inequalities
geometry
3D geometry
quadratics
Asymptote
analytic geometry
graphing lines

So I did my mock practice today, here are the problems I need counseling with and the "hide"s are what I did.

2. Find the value of $ |(2\minus{}i)(3\plus{}i)|$.

[hide]I FOILed and got $ 7\minus{}i$, but it was wrong.[/hide]

5. When factored, $ x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1 \minus{} y^3$ has one factor that is linear. Give this linear factor.

[hide]Well, somehow, synthetic division and checking for roots doesn't work.[/hide]

7. Solve the equation $ x^3 \minus{} 2x \minus{} 4\equal{}0$.

[hide]I got to $ x(x^2 \minus{} 2)\equal{} 4$ and got one of the answer: $ 2$ but didn't get the other one.[/hide]

8. Solve the inequality: $ \frac{x\plus{}1}{3\minus{}x}>4$.
[hide]
I got so much trouble with inequalities! I cross multiplied and get $ x>11/5$ and $ x \neq 0$, but I don't know how to tie it together so, it was wrong :(.[/hide]

10. Find the distance between two of the lines $ 3x\minus{}5y\plus{}8\equal{}0$ and $ 3x\minus{}5y\plus{}11\equal{}0$,

[hide]I put down $ 3$, but then realized that "distance" is defined as distance-line have to perpendicular both lines.[/hide]

16. Factor $ x^2 \minus{} y^2 \plus{} 8y \minus{} 16\equal{}0$

[hide]Kind of similar concept as $ 5.$ I did $ (x\plus{}y)(x\minus{}y) \plus{} 8y \minus{} 16$ and I stuck there.[/hide]

20. Solve the inequality: $ |x\minus{}3|<3$.

[hide]I solved and got $ x>0$ and $ x<6$, and seconds after the exam, I realized that it is $ (0,6)$ and not $ (\minus{}\infty,6)\cup(0,\infty)$. Could you summarize a little bit of solving inequalities with absolute value for me?[/hide]

All of them are pretty simple, but I only have time to do one of them right now:

[hide="Inequality + Absolute Value"]To solve inequalities with absolute values, you have to consider both the negative and positive cases. Solve each of these for $ x$, and then decide how to combine the two outcomes:

1. If $ x$ is between two numbers, then it can be written as $ a < x < b$ (possibly with different signs). This happens when the absolute value is less than some number.

2. If, however, $ x$ the possible values of $ x$ can not be written in one expression, it should be written $ x < a\quad\text{or}\quad x > b$. This happens when the absolute value is greater than some number.[/hide]

[hide="Solution to #20"]$ |x \minus{} 3| < 3$

$ x \minus{} 3 < 3,\quad x \minus{} 3 > \minus{} 3$

$ x < 6,\quad x > 0$

This can be written in one expression: $ \boxed{0 < x < 6}$.[/hide]

[hide="2"]You are correct in getting that it multiplies out to $ 7 \minus{} i$, but the question is asking for the absolute value (or modulus) of that.  For a complex number $ a \plus{} bi$, $ |a \plus{} bi| \equal{} \sqrt {a^2 \plus{} b^2}$.
So, $ |7 \minus{} i| \equal{} \sqrt {7^2 \plus{} ( \minus{} 1)^2} \equal{} \boxed{5\sqrt2}$.[/hide]

[hide="5"]$ x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1$ is the same as $ (x \plus{} 1)^3$, so the expression can be written as $ (x \plus{} 1)^3 \minus{} y^3$.
Using the formula for the difference of cubes, you get $ ((x \plus{} 1) \minus{} y)((x \plus{} 1)^2 \plus{} (x \plus{} 1)y \plus{} y^2)$.
Therefore, the linear term is $ \boxed{x \plus{} 1 \minus{} y}$.[/hide]

[hide="Hint for 16"]Try writing the equation as $ x^2\minus{}(y^2\minus{}8y\plus{}16)\equal{}0$.  See if you can get it from there.[/hide]

[hide="16"] $ x^2 \minus{} y^2 \plus{} 8y \minus{} 16 \equal{} 0$
$ x^2 \minus{} (y^2 \minus{} 8y \plus{} 16) \equal{} 0$
$ x^2 \minus{} [(y \minus{} 4)(y \minus{} 4)]$
Since this is the difference of squares:
$ (x \minus{} [y \minus{} 4])(x \plus{} [y \minus{} 4]) \equal{} (x \minus{} y \plus{} 4)(x \plus{} y \minus{} 4)$
YAY!!![/hide]

@i_like_pie
your summary is VERY helpful. I'm browsing through my algebra II textbook from the library, they left out the parts where it applies only to when absolute value is less than a number, and greater than a number.

These are pretty basic, but my school had IMP math program where they teach things in roundabout and wordy ways. It is good but not useful because many basic raw computation stuffs are left out.

[quote="jeez123"]
7. Solve the equation $ x^3 \minus{} 2x \minus{} 4 \equal{} 0$.

[hide]I got to $ x(x^2 \minus{} 2) \equal{} 4$ and got one of the answer: $ 2$ but didn't get the other one.[/hide]
[/hide][/quote]

There's actually 2 more
first divide the whole thing by x+2

then solve for x using the quadratic formula

[hide="7"]I think you meant dividing the whole thing by $ x \minus{} 2$

Well, I did that with synthetic division (where it actually worked :D), and got $ x^2 \plus{} 2x \plus{} 2$, then quadratic formula yields $ \boxed{ \minus{} 1 \pm i}$ as the other two!!![/hide]

remembering i_like_pie's method

[hide="8."]
$ \frac {x \plus{} 1}{3 \minus{} x} > 4$

$ x > 11/5$ or $ 2.2$
asymptote at $ x \equal{} 3$

So
number line 2.2 number line 3 number line

Checking $ x \equal{} 1$, the fraction will equal to $ 2/2$- not greater than $ 4$.
Checking $ x \equal{} 2.5$, the fraction will equal to $ 3.5/.5$- NOD
Checking $ x \equal{} 4$, the fraction will equal to $ 5/ \minus{} 1$ nope.

