ in $L(G)$ corresponding to edges $X,Y,Z$ in $G$,\r\n\tEach pair $(X,Y)$, $(Y,Z)$, $(Z,X)$ shares a vertex\r\n\tSo either $X,Y,Z$ share a vertex or $X,Y,Z$ are the edges of a triangle\r\nFor any vertex $V$ in $G$ with degree $d$,\r\n\tThere are $\\binom{d}{3}$ unordered triples of edges with common vertex $V$\r\nLet $t(G)$ be the number of triangles in $G$\r\nTherefore the $t(L(G))$ is\r\n\t$t(G)+\\sum{\\binom{d_{i}}{3}}$", "Solution_2": "Thanks for the reply. Actually, i knew the answer of the second question by trial and error, but could not have managed to prove it.Now it is clear by your explanaton. Also, after asking the first question, i noticed that it can be proved by getting use of the automorphism. Anyway, your help is appreciated." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $ S_N \\equal{} \\{X \\equal{} (n_1,n_2,...,n_k)|\\sum_i n_i \\equal{} N, n_1\\ge n_2\\ge ...\\ge n_k\\ge 1\\}$ set of Jung's diagram.\r\nLet $ F: S_N\\to S_n$ defined by for $ X \\equal{} (n_1\\ge n_2\\ge ...n_{l \\minus{} 1} > n_l\\ge ... n_m > n_{m \\plus{} 1} \\equal{} .. \\equal{} n_k \\equal{} 1)$, were $ n_{l \\minus{} 1} > k,n_l\\le k,n_m\\ge 2, n_{m \\plus{} 1} \\equal{} 1$ \r\n$ F(X) \\equal{} (n_1 \\minus{} 1,n_2 \\minus{} 1,...,n_{l \\minus{} 1},k,n_{l} \\minus{} 1,...,n_m \\minus{} 1).$ (all term $ n_i$ decrease $ n_i \\to n_i \\minus{} 1$ and add new term k)\r\nLet $ h(N) \\equal{} \\minus{} [\\frac {1 \\minus{} \\sqrt {8N \\plus{} 1}}{2}], t(N) \\equal{} \\frac {h(N)(h(N) \\plus{} 1)}{2} \\minus{} N.$\r\nLet $ F^{(n)}$ is n-th iteration of F ($ F^{(1)} \\equal{} F, F^{(n \\plus{} 1)} \\equal{} F(F^{(n)})$).\r\nDiagram X is in attractor if $ F^{(h(N))}(X) \\equal{} X$. Let $ L(X)$ is minimal distance to attractor $ F^{(h(N) \\plus{} L(N))}(X) \\equal{} F^{(L(N))}(X)$ but $ F^{(h(N) \\plus{} L(N) \\minus{} 1)}(X)\\not \\equal{} F^{(L(N) \\minus{} 1)}(X).$\r\n1. Prove: if $ X_1 \\equal{} (1,1,...,1)$ - (N times 1), then $ L(X_1) \\equal{} N \\plus{} 1 \\minus{} h(N).$ If $ N \\equal{} [(m^2 \\minus{} 1)/2],m \\equal{} h(N)$, then for any $ X\\in S_N$ $ L(X)\\le L(X_1)$.\r\n2. Prove that $ L(X)\\le h(N)(h(N) \\minus{} 1) \\ \\forall X\\in S_N$ and if $ N \\equal{} \\frac {m(m \\plus{} 1)}{2},m \\equal{} h(N), X_0 \\equal{} (m \\minus{} 1,m \\minus{} 1,m \\minus{} 2,m \\minus{} 3,...,2,1,1)$ \r\nthen $ L(X_0) \\equal{} h(N)(h(N) \\minus{} 1).$", "Solution_1": "What does $ n_i\\t0 n_i - 1$ mean?\r\n\r\nBesides, \"Jung\" should be \"Young\".\r\n\r\n Darij" } { "Tag": [ "geometry", "geometric transformation", "number theory" ], "Problem": "Hi. I'm searching for a good number theory book for intermediate solvers. If anybody has one please tell me. :|", "Solution_1": "[quote=\"\ufffd \u03b1\u03bd\u03b1\u03b3\u03b9\u03ce\u03c4\u03b7\u03c21\"]Hi. I'm searching for a good number theory book for intermediate solvers. If anybody has one please tell me. :|[/quote]\r\n\r\nuse engel or 103 problems in Number Theory by andreescu :)", "Solution_2": "Forget about Engel as concerns Number Theory.\r\nBut I can't say much until I know what is meant by \"intermediate solver\".\r\n\r\nDavid M. Burton, [i]Elementary Number Theory[/i], 1980\r\n\r\nis suitable for different levels, but I don't know whether it is freely avaliable online (there was a time when its German translation was online as PDF; I can send it to you if you wish).\r\n\r\n darij" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let X be a asset of 1+2^k points in the Euclidean plane. Prove there exist 3 points in X making a triangle whose largest angle has size at least 180*(1-1/k)", "Solution_1": "By Euclidean plane, what do you mean? An ordinary plane? :maybe:" } { "Tag": [], "Problem": "There is a new math symbol invented, it looks like this @\r\n\r\n2@=6\r\n3@=12\r\n5@=30\r\n7@=56\r\nwhat is 34@?", "Solution_1": "[hide=\"my answer\"]1190[/hide]", "Solution_2": "nice job gr8sk8r, you might be good enough to take a shot at the Relay race(click on my signature)", "Solution_3": "[hide]Yeah, I got that (1190) too! :) [/hide]", "Solution_4": "[hide]1190!! yay!![/hide]", "Solution_5": "in general,\r\n\r\nn@ = n(n+1)\r\nso 34@ = 34(35) or [color=cyan]1190[/color].", "Solution_6": "Yeah, pretty obvious.", "Solution_7": "[hide]so when a number multiplies by @, it is basically N(N+1)\nso 34*35=1190[/hide]", "Solution_8": "[hide]1190[/hide]", "Solution_9": "[hide]\n[tex]34@ = 34 \\times 35 = 1190[/tex]\n[/hide]", "Solution_10": "The pattern is, n@=n(n+1)\r\n\r\nSo 34@=34(34+1)=34*35=1190\r\n\r\n :lol:", "Solution_11": "[hide]Let x = any real number\n\nx@ = x(x+1)\n\nSo, 34@ = 34(35)\n\n(need a calculator for this :-D)\n\n[b]1190. There's your answer.[/b][/hide]", "Solution_12": "wow, i didn't think that many people would get it. Okay there is another new math symbol that looks like this: #\r\nif 5#=6\r\n 4#=4\r\n 3#=2\r\n 10#=16\r\nwhat is 32#", "Solution_13": "[quote=\"math92\"]wow, i didn't think that many people would get it. Okay there is another new math symbol that looks like this: #\nif 5#=6\n 4#=4\n 3#=2\n 10#=16\nwhat is 32#[/quote]\r\n\r\n[hide] $x$#$=2x-4$\n\n$32$#$=64-4=\\boxed{60}$[/hide]", "Solution_14": "*New*\r\n\r\nUse math92 @ symbol:\r\nIf I have n@@=5256, what is n?\r\n\r\nUse math92 # symbol:\r\nIf I have n###=2004, what is n?", "Solution_15": "[quote=\"10000th User\"]*New*\n\nUse math92 @ symbol:\nIf I have n@@=5256, what is n?\n\nUse math92 # symbol:\nIf I have n###=2004, what is n?[/quote]\r\n\r\n I dont quite understand, could you explain a little more thouroughly.", "Solution_16": "[hide]the first n=8\nthe second n=254[/hide]", "Solution_17": "[quote=\"G-UNIT\"][quote=\"10000th User\"]*New*\n\nUse math92 @ symbol:\nIf I have n@@=5256, what is n?\n\nUse math92 # symbol:\nIf I have n###=2004, what is n?[/quote]\n\n I dont quite understand, could you explain a little more thouroughly.[/quote]\r\nthat's strange to see that it is easy to proceed forward but then it is harder to understand when doing things backwards. for example, finding 3@@ is easy because 3@=12 and 3@@=12@=156, but finding n when n@@=156 is not understandable...", "Solution_18": "Let's check gunes's solutions, \r\n\r\nFirst one: does 8@@=5256?\r\n8@=8*9=72\r\n72@=72*73=5256. It checks. \r\n\r\nsecond one: does 254###=2004?\r\n254#=2(254)-4=504.\r\n504#=2(504)-4=1004.\r\n1004#=2(1004)-4=2004. It checks. \r\n\r\nWell done.", "Solution_19": "[hide]\n34*35=(30+4)(30+5)=900+150+120+20=1190\n\n[/hide]", "Solution_20": "[quote=\"math92\"]wow, i didn't think that many people would get it. Okay there is another new math symbol that looks like this: #\nif 5#=6\n 4#=4\n 3#=2\n 10#=16\nwhat is 32#[/quote]\r\n\r\n[hide]n#=2n-4\n\nSo 32#=2*32-4\n=60[/hide]", "Solution_21": "[quote=\"10000th User\"]*New*\n\nUse math92 @ symbol:\nIf I have n@@=5256, what is n?\n\nUse math92 # symbol:\nIf I have n###=2004, what is n?[/quote][hide=\"the first one\"]i used guess and check and got [b]8[/b] as my answer :lol: [/hide]", "Solution_22": "[quote=\"math92\"][quote=\"10000th User\"]*New*\n\nUse math92 @ symbol:\nIf I have n@@=5256, what is n?\n\nUse math92 # symbol:\nIf I have n###=2004, what is n?[/quote][hide=\"the first one\"]i used guess and check and got [b]8[/b] as my answer :lol: [/hide][/quote]\r\n :thumbup: \r\nNot too hard eh?\r\nI did this way: n@@=n@(n@+1)=5256 and find n@ is about 72.4. How to do it? OK see my square root calculate\r\n[code] 72.\n5256\n49\n 356 14_x_ missing digit is 2 (3 is too large)\n 284\n 7200 144_x_ missing digit is 4 (5 is a little too big)\n-----not important------[/code]\r\nSo n@ is about 72.4. if you check 72x73=72x70+72x3=5040+216=5256\r\nNow n@=n(n+1)=72 and you can check easy n=8\r\n\r\nmath92 please solve second one :D", "Solution_23": "[quote=\"10000th User\"]\nmath92 please solve second one :D[/quote]\r\ni'm working on it...", "Solution_24": "[quote=\"math92\"][quote=\"10000th User\"]\nmath92 please solve second one :D[/quote]\ni'm working on it...[/quote]\r\nHi math92\r\n[hide=\"hint\"]OK equation is n###=2004. You know # do this: double number and subtract 4. Now do 'backward' way. So add 4 and then divide by 2... this helps? ;)[/hide]", "Solution_25": "[hide]thanks alot for the hint. I got 254 as my answer. Is that correct?[/hide]", "Solution_26": "[quote=\"math92\"][hide]thanks alot for the hint. I got 254 as my answer. Is that correct?[/quote]\r\n :yup: :clap: :10: :flex: :jump: :thumbup: :wow:", "Solution_27": "whoa, do you do that everytime someone answers a problem? :P", "Solution_28": "[quote=\"math92\"]whoa, do you do that everytime someone answers a problem? :P[/quote]\r\nOnly for you because you are the most special middle school student here ;)", "Solution_29": "$n^2 + n$ is equal to $n(n+1)$", "Solution_30": "[hide]\n\nI started out by trying to get a pattern: 2@=6, 2*3=6; 3@=12, 3*4=12; the pattern is @=n+1.\n\nSo 34@ = 34*35 = $ \\framebox{1190} $[/hide]", "Solution_31": "[hide]\n\nI looked for a pattern and I got 2n - 4 is #.\n\nSo $ 32\\# = \\framebox{60} $[/hide]" } { "Tag": [ "function", "inequalities" ], "Problem": "let f:R->R be a functionwith the properties:\r\n1. f(x)f(x)<=1 x for any real\r\n2. f'(x)f'(x)+f\"(x)f\"(x)<=1 for any x real\r\n\r\nShow that f(x)f(x)+f'(x)f'(x)<=1 for any x real.", "Solution_1": "it was at procopiu contest at 12`th grade ? :)", "Solution_2": "Yes, didn't go to the contest because i don't know any phisics :D , could you give me an idea, please?", "Solution_3": "How come we can't have a point $x$ such that $f(x)=.9$, $f'(x)=.9$, and $f''(x)=-.1$? \r\n\r\n$f(x)f(x)=.81\\leq1$, $f'(x)f'(x)+f''(x)f''(x)=.82\\leq1$, but $f(x)f(x)+f'(x)f'(x)=1.62\\nleq 1$.\r\n\r\nOr am I misreading the question? I assumed by $f(x)f(x)$ you mean $f(x)$ times $f(x)$." } { "Tag": [], "Problem": "let A=(1,2,...,99) be a set.50 number are chosen from A,inwich the sum of each two number isnt equal to 99 or 100.\r\n\r\nprove that:\r\n\r\n the 50 chosen number should be : 50,51,...,99", "Solution_1": "[quote=\"ashegh\"]let A=(1,2,...,99) be a set.50 number are chosen from A,inwich the sum of each two number isnt equal to 99 or 100.\n\nprove that:\n\n the 50 chosen number should be : 50,51,...,99[/quote]\r\ni think if we suppose that the numbers are not 50,51,...,99 $\\rightarrow$ contradiction. :D", "Solution_2": "its a good fact...\r\n\r\nbut i think it doesnt help us too... :(", "Solution_3": "[hide]Suppose we choose an element $k$ ($k\\ne 99$) from $A$. We know that $99-k$ and $100-k$ cannot be chosen also. Therefore, the elements in $A$ can be paired up as: $(1,98), (2,97),\\ldots, (49,50), (99).$ We can only take one element from each set of parentheses. Thus, we must choose the element $99$ in order to end up with 50 numbers. But then we cannot choose 1 or else 99+1=100. Thus we have to choose 98. Similarily, we can't have 2. So we must take 97. This logic continues all the way down thus forcing us to choose $50,51,\\ldots 99.$[/hide]", "Solution_4": "nice solution joml88. :D" } { "Tag": [ "quadratics", "calculus", "calculus computations" ], "Problem": "\\[ x\\left( {\\frac{{dy}}\r\n{{dx}}} \\right)^2 \\plus{} (y \\minus{} x)\\frac{{dy}}\r\n{{dx}} \\minus{} y \\equal{} 0\\]", "Solution_1": "[quote=\"Xaenir\"]\n\\[ x\\left( {\\frac {{dy}} {{dx}}} \\right)^2 \\plus{} (y \\minus{} x)\\frac {{dy}} {{dx}} \\minus{} y \\equal{} 0\\]\n[/quote]\r\n\r\nIt is a quadratic in $ \\frac{dy}{dx}$\r\n\r\n$ \\Longrightarrow\\frac{dy}{dx}\\equal{}\\frac{x\\minus{}y\\pm\\sqrt{(y\\minus{}x)^2\\plus{}4xy}}{2}$\r\n\r\n $ \\frac{dy}{dx}\\equal{}\\frac{x\\minus{}y\\pm(x\\plus{}y)}{2}$\r\n \r\n $ \\frac{dy}{dx}\\equal{}x,\\minus{}y$\r\n \r\n $ y\\equal{}\\frac{x^2}{2}\\plus{}C$ and $ y\\equal{}e^{\\minus{}(x\\plus{}c)}$" } { "Tag": [ "inequalities", "trigonometry", "geometry", "inequalities open" ], "Problem": "In a triangle ABC prove that\r\n\r\n1+4CosACosBCosC >=2(CosACosB +CosBCosC +CosCCosA)", "Solution_1": "[quote=\"Michael Niland\"]In a triangle ABC prove that\n\n1+4CosACosBCosC >=2(CosACosB +CosBCosC +CosCCosA)[/quote]\r\n\r\nThis is equivalent to:\r\n\r\n$(\\cos A+\\cos B)^2 + (\\cos B + \\cos C)^2 + (\\cos C + \\cos A)^2 \\leq 3$\r\n\r\nFor acute triangle, it was proved at:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=14567\r\n\r\nThis is wrong for obtuse triangles.", "Solution_2": "Yes Fuzzylogic, you are right.\r\nThis inequality is only true for acute triangle and right triangle. If you rewrite it in the form :\r\n$ (cosA+cosB+cosC)^2 \\leq\\ ((sinA)^2+(sinB)^2+(sinC)^2) $\r\nIt is Walker's inequality.\r\n\r\nWalker's inequality has a nice form :\r\nIf H is the orthocenter of the non-obtuse triangle ABC then :\r\n$ (HA+HB+HC)^2 \\leq\\ AB^2+BC^2+CA^2 $\r\n\r\nTreegoner has a general inequality :\r\nFor non-obtuse triangle ABC and m,n,p be real number \r\n$(mcosA+ncosB+pcosC)^2 \\leq\\ (M(sinA)^2+N(sinB)^2+P(sinC)^2) $\r\n\r\n$M=p^2+n^2-m^2 $\r\n$N=m^2+p^2-n^2 $\r\n$P=n^2+m^2-p^2 $\r\n\r\nI remember that Treegoner also posted his theorem and proof on mathlinks recently." } { "Tag": [ "number theory", "prime factorization", "number theory unsolved" ], "Problem": "A positive integer $ m$ is called [i]good[/i] if there is a positive integer $ n$ such that $ m$ is the quotient of $ n$ by the number of positive integer divisors of $ n$ (including $ 1$ and $ n$ itself). Prove that $ 1, 2, \\ldots, 17$ are good numbers and that $ 18$ is not a good number.", "Solution_1": "[hide=\"Solution\"]If $ m$ is an odd prime, then we can let $ n \\equal{} 8m$.\nIf $ m \\equal{} 1$, $ n \\equal{} 1$\nIf $ m \\equal{} 2$, $ n \\equal{} 8$\nIf $ m \\equal{} 4$, $ n \\equal{} 36$\nIf $ m \\equal{} 6$, $ n \\equal{} 72$\nIf $ m \\equal{} 8$, $ n \\equal{} 80$\nIf $ m \\equal{} 9$, $ n \\equal{} 108$\nIf $ m \\equal{} 10$, $ n \\equal{} 180$\nIf $ m \\equal{} 12$, $ n \\equal{} 240$\nIf $ m \\equal{} 16$, $ n \\equal{} 128$\nLet the prime factorization of $ n$ be $ 2^{a_1}3^{a_2}5^{a_3}...p_n^{a_n}$\nNote that for any prime $ p$, $ \\frac {p^x}{x \\plus{} 1}$ is minimized when $ x \\equal{} 1$. So if $ m \\equal{} 18 \\equal{} 2 \\cdot 3^2$, then since $ 18|n$, $ n$ would have to be at least $ \\frac {2^1 \\cdot 3^2}{2 \\cdot 3} \\equal{} 3$. Thus, no prime $ k$ greater than $ 11$ can be in the prime factorization of $ n$, or we would have to multiply this by at least $ \\frac {k}{2}$ which would make the result greater than $ 18$. If $ 7$ or $ 11$ was in the prime factorization of $ n$, then one of the exponents on the other primes would have to be a multiple of $ 6$ or $ 10$, making the quotient too great. Thus, the quotient is of the form\n$ \\frac {2^{a_1}3^{a_2}5^{a_3}}{(a_1 \\plus{} 1)(a_2 \\plus{} 1)(a_3 \\plus{} 1)}$\nWe easily see $ a_3$ is either $ 0$ or $ 1$, or the fraction is too large. If it is $ 1$, then one of the other exponents must be $ 4$, and we can easily check that there is no solution. Thus, the quotient is of the form\n$ \\frac {2^{a_1}3^{a_2}}{(a_1 \\plus{} 1)(a_2 \\plus{} 1)}$.\nSince $ a_2$ is at least $ 2$, and also odd, it must be $ 3$, as $ 5$ would make the fraction too large. But we can easily check there is no solution in this case Thus, $ 18$ is not a good number.[/hide]\r\nBut I can't attain $ 14$ or $ 15$. Can anyone show how to get $ 14$ or $ 15$?", "Solution_2": "[quote=\"dgreenb801\"] But I can't attain $ 14$ or $ 15$. Can anyone show how to get $ 14$ or $ 15$?[/quote]\r\n\r\n$ 14\\equal{}\\frac{252}{18}\\equal{}\\frac{2^2}3\\frac{3^2}3\\frac 72$\r\n\r\n$ 15\\equal{}\\frac{360}{24}\\equal{}\\frac{2^3}{4}\\frac{3^2}3\\frac 52$" } { "Tag": [], "Problem": "20% of the male students at Carnage Middle School buy lunch in the cafeteria. The number of female students who buy lunch is 80% of the number of male students who buy lunch, but there are 20% more female students at Carnage than males. If 48 female students buy lunch, how many students attend Carnage Middle School?", "Solution_1": "[hide=\"Strange name for a school.\"]\nYou know that 48 females buy lunch. You also know that 48 is 80% of the number of male students who buy lunch, so there are 60 lunch-buying males.\n\n60 is 20% of the total males; therefore, there are 300 males attending the school.\n\nIf there are 20% more female students than male students, then there are 360 female students.\n\nThus, there are $ \\boxed{660 \\text{ students}}$ attending Carnage Middle School.[/hide]" } { "Tag": [ "modular arithmetic" ], "Problem": "Prove that for any prime p we can find an integer n such that n8 = 16 (mod p).", "Solution_1": "we must have $p|n^{8}-16=(n^{2}-2)(n^{2}+2)(n^{4}+4)$\r\nnow we know that if $p=8k+1,8k+7$ then there exist $a$ such that $a^{2}\\equiv2$ (mod p) , and if $p=8k+3,8k+1$ then there exist $a$\r\nsuch that $a^{2}\\equiv-2$ (mod p) . now we must prove the problem for $p=8k+5$\r\n$n^{4}+4=n^{4}+4n^{2}+4-4n^{2}=(n^{2}+2n+2)(n^{2}-2n+2)$ , now we must have $p|(n^{2}+2n+2)(n^{2}-2n+2)=[(n+1)^{2}+1][(n-1)^{2}+1]$\r\nnow we now that if $p=4k+1$ then there exist $a$ such that $a^{2}\\equiv-1$ (mod p) . now put $a=n+1$ and proof is complete[because $p=8k+5=4(2k+1)+1$]\r\n:)\r\nis my solution true ?", "Solution_2": "[quote=\"khashi70\"]is my solution true ?[/quote]\r\nWhy do You ask? You know it is true. \r\nFor instance to get decidibility of cogruence $x^{2}\\equiv-2 \\pmod{p}$ we have to culculate Legandre simbol\r\n\r\n$( \\frac{-2}{p})= (-1)^{\\frac{p-1}{2}+\\frac{p^{2}-1}{8}}=1 \\iff \\frac{(p-1)(p+5)}{8}$ is even $\\iff p\\equiv 1 \\pmod{8}\\vee p\\equiv 3 \\pmod{8}$.\r\n\r\nThe only missing case is $p=2$ which is trivial." } { "Tag": [ "inequalities", "floor function" ], "Problem": "$ a>1$ is not an integer.\r\n\r\ndefine $ f(n)\\equal{}\\frac{1}{n}\\lfloor{na}\\rfloor$.\r\n\r\nProve or disprove that $ f(n)>f(n\\plus{}1)$ for infinitely many $ n$.", "Solution_1": "in other topic it was written it should be known - could anyone give a link or a hint how to solve this problem ?\r\n\r\nmaybe it's not a good problem for \"Pre-Olympiad\" section?\r\nif so, please move it", "Solution_2": "[hide=\"Hint.\"]\n=====\n$ n\\equal{}\\frac{k}{a}$, $ k\\in \\mathbb{Z}_\\plus{}$.\n=====\n[/hide]", "Solution_3": "$ n\\in\\mathbb{N}$ , $ n$ is a positive integer (as usually $ n$ is) ;)\r\n\r\nso your hint is OK but only in the case $ a\\in\\mathbb{Q}$ :P\r\n\r\nwhat about irrational $ a$ ?", "Solution_4": "It's trivial to prove that if it's true for $ a_1,a_2$ so that $ a_1 < a_2$, then it's true for $ a_3\\in (a_1,a_2)$.\r\n\r\nThis hinges on showing that $ \\frac {1}{n}\\lfloor na_2\\rfloor > \\frac {1}{n \\plus{} 1}\\lfloor a_1(n \\plus{} 1)\\rfloor$, which is of course true because $ \\frac {1}{n \\plus{} 1}\\lfloor a_1(n \\plus{} 1)\\rfloor < \\frac {1}{n}\\lfloor na_1 \\rfloor \\le \\frac {1}{n}\\lfloor na_2 \\rfloor$, the last inequality being true because $ a_1 < a_2$.", "Solution_5": "[hide]Let $ a \\equal{} m \\plus{} \\alpha$ where $ m \\in \\mathbb{Z}$ is the floor of $ a$ and $ 0 \\leq \\alpha < 1$ is the fractional part. Then $ f(n) \\equal{} m \\plus{} \\frac{1}{n} \\lfloor n\\alpha \\rfloor$. Because $ \\alpha < 1$, there are infinitely many $ n$ such that $ \\lfloor n\\alpha \\rfloor \\equal{} \\lfloor (n\\plus{}1)\\alpha \\rfloor$ (*). Those $ n$ work.\n\n[hide=\"Proof of *\"]$ \\alpha < 1$ means $ \\alpha < 1 \\minus{} \\frac{1}{k}$ for some $ k$. Then $ \\lfloor n \\alpha \\rfloor > \\lfloor (n\\plus{}k)\\alpha \\rfloor \\minus{} k$, so there is one value in between $ n$ and $ n\\plus{}k$ where it didn't increase.[/hide][/hide]", "Solution_6": "perhaps [url=http://en.wikipedia.org/wiki/Hermite's_identity]Hermite's Identity[/url] may help.", "Solution_7": "um, except that it doesn't and has nothing to do with the problem except that they both contain floors?", "Solution_8": "Looks fairly helpful to me. It reinterprets $ f(n)$ as the average of $ \\lfloor a \\rfloor, \\lfloor a \\plus{} 1/n \\rfloor, \\ldots , \\lfloor a \\plus{} (n\\minus{}1)/n) \\rfloor$, which means you can analyze where the fractional part of $ a$ lies relative to $ 0, 1/n, 2/n \\ldots 1 \\minus{} 1/n$. I think that leads to a solution.", "Solution_9": "I admit I didn't see that and apologize for being hasty (and overly harsh, I guess), but that seems rather laborious and needlessly difficult." } { "Tag": [ "pen", "integration", "modular arithmetic", "number theory", "relatively prime", "Divisibility Theory" ], "Problem": "Find all positive integers $n$ that have exactly $16$ positive integral divisors $d_{1},d_{2} \\cdots, d_{16}$ such that $1=d_{1}54 $ d_8\\equal{}54,d_9\\equal{}54\\plus{}17\\equal{}71\\equal{}p$.\r\nTherefore all solutions are $ n\\equal{}2*3^7,n\\equal{}2*3^3*37,n\\equal{}2*3^3*71$.", "Solution_2": "Rust \r\n\r\nThe first solution is false because d8=54 and d9=81 then the difference is d9-d8=27, a contradiction.", "Solution_3": "Yes. You are rite. I don't chek $ d_9\\minus{}d_8\\equal{}17$ for these case.", "Solution_4": "Slightly different finish:\n\nSo as above, we get that $n=2 \\cdot 3^3 \\cdot p$ for some prime $p$.\n\nThen, we have that there exists $k$ with $k(k+17)=n=54p$\n\nConsidering modulo $27$ gives $k \\equiv 0, 10 \\pmod {27}$\n\nIn the former case, letting $k=27a$, we have $a(27a+17)=2p$. Since the two things are relatively prime, for there only to be prime factors of $p,2$ we must have $a=1,2$ (else there are at least 3 prime factors: a 2, and one further in each parenthetical part) $a=1$ yields $p=22$, which isn't prime, but $a=2$ yields $71$ which works.\n\nIn the latter case, letting $k=27a+10$ yields $(a+1)(27a+10)=2p$. Again, we have $a+1 \\leq 2 \\implies a \\leq 1$. But $a=1$ works, yielding $p=37$.\n\nThus the only possible working values of $n$ are $\\boxed{n=1998, 3834}$ (54 times 37 and 71 respectively). It can easily be verified that they work, so we're done.", "Solution_5": "@Mewto\n\n$k(k+17)=n=54p$ How did you know that $d_8,d_9$ don't have common prime divisors?" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "[i]Find all integer solutions x,y of the equations \n1)$(x+y^2)(x^2+y)=(x+y)^3$;\n2)$(x+y^2)(x^2+y)=(x-y)^3$\n3)$(x+y^3)(x^3+y)=(x+y)^4$\n4)$(x+y^3)(x^3+y)=(x-y)^4$[/i]\r\n[i]1) 45-th Swedish Mathematical Competition 2005, Final Round (Lund, November 19)\n2) Proposed by Dang Hung Thang in Vietnamese Mathematical Olympiad\n3) Australian Mathematical Competition 2004\n4) Just to fill the collection :) \nI solved only number 3). Help me :)[/i]", "Solution_1": "1)$(x+y^2)(x^2+y)=(x+y)^3$\r\n$\\Rightarrow xy(xy+1)=3xy(x+y)$or$x=0,y=0$\r\n$\\Rightarrow (x-3)(y-3)=8\\Rightarrow ...$\r\n2)Similar$\\Rightarrow x=0,y=0,(x-3)(y+3)=-10$\r\n3)$x=0,y=0$\r\n$\\Rightarrow x^2y^2+1=4x^2+6xy+4y^2$\r\n$\\Rightarrow (xy+1)^2=4(x+y)^2$\r\n$\\Rightarrow (x-4)(y-4)=15$or$(x+4)(y+4)=15\\Rightarrow ...$\r\n4)Similar 3)" } { "Tag": [ "Putnam", "geometry", "integration", "logarithms", "function", "college contests", "Putnam calculus" ], "Problem": "Fix an integer $ b \\geq 2$. Let $ f(1) \\equal{} 1$, $ f(2) \\equal{} 2$, and for each $ n \\geq 3$, define $ f(n) \\equal{} n f(d)$, where $ d$ is the number of base-$ b$ digits of $ n$. For which values of $ b$ does\n\\[ \\sum_{n\\equal{}1}^\\infty \\frac{1}{f(n)} \n\\]\nconverge?", "Solution_1": "jmerry just tried to link to this from the discussion of a 2008 problem, but the topic didn't exist. Now it does, and I'll go back and fix his link.", "Solution_2": "I realize that this is ancient history, but since I have an answer written up, I might as well post it. This is a harder problem then 2008 A4, which is the problem that got directed here.\r\n\r\n======================================\r\n\r\nFor the integer $ b\\ge 2,$ we have the sequence $ f(n)$ defined by $ f(1) \\equal{} 1, f(2) \\equal{} 2,$ and after that, $ f(n) \\equal{} nf(d),$ where $ d$ is the number of base-$ b$ digits of $ n.$ Then $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{f(n)}$ converges for $ b \\equal{} 2$ and diverges for $ b\\ge 3.$ As a lemma in our proof, we will need to estimate $ \\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{n}.$ On the one hand, picturing this sum as the sum of the areas of boxes shows that $ \\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{n} > \\int_{b^{k \\minus{} 1}}^{b^{k}}\\frac {1}{x}\\, dx .$\r\nHence, $ \\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{n} > \\ln b,$ and for $ b\\ge 3,\\ \\ln b > 1.$ On the other hand, since\r\n $ \\frac {1}{nb} \\plus{} \\frac {1}{nb \\plus{} 1} \\plus{} \\cdots \\plus{} \\frac {1}{nb \\plus{} b \\minus{} 1} < \\frac {1}{n},$ $ \\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{n}$ is a strictly decreasing function of $ k,$ and, in particular, for $ k\\ge 3,\\ \\sum_{n \\equal{} 2^{k \\minus{} 1}}^{2^{k} \\minus{} 1}\\frac {1}{n}\\le \\frac14 \\plus{} \\frac15 \\plus{} \\frac16 \\plus{} \\frac17 \\equal{} \\frac {319}{420},$ which is less than $ 1.$\r\nAssume that $ b\\ge 3.$ Then $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{f(n)} \\equal{} \\sum_{k \\equal{} 1}^{\\infty}\\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{f(n)} \\equal{} \\sum_{k \\equal{} 1}^{\\infty}\\frac {1}{f(k)}\\sum_{n \\equal{} b^{k \\minus{} 1}}^{b^{k} \\minus{} 1}\\frac {1}{n} > (\\ln b)\\sum_{k \\equal{} 1}^{\\infty}\\frac {1}{f(n)}.$\r\nThus $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{f(n)} > (\\ln b)\\sum_{n \\equal{} 1}^{\\infty}\\frac1{f(n)}.$ Since $ \\ln b > 1,$ this must mean that the sum diverges.\r\n\r\nNow assume that $ b \\equal{} 2.$ \r\n\r\n$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{f(n)} \\equal{} 1 \\plus{} \\frac12 \\plus{} \\frac16 \\plus{} \\sum_{n \\equal{} 4}^{\\infty}\\frac {1}{f(n)}$\r\n\r\n$ \\equal{} 1 \\plus{} \\frac12 \\plus{} \\frac16 \\plus{} \\sum_{k \\equal{} 3}^{\\infty}\\frac1{f(k)} \\sum_{n \\equal{} 2^{k \\minus{} 1}}^{2^{k} \\minus{} 1}\\frac1{f(n)}$\r\n\r\n$ \\equal{} 1 \\plus{} \\frac12 \\plus{} \\frac16 \\plus{} \\sum_{k \\equal{} 3}^{\\infty}\\frac1{f(k)}\\sum_{n \\equal{} 2^{k \\minus{} 1}}^{2^{k} \\minus{} 1}\\frac1n$\r\n\r\n$ < 1 \\plus{} \\frac12 \\plus{} \\frac16 \\plus{} \\left(\\frac {319}{420}\\right) \\sum_{k \\equal{} 3}^{\\infty}\\frac1{f(k)}.$\r\n\r\nSo, $ \\sum_{n \\equal{} 3}^{\\infty}\\frac1{f(n)} < \\frac16 \\plus{} \\left(\\frac {319}{420}\\right) \\sum_{k \\equal{} 3}^{\\infty}\\frac1{f(k)},$ or $ \\left(\\frac {101}{420}\\right)\\sum_{n \\equal{} 3}^{\\infty}\\frac1{f(n)} < \\frac16$ and the sum converges." } { "Tag": [ "trigonometry", "geometry" ], "Problem": "Solve for $x$ where $\\sin(20^{\\circ}+x)=2\\cos40^{\\circ}\\sin{x}$.", "Solution_1": "[hide=\"Solution\"]\\begin{eqnarray*}&& \\sin(20^\\circ+x)=2\\cos 40^\\circ\\sin x\\\\ &\\iff& \\sin 20^\\circ\\cos x+\\cos 20^\\circ\\sin x=2\\cos 40^\\circ\\sin x\\\\ &\\iff& \\sin 20^\\circ\\cos x=\\sin x(2\\cos 40^\\circ-\\cos 20^\\circ)\\\\ &\\iff& \\tan x=\\frac{\\sin 20^\\circ}{2\\cos 40^\\circ-\\cos 20^\\circ}\\\\ &\\iff& \\tan x=\\frac{\\sin 20^\\circ}{2\\cos (60^\\circ-20^\\circ)-\\cos 20^\\circ}\\\\ &\\iff& \\tan x=\\frac{\\sin 20^\\circ}{2(\\cos 60^\\circ\\cos 20^\\circ+\\sin 60^\\circ\\sin 20^\\circ)-\\cos 20^\\circ}\\\\ &\\iff& \\tan x=\\frac{\\sin 20^\\circ}{\\cos 20^\\circ+\\sqrt{3}\\sin 20^\\circ-\\cos 20^\\circ}\\\\ &\\iff& \\tan x={1\\over\\sqrt{3}}\\\\ &\\iff& x=30^\\circ+k\\cdot 180^\\circ, k\\in\\mathbb{Z}\\end{eqnarray*}[/hide]", "Solution_2": "[color=green]Different solution[/color]\r\n[hide]$\\sin(x+20^{o})=\\sin(x+40^{o})+\\sin(x-40^{o})$\n$\\sin(x+20^{o})-\\sin(x-40^{o})=\\sin(x+40^{o})$\n$2\\sin30^{o}\\cos(x-10^{o})=\\sin(x+40^{o})=\\cos(50^{o}-x)$\n$x-10^{o}=50^{o}-x+k.360^{o}$\n$x=30^{o}+k*180$\nThis equation comes from a famous geometry problem do you know it?[/hide]" } { "Tag": [ "arithmetic sequence", "algebra unsolved", "algebra" ], "Problem": "How many progressions with 4 elements have the elements in the set {1,2,...,n}? :)", "Solution_1": "[quote=\"xxxxtt\"]How many progressions with 4 elements have the elements in the set {1,2,...,n}? :)[/quote]\r\n\r\nYour problem is not very precise and many questions may be asked :\r\n- are you speaking about arithmetic progressions, geometric progressions or both ?\r\n- are you speaking about \"4 elements in progression\" or \"progressions with exactly 4 elements in the set\" ($ \\{2,3,4,5\\}$ would match the first, but not the second) ?\r\n- are you speaking about \"progressions\" or \"elements in progression\". In the first case $ \\{1,2,3,4\\}$ must be count twice (common difference $ \\plus{}1$ and common difference $ \\minus{}1$). In the second case, this set must be counted once.\r\n...\r\n\r\nIn the case where the problem is \"Count the number of distinct ordered subsets of positive integers $ \\{a,b,c,d\\}$ such that $ 1\\leq a0$, with $ s\\in\\{1,2,3\\}$, $ f(n)\\equal{}\\frac 32p(p\\minus{}1)\\plus{}ps$\r\n\r\nIMHO" } { "Tag": [ "geometry", "inequalities", "rearrangement inequality", "geometry unsolved" ], "Problem": "Prove :For all triangular we have:\r\n $ \\frac{1}{2Rr}\\leq \\frac{1}{a^2}\\plus{}\\frac{1}{b^2}\\plus{}\\frac{1}{c^2}$", "Solution_1": "We have $ \\frac {(a \\plus{} b \\plus{} c)r}{2} \\equal{} K \\equal{} \\frac {abc}{4R}$, where $ K$ is the area of the triangle. from this we get $ \\frac {1}{2Rr} \\equal{} \\frac {a \\plus{} b \\plus{} c}{abc} \\equal{} \\frac {1}{ab} \\plus{} \\frac {1}{bc} \\plus{} \\frac {1}{ca}$, so the inequality to prove is\r\n\\[ \\frac {1}{ab} \\plus{} \\frac{1}{bc} \\plus{} \\frac{1}{ca}\\leq \\frac{1}{a^2} \\plus{} \\frac {1}{b^2} \\plus{} \\frac {1}{c^2}\\]\r\nwhich is obvious by the Rearrangement Inequality (or also by AM-GM).\r\n\r\nequality occurs iff $ a \\equal{} b \\equal{} c$." } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "hey guys,\r\n\r\nanother year is over, so we have to do the yearly joke of course ;)\r\nand as usually there were some surprising changes in the German team.\r\nFirst the team and after this a few words.\r\n\r\nPeter Scholze (he did it this year^^, i think you know him)\r\nFriedrich Feuerstein (was at the IMO a few years ago but not in the last 3 years)\r\nGeorg Schr\u00f6ter (aka Naphthalin, first participation)\r\nStephan Neupert (also first one)\r\nLisa Sauermann (the first girl after about 5 years, first participation)\r\nMalte Lackmann (aka Malte Luck-Man, first one again)\r\n\r\nLisa Sauermann is just 14 years old and can particpate till 2011 so you can await something :P \r\nsurprisingly, christian sattler (aka NoName, golg medal last year) wasn't able to qualify for the IMO. also 2 guys who would have been in the team after the first 6 of 7 exams were \"thrown\" out...\r\nand also meru alagalingam who was in the successfull team last year ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88832]look here[/url]) couldn't qualify although he was 3rd in the pre-selection in december.\r\n\r\nbecause i am too old for this IMO, i wish the members of the team a lot of fun.\r\n\r\n :spam: :rotfl: \r\n\r\ndg", "Solution_1": "Just one question: Are you darij grinberg? :D", "Solution_2": "Can't believe Christian Sattler didn't make it into the team :o", "Solution_3": "[quote=\"NoName\"]Can't believe Christian Sattler didn't make it into the team :o[/quote]\r\n\r\nYeah, too bad that current students from LMU Munich cannot participate in their 14th year of school. :D", "Solution_4": "[quote=\"NoName\"]Can't believe Christian Sattler didn't make it into the team :o[/quote]I heard he was so noob that he couldn't solve $\\sum \\frac a{b+c}\\geq \\frac{3}{2}$. So, definitely the medal from last year was revoked by the IMO Committee and he was forbidden to ever participate in the IMO (including IMO 2009 in Germany, for which he will have to be in another country, not to disturb the students). \r\n\r\nIt is unfortunate that a good member of the AoPS-ML community has received such a harsh treatment, nevertheless we agree with the punishment in his case.", "Solution_5": "but this wasn't unforseeable. last year he was nearly disqualified from the IMO because he did some n00by things during the TST's. But I don't know whether I am allowed to tell this to you...\r\n\r\nNaphthalin", "Solution_6": "[quote=\"Valentin Vornicu\"][quote=\"NoName\"]Can't believe Christian Sattler didn't make it into the team :o[/quote]I heard he was so noob that he couldn't solve $\\sum \\frac a{b+c}\\geq \\frac{3}{2}$. So, definitely the medal from last year was revoked by the IMO Committee and he was forbidden to ever participate in the IMO (including IMO 2009 in Germany, for which he will have to be in another country, not to disturb the students). \n\nIt is unfortunate that a good member of the AoPS-ML community has received such a harsh treatment, nevertheless we agree with the punishment in his case.[/quote]\r\n\r\nWait...he got his medal revoked because he couldn't solve some inequality...?\r\n\r\nThen how did he get on the team in the first place? I don't actually know the rules for IMO, but it's not like I will even make it; I haven't even made it to the USAMO yet :P.", "Solution_7": "[quote=\"Zellex\"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]Unfortunately, yes :( \r\n\r\n[hide=\":whistling:\"]You do realize this thread is a hoax, right? :)[/hide]", "Solution_8": "[quote=\"Zellex\"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]\r\nI heard his medal was revoked, because he tried to achieve a second medal at the same IMO as \"GER7: Christian Stettle\".", "Solution_9": "[quote=\"Valentin Vornicu\"][quote=\"Zellex\"]Wait...he got his medal revoked because he couldn't solve some inequality...?[/quote]Unfortunately, yes :( \n\n[hide=\":whistling:\"]You do realize this thread is a hoax, right?[/hide][/quote]\r\n\r\nHow sad... wish him good luck next time :D", "Solution_10": "[quote=\"Valentin Vornicu\"]I heard he was so noob that he couldn't solve... [/quote]\r\n\r\nSorry for the stupid question but... What means [b]noob[/b]? (i never heard this before) \r\n\r\n$Tipe$", "Solution_11": "noob is an internet term. I think it developed from games. It basically transferred from \"newb\", who is a new player, or beginner.", "Solution_12": "[quote=\"Zellex\"]noob is an internet term. I think it developed from games. It basically transferred from \"newb\", who is a new player, or beginner.[/quote]\r\n\r\nand doesn't know so much --> makes many stupid mistakes", "Solution_13": "It actually comes from \"newbie\":http://en.wikipedia.org/wiki/Newbie\r\nThat punishment seems unfair to me." } { "Tag": [], "Problem": "Call the positive integer selfdivisible if it is divisible by each sum of its consecutive digits, in particular it is divisible by\r\neach of its digits. Prove that the set of selfdivisible integers is finite.", "Solution_1": "Isn't it infinite because all of the numbers consisting of only 1's are selfdivisible?\r\n\r\nExamples: $1$, $1111$, $1111111111111111111111111111$", "Solution_2": "[quote=\"rogue\"] if it is divisible [b]by each sum of its consecutive digits[/b], in particular it is divisible by\neach of its digits. [/quote]\r\n\r\nFor example, $1111$ is not divisible by $1+1$. \r\n\r\nAre we to assume that $0$ does not divide any integer?", "Solution_3": "[quote=\"t0rajir0u\"]\nAre we to assume that $0$ does not divide any integer?[/quote]\r\n\r\nYes.", "Solution_4": "i think its infinite....\r\n\r\n$24$, $2222 4$, $2222222 4$, etc are divisible by $2+2=4$ and $2+4=6$. Just keep adding three $2$'s to the beginning and it's always divisbile by 4 and 6.\r\n\r\nUnless you mean that the sums can e taken of three or more digits...", "Solution_5": "[quote=\"cincodemayo5590\"]i think its infinite....\n\n$24$, $2222 4$, $2222222 4$, etc are divisible by $2+2=4$ and $2+4=6$. Just keep adding three $2$'s to the beginning and it's always divisbile by 4 and 6.\n\nUnless you mean that the sums can e taken of three or more digits...[/quote]\r\n\r\n[b]Every[/b] set of consecutive digits. So, for example, $2 + 2 + 2 + 2 + 2 = 10$ does not divide $22222224$." } { "Tag": [ "geometry", "rectangle", "3D geometry" ], "Problem": "Consider a 6x6 grid.\r\nA)Find the number of squares of all sizes in this grid. Then generalize your result to an n x n grid.\r\nB) Find the number of rectangles of all sizes in this grid. Then generalize your result to an n x n grid.", "Solution_1": "Part b)\r\n\r\nFirst start by doing small values for $ n$,\r\n\r\nWhen $ n \\equal{} 1$ there are $ 1$ squares.\r\nWhen $ n \\equal{} 2$ there are $ 9$ squares.\r\nWhen $ n \\equal{} 3$ there are $ 36$ squares.