So $ \boxed{(\frac {11}{5}, 3)}$ YAY[/hide]

The last one:

[hide="10."] I got it in slope-intercept form (I don't know if that helps)

$ y\equal{}\frac{3}{5}x \plus{} \frac{11}{5}$ and
$ y\equal{} \frac{3}{5} x \plus{} \frac{8}{5}$.

And then...[/hide]


[quote="vidyamanohar"]$C_1, C_2, C_3$ are three circles.  Centre of $C_2$ lines on $C_1$.     $C_1$ and $C_2$ touch $C_3 $ internally.     $C_1$ and $C_2$ cut at A and B.  AB is extended to cut $C_3$ at X and Y.  $C_1$ and      $C_3$ touch at M.  M is joined to X and Y.  MX and MY cut $C_1$ at P and Q.  Prove that PQ is a tangent to $C_2$.

It is preferred to get the solution using coordinate geometry only.[/quote]

HHHAaa
It took this long by the time I have solved the problem and typed the solution.
Pls find the attachment for the solution.
Do post your opinion about the solution.
thanx


Find the number of zero at the end of the number
                      4^(5^6) + 6^(5^4)

[hide="dunno if this is right"]
rewrite the terms as $(5-1)^{5^{6}}+(5+1)^{5^{4}}$ and use binomial theorem to expand both.  The $-1$ and $+1$ at the end of each expansion will cancel.  The second to the last term in the first expansion is $\binom{5^{6}}{1}*5$, and the second to the last term in the second expansion is $\binom{5^{4}}{1}*5$.  If we factor $5^{5}$ out of the whole sum, we get $5^{5}(\dots+1)$.  The other factor will leave a modulo of 1 mod 5, (the dots term all have at least a factor of 5 in them). Since we have sufficient factors of 2 ($5^{4}$ of them), so we have $5$ zeros at the end of this number.
[/hide]

[quote="joml88"][quote="eryaman"]Hmm...I've simplified it to $4^{650}+4^{625}$, not sure if that helps...[/quote]

Eh, how did you do that?[/quote]Oops, nevermind.  I made an incredibly stupid arithmetic mistake that destroyed my method.  :blush:


but how did you get to the solution what method did you use?

One way is to [i]guess[/i] that you factorize it in $ \mathbb{Z}[x]$ as:

$ x^{7}+2x^{4}+1 = (x^{3}+Ax^{2}+Bx+D)(x^{4}+Ex^{3}+Fx^{2}+Gx+H)$

And compare coefficients. So for example $ GH=1$ since there are integers it means $ G=H=1$ or $ G=H=-1$ and use that to see what more you can deduce. So it is a very painful and boring  :sleeping: excercise.


Let $A$ and $B$ be sets of complex numbers such that for any $a \in A$ and $b \in B$ we have either $|a+b| = 1$ or $|a-b| = 1$. What is the largest possible value of $\min\{|A|, |B|\}$?

[quote="bthings_2000"]Is the magnitude of a set the number of elements in the set (cardinality) or the sum of the magnitudes of each of the elements in the set?[/quote]

The former.

I had trouble coming up with a rigorous proof of the result of this question, but the answer that paladin8 and I came up with was very convicing...

I am also struggling to find something rigorous.  I think[hide]the answer is probably 4.  I haven't tried letting 0 be one of the elements of one of the sets, but it seems unlikely that that will actually help.  The best I've done is to have 4 points in $A$ and 8 points in $B$ -- the values in $A$ are (for example) $\{\pm \frac{1}{4}, \pm \frac{3}{2}\}$, and the values in $B$ are the 8 intersection points formed by pairs of circles, one centered at $\pm \frac{1}{4}$ and the other centered at $\pm \frac{3}{2}$.[/hide]


Let $ABC$ be a triangle, and let $M$ be the midpoint of its side $AB$. Let a line through the point $M$ meet the lines $CA$ and $BC$ at the points $K$ and $L$ such that $CK = CL$. Denote by $D$ the foot of the perpendicular from the point $C$ to the line $AB$, and denote by $O$ the circumcenter of triangle $CKL$.

Prove that $OD = OM$.

[b]Problem.[/b] Let ABC be a triangle, and let M be the midpoint of its side AB. Let a line through the point M meet the lines CA and BC at the points K and L such that CK = CL. Denote by D the foot of the perpendicular from the point C to the line AB, and denote by O the circumcenter of triangle CKL.

Prove that OD = OM.

[i]Solution.[/i] Since CK = CL, the triangle KCL is isosceles. Hence, the angle bisector of its angle KCL is the symmetry axis of this isosceles triangle, and thus it is perpendicular to its base KL. In other words, the line KL is perpendicular to the angle bisector of the angle KCL. But the angle bisector of the angle KCL is simultaneously the angle bisector of the angle BCA. Thus, the line KL is perpendicular to the angle bisector of the angle BCA. Since this line KL passes through the point M, we thus can summarize: The line KL is the perpendicular to the angle bisector of the angle BCA through the point M.

Let the angle bisector of the angle BCA meet the circumcircle of triangle ABC at a point V (apart from the point C). Then, < VCA = < VCB. On the other hand, since the point V lies on the circumcircle of triangle ABC, we have < VCA = < VBA and < VCB = < VAB. Thus, < VCA = < VCB becomes < VBA = < VAB. This means that the triangle AVB is isosceles, so that AV = BV, and thus, the point V lies on the perpendicular bisector of the segment AB. Consequently, the orthogonal projection of the point V on the line AB is the midpoint M of the segment AB.