\r\n\r\nNow we can start seeing a pattern:\r\n$ 1 \\equal{} 1^3 \\equal{} 1^2$\r\n\r\n$ 1 \\plus{} 8 \\equal{} 1 \\plus{} 2^3 \\equal{} 9 \\equal{} (1 \\plus{} 2)^2$\r\n\r\n$ 9 \\plus{} 27 \\equal{} (1 \\plus{} 2^3) \\plus{} 3^3 \\equal{} 36 \\equal{} (1 \\plus{} 2 \\plus{} 3)^2$\r\n\r\nFrom this we can start to assume that for the $ n$ x $ n$ grid the amount of squares of all sizes is equal to the sum of the first $ n$ cubes or the equivalent, sum of the first $ n$ natural numbers, squared. So when $ n \\equal{} 6$, the amount squares of all sizes is $ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} 4^3 \\plus{} 5^3 \\plus{} 6^3 \\equal{} (1 \\plus{} 2 \\plus{} 3 \\plus{} 4 \\plus{} 5 \\plus{} 6)^2 \\equal{} 21^2 \\equal{} 441$.\r\n\r\nAnd for the $ n$ by $ n$ grid it's:\r\n\r\n$ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} \\cdots \\plus{} n^3 \\equal{} (1 \\plus{} 2 \\plus{} 3 \\plus{} \\cdots \\plus{} n)^2 \\equal{} \\left(\\dfrac{n(n \\plus{} 1)}{2}\\right)^2$\r\n\r\nEdit: oops thanks, tennisperson3.", "Solution_2": "Part b)\r\n\r\nYou have to choose 2 rows, for the edges of the rectangle, and there are $ n\\plus{}1$ rows. Same goes with the columns. Thus the answer is $ \\left(\\frac{n(n\\plus{}1)}{2}\\right)^2$. For 6 this gives 441.\r\n\r\ntonypr: Unfortunately, for part a, you actually did part b instead. \r\n\r\nPart a)\r\n\r\nFirst, we choose the side length for the square. This gives $ n$ choices, $ 1\\minus{}n$. If we choose $ k$ for this, the first $ n\\minus{}k\\plus{}1$ rows and first $ n\\minus{}k\\plus{}1$ columns can have the top corner. Thus we have $ (n\\minus{}k\\plus{}1)^2$ possibilities for a square of $ k$ length. Taking $ k$ from 1 to $ n$, we have $ n^2\\plus{}(n\\minus{}1)^2\\plus{}...\\plus{}1\\equal{}\\frac{n(n\\plus{}1)(2n\\plus{}1)}{6}$. For 6, this gives 91.", "Solution_3": "Chrispy might have been asking about a 6x6 grid of points, not of squares. \r\n\\[ .\\ .\\ .\\ .\\ .\\ .\\\\\r\n.\\ .\\ .\\ .\\ .\\ .\\\\\r\n.\\ .\\ .\\ .\\ .\\ .\\\\\r\n.\\ .\\ .\\ .\\ .\\ .\\\\\r\n.\\ .\\ .\\ .\\ .\\ .\\\\\r\n.\\ .\\ .\\ .\\ .\\ .\r\n\\]\r\n\r\nIt turns out that the formula for the number of squares (not necessarily integer side lengths) in this grid is $ \\frac{n^2(n^2\\minus{}1)}{12}$. We can prove this by considering the number of squares inscribed in a 1x1, 2x2, etc. and using induction. \r\n\r\nI don't know the answer to b) for such a grid of points.", "Solution_4": "If it's a grid of points then all you have to do is just subtract by one on the formula for part b, so it would be $ \\left(\\dfrac{(n \\minus{} 1)(n)}{2}\\right)^2$", "Solution_5": "That formula doesn't work for, for example, $ n\\equal{}3$.", "Solution_6": "For squares at least, it's not too hard to also count if they don't need to be orthogonal:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=155463\r\n\r\nLike ThinkFlow, I don't know what the answer should be for rectangles, but perhaps someone can extend the method from the thread I've linked?" } { "Tag": [ "LaTeX" ], "Problem": "Hello,\r\n\r\nHow could I make a footer with the page number and some text and how can I make the foot note and can I link the number in the foot note to the note down?\r\n\r\nThanks in advance", "Solution_1": "[quote=\"hasan4444\"]\nHow could I make a footer with the page number and some text[/quote][[LaTeX:Packages#fancyhdr]] which includes \\cfoot{\\thepage} which puts the page numbering the footer \n\n[quote=\"hasan4444\"]\nand how can I make the foot note and can I link the number in the foot note to the note down?[/quote]Put \\usepackage[hyperfootnotes]{hyperref} in preamble (see [url=http://www.tug.org/applications/hyperref/manual.html]hyperref manual[/url])" } { "Tag": [ "MATHCOUNTS", "geometry", "geometric transformation" ], "Problem": "To all of you going to nationals, wear a t-shirt with your username on it. I'll have DeeVee7 on mine. Just post what you will put so we can recognize each other.\r\n\r\nThanks, \r\nKaran Batra", "Solution_1": "I'll go write bob123 on my AOPS t-shirt :)", "Solution_2": "Bob123, are you from CA? I hope to see many AOPSers at nationals, 49 exactly with names on their shirts.", "Solution_3": "You guys can make 4 soccer teams and have a small championship :D :D", "Solution_4": "[quote=\"kchande\"]Bob123, are you from CA? I hope to see many AOPSers at nationals, 49 exactly with names on their shirts.[/quote]\r\n\r\nI'm #2 from CA... #1 is piggypi314, #3 is NoSoupForYou, #4 is aznness", "Solution_5": "I'll do a nametag.", "Solution_6": "I'll just tape it to my shirt or something like that.", "Solution_7": "[quote=\"Treething\"]I'll just tape it to my shirt or something like that.[/quote]\r\n\r\nY'know, that's a good idea...", "Solution_8": "You can put it anywhere as long as people can see it.\r\nBTW: not under a shirt", "Solution_9": "I will either be wearing my AoPS shirt with my username embroidered on it, or my Reader's Digest National Word Power Finalist shirt. :P \r\n\r\nBilly", "Solution_10": "will it have your username on it?\r\nit should", "Solution_11": "You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my \"Deny Everything\" DC Spy Museum shirt...or - forget it.", "Solution_12": "[quote=\"09husbri\"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my \"Deny Everything\" DC Spy Museum shirt...or - forget it.[/quote]\r\nGood idea (The forget it part)", "Solution_13": "[quote=\"09husbri\"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my \"Deny Everything\" DC Spy Museum shirt...or - forget it.[/quote]\r\n\r\nA seinfeld shirt? Keep up with the times man - Seinfeld is so 90's :D", "Solution_14": "...says \"NoSoupForYou\"... :lol: \r\n\r\nNo way, man! Seinfeld is [i]forever![/i]", "Solution_15": ":D", "Solution_16": "Alex, i like ur avatar. However, mine is better.", "Solution_17": "I agree, don't mess with cows!", "Solution_18": "Got milk?", "Solution_19": "Got rice?", "Solution_20": "Tienes arroz o leche? :D", "Solution_21": "No. Tengo arroz pero no tengo leche. No tengo gusto de la leche. Usted consigui\u00f3 pwned.\r\n\r\nOr in Dutch: Nr. Ik heb krullen maar ik heb geen melk. Ik houd van geen melk. U kreeg pwned.\r\n\r\nOr French: Non. Je prends le riz mais je ne prends pas le lait. Je n'aime pas le lait. Vous avez obtenu pwned.\r\n\r\nOr German: Nein. Ich habe Wellungen, aber ich habe nicht Milch. Ich mag nicht Milch. Sie erhielten pwned.\r\n\r\nMaybe Italian: No. Ho arricciature ma non ho latte. Non gradisco il latte. Avete ottenuto pwned. \r\n\r\nPerhaps Russian =): \u041d\u0435\u0442. \u042f \u0438\u043c\u0435\u044e \u0440\u0438\u0441 \u043d\u043e \u044f \u043d\u0435 \u0438\u043c\u0435\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u042f \u043d\u0435 \u043b\u044e\u0431\u043b\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u0412\u044b \u043f\u043e\u043b\u0443\u0447\u0438\u043b\u0438 pwned.\r\n\r\nOr Potuguese: N\u00e3o. Eu tenho ondas mas eu n\u00e3o tenho o leite. Eu n\u00e3o gosto do leite. Voc\u00ea come\u00e7ou pwned.\r\n\r\n\u03a6\u03cc\u03b2\u03bf\u03c2 \u03bc\u03b5\u03c4\u03b1\u03c6\u03c1\u03b1\u03c3\u03c4\u03ce\u03bd \u03bf\u03b9 \u03b4\u03b5\u03be\u03b9\u03cc\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bc\u03bf\u03c5! (Phear my translatoring skillz!)", "Solution_22": "[quote=\"NoSoupForYou\"][quote=\"09husbri\"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my \"Deny Everything\" DC Spy Museum shirt...or - forget it.[/quote]\n\nA seinfeld shirt? Keep up with the times man - Seinfeld is so 90's :D[/quote]\r\n\r\nLast I recall: \"No Soup for You!\" - Soup Nazi... from Seinfeld.... \r\nGOOD LUCK AT NATIONALS! I could've made it, but being from texas where Houston area schools pwn, it was a long shot... got 39th though, with a pathetic score (because of target round... missed four of those, did well on sprint) of 28! My lowest ever", "Solution_23": "[quote=\"shake9991\"][quote=\"NoSoupForYou\"][quote=\"09husbri\"]You can bet I'll be wearing my Seinfeld shirt with a picture of Kramer on it...or my United Nations T-shirt...or my chapter MathCounts shirt from 6th grade...or my \"Deny Everything\" DC Spy Museum shirt...or - forget it.[/quote]\n\nA seinfeld shirt? Keep up with the times man - Seinfeld is so 90's :D[/quote]\n\nLast I recall: \"No Soup for You!\" - Soup Nazi... from Seinfeld.... [/quote]\r\n\r\n*sigh* Does :D mean anything to you?", "Solution_24": "who knows\r\nit might", "Solution_25": "[quote=\"NoSoupForYou\"]No. Tengo arroz pero no tengo leche. No tengo gusto de la leche. Usted consigui\u00f3 pwned.\n\nOr in Dutch: Nr. Ik heb krullen maar ik heb geen melk. Ik houd van geen melk. U kreeg pwned.\n\nOr French: Non. Je prends le riz mais je ne prends pas le lait. Je n'aime pas le lait. Vous avez obtenu pwned.\n\nOr German: Nein. Ich habe Wellungen, aber ich habe nicht Milch. Ich mag nicht Milch. Sie erhielten pwned.\n\nMaybe Italian: No. Ho arricciature ma non ho latte. Non gradisco il latte. Avete ottenuto pwned. \n\nPerhaps Russian =): \u041d\u0435\u0442. \u042f \u0438\u043c\u0435\u044e \u0440\u0438\u0441 \u043d\u043e \u044f \u043d\u0435 \u0438\u043c\u0435\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u042f \u043d\u0435 \u043b\u044e\u0431\u043b\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u0412\u044b \u043f\u043e\u043b\u0443\u0447\u0438\u043b\u0438 pwned.\n\nOr Potuguese: N\u00e3o. Eu tenho ondas mas eu n\u00e3o tenho o leite. Eu n\u00e3o gosto do leite. Voc\u00ea come\u00e7ou pwned.\n\n\u03a6\u03cc\u03b2\u03bf\u03c2 \u03bc\u03b5\u03c4\u03b1\u03c6\u03c1\u03b1\u03c3\u03c4\u03ce\u03bd \u03bf\u03b9 \u03b4\u03b5\u03be\u03b9\u03cc\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bc\u03bf\u03c5! (Phear my translatoring skillz!)[/quote]\r\n :o Tell me you used some sort of electronic translator or a dictionary - or [i]soemthing[/i]. If you knew two or maybe even three of those languages, it would be (semi-)normal, but - whoa!\r\n\r\nPS Three years earlier and I could have given you the Hebrew translation, but I haven't been to Hebrew school in a loooong time.\r\n\r\nPPS Why am I telling this to somebody with a Nazi in his avatar? ;) Just kidding. You don't hate Jews, of course. You just don't give them soup.\r\n\r\nPPPS Don't take offense. I know you're not a Nazi.", "Solution_26": "[quote=\"09husbri\"][quote=\"NoSoupForYou\"]No. Tengo arroz pero no tengo leche. No tengo gusto de la leche. Usted consigui\u00f3 pwned.\n\nOr in Dutch: Nr. Ik heb krullen maar ik heb geen melk. Ik houd van geen melk. U kreeg pwned.\n\nOr French: Non. Je prends le riz mais je ne prends pas le lait. Je n'aime pas le lait. Vous avez obtenu pwned.\n\nOr German: Nein. Ich habe Wellungen, aber ich habe nicht Milch. Ich mag nicht Milch. Sie erhielten pwned.\n\nMaybe Italian: No. Ho arricciature ma non ho latte. Non gradisco il latte. Avete ottenuto pwned. \n\nPerhaps Russian =): \u041d\u0435\u0442. \u042f \u0438\u043c\u0435\u044e \u0440\u0438\u0441 \u043d\u043e \u044f \u043d\u0435 \u0438\u043c\u0435\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u042f \u043d\u0435 \u043b\u044e\u0431\u043b\u044e \u043c\u043e\u043b\u043e\u043a\u043e. \u0412\u044b \u043f\u043e\u043b\u0443\u0447\u0438\u043b\u0438 pwned.\n\nOr Potuguese: N\u00e3o. Eu tenho ondas mas eu n\u00e3o tenho o leite. Eu n\u00e3o gosto do leite. Voc\u00ea come\u00e7ou pwned.\n\n\u03a6\u03cc\u03b2\u03bf\u03c2 \u03bc\u03b5\u03c4\u03b1\u03c6\u03c1\u03b1\u03c3\u03c4\u03ce\u03bd \u03bf\u03b9 \u03b4\u03b5\u03be\u03b9\u03cc\u03c4\u03b7\u03c4\u03ad\u03c2 \u03bc\u03bf\u03c5! (Phear my translatoring skillz!)[/quote]\n :o Tell me you used some sort of electronic translator or a dictionary - or [i]soemthing[/i]. If you knew two or maybe even three of those languages, it would be (semi-)normal, but - whoa!\n\nPS Three years earlier and I could have given you the Hebrew translation, but I haven't been to Hebrew school in a loooong time.\n\nPPS Why am I telling this to somebody with a Nazi in his avatar? ;) Just kidding. You don't hate Jews, of course. You just don't give them soup.\n\nPPPS Don't take offense. I know you're not a Nazi.[/quote]\r\n\r\nlol, NoSoupForYou is jewish...\r\n\r\nBTW, Alex, where'd you get that picture, it looks cool :)", "Solution_27": "husbri: I used [url=http://www.worldlingo.com/en/products_services/worldlingo_translator.html]WorldLingo[/url].\r\n\r\nbob: What picture?", "Solution_28": "[quote=\"NoSoupForYou\"]husbri: I used [url=http://www.worldlingo.com/en/products_services/worldlingo_translator.html]WorldLingo[/url].\n\nbob: What picture?[/quote]\r\n\r\nyour avatar", "Solution_29": "google", "Solution_30": "USA TENNIS!!!" } { "Tag": [ "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "working on solving an unknown sequance equation...\r\ni have the (X,Y) exact curve ....but it seems that it's formula is unknown yet..\r\nneed ..assistance....(f(x)=????)", "Solution_1": "Looks like $y=Cx^{-p}$ for some $C,p>0$. Try to draw $\\ln y$ versus $\\ln x$. If you get a straight line, the dependence is, indeed, power." } { "Tag": [ "function", "Pascal\\u0027s Triangle" ], "Problem": "I've been wondering about this function recently:\r\n\r\nf(n,x,m) = the number of x's in the nth row of Pascal's Triangle mod m. That is, take each entry in Pascal's Triangle and reduce it modulo m, go to the nth row, and count how many occurences of the residue x there are. \r\n\r\nFor example, the 5th row of Pascal's Triangle, 1-5-10-10-5-1, when reduced mod 3, becomes 1-2-1-1-2-1. Then\r\n\r\nf(5,0,3) = 0\r\nf(5,1,3) = 4\r\nf(5,2,3) = 2\r\n\r\nI found some research out there on this when m = 2, but I was wondering how to go about determining this function for other moduli. I think there may just be a recurrence for it, or something.\r\n\r\nA similar challenging question: look at the so-called \"shallow diagonals\" or \"Fibonacci diagonals\" of Pascal's Triangle, and find the same type of information. That is, get a recursion or some sort of closed form for\r\n\r\ng(n,x,m) = the number of x's in the nth Fibonacci diagonal of Pascal's Triangle mod m. For example, the 7th diagonal, 1-5-6-1, when reduced modulo 4, gives us 1-1-2-1, and thus\r\n\r\ng(7,0,4) = 0\r\ng(7,1,4) = 3\r\ng(7,2,4) = 1\r\ng(7,3,4) = 0\r\n\r\nI am hoping someone knows how to determine at least some information about either of these functions, because I can't think of how to go about solving for them. I also think the Fibonacci diagonal one is harder, by the way.\r\n\r\nI would really appreciate any help - Thanks!", "Solution_1": "Let $p$ be a prime. Some useful identities:\r\nIf $p\\nmid k$, $\\binom{pn}{k}\\equiv 0\\mod p$\r\nIf $0\\le i,j0$ , then :\r\n\r\n$ \\sum_{cyc}\\frac{x^5}{yz}\\plus{}2\\sum_{cyc}\\frac{x^2y^2}{z}\\ge \\sum_{cyc}x^3\\plus{}\\sum_{sym}\\frac{x^4}{z}$", "Solution_1": "$ \\sum_{cyc}\\frac{x^{5}}{yz}\\plus{}2\\sum_{cyc}\\frac{x^{2}y^{2}}{z}\\ge\\sum_{cyc}x^{3}\\plus{}\\sum_{cyc}\\frac{x^{4}\\plus{}y^4}{z}$\r\n\r\n is equivalent to \r\n\r\n $ \\sum{(x^2\\minus{}yz)^2(x\\minus{}y)(x\\minus{}z)}\\plus{}\\sum{(yz)^2(x\\minus{}y)(x\\minus{}z)}\\ge{0},$\r\n\r\nit is obvious holds." } { "Tag": [ "number theory" ], "Problem": "Is there a way to do this problem [b]without dividing the entire thing out[/b]? No calculators are allowed.\r\n\r\nHow many times does the digit 8 appear in the first 2007 digits of the decimal expansion of 1/19?\r\n\r\n\r\nAns\r\n[hide]222[/hide]", "Solution_1": "The answer is incorrect, the decimal is 0.052631578947368421.... where the part I put keeps repeating so there will be an infinite number of 8s.", "Solution_2": "heh... oops... \r\nI meant the first 2007 digits of the decimal expansion.\r\n\r\nEDIT:\r\niniquitus, I said I need a way without long division, but thanks anyways...", "Solution_3": "Notice that $ \\frac{52631578947368421}{999999999999999999} \\equal{} \\frac{1}{19}$ by long division; therefore, in every 18 decimal places there are two 8's.\r\n\r\n$ 2007 \\div 19 \\equal{} 111 R9$, so there are $ 2\\times 111 \\plus{} 1 \\equal{} 223$ 8's??", "Solution_4": "Long division doesn't take that long; in any case, if you really wanted to do it quickly, you already know that the digits will repeat every $ 18$ digits because of [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url] and you can guess that approximately $ 2$ of those digits will be eights.", "Solution_5": "[quote=\"iniquitus\"]Notice that $ \\frac {52631578947368421}{999999999999999999} \\equal{} \\frac {1}{19}$ by long division; therefore, in every 18 decimal places there are two 8's.\n\n$ 2007 \\div 19 \\equal{} 111 R9$, so there are $ 2\\times 111 \\plus{} 1 \\equal{} 223$ 8's??[/quote]\r\n\r\nDon't you mean $ 2007 \\div 18 \\equal{} 111 R9$? by the way, I think this is correct.", "Solution_6": "yesh sorry :lol:" } { "Tag": [ "inequalities" ], "Problem": "Let $ x,y>0$ with $ xy\\geq 1$ with $ xy\\equal{}1$ if and only if $ x\\equal{}y$ and $ x\\geq y$ . Prove or disprove the following inequality \r\n\r\n$ 2x^2y^4\\plus{}x^3y^2\\geq 2x\\plus{}y^2$", "Solution_1": "I think that this inequality is not true. For example taking $ y \\equal{} \\frac {5}{4x}$ we land in an inequality:\r\n$ 2*(\\frac {5}{4})^{2}*\\frac {1}{x^{2}}\\geq \\frac {7}{16}x \\plus{} (\\frac {5}{4x})^{2}$ which is not true for sufficiently big x :wink:", "Solution_2": "If $ y\\equal{} \\frac{5}{4x}$ then $ xy \\equal{} \\frac{5}{4} \\neq 1$", "Solution_3": "Yes, but what the problem states is that:\r\nI)$ xy\\geq 1$\r\nII)$ x\\geq y$\r\nIII)$ (xy\\equal{}1)\\iff (x\\equal{}y)$\r\nAt least I understood it like that :wink:", "Solution_4": "[quote=\"polskimisiek\"]Yes, but what the problem states is that:\nI)$ xy\\geq 1$\nII)$ x\\geq y$\nIII)$ (xy \\equal{} 1)\\iff (x \\equal{} y)$\nAt least I understood it like that :wink:[/quote]\r\nI think the problem says\r\nIII) $ (xy \\equal{} 1)\\iff ((x \\equal{} y)$OR$ (x \\ge y)$.", "Solution_5": "[quote=\"archimedes1\"]\nI think the problem says\nIII) $ (xy \\equal{} 1)\\iff ((x \\equal{} y)$OR$ (x \\ge y)$.[/quote]That just wouldn't make since. I think the first interpretation is correct." } { "Tag": [ "quadratics" ], "Problem": "Hello to all! We have a very difficult homework and I don't know to answer it. I thing it is about Quadratic Equation. Here is the word problem:\r\n\r\n[b]Genelle and Eljen can paint a room together in 4 hours. Working alone, Genelle can paint the room in two hours less than Eljen can. Find how long it takes Eljen to paint the room alone.[/b]\r\n\r\nPlease, I am begging you\u2026 I want the answer and how to solve it\u2026\u2026 \r\nThank You Very much and your right answer could be greatly appreciated.\r\nThank You once again\u2026.", "Solution_1": "[hide=\"Solution\"]\nLet $ g$ be the rate at which Genelle paints and let $ e$ be the rate at which Eljen paints.\nLet $ t$ be the time it takes Eljen to paint the room.\n\nThen,\n$ 4(g\\plus{}e)\\equal{}1$\n$ (t\\minus{}2)g\\equal{}1$\n$ te\\equal{}1$\n\nSolving the equations,\n$ 4(\\frac{1}{t\\minus{}2}\\plus{}\\frac{1}{t})\\equal{}1$\n$ 4(t\\plus{}(t\\minus{}2))\\equal{}t(t\\minus{}2)$\n$ 8t\\minus{}8\\equal{}t^{2}\\minus{}2t$\n$ t^{2}\\minus{}10t\\plus{}8\\equal{}0$\n$ t\\equal{}5\\pm \\sqrt{17}$\n\nIf $ t\\equal{}5\\minus{} \\sqrt{17}$ then $ t\\minus{}2<0$\nThus, $ t\\equal{}5\\pm \\sqrt{17}$\n\nHope this helps...\n[/hide]", "Solution_2": "Looking back at the problem, $ 5\\minus{}\\sqrt{17}$ is obviously not a solution ($ t\\ge4$). Assuming mismatchtea is correct, the only solution is $ 5\\plus{}\\sqrt{17}$." } { "Tag": [ "AMC", "AIME", "AMC 10", "geometry", "email", "AMC 10 B", "trigonometry" ], "Problem": "The official scores have been released? Did everyone make it into AIME? :)", "Solution_1": "sadly, no.", "Solution_2": "well.. now that the cutoff is 110.. YES!! lol =)", "Solution_3": "have score reports come out? or do you just know the cutoff and assume that you made it?", "Solution_4": "123 on the 12s.\r\n\r\nI heard the cutoff for the 10's was 119?", "Solution_5": "[quote=\"dnbmathguy\"]123 on the 12s.\n\nI heard the cutoff for the 10's was 119?[/quote]\r\n\r\nNo, the cutoff for the AMC10 was 110.", "Solution_6": "It would be awesome if the cutoff really is 110.", "Solution_7": "What do you mean, \"if it is?\" It is.", "Solution_8": "No... They really did send out the official scores.. I got em yesterday!", "Solution_9": "ok. thanks. i just took tha amc-10 and i'm not allowed to discuss it right now but i don't think i made it.", "Solution_10": "made aime through amc-10A\r\n\r\n-140.5-", "Solution_11": "yea i made it through 10A too... 139.5\r\n\r\ni don't think i did as good on B though... was it just me or was 10B harder than 10A??", "Solution_12": "I think that the geometry problems (#24 and #20) on AMC 10B are harder than the geometry problems oan the A exam......", "Solution_13": "Think I did alright on the amc 12B\r\ni made it to the aime...120.5 on the 12A i think\r\nmaybe a couple of points higher on the 12B\r\n\r\nThe last few problems in 12B were definitely harder than the last ones on 12A", "Solution_14": "107.5 on 12B :'(, o well.", "Solution_15": "[quote=\"tetrahedr0n\"]amazing! the similarities continue mathfan: on 20, i took a ruler and measured the diagram, and my measurement yielded very close to the correct d. then i filled in e on my sheet like an idiot :(. I similarly brute-forced 24 and scored points on it. why is 25 easier than 24 i wonder?[/quote]\r\n\r\nInterestingly enough, I tried to do #20 with a ruler and diagram, but I didn't get anywhere so I just guessed C. I considered guessing D, but I figured I'd go with my first guess, which was C. Bah.\r\n\r\nAnd 25 was pretty easy compared to 20 and 24...hmm...", "Solution_16": "[quote=\"Syntax Error\"]I still don't remember! Someone post the question please.[/quote]\r\n\r\nIn triangle ABC, points D and E lie on BC and AC, respectively. If AD and BE interesect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?", "Solution_17": "what did you get on the 12 A mystic? though it was a lot easier i just did a lot worse", "Solution_18": "25 is often not as hard as the others -- two years ago, the AMC 12A had an immensely easy question 25, if you know anything about polynomials. It was a 1-line solution.", "Solution_19": "[quote=\"tetrahedr0n\"]amazing! the similarities continue mathfan: on 20, i took a ruler and measured the diagram, and my measurement yielded very close to the correct d. then i filled in e on my sheet like an idiot :(. I similarly brute-forced 24 and scored points on it. why is 25 easier than 24 i wonder?[/quote] \r\n\r\nthe AMC people probably ordered the problems in such way to fool students. notice that number 25 is on the last page, alone by itself........", "Solution_20": "I don't think they did that to fool students. In general, the AMC isn't out to fool you (although many people do get fooled, nonetheless). For instance, the answer choices often don't include answers that people could arrive at by making a stupid mistake. I know I've been saved before by not seeing my final answer as one of the choices...", "Solution_21": "[quote]For instance, the answer choices often don't include answers that people could arrive at by making a stupid mistake.[/quote]\r\n\r\nHow untrue (thinks back to AMC 12A #1, AMC 12B #20). :(", "Solution_22": "well, I took about 8 minutes (while in time trouble) trying to solve #20 without a clue until I turned my attention to #25 and solved it in about a minute... I really think that #25 should replace #20's position...", "Solution_23": "[quote=\"Fierytycoon\"]How untrue (thinks back to AMC 12A #1, AMC 12B #20). :([/quote]\r\n\r\n\r\nCompare to the exams from 15-20 years ago (I don't remember exactly when they switched they answer strategies). They hardly ever use answers that result primarily from simple errors like arithmetic mistakes, but generally more complicated subtle mistakes or fundamental misunderstandings.", "Solution_24": "In general they don't want someone to get a problem wrong because they made a very minor (and mathematically unimportant) error. If people get problems wrong, it should be because they are guessing or because they made more fundamental errors. The whole point of AMC12A #1 was to make sure you actually read the question rather than guess what it's asking. They want you to do that, and I think that's reasonable.", "Solution_25": "Are people saying that their teachers have received the official scores and list of AIME qualifiers by email, or just that the answers and cut-offs are available and people know their scores unofficially? My group has not received official scores, although AMC has an email address to send them to. Now I'm worrying that the answer sheets got lost in the mail...... :-(", "Solution_26": "[quote=\"texas137\"]Are people saying that their teachers have received the official scores and list of AIME qualifiers by email, or just that the answers and cut-offs are available and people know their scores unofficially? My group has not received official scores, although AMC has an email address to send them to. Now I'm worrying that the answer sheets got lost in the mail...... :-([/quote]\r\n\r\nI believe they're saying the former. My teacher received our scores via e-mail today at lunch. Considering that some people (e.g. Ragingg) got their scores several days back, it's likely that some scores are being sent out. You'll probably get yours soon.", "Solution_27": "[quote=\"mathfanatic\"]My teacher recieved our scores via e-mail today at lunch. Considering that some people (e.g. Ragingg) got their scores several days back, it's likely that some scores are being sent out. You'll probably get yours soon.[/quote]\r\n\r\nI remember back in November after the AMC 8 the scores were not all emailed on the same day. I don't know what the queueing protocol is, but there was a several day span for sending out official emails with AMC 8 score notifications. I wouldn't worry yet--but I might worry next week if I still hadn't seen an A score. \r\n\r\nI'm trying to calm my worries about a summer program application I submitted for my son, which was supposed to result in a confirmation of receipt postcard by now. I have to keep telling myself, \"No, it's not lost in the mail; it's just that the mail is real slow.\" \r\n\r\nAside to Mathfanatic: I recall that you desire to practice good grammar, so find your spelling mistake in the quoted text above, and correct it by editing your original post. I'll probably make a spelling mistake without noticing it in this post, just so we can tease each other about this issue. \r\n\r\nBest wishes to everyone who is still in suspense about scores.", "Solution_28": "In response to questions about emailing of school reports: The email reports are still going out. and probably will go out for a couple more (work) days. The very first paper reports were prepared on Friday, and I believe they went in the mail on Friday afternoon.\r\n\r\nHere is an an overview of the process: It takes several days for your answer sheets to even reach us after the contest. For the AMC A on Tues Feb 10, it was not until Friday Feb 13, that we had substantial numbers of contests in the building, and it was Tuesday, Feb 17 (remember Feb 16 was a federal holiday) that we finally had the majority. We are still receiving some A packages from distant parts of the globe. When I left the office on Friday, we had only received about 20 or so B packages.\r\n\r\nAfter receiving the packages, they are organized and grouped by states, and then opened and checked off the registration lists to determine which we have not received packages from. The teacher certification form is also checked to determine if all rules were followed and that no irregularities occurred. With 3000-4000 packages this takes several days.\r\n\r\nThe answer sheets are then scanned. Our scanner is very touchy, and we have had two (very expensive) visitis from the repairman, and that slowed us down some. We generally scan all results from a state together, and generally largest enrollment states first. The schools within each state are in random order. After scanning, we must still \"edit\" the data, since improperly filled in dots on the answer sheets leave gaps in our data. If a student does not bubble in the name data correctly, or uses an ink pen (not a lead pencil) then we must fill in the gaps. \r\n\r\nAfTempter editing, we run the data and start actual scoring. This year, this started on Monday Feb 23. We have to score a substantial number to determine the invitation levels. Only after all tht can we send the school reports by email, and of course, only if the school and contest manager provided an email address. Because of the chances of organization, scoring by state, the volume of editing, it seems we can get out about 600 reports by email per day. Therefore, it takes 5-8 (work) days to get all the reports completed by email. Packaging the paper reports is slow and painstaking because each package has individualized contents. \r\n\r\nWe make every effort to get your results to you as quickly as possible, if you have not received your results yet, rest assured that they are coming.\r\n\r\nSteve Dunbar\r\nAMC Director", "Solution_29": "to Steve Dunbar - thanks for posting this. I know I found it interesting and informative!" } { "Tag": [ "algebra", "polynomial", "complex numbers", "greatest common divisor", "IMC", "college contests" ], "Problem": "Let $p(z)$ be a polynomial of degree $n$, $n \\geq 1$ with complex coefficeints and $a_1, a_2,...,a_k$ be $k$, $k \\geq 2$ distinct complex numbers.\r\nProve that exist at least $(k-1)n+1$ complex numbers $z$ for which $p(z)$ is $a_1$ or $a_2$ or ... $a_k$.", "Solution_1": "The number of distinct roots of a polynomial $P$ is $n-\\deg(P,P')$, where $(f,g)$ denotes, as usual, the greatest common divisor of $f$ and $g$. This means that the number of distnct roots of $P+a_i,\\ i\\in\\overline{1,k}$ is $\\sum_{i=1}^k[n-\\deg(P+a_i,P')]=kn-\\sum_i\\deg(P+a_i,P')\\ (*)$.\r\n\r\nNow since $P+a_i$ are coprime, so are $(P+a_i,P')$, and since each one of them divides $P'$, their product must also divide $P'$, which has degree $n-1$. This means that $\\sum_i\\deg(P+a_i,P')\\le n-1$, so $(*)\\ge kn-(n-1)=(k-1)n+1$, as desired." } { "Tag": [ "geometry", "incenter", "circumcircle", "trigonometry", "inequalities", "geometry proposed" ], "Problem": "Let $ I$ and $ O$ denote respectively the incenter and the circumcenter of a triangle $ ABC$. Given that $ \\measuredangle AIO \\equal{} 90^{\\circ}$, prove that the area of the triangle $ ABC$ is less than $ \\frac {3\\sqrt 3}{4}AI^2$", "Solution_1": "$ \\angle AIO \\equal{} 90^\\circ \\ \\Longleftrightarrow \\ \\frac{r^2}{\\sin^2 {\\frac{A}{2}}} \\plus{} R^2 \\minus{} 2Rr \\equal{} R^2 \\ \\Longleftrightarrow \\ r \\equal{} 2R \\sin^2 {\\frac{A}{2}}$\r\n\r\n$ S <\\frac{3 \\sqrt{3}}{4} AI^2 \\ \\Longleftrightarrow \\ p r < \\frac{3 \\sqrt{3}}{4} \\frac{r^2}{\\sin^2 {\\frac{A}{2}}} \\ \\Longleftrightarrow \\ p < \\frac{3 \\sqrt{3}}{2} R$ that is know", "Solution_2": "Why?\r\n[quote=\"\u00ac[\u0192(Gabriel)\u00b3\u00b2\u00b9\u00ba]\u00bc\"]$ \\angle AIO \\equal{} 90^\\circ \\ \\Longleftrightarrow \\ \\frac {r^2}{\\sin^2 {\\frac {A}{2}}} \\plus{} R^2 \\minus{} 2Rr \\equal{} R^2 $[/quote]\nHow can solve this [quote]$ p < \\frac {3 \\sqrt {3}}{2} R$ that is know[/quote]", "Solution_3": "[quote=\"every\"]Why?\n[quote=\"\u00ac[\u0192(Gabriel)\u00b3\u00b2\u00b9\u00ba]\u00bc\"]$ \\angle AIO \\equal{} 90^\\circ \\ \\Longleftrightarrow \\ \\frac {r^2}{\\sin^2 {\\frac {A}{2}}} \\plus{} R^2 \\minus{} 2Rr \\equal{} R^2$[/quote][/quote]\nFor pytagora on $ \\triangle OIA$ : $ AO^2 \\equal{} OI^2 \\plus{}AI^2$\n\n[quote=\"every\"]How can solve this [quote]$ p < \\frac {3 \\sqrt {3}}{2} R$ that is know[/quote][/quote]\r\n\r\n$ p \\equal{} 4R \\cos \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2}$ so the inequality become: $ \\cos \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2} \\le \\frac {3 \\sqrt {3}}{8}$\r\n\r\nnow $ \\frac{a}{b\\plus{}b}\\equal{}\\frac{\\sin A}{\\sin B \\plus{} \\sin C} \\equal{} \\frac{2 \\sin \\frac{A}{2} \\cos \\frac{A}{2}}{2 \\sin \\frac{B\\plus{}C}{2} \\cos \\frac{B\\minus{}C}{2}}\\equal{}\\frac{\\sin \\frac{A}{2}}{\\cos \\frac{B\\minus{}C}{2}} \\ge \\sin \\frac{A}{2}$ with equality when $ B \\equal{} C$\r\n\r\nso $ \\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2} \\le \\frac{abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\le \\frac{1}{8}$ with equality when $ A\\equal{}B\\equal{}C$\r\n\r\nIt's know also $ \\sin^2 \\frac{A}{2} \\plus{} \\sin^2 \\frac{B}{2} \\plus{} \\sin^2 \\frac{C}{2} \\plus{} 2\\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2} \\equal{}1$ in every triangle.\r\n\r\nSo we obtain $ \\sin^2 \\frac{A}{2} \\plus{} \\sin^2 \\frac{B}{2} \\plus{} \\sin^2 \\frac{C}{2} \\ge \\frac{3}{4}$ with equality when $ A\\equal{}B\\equal{}C$.\r\n\r\nthat is equal to $ \\frac{9}{4} \\ge \\cos^2 \\frac{A}{2} \\plus{} \\cos^2 \\frac{B}{2} \\plus{} \\cos^2 \\frac{C}{2} \\ge 3 \\sqrt[3]{\\cos^2 \\frac{A}{2} \\cos^2 \\frac{B}{2} \\cos^2 \\frac{C}{2}}$ by AM-GM.\r\n\r\nso $ \\cos \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2} \\le \\frac{3\\sqrt{3}}{8}$ with equality when $ A\\equal{}B\\equal{}C$" } { "Tag": [ "parameterization", "Vieta", "quadratics", "algebra" ], "Problem": "Source: Bulgarian MO 1980\r\n1. For a real parameter $p \\neq 0$, let $x_1,x_2$ be the roots of the equation $x^2+px-\\frac{1}{2p^2}=0$. Prove that $x_1^4+x_2^4\\geq 2+\\sqrt{2}$.", "Solution_1": "Using Vieta's theorem, have $x_1+x_2=-p,x_1x_2=-\\frac{1}{2p^2}$ which yields:\r\n$x_1^2+x_2^2= (x_1+x_2)^2-2x_1x_2= p^2+\\frac{1}{p^2}$\r\nThen:\r\n$x_1^4+x_2^4= (x_1^2+x_2^2)^2-2x_1^2x_2^2$\r\n$= p^4+2+\\frac{1}{p^4}-\\frac{1}{2p^4}$\r\n$=2+p^4+\\frac{1}{2p^4}$\r\nNow let's prove that $p^4+\\frac{1}{2p^4} \\ge \\sqrt{2}$.\r\nWe have the ineq equivalent to:\r\n$p^4-\\sqrt{2}+\\frac{1}{2p^4} \\ge 0$\r\n$(p^2)^2-2*p^2*\\frac{1}{\\sqrt{2}p^2}+(\\frac{1} {\\sqrt{2}p^2})^2 \\ge 0$\r\n$(p^2-\\frac{1}{\\sqrt{2}p^2})^2 \\ge 0$\r\nTherefore $p^4+\\frac{1}{2p^4} \\ge \\sqrt{2}$ and:\r\n$x_1^4+x_2^4=2+p^4+\\frac{1}{2p^4} \\ge 2+\\sqrt{2}$\r\nQED", "Solution_2": "[quote=\"rem\"]Now let's prove that $p^4+\\frac{1}{2p^4} \\ge \\sqrt{2}$.\nWe have the ineq equivalent to:\n$p^4-\\sqrt{2}+\\frac{1}{2p^4} \\ge 0$\n$(p^2)^2-2*p^2*\\frac{1}{\\sqrt{2}p^2}+(\\frac{1} {\\sqrt{2}p^2})^2 \\ge 0$\n$(p^2-\\frac{1}{\\sqrt{2}p^2})^2 \\ge 0$\nTherefore $p^4+\\frac{1}{2p^4} \\ge \\sqrt{2}$[/quote]\r\nYou want to switch that around. To prove it, you take a true statement and then make it into the thing you want to prove.\r\nyou would say:\r\nNow let's prove that $p^4+\\frac1{2p^4}\\ge\\sqrt2$\r\nWe know $\\left(p^2-\\frac1{\\sqrt2p^2}\\right)^2\\ge0$\r\nso ${\\left(p^2\\right)}^2-2\\cdot p^2\\cdot\\frac1{\\sqrt2p^2}+{\\left(\\frac1 {\\sqrt{2}p^2}\\right)}^2\\ge0$\r\nthen $p^4-\\sqrt2+\\frac1{2p^4}\\ge0$\r\ntherefore $p^4+\\frac1{2p^4}\\ge\\sqrt2$.", "Solution_3": "You can also use the fact that every ineq is equivalent to the one below it.eg:\r\n\"$a^b+b^2=2ab$ is equivalent to $a^2-2ab+b^2 \\ge 0$ which is equivalent to $(a+b)^2 \\ge 0$\", which holds.QED\r\nSorry I didnot write the part \"the ineq is equivalent to\" in my solution. :blush: \r\nbtw the way \"the ineq is equivalent to\" is more effective, because if you are on a contest running out of time and have no time for rough work, you will use this method and hope you will arrive at the result insead of solving the ineq and doing it your way, which would take more time.\r\nBut if you have time, either way is fine. :)", "Solution_4": "OMG how close I was .......\r\nI gave up after I got $p^4+\\frac{1}{2p^4}\\geq \\sqrt{2}$.\r\nAnyway, thanks", "Solution_5": "is not $p^4 + \\frac{1}{2p^4} \\geq \\sqrt{2}$ by AMGM?", "Solution_6": "Here is my solution.\r\nSince $p\\neq 0,$\r\n$x^4=\\left(x^2+px-\\frac{1}{2p^2}\\right)\\left(x^2-px+p^2+\\frac{1}{2p^2}\\right)-\\left(p^3+\\frac{1}{p}\\right)x+\\frac{1}{4p^4}+\\frac{1}{2}.$\r\n\r\nSince $x_i\\ (i=1,2)$ is the solution of the given quadratic equation, we have $x_i^2+px_i-\\frac{1}{2p^2}=0.$ Thus\r\n$x_1^4+x_2^4=-\\left(p^3+\\frac{1}{p}\\right)(x_1+x_2)+2\\left(\\frac{1}{4p^4+\\frac{1}{2}}\\right)$\r\n$=-\\left(p^3+\\frac{1}{p}\\right)(-p)+2\\left(\\frac{1}{4p^4}+\\frac{1}{2}\\right)\\ \\because{x_1+x_2=-p}$\r\n$=p^4+\\frac{1}{2p^4}+2.$\r\nAs sen pointed, we can use $A.M.-G.M.