Let K’ and L’ be the orthogonal projections of the point V on the lines CA and BC. By the Simson line theorem, since the point V lies on the circumcircle of triangle ABC, the orthogonal projections L’, K’ and M of this point V on the sidelines BC, CA and AB of triangle ABC lie on one line (namely, the Simson line of the point V with respect to the triangle ABC). In other words, the line K’L’ passes through the point M.

Now, remember that the point V lies on the angle bisector of the angle BCA, and that the points K’ and L’ are the orthogonal projections of this point V on the lines CA and BC. Thus, it follows from symmetry that the points K’ and L’ are symmetric to each other with respect to the angle bisector of the angle BCA. Hence, the line K’L’ is perpendicular to the angle bisector of the angle BCA. Since we know that this line K’L’ passes through the point M, it follows that the line K’L’ is the perpendicular to the angle bisector of the angle BCA through the point M. But we have previously shown that the line KL is the perpendicular to the angle bisector of the angle BCA through the point M. Thus, the lines K’L’ and KL are one and the same line. Consequently, the points $K=KL\cap CA$ and $K^{\prime}=K^{\prime}L^{\prime}\cap CA$ coincide; similarly, the points L and L’ coincide.

Since we have defined the points K’ and L’ as the orthogonal projections of the point V on the lines CA and BC, we thus obtain that the points K and L are the orthogonal projections of the point V on the lines CA and BC. Thus, < CKV = 90° and < CLV = 90°. This implies that the points K and L lie on the circle with diameter CV. In other words, the circumcircle of triangle CKL is the circle with diameter CV. Consequently, the circumcenter O of triangle CKL is just the center of the circle with diameter CV, thus the midpoint of the segment CV.

Let T be the midpoint of the segment DM.

Since the point V lies on the perpendicular bisector of the segment AB, we have $VM\perp AB$ (since the point M is the midpoint of the segment AB). Also, clearly $CD\perp AB$, since the point D is the foot of the perpendicular from the point C to the line AB. Hence, the lines VM and CD are both perpendicular to the line AB, so that the quadrilateral VMDC is a trapezoid with its bases VM and CD both perpendicular to the line AB. Thus, the line joining the midpoints O and T of its legs CV and DM must also be perpendicular to the line AB. In other words, we have $OT\perp AB$, so that the point O lies on the perpendicular to the line AB at the point T. But since the point T is the midpoint of the segment DM, the perpendicular to the line AB at the point T is the perpendicular bisector of the segment DM. Thus, the point O lies on the perpendicular bisector of the segment DM, and we get OD = OM, completing our solution.

Sorry for the length of the above solution – it’s the first that came into my mind and I didn’t have time to look for a simpler one or to optimise the above writeup. Nice problem, by the way.

  darij

Just see the Isosceles trapezoid $ EFDM$ .

Since $MA \equal{} MB$ and $CK \equal{} CL,$ by Menelaus' theorem for $\triangle ABC$ cut by $\overline{ KML},$ we get $ AK \equal{} BL.$  Thus, $E \equiv \odot(CKL) \cap \odot(ABC) \not\equiv C$ is the center of the rotation carrying $AK$ into $BL,$ and due to $EA \equal{} EB,$ then $E$ is the midpoint of the arc $AB$ of $\odot(ABC).$

Clearly $E$ lies on the internal bisector of $\angle ACB$ and since $CE$ is the circumdiameter of $\triangle CKL$ perpendicular to $ KL,$ it follows that $O$ is the midpoint of $CE.$ Moreover, if $A',B'$ denote the midpoints of $CB,CA,$ then $O$ is the intersection of $CE$ with the perpendicular bisector of  segment $A'B',$ because the positive homothety with center $C$ and factor $\frac {_1}{^2}$ transforms $A,B$ into $A',B',$  $E$ into $O$ and  the perpendicular bisector  $EM$ of $ AB$ into the perpendicular bisector of $A'B'.$ Now, it remains to note that $A'B'DM$ is an isosceles trapezoid, in which $O$ lies on its symmetry axis. Indeed,  $\triangle MA'B'$ and  $ \triangle DB'A'$ are symmetric about the perpendicular bisector of $A'B',$ hence $OD \equal{} OM.$

Let $ P$ be the intersection the line passing through $ K$ perpendicular to $ CK$ and the line passing through $ L$ perpendicular to $ CL$. Since $ \angle CKP \plus{} \angle CLP \equal{} 90^o \plus{} 90^o \equal{} 180^o$, we have that $ CKPL$ is cyclic. Furthermore, the line $ KL$ is Simson Line for $ P$. The line $ KL$ also contains $ M$, which lies on $ AB$. This implies $ PM$ is perpendicular to $ AB$ (by the property of the Simson Line). Since $ CD$ is also perpendicular to $ AB$, we have that $ CD \parallel{} MP$.

But the quadrilateral $ CKPL$ is cyclic, whose corresponding circle has center $ O$ (since $ O$ is the circumcenter of $ \Delta CKL$). Since $ \angle CKP \equal{} 90^o$, $ CP$ is a diameter of this circle, which implies $ C, O, P$ are collinear and $ |OC| \equal{} |OP|$. Since $ CD \parallel{} MP$, we conclude that $ O$ is equidistant to the lines $ CD$ and $ MP$. This combined with the fact that $ MD$ is perpendicular to $ CD$ and $ MP$ implies $ |OD| \equal{} |OM|$, as desired.

Take A' on (CB with CA' = CA, call V the middle of the arc AB not containing C on (ABC); obviously, CV is the angle bisector of <ACB and, due to symmetry, A'V = AV ( 1 ), but AV = BV, so A'V = BV ( 2 ). It's obvious also that AA' is perpendicular to CV, hence AA'||ML and, M being the middle of AB, L is the middle of A'B; the triangle A'VB being isosceles - see relation (2) - it follows that VL is its altitude. Analogously prove VK and AC perpendicular, hence CKVL cyclic, and the circle of diameter CV is its circumcircle. DM being the orthogonal projection of the segment CV on AB, its extremities, D and M are equally apart of the middle of CV.