$ inequality." } { "Tag": [ "algebra", "polynomial", "real analysis", "real analysis unsolved" ], "Problem": "Define the trigonometric polynomials $ P_0=Q_0=1, P_{n+1}(t)=P_n(t)+e^{i\\cdot 2^nt}Q_n(t), Q_{n+1}(t)=P_n(t)-e^{i\\cdot 2^n t}Q_n(t)$. Also, define $ f_n=\\frac{P_n-P_{n-1}}{2^n}$ and finally $f=\\sum_{n\\geq 1}{f_n}$. Prove that even if $f$ is $1/2$ holderian the partial Fourier series of $f$ does not converge normally (it can be proved-I posted it some time ago-that if $f$ is $k>1/2$ holderian then the Fourier series converges normally).", "Solution_1": "Rudin-Shapiro polynomials", "Solution_2": "Ok, and a proof? ;)" } { "Tag": [ "inequalities", "function", "calculus", "derivative", "inequalities unsolved" ], "Problem": "if $ a,b,c>0$ then \r\n\r\n$ \\sum_{cyc}\\left(\\frac{a^2}{b^2\\plus{}c^2}\\right) \\ge \\sum_{cyc}\\left(\\frac{a}{b\\plus{}c}\\right)$", "Solution_1": "[quote=\"Evariste-Galois\"]if $ a,b,c > 0$ then \n\n$ \\sum_{cyc}\\left(\\frac {a^2}{b^2 \\plus{} c^2}\\right) \\ge \\sum_{cyc}\\left(\\frac {a}{b \\plus{} c}\\right)$[/quote]\r\nThe inequality is equivalent to $ \\sum \\frac {ab(a \\minus{} b)^2(a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} ab \\plus{} bc \\plus{} ca)}{(b^2 \\plus{} c^2)(a^2 \\plus{} c^2)(b \\plus{} c)(a \\plus{} c)} \\geq 0$ which is obviously true !", "Solution_2": "[quote=\"rachid\"][quote=\"Evariste-Galois\"]if $ a,b,c > 0$ then \n\n$ \\sum_{cyc}(\\frac {a^2}{b^2 + c^2}) \\ge \\sum_{cyc}(\\frac {a}{b + c})$[/quote]\nThe inequality is equivalent to $ \\sum \\frac {ab(a - b)^2(a^2 + b^2 + c^2 + ab + bc + ca)}{(b^2 + c^2)(a^2 + c^2)(b + c)(a + c)} \\geq 0$ which is obviously true ![/quote]\r\n\r\nyes that's it Rachid :lol: \r\n\r\nme I stoped here :\r\n\r\n$ \\sum_{cyc}(ab(a - b)(\\frac {1}{(b^2 + c^2)(b + c)} - \\frac {1}{(a^2 + c^2)(a + c)}))\\ge 0$\r\n\r\nWhich is true obviously true !! because $ (a - b)$ and $ (\\frac {1}{(b^2 + c^2)(b + c)} - \\frac {1}{(a^2 + c^2)(a + c)}))$ have the same sign !!", "Solution_3": "Old problem of Vasc.", "Solution_4": "the function $ f(x) \\equal{} \\sum\\frac {a^x}{b^x \\plus{} c^x}$ is increasing because of th first derrivative so....................." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Given a triangle ABC, let the internal and the external angle bisectors of the angle CAB meet the circumcircle of triangle ABC at the points L and M, respectively (with $L \\neq A$ and $M \\neq A$, of course). Then let P and Q be the reflections of the point A in the points M and L. The circumcircles of triangles AMQ and ALP intersect each other at the point A and at some other point, which we call S. Show that AS || BC.\r\n\r\n Darij", "Solution_1": "Actually a little bit more general result holds. If P and Q are points so that AM/AP = AL/AQ then AS//BC.\r\n\r\nProof.\r\n\r\nAL and AM are orthogonals, therefore ML, PL and MQ are diameters, i.e. O_1 (center of c(PAL)) and O_2 (center of c(MAQ)) stay on PL and MQ respectively.\r\n\r\nGiven that ML//PQ (from ipothesys AM/AP = AL/AQ) then O_1O2//ML. But ML is orthogonal to BC, by simmetrie, and O_1O_2 is orthogonal to AS also, then the thesis.", "Solution_2": "The problem actually holds for any $ P$ and $ Q$ | $ P \\in AM$, $ L \\in AL$ and $ PQ$ || $ ML$, we shall get $ \\triangle OO_1O_2 \\sim \\triangle AML$ and $ OO_1 \\parallel AM$ , $ OO_2 \\parallel AL$, hence $ O_1O_2 \\parallel ML$, done.\r\n\r\nObservation: $ PL$ and $ MQ$ concur on $ OA$.\r\n\r\nBest regards\r\nsunken rock" } { "Tag": [ "geometry", "rectangle", "3D geometry", "search", "number theory", "prime numbers" ], "Problem": "I am taking a college math class to become an elementary school teacher. I am currently working on an assignment and am struggling with finding the \"shortcut\"\r\nHere is the question:\r\n\r\n1. Which numbers less than 100 have an odd number of factors? Explain how rectangles help to figure out the answer.\r\n2. Which numbers less than 100 have EXACTLY three factors? Explain.\r\n3. Explain how to find all numbers less than 1000 with exactly three factors. ( I just have to explain it, not write out the numbers themselves)\r\n4. Which numbers less than 100 have EXACTLY four factors? Explain.\r\n5. Explain how to find all numbers less than 1000 with exactly four factors. (same as #3)\r\n\r\nI know how to solve this the long way by writing out all the numbers, but what is the trick here? Help!", "Solution_1": "Sorry mine is without rectangles.\r\n1) Those are only squares, if you look at formula for the number of factors: $ d(n)\\equal{}(x_1\\plus{}1)(x_2\\plus{}1)...(x_k\\plus{}1)$ if $ n\\equal{}p_1^{x_1}p_2^{x_2}...p_k^{x_k}$ where $ p_1,p_2,...p_k$ are different primes, which then gives you that all $ x_1,x_2,...,x_n$ are even.\r\n2) The formula above gives you those are squares of primes\r\n3) = 2)\r\n4) Again, the formula above gives you cubes of the primes, or product of 2 different primes\r\n5) = 4)", "Solution_2": "1. The numbers that have an odd number of factors are the ones that are the squares. This is because the square root of these numbers doesn't have a different number to multiply with to get the square. \r\n\r\n2. The numbers that have three factors are the ones that have the factors; $ 1, x,$ and $ x^2$. $ x$ is prime though, because otherwise we could factor out $ x$ and get more factors. The numbers that are prime and less than ten are 2, 3, 5, and 7. The squares of these are 4, 9, 25, and 49. \r\n\r\n3. Well we can start off by finding the largest perfect square less than 1000. This is 961 which is $ 31^2$. So we are looking for the primes less than or equal to 31. Once we have found these primes, then we have to square them to get our answers. \r\n\r\n4. Alright, this is a tricky problem. So I now think it is time to introduce a good trick of finding the number of factors of a number. First you factor the number into the form $ 2^a 3^b 5^c....$. We do this, because we want to know the exponents of the primes. Now this is how we find the factors; \r\n\r\nWe add one to all the exponents of the primes, and then we multiply all these numbers together to get; \r\n\r\n$ (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1)...$. This is the number of factors. So there are four factors so; \r\n\r\n$ (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1)... \\equal{} 4$\r\n\r\n$ 4$ can factor into $ (1)(4)$, or $ (2)(2)$. For the first case, we know that if the exponent of one of the primes is $ x$, then \r\n$ x \\plus{} 1 \\equal{} 1$, and if $ y$ is the exponent of one of the primes, then $ y \\plus{} 1 \\equal{} 4$. So $ x \\equal{} 0$, and $ y \\equal{} 3$. If we look at $ x$, we notice that anything to the power of $ 0$ except $ 0$ is 1. We don't need to worry about $ 0$ though, because $ 0$ is composite. So that means any number that has four factors has to be a perfect cube, if this is the case. Now lets look at the other case. $ (2)(2)$. $ x \\plus{} 1 \\equal{} 2$, $ y \\plus{} 1 \\equal{} 2$. So $ x \\equal{} y \\equal{} 1$. So this means that the number can be expressed as; \r\n\r\n$ a^1 b^1$ or $ c^3$ if it has four factors. So it is either the product of two distinct prime numbers, or the cube of a prime number. \r\n\r\n\r\n5. To find all the numbers less than 1000 that have four factors, you have to search for a while by plugging primes into \r\n$ a^1 b^1$, and $ c^3$. \r\n\r\n\r\nHope that helped." } { "Tag": [ "calculus", "integration", "logarithms", "limit", "function", "algebra", "domain" ], "Problem": "find :\r\n$ p.v. \\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 \\minus{} x)^{3}}dx$", "Solution_1": "hello, $ \\int_{0}^{\\infty}\\frac{(\\ln(x))^2}{(1\\minus{}x)^3}dx\\equal{}\\frac{1}{3}\\pi^2\\plus{}\\pi$.\r\nSonnhard.", "Solution_2": "Hello Sonnhard.\r\nWhy do you think so? My calculations shows that answer is $ \\pi^{2}$. (but I'm not sure, I solved it , but my solution is very big since all complex analysis lays there :( :( )", "Solution_3": "Oops... my calculation shows that the answer is $ \\frac {\\pi^2}{3}$.\r\n\r\n[b]Step 1)[/b] $ PV \\int_{0}^{\\infty} \\frac {\\ln^2 x}{(1 - x)^3} \\, dx = PV \\int_{\\mathbb{R}} \\frac {t^2 e^{t}}{(1 - e^t )^3} \\, dt$.\r\n\r\n[hide=\"Solution\"]Put $ I_{\\epsilon} = \\int_{0}^{1 - \\epsilon} + \\int_{1 + \\epsilon}^{\\infty} \\frac {\\ln^2 x}{(1 - x)^3} \\, dx = \\int_{ - \\infty}^{\\ln(1 - \\epsilon)} + \\int_{\\ln(1 + \\epsilon)}^{\\infty} \\frac {t^2 e^{t}}{(1 - e^t )^3} \\, dt.$\n\nIf we let $ f(t) = \\frac {t^2 e^{t}}{(1 - e^t )^3}$, then $ \\lim_{t\\to 0} t f(t) = 1$. So $ f(t) \\approx \\frac {1}{t}$ near $ t = 0$. Note that\n\n$ I_{\\epsilon} = \\int_{\\mathbb{R} - ( - \\epsilon, \\epsilon)} f(t) \\, dt + E(\\epsilon)$\n\nwhere $ E(\\epsilon) = \\int_{ - \\epsilon}^{\\ln(1 - \\epsilon)} + \\int_{\\ln(1 + \\epsilon)}^{\\epsilon} f(t) \\, dt$. Since $ |E(\\epsilon)| \\leq C \\ln \\left| \\frac {\\ln(1 - \\epsilon)}{\\ln(1 + \\epsilon)} \\right|$ for sufficiently small $ t$, $ E(\\epsilon) \\to 0$ as $ \\epsilon \\to 0$. This proves the claim.[/hide]\r\n\r\n[b]Step 2)[/b] As you can confirm,\r\n\r\n$ I = PV \\int_{\\mathbb{R}} \\frac {t^2 e^{t}}{(1 - e^t )^3} \\, dt = \\int_{0}^{\\infty} \\frac {t^2 (e^{ - t} - e^{ - 2t})}{(1 - e^{ - t})^3} \\, dt.$\r\n\r\nNote that $ \\frac {1}{(1 - x)^3} = \\sum_{n = 1}^{\\infty} \\frac {n(n + 1)}{2} x^{n - 1}$ for $ |x| < 1$. Then\r\n\r\n\\begin{eqnarray*} I & = & \\sum_{n = 1}^{\\infty} \\frac {n(n + 1)}{2} \\int_{0}^{\\infty} t^2 (e^{ - nt} - e^{ - (n + 1)t}) \\, dt \\\\\r\n& = & \\sum_{n = 1}^{\\infty} n(n + 1) \\left( \\frac {1}{n^3} - \\frac {1}{(n + 1)^3} \\right) \\\\\r\n& = & \\sum_{n = 1}^{\\infty} \\left( \\frac {1}{n} + \\frac {1}{n^2} - \\frac {1}{n + 1} + \\frac {1}{(n + 1)^2} \\right) \\\\\r\n& = & 2\\zeta(2) \\\\\r\n& = & \\frac {\\pi^2}{3}. \\end{eqnarray*}", "Solution_4": "It looks like true :lol: , I'll try to check my solution and (also your).", "Solution_5": "The integral equals $ \\int_{0}^{\\infty }{\\frac{x^{2}e^{x}}{\\left( 1\\minus{}e^{x} \\right)^{2}}\\,dx},$ which gives sos440's result.", "Solution_6": "hello, yes, my result is wrong, the result $ \\frac{\\pi^2}{3}$ is correct.\r\nSonnhard.", "Solution_7": "here is what I do:\r\nConsider function :\r\n$ f(z) = \\frac {(\\ln(z) - \\pi i)^{3}}{(1 - z)^{3}}$\r\nand integrate it over the contour .\r\nnote that :\r\n$ \\int_{\\gamma}f(z) = - 3\\pi^{3}i - 6\\pi^{2}$ by residue .\r\non big and small cicrcle integral is equal to $ 0$, \r\nlet $ A$ will be upper contour and $ B$ will be down .\r\nthus we have \r\n$ \\int_{A}\\frac {(\\ln(z) - \\pi i)^{3}}{(1 - z)^{3}} + \\int_{B}\\frac {(\\ln(z) + \\pi i)^{3}}{(1 - z)^{3}}$\r\n=\r\n$ \\int_{A}\\frac {(\\ln(z) - \\pi i)^{3}}{(1 - z)^{3}} - \\int_{A}\\frac {(\\ln(z) + \\pi i)^{3}}{(1 - z)^{3}}$\r\n=\r\n$ \\int_{A}\\frac { - 6i\\pi\\ln^{2}z + 2i\\pi^{2}}{(1 - z)^{3}}$ \r\n$ \\int_{A}\\frac {2i\\pi^{2}}{(1 - z)^{3}} = - i\\pi^{2}$ standart integral.\r\nand ${ - 6i\\pi\\int_A}\\frac {\\ln^{2}z}{(1 - z)^{3}}dz = p.v - 6i\\pi\\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 - x)^{3}}dx - 6i\\pi\\int_{M}\\frac {\\ln^{2}z}{(1 - z)^{3}}dz$\r\nwhere $ M$ is small half circle.\r\nthen :\r\n$ \\int_{M}\\frac {\\ln^{2}z}{(1 - z)^{3}}dz = i \\pi res_{z = 1}\\frac {\\ln^{2}z}{(1 - z)^{3}} = i\\pi$ by half residue \r\nthus :\r\n $ p.v - 6i\\pi\\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 - x)^{3}}dx = - 3\\pi^{3}i + i\\pi^{2} - 6\\pi^{2} - 6\\pi^{2} = - 2\\pi^{3}i - 12\\pi^{3}$\r\nand \r\n$ \\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 - x)^{3}}dx = \\frac {\\pi^{2}}{3} - 2i\\pi^{3}$\r\nreal part is $ \\frac {\\pi^{2}}{3}$ it's answer yes I agree with sos440, but Why I have $ Im$ ?? why $ - 2i\\pi^{3}$ ??", "Solution_8": "nice integral! i try to come up with another solution", "Solution_9": "[quote=\"Extremal\"]find :\n$ p.v. \\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 \\minus{} x)^{3}}dx$[/quote]\r\n\r\nhold on, i see people are posting solutions, but i don't think the integral even exists.\r\n\r\nwhen you have an integral like $ \\int_0^\\infty f(x)\\;\\ln^2\\,{x}\\;dx$\r\n\r\nthis integral will exist only so long as $ f(z)$ has a double pole at $ z \\equal{} 1$, that's because the principal branch of $ \\ln\\,z$ has a simple $ 0$ at $ z \\equal{} 1$. this integral still has a meaning and we can evaluate it in the complex domain by taking a detour at $ z \\equal{} 1$ using a small semicircle.\r\n\r\nbut what you have is $ f(z)$ with a [u]triple[/u] pole at $ z \\equal{} 1$, and the numerator is only $ \\ln^{2}\\,z$, so the integral will blow up. \r\n\r\nyes, one can do\r\n$ \\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 \\minus{} x)^{2}}dx$\r\nor $ \\int_{0}^{\\infty}\\frac {\\ln^{3}x}{(1 \\minus{} x)^{3}}dx$\r\n\r\nbut not $ \\int_{0}^{\\infty}\\frac {\\ln\\,x}{(1 \\minus{} x)^{2}}dx$\r\nor $ \\int_{0}^{\\infty}\\frac {\\ln^{2}x}{(1 \\minus{} x)^{3}}dx$", "Solution_10": "Hello Misan. I agree with you about pole, Therefore I wrote $ P.V.$ this mean principal value of integral :) .\r\nor more generaly it's same as to find principal value of \r\n$ \\lim_{\\epsilon \\to 0} \\int_{\\mathbb{R}^{\\plus{}}\\minus{}(1\\minus{}\\epsilon, 1\\plus{}\\epsilon)} ...$", "Solution_11": "[quote=\"Extremal\"]\nreal part is $ \\frac {\\pi^{2}}{3}$ it's answer yes I agree with sos440, but Why I have $ Im$ ?? why $ - 2i\\pi^{3}$ ??[/quote]\r\n\r\nKnow what, I really feel I have to apologize for injecting so much complex analysis in here but it's a beautiful problem and sides, Extremal's having problems with it and well . . . I'm helping him. :D \r\n\r\nFirst let $ log(z) = ln(r) + i\\Theta,\\; 0\\leq \\Theta < 2\\pi$. This is a branch of $ \\text{log}$ with the branch cut along the positive real axis. Now, referring to the diagram below, the contour $ \\Gamma$ encloses no poles and so:\r\n\\[ \\int\\limits_{\\Gamma}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz = 0\r\n\\]\r\nAs Extremal stated above, the integral along the outer circular contour goes to zero as this contour is expanded and likewise goes to zero as the contour around the origin decreases. So we're left with:\r\n\\[ \\int\\limits_{A}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz + \\int\\limits_{B}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz = 0\r\n\\]\r\nNow:\r\n\\[ \\int\\limits_{A}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz = \\text{P.V.}\\int_0^{\\infty}\\frac {(\\ln(x) - \\pi i)^3}{(1 - x)^3}dx + \\lim_{\\rho\\to 0}\\int\\limits_{Green}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz\r\n\\]\r\n\r\n\\[ \\int\\limits_{B}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz = \\text{P.V.}\\int_{\\infty}^{0}\\frac {(\\ln(x) + \\pi i)^3}{(1 - x)^3}dx + \\lim_{\\rho\\to 0}\\int\\limits_{\\text{Blue}}\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}dz\r\n\\]\r\nWe can use the Residue Theorem to calculate the integral over the Green and Blue contours:\r\n\\begin{align*} \\lim_{\\rho\\to 0}\\int\\limits_{Green}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz & = - \\pi i\\mathop\\text{Res}\\limits_{z = 1}\\left\\{\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}\\right\\} \\\\\r\n& = - \\pi i\\bigg(3\\pi i - 3\\pi^2/2\\bigg) = 6\\pi^2 + 3\\pi^3 i/2 \\end{align*}\r\n\r\n\\begin{align*} \\lim_{\\rho\\to 0}\\int\\limits_{Blue}\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}dz & = - \\pi i\\mathop\\text{Res}\\limits_{z = 1}\\left\\{\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}\\right\\} \\\\\r\n& = - \\pi i\\bigg( - 3\\pi i - 3\\pi^2/2\\bigg) = - 6\\pi^2 + 3\\pi^3 i/2 \\end{align*}\r\ntherefore:\r\n\\[ \\lim_{\\rho\\to 0}\\int\\limits_{Green}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz + \\lim_{\\rho\\to 0}\\int\\limits_{Blue}\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}dz = 3\\pi^3 i\r\n\\]\r\nCombining this result with the two principal-valued integrals:\r\n\\[ \\text{P.V.}\\int_0^{\\infty}\\frac {(\\ln(x) - \\pi i)^3}{(1 - x)^3}dx - \\text{P.V.}\\int_{0}^{\\infty}\\frac {(\\ln(x) + \\pi i)^3}{(1 - x)^3}dx + 3\\pi^3 i = 0\r\n\\]\r\nor:\r\n\\[ \\text{P.V.}\\int_0^{\\infty}\\frac {(\\ln(x) - \\pi i)^3 - (\\ln(x) + \\pi i)^3}{(1 - x)^3}dx = - 3\\pi^3 i\r\n\\]\r\nsimplifying:\r\n\\[ \\text{P.V.}\\int_0^{\\infty}\\frac {2\\pi^3 i - 6\\pi i \\ln^2(x)}{(1 - x)^3}dx = - 3\\pi^3 i\r\n\\]\r\nThen finally:\r\n\\[ \\text{P.V.}\\int_0^{\\infty}\\frac {2\\pi^3 i}{(1 - x)^3}dx - \\text{P.V.}\\int_0^{\\infty}\\frac {6\\pi i\\ln^2(x)}{(1 - x)^3}dx = - 3\\pi^3 i\r\n\\]\r\n\r\n\\[ - \\pi^3 i - \\text{P.V.}\\int_0^{\\infty}\\frac {6\\pi i\\ln^2(x)}{(1 - x)^3}dx = - 3\\pi^3 i\r\n\\]\r\n\r\n\\[ \\text{P.V.}\\int_0^{\\infty}\\frac {\\ln^2(x)}{(1 - x)^3}dx = \\frac {2\\pi^3 i}{6\\pi i} = \\frac {\\pi^2}{3}\r\n\\]", "Solution_12": "yes, :) . I get $ Im$ maybe I don't calculate right half residue.", "Solution_13": "[quote=\"shawtoos\"]\nWe can use the Residue Theorem to calculate the integral over the Green and Blue contours:\n\\begin{align*} \\lim_{\\rho\\to 0}\\int\\limits_{Green}\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}dz & = - \\pi i\\mathop\\text{Res}\\limits_{z = 1}\\left\\{\\frac {(\\text{log}(z) - \\pi i)^3}{(1 - z)^3}\\right\\} \\\\\n& = - \\pi i\\bigg(3\\pi i - 3\\pi^2/2\\bigg) = 6\\pi^2 + 3\\pi^3 i/2 \\end{align*}\n\n\\begin{align*} \\lim_{\\rho\\to 0}\\int\\limits_{Blue}\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}dz & = - \\pi i\\mathop\\text{Res}\\limits_{z = 1}\\left\\{\\frac {(\\text{log}(z) + \\pi i)^3}{(1 - z)^3}\\right\\} \\\\\n& = - \\pi i\\bigg( - 3\\pi i - 3\\pi^2/2\\bigg) = - 6\\pi^2 + 3\\pi^3 i/2 \\end{align*}\n[/quote]\r\n\r\nI think this is a problem (not to mention combining two principal-value integrals further down in the analysis but I digress). The particular theorem used to make those equalities is only valid for a simple pole and those aren't simple. I don't know why everything falls into place; something ain't right I think.", "Solution_14": "just so you know, the contour you are using will not work, [u]where[/u] is your branch cut ? you cannot have your branch cut on $ [0,\\infty)$ 'cause you have a pole there. it has to be defined somewhere else e.g. $ ( \\minus{} \\infty,0],\\quad\\text{or}\\quad ( \\minus{} \\infty\\imath, 0\\imath],\\quad\\text{or}\\quad [0\\imath, \\infty\\imath)$", "Solution_15": "I see. therefore you need to choose such contour as mine. Choose my contour and then will be everything good. :) you will have simple pole.", "Solution_16": "no, your contour will not work. your contour is no different from the one that shawtoos used. in fact, there is no contour on this planet that will work for the $ f(z)$ you have posed to solve the integral.", "Solution_17": "No misan, I don't need brunch cut since I integrate it on countour and therefore I consider my function In area (inside my contour) .\r\nI know that there is pole on $ [0,\\infty]$ maybe I must explain well every my step.. if you try to write everything well then will be good.", "Solution_18": "Here is red area where I consider my function .I'm sorry about my picture, since it not looks good. :blush: )", "Solution_19": "For integrals like this, the standard contour is real axis from -R to R with two little semicircle around x = 0 and x= 1, and with a big semicircle around x = 0 with radius R. All semicircles are on the upper half plane.", "Solution_20": "extremal, look at your contour, you should get\r\n$ \\oint _{|z| \\equal{} \\delta,\\delta\\to 0}\\;f(z)\\;dz\\; \\plus{} \\;\\int_0^\\infty \\frac {(\\ln\\,x \\minus{} \\pi i)^{3}}{(1 \\minus{} x)^{3}} \\;dx$\r\n\r\n$ \\plus{} \\oint_{|z| \\equal{} R,R\\to\\infty}\\;f(z)\\; dz\\; \\plus{} \\int_\\infty^0 \\frac {(\\ln(x\\;e^{2\\pi i}) \\minus{} \\pi i)^{3}}{(1 \\minus{} x\\;e^{2\\pi i})^{3}}\\;e^{2\\pi i}\\;dx \\quad \\equal{} \\quad \\minus{} 3\\pi^{3}i \\minus{} 6\\pi^{2}$\r\n\r\n (contributions from the $ 2$ semi-circles at $ z \\equal{} 1$ cancel each other out...)\r\n\r\nnow look at the $ 4$ terms on the LHS, clearly one of them does [b]not[/b] go to $ 0$.... can you see which one ?", "Solution_21": "I'd like to better understand this problem. I'm not trying to force-fit it to my whims but only to understand precisely what the problem is. Letting $ \\displaystyle f_1(x)\\equal{}\\frac{(\\ln(x)\\minus{}\\pi i)^3}{(1\\minus{}x)^3}$ and $ \\displaystyle f_2(x)\\equal{}\\frac{(\\ln(x)\\plus{}\\pi i)^3}{(1\\minus{}x)^3}$ then without regard to its relevance to the above problem, does the following limit exists:\r\n\r\n\\[ \\lim_{\\epsilon\\to 0}\\left\\{\\int_0^{1\\minus{}\\epsilon} f_1(x)dx\\plus{}\\int_{1\\plus{}\\epsilon}^{\\infty}f_1(x)dx\\plus{}\\int_{\\infty}^{1\\plus{}\\epsilon} f_2(x)dx\\plus{}\\int_{1\\minus{}\\epsilon}^0 f_2(x)dx\\right\\}\\]\r\n\r\nIt looks to me the limit does exists in the Principal Value sense and approaches $ \\minus{}3\\pi^3 i$, that is, the unbounded terms, while approaching infinity and minus infinity, nevertheless cancel. Is this not true?\r\n\r\nIf it is true, then I believe an argument could be made to solve the original integral using the contour I posted.", "Solution_22": "I understand. I must explain every my step more clearly (I do it only once more, since Must prepare to my exams , and hence I have no free much time.)\r\nConsider contour such as on picture (please Look at picture well :( ).\r\n\r\nMy picture is not very well (since I can't draw it well) I note that on the picture points $ (A,B,C,D,H,F,E,G)$ lie on same line $ (0,\\infty)$\r\nbut also note that $ A \\neq H$ but they are very close, such that $ dist (A,H) \\equal{} 0$ :lol: .\r\nalso for points $ (B,F)$ , $ (C,E)$ and $ (D,G)$. OK.\r\nNow I try to explain other things on my contour.\r\n $ |AB| \\equal{} |HF|$ or dist. is equal. \r\n$ |CD| \\equal{} |EG|$\r\nlet's point $ 1$ called point $ 1$ :lol: and also point $ 0$ called point $ 0$.\r\n\r\nand fix some small $ \\epsilon > 0$ and big $ R > 0$ say :\r\n$ |1F| \\equal{} |1E| \\equal{} |1B| \\equal{} |1C| \\equal{} |0H| \\equal{} |0A| \\equal{} \\epsilon$\r\nalso\r\n$ |0D| \\equal{} |0G| \\equal{} R$\r\nevery one understand that $ R$ is radius of big corcle , and $ \\epsilon$ is radius of small circles.\r\nnow I can \r\n consider function :\r\n\r\n\r\n$ f(z) \\equal{} \\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}$ where $ \\lim_{\\delta \\to 0} \\ln (1 \\plus{} \\delta\\i) \\equal{} 0$ (or argument on upper line is equal to $ 0$.)\r\n in area of contour it's [b]well defined[/b] (I think every one understand: what does it mean \"well defined\". ) function .\r\nlets' called this contour $ L$ and integrate on this contour our function $ f(z)$\r\n$ \\int\\limits_{L}f(z)dz$ \r\nfrom complex analysis and residue theorem it's wellknown that integral will be equal to :\r\n\\[ \\int\\limits_{L}f(z)dz \\equal{} 2 \\pi i Rez _{z \\equal{} 1}f(z)\r\n\\]\r\nsince only point $ 1$ is in area of our contour. OK>\r\nNow consider our integral by next way (as sum of the contours.)\r\n\\[ \\int\\limits_{L}f(z)dz \\equal{} \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int\\limits_{Circle(D,G)}f(z)\r\n\\]\r\n\r\n\\[ \\plus{} \\int\\limits_{G,E,F,H}\\frac {(\\ln(z) \\plus{} 2\\pi i \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int\\limits_{circle(H,A)}f(z)\r\n\\]\r\nfirst simple question: why I have :\r\n\\[ \\int\\limits_{G,E,F,H}\\frac {(\\ln(z) \\boxed{ \\plus{} 2\\pi i} \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nAnswer: since function $ ln (z) \\equal{} \\ln |z| \\plus{} i arg(z)$\r\nhence after contour $ Circle(D,G)$ argument of $ z$ will be $ 2\\pi i$ hence I have $ \\plus{} 2\\pi i$\r\nOK.\r\nNow I use something simple, I say that : if I change orientation of contour then there will changed sign, or\r\n\\[ \\int\\limits_{G,E,F,H}\\frac {(\\ln(z) \\plus{} 2\\pi i \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} 2\\pi i \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{}\r\n\\]\r\n\r\n\\[ \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\n.\r\nOk, Now let's note that\r\n\\[ \\int\\limits_{Circle(D,G)}f(z) \\to 0\r\n\\]\r\nas $ R \\to \\infty$ or more exact it's something $ \\equal{} O(\\frac {\\ln^{3} R}{R^{2}})$ for big $ R$\r\nnow note that :\r\n\\[ \\int\\limits_{circle(H,A)}f(z) \\equal{} O(\\epsilon \\ln^{3} \\epsilon)\r\n\\]\r\nfor small $ \\epsilon$\r\n\r\n(also note that $ \\lim_{\\epsilon \\to 0} O(\\epsilon \\ln^{3} \\epsilon) \\equal{} 0$)\r\nfinally we have :\r\n\\[ .... \\equal{} \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} O(\\frac {\\ln R}{R^{2}}) \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} O(\\epsilon \\ln \\epsilon)\r\n\\]\r\nLet's denote by\r\n\\[ W(\\epsilon,R) \\equal{} O(\\epsilon \\ln^{3} \\epsilon) \\plus{} O(\\frac {\\ln^{3} R}{R^{2}})\r\n\\]\r\nand note that \r\n$ \\lim\\limits_{\\epsilon \\to 0, R \\to \\infty } W(\\epsilon,R) \\equal{} 0$\r\nOK.\r\nor we have\r\n\\[ \\equal{} \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} W(\\epsilon,R)\r\n\\]\r\nOK.\r\nNow ()at first consider :\r\n\\[ \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nthis integral .\r\nNow I divide contour $ (A,B,C,D)$ by parts such that :\r\n\\[ \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\int\\limits_{A,B}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int\\limits_{circle(B,C)}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int\\limits_{C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nnote that \r\n$ |AB| \\equal{} (\\epsilon,1 \\minus{} \\epsilon)$ and $ |CD| \\equal{} ( 1 \\plus{} \\epsilon,R)$\r\nthus we have\r\n\\[ \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon}\\frac {(\\ln(x) \\minus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\plus{} \\int\\limits_{circle(B,C)}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int_{1 \\plus{} \\epsilon}^{R}\\frac {(\\ln(x) \\minus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx\r\n\\]\r\nsimilary for:\r\n\\[ \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\minus{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon}\\frac {(\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\minus{} \\int\\limits_{circle(F,E)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int_{1 \\plus{} \\epsilon}^{R}\\frac {(\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx\r\n\\]\r\nafter summation this two results we will have :\r\n\\[ \\int\\limits_{A,B,C,D}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int\\limits_{H,F,E,G}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} W(\\epsilon,R) \\equal{}\r\n\\]\r\n\r\n\\[ \\equal{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon}\\frac {(\\ln(x) \\minus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\plus{} \\int\\limits_{circle(B,C)}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\plus{} \\int_{1 \\plus{} \\epsilon}^{R}\\frac {(\\ln(x) \\minus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\minus{}\r\n\\]\r\n\r\n\\[ \\minus{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon}\\frac {(\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\minus{} \\int\\limits_{circle(F,E)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int_{1 \\plus{} \\epsilon}^{R}\\frac {(\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx\r\n\\]\r\nNote that\r\n\\[ \\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}} \\equal{} \\frac {(\\ln^{3}(z) \\minus{} 3\\ln^{2}(z) \\pi i \\minus{} 3\\pi^{2}\\ln (z) \\plus{} i\\pi^{3} )}{(1 \\minus{} z)^{3}}\r\n\\]\r\n\r\n\\[ \\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}} \\equal{} \\frac {(\\ln^{3}(z) \\plus{} 3\\ln^{2}(z) \\pi i \\minus{} 3\\pi^{2}\\ln (z) \\minus{} i\\pi^{3} )}{(1 \\minus{} z)^{3}}\r\n\\]\r\ntherefore :\r\n\\[ \\int_{\\epsilon}^{1 \\minus{} \\epsilon}\\frac {(\\ln(x) \\minus{} \\pi i )^{3} \\minus{} (\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\equal{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\frac { \\minus{} 6\\ln^{2}(z) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3} }dx\r\n\\]\r\nand\r\n\\[ \\int_{1 \\plus{} \\epsilon}^{R}\\frac {(\\ln(x) \\minus{} \\pi i )^{3} \\minus{} (\\ln(x) \\plus{} \\pi i )^{3}}{(1 \\minus{} x)^{3}}dx \\equal{} \\int_{1 \\plus{} \\epsilon}^{R} \\frac { \\minus{} 6\\ln^{2}(z) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx\r\n\\]\r\nfinally we obtain :\r\n\\[ \\equal{} \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\frac { \\minus{} 6\\ln^{2}(x) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\frac { \\minus{} 6\\ln^{2}(x) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx \\plus{} \\int\\limits_{circle(B,C)}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int\\limits_{circle(F,E)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nsince differenc between points $ (B,F)$ and also $ (C,E)$ is equal to $ 0$ then I can say that :\r\n\\[ \\minus{} \\int\\limits_{circle(F,E)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\minus{} \\int\\limits_{circle(B,F)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nand therefore :\r\n\\[ \\int\\limits_{circle(B,C)}\\frac {(\\ln(z) \\minus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\minus{} \\int\\limits_{circle(F,E)}\\frac {(\\ln(z) \\plus{} \\pi i )^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\int\\limits_{circle(B,C)}\\frac { \\minus{} 6\\ln^{2}(z) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} z)^{3}}dz\r\n\\]\r\nnote that :\r\n\\[ \\int\\limits_{circle(B,C)}\\frac { \\minus{} 6\\ln^{2}(z) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\equal{} \\int\\limits_{circle(B,C)}\\frac { \\minus{} 6\\ln^{2}(z) \\pi i}{(1 \\minus{} z)^{3}}dz \\plus{} \\int\\limits_{circle(B,C)}\\frac {2i\\pi^{3}}{(1 \\minus{} z)^{3}}dz \\equal{} \\minus{} \\pi i Rez_{z \\equal{} 1} \\frac { \\minus{} 6\\ln^{2}(z) \\pi i}{(1 \\minus{} z)^{3}} \\plus{} i\\pi^{3}\\frac {1}{(1 \\minus{} z)^{2}}|_{1 \\minus{} \\epsilon}^{1 \\plus{} \\epsilon} \\equal{} 6\\pi^{2}\r\n\\]\r\nafter note that :\r\n\\[ \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\frac { \\minus{} 6\\ln^{2}(x) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\frac { \\minus{} 6\\ln^{2}(x) \\pi i \\plus{} 2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx \\equal{} \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\plus{} \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac {2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx\r\n\\]\r\nnote that\r\n\\[ \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac {2i\\pi^{3}}{(1 \\minus{} x)^{3}}dx \\equal{} \\minus{} \\frac {i\\pi^{3}}{(1 \\minus{} \\epsilon)^{2}}\r\n\\]\r\nthus we have :\r\n\\[ \\equal{} \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\minus{} \\frac {i\\pi^{3}}{(1 \\minus{} \\epsilon)^{2}}\r\n\\]\r\nfinally we get :\r\n\\[ \\int_{L}f(z)dz \\equal{} 6\\pi^{2} \\plus{} \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\minus{} \\frac {i\\pi^{3}}{(1 \\minus{} \\epsilon)^{2}} \\plus{} W(\\epsilon,R) \\equal{} 2 \\pi i Rez _{z \\equal{} 1}f(z)\r\n\\]\r\nNow let $ R\\to \\infty$ and $ \\epsilon \\to 0$ then\r\n\\[ \\left( \\int_{\\epsilon}^{1 \\minus{} \\epsilon} \\plus{} \\int_{1 \\plus{} \\epsilon}^{R} \\right) \\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\minus{} \\frac {i\\pi^{3}}{(1 \\minus{} \\epsilon)^{2}} \\plus{} W(\\epsilon,R) \\to P.V. \\int_{0}^{\\infty}\\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\minus{} i\\pi^{3}\r\n\\]\r\nand finally :\r\n\\[ \\int_{L}f(z)dz \\equal{} P.V. \\int_{0}^{\\infty}\\frac { \\minus{} 6\\ln^{2}(x) \\pi i}{(1 \\minus{} x)^{3}}dx \\minus{} i\\pi^{3} \\plus{} 6\\pi^{2} \\equal{} 2 \\pi i Rez _{z \\equal{} 1}f(z)\r\n\\]\r\nthus we obtain\r\n\\[ P.V \\int_{0}^{\\infty}\\frac {\\ln^{2}(x)}{(1 \\minus{} x)^{3}}dx \\equal{} \\frac {2 \\pi i Rez _{z \\equal{} 1}f(z) \\plus{} i\\pi^{3} \\minus{} 6\\pi^{2}}{ \\minus{} 6\\pi i} \\equal{} \\frac {\\pi^{2}}{3}\r\n\\]\r\n:sleeping:", "Solution_23": "Isn't the solution (pi^2)/3 + pi ??", "Solution_24": "I yield. With some effort on my part I was able to work through sos440's solution above. It's quite elegant I believe (where's the emotiocon for \"we're not worthy\"?) I'm sorry for cluttering-up this thread and the forum with mumbo-jumbo complex analysis. Sorry Extremal and I'm sure you'll do fine on your test :)", "Solution_25": "[quote=\"shawtoos\"]... I'm sure you'll do fine on your test :)[/quote]\r\nthank you shawtoos :roll: \r\nsometimes explanations looks very hard (or very big)... in this situation it easy to see solution then to try explain . :|" } { "Tag": [ "search", "function", "LaTeX", "number theory", "number theory unsolved" ], "Problem": "If $a,b,c$ are positive integers and\r\n$05 threads, I merged them now.", "Solution_5": "I see. I hope in the future I'll be a better searcher. :D \r\n\r\nBut, in Myth's solution, case 2, why is $\\frac{b^{2}-d}{a} u'x \\equal{} \\frac{u^4\\minus{}u^2}{2u^2\\minus{}2u\\minus{}2} \\equal{}> \\frac{2u^2\\minus{}2u\\minus{}2}{u^2(u\\minus{}1)(u\\plus{}1)}du \\equal{} \\frac{dx}{x}$\r\n\r\n$ \\equal{}> \\left(\\frac{2}{u} \\plus{} \\frac{2}{u^2} \\plus{} \\frac{1}{1\\minus{}u} \\minus{} \\frac{1}{1\\plus{}u}\\right)du \\equal{} \\frac{dx}{x}$\r\n\r\n$ \\equal{}> \\ln\\left|\\frac{u^2}{1\\minus{}u^2}\\right| \\minus{} \\frac{2}{u} \\equal{} \\ln|Cx|$.\r\n\r\nNow make u = y/x to get\r\n\r\n$ \\ln\\left|\\frac{y^2}{x^2\\minus{}y^2}\\right| \\minus{} \\frac{2x}{y} \\equal{} \\ln|Cx|$." } { "Tag": [ "limit", "trigonometry", "logarithms" ], "Problem": "Sa se calculeze:\r\n \r\n $\\lim\\rightarrow{x\\to {0}} \\frac{1-\\ln(e+\\sinh)\\cdot \\ln(e+\\sin2x)\\cdot...\\cdot \\ln(e+\\sin nx)}{x}$ ;)", "Solution_1": "Ah ce problema urata! :( \r\nOricum, o chestiune de scriere: pune \\ln, \\lim, \\sin ca sa iasa functiile frumos ;) si \\cdots pentru 3 puncte centrale (inmultire continua). \r\n\r\nUite varianta \"buna\" :P\r\n[quote=\"cezar lupu\"]Sa se calculeze:\n \\[ \\lim_{x\\to 0} \\frac{1-\\ln {(e+\\sin x)}\\cdot \\ln {(e+\\sin {2x})}\\cdots \\ln {(e+\\sin {nx})}}{x}. \\][/quote]", "Solution_2": "Ok,mersi de indicatie. Apropo limita cum se scrie? :?", "Solution_3": "[quote=\"cezar lupu\"]Ok,mersi de indicatie. Apropo limita cum se scrie? :?[/quote]Vezi ca am editat postul mai sus ... :roll:", "Solution_4": "[quote=\"Valentin Vornicu\"]Ah ce problema urata! :( \nOricum, o chestiune de scriere: pune \\ln, \\lim, \\sin ca sa iasa functiile frumos ;) si \\cdots pentru 3 puncte centrale (inmultire continua). \n\nUite varianta \"buna\" :P\n[quote=\"cezar lupu\"]Sa se calculeze:\n \\[ \\lim_{x\\to 0} \\frac{1-\\ln {(e+\\sin x)}\\cdot \\ln {(e+\\sin {2x})}\\cdots \\ln {(e+\\sin {nx})}}{x}. \\][/quote][/quote]\r\n\r\nAdaugand 2n-2 termeni (1/2 cu plus si 1/2 cu minus), iar apoi aplicad l'Hospital, avem:\r\n\r\n$\\lim_{x\\to 0} \\frac{1-\\ln {(e+\\sin x)}\\cdot \\ln {(e+\\sin {2x})}\\cdots \\ln {(e+\\sin {nx})}}{x} = \\lim_{x\\to 0} \\frac{\\sum_{k=0}^{n} [\\prod_{l=1}^{k-1} \\ln (e + \\sin lx)][1 - \\ln(e+\\sin kx)]}{x} = \\sum_{k=0}^{n} \\lim_{x\\to 0} \\frac{1- \\ln (e + sin kx)} {x} = \\sum_{k=0}^{n} \\lim_{x\\to 0} \\frac{k \\cos kx}{e + \\sin kx}= \\sum_{k=0}^{n} \\frac{k} {e} = \\frac{n (n+1)} {2 e}$" } { "Tag": [ "AMC", "AMC 8" ], "Problem": "If $ \\angle A\\equal{}20^\\circ$ and $ \\angle AFG\\equal{}\\angle AGF$, Then how many degrees is $ \\angle B\\plus{}\\angle D$?\n[asy]/* AMC8 2000 #24 Problem */\npair A=(0,80), B=(46,108), C=(100,80), D=(54,18), E=(19,0);\ndraw(A--C--E--B--D--cycle);\nlabel(\"$A$\", A, W);\nlabel(\"$B$\", B, N);\nlabel(\"$C$\", shift(7,0)*C);\nlabel(\"$D$\", D, SE);\nlabel(\"$E$\", E, SW);\nlabel(\"$F$\", (23,43));\nlabel(\"$G$\", (35, 86));[/asy]", "Solution_1": "\\begin{eqnarray*}\r\n\\angle{B}+\\angle{D}\r\n& = & \\angle{AFB} \\\\\r\n& = & \\angle{AFG} \\\\\r\n& = & \\frac{180^\\circ-\\angle{A}}{2} \\\\\r\n& = & \\frac{180^\\circ-20^\\circ}{2} \\\\\r\n& = & \\frac{160^\\circ}{2} \\\\\r\n& = & \\boxed{80^\\circ}\r\n\\end{eqnarray*}" } { "Tag": [], "Problem": "Nancy spent 3 hours painting her dormitory room. Her roommate, Marcia, didn't like the color. So Marcia did the job over again in 5 hours. Working together, how long would it have taken to paint the room? \r\nThe answer is 9/8, but can someone show me how you get it?", "Solution_1": "Ok, Nancy spent 3 hours on the job, and Marcia spent 5 hours on the same job. \r\n\r\nSo, in 1 hour, Nancy completed $\\frac{1}{3}$ of the total job, and in 1 hour, Marcia completed $\\frac{1}{5}$ of that same job. So in 1 hour, they can complete $\\frac{1}{3} + \\frac{1}{5}$ of the entire job. Now, since the entire job is $1j$, and each hour they complete $\\frac{8}{15}j$ of the job, they will get it done in $\\frac{1j}{\\frac{8}{15}j}$, which is $\\frac{15}{8}$, not $\\frac{9}{8}$.", "Solution_2": "[hide]\n$Rate * Time = Jobs$\n$R_N(3) = 1$\n$R_N = \\frac{1}{3}$\n$R_M(5) = 1$\n$R_M = \\frac{1}{5}$\n$\\frac{1}{3}T + \\frac{1}{5}T = 1$\n$\\frac{8}{15}T = 1$\n$T = \\frac{1}{\\frac{8}{15}}$\n$T = \\frac{15}{8}$\n[/hide]" } { "Tag": [ "inequalities", "MIT", "college", "inequalities proposed" ], "Problem": "Let $ a_{1},a_{2},...,a_{n}$ be real numbers. Prove that\r\n\\[ 1+a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+(a_{1}+a_{2}+...+a_{n})^{2}\\ge \\]\r\n\r\n\\[ \\ge \\frac{2}{\\sqrt{n+1}}(|a_{1}|+|a_{2}|+...+|a_{n}|+|a_{1}+a_{2}+...+a_{n}|). \\]\r\n\r\n\r\nFind the best constant that can replace for $ \\frac{2}{\\sqrt{n+1}}$?", "Solution_1": "$ \\frac{2}{\\sqrt{n+1}}(/a_{1}/+/a_{2}/+...+/a_{n}/+/a_{1}+a_{2}+...+a_{n}/)=<\\frac{2}{\\sqrt{n+1}}\\sqrt{(1+^{2}+...+1^{2})(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=\\frac{2}{\\sqrt{n+1}}\\sqrt{(n+1)(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=$\r\n$ =2\\sqrt{a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}}=<1+a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}$\r\n\r\n\r\n;", "Solution_2": "[quote=\"zaya_yc\"]$ \\frac{2}{\\sqrt{n+1}}(/a_{1}/+/a_{2}/+...+/a_{n}/+/a_{1}+a_{2}+...+a_{n}/)=<\\frac{2}{\\sqrt{n+1}}\\sqrt{(1+^{2}+...+1^{2})(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=\\frac{2}{\\sqrt{n+1}}\\sqrt{(n+1)(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=$\n$ =2\\sqrt{a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}}=<1+a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}$\n\n\n;[/quote]\r\n\r\nCould you explain why\r\n\r\n\\[ 2\\sqrt{a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}}\\le 1+a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}?\\]", "Solution_3": "[quote=\"hungkhtn\"][quote=\"zaya_yc\"]$ \\frac{2}{\\sqrt{n+1}}(/a_{1}/+/a_{2}/+...+/a_{n}/+/a_{1}+a_{2}+...+a_{n}/)=<\\frac{2}{\\sqrt{n+1}}\\sqrt{(1+^{2}+...+1^{2})(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=\\frac{2}{\\sqrt{n+1}}\\sqrt{(n+1)(a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2})}=$\n$ =2\\sqrt{a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}}=<1+a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}$\n\n\n;[/quote]\n\nCould you explain why\n\\[ 2\\sqrt{a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}}\\le 1+a_{1}^{2}+...+a_{n}^{2}+(a_{1}+...+a_{n})^{2}? \\]\n[/quote]\r\n\r\njust an am-gm inequality on $ 1$ and $ a_{1}^{2}+\\cdots+a_{n}^{2}+(a_{1}+\\cdots+a_{n})^{2}$.", "Solution_4": "OK, you are right! nice solution, zara_yc and sung. Actually, the following stronger inequality holds for all real numbers $ a_{1},a_{2},...,a_{n}$\r\n\r\n\r\n\\[ 1+a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+(a_{1}+a_{2}+...+a_{n})^{2}\\ge\\]\r\n\r\n\\[ \\ge k(|a_{1}|+|a_{2}|+...+|a_{n}|+|a_{1}+a_{2}+...+a_{n}|)\\]\r\n\r\nin which\r\n\r\n$ k=\\frac{2}{\\sqrt{n+1}}$ if $ n$ is odd and $ k=\\frac{2\\sqrt{n+1}}{\\sqrt{n(n+2)}}$ if $ n$ is even.\r\n\r\n@ sung. I hear that you will come to MIT this year? Congrat!!!", "Solution_5": "[quote=\"hungkhtn\"]\n@ sung. I hear that you will come to MIT this year? Congrat!!![/quote]\r\n\r\nwell i'm now undergrad student of MIT.. i'm gonna be a junior this semester. :)", "Solution_6": "@ sung. Anyway, this is a very good new! Congratulation! Hope we can meet in US! You may be not older than me but this year I just become freshman.\r\n\r\nFor my problem, actually, the solution is very easy. Using\r\n\r\n\\[ x^{2}+\\frac{1}{n+1}\\ge \\frac{2x}{\\sqrt{n+1}}.\\]\r\n\r\nLet $ x=a_{1},a_{2},...,a_{n},a_{1}+a_{2}+...+a_{n}$ then summing up, we are done.\r\n\r\n\r\nActually, in this solution, according to the case of equality, we realize that the equality holds if and only if $ n$ is odd. For $ n$ is even, and $ k=\\frac{2}{\\sqrt{n+1}}$ the equality can not hold. I found the best constant is $ \\frac{2\\sqrt{n+1}}{\\sqrt{n(n+2)}}$. I hope someone can find a solution to this case ($ n$ is even)." } { "Tag": [ "counting", "distinguishability" ], "Problem": "A set of 25 square blocks is arranged into a 5 x 5 square. How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?\r\n\r\n[b](A)[/b] 100 [b](B)[/b] 125 [b](C)[/b] 600 [b](D) [/b]2300 [b](E)[/b] 3600", "Solution_1": "Does order matter?\r\n\r\n[hide=\"If so\"]We choose an arbitrary block, then remove that row and column. We repeat the process, then repeat it again, so the answer is $ 25\\cdot 16\\cdot 9 \\equal{} 3600\\Rightarrow E$.[/hide]\n\n[hide=\"If not\"]$ 3600/3!\\equal{}600\\Rightarrow C$[/hide]", "Solution_2": "well since it says blocks im going to guess that you have to divide by 6 since blocks are indistinguishable from another block." } { "Tag": [ "MATHCOUNTS", "geometry", "ratio", "Pythagorean Theorem", "similar triangles", "congruent triangles", "number theory" ], "Problem": "This is for a list of topics that are frequently used in Mathcounts, or even those that don't appear that often.\r\n\r\nActually, I thought of making this thread because I had a question: Are logs used in MC?", "Solution_1": "[quote=\"star99\"]This is for a list of topics that are frequently used in Mathcounts, or even those that don't appear that often.\n\nActually, I thought of making this thread because I had a question: Are logs used in MC?[/quote]\r\nlogs are never ever used in mathcounts. Besides, any problems with logs can be rewritten with exponents so there.", "Solution_2": "MathCounts is pretty much about applying the simplest of concepts (like the pythagorean theorem or ${n\\choose r}$) in creative ways to solve problems quickly. It's alot about finding patterns, using tricks, shortcuts, etc.\r\n\r\nSome topics:\r\n\r\nGeometry-Similar Triangles, Pythagorean Theorem, Congruent Triangles (every once in a while), Polygons, Circles (area+circumference)\r\nNumber Theory-Modular Arithmetic, Divisibility (like the two smallest prime factors of $2^{1024}$), Primes, some other stuff that I can't quite allow to come to my head right now\r\nCombinatorics-Combinations, Permutations, Sets\r\nAlgebra-Any creative problems that somehow end up in setting up equations, that's another thing that Mathcounts stresses (d=rt, stuff like that)", "Solution_3": "[quote=\"13375P34K43V312\"]MathCounts is pretty much about applying the simplest of concepts (like the pythagorean theorem or ${n\\choose r}$) in creative ways to solve problems quickly. It's alot about finding patterns, using tricks, shortcuts, etc.\n\nSome topics:\n\nGeometry-Similar Triangles, Pythagorean Theorem, Congruent Triangles (every once in a while), Polygons, Circles (area+circumference)\nNumber Theory-Modular Arithmetic, Divisibility (like the two smallest prime factors of $2^{1024}$), Primes, some other stuff that I can't quite allow to come to my head right now\nCombinatorics-Combinations, Permutations, Sets\nAlgebra-Any creative problems that somehow end up in setting up equations, that's another thing that Mathcounts stresses (d=rt, stuff like that)[/quote]\r\ncongruent triangles are the same as similar triangles with a ratio of 1.", "Solution_4": "Well SORRY, but congruent triangles are a little easier to deal with.", "Solution_5": "[quote=\"13375P34K43V312\"]Well SORRY, but congruent triangles are a little easier to deal with.[/quote]\r\nI was only trying to shorten the list :(" } { "Tag": [ "percent", "LaTeX" ], "Problem": "Dr. Phil decides to donate 500,000 dollars to the local oil monopoly. 75% of it goes to the CEO before he tells anybody about the donation. The rest of it goes to the company.\r\n\r\na) How much goes to the company?\r\n\r\n60% of it goes to the marketing section, and the rest is equally divided between the employee budget and the oil mining department.\r\n\r\nb) How much of it goes to the oil mining department?\r\n\r\nOut of generosity, the employee budget people decide to give 50% of what they got back to Dr. Phil.\r\n\r\nc) What percent of Dr. Phil's donation get backs to him?", "Solution_1": "[quote=\"Treething\"]Dr. Phil decides to donate 500,000 dollars to the local oil monopoly. 75% of it goes to the CEO before he tells anybody about the donation. The rest of it goes to the company.\n\na) How much goes to the company?\n\n60% of it goes to the marketing section, and the rest is equally divided between the employee budget and the oil mining department.\n\nb) How much of it goes to the oil mining department?\n\nOut of generosity, the employee budget people decide to give 50% of what they got back to Dr. Phil.\n\nc) What percent of Dr. Phil's donation get backs to him?[/quote]\r\n\r\n[hide]\na) If 75% goes to the CEO, then 25% goes to the company. 500,000*(25/100)=125,000. 125,000 goes to the company.\n\nb) If 40% goes to the marketing section, then 60% is equally divided between the emplyee budget and the oil mining department. 125,000*(60/100)=75,000. Because the 75,000 is equally divided, you need to divided 75,000 by 2. 75,000/2=37,500. 37,500 goes to the oil mining department.\n\nc) Haha, funny.\n\n...Wait, it is a joke, right? :? \n[/hide]\r\n\r\nThis is good practice!\r\n\r\nEdit: Wow, I am confused. Some of my answer gets cut out because of some Latex error. Could a mod fix this, please? I don't know Latex... ;)", "Solution_2": "[hide=\"a\"]125000[/hide]\n[hide=\"b\"]25000[/hide]\n[hide=\"c\"]12500[/hide]\r\n\r\nis that all they get? :(", "Solution_3": "a\r\n[hide]\n500000/4=125000\n>>125000<<[/hide]\n\nb\n[hide]\n125000*.4=50000\n50000/2=25000\n>>25000<<[/hide]\n\nc\n[hide]\n25000/2=12500\n12500/500000\n0.025\n>>2.5%<<[/hide]\r\n\r\nErr, shinwoo, didn't the problem say \"percent\"?\r\n\r\nMan, Dr. Phil made several bad mistakes here.\r\n1:Its an OIL company! If he really needed to make a donation, why did it have to be to an OIL company?!\r\n2:Not only is it an oil company, but it's also an oil MONOPOLY!!! It's already got enough of the market, it doesn't need anymore!\r\n3:Never donate money to an oil monopoly unless you personnaly know the CEO, especially if the CEO is a CURRUPT one!!!\r\n\r\nAt least the employee branch was kind enough to bive back some money... :(", "Solution_4": "Hey, Dr. Phil's corrupt too. See the previous topic about him...\r\n\r\nOompa Loompa doompa dee dee-aaagh!!!", "Solution_5": "[quote=\"SoccerBrainy40\"]\nEdit: Wow, I am confused. Some of my answer gets cut out because of some Latex error. Could a mod fix this, please? I don't know Latex... ;)[/quote]\r\n\r\nSorry 'bout that - our message board takes anything between dollar signs and runs LaTeX (a typesetting program that lets you do math stuff) on it. Your post had dollar signs in it because the problem was about dollars. I took the dollar signs out of your post, so it looks ok now.", "Solution_6": "Whoaps Dazy. :D \r\n\r\nthanks for correcting me...\r\n\r\nyou know the answer anyway so I'll just say the answer\r\n\r\ndwx's answer is correct" } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "in how many ways two bishops can be placed on on a chess board so that they will not attack each other.", "Solution_1": "The only way they can attack each other is if they are on the same diagonal line ;)\r\nthis can be chosen in exactly\r\n\\[ 2\\left(2\\sum_{i\\equal{}2}^7\\binom{i}{2}\\plus{}\\binom{8}{2}\\right)\\] ways. So your answer is\r\n\\[ \\binom{64}{2}\\minus{}4\\sum_{i\\equal{}2}^7\\binom{i}{2}\\minus{}2\\binom{8}{2}\\equal{}\\binom{64}{2}\\minus{}2\\binom{9}{3}\\minus{}2\\binom{8}{3}\\equal{}\\boxed{1736}\\]", "Solution_2": "The answer is completely Correct", "Solution_3": "[quote=\"Albanian Eagle\"]The only way they can attack each other is if they are on the same diagonal line ;)\nthis can be chosen in exactly\n\\[ 2\\left(2\\sum_{i \\equal{} 2}^7\\binom{i}{2} \\plus{} \\binom{8}{2}\\right)\\]\nways. So your answer is\n\\[ \\binom{64}{2} \\minus{} 4\\sum_{i \\equal{} 2}^7\\binom{i}{2} \\minus{} 2\\binom{8}{2} \\equal{} \\binom{64}{2} \\minus{} 2\\binom{9}{3} \\minus{} 2\\binom{8}{3} \\equal{} \\boxed{1736}\\]\n[/quote]\r\n\r\nI dont understand. How it can be possible? If on chessboars there are 64*63 (4 032) ways, how the bishops can be placed. If the first bishop will be on first row, than the second cant be only on 7 cells (it means, that here is 58 possible cells). Most cells in danger will be, if the first bishop will be in the center. Than the second bishop cant be on 13 cells (40 is possible). Try it and calc.\r\n\r\nIf I divide your solution by 64, I get average about 27. How it can be true, if minimum of possible cells can be 40? Something is wrond, hey?", "Solution_4": "[quote=\"thco\"]If on chessboars there are 64*63 (4 032) ways[/quote] There are $ \\binom{64}{2} \\equal{} \\frac{64 \\cdot 63}{2} \\equal{} 2016$ ways to place two identical bishops on different squares of a chessboard.", "Solution_5": "[quote=\"JBL\"][quote=\"thco\"]If on chessboars there are 64*63 (4 032) ways[/quote] There are $ \\binom{64}{2} \\equal{} \\frac {64 \\cdot 63}{2} \\equal{} 2016$ ways to place two identical bishops on different squares of a chessboard.[/quote]\r\nWell, its better. From problem it isnt evident, that they are identical." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "with the digits $ 1\\minus{}9$ construct two numbers with maximal and minimal product", "Solution_1": "no one interested?" } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "inequalities", "triangle inequality" ], "Problem": "Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.", "Solution_1": "[hide=\"Hint\"]The triangle inequality still applies to the faces.[/hide]\n\n[hide=\"Solution\"]Let the tetrahedron be $ ABCD$. Suppose the three edge lengths from every vertex do not form the sides of a triangle. WLOG $ AB$ is the maximum length edge. Then $ AB > AC \\plus{} AD$ and $ AB > BC \\plus{} BD$. Summing these gives $ 2AB > AC \\plus{} AD \\plus{} BC \\plus{} BD$. But by the triangle inequality on the faces $ ABC$, $ ABD$, we have $ AB < AC \\plus{} BC$ and $ AB < AD \\plus{} BD$. Summing these gives $ 2AB < AC \\plus{} AD \\plus{} BC \\plus{} BD$. Contradiction.[/hide]" } { "Tag": [ "ARML", "email", "AMC", "AMC 12", "AIME" ], "Problem": "[code]Greetings,\n \nWhen asked to assemble a team of mathematically talented high students to compete in the 2009 National ARML Competition, you guys were the first on my mind. I know that you have grown mathematically and enjoy the company of highly capable young people. Certainly, the ARML Competition brings together the best math students in the country and is an experience that one never forgets.\n \nI have attached your invitation and the application (in Word) for the 2009 Team. Please feel free to download the application and share with another mathematically talented student at your school.\n \nPlease let me know as soon as possible if you or a friend would like to join us. My email address is [color=red]pottergene @ msn.com[/color]. ) ([color=red]take out spaces[/color])\n \nBest regards,\nGene Potter\nSecretary-Treasurer\nMO-ARML\n***-***-**** ([color=red]phone number blanked out[/color])[/code]\r\nKS does not have an ARML team, as far as I know (being a KSan), and the closest ARML team is MO ARML. PM me your email if you would like this email forwarded. MO ARML has sent two teams and a few alternates, we go to Iowa by bus on Thurs Mar 28th, stay two nights, and leave after the competition on Sat Mar 30th. \"The total cost of the trip is 200 dollars per student\". It's fun and you meet many interesting people :) .\r\n\r\nhttp://math.missouristate.edu/MissouriARML.htm", "Solution_1": "Attachments:", "Solution_2": "I probably won't go this year simply because I'm not prepared well enough. What are some good resources for ARML training?", "Solution_3": "AoPS v2, of course. Old ARML tests (the last few years are online), the old ARML marathons and practice series. And other contests, AMC12, AIME (though they have different formats, it's the same math concepts), local Great Plains Math League meets (throughout the school year, the tests are modeled after ARML, though significantly easier)." } { "Tag": [ "conics", "hyperbola", "geometry", "incenter", "circumcircle", "analytic geometry", "geometric transformation" ], "Problem": "Prove that for all $M$ lie on the Feuerbach hyperbola of triangle $ABC$ with incenter $I$ and circumcenter $O$ then the cevian quotient $M/I$ lie on line $IO$", "Solution_1": "We prove a stronger result: M/I lies on OI if and only if M lies on the Feurebach hyperbola of ABC. Using barycentric coordinates WRT ABC, we have\n\n$O(a^2S_A:b^2S_B:c^2S_C) \\ , \\ I(a:b:c) \\ , \\ M(x:y:z) \\ \\Longrightarrow$\n\n$M/I \\left (a \\left ( \\frac{b}{y}+\\frac{c}{z}-\\frac{a}{x} \\right) :b \\left ( \\frac{c}{z}+\\frac{a}{x}-\\frac{b}{y} \\right):c \\left ( \\frac{a}{x}+\\frac{b}{y}-\\frac{c}{z} \\right) \\right)$\n\nHence, $O,I, M/I$ are collinear if and only if \n\n$\\left [\\begin {array}{ccc} a & b & c\\\\ a^2S_A & b^2S_B& c^2S_C\\\\ a \\left ( \\frac{b}{y}+\\frac{c}{z}-\\frac{a}{x} \\right)& b \\left ( \\frac{c}{z}+\\frac{a}{x}-\\frac{b}{y} \\right)&c \\left ( \\frac{a}{x}+\\frac{b}{y}-\\frac{c}{z} \\right) \\end{array}\\right]=0$ \n\n$\\Longleftrightarrow a(bS_B-cS_C)yz+b(cS_C-aS_A)zx+c(aS_A-bS_B)xy=0,$\n\nwhich is the equation of the Feuerbach hyperbola of ABC, i.e. isogonal conjugate of IO.", "Solution_2": "[b]Lemma:-[/b]\n\nGiven a triangle, $ ABC $ and a point $ P $, the locus of all points $ Q $ so that the ceva quotient $ Q/P $ lies on a fixed line is a hyperbola passing through $ A,B,C $.\n\n[b]Proof:-[/b]\n\nIf we take $ P $ to the centroid of $ ABC $ by a projective transformation, then $ Q/P $ becomes the anti-complement of the isotomic point of $ Q $ wrt $ ABC $. So if it moves on a line, then $ Q $ moves on a conic passing through $ A,B,C $ since isotomic conjugate of a circumconic is a line.\n\n[b]Back to Main Proof:-[/b]\n\nIn this problem, clearly, $ I/I=I $ and if $ G_e $ is the Georgenne point of $ ABC $, then clearly, $ G_e/I $ is the center of homothety of intouch triangle and excentric triangle of of $ ABC $, which lies on $ OI $. So the locus of $ Q $ is the conic passing through $ A,B,C,I,G_e $ which is the Feuerbach hyperbola.", "Solution_3": "Nice proof, but there are some assertions that are not necessarily true.\n\n[quote=\"RSM\"] Note that, the ceva quotient $Q/P$ and $P/Q$ are the same. If we take $P$ to the centroid of $ABC$ by a projective transformation, then $Q/P$ becomes the complement of the isotomic point of $Q$ wrt $ABC.$ So if it moves on a line, then $Q$ moves on a hyperbola passing through $A,B,C$ since isotomic conjugate of a hyperbola is a line.\n[/quote]\nIn general, Q/P and P/Q are not the same point and the isotomic conjugate of a line is not necessarily a hyperbola. This depends on whether the line meets the Steiner circumellipse of the reference triangle at two, one or any point.\n\n[quote=\"RSM\"]$I/I=I$ and if $G_e$ is the Georgenne point of $ ABC,$ then clearly, $G_e/I$ is the centroid of the intouch triangle of $ ABC,$ which lies on $OI.$ [/quote]\nGe/I is not the centroid of the intouch triangle but the Isogonal Mittenpunkt, i.e. exsimilicenter of the incircle and the circumcircle of the excentral triangle.", "Solution_4": "Thank you for pointing out the mistakes. I have edited the proof.\n\nActually I messed up with the definitions. I took the following definition as the definition of ceva-quotient:-\n\nIf $ AP,BP,CP $ intersects cevian triangle of $ Q $ at $ A',B',C' $, then perspective center of the cevian triangle of $ Q $ and $ A'B'C' $ is $ Q/P $.\n\nIn this definition $ P/Q $ and $ Q/P $ are the same. But I searched for the definition in google just and saw that its different." } { "Tag": [], "Problem": "Express the following as a common fraction.\n\n\\[ \\dfrac{1\\plus{}3\\plus{}5\\plus{}\\ldots\\plus{}19}{2\\plus{}4\\plus{}6\\plus{}\\ldots\\plus{}20}\\]", "Solution_1": "the sum 1+3+5....+19 is the sum of the first 10 odd positive integers. The sum of the first n odds is n^2. \r\n\r\nthe sum 2+4+6....20 is the sum of the first 10 even positive integers. The sum of the first n evens is n^2+n.\r\n\r\nthus the sum of the first 10 odds divided by the sum of the first 10 evens is 10^2/(10^2+20)=100/110=10/11", "Solution_2": "For the even ones, you could notice each number is $ (1\\plus{}1), (3\\plus{}1), (5\\plus{}1)$, etc. So the sum is $ 100 \\plus{} 10$, or $ 110$.\r\n\r\nThus, the answer is $ \\boxed{\\frac{10}{11}}$." } { "Tag": [], "Problem": "where can i find some past bacalaureat exams from all over the world in order to preapare for it .\r\nthanks :wink:", "Solution_1": "This [url=http://subiecte2007.edu.ro/bacalaureat/subiecte/proba_D/filiera_teoretica/profil_real/specializarea_matematica_informatica/matematicaM1_1/matematicaM1_1/varianta_001.html]link[/url] gives you this years proposed problems for the Romanian one.", "Solution_2": "$10^{n}$ thanks" } { "Tag": [], "Problem": "Hi All.\r\n\r\nYesterday I made an experiment on the synthesis of $ \\text{KMnO}_4$. I resulted in some aciform crystal. But some of my classmates got lamellar crystal. So can anyone please tell me under what conditions we can get the two kinds of crystals?\r\n\r\nThanks a lot!", "Solution_1": "hey i am just a beginner in this field could you first tell me what lamellar crystal and aciform crystal are??? :maybe:", "Solution_2": "[quote=\"shobber\"]Hi All.\n\nYesterday I made an experiment on the synthesis of $ \\text{KMnO}_4$. I resulted in some aciform crystal. But some of my classmates got lamellar crystal. So can anyone please tell me under what conditions we can get the two kinds of crystals?\n\nThanks a lot![/quote]\r\ndid u actually synthesis kmno4 frm simpler starting materials or did u just grow a crystal? at least dats wat i did, but since i grew them at home, one cant expect to actually synthesise them. and i havent grown kmno4 but a few other crystals like alum and rochelle salt.", "Solution_3": "In the experiment, we synthesised KMnO4 from KClO3, KOH, MnO2 and CO2." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Solve the system:\r\n$ 2^x\\plus{}y\\equal{}1$\r\n$ 2^y\\plus{}z\\equal{}1$\r\n$ 2^z\\plus{}x\\equal{}1$", "Solution_1": "$ x<0$ $ \\implies$ $ 2^{x}<1$ $ \\implies$ $ y>0$ $ \\implies$ $ 2^{y}>1$ $ \\implies$ $ z<0$ $ \\implies$ $ 2^{z}<1$ $ \\implies$ $ x>0$ Contradiction!\r\n$ x>0$ $ \\implies$ $ 2^{x}>1$ $ \\implies$ $ y<0$ $ \\implies$ $ 2^{y}<1$ $ \\implies$ $ z>0$ $ \\implies$ $ 2^{z}>1$ $ \\implies$ $ x<0$ Contradiction!\r\nonly solution $ x\\equal{}0$ $ \\implies$ $ y\\equal{}z\\equal{}0$ \r\n$ \\{(0,0,0)\\}$ :wink:" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Let A, B - two matrix, nilpotent of degree r, such that AB = BA. Why matrix \r\nA + B is nilpotent of degree r, too?\r\n\r\nI know that it is something very simple, but I have though about it a lot and cannot prove. I already know, that A + B in degree 2r is zero: AB = BA and we can use Newton binomial, so in all element of decomposition of (A+B)^2r there is A or B in degree at least r. But why it is zero already in degree r?", "Solution_1": "Wow, this time I am here before the White Wolf.\r\n\r\nFirst, let me note that your proof of $ \\left(A \\plus{} B\\right)^{2r} \\equal{} 0$ actually shows the stronger assertion $ \\left(A \\plus{} B\\right)^{2r \\minus{} 1} \\equal{} 0$. Of course, if we are talking about $ n\\times n$ matrices over a field, then this strengthens to $ \\left(A \\plus{} B\\right)^{\\mathrm{min}\\left\\{2r \\minus{} 1,n\\right\\}} \\equal{} 0$.\r\n\r\nImproving the exponent beyound this seems challenging. But we cannot generally have $ \\left(A \\plus{} B\\right)^r \\equal{} 0$. In fact, if we would, then, for $ r \\equal{} 2$,\r\n\r\n[color=green][b](1)[/b][/color] any two matrices $ A$ and $ B$ satisfying $ AB \\equal{} BA$ and $ A^2 \\equal{} B^2 \\equal{} 0$ would satisfy $ \\left(A \\plus{} B\\right)^2 \\equal{} 0$\r\n\r\nand thus also $ AB \\equal{} 0$ (at least over a field of characteristic $ \\neq 2$, since $ AB \\equal{} \\frac12\\left(\\left(A \\plus{} B\\right)^2 \\minus{} A^2 \\minus{} B^2\\right)$), what contradicts Example 2.1 (the $ E_{14}$ part) of the article http://www.emis.de/journals/ELA/ela-articles/articles/vol16_pp237-247.pdf . Note that this counterexample is for $ n \\equal{} 4$, while [color=green][b](1)[/b][/color] is still true for $ n \\equal{} 3$ (as follows from the end of page 5 in the above-mentioned article). Maybe this generalizes...\r\n\r\n darij", "Solution_2": "[quote=\"darij grinberg\"]Wow, this time I am here before the White Wolf.\n\nFirst, let me note that your proof of $ \\left(A \\plus{} B\\right)^{2r} \\equal{} 0$ actually shows the stronger assertion $ \\left(A \\plus{} B\\right)^{2r \\minus{} 1} \\equal{} 0$. Of course, if we are talking about $ n\\times n$ matrices over a field, then this strengthens to $ \\left(A \\plus{} B\\right)^{\\mathrm{min}\\left\\{2r \\minus{} 1,n\\right\\}} \\equal{} 0$.\n darij[/quote]\n\nYes, I already have seen it myself. \n\n[quote=\"darij grinberg\"]But we cannot generally have \\left(A + B\\right)^r = 0. In fact, if we would, then, for r = 2,\n\n(1) any two matrices A and B satisfying AB = BA and A^2 = B^2 = 0 would satisfy \\left(A + B\\right)^2 = 0\n\nand thus also AB = 0 (at least over a field of characteristic \\neq 2, since AB = \\frac12\\left(\\left(A + B\\right)^2 - A^2 - B^2\\right)), what contradicts Example 2.1 (the E_{14} part) of the article http://www.emis.de/journals/ELA/ela-articles/articles/vol16_pp237-247.pdf . Note that this counterexample is for n = 4, while (1) is still true for n = 3 (as follows from the end of page 5 in the above-mentioned article). Maybe this generalizes... [/quote]\r\nThank you very much!" } { "Tag": [], "Problem": "Prove for any $ x,y,z\\geq 0$:\r\n$ 8(x^3\\plus{}y^3\\plus{}z^3)^2 \\geq 9(x^2\\plus{}yz)(y^2\\plus{}zx)(z^2\\plus{}xy)$", "Solution_1": "nice ineq :) \r\nhere's my solution:\r\n\r\n[hide]we have to prove \n$ 8(x^3 + y^3 + z^3)^2 >= 9*(x^2 + yz)(y^2 + xz)(z^2 + xy)$\n$ < = >$\n$ 2 (x^3 + y^3 + z^3)^{\\frac{2}{3}} >= \\frac{3}{3^{1/3}}* \\sqrt[3]{(x^2 + yz)(y^2 + xz)(z^2 + xy)}$\nfrom AM-GM we know\n$ 3\\sqrt[3]{(x^2 + yz)(y^2 + xz)(z^2 + xy)} <= x^2 + yz+ y^2 + xz + z^2 + xy <= 2(x^2 + y^2 + z^2)$\nso now we have to prove:\n${ 2(x^3 + y^3 + z^3)^{2/3} >=\\frac{2}{3^{1/3}} (x^2 + y^2 + z^2)}$ or\n${ (x^3 + y^3 + z^3)^{2/3} >=\\frac{1}{3^{1/3}} (x^2 + y^2 + z^2)}$\nfrom the power mean ineq we know that \n$ \\frac{(x^3 + y^3 + z^3)^{1/3}}{3^{1/3}} >= \\frac{(x^2 + y^2 + z^2)^{1/2}}{3^{1/2}}$\n$ < = >$\n$ \\frac{(x^3 + y^3 + z^3)^{2/3}}{3^{2/3}}>= \\frac{x^2 + y^2 + z^2}{3}$\n$ < = >$\n${ (x^3 + y^3 + z^3)^{2/3} >=\\frac{1}{3^{1/3}}(x^2 + y^2 + z^2)}$ \nQED[/hide]", "Solution_2": "[quote=\"balan razvan\"]Prove for any $ x,y,z\\geq 0$:\n$ 8(x^3 \\plus{} y^3 \\plus{} z^3)^2 \\geq 9(x^2 \\plus{} yz)(y^2 \\plus{} zx)(z^2 \\plus{} xy)$[/quote]\r\n$ 24(x^3\\plus{}y^3\\plus{}z^3)^2\\geq8(x^2\\plus{}y^2\\plus{}z^2)^3\\equal{}(x^2\\plus{}y^2\\plus{}z^2\\plus{}x^2\\plus{}y^2\\plus{}z^2)^3\\geq$\r\n$ \\geq(x^2\\plus{}y^2\\plus{}z^2\\plus{}xy\\plus{}xz\\plus{}yz)^3\\geq27(x^2\\plus{}yz)(y^2\\plus{}xz)(z^2\\plus{}xy).$ :)" } { "Tag": [ "linear algebra", "matrix", "function", "calculus", "derivative", "algebra", "polynomial" ], "Problem": "I've seen a proposition today: \r\nLet $A$ be an $n\\times n$ matrix and $p(x)$ be any polynomial. If $\\lambda$ is an eigenvalue of $A$, then $p(\\lambda)$ is an eigenvalue of $p(A)$.\r\n\r\nIt is easy to prove while $A$ is diagonalizable, BUT how about the general cases? :(", "Solution_1": "I think it's still true . . . A is similar to its Jordan form (actually, I don't think we even need that - you could note that it's similar (unitarily, in fact) to an upper triangular matrix); now, if $A = C^{-1}UC$, where $U$ is upper triangular, we know $p(A) = C^{-1}p(U)C$. Now, if $\\lambda$ is an eigenvalue, that means it is one of the diagonal elements of $U$, and we seek to prove that $p(\\lambda)$ is one of the diagonal elements of $p(U)$. Write $U = D + T$, where $D$ is diagonal and $T$ is strictly upper triangular. Then, I think that the powers of $T$ will also be strictly upper triangular and will remain that way if multiplied by diagonal matrices, so the diagonal elements of $p(U)$ should be the same as $p(D)$, which in turn are simply the images of the individual diagonal elements under $p$, which suffices to prove the claim.", "Solution_2": "where did you see this proposition?", "Solution_3": "[quote=\"uranium\"]Let $A$ be an $n\\times n$ matrix and $p(x)$ be any polynomial. If $\\lambda$ is an eigenvalue of $A$, then $p(\\lambda)$ is an eigenvalue of $p(A)$. [/quote]\r\nFirst of all, let us remind that $\\lambda$ is an eigenvalue of a $n \\times n$ matrix $A$ , if and only if there exists a $n\\times 1$ non-zero matrix $X$ such that \r\n\\[ (1)\\; \\; \\; \\; \\begin{array}{|c|} \\hline AX=\\lambda X \\\\ \\hline \\end{array} \\]\r\nAssume (1) as verified. One finds $A^2X=A\\left(\\lambda X\\right) = \\lambda(AX)=\\lambda^2X$ and in a similar manner one proves that $(2)\\; \\; \\; \\; \\; \\; A^kX=\\lambda^kX$ for $k\\in \\{0,1,...\\} .$ Let $p(z)=\\sum\\limits_{k=0}^M c_k z^k$ with complex coefficients. From (2) it follows\r\n \\[ p(A)X: =\\left(\\sum_{k=0}^Mc_kA^k\\right)X = \\sum\\limits_{k=0}^M c_k\\left(A^kX\\right) = \\left(\\sum\\limits_{k=0}^Mc_k\\lambda^k\\right)X=p(\\lambda)X \\; \\]\r\nwith $X$ a non-zero $n\\times 1$ -matrix. But according to (1) this means that $p(\\lambda)$ is an eigenvalue of the matrix $p(A) .$ \r\n\r\n\r\n[b]Remarks:[/b]\r\n[b]1)[/b] Of interest is to observe that if $A$ is non-singular ( that is $\\det(A)\\ne 0$) having the spectrum (i.e. the set of eigenvalues) \r\n$\\sigma_A: =\\{\\lambda_1,...,\\lambda_n\\}$ , then ( using $\\det(A)=\\lambda_1\\cdots\\lambda_n\\ne 0$ ) the spectrum of $A^{-1}$ is $\\sigma_{A^{-1}}=\\left\\{\\frac{1}{\\lambda_1},...,\\frac{1}{\\lambda_n}\\right\\}.$ To prove this observe that \r\n$P_A(\\lambda): =\\det(A-\\lambda\\cdot I)=\\det\\left( A(I-\\lambda\\cdot A^{-1}) \\right)=(-1)^n\\lambda^n\\det(A)\\det\\left(A^{-1}-\\frac{1}{\\lambda}\\cdot I\\right)$. Therefore $P_A(\\lambda)=0$ iff $\\lambda^nP_{A^{-1}}\\left(\\frac{1}{\\lambda}\\right) = 0 .$\r\n In fact this imply that equalities (2) remain true also for $k\\in \\{-1,-2,\\cdots\\}.$\r\n[b]2)[/b] Further assume that $A$ is arbitrary , having the spectrum $\\sigma_A$ with the eigenvalues\r\n\\[ \\begin{array}{l} \\underbrace{\\lambda_1,\\lambda_1,\\ldots,\\lambda_1}_{m_1} \\\\ \\underbrace{\\lambda_2,\\lambda_2,\\ldots,\\lambda_2}_{m_2} \\\\ \\vdots \\\\ \\underbrace{\\lambda_j,\\lambda_j,\\ldots,\\lambda_j}_{m_j} \\\\ \\end{array} \\; \\; \\; ,\\; \\; \\; \\; \\begin{array}{|c|} \\hline m_1+m_2+\\ldots+m_j=n\\\\ \\hline \\end{array}\\; , \\]\r\nwhere $\\{\\lambda_1,\\ldots,\\lambda_j\\}$ are mutual distinct, i.e. $\\lambda_{\\alpha}\\ne \\lambda_{\\beta}$ for $1\\le \\alpha \\ne \\beta \\le j .$\r\nAssume further that $D\\subset {\\mathbb C}$ and $f$$: D\\to {\\mathbb C}$ is a function satisfying following conditions :\r\ni) $\\sigma_A\\subset D$,\r\nii) the derivatives $f^{(m_1-1)}(\\lambda_1),\\ldots ,f^{(\\lambda_j-1)}(\\lambda_j)$ exist.\r\n[b] Then the matrix (=matrix-function) $f(A)$ may be defined ![/b] \r\n\r\n[b]Proposed questions[/b]\r\nQ1) Try to find how $f(A)$ is defined .\r\nQ2) Prove that if $\\lambda$ is an eigenvalue of $A$ , then $f(\\lambda)$ is an eigenvalue of the matrix $f(A)$ . \r\nQ3) If $A=\\left(\\begin{array}{cc} a & b\\\\ c & d \\end{array}\\right)$ has the spectrum $\\sigma_A: =\\{\\lambda_1,\\lambda_2\\}\\subset {\\mathbb R},$ find the matrix $\\sin(A) .$ \r\nQ4) Consider the [b]\"Fibonacci matrix\"[/b] $F: =\\left(\\begin{array}{cc} 1 & 1\\\\ 1& 0 \\end{array}\\right)$ . Find following matrices $F^{2006} \\; ,\\; \\; e^{F}\\; .$ Prove that $\\det\\left(f(F)\\right)=f(\\lambda_1)f(\\lambda_2)$ \r\nwhere $\\lambda_1, \\lambda_2$ are the roots of equation $\\lambda^2 -\\lambda-1 = 0 .$\r\nQ5) If $\\Gamma$ denotes Gamma -function , for which matrices $A$ the matrix $\\Gamma(A)$ is [i]well-defined[/i] ? \r\nQ6) Verify that if $A$ is as in Q3) , then \r\n\\[ f(A)=\\alpha(f)\\cdot A +\\beta(f)\\cdot I \\; \\; \\; , \\; \\; \\; I: =\\left(\\begin{array}{cc} 1 &0\\\\ 0 & 1 \\end{array}\\right) \\]\r\nwhere $\\alpha(f)=\\left\\{\\begin{array}{ccl} \\displaystyle \\frac{f(\\lambda_2)-f(\\lambda_1)}{\\lambda_2-\\lambda_1} &, & \\lambda_2\\ne \\lambda_1\\\\ \\displaystyle f^{\\prime}(\\lambda_1) &, & \\lambda_2=\\lambda_1 \\end{array}\\right.\\;$ and $\\beta(f)=\\left\\{\\begin{array}{ccl} \\displaystyle \\frac{\\lambda_2f(\\lambda_1)-\\lambda_1f(\\lambda_2)}{\\lambda_2-\\lambda_1} & , & \\lambda_2\\ne \\lambda_1\\\\ \\displaystyle f(\\lambda_1) -\\lambda_1f^{\\prime}(\\lambda_1) &, & \\lambda_2=\\lambda_1 \\end{array}\\right.\\; .$\r\nQ7) Justify following assertion:\r\n [b] If $A$ is a $n\\times n$ matrix , $(n\\ge 2)$ , then \"the more complicated matrix- functions\" are the matrix-polynomials of degree $\\le n-1 .$ [/b]\r\n\r\n[b]Some references[/b] (of course, the author is a Russian mathematician) :\r\n[1]F.R.Gantmacher , [i] The Theory of Matrices, Vol. I [/i], AMS Chelsea Publishing,(translated by K.A.Hirsch), 1959,1960,1970,1998) .\r\n[2]F.R.Gantmacher , [i] The Theory of Matrices, Vol. II [/i], AMS Chelsea Publishing,(1959,1987,1989) .", "Solution_4": "Aha, nice proof, nice remarks!\r\nThanks a lot! :lol:" } { "Tag": [ "inequalities", "number theory proposed", "number theory" ], "Problem": "Suppose $ m,n$ are psotive integrs such that $ m > 1$ and $ 2^{(2m \\plus{} 1)}\\ge n^2$. Prove that $ 2^{(2m \\plus{} 1)} \\ge n^2 \\plus{} 7$.", "Solution_1": "Since $ 2^{2m\\plus{}1}$ cannot be a square, $ 2^{2m\\plus{}1}>n^{2}$. Since a perfect square is of the form 4k or 4k+1, if we could prove that $ 2^{2m\\plus{}1}\\minus{}3$ and $ 2^{2m\\plus{}1}\\minus{}4$ cannot be perfect squares, then we are done.\r\nAssume $ 2^{2m\\plus{}1}\\minus{}3\\equal{}n^{2}$ which implies $ 4(2^{2m\\minus{}1}\\minus{}1)\\equal{}(n\\minus{}1)(n\\plus{}1)$\r\nThis is impossible since RHS is divisible by 8 ( Since n is odd, one of n-1 and n+1 is got to be divisible by 4)\r\nAssume $ 2^{2m\\plus{}1}\\minus{}4\\equal{}n^{2}$ which implies $ 2^{2m\\minus{}1}\\minus{}1\\equal{}(n/2)^{2}$ which is impossible since LHS is of the form 4k+3" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "For polynom P(x) with integer coefficient P(a)=1 P(b)=2 P(c)=3\r\nwhere a, b, c. prove that b is between a and b.", "Solution_1": "lemma:\r\nfor all integers $ x$ and $ y$ we have $ x \\minus{} y$ divide $ p(x) \\minus{} p(y).$\r\n\r\nso we have $ b \\minus{} a |p(b) \\minus{} p(a) \\equal{} 1$ so $ b \\minus{} a \\in${$ \\minus{} 1,1$} and $ c \\minus{} b \\in${$ \\minus{} 1,1$}.\r\nif we suppose that $ (b > a$ and $ b > c)$ we have $ b \\minus{} a \\equal{} 1$ and $ c \\minus{} b \\equal{} \\minus{} 1$ so $ a \\equal{} c$,which it is a contradiction .\r\nand with the same way if $ (b < a$ and $ b < c) a \\equal{} c.$" } { "Tag": [ "trigonometry", "absolute value" ], "Problem": "Complex Zeta, in June you solved one of my questions, but your solution was over my head. However, I've never seen your method before, and I would like to see it again. \r\n\r\nThe question was sin1sin3sin5...sin179\r\n\r\nIf you would explain it again I would appreciate it. Thanks.", "Solution_1": "Ok. What I did was I evaluated sin(1)*sin(2)*sin(3)*...*sin(179) and sin(2)*sin(4)*sin(6)*...*sin(178) and then divided.\r\n\r\nInstead, let's do sin(180/n)*sin(360/n)*...*sin(180*(n-1)/n). So let's consider x^n-1=0. These are the vertices of a regular n-gon on the unit circle. Let's remove the root x=1 because that's not one of the points we need to consider. Thus we are left with x^(n-1)+x^(n-2)+...+x+1. But if we let w=e^(i*pi/n), then x^(n-1)+x^(n-2)+...+x+1 = (x-w)(x-w^2)(x-w^3)...(x-w^(n-1)). Let's now let x=1 and take absolute values of both sides. The absolute value of the LHS is n. The absolute value of the RHS is the product of the absolute value of each term: |1-w|*|1-w^2|...|1-w^(n-1)|.\r\n\r\nNow recall (or prove) the identity |1-e^(i*theta)|=2*sin(theta/2). Thus we get n=2*sin(180/n)*2*sin(360/n)*...*2*sin(180*(n-1)/n). Thus sin(180/n)*sin(360/n)*...*sin(180*(n-1)/n)=n/2^(n-1). Now let n=180 and 90, respectively, and then divide." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "[i]Suppose $ G$ is a connected graph and $ f : G \\to G$ is an isomorphism. Prove that there is a cycle $ v_0 \\minus{} \\cdots \\minus{} v_n \\minus{} v_0$ such that $ f(\\{v_0, \\ldots, v_n\\}) \\equal{} \\{v_0, \\ldots, v_n\\}$.