Best regards,
sunken rock

[quote="Herodatviet"][color=darkred]Let $ ABC$ be a triangle, and let $ M$ be the midpoint of $ [AB]$ . A line through $ M$ meet $ CA$ , $ CB$ at $ K$ , $ L$ respectively such that

$ CK \equal{} CL$ . Denote $ D\in AB$ for which $ CD\perp AB$ and the circumcircle $ C(R,\rho )$ of $ \triangle KCL$ . Prove that $ RD \equal{} RM$ .[/color] [/quote]

[color=darkblue][b][u]Proof I (synthetic)[/u].[/b] Suppose w.l.o.g. that $ b < a$ . In this case $ A\in (CK)$ and $ L\in (CB)$ . Let $ w \equal{} C(O,R)$ be the circumcircle of $ \triangle ABC$

with the diameter $ [SN]$ ,where $ M\in SN$  and the line $ AB$ separates the points $ S$ , $ C$ . Construct $ \left\|\begin{array}{c} K'\in CA\ ;\ SK'\perp CA \\
\ L'\in CB\ ;\ SL'\perp AB\end{array}\right\|$ . Observe that

$ \left\|\begin{array}{c} CK' \equal{} CL' \\
\ SK' \equal{} SL' \\
\ SA \equal{} SB\end{array}\right\|\implies$ the right triangles $ K'AS$ and $ L'BS$ are equivalently. Thus, $ K'A \equal{} L'B$ and $ \frac {K'A}{K'C}\cdot \frac {L'C}{L'B}\cdot\frac {MB}{MA} \equal{} 1$ . Therefore,

$ M\in K'L'$ and $ CK' \equal{} CL'\ \implies\ K'\equiv K$ , $ L'\equiv L$ and the quadrilateral $ CKSL$ is a kite inscribed in a circle with the diameter $ [CS]$ .

In conclusion, $ RC \equal{} RS \equal{} \rho$ and $ RD \equal{} RM$ because $ CD\parallel MS\perp AB$ .[/color]

[color=darkblue][b][u]Proof II (metric)[/u].[/b] Suppose w.l.o.g. that $ b < a$ . In this case $ A\in (CK)$ and $ L\in (CB)$ . Denote the midpoint $ X$ of $ [KL]$ , i.e. $ CX\perp KL$ and the second intersection $ S$ of the line  $ CX$ with the circumcircle $ C(O,R)$ of $ \triangle ABC$ , i.e. $ SM\perp AB$ . Apply the [u][b]Menelaus[/b]' theorem[/u] to the transversal $ \overline {KML}$ and $ \triangle ABC$ : $ \frac {KA}{KC}\cdot\frac {LC}{LB}\cdot\frac {MC}{MB} \equal{} 1\implies$ $ KA \equal{} LB$ . Since $ CK \equal{} CL$ obtain $ KA \plus{} b \equal{} a \minus{} LB$ $ \implies$  $ KA \equal{} LB \equal{} \frac {a \minus{} b}{2}$ and $ CK \equal{} CL \equal{} \frac {a \plus{} b}{2}$ . Thus, $ CK \equal{} 2\rho\cos\frac C2$ $ \implies$ $ \rho \equal{} \frac {a \plus{} b}{4\cos\frac C2} \equal{}$ $ R\cos\frac {A \minus{} B}{2} \equal{}$ $ \frac {CS}{2}$ . Therefore, $ CR \equal{} RS \equal{} \rho$ , i.e. the quadrilateral $ CKSL$ is a kite with $ KC\perp KS$ and its circumcenter $ R$ is the midpoint of the diagonal $ [CS]$ in the trapezoid $ CDSM$ , where $ CD\parallel MS\perp AB$ . In conclusion, $ FD \equal{} FM$ . [/color]

WLOG we can assume that $AC>BC$. Let $P$ be the point diametrically opposite to $C$ on the circumcenter of $\triangle CKL$. Then $\angle PKC=\angle PLC=90^{\circ}$. Also let $E,N,F$ be the feet of the perpendiculars from $A,P,B$ respectively to the line $KL$. $\triangle AEM=\triangle BFM$ as $AM=MB$, so $AE=BF$. $\angle EKA=\angle LKC=\angle KLC=\angle FLB$, so $\triangle AEK=\triangle BFL$ and $EK=FL$.
Let a circle built on $AP$ as on a diameter intersects $KL$ at $M_1$ besides $K$. If $Q$ is the projection of the circle's center on $EN$, then $KQ=QM_1$. Moreover, $EQ=QN$ because $AENP$ is a trapezoid. That's why $EK=M_1N$. Similarly, if a circle built on $PB$ as on a diameter intersects $KL$ at $M_2$ besides $L$, then $M_2N=FL$. We have: $M_1N=EK=FL=M_2N$, so $M_1=M_2$. But $\angle AM_1P=\angle PM_2B=90^{\circ}$. Therefore, $AM_1B$ is a straight line. As $M_1$ lies both on $EL$ and on $AB$, it is therefore the point of intersection of these lines and $M_1=M$.
Conclusion: $\angle AMP=90^{\circ}$, $O$ is the midpoint of $PC$, so its projection on $MD$ is the midpoint of latter, that's why $OD=OM$.


inequalities
integration
inequalities unsolved

Please deduce $1/(k+1)$ $\leq$ $(1^k+2^k+3^k+...+p^k)/p^{k+1}$ $\leq$ $[(1+1/p)^{k+1} - (1/p)^{k+1}]/(k+1)$
by $nm^{n-1}\leq (m+1)^n-m^n\leq n(m+1)^{n-1}$,
where k,m,n,p are positive integers

P.S.:sorry that I wrote it wrongly...

Are you sure that $nm^{n-1}\geq (m+1)^n-m^n\geq n(m+1)^{n-1}$?

I think that $nm^{n-1}\leq (m+1)^n-m^n\leq n(m+1)^{n-1}$.