\n\n[color=brown](A cycle does not pass through the same vertex twice. A cycle can be of length 0, (a vertex) and of length 1 (an edge))[/color] [/i]\r\n\r\n[color=olive] [b]Special Case 1.[/b] If $ G$ has no cycles, (i.e. if $ G$ is a tree) then $ n$ must be $ 0$ or $ 1$, so either $ f(v) \\equal{} v$ for some $ v$ or $ f(u) \\equal{} v$ and $ f(v) \\equal{} u$ for some $ u,v$.[/color]\r\n\r\n[color=olive] [b]Special Case 2.[/b] If a graph is axially symmetric, then it contains a cycle that is axially symmetric via the same axis.[/color]", "Solution_1": "I think this is not true. \r\n\r\nConsider the following graph, with vertex set $ \\{A,B,C,V_1,V_2,V_3,V_4,V_5,V_6\\}$ and edge set $ \\{AV_1,BV_1,AV_4,BV_4,BV_2,CV_2,BV_5,CV_5,CV_3,AV_3,CV_6,AV_6\\}$. \r\nNow consider the following automorphism:\r\n\r\n$ A \\rightarrow B$ \r\n$ B \\rightarrow C$\r\n$ C \\rightarrow A$\r\n\r\n$ V_1 \\rightarrow V_2$ \r\n$ V_2 \\rightarrow V_3$\r\n$ V_3 \\rightarrow V_4$\r\n$ V_4 \\rightarrow V_5$\r\n$ V_5 \\rightarrow V_6$\r\n$ V_6 \\rightarrow V_1$\r\n\r\nIt's easy to verify that there is no fixed cycle:\r\n\r\nThere is no fixed cycle of order $ 0$ or $ 1$.\r\nThis graph is bipartite, so a cycle must contain at least a vertex from $ \\{A,B,C\\}$ and a vertex from $ \\{V_1,V_2,V_3,V_4,V_5,V_6\\}$. \r\nBut a fixed cycle that contains a vertex from one set must contain the whole set.\r\nSo if there is a fixed cycle it must contain the whole vertex set. This cycle must alternate between vertices of each group. But $ |\\{A,B,C\\}|\\neq |\\{V_1,V_2,V_3,V_4,V_5,V_6\\}|$, so there is no fixed cycle." } { "Tag": [ "MATHCOUNTS", "AMC", "AIME" ], "Problem": "find the sum of the first thirty termsm of the sequence\r\n\r\n1,5,6,11,17,28...\r\nif the 30th term is 2,888,956 and the 31st term is 4,674,429\r\n\r\ndont use a calculator, this was 1984 and they didnt have calculators... :P \r\n\r\nand...\r\nfind a general formula for the sum of the first n fibbonacci numbers \r\n1,1,2,3,5,8,13,.....", "Solution_1": "first n fibonacci numbers is F(n+2)-1.\r\n\r\nYour first problem is 30th term+31st term-5.", "Solution_2": "[quote=\"mathking123\"]find the sum of the first thirty termsm of the sequence\n\n1,5,6,11,17,28...\nif the 30th term is 2,888,956 and the 31st term is 4,674,429\n\ndont use a calculator, this was 1984 and they didnt have calculators... :P \n\nand...\nfind a general formula for the sum of the first n fibbonacci numbers \n1,1,2,3,5,8,13,.....[/quote]\r\n+4,+1,+5,+6,+11?\r\nI don't see a pattern.", "Solution_3": "think about recursion :wink:", "Solution_4": "Finding a general formula for the Fibonacci numbers is above mathcounts level. This is known as Binet's formula: google it.", "Solution_5": "thanks everyone!\r\n\r\n@firecricket\r\n\r\ndoes that formula work even if the sequence does not start with 1,1,2,3,5......?", "Solution_6": "1,3,4,7,11...\r\n\r\ndoesn't seem to work for this one", "Solution_7": "alanchou, thanks for trying lol i was too lazy :lol: \r\n\r\nehh, @firecricket, where did the 30th term +31st term-5 come from??", "Solution_8": "you just notice the pattern after a while. \r\n\r\nif you do want to actually find out why that works, go ahead though", "Solution_9": "Nah, or you could use this amazing sequence technique (your way is better for mathcounts though, if you are good at finding patterns)\r\n\r\n$ F(n) \\plus{} F(n \\plus{} 1) \\equal{} F(n \\plus{} 2)$ where $ n$ is $ >\\equal{} 1$ is any formula in this shape.\r\n\r\nSo basically\r\n\r\n$ F(1) \\plus{} F(2) \\equal{} F(3)$\r\n$ F(2) \\plus{} F(3) \\equal{} F(4)$\r\n$ F(3) \\plus{} F(4) \\equal{} F(5)$\r\n....\r\n$ F(30) \\plus{} F(31) \\equal{} F(32)$\r\n\r\nnow add up ALL terms\r\n\r\nWe get $ F(1) \\plus{} 2[F(2) \\plus{} F(3) \\plus{} F(4)... \\plus{} F(30)] \\plus{} F(31) \\equal{} F(3) \\plus{} F(4) \\plus{} ...F(32)$\r\nNow cancel out common terms from both sides\r\nSo we now have\r\n$ F(1) \\plus{} 2F(2) \\plus{} F(3) \\plus{} F(4)... \\plus{} F(30) \\equal{} F(32)$\r\nWe are almost done\r\nSo we now have\r\n$ F(1) \\plus{} F(2) \\plus{} F(3)... \\plus{} F(30) \\equal{} F(32) \\minus{} F(2)$\r\n\r\nFor the first problem the sum of the first 30 terms is $ F(32) \\minus{} F(2) \\equal{} F(30) \\plus{} F(31) \\minus{} F(2) \\equal{}$what firecricket said $ \\equal{} 2,888,956 \\plus{} 4,674,429 \\minus{} 5$ since $ F(30)$, $ F(31)$, and $ F(2)$ are given.\r\n\r\nThus for any general sequence when $ F(n) \\plus{} F(n \\plus{} 1) \\equal{} F(n \\plus{} 2)$, the sum of the first $ n$ terms is denoted by $ F(n \\plus{} 2) \\minus{} F(2)$ where it starts at $ n \\equal{} 1$\r\n\r\nSpecifically, the sum of the first $ n$ terms in the fibonacci sequence is $ F(n \\plus{} 2) \\minus{} 1$ where $ F(1) \\equal{} 1$, $ F(2) \\equal{} 1$, $ F(3) \\equal{} 2$...\r\n\r\nEDIT: Dont believe how awesome this technique is, all you mathcounts people can now solve an AIME number 11 in 1 minute.\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=464978#p464978", "Solution_10": "OK THANKS!", "Solution_11": "u should memorize $ F_{n \\plus{} 2} \\minus{} 1$ is the sum of the fibonacci terms up to the nth term.", "Solution_12": "[quote=\"toadoncart\"]u should memorize it: $ F_{n \\plus{} 2} \\minus{} 1$ is the sum of the fibonacci terms up to the nth term.[/quote]\r\n\r\neven better: understand it, prove it. :)" } { "Tag": [ "inequalities" ], "Problem": "At one point in the solution of an inequality, I saw the following expression:\r\n\r\n[tex]\r\n\\displaystyle xyz \\Sigma_{cyclic} (x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz) \\leq \\\\\r\n(x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz)(z^{3}+x^{3}+xyz)\r\n[/tex]\r\n\r\nIs there an easy way to simplify the above to the below without have to multiply out the expressions?\r\n\r\n[tex]\r\n\\displaystyle \\Sigma_{symmetric} x^6 y^3 \\geq \\Sigma_{symmetric} x^5 y^2 z^2\r\n[/tex]\r\n\r\nThanks.", "Solution_1": "[quote=\"Fierytycoon\"]At one point in the solution of an inequality, I saw the following expression:\n\n[tex]\n\\displaystyle xyz \\Sigma_{cyclic} (x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz) \\leq \\\\\n(x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz)(z^{3}+x^{3}+xyz)\n[/tex]\n\nIs there an easy way to simplify the above to the below without have to multiply out the expressions?\n\n[tex]\n\\displaystyle \\Sigma_{symmetric} x^6 y^3 \\geq \\Sigma_{symmetric} x^5 y^2 z^2\n[/tex]\n\nThanks.[/quote]\r\n\r\nFirst, I am better off with symmetric notation (not cyclic) so I suggest multiplying the whole thing by 2 and changing the \"cyclic\" by \"symmetric\".\r\n\r\nSecond, in a symmetric sum, you can freely slide symmetric expressions in and out of the sigma (for example, you can put the xyz inside the expression.\r\n\r\nThird, inside a symmetric sum, you can replace the variables in any term with other variables - for example, if you have x^2*y+2z^2*y, you can replace that by 3x^2*y. \r\n\r\nNext, to convert the right the right hand side to a sym sum, cound the number of \"like terms\" and divide by 6.\r\n\r\nFor this example we have (remember we multiplied the whole thing by 2):\r\n\r\n6 terms of the form x^3y^3z^3\r\n12 terms of the form x^6y^3\r\n6 terms of the form x^7yz\r\n18 terms of the form x^4y^4z\r\n12 terms of the form x^5y^2z^2\r\n\r\nChecking, 6+12+6+18+12=54, which is the total number of terms\r\nthus the sym sum of the RHS is\r\nsym(x^3y^3z^3+2x^6y^3+x^7yz+3x^4y^4z+2x^5y^2z^2).\r\n\r\nThe LHS is easier since it is already in symmetric sum notation; just multiply out xyz(x^3 + y^3 +xyz)(y^3+z^3+xyz ) and replace terms using the third \"trick\" above.", "Solution_2": "Thanks for all of the tips Celeborn.\r\n\r\nI'm just curious: how exactly did you find the following data? The task would personally take me a while. What kind of combinatorial counting argument did you use?\r\n\r\n\r\n[quote]For this example we have (remember we multiplied the whole thing by 2): \n\n6 terms of the form x^3y^3z^3 \n12 terms of the form x^6y^3 \n6 terms of the form x^7yz \n18 terms of the form x^4y^4z \n12 terms of the form x^5y^2z^2[/quote]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ are positive real number such that $abc=1$. Prove that:\r\n$6\\sqrt[3]{(a^{3}+1)(b^{3}+1)(c^{3}+1)}\\ge a^{2}+b^{2}+c^{2}+9$", "Solution_1": "Wrong, try $a=1, b=1/10, c=10$ :P" } { "Tag": [ "geometry", "3D geometry", "prism", "ratio" ], "Problem": "The scale factor of two similar prisms is 3 : 4. The surface area and volume of the smaller prism are 63 cm^2 and 54 cm^3, respectively. Find the surface area and volume of the larger prism.", "Solution_1": "What kind of prism? Rectangular? Pyramidal? Parallelpiped?", "Solution_2": "It doesn't matter; ratio of surface area of two similar figures is equal to square of ratio of their lengths, and ratio of volume of two similar figures is equal to the cube of the ratio of their lengths. Thus $ 63 \\left(\\frac 43 \\right)^2 \\equal{} 112, 54 \\left(\\frac 43 \\right)^3 \\equal{} 128$." } { "Tag": [], "Problem": "Ok...I'm not sure how to prove using an Inductive step..Can someone help me please\r\n\r\n\r\nOk I have 3 cases I need to prove with this step.\r\n\r\nCase1\r\n\r\nn^2 <= n!\r\n\r\nCase2\r\n\r\nn^2 <= 2^n\r\n\r\nCase3\r\n\r\n2^n <= n!\r\n\r\nPlease help...I need to figure this out so I can use it in a program im writing for work...Thank you", "Solution_1": "[hide=\"Hint for Case 2.\"]To go from $n^2$ to $(n+1)^2$, you add $2n+1$, and to go from $2^n$ to $2^{n+1}$ you add $2^n$... if $n$ is at least 2, which is larger?[/hide]\n\n[hide=\"Hint for Case 3.\"]To go from $2^n$ to $2^{n+1}$, you multiply by 2, and to go from $n!$ to $(n+1)!$, you multiply by $n+1$... if $n$ is at least 2, which is larger?[/hide]\r\n\r\nCase 1 follows from Cases 2 and 3. In fact, it is easier to prove Cases 1 and 3 directly than to prove them by induction... see if you can do so." } { "Tag": [ "function", "vector", "trigonometry", "algebra", "polynomial", "advanced fields", "advanced fields theorems" ], "Problem": "Hi all,\r\nI am here for the first time and thank all of you who want to help me.\r\nI have few problems that I do not know how to solve.\r\nthis is one of them:\r\n\r\n(\u03b3\u00b2u(x,y)) / (\u03b3x\u03b3y)=0\r\n\r\nFind all solutions u(x,y) to this equations. How does this compare with an ODE;\r\n\r\n(d\u00b2y)/(dx\u00b2)=0", "Solution_1": "Did you mean $\\frac{\\partial^{2}u(x,y)}{\\partial x\\,\\partial y}=0$?\r\nIf so, recall that $\\frac{\\partial^{2}u(x,y)}{\\partial x\\,\\partial y}=\\frac{\\partial}{\\partial y}\\left(\\frac{\\partial u(x,y)}{\\partial x}\\right)$ and think of what it means that $\\frac{\\partial f(x,y)}{\\partial y}=0$ first.", "Solution_2": "Yes I got to that step but I have trouble of solving it..", "Solution_3": "The solutions of $\\frac{df(y)}{dy}=0$ are simply constants.\r\n\r\nNow, what if we want to make this a function of another variable as well? Then $\\frac{\\partial f(x,y)}{\\partial y}=0.$ should be constant [i]as far as $y$ is concerned.[/i]\r\n\r\nWhat should that mean? \r\n\r\n(There are some issues of \"regularity\" that I'd suggest we not address right away.)", "Solution_4": "[quote=\"Kent Merryfield\"]The solutions of $\\frac{df(y)}{dy}=0$ are simply constants.\n\nNow, what if we want to make this a function of another variable as well? Then $\\frac{\\partial f(x,y)}{\\partial y}=0.$ should be constant [i]as far as $y$ is concerned.[/i]\n\nWhat should that mean? \n\n(There are some issues of \"regularity\" that I'd suggest we not address right away.)[/quote]\r\n\r\nYes, but can we find s specific solutions? if we can how many we have?", "Solution_5": "[quote=\"mina2007\"]Yes, but can we find s specific solutions? if we can how many we have?[/quote]\r\nYou do realize that it is typical for linear homogeneous PDE's to have a solution space which is an infinite dimensional vector space? \"How many\" is not a good question.\r\n\r\nYou want a specific solution? How about this one:\r\n\r\n$f(x,y)=\\frac{\\sin^{2}(\\cos(e^{12x}))+\\tan^{-1}(x^{3})}{(1+x+x^{2})^{\\frac{7}{13}}}+y^{9}-3y^{4}+3$", "Solution_6": "[quote=\"Kent Merryfield\"][quote=\"mina2007\"]Yes, but can we find s specific solutions? if we can how many we have?[/quote]\nYou do realize that it is typical for linear homogeneous PDE's to have a solution space which is an infinite dimensional vector space? \"How many\" is not a good question.\n\nYou want a specific solution? How about this one:\n\n$f(x,y)=\\frac{\\sin^{2}(\\cos(e^{12x}))+\\tan^{-1}(x^{3})}{(1+x+x^{2})^{\\frac{7}{13}}}+y^{9}-3y^{4}+3$[/quote]\r\n\r\nThe problem says to find all solutions to this equations and also how many are there? thats why I can not solve it but thanks for trying", "Solution_7": "I repeat, \"how many?\" is the wrong question. Do you understand why the ludicrous function I wrote down (and I got tired of the game or I would have made it even more ridiculous) is a solution?", "Solution_8": "[quote=\"Kent Merryfield\"]I repeat, \"how many?\" is the wrong question. Do you understand why the ludicrous function I wrote down (and I got tired of the game or I would have made it even more ridiculous) is a solution?[/quote]\r\n\r\nI am sorry, this is my first time in PDE so I am not really familiar with those things,I am trying to learn..\r\nI do not understand why is than function a solution..if u can explain me that would be perfect...thank you so much", "Solution_9": "Let's start with an ODE $\\frac{df(x)}{dx}=0$. Then every constant function is a solution, so you have infinitely many of them. The same story happens with your equation though the solutions are not only constants now (every constant function is still a solution, though).\r\n\r\nTo check that Kent's function is a solution, just differentiate it and see what you get ;). Then think what (if anything) is special about his terrible function (Hint: it consists of two parts: a horrible fraction and a polynomial; what can you say about each of them?).\r\n\r\nFirst time is always hard with anything :). Just forget it is \"PDE\" and think about the problem one variable a time.", "Solution_10": "[quote=\"fedja\"]Let's start with an ODE $\\frac{df(x)}{dx}=0$. Then every constant function is a solution, so you have infinitely many of them. The same story happens with your equation though the solutions are not only constants now (every constant function is still a solution, though).\n\nTo check that Kent's function is a solution, just differentiate it and see what you get ;). Then think what (if anything) is special about his terrible function (Hint: it consists of two parts: a horrible fraction and a polynomial; what can you say about each of them?).\n\nFirst time is always hard with anything :). Just forget it is \"PDE\" and think about the problem one variable a time.[/quote]\r\n\r\nThank you so much :) you helped me...I understand this problem now...\r\nI also posted one more problem that I dont know about...if you can see maybe you know waht is about...i will go to study now but tomorrow I will be here and i will try to do it...i can see that you are good in it...thanks again", "Solution_11": "You are welcome :). One more rule: don't quote the last post in the thread entirely; you are replying to it by default. Use the quote option for specific parts you want to comment on." } { "Tag": [ "AMC", "AIME" ], "Problem": "How'd you do?\r\nFor reference, these are the (official, from [url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/07-AIMEanswers.shtml[/url]) answers:\r\n\r\n1. 83\r\n2. 52\r\n3. 15\r\n4. 105\r\n5. 539\r\n6. 169\r\n7. 477\r\n8. 030\r\n9. 737\r\n10. 860\r\n11. 955\r\n12. 875\r\n13. 080\r\n14. 224\r\n15. 989", "Solution_1": "Woah! I just took the AIMEII today and those answers gave me a bit of a scare...", "Solution_2": "I totally screwed up on it, made like 50 mistakes, and only got 3 correct", "Solution_3": "This year's AIME is more difficult." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "$ a$, $ b$, $ x$, $ y$ are positive real numbers and $ 1\\geq a^{5}\\plus{}b^{5}$, $ 1\\geq x^{5}\\plus{}y^{5}$. Proove that\r\n\\[ a^{2}x^{3}\\plus{}b^{2}y^{3}\\leq1\\]", "Solution_1": "Use AM-GM to find that \r\n$ 5a^{2}x^{3}\\leq 2a^{5}\\plus{}3x^{5}$ and $ 5b^{2}y^{3}\\leq 2b^{5}\\plus{}3y^{5}$" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "logarithms", "geometry", "number theory proposed" ], "Problem": "(i) Let $f,g \\in \\mathbb{Z}[x]$ be monic irreducible polynomials such that, for every $n$, the sets of prime divisors of $f(n)$ and $g(n)$ coincide. Show that $f \\equiv g$.\r\n\r\n(ii)* Is this necessarily true if we only assume that $f(n)$ and $g(n)$ have the same set of prime divisors for [i]infinitely many[/i] $n$?\r\n\r\nTo me, (ii) is open, though I believe that the answer is 'yes.' But give (i) a try, it is not hard at all! ;) \r\n\r\n--Vesselin", "Solution_1": "If we assume that they do not coincide, then they must be coprime. We can find integral polynomials $u,v$ s.t. $uf+vg$ is a constant $C$. This, together with the condition on the sets of prime factors, implies that for all $n$, the prime factors of $f(n)$ are among those of $C$, and this contradicts the well-known fact (discussed on the forum before, I believe) that there are infinitely many primes which divide some $f(n)$.", "Solution_2": "Ok, using heavy artilery (vess likes it, I like it too, so everyone is happy) one can prove the second assertion of the problem. Indeed, after some results of Schinzel, Mahler and many others that I forgot one can show (don't do this at home!) that for any polynomial $f$ there exists a constant $C>0$ such that for all sufficientely large $x$ the greatest prime factor of $f(x)$ is greater than or equal to $ C ln ln x$. I remember P\u00eferre gave me some more precise information in a private message about this estimation, but I lost the message. Pierre, could you please give some references about these estimations?", "Solution_3": "I do not remember what I gave you as results ;) , but it seems to me that we must add some hypothesis (otherwise the asserted result it trivially false for $f(x) = x^2$). But maybe is $f(x)$ assumed to be irreducible?\r\n\r\nAnyway, all I have from now is the following :\r\n\r\nLet $D(n)$ be the greatest prime divisor of $n$.\r\n\r\nLet $f(x)$ be a polynomial with integer coefficients, having at least two distinct roots (I assumed they are supposed to be integers, am I wrong?). Then there is an effectively computable constant $c>0$ such that $D(f(x)) > c \\log \\log (x)$ for every integer $x>3.$\r\n\r\nReference :\r\nP. Erd\u00f6s, T.N. Shorey, On the greatest prime factor of $2^p-1$ for a prime $p$ and other expressions, Acta Arithmetica, 30 (1976), p. 257-265.\r\n\r\nPierre.", "Solution_4": "I found a reference of trust: Erdos. He says that it has been proved that if $P$ is any irreducible polynomial with integer coefficients and degree greater than 1, then there exists $ c=c(f)>0$ such that the greatest prime factor of $ f(n)$ is greater than $ clnlnn$ for all $n$. Thus for our problem it remains the case when one of the polynomials is of degree 1.", "Solution_5": "It is really cool. In particular, it killes all problems like Ramanudjan $x^2+7=2^k$ and so on.", "Solution_6": "Yes, I find this area in mathematics absolutely incredible. Too bad we know practically nothing in this field. :(" } { "Tag": [ "algebra", "polynomial" ], "Problem": "Descartes rule of signs is a precalc and sometimes even advanced algebra topic. Anyway I've seen it before and have no idea why it works. It allows you to find out how many real negative/positive roots there are by looking at how often the sign of the coefficients in a polynomial changes. For example if you had 2 sign changes such as : x^5-x^3+2x, you would have 2 negative/positive real roots. Why does this work? (Note: terms such as 0x^4 are ignored.)", "Solution_1": "it is not an easy proof but, [url=http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF]http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF[/url]\r\n\r\nand it starts on the 16th page", "Solution_2": "[quote=\"Scelesta\"]Descartes rule of signs is a precalc and sometimes even advanced algebra topic. Anyway I've seen it before and have no idea why it works. It allows you to find out how many real negative/positive roots there are by looking at how often the sign of the coefficients in a polynomial changes. For example if you had 2 sign changes such as : x^5-x^3+2x, you would have 2 negative/positive real roots. Why does this work? (Note: terms such as 0x^4 are ignored.)[/quote]\r\nif it works i should say it is very incredible and cool.really really cool. :!:", "Solution_3": "reading over your post, your are close to the rule, but with the variation of sign, it is that number minus a mulitple of 2, if there is 3 variations in sign, that means that there could be 3 or 1 positive real roots, and if you do f(-x), then you can find the negative roots, (this follows directly from the other one since you are making the negative roots positive)" } { "Tag": [ "trigonometry", "inequalities", "geometry unsolved", "geometry" ], "Problem": "Prove that in any triangle $ABC$,\r\n\\[ 0 < \\cot { \\left( \\frac{A}{4} \\right)} - \\tan{ \\left( \\frac{B}{4} \\right) } - \\tan{ \\left( \\frac{C}{4} \\right) } - 1 < 2 \\cot { \\left( \\frac{A}{2} \\right) }. \\]", "Solution_1": "$\\tan(\\frac{\\pi}{4}-\\frac{A}{4})=\\tan(\\frac{B}{4}+\\frac{C}{4})$\r\n$\\frac{\\cot\\frac{A}{4}-1}{\\cot\\frac{A}{4}+1}=\\frac{\\tan\\frac{B}{4}+\\tan\\frac{C}{4}}{1-\\tan\\frac{B}{4}\\tan\\frac{C}{4}}$\r\nSo $\\cot\\frac{A}{4}-\\tan\\frac{B}{4}-\\tan\\frac{C}{4}-1=(1-\\frac{1-\\tan\\frac{B}{4}\\tan\\frac{C}{4}}{1+\\cot\\frac{A}{4}})(\\cot\\frac{A}{4}-1)$\r\n$=\\frac{\\cot\\frac{A}{4}+\\tan\\frac{B}{4}\\tan\\frac{C}{4}}{1+\\cot\\frac{A}{4}}(\\cot\\frac{A}{4}-1)$\r\n$=\\frac{1+\\tan\\frac{A}{4}\\tan\\frac{B}{4}\\tan\\frac{C}{4}}{1+\\tan\\frac{A}{4}}(\\cot\\frac{A}{4}-1)$\r\nSince $\\tan\\frac{A}{4},\\tan\\frac{B}{4},\\tan\\frac{C}{4}\\in(0,1)$\r\nSo it's obvious that's larger than 0.\r\nAnd $2\\cot\\frac{A}{2}=\\frac{\\cot^2\\frac{A}{4}-1}{\\cot\\frac{A}{4}}$\r\nSo we just need to prove\r\n$\\frac{1+\\tan\\frac{A}{4}\\tan\\frac{B}{4}\\tan\\frac{C}{4}}{1+\\tan\\frac{A}{4}}<\\frac{\\cot\\frac{A}{4}+1}{\\cot\\frac{A}{4}}=1+\\tan\\frac{A}{4}$\r\nEasy to see left side is smaller than 1 and the right side larger than 1.", "Solution_2": "[quote=\"Rushil\"]Prove that in $ \\triangle\\ ABC$ there is the relation $ \\cot \\frac A4 \\minus{} \\tan\\frac B4 \\minus{} \\tan\\frac C4 \\minus{} 1 < 2 \\cot\\frac A2$ .[/quote]\r\nDenote $ \\tan \\frac A4 \\equal{} x\\ ,\\ \\tan \\frac B4 \\equal{} y\\ ,\\ \\tan\\frac C4 \\equal{} z$ and $ S \\equal{} y \\plus{} z$ , $ P \\equal{} yz$ . . Observe that $ \\frac {\\pi}{4} \\minus{} \\frac A4 \\equal{} \\frac B4 \\plus{} \\frac C4\\implies$\r\n\r\n$ \\tan\\left(\\frac {\\pi}{4} \\minus{} \\frac A4\\right) \\equal{} \\tan\\left(\\frac B4 \\plus{} \\frac C4\\right)\\implies \\frac {1 \\minus{} x}{1 \\plus{} x} \\equal{} \\frac {S}{1 \\minus{} P}\\implies x \\equal{} \\frac {1 \\minus{} S \\minus{} P}{1 \\plus{} S \\minus{} P}$ . The proposed inequality is \r\n\r\nequivalently with $ \\frac 1x < S \\plus{} 1 \\plus{} 2\\cdot\\frac {1 \\minus{} x^2}{2x}\\Longleftrightarrow x < S \\plus{} 1$ what is truly because $ 0 < x < 1 < S \\plus{} 1$ .\r\n\r\n[b]Remark.[/b] $ \\frac {\\pi}{4} \\minus{} \\frac A4 \\equal{} \\frac B4 \\plus{} \\frac C4\\Longleftrightarrow$ $ \\frac {1 \\minus{} x}{1 \\plus{} x} \\equal{} \\frac {y\\plus{}z}{1 \\minus{} yz}\\Longleftrightarrow$ $ x\\plus{}y\\plus{}z\\plus{}xy\\plus{}yz\\plus{}zx\\equal{}1\\plus{}xyz\\Longleftrightarrow$ \r\n\r\n$ (1\\minus{}x)(1\\minus{}y)(1\\minus{}z)\\equal{}2(1\\minus{}x\\minus{}y\\minus{}z)\\implies$ $ \\boxed {x\\plus{}y\\plus{}z<1}$ . This solution is similarly with upper [b]Jin[/b]'s proof." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that $ p(n)\\equal{}(a\\plus{}0.5)^{n}\\plus{}(b\\plus{}0.5)^{n}$ is an integer for only finitely many $ n$\r\nin genral is there an approach to solve problems of this kind that is\r\n\"Prove that there are infinitely many...\"\r\n\"Prove that there are finitely many...\"\r\n\"Prove that for every...\"", "Solution_1": "As much as my knowledge and intuition goes, there is no general way to . Probably with a bit of elaboration, one could reduce it to Hilbert's 10th problem to make it more formal.\r\n\r\nBut the problem you mention was posted before. A usefull way to attack this one is to write it as $ 2^n | u^n \\plus{}v^n$ for odd $ u,v$ (and then to proceed further).", "Solution_2": "General [i]solutions[/i], no. General strategies? Maybe contradiction in the finite case. In the infinite case, an explicit construction usually works better than a non-constructive proof." } { "Tag": [ "complex analysis", "function", "calculus", "integration", "logarithms", "complex analysis unsolved" ], "Problem": "Let f be a holomorphic function inside the circle ${|z| < 1}$ and continuous up to $|z| \\leq 1$ Let M be the supremum of $|f|$ on $|z| \\leq 1$. Let L be the intersection of $|z| \\leq 1$ and $Re[z] = 1/2$ Let m be the supremum of f on L. Show that $|f(0)|^{3}\\leq m*M^{2}$. The hint given is: consider the product of some functions obtained from f.\r\n\r\nI expect you're going to have to use the Cauchy integral formula, and the standard method of bounding (supremum norm * Length), but I can't figure out the appropriate functions. \r\n\r\nI was thinking something like: $(f^{2}/m)^{1/3}*(f^{2}/m)^{2/3}*m/f$ since $m/M < 1$ on the unit circle.\r\n\r\nbut you get $|f(0)| \\leq (M^{2}/m)^{\\frac{2}{3}}*(M^{2}/m)^{\\frac{1}{3}}*1 \\implies |f(0)|^{3}\\leq M^{4}/m^{2}*M^{2}/m^{2}= M^{6}/m^{3}$, which seems close, but doesn't do me good.", "Solution_1": "You meant $|f(0)|^{3}\\le mM^{2}$, right? \r\nConsider $F(z)=f(z)f(\\zeta z)f(\\zeta^{2}z)$, where $\\zeta^{3}=1$. Apply the maximum principle to $F$ on a triangle inscribed into the circle.", "Solution_2": "You are correct. I still don't understand your solution. I understand that $F(z)=f(z)f(\\zeta z)f(\\zeta^{2}z) \\implies |F(z)|=|f(z)f(\\zeta z)f(\\zeta^{2}z)|\\leq |f(z)f(\\zeta z)f(\\zeta^{2}z)|$ on the bdy of the triangle. So $|F(0)| \\leq |f(z)f(\\zeta z)f(\\zeta^{2}z)|$ on the bdy of the triangle. However, why would the maximum on the boundary equal $M^{2}m$? Am I missing an intermediate step, somewhere?\r\n\r\nSorry if this question seems trivial.", "Solution_3": "I meant the triangle with vertices $-1$ and $1/2\\pm i\\sqrt{3}/2$. On each side of this triangle one of the functions $f(z)$, $f(\\zeta z)$, $f(\\zeta^{2}z)$ is bounded by $m$. The other two are bounded by $M$. This is where $M^{2}m$ comes from.", "Solution_4": "Generalization: Let $r(\\theta) = |f(e^{i \\theta})|$. Is it true that \r\n\\[|f(0)| \\le \\exp \\left( \\frac{1}{2 \\pi}\\int_{0}^{2 \\pi}\\ln r(\\theta) d \\theta \\right) \\quad ?\\]", "Solution_5": "You are absolutely right! :thumbup: The function $u(z)=\\log |f(z)|$ is [url=http://en.wikipedia.org/wiki/Subharmonic_function]subharmonic[/url], which means that it has the sub-mean value property: $u(z_{0})\\le \\frac{1}{2\\pi}\\int_{0}^{2\\pi}u(z_{0}+re^{i\\theta})\\,d\\theta$ for any $r>0$ (provided that the circle with center $z_{0}$ and radius $r$ lies in the region where $u$ is subharmonic)." } { "Tag": [ "LaTeX" ], "Problem": "When writing negatives, I usually use the minus sign. The only problem is that it turns out big, like $ \\minus{}22$. Is there a way to get a smaller negative sign?", "Solution_1": "You can make the text smaller for that particular symbol.", "Solution_2": "It doesn't work on $ \\text{\\LaTeX}$, unless I'm doing something wrong.", "Solution_3": "You could use the text version which gives $ \\text{\\minus{}}22$ instead of $ \\minus{}22$. The minus sign is a little low so you could raise it using $ \\raisebox{0.3ex}{\\text{\\minus{}}} 22$. In a document you could use \\newcommand{\\minus}{\\raisebox{0.2ex}{\\text{-}}} so that you could write \\minus 22", "Solution_4": "Like this?\r\n\r\n$ \\text{\\minus{}22}$\r\n\r\nThanks! :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Consider the points $P$ =$(0,0)$,$Q$ = $(1,0)$, $R$= $(2,0)$, $S$ =$(3,0)$ in the $xy$-plane. Let $A,B,C,D$ be four finite sets of colours(not necessarily distinct nor disjoint). In how many ways can $P,Q,R$ be coloured bu colours in $A,B,C$ respectively if adjacent points have to get different colours? In how many ways can $P,Q,R,S$ be coloured by colours in $A,B,C,D$ respectively if adjacent points have to get different colors?", "Solution_1": "[quote=\"Rushil\"]Let $A,B,C,D$ be four finite sets of colours(not necessarily distinct nor disjoint).[/quote]\r\n\r\nWhat do you mean? You can give more than one color to each point?\r\n\r\nPierre.", "Solution_2": "I think its clear that one point gets one colour but $A,B,C,D$ might not be distinct." } { "Tag": [], "Problem": "Caculate:\r\n$ S\\equal{}\\frac{1}{1!2003!}\\plus{}\\frac{1}{3!2001!}\\plus{}...\\plus{}\\frac{1}{2001!3!}\\plus{}\\frac{1}{2003!1!}$", "Solution_1": "[hide=\"solution\"]\nlet \n$ S \\equal{}\\frac{1}{1!2003!}\\plus{}\\frac{1}{3!2001!}\\plus{}\\cdots\\plus{}\\frac{1}{2001!3!}\\plus{}\\frac{1}{2003!1!}$\n$ S \\equal{}\\frac{1}{2004!}(\\frac{2004!}{1!2003!}\\plus{}\\frac{2004!}{3!2001!}\\plus{}\\cdots\\plus{}\\frac{2004!}{2001!3!}\\plus{}\\frac{2004!}{2003!1!})$\n$ S \\equal{}\\frac{1}{2004!}(\\sum_{k \\equal{} 0}^{1001}\\dbinom{2004}{2k\\plus{}1})$\n$ S \\equal{}\\frac{2^{2003}}{2004!}$\n[/hide]" } { "Tag": [], "Problem": "How many digits are in the value of the following expression?\r\n$2^{2007}\\times5^{1957}\\div4^{28}$\r\nI keep getting 1957, but they say 1956. \r\n\r\nWho is right?", "Solution_1": "They are, check with a calculator, answer is $1.5625 * 10^{1955}$", "Solution_2": "[hide]$\\frac{2^{2007} \\cdot 5^{1957}}{ 4^{28}}=\\frac{10^{1957}\\cdot 2^{50}}{2^{56}}$\n$=\\frac{1}{2^6} \\cdot 10^{1957}$\n$\\frac{1}{10} \\cdot 10^{1957} > \\frac{1}{64} \\cdot 10^{1957}> \\frac{1}{100} \\cdot 10^{1957}$\nThere are 1956 digits.[/hide]", "Solution_3": "I get 1956, what they said :? I think you are overcounting", "Solution_4": "Heres what I did\r\n\r\n$4^{28}=2^{56}$\r\n\r\nSo the expression becomes\r\n\r\n$2^{1951}5^{1957}$\r\n\r\nThat is $10^{1951}5^6$.\r\n$5^6$ has 5 digits while $10^{1951}$ has 1952. Add to get 1957", "Solution_5": "Look at pianoforte's solution. It looks like what I did except much neater. You are overcounting 1, because you should subtract one when you add the 6 and 1951-example-one digit number times two digit number (1 and 10) If you add 1 digit plus 2 digits you get three digits-but in reality you get 1*10=10=2 digits. ;)", "Solution_6": "Nah, I figured out what he did wrong. $10^{1951}$ has 1951 digits, not 1952 digits :rotfl:", "Solution_7": "um no. 10^1 has 2, 10^2 has 3,...\r\nWhat I did wrong (just realized) is that when multiplying by 5^6, one of those digits takes the place of the 1 in 10^1951", "Solution_8": "[quote=\"biffanddoc\"]um no. 10^1 has 2, 10^2 has 3,...\nWhat I did wrong (just realized) is that when multiplying by 5^6, one of those digits takes the place of the 1 in 10^1951[/quote]\r\nPfft.. I was thinking about number of 0s, oops.", "Solution_9": "[quote=\"biffanddoc\"]Heres what I did\n\n$4^{28}=2^{56}$\n\nSo the expression becomes\n\n$2^{1951}5^{1957}$\n\nThat is $10^{1951}5^6$.\n$5^6$ has 5 digits while $10^{1951}$ has 1952. Add to get 1957[/quote]\r\n\r\nThe thing is, the first digit of 10^1951 is 1, so when it is multiplied by 5^6, you have to take away one digit." } { "Tag": [ "percent" ], "Problem": "A mixture of 40 liters of paint is 30 percent red tint, 15 percent yellow tint, and 50 percent water. If 10 liters of yellow tint are added to the original mixture, what is the percent of yellow tint in the new mixture?", "Solution_1": "[hide]\n$ \\frac {40 \\cdot .15 \\plus{} 10}{50} \\equal{} \\frac {8}{25}$.\n[/hide]", "Solution_2": "Wait a minute...\r\n[quote=\"sharkman\"]A mixture of 40 liters of paint is 30 percent red tint, 15 percent yellow tint, and 50 percent water.[/quote]\r\nThat only adds up to 95 percent." } { "Tag": [ "limit", "algebra proposed", "algebra" ], "Problem": "Given sequence $\\{x_{k}\\}_{1}^\\infty$ ,$x_{k}=\\sum_{i=1}^{k}\\frac{k}{(k+1)!}$\r\nfind the limit\r\n$\\lim_{n\\rightarrow+\\infty}\\sqrt[n]{x_{1}^{n}+x_{2}^{n}+...x_{2007}^{n}}$", "Solution_1": "Note : $\\lim_{n\\to\\infty}\\sqrt[n]{a_{1}^{n}+a_{2}^{n}+...+a_{k}^{n}}=\\max{(a_{1},a_{2},...,a_{k})}$", "Solution_2": "[quote=\"N.T.TUAN\"]Note : $\\lim_{n\\to\\infty}\\sqrt[n]{a_{1}^{n}+a_{2}^{n}+...+a_{k}^{n}}=\\max{(a_{1},a_{2},...,a_{k})}$[/quote]\r\n\r\nCan you give a short proof of that or a hint?", "Solution_3": "Yes, because $\\max{(a_{1},a_{2},...,a_{k})}\\leq\\sqrt[n]{a_{1}^{n}+...+a_{k}^{n}}\\leq\\max{(a_{1},a_{2},...,a_{k})}\\sqrt[n]{k}$.", "Solution_4": "$\\max{(a_{1},a_{2},...,a_{k})}=?$ :D", "Solution_5": "\\[\\max{(x_{1},x_{2},...,x_{2007})}=x_{2007}=\\sum_{k=1}^{2007}\\frac{k}{(k+1)!}=\\sum_{k=1}^{2007}(\\frac{1}{k!}-\\frac{1}{(k+1)!})=\\\\1-\\frac{1}{2008!}\\]\r\n. Done! :D" } { "Tag": [ "limit", "integration" ], "Problem": "Find:\r\n$\\mathop {\\lim }\\limits_{n \\to \\infty } \\left( {2n + 1} \\right)\\int\\limits_0^1 {x^n e^x dx} $", "Solution_1": "hey metru\r\nnice problem but maybe u should read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]this[/url] first\r\nand then post ur problem again [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]here[/url]!\r\n\r\npeeta", "Solution_2": "But can somebody at least solve the problem???", "Solution_3": "he did the right thing to do:\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45226]www.artofproblemsolving.com/Forum/viewtopic.php?t=45226[/url]" } { "Tag": [ "geometry", "rectangle", "rotation", "geometric transformation", "AMC" ], "Problem": "Rectangle $ PQRS$ lies in a plane with $ PQ = RS = 2$ and $ QR = SP = 6$. The rectangle is rotated $ 90^\\circ$ clockwise about $ R$, then rotated $ 90^\\circ$ clockwise about the point that $ S$ moved to after the first rotation. What is the length of the path traveled by point $ P$?\r\n\r\n${ \\textbf{(A)}\\ (2\\sqrt3 + \\sqrt5})\\pi \\qquad \\textbf{(B)}\\ 6\\pi \\qquad \\textbf{(C)}\\ (3 + \\sqrt {10})\\pi \\qquad \\textbf{(D)}\\ (\\sqrt3 + 2\\sqrt5)\\pi \\\\ \\textbf{(E)}\\ 2\\sqrt {10}\\pi$", "Solution_1": "[hide]The first rotation moves $ P$ one-fourth of the circumference of the circle with radius $ PR \\equal{} 2\\sqrt {10}$. Thus, $ P$ moves $ \\frac {1}{4}*4\\sqrt {10}*\\pi \\equal{} \\sqrt {10}\\pi$. We then rotate $ 90^\\circ$ around the new position of $ S$. The distance $ P$ moves is simply one-fourth of the circumference of the circle with radius $ SP \\equal{} 6$. Thus, we have that our answer is $ \\boxed{(3 \\plus{} \\sqrt {10})\\pi\\equal{}C}$[/hide]", "Solution_2": "My explanation\r\n\r\nThe rotations are 1/4 circumference of two different circles because 90 degrees is one-fourth of a whole circle's degree measure of 360. The first circle will have a radius of the length of rectangle and the second cirlce will have a radius of the diagonal of the rectangle." } { "Tag": [], "Problem": "If $ y \\minus{} x \\equal{} 7$, what is the value of $ x \\minus{} y$?", "Solution_1": "We know that $ x\\minus{}y\\equal{}\\minus{}(y\\minus{}x)$. Since we know that $ y\\minus{}x\\equal{}7$, we can substitute and get $ x\\minus{}y\\equal{}\\boxed{\\minus{}7}$" } { "Tag": [ "inequalities", "logarithms", "search", "inequalities proposed" ], "Problem": "Let $ p \\geq \\log{(3/2)} / \\log{2}$ be an arbitrary real number. Then, for all triangles with sidelengths $ a$, $ b$, $ c$ and semiperimeter $ s$, prove that\r\n\\[ \\left(\\frac {s \\minus{} a}{a}\\right)^{p} \\plus{} \\left(\\frac {s \\minus{} b}{b}\\right)^{p} \\plus{} \\left(\\frac {s \\minus{} c}{c}\\right)^{p}\\geq \\frac {3}{2^{p}}.\r\n\\]", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=1744010150&t=76277" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\\plus{}b\\plus{}c \\equal{} 1$.\r\nProve that\r\n\\[ \\sum_{cyc} \\left(a \\plus{} \\frac{1}{a}\\right)^2 \\ge \\frac{100}{3}\r\n\\]\r\nand\r\n\\[ \\sum_{cyc} \\left(a \\plus{} \\frac{1}{a}\\right)\\left(a \\plus{} \\frac{1}{b}\\right) \\ge \\frac{100}{3}\r\n\\]\r\n\r\nI proved it by a technical method. \r\nBut it doesn't seem to be....elegant.