I give you a hint.

$\frac{1}{k+1} \left\{\left(1+\frac{1}{p}\right)^{k+1}-\left(\frac{1}{p}\right)^{k+1}\right\}=\int_{\frac{1}{p}}^{1+\frac{1}{p}} x^k dx$ and,

$\frac{1}{p^{k+1}}(1^k+2^k+\cdots +p^k)=\frac{1}{p} \left\{\left(\frac{1}{p}\right)^k+\left(\frac{2}{p}\right)^k+\cdots +\left(\frac{p}{p}\right)^k\right\}$,thus

You are to imagine ...... :)


function
calculus
algebra
polynomial
calculus computations

Find the equation of the line tangent to the function at the given point.

$ f(x)\equal{}10x^2$ at $ x\equal{}6$

Solution without using calculus:

If $ y \equal{} l(x)$ is the tangent line, then $ 10x^2\equal{}(x \minus{} 6)^2*10\plus{}l(x)$.

[quote="ben_b"]Find the equation of the line tangent to the function at the given point.

$ f(x) = 10x^2$ at $ x = 6$[/quote]

$ \text{General equation of line: }(y-y_1)=m(x-x_1)
\\
\\
m= f'(x)|_{x=6}$

[quote="kunny"]Solution without using calculus:

If $ y \equal{} l(x)$ is the tangent line, then $ 10x^2 \equal{} (x \minus{} 6)^2*10 \plus{} l(x)$.[/quote]
Can this be proved without calculus? Is there any generalization to this?

Let $ P(x)$ be the polynomial with degree $ n\geq 2$. The tangent line $ y\equal{}l(x)$ of the curve $ y\equal{}P(x)$ at the point $ (a,\ P(a))$ can be found by

$ P(x)\minus{}l(x)\equal{}(x\minus{}a)^2Q(x)\Longleftrightarrow P(x)\equal{}(x\minus{}a)^2Q(x)\plus{}l(x)$.

$ Q(x)$ is the quotient, in this case $ Q(x)$ is the polynomial with degree $ n\minus{}1\ (n\geq 2)$.


modular arithmetic
inequalities
number theory
Diophantine equation
number theory proposed

Solve in natural numbers the following equation: \[n^{3}-5n+10=2^{k}.\]

the only solution is $n=2$
for $n\geq 3$ we have :
$n^{3}-5n+10=2^{k}$ iff $n(n-2)(n+2)=8(2^{k-3}-1)$ but if $n(n-2)(n+2)$ is even then it is divisible by at least $16$
 so $k=3$ and $n=2$ is the only solution.

The "Albanian Eagle" has the right answer to this problem, but his/her solution is incorrect! The Diophantine equation

$(1)\;\; n^{3}-5n+10 \;=\; 2^{k}$

is [u]not[/u] equivalent to $n(n-2)(n+2) = 8(2^{k-3}-1).$ In fact (1) is equivalent to $(n-2)(n^{2}+2n-1) = 8(2^{k-3}-1).$

One way to solve equation (1), is to compute $2^{k}$ and $n^{3}-5n+10 = f(n)$ modulo 7. Writing $k = 3m+r$ where $m$ and $0 \leq r \leq 2$ are integers, we obtain

$(2) \;\; 2^{k}\;=\; 2^{3m+r}\;=\; 8^{m}\cdot 2^{r}\; \equiv \; 2^{r}\pmod{7}.$

Futhermore

$\begin{tabular}{|c|c|}\hline n & f(n) mod 7 \\ \hline 0 & 3 \\ \hline 1 & 6 \\ \hline 2 & 1 \\ \hline 3 & 1 \\ \hline 4 & 5 \\ \hline 5 & 5 \\ \hline 6 & 0 \\ \hline \end{tabular}$

This table combined with (1) and (2) gives $r=0$. Hence $2^{k}= 2^{3m}$, so

$(3) \;\; 5(2-n) \;=\; (2^{m})^{3}-n^{3}$

according to (1). We observe that $n=1$ and $n=2$ in (1) gives $2^{k}= 6$ and $2^{k}= 8$ respectively. Thus there is one solution of (1) when $n \leq 2,$ namely $(n,k) = (2,3).$ For $n \geq 3$ we get $2^{m}< n$ by (3), hence

$5(2-n) \;=\; (2^{m})^{3}-n^{3}\; \leq \; (n-1)^{3}-n^{3}$

i.e.

$3n^{2}-8n+11 \; \leq \; 0,$

which is impossible. Therefore the only solution of the Diophantine equation (1) is $(n,k) = (2,3).$

thanks for pointing out my mistake! :blush:
[I shouldn't rush when posting solutions...]

anyway, you also have a small mistake in the end of your proof !
you should have $5(2-n)=(2^{m})^{3}-n^{3}$ so the inequality is incorrect. :)
but we have $5(n-2)=n^{3}-(2^{m})^{3}\geq n^{3}-(n-1)^{3}$ which implies
$8n\geq 3n^{2}+11$ which is imposible (it's almost the same thing :P )

Thank you for pointing out my sign error, Albanian Eagle! I have now corrected this error in accordance with your suggestion.


algebra
polynomial
parameterization
superior algebra
superior algebra unsolved

My teacher had one problem and all of us can solve it  :huh:
Can anyone help us?
 p is a prime.
 Evaluate the product of all Z/p polynomial of order n with the highest parameter is 1  .

I've tried to solve it many times, but make no progress anymore. So I just post my results, perhaps someone can complete them.

Denote $A(n)$ the product of all monic polynomials of degree $n$ over $\mathbb{F}_{p}$, and \[B(n) = \prod_{i=0}^{n-1}(X^{p^{n}}-X^{p^{i}})\] I'm very sure that $A(n) = B(n)$, but I can't prove it for every $n$. We have the recursion \[B(1) = X^{p}-X ,~ B(n) = (X^{p^{n}}-X) B(n-1)^{p}\] So it is enough to show $A(n) = (X^{p^{n}}-X)*A(n-1)^{p}$.

Conerning the structure of the polynomial $B(n)$: It is monic of degree $np^{n}$, consists of $2^{n}$ monomials with coefficients $\pm 1$, which appear in sequences with lengths $\binom{n}{k}$. Explicitely: \[B(n) = \sum_{\stackrel{0 \leq k \leq n}{0 \leq t_{1}< ... < t_{k}\leq n-1}}(-1)^{k}X^{(n-k)p^{n}+p^{t_{1}}+...+p^{t_{k}}}\] For example, \[B(3) = X^{3p^{3}}-X^{2p^{3}+p^{2}}-X^{2p^{3}+p}-X^{2p^{3}+1}+X^{p^{3}+p^{2}+p}+X^{p^{3}+p^{2}+1}+X^{p^{3}+p+1}-X^{p^{2}+p+1}\]


A game is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players $ A$, $ B$, and $ C$ start with $ 15$, $ 14$, and $ 13$ tokens, respectively. How many rounds will there be in the game?

$ \textbf{(A)}\ 36 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 38\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 40$

[hide]my initial guess was 42, but then 13 will end the game sooner than 14 and 15, so it will take 13+13+13 or 39 rounds (for each player will give one token to the other), so D[/hide]

[hide]
If you write out the first few (3~4) turns, you see that every 3 turns, starting with $n,n-1,n-2$, the # of coins becomes $n-1,n-2,n-3$.  So I would say $3\cdot 13=39$ ($D$).
[/hide]

[hide]Every 3 turns, each player has 1 less token. Player C has the fewest number of tokens, 13, so it will take $3\cdot13=39$ turns.[/hide]

[hide]

heh doesnt matter if 13 has the least cuz the player with 15 will lose 3 first, for example the first couple turns are

15, 14, 13
12, 15, 14
13, 12, 15
14, 13, 12

and etc, hence it is easy to see that the player with 15 will drop to 0 first, and the last move will be

3 2 1
0 3 2

Each turn a player discards one token, and since we start off with 42 tokens and end up with 5, there will be 37 moves, hence the answer is $B$.

[/hide]

What's the answer?  I used brute force and got the same answer as gotztahbeazn.

i think the answer is B

apprentice, ccy, and catcurio all missed something vital that gotztahbeazn mentioned
they calculated how many turns it would take for the player with 13 tokens to run out, when in actuality the player with 15 tokens will have to give up 3 on his last turn, and thus ending the game

this is similar to the snail up a wall question (or frog in a well, slight variations...)
the snail wants to crawl up a 100 cm wall
during the day, he climbs 3 cm
at night, he rests and slides back down 2 cm

obviously, the apparent answer would be it takes him 100 days, but after careful consideration, one can see that it takes 98 days, because at the end of 98 days, the snail has reached to top and won't slide down

this same idea is used in this problem

therefore, the answer would be B. 37

[quote="vallon22"]this is similar to the snail up a wall question (or frog in a well, slight variations...)
the snail wants to crawl up a 100 cm wall
during the day, he climbs 3 cm
at night, he rests and slides back down 2 cm
[/quote]That was exactly what I was thinking of. Yeah, the answer is B.

well lets take the pattern of the 13,14,15
if you right the first four of each down you will see a pattern

13,14,15,12

and it will go on as 13,14,11,12,13

so for 13 there are three up then it goes to 12 three then 11 then three then 10.....three then 1 then 0

so there are 39 before it hits 0. on the 40th round the  game ends. so for 13 there are 40 rounds

now lets do 14

14,15,12,13,14,11 so there are 12 sets of three before it hits 0 +2

that would be 12x3+2+1

there is a 1 there because the zero will count as the last round

so it is 36+2+1=39

so there are 39 rounds for 14

now lets do 15

the pattern is:15,12,13,14,11

so in this pattern there are 12 groups of three +1 because there is a 15 and you add one more because zer ounts as a round

so it is 36 +1+1=38

38 is the lowest number of rounds that you get so the answer is 38

you cannot forget the zero because all the patterns 1,2,3 and then no one counts the finishing round when someone gets zero.the question is asking how many rounds are there in the game not how many rounds before the game finishes.

so the correct answer I think is 38

i'm pretty sure the answer is 37

let's figure out how to get down from 15, 14, 13 to 3, 2, 1 respectively

i'm gonna list out the first few rounds

15, 14, 13
12, 15, 14
13, 12, 15
14, 13, 12

so we see that for every three rounds, everyone loses one token

to get from 15, 14, 13 to 3, 2, 1, there must be 12 of these 3-round groups
that means it will take 36 rounds to get down to 3, 2, 1

from there, one additional round will bring our values to 0, 3, 2, and thus ending the game

so the answer is 37

@calculatorwiz i think you included the very first setup (15, 14, 13) as one round
i believe the question implies that the first round is after the first move (12, 15, 14)

oh my bad I am sorry I was countng the starting as a round. sorry to make you do all that work. my bad


superior algebra
superior algebra unsolved

I am searching this problem : prove that $ L$, the maximal unramified extension of a local field $ K$ is not complete. Do you have ideas? Thanks for advance.

I think there is a Cauchy sequence like $ \sum_{i \equal{} 1}^n p^i\zeta_{\ell^i}$ which does not converge. Perhaps Krasner's lemma is useful.

BTW: is it true that every infinite algebraic extension of a local field is not complete?


I need help with the following problem:
Find all integers such that they are equal to the sum of the factorials of their digits.

This seems tough to me.  My first step would be to find the maximal number of digits such a number could have and perhaps think about the rightmost digit.

Is this question same as:

\[abc = a!+b!+c!\]

Supposing that a,b, and c each represent the digt?