\r\nI hope to show your elegant proof.", "Solution_1": "for the first one,\r\nBy Cauchy,\r\n$ S\\equal{}\\sum_{cyc} \\left(a \\plus{} \\frac{1}{a}\\right)^2 \\ge \\frac{1}{3}\\left(1 \\plus{} \\sum \\frac{1}{a}\\right)$\r\n$ (a\\plus{}b\\plus{}c)\\left(\\sum \\frac{1}{a}\\right)\\geq 9 \\Leftrightarrow \\sum \\frac{1}{a} \\geq 9$\r\nthen $ S\\geq \\frac{100}{3}$ equality hold if $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$\r\nfor the second one,\r\nwe have,\r\n$ S\\equal{}\\sum\\left(a\\plus{}\\frac{1}{a}\\right)\\left(a\\plus{}\\frac{1}{b}\\right)\\equal{}\\sum a^2 \\plus{}\\sum \\frac{a}{b}\\plus{} 3\\plus{}\\frac{1}{abc}$\r\nBy AM-GM $ abc\\leq \\frac{(a\\plus{}b\\plus{}c)^3}{27}\\Leftrightarrow \\frac{1}{abc}\\geq \\frac{1}{abc}$ (1)\r\nand also $ \\sum \\frac{a}{b} \\geq 3$ (2)\r\nBy Cauchy, $ \\sum a^2 \\geq \\frac{1}{3}(a\\plus{}b\\plus{}c)^2\\equal{}\\frac{1}{3}$ (3)\r\nfrom (1) and (2) and (3), we get,\r\n$ S\\geq \\frac{1}{3}\\plus{}3\\plus{}27\\plus{}3\\equal{}\\frac{100}{3}$ equality hold if $ a\\equal{}b\\equal{}c\\equal{}\\frac{1}{3}$", "Solution_2": "Thank you for your proof.\r\n\r\nMy proof\r\n$ \\sqrt[3]{abc} \\le \\frac{a+b+c}{3}=\\frac{1}{3}$.\r\nSo, $ abc \\le 27$.\r\n$ 3(a^2+b^2+c^2)-(a+b+c)^2=(a-b)^2+(b-c)^2+(c-a)^2 \\ge 0$.\r\nSo, $ a^2+b^2+c^2 \\ge \\frac{1}{3}(a+b+c)^2 = \\frac{1}{3}$.\r\n\r\n[b]first[/b]\r\n\\begin{eqnarray*}\r\n\\sum \\left(a + \\frac{1}{a}\\right)^2\r\n& = & \\sum a^2 + \\sum \\frac{1}{a^2} + 6 \\\\\r\n& \\ge & \\frac{1}{3} + 3\\sqrt[3]{\\frac{1}{(abc)^2}} + 6 \\\\\r\n& \\ge & \\frac{1}{3} + 3\\sqrt[3]{27^2} + 6 \\\\\r\n& = & \\frac{1}{3} + 27 + 6 \\\\\r\n& = & \\frac{100}{3}\r\n\\end{eqnarray*}\r\n\r\n[b]second[/b]\r\n\\begin{eqnarray*}\r\n\\sum \\left(a + \\frac{1}{a}\\right)\\left(a + \\frac{1}{b}\\right)\r\n& = & \\sum a^2 + \\sum \\frac{a}{b} + \\sum \\frac{1}{ab} +3 \\\\\r\n& \\ge & \\frac{1}{3} + 3 + 3\\sqrt[3]{\\frac{1}{(abc)^2}} + 3 \\\\\r\n& \\ge & \\frac{1}{3} + 3 + 3\\sqrt[3]{27^2} + 3 \\\\\r\n& = & \\frac{1}{3} + 3 + 27 + 3 \\\\\r\n& = & \\frac{100}{3}\r\n\\end{eqnarray*}\r\n\r\n\r\nSecond is proof similar to me.\r\nI felt that the proof of first was simple and excellent.\r\n\r\nAnybody, is there another proof?", "Solution_3": "Here is my solution:\r\n\r\nQ1. For $ a > 0$, we have $ \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\plus{} \\frac {160}{3}a \\minus{} \\frac {260}{9} \\equal{} \\frac {(3a \\minus{} 1)^2}{9a^2}(a^2 \\plus{} 54a \\plus{} 9)\\geq 0$\r\n\r\n$ \\Longleftrightarrow \\left(a \\plus{} \\frac {1}{a}\\right)^2\\geq \\minus{} \\frac {160}{3}a \\plus{} \\frac {260}{9}.$\r\n\r\n$ \\therefore \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\plus{} \\left(b \\plus{} \\frac {1}{b}\\right)^2 \\plus{} \\left(c \\plus{} \\frac {1}{c}\\right)^2\\geq \\minus{} \\frac {160}{3}(a \\plus{} b \\plus{} c) \\plus{} \\frac {260}{9}\\cdot 3.$, \r\n\r\nyielding $ \\therefore \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\plus{} \\left(b \\plus{} \\frac {1}{b}\\right)^2 \\plus{} \\left(c \\plus{} \\frac {1}{c}\\right)^2\\geq \\frac {100}{3}.$\r\n\r\nAlternative Solution:\r\n\r\nLet $ f(a) \\equal{} \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\plus{} \\frac {160}{3}a \\minus{} \\frac {260}{9}\\ (a > 0)$,\r\n\r\n$ f'(a) \\equal{} \\frac {1}{3}(3a \\minus{} 1)(a^3 \\plus{} 27a^2 \\plus{} 9a \\plus{} 3)$, yielding $ f(a)\\geq f\\left(\\frac {1}{3}\\right) \\equal{} \\frac {100}{9}$, the rest is easy.", "Solution_4": "I found simpler proof !!\r\n\r\nWe use Jensen ineq.\r\n$ \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\ (a > 0)$ is convex function.\r\nSo, $ \\frac {\\sum \\left(a \\plus{} \\frac {1}{a}\\right)^2}{3} \\ge \\left(\\frac {\\sum a}{3} \\plus{} \\frac {1}{\\frac {\\sum a}{3}}\\right)^2 \\equal{} \\left(\\frac {1}{3} \\plus{} 3\\right)^2 \\equal{} \\frac {100}{9}$.\r\nThus, $ \\sum \\left(a \\plus{} \\frac {1}{a}\\right)^2 \\ge \\frac {100}{3}$." } { "Tag": [ "inequalities", "3-variable inequality", "cyclic inequality", "square root inequality" ], "Problem": "Find the smallest constant $ k$ such that\r\n\r\n$ \\frac {x}{\\sqrt {x \\plus{} y}} \\plus{} \\frac {y}{\\sqrt {y \\plus{} z}} \\plus{} \\frac {z}{\\sqrt {z \\plus{} x}}\\leq k\\sqrt {x \\plus{} y \\plus{} z}$\r\n\r\nfor all positive $ x$, $ y$, $ z$.", "Solution_1": "The correct answer is $ k=\\frac54$.", "Solution_2": "This inequality is beautiful and very difficult to solve.\r\nI woul like to know if you Fedor created it or you found it on a mathematical review or it is an olympiad problem.\r\n\r\nThank you.", "Solution_3": "I agree that the inequality is extremely difficult and beautiful. It is an inequality that you remember after you had solved it. I saw it in Crux and I think (but I'm not sure) that it was given also at an american contest.", "Solution_4": "Unfortunately, yesterday was a very bad day for me. I will post a detailed solution of this problem and I hope today I'm not so stupid.\r\n As in the above \"solution\", we have to prove that if a^2,b^2,c^2 are the sides of o triangle( even degenerate), then:\r\n a+b+c+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/(2sqrt2)*sqrt(a^2+b^2+c^2).\r\n\r\nProposition:\r\n The above inequality is veryfied when b<=c<=a<=sqrt(b^2+c^2).\r\n\r\n Proof:\r\n It is easy to prove that 5/(2sqrt2)*sqrt(a^2+b^2+c^2)>=5/4*(a+sqrt(b^2+c^2)). Thus, we have to prove that (a+b+c)+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/4(a+sqrt(b^2+c^2)), which reduces (after clearing denominators) to proving that f(a)<=0, where \r\n f(x)=4x^3(c-b)-x^2bc+x(4b^3+4b^2c+4bc^2-4c^3-5bcsqrt(b^2+c^2))+4bc(c^2-b^2).\r\n We know that b<=c<=a<=sqrt(b^2+c^2). If b=c, proving that f(a)<=0 reduces to ab^2(a-b+(5sqrt2-7)b)>0, which is obvious because a>=b.\r\n So, we can suppose that b=0 because c>=b. It is trivial to prove that f(c)<=0 (it reduces to 5sqrt(b^2+c^2)>=4b+3c, which is Cauchy and that f(sqrt(b^2+c^2))<=0 (which reduces to bc(sqrt(b^2+c^2)-2b)^2>=0.\r\n So f has a negative (possible 0) root, a root between 0 and c and another one which is at least sqrt(b^2+c^2). Consequently, f doesn't change sign in the interval [c, sqrt(b^2+c^2)]. Since it isn't positive in c and sqrt(b^2+c^2) and c<=a<=sqrt(b^2+c^2), we must have f(a)<=0.\r\n Now let's solve the problem. One can easily see that we can suppose that c>=b (if not, we interchange b and c and the LHS will become smaller, while the RHS will remain the same). Now, if a<=b<=c, then we apply the proposition for the triplet (c,a,b) (this means that we have sqrt(b^2+a^2)>=c>=b>=a and we can appply the proposition) and if b<=a<=c ,then we apply the proposition for the triplet (c,b,a).", "Solution_5": "Congratulations, Harazi, your solution is now correct. I verified it in detail. Is very nice solution for this hard problem. I also came to idea to change from (x,y,z) to (a,b,c), but can't go further. The case (a>=c>=b) is really main case in this inequality. The idea f(c) <= 0, f(sqrt(b^2+c^2)) <=0 is nice. \r\n\r\nNamdung", "Solution_6": "[quote=\"hitek\"]Let $x,y,z$ be real non-negative numbers, prove that\n\\[ \\frac x{\\sqrt{x+y}}+\\frac y{\\sqrt{y+z}}+\\frac z{\\sqrt{z+x}}\\leq\\frac54\\sqrt{x+y+z}. \\]\n[/quote]\r\nThis was suggested by Walther Janous's problem Crux 1366 [1989:271].\r\nThere can be someone it knows? What there was, in Crax 1366? Thank you.", "Solution_7": "\\[ \\frac{a}{\\sqrt{a+b}}+\\frac{b}{\\sqrt{b+c}}+\\frac{c}{\\sqrt{c+a}}\\le \\frac{5}{4}\\sqrt{a+b+c} \\]\r\n\r\nHere's my solution\r\n\r\nNotice that if one of three number $a,b,c$ equal $0$ then we are done. Otherwise, let \r\n\\[ x=\\sqrt{\\frac{a+b}{2}}, y=\\sqrt{\\frac{a+c}{2}}, z=\\sqrt{\\frac{b+c}{2}},k=\\frac{5\\sqrt{2}}{4} \\]\r\nRewrite the problem into the form\r\n\\[ (x+y+z)+(x-y)(x-z)(z-y)(\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{zx}) \\le k\\sqrt{x^2+y^2+z^2} (*) \\]\r\nClearly, we only need to solve problem incase $x \\ge y \\ge z$\r\nNotice that for all $t \\ge 0$ then\r\n\\[ k\\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \\ge k\\sqrt{x^2+y^2+z^2}+3t \\] \r\nFrom this result, we deduce that if we decrease all $x,y,z$ by a number $t \\le \\min(x,y,z)$ then the expression\r\n\\[ (x-y)(x-z)(z-y)(\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{zx}) \\]\r\nis increased but the expression\r\n\\[ k\\sqrt{x^2+y^2+z^2}-(x+y+z) \\]\r\nis decreased. So after choosing a number $t$ for which $x-t,y-t,z-t$ is three lengths of a right triangle, we only need to consider the first problem incase $abc=0$, which was proved as above.", "Solution_8": "This is Crux Mathematicorum, volume 17,#1, January 1991, problem 1490. Proposed by J.Garfunkel. Search in the net. There you will find the solution!\r\n Or see here post #9:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Crux&t=1459\r\nbut it is very hard to read.", "Solution_9": "I think this would help", "Solution_10": "[quote=\"Fedor Bakharev\"]Find the smallest constant $k$ such that \n$\\frac{x}{\\sqrt{x+y}}+\\frac{y}{\\sqrt{y+z}}+\\frac{z}{\\sqrt{z+x}}\\leq k\\sqrt{x+y+z}$\nfor all positive $x,y,z$.[/quote]\r\nThis was proposed by Jack Garfunkel, see:\r\nProblem $1490^{*}$, [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1989, No.9, p.270.\r\n\r\n$\\sum_{cyc}\\frac{x}{\\sqrt{(x+y)(x+y+z)}}$\r\n\r\n$\\leq\\sum_{cyc}\\frac{425x^{2}+9y^{2}+81z^{2}-54yz+810zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}=\\frac{5}{4}$,\r\n\r\nwith equality if $x=3,y=1,z=0$.", "Solution_11": "Any proof to go with it? Is it just another computer proof... that's a really nice identity/inequality if you can prove it. What is the equality case for\r\n\\[\\frac{425x^{2}+9y^{2}+81z^{2}-54yz+8101zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}\\ge\\frac{x}{\\sqrt{(x+y)(x+y+z)}}\\]\r\nIs equality at $x=3y,z\\to 0$?\r\n\r\nedit. oops didn't see your equality case... but it is in fact for any $x=3y$, right?", "Solution_12": "And now, here is my solution for it\r\n\\[LHS^{2}\\le \\sum a(5a+b+9c)\\cdot \\sum \\frac{a}{(a+b)(5a+b+9c)}=5(a+b+c)^{2}\\cdot\\sum\\frac{a}{(a+b)(5a+b+9c)}\\]\r\nHence, it suffices to prove that\r\n\\[(a+b+c)\\cdot\\sum\\frac{a}{(a+b)(5a+b+9c)}\\le \\frac{5}{16}\\]\r\nBut we have\r\n\\[RHS-LHS=\\frac{\\sum ab(a+b)(a+9b)(a-3b)^{2}+243\\sum a^{3}b^{2}c+835\\sum a^{3}bc^{2}+232\\sum a^{4}bc+1230a^{2}b^{2}c^{2}}{16\\prod (a+b) \\cdot \\prod (5a+b+9c)}\\ge 0 \\]\r\nWe are done. :)\r\n\r\n@: I'm looking for you idea. :)", "Solution_13": "[quote=\"hungkhtn\"]\\[ k\\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \\ge k\\sqrt{x^2+y^2+z^2}+3t \\] \nFrom this result, we deduce that if we decrease all $x,y,z$ by a number $t \\le \\min(x,y,z)$ then the expression\n\\[ (x-y)(x-z)(z-y)(\\frac{1}{xy}+\\frac{1}{yz}+\\frac{1}{zx}) \\]\nis increased but the expression\n\\[ k\\sqrt{x^2+y^2+z^2}-(x+y+z) \\]\nis decreased. [/quote]\n\nI am sorry,I think your solution is not very correct.Because \n\\[ k\\sqrt{x^2+y^2+z^2}-(x+y+z) \\]\nmay be not decreased when $x,y,z$ reduce be a number $t \\le \\min(x,y,z)$ .", "Solution_14": "Dear Zhaobin,\n i think the idea by mv method is correct,though it is not detailed.", "Solution_15": "[quote=\"Fedor Bakharev\"]Find the smallest constant $ k$ such that\n\n$ \\frac {x}{\\sqrt {x \\plus{} y}} \\plus{} \\frac {y}{\\sqrt {y \\plus{} z}} \\plus{} \\frac {z}{\\sqrt {z \\plus{} x}}\\leq k\\sqrt {x \\plus{} y \\plus{} z}$\n\nfor all positive $ x$, $ y$, $ z$.[/quote]\nFor $x=3$, $y=1$ and $z\\rightarrow0^+$ we get $k\\geq\\frac{5}{4}$.\nWe'll prove that:\n\\[ \\frac {x}{\\sqrt {x \\plus{} y}} \\plus{} \\frac {y}{\\sqrt {y \\plus{} z}} \\plus{} \\frac {z}{\\sqrt {z \\plus{} x}}\\leq \\frac{5}{4}\\sqrt {x \\plus{} y \\plus{} z}\\]\nfor all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\\neq0$.\nBy Cauchy-Schwarz $\\left(\\sum_{cyc}\\frac {x}{\\sqrt {x \\plus{} y}}\\right)^2\\leq\\sum_{cyc}\\frac{x(2x+4y+z)}{x+y}\\sum_{cyc}\\frac{x}{2x+4y+z}$.\nId est, it remains to prove that\n$\\sum_{cyc}\\frac{x(2x+4y+z)}{x+y}\\sum_{cyc}\\frac{x}{2x+4y+z}\\leq\\frac{25(x+y+z)}{16}$, which is\n$\\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$\n$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\\geq0$, which is \n$\\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$\n$+\\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\\geq0$, which is obvious.", "Solution_16": "Can we let a=sqrt(x+y),b=sqrt(y+z),c=sqrt(z+x),and then use 2c cosB to replace (a^2-b^2+c^2)/a...and so on(it must form a triangle because x,y,z are positive).But I don't know how to do it then.", "Solution_17": "[quote=\"hitek\"]Let $x,y,z$ be real non-negative numbers, prove that\n\\[ \\frac x{\\sqrt{x+y}}+\\frac y{\\sqrt{y+z}}+\\frac z{\\sqrt{z+x}}\\leq\\frac54\\sqrt{x+y+z}. \\]\n[/quote]\nI think I've got a new solution: \nLet $x+y=c^2,y+z=a^2,z+x=b^2$, with $a,b,c>0$.Wlog, we suppose $a\\ge b,c$.Then we have :\n$x=\\frac{c^2+b^2-a^2}{2}$,$y=\\frac{a^2-b^2+c^2}{2}$, $z=\\frac{a^2+b^2-c^2}{2}$.\nThen, we just have to prove that :\n$$\\frac{c^2+b^2-a^2}{2}+\\frac{a^2-b^2+c^2}{2}+\\frac{a^2+b^2-c^2}{2}\\le\\frac{5}{2\\sqrt{2}}\\sqrt{a^2+b^2+c^2}, $$i.d\n$$(a+b+c)+\\frac{(a+b+c)(a-b)(a-c)(c-a)}{abc}\\le\\frac{5}{2\\sqrt{2}}\\sqrt{a^2+b^2+c^2}. $$\nBut $$\\sqrt{a^2+b^2+c^2}\\ge\\frac{a+\\sqrt{b^2+c^2}}{\\sqrt{2}} ;$$\nSo we just have to prove that $f(a)\\le 0$, where : \n$$f(a)=4a^3(c-b)-a^2bc+a(4b^3+4b^2c-4c^3-5bc\\sqrt{b^2+c^2})+4bc(c^2-b^2)$$, where $a$ is between $max(a,b)$ and $\\sqrt{a^2+b^2}$. Then we just have to prove the inequality when $b\\le c$, i.d when $b=c$ and $b0$,\nand $$f(c)=-bc^2(5\\sqrt{b^2+c^2}-4b-3c)\\le 0. $$\nWe also have : $$f(\\sqrt{b^2+c^2})=2bc(4b\\sqrt{b^2+c^2}-5b^2-c^2)=-2bc(\\sqrt{b^2+c^2}-2b)^2\\le 0. $$\nThus, $f$ admits three real roots (a negative root, a root in $[0,c]$ and a root $\\ge\\sqrt{b^2+c^2}$. Then, $f$ is negative in $[c,\\sqrt{b^2+c^2}]$.\nWe have $f(a)=0$ if $\\sqrt{b^2+c^2}=2b$ and $a=\\sqrt{b^2+c^2}$, i.e the vector $(a,b,c)$ is proportional to $(2,1,\\sqrt{3})$. In this case, the inequality $$\\frac{a+\\sqrt{b^2+c^2}}{\\sqrt{2}}\\le\\sqrt{a^2+b^2+c^2} $$ is an equality. In conclusion, we have an equality iff vectors $(x,y,z)$ and $(0,3,1)$ are proportional.", "Solution_18": "[quote=Fedor Bakharev]Find the smallest constant $ k$ such that\n\n$ \\frac {x}{\\sqrt {x \\plus{} y}} \\plus{} \\frac {y}{\\sqrt {y \\plus{} z}} \\plus{} \\frac {z}{\\sqrt {z \\plus{} x}}\\leq k\\sqrt {x \\plus{} y \\plus{} z}$\n\nfor all positive $ x$, $ y$, $ z$.[/quote]\n\n", "Solution_19": "[quote=\"hitek\"]Let $x,y,z$ be real non-negative numbers, prove that\n\\[ \\frac x{\\sqrt{x+y}}+\\frac y{\\sqrt{y+z}}+\\frac z{\\sqrt{z+x}}\\leq\\frac54\\sqrt{x+y+z}. \\]\n[/quote]\nLet $a,b,c$ be real non-negative numbers, prove that\n\\[ \\frac{\\sqrt{2}}{2}\\left(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\right)\\leq\\frac a{\\sqrt{a+b}}+\\frac b{\\sqrt{b+c}}+\\frac c{\\sqrt{c+a}}\\leq \\frac54\\sqrt{a+b+c}. \\]\n[url=https://artofproblemsolving.com/community/c6h56071p347277]here[/url]\n[b]Walther Janous\nJack Garfunkel[/b]", "Solution_20": "Just try to graph this problem. WLOG let $x+y+z =1$ and all variables are positive (because we are looking for max). Use spherical coordinate to parameterize $x,y$ and $z$: \n$ z=\\sin ^2(\\theta ); y=(\\cos (\\theta ) \\cos (\\phi ))^2; x=(\\cos (\\theta ) \\sin (\\phi ))^2$\n$\\text{Plot3D}\\left[\\frac{x}{\\sqrt{x+y}}+\\frac{z}{\\sqrt{x+z}}+\\frac{y}{\\sqrt{y+z}},\\left\\{\\phi ,-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right\\},\\{\\theta ,0,2 \\pi \\},\\text{PlotPoints}\\to 50\\right]$\nFind one maximum point by: \n$\\text{FindMaximum}\\left[\\frac{x}{\\sqrt{x+y}}+\\frac{z}{\\sqrt{x+z}}+\\frac{y}{\\sqrt{y+z}},\\{\\{\\phi ,0\\},\\{\\theta ,2 \\pi \\}\\}\\right]$ $\\Rightarrow 1.25$\n\n", "Solution_21": "Let $ x,y,z> 0.$ Prove that \n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{2y}{\\sqrt{y+z}}+\\frac{z}{\\sqrt{z+x}} \\leq \\frac{7}{8}\\sqrt{x+y+z}$$\n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{2y}{\\sqrt{y+z}}+\\frac{2z}{\\sqrt{z+x}} \\leq \\frac{5}{2}\\sqrt{x+y+z}$$\n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{4y}{\\sqrt{y+z}}+\\frac{4z}{\\sqrt{z+x}} \\leq 5\\sqrt{x+y+z}$$\n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{y^2}{\\sqrt{y+z}}+\\frac{z}{\\sqrt{z+x}} \\leq \\frac{5}{4}\\sqrt{x+y+z}$$", "Solution_22": "[quote=sqing]Let $ x,y,z> 0.$ Prove that \n\n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{y^2}{\\sqrt{y+z}}+\\frac{z}{\\sqrt{z+x}} \\leq \\frac{5}{4}\\sqrt{x+y+z}$$[/quote]\nIt's wrong. Try $x=z=1$ and $y=100.$\n\n", "Solution_23": "[quote=sqing]Let $ x,y,z> 0.$ Prove that \n$$ \\frac{x}{\\sqrt{x+y}}+\\frac{2y}{\\sqrt{y+z}}+\\frac{z}{\\sqrt{z+x}} \\leq \\frac{7}{8}\\sqrt{x+y+z}$$\n[/quote]\nIt's wrong for $y\\rightarrow+\\infty.$\n\n" } { "Tag": [ "factorial", "number theory", "prime factorization" ], "Problem": "Hey guys, how do you solve this kind of problem?\r\n\r\nHow many factors has (10!)(9!)(8!)(7!)(6!)(5!)(4!)(3!)(2!)(1!) :blush:", "Solution_1": "[quote=\"guile\"]Hey guys, how do you solve this kind of problem?\n\nHow many factors has (10!)(9!)(8!)(7!)(6!)(5!)(4!)(3!)(2!)(1!) :blush:[/quote]\r\n[hide]$(10!)(9!)(8!)(7!)(6!)(5!)(4!)(3!)(2!)(1!)=(2^8\\cdot 3^4 \\cdot 5^2\\cdot 7)(2^7\\cdot 3^4 \\cdot 5\\cdot 7)(2^7\\cdot 3^2 \\cdot 5\\cdot 7)(2^4\\cdot 3^2 \\cdot 5\\cdot 7)(2^4\\cdot 3^2 \\cdot 5)(2^3\\cdot 3 \\cdot 5)(2^3\\cdot 3)(3\\cdot 2)(2)=2^{38}\\cdot 3^{17}\\cdot 5^{7}\\cdot 7^{4}$ which has $(38+1)(17+1)(7+1)(4+1)=\\boxed{28080}$ factors.[/hide]", "Solution_2": "Woah, I got [hide]19656[/hide] :blush:", "Solution_3": "generally,no. of divisors of a number suppose prime factorised in the form $2^a3^b5^c$ is given by $(a+1)(b+1)(c+1)$.So do this in the question...(yeah what chinesechess did)", "Solution_4": "Actually, even more generally, if you have :\r\n$n = \\prod_{i=1}^l p_i^{a_i}$, ($p_i \\in \\mathbb{P}$, $p_i \\ne p_j$, $a_i \\in \\mathbb{N}$) \r\nthen, the number of divisors is :\r\n$d(n) = \\prod_{i=1}^l (a_i + 1)$.\r\n\r\n:)", "Solution_5": "[quote=\"easyas3.14159...\"]Woah, I got [hide]19656[/hide] :blush:[/quote]\r\n\r\nWhat was your prime factorization?", "Solution_6": "I also got $28080$ factors.\r\nLooks like easyaspi used $(38+1)(17+1)(7)(4)$ instead of $(38+1)(17+1)(7+1)(4+1)$ :D", "Solution_7": "What set is $\\mathbb{P}$? Prime numbers?", "Solution_8": "Yes DanK. \r\n$\\mathbb{P}$ is indeed the set of the prime numbers. :)", "Solution_9": "[quote=\"236factorial\"][quote=\"easyas3.14159...\"]Woah, I got [hide]19656[/hide] :blush:[/quote]\n\nWhat was your prime factorization?[/quote]Woah, 236factorial is right. If you don't show your work, there is no way for us to point out what you did wrong so that you can correct and improve... :roll:", "Solution_10": "[quote=\"mathmanman\"]Actually, even more generally, if you have :\n$n = \\prod_{i=1}^l p_i^{a_i}$, ($p_i \\in \\mathbb{P}$, $p_i \\ne p_j$, $a_i \\in \\mathbb{N}$) \nthen, the number of divisors is :\n$d(n) = \\prod_{i=1}^l (a_i + 1)$.\n\n:)[/quote]\r\n\r\nJust one thing. Why is $p_i \\neq p_j$ in there? There isn't even a $p_j$ in your thing...", "Solution_11": "Just to say that the primes are pairwise distinct. ;)", "Solution_12": "I got $\\boxed{28080}$.\r\n\r\n[hide=\"my soln\"]\n$10!=2*5*3^2*2^3*7*2*3*5*2^2*3*2$\n$9!=3^2*2^3*7*2*3*5*2^2*3*2$\n$8!=2^3*7*2*3*5*2^2*3*2$\n$7!=7*2*3*5*2^2*3*2$\n$6!=2*3*5*2^2*3*2$\n$5!=5*2^2*3*2$\n$4!=2^2*3*2$\n$3!=3*2$\n$2!=2$\n\nThis comes out to be(unless i counted wrong) $2^{38}*5^{7}*3^{17}*7^{4}$.\nSo the number of factors are $(38+1)(7+1)(17+1)(4+1)=\\boxed{28080}$.[/hide]", "Solution_13": "[quote=\"lkryptonicl\"]I got $\\boxed{27360}$.[/quote]\r\nDid you get $2^{37}$? Or did you forget to add $1$? You should get $(38+1)(17+1)(7+1)(4+1)$.", "Solution_14": "yea i counted wrong", "Solution_15": "Ok now you know how many factors it has right? So now can you find the sum of all of the factors?", "Solution_16": "[hide]\n${\\displaystyle\\sum_{i=0}^{38}2^i}*{\\displaystyle\\sum_{j=0}^{17}3^j}*{\\displaystyle\\sum_{k=0}^{7}5^k}*{\\displaystyle\\sum_{l=0}^{4}7^l}$ :D [/hide]", "Solution_17": "[hide=\"sum of factors i think\"]\n\n$(38+2)(7+5)(17+3)(4+7)=105600$[/hide]", "Solution_18": "Hmm... the sum of the factors of 8 acccording to:\r\nme: $2^0+2^1+2^2+2^3=1+2+4+8=15$\r\nyou: $3+2=5$\r\n$1+2+4+8=\\boxed{15}$ :D \r\nBut does anyone know the formula for product of factors?", "Solution_19": "Sum of factors is... [hide]\n${\\displaystyle\\sum_{i=0}^{38}2^i}*{\\displaystyle\\sum_{j=0}^{17}3^j}*{\\displaystyle\\sum_{k=0}^{7}5^k}*{\\displaystyle\\sum_{l=0}^{4}7^l}=(2^{39}-1)(193710244)(97656)(2801)=2.912959587\\times10^{28}$ :D \n[/hide]", "Solution_20": "You think that the sum of the factors of that expression is 15?", "Solution_21": "The sum of the factors of a number who's prime factorization is\r\n\r\n$p_1^{e_1} * p_2^{e_2} * \\ldots * p_n^{e_n}$\r\n\r\nis\r\n\r\n$(p_1^0 + p_1^1 + ... + p_1^{e_1}) * (p_2^0 + p_2^1 + \\ldots + p_2^{e_3})*...*(p_n^0 + p_n^2 + ... + p_n^{e_n})$\r\n\r\nThe number of factors is \r\n\r\n$(e_1 + 1)*(e_2 + 1)*\\ldots*(e_n + 1)$", "Solution_22": "[quote=\"lkryptonicl\"]You think that the sum of the factors of that expression is 15?[/quote]\r\n\r\nUmm...I was trying to prove that my method was correct. :P So I used an example of 8. And for the sum of the expression, look at my other post. :D", "Solution_23": "there is a more efficient way of doing this problem..., what if it were up to 30!", "Solution_24": "[quote=\"Altheman\"]there is a more efficient way of doing this problem..., what if it were up to 30![/quote]\r\n\r\nYou mean number of divisors.Then I can give u 5 options [hide=\" :P \"]This ones on MIMC[/hide]", "Solution_25": "And the product is $((10!)(9!)(8!)(7!)(6!)(5!)(4!)(3!)(2!)(1!))^{\\frac{28080}{2}}=(6.6586\\times10^{27})^{14040}$ :D", "Solution_26": "The product of the divisors of $n$ is $n^x$, where $x=\\frac{\\text{number of divisors}}{2}$.", "Solution_27": "[quote=\"lotrgreengrapes7926\"]Hmm... the sum of the factors of 8 acccording to:\nme: $2^0+2^1+2^2+2^3=1+2+4+8=15$\nyou: $3+2=5$\n$1+2+4+8=\\boxed{15}$ :D \nBut does anyone know the formula for product of factors?[/quote]\r\nThe product of all factors of $n$ would just be $n^{k/2}$ where $k$ is the number of factors.", "Solution_28": "My formulas looked prettier :P" } { "Tag": [ "calculus", "derivative", "function", "integration", "calculus computations" ], "Problem": "I Created this Little problem, thought you would like to see it.\r\n\r\n[color=blue]Problem[/color] Find a function like $f: R \\longrightarrow R$ such that $f'$ is periodic but $f$ is not. :)", "Solution_1": "something like: f(x)=sin(x)+x, f'(x)=cos(x)+1 ?", "Solution_2": "Assume $f$ satisfies some reasonable regularity condition - $C^{1}$ ought to be good enough - and suppose that $f'$ is periodic of period $P$ for some $P>0.$\r\n\r\nClaim: $f$ is periodic of period $P$ if and only if $\\int_{a}^{a+P}f'(x)\\,dx=0$ for any $a.$\r\n\r\nThe connection to rotceh's example should be clear.\r\n\r\n[Note: I don't mean for the phrase \"periodic of period $P$\" to preclude being periodic with a shorter period, or even being constant.]" } { "Tag": [ "IMO Shortlist" ], "Problem": "Not really an IIT oriented problem. Its actually from a programming contest. I wrote the program, will post it if you want but anybody got the logic?? Simple math.\r\n\r\n As we all know caterpillars love to eat leaves. Usually, a caterpillar sits on leaf, eats as much of it as it can (or wants), then stretches out to its full length to reach a new leaf with its front end, and finally \"hops\" to it by contracting its back end to that leaf.\r\n\r\nWe have with us a very long, straight branch of a tree with leaves distributed uniformly along its length, and a set of caterpillars sitting on the first leaf. (Well, our leaves are big enough to accommodate upto 20 caterpillars!). As time progresses our caterpillars eat and hop repeatedly, thereby damaging many leaves. Not all caterpillars are of the same length, so different caterpillars may eat different sets of leaves. We would like to find out the number of leaves that will be undamaged at the end of this eating spree. We assume that adjacent leaves are a unit distance apart and the length of the caterpillars is also given in the same unit.\r\n\r\nFor example suppose our branch had 20 leaves (placed 1 unit apart) and 3 caterpillars of length 3, 2 and 5 units respectively. Then, first caterpillar would first eat leaf 1, then hop to leaf 4 and eat it and then hop to leaf 7 and eat it and so on. So the first caterpillar would end up eating the leaves at positions 1,4,7,10,13,16 and 19. The second caterpillar would eat the leaves at positions 1,3,5,7,9,11,13,15,17 and 19. The third caterpillar would eat the leaves at positions 1,6,11 and 16. Thus we would have undamaged leaves at positions 2,8,12,14,18 and 20. So the answer to this example is 6.\r\n\r\nSo given the no of leaves, no of caterpillars and each caterpillars length.. Give me an algo to find. Its really easy. Maybe write a CPP program if you are interested.", "Solution_1": "there was another amazing question.. i think this was an IMO shortlist or something.. also found it a few days back in titu andrescu and razwan gelca.. Check it out -\r\n\r\nA chameleon repeatedly rests and then catches a fly. The first rest is for a period of 1 minute. The rest before catching the fly 2n is the same as the rest before catching fly n. The rest before catching fly 2n+1 is 1 minute more than the rest before catching fly 2n. How many flies does the chameleon catch before his first rest of 9 minutes? How many minutes (in total) does the chameleon rest before catching fly 98? How many flies has the chameleon caught after 1999 total minutes of rest?\r\n\r\nthe solution eventually lies in functional equations. check it out. cool function." } { "Tag": [ "USAMTS" ], "Problem": "Does USAMTS notify you when they have registered you via e-mail?", "Solution_1": "You should go ask this in the USAMTS forum. (Or a moderator could be generous and move it there.)", "Solution_2": "Or you could just go on their website :)", "Solution_3": "USAMTS has a forum? Oops.", "Solution_4": "USAMTS has a forum on this website:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/index.php?f=164\r\n\r\n(It's under the category Contests & Programs.)", "Solution_5": "Why do you need to be notified? Just register, and ... :idea: you're registered!", "Solution_6": "I like to know that they registered me and the mail that I sent didn't magically disappear. ;)", "Solution_7": "Log onto the USAMTS website and click on \"My USAMTS\" -- if you're registered you will see a blank score sheet for Year 17.", "Solution_8": "Ok, my problems are solved." } { "Tag": [ "trigonometry", "symmetry", "geometry", "circumcircle", "power of a point" ], "Problem": "In the diagram, the smaller circle is inscribed in equilateral $ \\triangle{EFG}$ and $ \\triangle{EFG}$ is inscribed in the larger circle. B and C are points of tangency and are also points on chord $ AD$. If AB=2, then BC =?\r\n\r\nA. $ \\sqrt {5} \\minus{} 1$\r\nB. $ 2$\r\nC. $ 3$\r\nD. $ \\sqrt {5} \\plus{} 1$\r\nE. $ 4$", "Solution_1": "[quote=\"hohoho11\"]In the diagram, the smaller circle is inscribed in equilateral $ \\triangle{EFG}$ and $ \\triangle{EFG}$ is inscribed in the larger circle. B and C are points of tangency and are also points on chord $ AD$. If AB=2, then BC =?\n\nA. $ \\sqrt {5} \\minus{} 1$\nB. $ 2$\nC. $ 3$\nD. $ \\sqrt {5} \\plus{} 1$\nE. $ 4$[/quote]\r\n\r\nIs what he meant to say again.", "Solution_2": "[hide=\"Answer\"]D[/hide]How would you do this? Power of a Point? I'm dumb...", "Solution_3": "solution for dumb people like me???", "Solution_4": "[hide=\"I think my solution is too complicated...\"]We can split the triangle up into 6 congruent right triangles. The hypotenuse of each of them is equal to the radius of the large circle. Let the radius of the small circle be $ r$. Using right triangles, we find that $ BC=2r\\sin60^\\circ=r\\sqrt{3}$, so once we find $ r$, we will be able to determine the length of $ BC$. Also, we can use $ r$, with right triangles, to determine that the radius of the large circle is $ \\frac{r}{\\cos 60^\\circ}=2r$\nNow, we let the circumcenter/incenter of the triangle be $ O$, the midpoint of $ BC$ be $ M$, and the point diametrically opposite $ E$ to be $ H$. By Power of a Point, we have:\n\\[ \\&EM \\cdot MH=AM \\cdot DM \\\\\n\\&\\Rightarrow \\left(EO-MO\\right)\\left(EO+MO\\right)=\\left(AB+BM\\right)^2 \\\\\n\\&\\Rightarrow EO^2-MO^2=\\left(AB+BM\\right)^2\\]Plugging in numbers, we have:\n\\[ \\left(2r\\right)^2-\\left(r\\cos 60^\\circ\\right)^2=\\left(2+r\\sin 60^\\circ\\right)^2\\]Solving this complicated equation, we find that $ r=\\frac{\\sqrt{15}+\\sqrt{3}}{3}$.\nThen, $ BC=\\sqrt{3} \\cdot \\frac{\\sqrt{15}+\\sqrt{3}}{3}=\\sqrt{5}+1 \\Rightarrow \\boxed{D}$.[/hide]", "Solution_5": "[quote=\"hohoho11\"]In the diagram, the smaller circle is inscribed in equilateral $ \\triangle{EFG}$ and $ \\triangle{EFG}$ is inscribed in the larger circle. B and C are points of tangency and are also points on chord $ AD$. If AB=2, then BC =?\n\nA. $ \\sqrt {5} \\minus{} 1$\nB. $ 2$\nC. $ 3$\nD. $ \\sqrt {5} \\plus{} 1$\nE. $ 4$[/quote]\r\n[hide]Since $ \\triangle EFG$ is equilateral, we can exploit symmetry here. Let the radius of the incircle be $ a$, so radius of the circumcircle is $ 2a$. Hence the side length of the triangle is $ 2\\sqrt {3}a$. Let $ BC \\equal{} x$.\n\nUsing power of a point on the larger circle, $ 2(2 \\plus{} x) \\equal{} 3a^2$. \n\nSince the centers of the circles $ O$ lie on the centroid, $ OX \\equal{} \\frac {a}{2}$ and $ AX \\equal{} \\sqrt {4a^2 \\minus{} \\frac {a^2}{4}} \\equal{} \\frac {\\sqrt {15}a}{2}$. This means $ \\frac {\\sqrt {15}a}{2} \\minus{} 2 \\equal{} \\frac {x}{2}$. From these two equations, we find $ x \\equal{} \\sqrt {5} \\plus{} 1$.[/hide]", "Solution_6": "It is easy to see that $ BC$ is half the length of the side of the larger equilateral triangle and $ B$ bisects $ GE$, a side. So $ CB\\equal{}GB\\equal{}EB$. By symmetry, $ CD\\equal{}AB\\equal{}2$. By power of a point, \\[ AB\\cdot BD\\equal{}BG\\cdot EB\\implies 2\\cdot (2\\plus{}x)\\equal{}x^2\\implies x\\equal{}\\sqrt{5}\\plus{}1\\]", "Solution_7": "Wow, how'd I miss that...I wasted so much time on my convoluted solution. lol" } { "Tag": [], "Problem": "Which is the better Power Ranger Theme?\r\n\r\nNINJA STORM\r\n[youtube]wWawGV7MXLU[/youtube]\r\n\r\nMYSTIC FORCE\r\n[youtube]GNw0LoNcosc[/youtube]", "Solution_1": "...... Could they make it more exciting? :P", "Solution_2": "power what?" } { "Tag": [ "modular arithmetic" ], "Problem": "Let $a, n$ be positive integers such that $(a, n) = 1$. \r\n\r\nShow that $n | \\phi (a^{n}-1)$.", "Solution_1": "Was no one able to solve this? :o \r\n\r\nI suppose I will have to provide a [hide=\"hint\"] Consider residues $\\bmod (a^{n}-1)$. In other words, work in $\\mathbb{Z}_{ a^{n}-1 }$. [/hide]\n\nAnd perhaps [hide=\"another\"] More specifically, consider the residues left by the powers of $a$. [/hide]", "Solution_2": "For t FC = 2x/3 and FB = x/3\r\nFD = FC - DC = x/6\r\n\r\n=> BF = 2FD\r\nthese are the bases of the two triangles BFE and FDE\r\nthr altitudes are equal..\r\n...\r\n\r\n6th part:\r\nAFC and BFE are similar\r\nand AC/BE = 2\r\n=> area AFC / area BFE = 2 sq. = 4\r\nuse this + 5th part..", "Solution_2": "this is ncert question if m not mistaken... ninth class", "Solution_3": "Yes very much the same. And what muks has done is what the hint given {though I don't kow whether both coincide}", "Solution_4": "You are right.This is frm NCERT 9. Let's see..", "Solution_5": "Thank you.", "Solution_6": "oops ! i am going too face this ncert exam :(" } { "Tag": [ "function", "calculus", "integration", "college contests" ], "Problem": "Let be given continous function f(x) on [0,+oo), f takes real postitive value. Show that function\r\n\r\n F(x) = Integral{0_x} tf(t)dt / Integral{0_x} f(t)dt\r\n\r\nis increasing on [0, +oo).", "Solution_1": "$F(x)=\\frac {\\int_0^x tf(t)dt}{\\int_0^xf(t)dt}$\r\nF is derivable.\r\n$F'(x)=\\frac{xf(x)\\cdot \\int_0^xf(t)dt-\\int_0^xtf(t)dt\\cdot f(x)}{(\\int_0^xf(t)dt)^2}$.\r\nF is increasing iff F'>0 iff $x\\cdot \\int_0^xf(t)dt-\\int_0^xtf(t)dt>0$ iff $\\int_0^x(x-t)\\cdot f(t)dt>0$. The last is true because f is a positive valued function.\r\n\r\nNot a brain cracking contest isn't it?", "Solution_2": "Yes, as usual, VUMC problems are easy. Only one or two (of 5 or 6) problems are a little bit harder." } { "Tag": [ "floor function" ], "Problem": "Can anyone help me understand problem 476 in AoPS volume 1\r\n\r\nI have the answers and solutions but I still don't get it", "Solution_1": "Can you post the problem?", "Solution_2": "Find the greatest n for which 12^n evenly divides 20 factorial.", "Solution_3": "$ 12\\equal{}2^2\\cdot 3$. Try finding the amount of powers of 2 and 3 in $ 20!$", "Solution_4": "i did. But problem is that it has to be in a relationship between 2^2 and 3, which makes it complicated", "Solution_5": "Well, try this: You might notice that in $ a^n$, there are $ \\left\\lfloor \\frac{n}{x} \\right\\rfloor$ powers of $ a^x$", "Solution_6": "[hide]Well, there is one factor of 3 in 3, 6, 12, and 15 and two factors of 3 in 9 and 18 for a total of 4(1) +2(2) = 8. There is one factor of 2 in 2, 6, 10, 14, and 18 two factors of 2 in 4, 12, and 20, 3 factors in 8, and 4 factors in 16 for a total of 5(1) + 2(3) + 3(1) + 4(1) = 18. Thus n=8 since 12^8 = 2^16 * 3^8.[/hide]", "Solution_7": "hmmm\r\nI still don't get, what you want me to do", "Solution_8": "In general, if $ a$ is a factor of $ n!$, $ x\\equal{}\\sum_{k\\equal{}1}^{\\infty}\\left\\lfloor\\frac{n}{a^k}\\right\\rfloor$ is the largest $ x$ such that $ a^x|n!$.\r\n\r\n$ 12\\equal{}2^2\\cdot3$.\r\n$ 2^x|20!\\implies x\\equal{}\\left\\lfloor\\frac{20}{2}\\right\\rfloor\\plus{}\\left\\lfloor\\frac{10}{2}\\right\\rfloor\\plus{}\\left\\lfloor\\frac{5}{2}\\right\\rfloor\\plus{}\\left\\lfloor\\frac{2}{2}\\right\\rfloor\\equal{}10\\plus{}5\\plus{}2\\plus{}1\\equal{}18$\r\nTherefore, $ 2^{18}|20!$.\r\nNow, consider:\r\n$ 3^x|20!\\implies x\\equal{}\\left\\lfloor\\frac{20}{3}\\right\\rfloor\\plus{}\\left\\lfloor\\frac{6}{3}\\right\\rfloor\\equal{}6\\plus{}2\\equal{}8$\r\n\r\nSo $ 2^{18}\\cdot3^8\\equal{}2^{5}\\cdot\\boxed{12^{8}}$.\r\n\r\nThis is pretty much what crazypianist1116 said, but in a nice summation formula.", "Solution_9": "I don't see what's complicated. Just write down $ 20\\cdot19\\cdot18\\cdot17\\cdots2$, then $ 20\\equal{}5\\cdot2^2, 19\\equal{}19, 18\\equal{}3^2\\cdot2$ etc. Count all $ 2's$ and $ 3's$ (18 and 8) and you get $ n_{max}\\equal{}8$ (why not $ 18: 2\\equal{}9$? simple, there are no nine $ 3's$). Others showed nicer solutions but you could have done this if nothing else worked for you.", "Solution_10": "[quote=\"Flame\"] (why not $ 18: 2 \\equal{} 9$? simple, there are no nine $ 3's$)..[/quote]\r\n\r\nAh, so that's what the solution was talking about.\r\nThe solution said something like that but now I understand it" } { "Tag": [ "ceiling function", "logarithms" ], "Problem": "There is a house which has $N$ floors and there are $n$ eggs. The all of them are identical on their durability. It is known that the egg will be crashed if we shall dump it from the $d$-th floor or more higher one and will preserve their durability when damped from some lower floor. How many attempts is required to define this number $d$ in any case? \r\nThe egg would be crashed if we shall dump it from $N$-floor surely. For instance, if $n=1$ then evidently we need $N-1$ ettemps.", "Solution_1": "I don't see why $n$ matters...the question only has to do with the number of floors, $N$. I suppose an upper bound is the number of digits in the base-2 representation of N. Pick the middle floor, then depending on what happens, pick the middle of the bottom half or the top half, etc. each time you find out one of the base-2 digits.", "Solution_2": "$n$ does indeed matter. If you break all of your eggs, you have failed the experiment. The example given is if $n=1$, you have to start at the bottom floor and work your way up to be sure. If $N$ was 5, and $d$ was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether $d$ was 1 or 2.", "Solution_3": "Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). However, I still don't see why $n$ matters. Suppose in the example that we are given $n=N-1$, then we only need one \"try,\" but we still drop the egg $N-1$ times. That is, the answer or method does not have anything to do with n.", "Solution_4": "$n$ matters because if you have less eggs than you need for a specific optimum strategy (that is, $n = \\infty$), then you need to use the eggs you already have \"more carefully.