Maybe I'm not expert at these stuffs but I do know some I think

1 ---> 1 = 1! = 1

2 ---> 2 = 2! = 2

Hmm.. I at least got 2 solutions. Better than nothing. :lol:

Only thing I can say to you is that the number is going to be quite hard to find.

Isn't this too hard for Intermediate if JBL thinks it's tough?

Maybe I misunderstood the question, but I believe the only two solutions are 1 and 2.

[hide="A non trivial solution"]
145
[/hide]

Oberservations:
[list] 1. Since $9! = 362880, so it cannot be more than 7 digits.
2. Ignoring 1s and 2s, the sum of the digits must be 0 (mod 3).[/list]

As already pointed out There are only 4 such numbers  1,2,145, and  40585.
For more info google on Factorion.  eg  [url=http://icl.pku.edu.cn/yujs/MathWorld/math/f/f022.htm]here[/url]


geometry
parallelogram
ratio
geometric transformation
homothety
projective geometry
geometry proposed

We have to pairs of parallel lines (not all four lines are parallel), and a circle in the plane. Construct the circle's center using only a ruler.

As firts insigth I will give only the "strategy" of solution. Later on I will try to provide a complete proof.

Having two parallel we can construct a new parallel from a point on the plane and also we can find the midpoint of a segment which is parallel wrt the two given parallel.

Using this fact we can construct two chords of the circle parallel wrt one of the couple of given parallels and their midpoints.

The line through the midpoints of those two chords is a diameter of the circle.

The requested center is the intersection of two diameter.

PS
what is G.M.?

G.M. is Gazeta Matematica, a Romanian magazine.

I have a similar proof, but different. You don't need to post all the boring details, if you don't want to :).

Here is my solution. It is slightly different from that of Sprmnt21 (actually I am not familiar with the usual ruler constructions):

Let our two pairs of parallel lines form a parallelogram ABCD. We can construct the diagonals AC and BD of this parallelogram; these diagonals intersect at a point M, which is the midpoint of both of them.

Now, we have:

[b]Lemma 1.[/b] If P is an arbitrary point in the plane, then we can construct the parallels to the lines AC and BD through the point P.

[i]Proof.[/i] Let D' a point lying on the line PD and different from the point D, and let the line BD' meet the line PM at a point M'. Now, using the complete quadrilateral theorem, we can construct the harmonic conjugate I of the point M' with respect to the segment BD'. We will succeed to prove Lemma 1 if we show that the line PI is parallel to the line BD; for then, with the line PI, we will have constructed the parallel to the line BD through the point P, and the construction of the parallel to the line AC through the point P will be similar.

So we must prove that the line PI is parallel to the line BD. In fact, if we call I' the (possibly infinite) point of intersection of the lines PI and BD, then the cross ratio theorem yields that the cross ratios (BD'; M'I) and (BD; MI') are equal. But the points M' and I divide the segment BD' harmonically; hence, it follows that the points M and I' divide the segment BD harmonically, and since the point M is the midpoint of the segment BD, this yields that the point I' is an infinte point, and the line PI is parallel to the line BD, as claimed. Thus, Lemma 1 is proven.

Now take two points P and Q on the circle. After Lemma 1, we can construct the parallels to the line AC through these points P and Q; let these parallels meet the circle again at the points P' and Q', respectively. If we call K and L the midpoints of the segments PP' and QQ', then the line KL passes through the center of the circle (since the midpoint of a chord lies on the perpendicular to the chord from the center of the circle, and thus, the midpoints of two parallel chords lie on the common perpendicular to the two chords from the center of the circle). Hence, if we will succeed to construct the points K and L, then we will obtain a line through the center of the circle; we can also obtain another line through the center of the circle if we construct not the parallels to the line AC through the points P and Q, but the parallels to the line BD through the points P and Q. Thus, we will have constructed two lines through the center of the circle, and consequently, we will be able to construct the center itself.

Hence, it remains to construct the points K and L. I will only construct the point K here; the point L is analogous.

Using Lemma 1, we can draw the parallel to the line BD through the point P. Let X be a point on this parallel. Again using Lemma 1, we can draw the parallel to the line AC through the point X, and the parallel to the line BD through the point P'. These two parallels intersect at a point Y. Let the lines PY and P'X intersect at a point Z. Since PP' || AC and XY || AC, we have PP' || XY; since PX || BD and P'Y || BD, we have PX || P'Y. Hence, the quadrilateral PXYP' is a parallelogram; it follows that the point of intersection Z of its diagonals PY and P'X is the midpoint of these two diagonals. Hence, we have showed that the point Z is the midpoint of the segment P'X. On the other hand, the point K (which we want to construct) is the midpoint of the segment PP'. For this reason, ZK || XP. Since XP || BD, we thus have ZK || BD. Hence, the point K lies on the parallel to the line BD through the point Z. In other words, we can construct the point K as the meet of the parallel to the line BD through the point Z with the line PP'. Finish.

  Darij

[quote="grobber"]
I have a similar proof, but different. You don't need to post all the boring details, if you don't want to :).
[/quote]

 In this case the effort is not so hard  :) .

Tool 1.  How to find the midpoint of a segment, given  a line parallel wrt the segment.

Proof

Let R"S" the given segment and p' the parallel to p"=R"S". Let L be a point on the plane of p' and p" and let r and s the two lines through L passing through R" and S" respectively. let's call R'=r^p', S'=s^p', N=S"^R' and M=LN^p".
By Thales it holds that LR'/R'R"=LS'/S'S". Then by Ceva it follows that M is the midpoint of R"S".

Tool 2. Given a segment and its midpoint, find a line parallel to the line containing the segment through a given point.

Proof

We can go backward to the proof of tool 1, given R", M, S" and, for instance, S'. Take L on s=S'S", if N=LM^R"S' then R'=S"N^LR". The line p'=R'S' is parallel to p"=R"S".

Now using the two tools we can solve the problem.

[quote="sprmnt21"]Tool 1.  How to find the midpoint of a segment, given  a line parallel wrt the segment.