\"\r\n\r\nFor example, if $n = 1$ the only permissible strategy is to drop from the first floor and up until the egg breaks, since you only have one egg to work with.\r\n\r\nEdit: \r\n\r\n[quote=\"Phelpedo\"]Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). [/quote]\r\n\r\nThere is.", "Solution_5": "[b]2 Moderators[/b]. \r\n\r\nPossibly [b]Combinatorics Proposed & Own Problems[/b]\r\n is more nice place for this problem?", "Solution_6": "I find the wording to be confusing.", "Solution_7": "[quote=\"t0rajir0u\"]$n$ matters because if you have less eggs than you need for a specific optimum strategy (that is, $n = \\infty$), then you need to use the eggs you already have \"more carefully.\"\n\nFor example, if $n = 1$ the only permissible strategy is to drop from the first floor and up until the egg breaks, since you only have one egg to work with.\n[/quote]\nFrom your examples I still can't tell the difference. If you still know the results from failed experiments, then it's the same as if they hadn't failed and you had that many eggs to begin with. If you don't know results from failed experiments...then the problem becomes pretty much impossible. \n\n[quote=\"besttate\"]\nIf N was 5, and d was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether d was 1 or 2.\n\n[/quote]\n\nIf d is 2 and I drop it from the 3rd floor, I am still clueless whether it is 1 or 2. I can determine the answer by dropping from the 1st floor and then the 2nd.\n\nIf d is 1, then I do the exact same thing with the next egg...\n\n\n[quote=\"t0raji0u\"]\n[quote=\"Phelpedo\"]Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). [/quote]\n\nThere is.[/quote]\r\nI meant I thought there was only one floor (as opposed to one and all higher ones).", "Solution_8": "hmmm... perhaps the best strategy is to drop the first one at floor N/2, then weather it breaks or not we use the next drop at floor N/4 or 3N/4 respectively. Each time we split the amount left to check in half, then we use the last egg to start from the least open value and work our way up. So given n and N we know that we can split N in half n-1 times worst case senario (if they all break each time), and then the floor d can be right underneath the upper bound of what we have to check with our last egg so thats at most N/(2^(n-1)) - 1 we have to check with the last one. So thats n + N/(2^(n-1) - 2 total drops?", "Solution_9": "[b]nate:[/b] Why does your number of tries increase with larger $n$? It should decrease as $n$ approaches infinity, and the best-case scenario should be $\\lceil \\log_{2}N \\rceil$. \r\n\r\n[quote=\"Phelpedo\"][quote=\"besttate\"]\nIf N was 5, and d was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether d was 1 or 2.\n\n[/quote]\n\nIf d is 2 and I drop it from the 3rd floor, I am still clueless whether it is 1 or 2. I can determine the answer by dropping from the 1st floor and then the 2nd.\n\nIf d is 1, then I do the exact same thing with the next egg...[/quote]\r\n\r\nLet's go off of this case and make it very simple. Say $N = 3$. \r\n\r\nNow, let's look at what our strategy is like if $n = 2$. When $n = 2$ we can afford to drop off of the second floor. If egg 1 breaks, we can drop off of the first floor to determine whether the floor in question is $2$ or $1$. If egg 1 does not break, we already know it must be the third floor. Thus we require $\\boxed{2}$ attempts at most. \r\n\r\nThis strategy fails when $n = 1$ because if we break on the second floor we have no way of discovering whether the floor in question is $2$ or $1$, so we need to drop from the first floor up, which requires $\\boxed{3}$ attempts at most.\r\n\r\nHence the worst-case number of required attempts varies depending on $n$.", "Solution_10": "We drop from every $n$-th floor. When it breaks we use our $n-1$ remaining eggs to inspect the floors between the floor we just were at and the last known good floor. This gives $N/n+n-1$ as the number of tries.", "Solution_11": "As I said, the optimal case when $n > log_{2}N$ should give an answer of $\\lceil log_{2}N \\rceil$ by binary search.", "Solution_12": "[quote=\"t0rajir0u\"]\n\nNow, let's look at what our strategy is like if $n = 2$. When $n = 2$ we can afford to drop off of the second floor. If egg 1 breaks, we can drop off of the first floor to determine whether the floor in question is $2$ or $1$. If egg 1 does not break, we already know it must be the third floor. Thus we require $\\boxed{2}$ attempts at most. \n\nThis strategy fails when $n = 1$ because if we break on the second floor we have no way of discovering whether the floor in question is $2$ or $1$, so we need to drop from the first floor up, which requires $\\boxed{3}$ attempts at most.\n\nHence the worst-case number of required attempts varies depending on $n$.[/quote]\r\n\r\nIf we are allowed to drop from the first floor then the second, why can't we just drop from the second and then the first (like in the n=2 case).", "Solution_13": "[quote=\"Phelpedo\"][quote=\"t0rajir0u\"]Hence the [b]worst-case[/b] number of required attempts varies depending on $n$.[/quote]\n\nIf we are allowed to drop from the first floor then the second, why can't we just drop from the second and then the first (like in the n=2 case).[/quote]\r\n\r\nIn the worst case, $d = 1$ and dropping from the second floor would break the egg, leaving us unsure whether $d = 1$ or $d = 2$.", "Solution_14": "So we drop it from one...that's only two attempts (same as d=2) according to your previous post.", "Solution_15": "Ah. You're correct there. But let's re-analyze for $N = 5$.\r\n\r\nWith $n = 1$ the worst-case scenario comes if $d = 5$, in which case it will require $\\boxed{4}$ tries (on floors $1, 2, 3, 4$) to ascertain the correct floor. We will need to drop starting from the first floor. \r\n\r\nWith $n = 2$, however, we can drop at the third floor. If it breaks, we test the first and second floors for a total of three tries. If it does not break, we test the fourth floor for a total of two tries. Hence we require $\\boxed{3}$ tries in this case. \r\n\r\n(I'm pretty sure this is correct this time.)" } { "Tag": [ "analytic geometry", "email" ], "Problem": "can anyone suggest good article on coordinate system.......A book or a good note will do.one may email me any article they possess.........", "Solution_1": "hello, here you will find an overview\r\nhttp://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys_f.html\r\nSonnhard." } { "Tag": [], "Problem": "The Gnollish language consists of 3 words, ``splargh,'' ``glumph,'' and ``amr.'' In a sentence, ``splargh'' cannot come directly before ``glumph''; all other sentences are grammatically correct (including sentences with repeated words). How many valid 3-word sentences are there in Gnollish?", "Solution_1": "[hide=\"There is a much faster way then this, but...\"]splargh amr glumph\nsplargh amr amr\nsplargh amr splargh\nsplargh splargh amr\nsplargh splargh splargh\namr amr amr\namr amr splargh\namr amr glumph\namr glumph splargh\namr glumph amr\namr glumph glumph\namr splargh amr\namr splargh splargh\nglumph amr glumph\nglumph amr splargh\nglumph amr amr\nglumph glumph amr\nglumph glumph glumph\nglumph glumph splargh\nglumph splargh amr\nglumph splargh splargh[/hide]\r\n\r\nI got 20.", "Solution_2": "there are total $ 3^3$ ways not regarding grammar mistakes.\r\n\r\nand there are 2*3=6 grammatically incorrect sentences.\r\n\r\nso 27-6=21\r\n\r\ni got 21.", "Solution_3": "Yeah, me too. I thought 8 + 8 + 5 = 20. :P" } { "Tag": [ "linear algebra", "matrix", "advanced fields", "advanced fields unsolved" ], "Problem": "Max logdet(S)\r\nst.(1)LMI>=0 , LMI=[S,Phi*S,Phi1*S,Phi2*S;(Phi*S)',S-r*S,0,0;(Phi1*S),0,r*S,0;\r\n\r\n(Phi2*S)',0,0,r*S]; \r\n(2)S>0; \r\n(3)0R* such that\r\nf(x*f(y))=f(x*y)+x;for all x,y in R* :( \r\n\r\n :( \r\n\r\n :)", "Solution_1": "This was our TST problem.\r\nPut x=f(z),f(f(z)f(y))=f(f(z)y)+f(z)=f(zy)+f(z)+y=f(zy)+f(y)+z\r\nhence f(x)=x+c.subsituting in original equation we'll get f(x)=x+1", "Solution_2": "I don't understand your last step: why $f(zy)+f(z)+y=f(zy)+f(y)+z$?", "Solution_3": "change y and z 's position in the equality f(f(z)f(y))=f(f(z)y)+f(z).\r\nI think it is a useful way to let the equality have symmetry.", "Solution_4": "Oh, I see, thanks [b]zhaobin[/b]. :)", "Solution_5": "you put $x=f(z)$ how do you guarantee that $z$ exist?", "Solution_6": "For all values of z there exist x,and the last equation f(zy)+f(z)+y=f(zy)+f(y)+z holds for any y,z element of R*", "Solution_7": "[quote=\"hardsoul\"]For all values of z there exist x,and the last equation f(zy)+f(z)+y=f(zy)+f(y)+z holds for any y,z element of R*[/quote]\r\nWe must prove that for every $x$ where exists $z$ such that $f(z)=x$ and not x such that...", "Solution_8": "Why?", "Solution_9": "I agree with Hardsoul.\r\n\r\nIf f:R*-->R*with $x$, $y$, $z$ in R*, then\r\n$f(xf(y))=f(xy)+x$\r\nimplies\r\n$f(f(z)f(y))= f(zf(y)) + f(y) = f(yf(z)) + f(z)$\r\nwhich implies \r\n$f(zy) + f(z) + y = f(zy) + f(y) + z$\r\nwhich implies\r\n$f(z) - z = f(y) - y$\r\nwhich can only be true if\r\n$ f(x) = x + c$.\r\nThus, whenever the given identity is valid, the function must be in this form. We can quickly verify that the only $c$ which works is $1$.\r\n\r\nSorry if it just seems as if I am repeating everything. I just wanted to sort out the logic.", "Solution_10": "I don't agree:\r\nLet A be the image f(R).\r\nWhy is x in A ?\r\n\r\n(It's the same thing in IMO 1999 6/ )", "Solution_11": "[quote=\"hardsoul\"]Why?[/quote]\r\nBecause you used in your solution $x=f(z)$ for [b]every[/b] x, but how do you know that for every x there exist z such that $f(z)=x$??? You must prove that f is sjurective.\r\n\r\nP.S. I wrote the same as Igor just in other words.", "Solution_12": "Put $y = \\frac{1}{x}$ so that $f(xf(\\frac{1}{x})) = f(1) + x$. $f(1)$ is constant, so as x varies over R, so does $f$ of something.", "Solution_13": "[quote=\"Renshaw\"]Put $y = \\frac{1}{x}$ so that $f(xf(\\frac{1}{x})) = f(1) + x$. $f(1)$ is constant, so as x varies over R, so does $f$ of something.[/quote]\r\nYou have missed something: $f(xf(\\frac{1}{x})) = f(1) + x$ , $x$ doesn't varies over $\\mathbb{R}$ , $x$ varies over $\\mathbb{R^{+}}$, so you can't get values which are less than $f(1)$.", "Solution_14": "I had thought that \"x in R*\" just meant x was real and not zero. Does the R* actually mean $\\mathbb{R}^+$?", "Solution_15": "Actually, nhat wote $\\mathbb R^*$, and that's usually the notation for $\\mathbb R\\setminus\\{0\\}$, so we're not only working with positive numbers, but with all non-zero numbers. In this case, the surjectivity of the function is pretty clear, since $f(xy)+x$ can take any value we want it to take (which is just what Renshaw did, put into words).", "Solution_16": "I still don't think it is necessary that we prove $f$ to be surjective.\r\n\r\n[quote]\nPut x=f(z),f(f(z)f(y))=f(f(z)y)+f(z)=f(zy)+f(z)+y=f(zy)+f(y)+z\nhence f(x)=x+c.subsituting in original equation we'll get f(x)=x+1\n[/quote]\r\n\r\nThe point is that we don't care whether or not $f(z)$ gives each possible value of $x$. The identity $f(xf(y))=f(xy)+x$ will hold if we put $f(z)$ in for $x$, because the range of $f$ is a subset of the range of $x$. This is more of a \"plugging in\" than a substitution, for we plan not to let $x$ vary, but $z$. \r\n\r\nThen we use the whole symmetry thing, and arrive at $f(z) - z = f(y) - y$. Since both $z$ and $y$ are both free to vary thoughout the enitre domain (R* or whatever), this identity can only hold if $f(x) - x = c $. The original identity then gives $c = 1$.\r\n\r\nI think we were all just confused about what it means to \"put\" f(z) = x.", "Solution_17": "Igor wrote:\r\n\"I don't agree: \r\nLet A be the image f(R). \r\nWhy is x in A ? \"\r\nDear Igor x can take any value of R*(positive reals) so why can't it take all the values that f(A) can take?\r\nThe set of the range of the f(A) is not greater than all R*.\r\nArmo wrote:\r\n\"Because you used in your solution for every x, but how do you know that for every x there exist z such that ??? You must prove that f is sjurective. \"\r\nDear Armo I'm not trying to prove that f is surjective in the subsitution x=f(z) I'm just putting those values that f(z) can take.And in the equation f(xf(y))=f(f(z)f(y)) for any z we can take x such that x=f(z).\r\nI wrote the same as Renshaw wrote.", "Solution_18": "Igor wrote:\r\n\"I don't agree: \r\nLet A be the image f(R). \r\nWhy is x in A ? \"\r\nDear Igor x can take any value of R*(positive reals) so why can't it take all the values that f(A) can take?\r\nThe set of the range of the f(A) is not greater than all R*.\r\nArmo wrote:\r\n\"Because you used in your solution for every x, but how do you know that for every x there exist z such that ??? You must prove that f is sjurective. \"\r\nDear Armo I'm not trying to prove that f is surjective in the subsitution x=f(z) I'm just putting those values that f(z) can take.And in the equation f(xf(y))=f(f(z)f(y)) for any z we can take x such that x=f(z).\r\nI wrote the same as Renshaw wrote.", "Solution_19": "Sorry guys, I thought R* means $\\mathbb{R^{+}} $. And this is another reason to write your texts using LaTeX. :mad:", "Solution_20": "Never mind.I'm not very good at explaining,may be that's why I always got low points at contests . :)" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "congruent triangles" ], "Problem": "In a triangle $ABC$, $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\\angle A$. Suppose $\\angle B = 2 \\angle C$ and $CD =AB$. prove that $\\angle A = 72^{\\circ}$.", "Solution_1": "[hide=\"My solution\"]Since $AD$ is bisector of $\\triangle ABC$ we have\n\n$\\frac{AB}{AC} = \\frac{BD}{DC} \\Rightarrow$\n\n$AB \\cdot DC = BD \\cdot AC$\n\n$AB^2 = BD \\cdot AC$\n\nHence the segment $BA$ is tangent from $B$ to the circumcircle of $\\triangle ADC$\n\nSo we have $\\angle DCA = \\angle DAB \\Rightarrow \\widehat A = 2\\widehat C$\n\n$\\widehat A + \\widehat B + \\widehat C = 180 \\Rightarrow$\n\n$2\\widehat C + 2\\widehat C + \\widehat C = 180 \\Rightarrow \\widehat C=36 \\Rightarrow \\widehat A=72$[/hide]", "Solution_2": "Surprisingly, this question is a part of our high school syllabus.", "Solution_3": "i don't have the time to type out a full solution, but another way goes like this\r\n\r\n[hide]draw biscector of $B$, this intersect $AD$ at $I$, and $AC$ at $E$, let $BAD=y$, and $C=x$\n\n$C=EBC$, so $CE=EB$, using SAS congurency, $\\triangle ABE=\\triangle DCE$ (1) \nthen $AE=ED$ by (1) so $EAD=EDA=y$,\n\n$ECD=EAB=2y$ by (1)\n\nlooking at $\\triangle ADC$, $180=4y+x$, \nlooking at $\\triangle ABC$, $180=2y+3x$, \nsubtracting and simplifying gives $y=x$, so $180=5y \\implies y=36 \\implies \\angle A =72^{\\circ}$[/hide]", "Solution_4": "Altheman's solution is clear but I have a doubt about pontios' solution. My doubt is about the logical deduction that follows in the following lines:\r\n\r\n[quote=\"pontios\"]$ AB^2 \\equal{} BD \\cdot AC$\n\nHence the segment $ BA$ is tangent from $ B$ to the circumcircle of $ \\triangle ADC$[/quote]\r\nWhat I understand is that the power of a point at $ B$ is $ AB^2\\equal{}BD\\cdot BC$. How can he deduce that's the circumcircle of $ \\triangle ADC$ if it is not given that $ AC\\equal{}BC$?", "Solution_5": "[quote=\"thundercoder\"]Surprisingly, this question is a part of our high school syllabus.[/quote]\r\nlol do the they still have this in the CBSE syllabus book? It used to be there when I was in india..", "Solution_6": "[quote=\"10000th User\"][quote=\"pontios\"]$ AB^2 \\equal{} BD \\cdot AC$\n\nHence the segment $ BA$ is tangent from $ B$ to the circumcircle of $ \\triangle ADC$[/quote]\nWhat I understand is that the power of a point at $ B$ is $ AB^2 \\equal{} BD\\cdot BC$. How can he deduce that's the circumcircle of $ \\triangle ADC$ if it is not given that $ AC \\equal{} BC$?[/quote]\r\n\r\nThis proposition is true. Just work backwards through the standard proof for power of a point for tangent-secant.", "Solution_7": "[quote=\"Altheman\"][quote=\"10000th User\"][quote=\"pontios\"]$ AB^2 \\equal{} BD \\cdot AC$\n\nHence the segment $ BA$ is tangent from $ B$ to the circumcircle of $ \\triangle ADC$[/quote]\nWhat I understand is that the power of a point at $ B$ is $ AB^2 \\equal{} BD\\cdot BC$. How can he deduce that's the circumcircle of $ \\triangle ADC$ if it is not given that $ AC \\equal{} BC$?[/quote]\n\nThis proposition is true. Just work backwards through the standard proof for power of a point for tangent-secant.[/quote]I just tried following his steps and unfortunately, I am unable to see why $ AB^2 \\equal{} BD \\cdot AC$ implies the power of $ B$ with respect to the circumcircle of $ \\triangle ADC$.\r\n\r\nI can see it if it is $ AB^2 \\equal{} BD\\cdot BC$, but not from $ AB^2 \\equal{} BD \\cdot AC$...", "Solution_8": "Okay nevermind, I read it as the power of a point theorem...i guess you are right then...", "Solution_9": "[quote=\"10000th User\"]Altheman's solution is clear but I have a doubt about pontios' solution. My doubt is about the logical deduction that follows in the following lines:\n\n[quote=\"pontios\"]$ AB^2 \\equal{} BD \\cdot AC$\n\nHence the segment $ BA$ is tangent from $ B$ to the circumcircle of $ \\triangle ADC$[/quote]\nWhat I understand is that the power of a point at $ B$ is $ AB^2 \\equal{} BD\\cdot BC$. How can he deduce that's the circumcircle of $ \\triangle ADC$ if it is not given that $ AC \\equal{} BC$?[/quote]\r\n\r\noops! you're right! It should be \r\n$ AB^2 \\equal{} BD \\cdot BC \\equal{}>$ [the segment $ BA$ is tangent from $ B$ to the circumcircle of $ \\triangle ADC$]\r\n\r\nI don't remember what was my thought (it's more than 2 years..)\r\nI guess that I misread and confused about $ AC$ and $ BC$, I'm sorry :oops:", "Solution_10": "[hide]CAN BE DONE USEING SINE LAW[/hide]", "Solution_11": "Working on [b]Altheman[/b]'s little bit:\n$\\triangle CDE\\cong\\triangle BAE$ means that a rotation about $E$ will map the two congruent triangles, i.e. $\\angle BEC=\\angle AED$; with $DE=AE, BE=CE$ we see that, actually $\\triangle BEC\\sim\\triangle AED\\implies y=x$, hence $5y=180^\\circ\\iff\\hat A=72^\\circ$.\n\nBest regards,\nsunken rock", "Solution_12": "Sorry for reviving this old thread but I have a different solution.\n\nLet $\\angle BAD = y$ and $\\angle ACB = x$, the bisector of $\\angle B$ meet $AC$ at $E$. \n\nNote that $\\Delta ABE \\cong \\Delta DCE$. Therefore, $\\angle EDC = 2y \\implies \\angle EDB = 180-2y \\implies ABDE$ is cyclic. Considering segment $DE$, we get $x=y$. \n\nThus, $\\angle A + \\angle B + \\angle C = 180^o \\implies 2y+2x+x=180^o \\implies \\angle A = 2y = 72^o$", "Solution_13": "By angle bisector theorem ...... it directly implies $\\Delta ABC \\sim \\Delta BDA \\implies \\angle A=72^{\\circ} $", "Solution_14": "Note that $\\angle B =2C$,$\\angle A=180-3C$,$\\angle ADB = 90-\\frac{C}{2}$.\n\nUsing Sin rules in $\\triangle ADB$ ,\n\n$\\frac{AB}{\\cos \\frac{C}{2}} =\\frac{AD}{\\sin 2C}$.\n\n\nUsing sin rules in $\\triangle ADC$ we have ,\n\n$\\frac{AD}{\\sin C} = \\frac{CD}{\\cos \\frac{3C}{2}}$.\n\n\nWe get the following results from two above equation,\n\n$\\cos \\frac{C}{2} .\\sin {C} = \\sin {2C}.\\cos \\frac{3C}{2}$.\n\nGet,$\\sin \\frac{3C}{2} =\\sin \\frac{7C}{2}$.\n\nGet,$C=36$.\n\nSo,$A=72$.\n\n", "Solution_15": "[quote=problemsolve_AoPS]Sorry for reviving this old thread but I have a different solution.\n\nLet $\\angle BAD = y$ and $\\angle ACB = x$, the bisector of $\\angle B$ meet $AC$ at $E$. \n\nNote that $\\Delta ABE \\cong \\Delta DCE$. Therefore, $\\angle EDC = 2y \\implies \\angle EDB = 180-2y \\implies ABDE$ is cyclic. Considering segment $DE$, we get $x=y$. \n\nThus, $\\angle A + \\angle B + \\angle C = 180^o \\implies 2y+2x+x=180^o \\implies \\angle A = 2y = 72^o$[/quote]\n\nThat's really mindblowing bro..." } { "Tag": [ "induction" ], "Problem": "\"A horizontal and a vertical line intersect in exactly one point. Prove by induction that given n >= 1 horizontal\r\nlines and n vertical lines, that the total number of intersections between horizontal and vertical lines is never\r\nmore than n^2.\"\r\n\r\nInduction seems to be my weak point :( I can never figure it out,", "Solution_1": "Given $ n$ horizontal and $ n$ vertical lines, add an extra horizontal line. How many extra intersections do you get? Now add a vertical line. Same thing." } { "Tag": [ "algebra", "polynomial", "function", "college", "complex numbers" ], "Problem": "[i]Let f(x)=c[size=75]k[/size]x^k+...+c[size=75]0[/size] be a polynomial with c[size=75]k[/size] and c[size=75]0[/size] unequal to 0. Factor f(x) over the complex numbers as f(x)= c[size=75]k[/size](x-r1)^(m1) .(x-r2)^(m2)...(x-rn)^(mn), where the r sub n's are distinct nonzero complex numbers and m sub n's are positive integers. Then a sequence {a[size=75]n[/size]} satisfies the linear recurrence with characteristic polynomial f(x) if and only if there exists polynomials g1(n), g2(n), g3(n)...,gn(n) (numbers following g are subbed) with deg gi :le: mi-1 for i=1,2,...n such that\na[size=75]n[/size]=g1(n)(r1)^n+.....g[size=75]n[/size](n)(r sub n)^n\nfor all n. phew![/i]\r\n\r\njust in case you were wondering whether this was a pointless challenge or a question, i was just wondering if this theorem had a name (because its quite useful and i was doing some proof-type questions with it).", "Solution_1": "It's just called the characteristic equation.", "Solution_2": "oh...its that simple? thanks!", "Solution_3": "Once again, Simon takes the cake. Or the cheese. Eh, let's rephrase that:\r\n\r\n[b]Once again, Simon takes the cheesecake![/b]\r\n\r\nJust to make this a meaningful post, what is a characteristic polynomial?", "Solution_4": "If you have a linear recursion a_n+c_0*a_(n-1)+c_1*a_(n-2)+...=0, then the characteristic polynomial is x^n+c_0*x^(n-1)+c_1*x^(n-2)+...", "Solution_5": "i have another question about recurrences - what happens when the recurrence isnt linear? for example, what would happen if in some sequence {a[size=59]n[/size]}, a[size=59]n+k[/size] is multiplied or exponentiated or divided or logarithmed by n , instead of a constant? how would i find a general form for it then? would its characteristic polynom. be defined differently?", "Solution_6": "You can't (as far as I know) use characteristic polynomials in that case. You can use generating functions sometimes, but you'll probably have to solve a differential equation at the end. If you know how to do that, great. Otherwise, you're in a bit of trouble.", "Solution_7": "so, how would you find a formula for a[size=75]2n[/size]=a[size=75]n[/size]+n? the uc-berkely mathcircle packet talks about these inhomogenuous ones...but is there are another method?", "Solution_8": "There are infinitely many functions f: Z to R (or C or Z or whatever you want) satisfying that because you don't have any rules about the odd terms.", "Solution_9": "suppose we drag a constant terms behind a homogenous relation, something like:\r\na_n=a_(n-1)+a_(n-2)+k\r\nhow can i find an explicit formula?", "Solution_10": "Look at the first few terms of that particular example. I think it is fairly clear what happens, no? Just look for the pattern.", "Solution_11": "if we just look at the coefficient of k, we get:\r\n1,2,4,7,12,20,33,54,88,143,...\r\ni dont see the pattern...", "Solution_12": "Check out the first differences. (That wasn't really as obvious as I made it sound, sorry. But looking at differences is often a useful way of unlocking the pattern in a sequence.)", "Solution_13": "so the nth term would look something like:\r\nu_n=a*(r1)^n+b*(r2)^n+(phi)^(n+1)+(phi-1)^(n+1)\r\n??", "Solution_14": "Well, if we ignore the k part for a bit, we get a Fibonacci-type sequence, which is given by (a*:phi:^n + b*(-:phi:)^(-n)) for some real numbers a and b. Then, the k will be entirely independent. The coefficient of the k terms is partial sums of sequences of Fibonacci numbers, which you can evaluate using Binet's formula, or knowing the identity F(0) + F(1) + ... + F(n) = F(n + 2) - 1. So in the case like that where the k can be summed independantly, it isn't bad at all. It's when you start mixing things in together that it becomes difficult." } { "Tag": [ "function", "AoPSwiki", "articles", "search" ], "Problem": "I need to learn more about generating functions. I have read the AopsWiki article on it, but it doesn't say much more than what it is. What are some good sites to learn more about them? Some example questions, explanations, solutions to example questions, etc. would also be helpful. Thanks in advance!", "Solution_1": "i am not sure if it is useful for you \r\n\r\n enter here http://math.haiyanh.com/articles/generating-functions.html", "Solution_2": "Got this off of a google: http://en.wikibooks.org/wiki/HSE_Counting_and_Generating_functions\r\n\r\nI only skimmed over it, so I'm not absolutely sure of its accuracy/reliability. But it did cover a lot of stuff.\r\n\r\nOne good problem to do when you think you've finally gotten some mastery of generating functions is to prove that the number of ways to partition a number into distinct parts is the same as the number of ways to partition it into odd parts. http://mathworld.wolfram.com/PartitionFunctionP.html has more about the partition function, and this problem is briefly discussed if you scroll to a bit above the middle of the page.", "Solution_3": "Go search the web or the forum for \"generatingfunctionology,\" a book by the mathematician Herbert Wilf that he makes available for free online.", "Solution_4": "this time i am sure it will be useful.\r\n\r\n[size=150][b]the book for Generating Functions[/b][/size]\r\n\r\ndownload PDF here http://math.haiyanh.com/articles/generating-functions.html" } { "Tag": [ "Ross Mathematics Program" ], "Problem": "I will be applying to the OSU Ross Program this year.\r\nHowever, I don't really know I'll be accepted or not...\r\nI am solving the problems and can't solve two of them, but I'm still trying...\r\n\r\nSo, does anybody know how much people Ross accepts? \r\nBecause I will make other plans, if it's hard to get in.\r\nAlso, is there a date that the program announces people accepted?\r\n\r\nThanks.", "Solution_1": "I am an alum of Ross, and mathematically speaking, it was one of the most amazing experiences that I have had.\r\n\r\nOn the application it says that you don't have to solve the problems. Perhaps as important as actually presenting solutions is describing your thought process and demonstrating your ability to explore unknown and difficult math problems.\r\n\r\nI believe there is no set number of people that Ross accepts, though the program may try to prevent enormous amounts of people from going. The rolling admission means that they tell you whether or not you were accepted once they receive your application, though it is probably beneficial to apply early in case the program fills up.", "Solution_2": "[quote=\"Myself\"]I am an alum of Ross, and mathematically speaking, it was one of the most amazing experiences that I have had.[/quote]\r\n\r\nThis is true - I was a first year last year and now I know quite a bit of number theory. I'm also better at proving things and exploring mathematical topics due to Ross.\r\n\r\nI believe that you'll get in - I remember not solving about one and a half problems last year.\r\n\r\nDo apply early as you can though - I applied late April last year and got in but who knows this year.\r\n\r\nI hear there about 30-40 people going this summer.", "Solution_3": "Keep working steadily on the problems, and show your steps and your thought processes. Ross is very worthwhile; it's a great program for learners who are serious about learning more math." } { "Tag": [ "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "Show that if $ R$ is a commutative ring and $ N \\equal{} (a_1, \\cdots, a_m)$ where the $ a_i$ are nilpotent then $ N$ is a nilpotent ideal.", "Solution_1": "I haven't thought this one through fully. Doesn't this pretty much follow from the fact that the sum of nilpotent elements is nilpotent [use \r\nthat $ a^m\\equal{}0$, $ b^n\\equal{}0$ gives $ ra\\equal{}0$ for all $ r$ and $ (a\\plus{}b)^{m\\plus{}n}\\equal{}0$]. So, \r\nit seems to me if $ a_i^{n_i} \\equal{} 0$, we should think $ I^N\\equal{}0$ where $ N \\equal{} \\sum n_i$?", "Solution_2": "Recall that the ideal $ I^n$ consists of finite sums of terms of the form $ x_1 \\cdots x_n$ where the $ x_i \\in I$. Of course, an ideal is nilpotent if there exists a positive integer $ n$ so that $ I^n \\equal{} 0$. This is trickier than it appears because if $ I \\equal{} (a_1, \\cdots, a_m)$ then each of the above $ x_i$ is an $ R$-linear combination of the nilpotent $ a_i$.", "Solution_3": "Try induction on $ n$ and use the binomial formula on $ (x\\plus{}y)^k\\equal{}\\dots$.", "Solution_4": "I have tried that, but the method you're suggesting isn't quite enough. Try applying it formally to the situation I just described and you'll see that the expressions get fantastically complicated. It makes me wonder if there's a less brute-force approach.\r\n\r\nThe approaches suggested thus far would be appropriate when showing that the nilradical is an ideal (i.e. $ R$-linear combinations of nilpotent elements are nilpotent) but this is a little more difficult. I should stress also that the ideal is finitely generated, which I think allows you to put an upper bound on the nil exponents of elements of the ideal.", "Solution_5": "OK, the trick I have in mind helps you to perform the induction and to show that the nilradical is an ideal: Let $ x^n\\equal{}0$ and $ y^m\\equal{}0$. Then $ (x\\plus{}y)^{m\\plus{}n}\\equal{}0$ because the expansion of $ (x\\plus{}y)^{m\\plus{}n}$ is a sum of terms of the form $ x^ky^{m\\plus{}n\\minus{}k}$ in which the former vanishes for $ k\\ge n$ and the latter for $ k1$, then $ I_m\\equal{}I_{m\\minus{}1}\\plus{}Ra_m$. Now show that, for $ a\\in I_{m\\minus{}1}$, $ b\\in R$, $ (a\\plus{}ba_m)^{N_{m}}\\equal{}0$ with the trick above. ... done. This is not too tricky. ;)\r\n\r\nNote that basically the following is sufficient for the situation: Let $ J_1, J_2$ be nilpotent ideals in a commutative ring, i.e. $ J_i^{n_i}\\equal{}0$, then $ (J_1\\plus{}J_2)^{n_1\\plus{}n_2}\\equal{}0$.", "Solution_9": "Wonderful, I see it now. I do think that's a pretty subtle induction argument, but you're right, the above trick applies with $ m \\equal{} N_{m\\minus{}1}$ and $ n \\equal{} n_m$." } { "Tag": [ "probability", "articles" ], "Problem": "i wanted to know what u guys think since there seems to be a controversy \r\ni believe dumbledore asked snape to kill him", "Solution_1": "i think everything is the way it seems for once. I think snape wanted to kill dumbledore and succeded", "Solution_2": "Not dead. And I still think Snape is \"good\". \r\n\r\nUh...Dumbledore's \"death\" seemed too stupid on his part for it to be realistic. I don't know...that's just my opinion.", "Solution_3": "or maybe...dumbledore told snape that his position as volde's right=hand man would be of much more importance than dumble's own life...and that if given a choice b/w one...choose spyness\r\n\r\nbut either way...snape didn't like dumble too much lol...cuz didn't it say in book 5 that you needed to actually want to hurt somebody to use cruciatus? so you'd think it'd be the same for avada...specially since it prolly requires a lot more power\r\n\r\nso in the end...snape did what he wanted...but i think he's still a good guy at heart...(saving harry from certain death at the end of book 6...saving harry from quirrell in book 1...bringing harry back from the forest stead of killing him in book 3...errr ignore book 5 lol)", "Solution_4": "i said this in the hbp thread too, but snape is always loyal to dumbledore, he is only loyal to dumbledore; dumbledore does not fear death like other people/wizards; dumbledore mentioned repeatedly that harry's life is worth much more than his own in the chapter before his death.\r\n\r\nyou need to be angry/filled with hatred to perform cruciatus or avada kedavra. Snape WAS very angry - at having to follow dumbledore's wishes out of loyalty but as a result be branded as a traitor and a [i]coward[/i]; also probably at harry for his ignorance and inability to master occlumency and silent spell-casting.", "Solution_5": "very nicely put L_Li\r\n\r\n :omighty:", "Solution_6": "Dumbledore wanted to use Snape as a spy - most certain. Otherwise, the book has no room to um... expand. You know.", "Solution_7": "why the unbreakable vow tho?", "Solution_8": "My thoughts:\r\n\r\n1. Dumblebore isn't dead.\r\n2. Snape is a good person.\r\n3. Snape doesn't want to kill Dumbledore.\r\n\r\nNo I haven't read the book but that's what I have in mind.", "Solution_9": "Who is dumbledore??? :?", "Solution_10": "if anyone hasn't read the book yet, this title sort of gives away who \"dies\" ... ;)", "Solution_11": "[quote=\"Derek\"]My thoughts:\n\n1. Dumblebore isn't dead.\n2. Snape is a good person.\n3. Snape doesn't want to kill Dumbledore.\n\nNo I haven't read the book but that's what I have in mind.[/quote]\r\n\r\noh ok so is this what the other party consists of? people who havent even READ THE BLOODY BOOK?!\r\nif you actually read the two chapters leading up to his death, you'll be more convinced that dumbledore really died and that there's really no reason for rowling to want him to be secretly alive anyway.", "Solution_12": "go to [url=http://dumbledoreisnotdead.com/introduction.html]dumbledoreisnotdead.com[/url] thats all i have to say...", "Solution_13": "but tricking voldemort into actually not killing dumby is pretty hard even with skilled occlumency", "Solution_14": "I think we just spoiled the ending of the book for a lot of peoples.", "Solution_15": "He is not actually [i]bad[/i], only a bit misled. If Harry had a troubled childhood, Voldy had double that. We must understand his feelings.", "Solution_16": "Actually I think he is evvil, cause he's villain..... or he has a seriously messed up mind, which I think could be the case just as well", "Solution_17": "Well if you did read that site duimbledoreisnotdead.com,ou'll change your mind and say dumby's alive,read each and every thing.\r\n\r\nI fet some of the things had been stolen from LORD OF THE RINGS.", "Solution_18": "I have better work thank you!\r\nI know the kind of rubbish that is posted in these sites. \r\nBut the picture of Dumbledore is nice! :P", "Solution_19": "I agree that Dumbledore must be alive or Rowling would be booed on her 7th book...somehow he's alive. If he's not, I admit I am wrong", "Solution_20": "But isn't he technically dead since his picture got hung up in the office?", "Solution_21": "That gives me an idea... maybe harry will find a way to get dumbledore out of the portrait and dumbeldore will come back. But I still go with th idea of a horcrux.", "Solution_22": "He is technically dead in more ways than one. But vicseb could have a whole team of ex-Hogwarts headmasters like that!", "Solution_23": "i have read in a mugglenet article that the water of the lake might have been draught of living death and dd might b sleeping in his tomb\r\nit didnt make much sense", "Solution_24": "Yes Snape's Avada Kedavra must put him to sleep somehow.", "Solution_25": "Yeah, but he fell out of the tower (which must've/could've? killed Dumbledore). However, you have to WANT to kill someone to do the Avada Kedavra curse. Maybe Snape just wanted to put Dumbledore into a swoon... I remember something in the fourth book about Moody saying that you have to feel rage. Maybe Snape \"tried\" to feel rage so that he could just make Dumbledore faint... :ninja:", "Solution_26": "But wizards like Flitwick, McGonagall, Slughorn, Lupin, Moody and the others are more powerful than Snape and would have recognized a sign of life in A.D.\r\n\r\nOr maybe they did? Maybe the funeral was just to deceive the Deah Eaters?", "Solution_27": "i say only RAB knows wats goin on", "Solution_28": "and only JKR knows who's RAB so let the matter rest", "Solution_29": "genius turned to madness due to the potion (or poison) and asked snape to kill him" } { "Tag": [ "induction" ], "Problem": "Find\r\n\r\n$ \\sum_{x_{1}\\equal{} 1}^{6}\\sum_{x_{2}\\equal{} 1}^{x_{1}}\\sum_{x_{3}\\equal{} 1}^{x_{2}}...\\sum_{x_{9}\\equal{} 1}^{x_{8}}1$\r\n\r\n[hide=\"hint\"]Balls and urns in disguise[/hide]", "Solution_1": "$ \\sum_{a\\equal{}1}^{b}1\\equal{}b$\r\n$ \\sum_{b\\equal{}1}^{c}b\\equal{}\\frac{1}{2}c (c\\plus{}1)$\r\n$ \\sum_{c\\equal{}1}^{d}\\frac{1}{2}c (c\\plus{}1)\\equal{}\\frac{1}{6}d (d\\plus{}1) (d\\plus{}2)$\r\n$ \\sum_{d\\equal{}1}^{e}\\frac{1}{6}d (d\\plus{}1) (d\\plus{}2)\\equal{}\\frac{1}{24}e (e\\plus{}1) (e\\plus{}2) (e\\plus{}3)$\r\n\r\n$ \\ldots$\r\n\r\nBy induction, we can afirm:\r\n\r\n$ S \\equal{}\\sum_{x_{1}\\equal{}1}^{s}\\sum_{x_{2}\\equal{}1}^{x_{1}}\\ldots\\sum_{x_{j}\\equal{}1}^{x_{j\\minus{}1}}1\\equal{}\\frac{1}{j!}\\prod_{i\\equal{}0}^{j\\minus{}1}(s\\minus{}i)$\r\n\r\nfor your problem $ j\\equal{}9,s\\equal{}6$\r\n\r\nAnd by simple inspection of the formula we can afirm that $ S\\equal{}0$ for $ s 0$ is rational and that $sin(x)$ is rational then $\\frac x {\\pi} = 0, \\frac 1 2 , \\frac 1 6$.\r\nIt follows that the solutions of the given equation are $n=0,1,2$.\r\n\r\nPierre." } { "Tag": [ "LaTeX" ], "Problem": "You've most likely seen the \\mathbb-styled 1 they often denote for the identity map.\r\n\r\nHow do you write it? $ \\mathbb{1}$ seems to produce something different, so I wondered.", "Solution_1": "See [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LATEX Symbol List[/url] page 65. Hence\r\n[code]\\usepackage{dsfont} or if you prefer \\usepackage[sans]{dsfont}\n...\n$\\mathds{1}$[/code]" } { "Tag": [ "quadratics", "complex numbers" ], "Problem": "I know that this has been posted before, but how do you solve this?\r\n\r\n$a + b = 6$\r\n$ab = 14$\r\n\r\nor something of the sort.", "Solution_1": "Say that a=6-b. Then you have (6-b)b=14, 6b-b^2=14. b^2-6b+14=0, then just apply quadratic formula.", "Solution_2": "I'm guessing you want the values of $a$ and $b$.\r\n\r\n$a + b = 6$\r\n\r\n$a = 6 - b$\r\n\r\n$(6-b)b = 14$ \r\n\r\n$6b-b^2=14$\r\n\r\n$b^2-6b+14=0$\r\n\r\nThe roots of this equation involve Complex Numbers so it shouldnt be in this forum, but the roots are $3 + i\\sqrt5$ and $3 - i\\sqrt5$", "Solution_3": "Ok, the first one works, although I prefer completing the square.", "Solution_4": "Oh, yeah", "Solution_5": "[quote=\"4everwise\"]I know that this has been posted before, but how do you solve this?\n\n$a + b = 6$\n$ab = 14$\n\nor something of the sort.[/quote]\r\n\r\nI've devised my own little method for this kind of thing. Set a=b, so a=3, b=3, then do ab=9. Take 14-9=5, so you make $a=3+i\\sqrt{5}$ and b=6-a\r\nThe part you add to a is the square root of -(ab-a-b)... May sound complicated, but it's simple to me and it works... I like making up my own methods for doing problems because:\r\na) They're faster (to me)\r\nb) They're more understandable (to me)\r\nc) Everyone likes doing things the way they want to do it... (anyone denying this is werid :))" } { "Tag": [ "geometry", "calculus", "percent" ], "Problem": "As a member of Art of Problem Solving Forums, you must be very good at math, so your math grade at school is obvious to be really good. \r\n\r\nBut how do you do in other subjects? Such as:\r\n\r\n-Science (Chemistry, Physics, Biology, Botany...)\r\n\r\n-English (Language Arts, Writing, Critical Reading...)\r\n\r\n-History (Social Studies...)\r\n\r\n-Foreign Language (French, Latin, Spanish...)\r\n\r\n-Fine Arts (Music, Art..)\r\n\r\n\r\n[hide=\"My Personal Profile\"]\n[b]Grade Averages[/b]\nScience- A+\nEnglish- A-/B+\nHistory- B+/A-\nForeign Language- A+\nFine Arts- A+\n\nAs you can see, my weaknesses are English and History... :blush:\n[/hide]", "Solution_1": "mid quarter grades were (i'm in eigth grade)\r\n\r\n[hide]tech ed, A-\nstrings, A\nsocial studies, A\nscience, A\ncommunication arts, A\n\ngeometry, 97.6\n\nTech ed's my only bad subject, but only because I miss alot of class because of geometry[/hide]", "Solution_2": "i'm good at most subjects. through 3 years of high school, i've only made one B - that was in English 3 H. i should make all A's this year as well. english is my weakest subject. i like sciences and history a lot. took 3 years of french and stopped as i quit enjoying the class. never took any fine arts classes.", "Solution_3": "I consider myself to be pretty good at.. most subjects:\r\n[hide]Last Year Final Grades:\nEnglish: A (I think i have a great vocab)\nScience: A+ (easy class)\nSocial Studies: B+ (messed up one semester)\nComposition: B (didnt hand in 1 book report.. would explain)\nComputers: A (but computers in my school isnt learning bout them, its just doing some projects for other teachers mostly. Otherwise, my friends come to me for tech help.)\nMath: A - but i guess thats not another subject\nForeign Language (Hebrew): A\nI cant think of any other subjects right now.\n\nThis year so far i have a :\n\n100 in foreign language (hebrew)\n104 (Religious test)\n87 (religious)\n95 (math - but it was a tough test, highest class, and was highest grade in the grade (half the grade failed it)\nthats it so far i think.[/hide]", "Solution_4": "LOL, why is it that everyone who has posted so far has failed to include there Physical Education marks? I can proudly say the majority of my P.E. marks have been A's. :lol:", "Solution_5": "Well i only had to take one PE class in high school, but i did get an A. The sciences are very good subjects for me (chem is almost better than math), although i do not enjoy the memorization required for biology. History/Geography are also very good classes for me and I enjoy learning more about such topics. English is not class i am particularly talented in, but i work harder in it than in other classes (not that i work in any class, but anyway). I performed well in foreign languages (French), but the teacher had some mental problems (very stressed out, unbalanced) and i decided to not pursue it further. Visual arts are my second passion. I even spend time outside of class drawing for my own enjoyment, but other arts are not my favourite subjects.", "Solution_6": "[quote=\"G-UNIT\"]LOL, why is it that everyone who has posted so far has failed to include there Physical Education marks? I can proudly say the majority of my P.E. marks have been A's. :lol:[/quote]\r\n\r\nI didn't put it because my school doesn't give a grade for Physical Education.\r\n\r\nI don't know about others. :(", "Solution_7": "I didnt think that PE was really important to mention... my school's PE is easy.. i got an A - but im also on the basketball team and wrestling team.. soo...", "Solution_8": "i think that people haven't mentioned PE becasue the class can't be graded on physical/athetic abilities so it must be based on participation. since (almost) everyone participates, everyone gets A's in PE. since everyone get's A's, it's useless to mention.", "Solution_9": "I just love how everybody loves to inflate their own grades. Makes you feel very warm inside. And I also love excuses for \"low\" grades. Do we really care if everyone else flunked the class? I dunno. But still, it looks better when you try to impress other people.\r\n\r\nme last year-\r\nbc- A+\r\neng- A \r\nspanish- A\r\nAP World- A\r\nPhysics- A\r\nEcon (macro + micro aps)- A, A+\r\nAP Bio- A- (which was with the hardest teacher in the entire school and half the class got Cs or lower, and I got a 98 on the final when the mean score was a 72)\r\nAP Suckup- A", "Solution_10": "[hide=\" Shameless boosting of self\"]Physics AP- A\nChemistry AP- A\nBusiness Law- A\nSpanish 3- A\nComposition- A\n\nTo tell you the truth, my hardest class is Chemistry AP, even though its a science. Its just too confusing for me.[/hide]", "Solution_11": "I personally hate chemistry.\r\n\r\nAnd maybe if it made sense I'd like it.", "Solution_12": "Science = no problem\r\nEnglish = yes lots of problems.\r\nHistory - I thought I sucked but I did not.\r\nJapanese - We don't do anything.\r\nArt - I am terrible.\r\nPE - I'm ok.", "Solution_13": "Everything is good to an extent.... but ENGLISH!!! ROAR! :mad: Ever since high school... that subject... will be my doom....", "Solution_14": "I got along fine with all of my subjects, at least up till now\r\nExcept for the Comprensive Reading standardized test... I always in the B range\r\nMy Gym grade is unsteady too.", "Solution_15": "[quote=\"yif man12\"]I just love how everybody loves to inflate their own grades. Makes you feel very warm inside. And I also love excuses for \"low\" grades. Do we really care if everyone else flunked the class? I dunno. But still, it looks better when you try to impress other people.\nAP Bio- A- ([b]which was with the hardest teacher in the entire school and half the class got Cs or lower, and I got a 98 on the final when the mean score was a 72[/b])\n[/quote]", "Solution_16": "I did bad in english one year, history one year, and this class which was basically dissecting films like they were books, i did bad in that class.", "Solution_17": "[quote=\"bleumoose\"][quote=\"yif man12\"]I just love how everybody loves to inflate their own grades. Makes you feel very warm inside. And I also love excuses for \"low\" grades. Do we really care if everyone else flunked the class? I dunno. But still, it looks better when you try to impress other people.\nAP Bio- A- ([b]which was with the hardest teacher in the entire school and half the class got Cs or lower, and I got a 98 on the final when the mean score was a 72[/b])\n[/quote][/quote]\r\n\r\nuhh... what's the point of that? Because I'm pretty sure I made it so that others would not take my grades seriously. Those are all made up. Well, not all made up. But partially made up. I keep this kind of info to myself. So guess away. Maybe one of them's right, maybe 6. Who knows? I know I do.", "Solution_18": "GPA: 4.0 ... says it all, doesn't it? or perhaps you need AP Test Average: 5 :D of course only one of those was not math or science...\r\n\r\nOf course I don't consider myself a history or english expert by ANY means... I'm just good at snatching those A's.", "Solution_19": "Last year i earned a 4.0 (weighted and unweighted, no honors or AP classes cuz im a noob D:) during 2nd semester with Band, Teachers Aide (I should have taken German 3 honors but German is THAT boring that i couldn't do another year), Language Arts 3, US History, Algebra 2, and Physics.\r\n\r\nBack in 7th grade i was in the honors track for both Language Arts (English) and Math. I excelled in honors language arts but struggled in honors 7th grade math (honors pre-algebra). Eventually i fell so behind that i stopped trying or caring. I passed with a C- but i needed a B- or higher to go onto Algebra 1 or an A- or higher to go onto Algebra 1 honors. So in 8th grade i continued to excel in honors language arts and i did very well in a second year of honors pre-algebra with a different teacher (thank god). So if you judge my math skill by what level of math i am taking as a senior (Trigonometry/Math Analysis....normal schools call it Precalculus), you'd probably say i am average (or possibly a little bit remedial since they are pushing back Algebra 1 to 8th grade so rapidly). However, in Algebra 1 in high school, i started to realize that i liked math and that i hated language arts so i took regular language arts instead because honors was just a pain in the ass and i couldn't stand to read as much as they expected you to. My math grades show how i gained an interest in math:\r\n\r\nAlgebra 1: B -\r\nGeometry: B+\r\nAlgebra 2: A\r\nTrig/math analysis and AP Stats: Hopefully A's :D\r\n\r\nSo even though my math level is average and most of my peers who were in my 7th grade math class are in Calculus BC right now, i'd say math is my best subject. I get A's in everything else but it's because college prep classes at my school are a joke. For example, i had a really easy chemistry teacher in 10th grade and i got A's both semesters but since she was easy, i didn't learn anything. All i remember from chemistry is how to balance basic equations and HCLBR FINO (Huckleberry fino....some gay pneumonic device for remembering the diatomic elements, H, Cl, Bf, F, I, N, O). Too bad i forgot what a diatomic element is!!!!! \r\n\r\nMan i wish i tried harder in 7th grade honors pre-algebra D:", "Solution_20": "GPA: 4.0 Unweighted.\r\n\r\nI guess that's pretty good.\r\n\r\nI love english. It kind of contradictory to like english almost as much as math... but I do. I'm on our Newspaper Club, I'm a Literary Magazine Editor, and I do Speech and Debate.\r\n\r\nPre-Calculus A+\r\nPhysics A\r\nChemistry A+ \r\nEnglish A+\r\nSpanish A+\r\nAP European History A", "Solution_21": "Just look at Bertrand Russel!", "Solution_22": "[hide=\"grades as of right now\"]\nComputers: A+\nSocial Studies: A like 98%\nScience: A+\nMath:A+\nIndustrial Tech: A \nGym: A (all based on effort and wearing your gym clothes)\nEnglish Language Arts: A+\nI'm in 7th, and my school is really easy. :)\n[/hide]", "Solution_23": "My grades are all right. I like reading, but not writing, at least not to the school prompts.", "Solution_24": "grades:\r\n\r\nAP Human Geography: 96\r\nAP Biology: 98\r\nHonors Algebra II/Trig: 104\r\nHonors English 9: 98\r\nSpanish I: 98\r\nP.E. : A (don't know the percent)\r\n\r\ni really like bio...i hate spanish, human geography is really boring, english is easy,p.e. is basically wear p.e. clothes and participate[/hide]", "Solution_25": "Afrikaans 90%\r\nEnglish 96%\r\nMaths 100%\r\nAdditional maths 100%\r\nScience 93%\r\nAccountancy 99%\r\nMusic 95%\r\nMusic second instrument 90%\r\nHistory 100%\r\n\r\nIn our country we have only 6 subjects, but I take 9, lol :rotfl: .\r\nFunny that I do better in English than Afrikaans, since I am actually Afrikaans speaking.\r\n\r\nThe government wants to change the educational system in South Africa. In the new system, everyone who has 70%+ will get a \"4\", 40-69% will get a \"3\", 35-39% = \"2\" and less than 35% = \"1\". This will be the only mark that you receive. There will be no way for you to know whether you get 70% or 100% - you will get a 4. This is really stupid, isn't it? It sucks! :(", "Solution_26": "Im pretty good at other subjects, even though Im just a freshman and school started like a month ago. At my school we have a block schedule, so Im only taking 4 classes now, and next semester I'll take the other 4 classes but my grades right now are-\r\n\r\nHonors Algebra II- A (greater than 95%)\r\nHonors Biology- A(close to B...around 94%)\r\nHonors Symphonic Band- A(Around 95%)\r\nHonors Pre-Calc- A(Around 95%)\r\n\r\nNext semester I will taking Honors English, Honors World History, Honors Symphonic Band II, and P.E. (not honors)", "Solution_27": "I have 98+ in mathematics and an average of 95+ in all of my subjects except my currect science class because they are talking about biology and stuff (bad memory causes a drop in marks). Fortunately, I'll have physics, and all maths next semester, so I'll be good for next semester. Besides, doign math in boring classes in fun. :lol: \r\n\r\nMasoud Zargar", "Solution_28": "i rule at science!\r\nnot good at LA or SS, tho\r\n :blush:", "Solution_29": "I used to be good last year in subjects like social,hindi computer but now that feeling of mugging stuff(Even in bio) is just fading away...I\"m still good at english and Science.\r\n\r\nI am good at math but i dont know why i fluctuate so much in xams coz of calculations n stuff,i get 98.5,then 81,then 94....li just finished my xams today...lets see what happens", "Solution_30": "[hide=\"Current Grades\"]\nHonors French IV: A\nAP Calc BC: A\nHonors Orchestra: A\nHonors Chemistry:A+\nAncient Philosophy: A\nHonors British Lit: A\nHonors World History: B+\n [/hide]", "Solution_31": "my grades for my first quarter... midterm for other classes\r\n\r\nAP US History - B\r\nAP Language (English) - B\r\nAP Calculus AB - A\r\nAP Calculus BC - A\r\nAP Spanish - A\r\nTheory of Knowledge - A\r\nIB Philosophy - A", "Solution_32": "maths: 10\r\nenglish(my 2nd language): 9 \r\nchemistry: 8\r\ngerman(my3d language): 8\r\nlatin:7\r\nhistory:7\r\nitalian:7\r\nin italy the marks are\r\nunder 6: bad\r\n6 good\r\n7 better than 6\r\n8 very good\r\n9outstanding\r\n10 more tha outstanding", "Solution_33": "[hide=\"my current grades\"]\nHonors Geometry: A\nEnglish: A-\nSocial Studies: A+\nScience: A+\nGerman: A+\nStrings (violin): A+[/hide]", "Solution_34": "I'm really good at Math, French, and Science (my 3 honors). I'm pretty good at History/Social Studies, only not in honors cuz I don't wanna have way too much hw, and ok at writing (English), though I have been improving my writing more recently. And in English, I'm really good at grammar.\r\n\r\nAlso, I'm a freshman in HS, and I've found it's easier to get A's at HS. In JH I had various B's here and there, mainly in English. Now at HS it's all A's.\r\nWell, in French and Math I seemed to be the best person in my grade, but not as much in Science...now in HS science I (and various others) are taking biology, and I'm beating most ppl on our tests :lol: \r\n\r\nI especially hate projects though....95.5 test average in bio......got a 75 on a term project :(. I also forget to do homework a lot....got a couple B+ in french b/c of it.\r\n\r\nIn Math (all honors):\r\nAlgebra 1 was a joke, easy A+\r\nBut in geometry it was hard.....only a B\r\nNow Algebra 2 is really easy (said to be hardest honors class at my school)", "Solution_35": "[hide=\"estimated grades\"]\nenglish: 102\nscience: 97\nsocial studies: 96\nspanish: 100 \n\nThat's basically it.. there aren't a lot of grades for me [/hide]", "Solution_36": "[quote=\"pkothari13\"]Everything is good to an extent.... but ENGLISH!!! ROAR! :mad: Ever since high school... that subject... will be my doom....[/quote]We are in the same boat, that subject... will be my doom too.... \r\n[hide=\"click here to find out why\"]I got a C on my recent English test :([/hide]", "Solution_37": "[hide]geometry: A-\nreading language arts: A\nsocial studies: A+\nscience: A\nPE: A\nFrench: A\nmath elective: A\n\nI am surprised at my math grade......but I bombed a test, so it shows :( [/hide]", "Solution_38": "I'm good in everything except French, which is very hard, and the smartest class I'm in short of my math.", "Solution_39": "Well pretty good, I got clear A+'s in Math, Science and social science and English and barely scraped an A+ in language(Sanskrit), (I didn't deserve it!)\r\n Ashwath", "Solution_40": "GPA: 4.0\r\ncurrently taking math and science classes at a community college, cause i'm done with HS. :) \r\n\r\nmy science (ESPECIALLY Chemistry) is lot stronger than my math, my math is strong also\r\n\r\ntook the SATII this Oct for first time, Math: 800, Chem:800, Physics:770" } { "Tag": [ "linear algebra", "matrix", "vector", "algebra", "polynomial" ], "Problem": "I have a problem here:\r\n\r\nLet matrix A = uv^T is a column times a row (a rank-1 matrix).\r\nNote: v^T means transpose of v\r\n\r\n1)By multiplying A times u, show that u is an eigenvector. What is its eigenvalue.\r\n\r\n2) What are the other eigenvalues of A (and why?)", "Solution_1": "A = uv^T is a rank-1 matrix, so how do we show that u is an eigenvector if multiplying A times u?\r\n\r\nWe have Au = Lu, (A - LI)u = 0, where L represents lamda.\r\n\r\nso how to show that det(A - LI) = 0 ?", "Solution_2": "You don't want to do that (look at the determinant). Instead, focus directly on the definition of eigenvector. Use matrix notation and the associativity of matrix multiplication.\r\n\r\n$Au=(uv^{T})u=u(v^{T}u).$\r\n\r\nNow, $v^{T}u$ is a $1\\times 1$ matrix - that is, a scalar. To be exact, it's the inner product of $u$ and $v$. Since it is scalar, it can be written on either side of the vector it's multiplying. Hence:\r\n\r\n$Au=(v^{T}u)u$ or $Au=\\lambda u,$ where $\\lambda = v^{T}u=u\\cdot v.$\r\n\r\nBeyond that, the fact that $A$ is a rank-1 matrix means that zero is an eigenvalue of it of multiplicity $n-1.$ So the eigenvalues of $A$ are $n-1$ copies of zero, and $u\\cdot v.$ The characteristic polynomial of $A$ is $x^{n}-(u\\cdot v)x^{n-1},$ and the minimal polynomial is $x^{2}-(u\\cdot v)x.$", "Solution_3": "An interesting special case - what would happen if $u$ and $v$ were orthogonal (without either one being zero)? Try it with $u=\\begin{bmatrix}1\\\\1\\end{bmatrix}$ and $v=\\begin{bmatrix}1\\\\-1\\end{bmatrix}$ just to get the idea.\r\n\r\n[hide=\"The results.\"]You get a rank-1 nilpotent matrix - in particular, you get a rank-1 matrix $A$ such that $A^{2}=0.$ Of course, all of its eigenvalues would be zero.[/hide]" } { "Tag": [ "vector", "calculus" ], "Problem": "I need a balancing equations program for my TI-89 Titanium.\r\n\r\nDoes anyone know of a program I could download (from like ticalc.org or something)?\r\n\r\n\r\nSimple Equations like:\r\n\r\nC3H8 + O2 ---> CO2 + H2O\r\nAnswer would be\r\nC3H8 + 5O2 ---> 3CO2 + 4H2O\r\n\r\n3 Carbons\r\n8 Hydrogens\r\n10 Oxygens\r\n\r\n\r\nDoes anyone know of a program I could download (from like ticalc.org or something?", "Solution_1": "The only I have seen that did that, did not give you a direct answer, it just told you how to set it up.", "Solution_2": "uhhhh just use matrices and the auto-solve thingy\r\n\r\n\r\nalso gonna point out that you make a program pretty easily if you use vectors in n dimensions, where n is the number of elements you're using (hey, an application of something i learned in multivariable calculus)" } { "Tag": [ "AMC", "AIME", "Putnam" ], "Problem": "Have you seen the web site [url]http://www.kalva.demon.co.uk[/url] by John Scholes? It's got a really nice collection of problems from the AIME, US Olympiads, International Olympiads, Putnam, and many more sources. Anyway, I noticed that many of the pages there say that \"to avoid possible copyright problems, I have changed the wording, but not the substance, of the problems\". Is that legal? I would have thought that just changing the wording is not enough to avoid copyright infringement. Just wondering...", "Solution_1": "My guess would be that he's a little bit iffy on the copyright question, but the math contest folk don't seem particularly likely to go suing people . . .\r\n\r\nI bet he is violating the copyright laws, actually. Because if you write a book similar enough to another work, that can be a violation. So odds are good he's breaking the law.\r\n\r\nIt is a great site. I should go practice for the Putnam . . .", "Solution_2": "Joel, I don't think that's quite right. It is not possible to patent mathematical expressions and formulas (circa copyright laws in the US, don't know about elsewhere)\r\n\r\nKalva has specifically modified the wording of each question as to change the actual statement of the problem but not the spirit of the problem. It is well known in the competition circles that everyone \"borrows\" everyone else's problems from time to time, and it would be almost impossible to find a problem that one cannot find prior art for.", "Solution_3": "If you can copyright words, you can copyright math. And if you went through Harry Potter, carefully changing the wording of every sentance while keeping the 'spirit' intact, that would probably be copyright infringement, too. But, as I said, no one is going to bother about it anyways.", "Solution_4": "[quote=\"JBL\"]If you can copyright words, you can copyright math.[/quote]\r\n\r\nWhat one can actually copyright is termed \"expression\" by lawyers. Once I looked it up, after someone asked on the Internet whether cooking recipes are subject to copyright. The answer, which amazed me at the time, is no. The legal idea is that a mere recitation of facts doesn't have enough \"expression\" to be deserving of copyright protection. So a cook can write rhapsodically about how delicious food is, and all of those paragraphs might be protected by copyright, but the recipes in the cook's cookbook might not be. Moreoover, there are exceptions to general copyright protection related to the intended use of the copied material, and someone who adapts a FEW problems from a published source, for NON-PROFIT purposes, not in verbatim form, and credits the original source (which keeps readers interested in buying the original source, thus not destroying its market) will probably be found to have fit within the \"fair use\" doctrine, an explicit statutory exception in the United States copyright law. \r\n\r\nAll that having been said, I BUY sets of old problems from the AMC people at list price rather than photocopying them, because I think they ought to get fair value for their efforts in putting together contest programs. And I buy a lot of problem books--and notice that many problems recur in several of those books. I hope everyone here agrees that math problem books are, to quote Marcus Aurelius in a closely related context, something \"on which money should be eagerly spent.\"" } { "Tag": [], "Problem": "What non-zero, real value of $ x$ satisfies $ (5x)^4\\equal{} (10x)^3$? Express\nyour answer as a common fraction.", "Solution_1": "Divide by 5x ^3 we get 5x=2^3 so x=8/5" } { "Tag": [ "MIT", "college", "Duke", "Princeton" ], "Problem": "What colleges did you all end up deciding on? A combination of financial aid concerns and an awesome CPW experience convinced me to choose MIT.", "Solution_1": "Congratulations! I'm sure you'll be very happy here. :)", "Solution_2": "$ \\frac{E}{c^{2}} \\sqrt{\\minus{}1} \\frac{PV}{nR} !$", "Solution_3": "I'll be at MIT as well. See you guys there!", "Solution_4": "I'm going to Duke, but MIT is a good choice too. :wink:", "Solution_5": "I could be wrong but, jonathanchou711, I thought you were in Middle School right now...", "Solution_6": "I will be heading off to Cornell University for engineering. i look forward to it more than anything right now.", "Solution_7": "I'm going to Princeton University :D", "Solution_8": "$ \\text{Financial Aid} \\bigcap \\text{Good Weather} \\Rightarrow \\mathbb{CALTECH}$", "Solution_9": "Will also be at MIT.", "Solution_10": "University of Cambridge, UK", "Solution_11": "[quote=\"LightEntropy\"]I will be heading off to Cornell University for engineering. i look forward to it more than anything right now.[/quote]\r\nSo is my friend Brett", "Solution_12": "I'm going to Princeton. :)", "Solution_13": "I'm going to Princeton!", "Solution_14": "Hah, I'm jealous of all of you. (just found this thread linked through the other one)\r\n\r\nI'm taking a gap year (and hopefully going to college after that), to a good degree due to financial concerns.\r\n\r\nAnyways, I think this will be a good move for me in general, but I have to wonder if I'm literally the only USAMOan this year not heading to college (anyone know?). (Actually, I'm curious in general about how math competition results correlate with other things in life, like sports and language skills. A topic for the round table, I suppose)" } { "Tag": [], "Problem": "$ 3^7 \\minus{} 3^6 \\equal{} N\\cdot3^6$. Find $ N$.", "Solution_1": "Factoring yields $ 3^6(3\\minus{}1)$, which takes the form of $ N\\cdot 3^6$. Therefore, N = 3-1 =2." } { "Tag": [], "Problem": "Are there any programs for the Ti84+SE so that I can directly write in assembly code on the calculator?\r\nI've searched on ticalc.org, and found ASM editor and on calc asm editor, etc., but none of them works well on my calculator.\r\nAlso I'd like to avoid opcoding if possible, or perhaps try it only after I get better with asm.\r\nCould anyone give me some advice on where I can find such programs?\r\nthx.", "Solution_1": "You can always look for emulators for ASM calculator.s" } { "Tag": [], "Problem": "I saw this problem somewhere but it didnt give an answer so I want to make sure my solution is correct.\r\nHow many ways are there to get from the point (0,0) to (4,4) in 10 or less steps if for every step you can go either one unit north, south, west or east. If you reach (4,4), you are done.\r\n[hide=\"I got\"]I got 2310, because it has to be 8 or 10 moves. Ofr 8 moves, its just 8C4, for 10 moves it can be 5N,1S,4E or 5E,1W,4N. FOr both of those combinations, there are 10!/(4!5!), but some of those you would already reach (4,4) before the 10 steps and there are 8C4*2 of them so (10!/(4!5!)-8C4*2)*2+8C4=2310. I dont think I did anything wrong but i just want to be sure[/hide]", "Solution_1": "[hide=\"solution\"]It will either take 8 or 10 steps to get there. \n\nIf he takes 8 steps, its just ${8\\choose 4}=\\frac{8\\cdot 7\\cdot 6\\cdot 5}{4\\cdot 3\\cdot 2}=70$.\n\nIf he takes 10 steps he can go 5 west, 4 north and 1 east, or 5 north, 4 west and 1 south. \n\nSo $2\\cdot{10\\choose 5}\\cdot 5=10\\cdot \\frac{10\\cdot 9\\cdot 8\\cdot 7\\cdot 6}{5\\cdot 4\\cdot 3\\cdot 2}=2520$\n\n$2590$\n\nI think I calculated one of the combanatorics wrong. :maybe: [/hide]\r\n\r\nI don't see anything wrong with your solution", "Solution_2": "Wait, jli, do you have to subtract something frmo your second case (10 moves) because in some of those cases for 10 moves, you would already have raeched (4,4), so you dont have to go anymore (for example if you go NNNNEEEEWE, you will already have reached (4,4) in 8 moves, but when you counted the ones for the 10 moves, you also incorperated that into the solution so you overcounted)\r\n\r\nI was just wondering because everyone else got like really small answers (like 144, 210...)", "Solution_3": "ArtofOwna, you are correct (and you correctly point out jli's error)." } { "Tag": [], "Problem": "So it turns out the MOP is June 12th through July 2nd. New York Regents exams are going to be held from June 16th to June 23rd. Some teachers will undoubtedly be giving final exams in June also. Surely this issue has come up for NYC kids in the past. How have people dealt with this?", "Solution_1": "The year I went, the timing was such that my last regents exam was on the same day I left and I got into Nebraska around midnight, so I didn't have that problem. You should start talking to the school (Danny Jaye would be a good first bet) and to the MOP organizers right now about the opportunity to take the test, proctored, at MOP.", "Solution_2": "I wish I had the same issue.\r\n\r\nbleh, I'm graduating.", "Solution_3": "How will your school let you miss so much school w/o failing your grade? \r\n\r\nNext year if I make MOP (8th grade this year), I'd have the same problem.", "Solution_4": "Regents exams take place after classes end, so you only have to come in if you have exams to take.", "Solution_5": "Hooyoung ... my year, Danny Jaye had faxed them my history regents to Nebraska, where I took it ... 1 hr later than ny time \r\n\r\nKD", "Solution_6": "Thanks for the responses. I will talk to Mr. Jaye tomorrow and try to work something out.", "Solution_7": "i second geehoon :(\r\nso is hooyoung the only NYC perosn going? if so, congrats to hooyoung:)\r\notherwise, congrats to all who made MOP", "Solution_8": "Danny Zhu is going also, but i'm not surprised of course.", "Solution_9": "yea but didnt zhu not make it last year?", "Solution_10": "wow this thread is like a year old...\r\n\r\nstop bumping old threads!" } { "Tag": [ "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Humanitarian organization has to deliver 18 tons of humanitarian aid which is distributed in n containers, where n\u226535. They have 7 trucks each that can carry infinite number of containers whose total mass does not exceed 3 tons. It is known that these trucks can carry any 35 containers, from given n containers, in one tour. Prove that these trucks can carry all n containers in one tour.", "Solution_1": "It realy looks so easy but the complexity of its definition makes it nonattractive and tough.", "Solution_2": "Anybody clever enough to solve? :maybe:", "Solution_3": "It seems to me that this is a simple induction on $ n$. It obviously holds for $ n \\equal{} 35$. \r\nSuppose it holds for some $ n\\geq35$ and let's prove that it holds for $ n \\plus{} 1$. Choose\r\nthe least container with mass $ t\\leq18/(n \\plus{} 1)$. All the other containers can be put on \r\n7 trucks by hypothesis. At least one truck has free space for $ \\geq3 \\minus{} \\frac {18 \\minus{} t}{7} \\equal{} (3 \\plus{} t)/7$\r\ntones and this is not less than $ t$, because $ t\\leq(3 \\plus{} t)/7$ is equivalent with $ t\\leq1/2$.\r\nSo we can put the last container in that truck." } { "Tag": [], "Problem": "\u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bd \u03cc\u03bb\u03bf\u03b9 \u03bf\u03b9 \u03cc\u03c1\u03bf\u03b9 \u03c4\u03b7\u03c2 $ (a_{n})$ \u03bc\u03b5 $ a_{n}\\equal{} 1!\\plus{}2!\\plus{}3!\\plus{}...\\plus{}n!$ \u03ba\u03b1\u03b9 $ n\\geq1$ \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b3\u03c9\u03bd\u03b1 \u03b1\u03ba\u03b5\u03c1\u03b1\u03af\u03c9\u03bd.", "Solution_1": "\u0394\u03b5\u03c2 \u03b5\u03b4\u03ce: :wink: \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=199152", "Solution_2": "[quote=\"Protonios\"]\u0394\u03b5\u03c2 \u03b5\u03b4\u03ce: :wink: \n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=199152[/quote]\r\n\r\n\u0394\u03b5\u03bd \u03c4\u03bf \u03c0\u03c1\u03cc\u03c3\u03b5\u03be\u03b1 :blush: \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c5\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 protonios :)" } { "Tag": [], "Problem": "Find the coefficient of the $a^{2}bc$ term in the expansion of $(a+2b+3c)^{4}$.", "Solution_1": "[quote=\"xxxxx\"]Find the coefficient of the $a^{2}bc$ term in the expansion of $(a+2b+3c)^{4}$.[/quote]\r\n\r\n$(a+2b+3c)(a+2b+3c)(a+2b+3c)(a+2b+3c)$\r\n\r\n[hide]There is ${4\\choose 2}=6$ ways to choose the $a^{2}$. There are 2 ways left to choose the $b$ term but the coefficient in fron of all the $b$'s is 2 so $2\\times 2=4$. There is only one way to choose the $c$ term but the coefficient in front of all the $c$'s is 3 so $3\\times 1=3$. Therefore, $6\\times 4\\times 3=72$ i think.[/hide]", "Solution_2": "[hide=\"answer\"]72[/hide]\r\n\r\nI'm quite sure the answer is right. Can anyone write the binomial expasion equation for this please? (using combinatorics I think?)" } { "Tag": [ "induction", "modular arithmetic" ], "Problem": "To proof for any natural number n that:\r\n$ a)323/a_{n}\\equal{}20^{2n}\\plus{}16^{2n}\\minus{}3^{2n}\\minus{}1$\r\n\r\n$ b)24/a_{n} \\equal{} n^6 \\minus{} 3n^5 \\plus{} 6n^4 \\minus{} 7n^{3} \\plus{} 5n^{2} \\minus{} 2n$", "Solution_1": "Hi barcelona. Could you retype the first bit? Thanks.", "Solution_2": "It's clear now.", "Solution_3": "Hi barcelona.\r\n\r\nI've got to go but I guess both will succumb to atttack by mathematical induction, the second one since its factors seem to be \r\n\r\n$ (n\\minus{}1)n(n^{2}\\minus{}n\\plus{}1)(n^{2}\\minus{}n \\plus{} 2)$.\r\n\r\n(By the way, I think \\mid gives the divides sign: $ \\mid$).", "Solution_4": "a) Of course it will succumb to induction, but I much rather prefer the following.\r\n\r\n$ 323\\equal{}17\\cdot 19$. Now, $ 20^{2n} \\plus{} 16^{2n} \\minus{} 3^{2n} \\minus{} 1 \\equiv 3^{2n} \\plus{} (\\minus{}1)^{2n} \\minus{} 3^{2n} \\minus{} 1 \\equiv 0 \\pmod{17}$, and $ 20^{2n} \\plus{} 16^{2n} \\minus{} 3^{2n} \\minus{} 1 \\equiv 1^{2n} \\plus{} (\\minus{}3)^{2n} \\minus{} 3^{2n} \\minus{} 1 \\equiv 0 \\pmod{19}$.", "Solution_5": "For question (a) I also like to look at two different ways of \"factoring\" the problem. I am not quite factoring but breaking it down to two differences of squares that can be factored, to show that 17 divides the number. Then rearranging the terms and factoring two different differences of squares, to show that 19 divides it. Like so:\r\n\r\n$ (20^{n}\\minus{}3^{n})(20^{n}\\plus{}3^{n}) \\plus{} (16^{n}\\plus{}1^{n})(16^{n}\\minus{}1^{n})$\r\n\r\nWhich shows that 17 is a factor.\r\n\r\nor\r\n\r\n$ (20^{n}\\minus{}1^{n})(20^{n}\\plus{}1^{n}) \\plus{} (16^{n}\\plus{}3^{n})(16^{n}\\minus{}3^{n})$\r\n\r\nWhich shows that 19 is a factor. In both cases you need to be careful that you make it clear what happens when n is odd vs when n is even. Though that's not too tough.\r\n\r\nHence, their product 323 is a factor.\r\n\r\nHope that makes sense." } { "Tag": [ "calculus", "integration", "logarithms", "function", "limit", "geometry", "geometric transformation" ], "Problem": "Here is an interesting integral formula:\r\n\r\nLet $a$ and $b$ be real positive numbers such that $a