Proof

Let R"S" the given segment and p' the parallel to p"=R"S". Let L be a point on the plane of p' and p" and let r and s the two lines through L passing through R" and S" respectively. let's call R'=r^p', S'=s^p', N=S"^R'[/quote]

I think this should be N = S"R' /\ S'R".

[quote="sprmnt21"]and M=LN^p".
By Thales it holds that LR'/R'R"=LS'/S'S". Then by Ceva it follows that M is the midpoint of R"S".[/quote]

In fact, you don't even need Ceva to show this. Call U and V the midpoints of the segments R'S' and R"S". Then, since the points L, R' and R" are collinear, since the points L, S' and S" are collinear, and since R"S" || R'S', there exists a homothety with center L mapping the points R' and S' to the points R" and S". This homothety then must also map the midpoint U of the segment R'S' to the midpoint V of the segment R"S". Since the center of the homothety is L, we can conclude that the points L, U and V lie on one line. Similarly, the points N, U and V lie on one line. Thus, all four points L, N, U and V lie on one line, what implies that the point U lies on the line LN. Thus, U = LN /\ p" = M, so that the point M is the midpoint of the segment R"S".

[quote="sprmnt21"]Tool 2. Given a segment and its midpoint, find a line parallel to the line containing the segment through a given point.

Proof

We can go backward to the proof of tool 1, given R", M, S" and, for instance, S'. Take L on s=S'S", if N=LM^R"S' then R'=S"N^LR". The line p'=R'S' is parallel to p"=R"S".[/quote]

Nice constructions, certainly much easier than mine. Thank you!

  Darij

[quote="darij grinberg"]

In fact, you don't even need Ceva to show this.

...

Darij[/quote]

Yes.

Indeed (at the last end) only Thales needs.

If p is the parallel through L at p'//p" and we denote R=S"R'^p and S=R"S'^p, we have that LS/R"S"=LS'/S'S" and LR/R"S"=LR'/R'R" from this, as LS'/S'S"=LR'/R'R", it follows that LS/R"S"=LR/R"S" i.e. LS=LR and from this is easy to conclude that MR"=MS".

But actually, the above argument is one of the way to prove Ceva's theorem.

I would like to comment in a general way your remark (I hope my english will work).

I usually think about the powerfull theorems (like Ceva, Pascal, Desargues, and so on) as a shorten way to comunicate a concept.

I think at that as a macro(high-level) language respect a  machine(low-level) language in the computer field.


topology
limit
combinatorial geometry
advanced fields
advanced fields unsolved

$E$ a Banach space. Prove that if $K$ is a weak compact set, then so is its closed convex hull.

I don’t seem to be using the completeness of $E$. I other words, I believe this holds in any real normed space :?.

The closed convex hull $\overline{\mbox{co}K}$ of $K$ is, of course, (strongly) closed and convex :). In any normed space, this implies that it is also weakly closed. Next, notice that for any functional $f\in E^{*}$ (the dual of $E$), $f(K)$ is compact, and hence contained in some compact interval $I_{f}$. We also have $f\left(\overline{\mbox{co}K}\right)\subset I_{f}$, so in the weak topology the set $\overline{\mbox{co}K}$ can be identified with a closed subset of the compact topological product $\prod_{f\in E^{*}}I_{f}$. This implies that it’s compact, and we’re done.

Is there anything wrong with this?

It works. $E^{*}$ is complete even when $E$ isn't.

I'm afraid this proof doesn't work. Completeness is required.

One of my professors produced both a counterexample (in an
incomplete Normed Space) and an explanation of where the
proof fails.

Here's his Counterexample:

Let $E$ be the set of all "finite" sequences in $\ell^{2}$, that is
$E = \{ \langle x_{n}\rangle \in \ell^{2}: \exists N$ s.t. $x_{n}$ = 0 when n$\textgreater$N $\}$.
Let $E$ inherit the norm topology from $\ell^{2}$. $E$ is not a complete space.

Let $e_{n}$ = $\langle$0, $\dots$ 0,1,0 $\dots 0\rangle$, with a 1 in the n'th place.

Let $K$ = $\{$ $\frac{1}{2}e_{1}$ , $\frac{1}{4}e_{2}$ , $\frac{1}{8}e_{3}$ , $\frac{1}{16}e_{4}$ , $\dots \}$ $\cup$ $\{$ 0 $\}$
$K$ is strongly compact, and thus weakly compact.

If the proof above is correct, $\overline{co(K)}^{E}$ ( the closure of $co(K)$ in $E$)
would be weakly compact in $E$. So it's image under the inclusion
$E \hookrightarrow \ell^{2}$ would be weakly compact in $\ell^{2}$, and hence weakly closed $\ell^{2}$,
 and hence strongly closed $\ell^{2}$.

But it's not strongly closed $\ell^{2}$:   Let $x_{n}= \langle \frac{1}{4}, \frac{1}{16}, \dots , \frac{1}{4^{n}},0,0, \dots \rangle$.

Then $x_{n}= \frac{1}{2^{n}}\bullet 0+\sum_{k=1}^{n}\frac{1}{2^{k}}( \frac{1}{2^{k}}e_{k})$,  so $x_{n}\in co(K)$

Let $x = \langle \frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \frac{1}{256}, \dots \rangle$. Then $x = \lim x_{n}$ in $\ell^{2}$.

But x is not in $E$, so it's not in $\overline{co(K)}^{E}$.
Thus $\overline{co(K)}^{E}$ is not strongly closed in $\ell^{2}$.

I can see where the "proof" fails (and it was a pretty dumb mistake :)). I'll see what I can do when I have more time.

I don't think it's dumb at all. I think it's very subtle. I didn't see it - that's why I asked my professor about it.

It's interesting that this method works so easily in the standard proof of the Banach-Alaoglu Theorem. I am wondering if it can be "fixed", in this case, when $E$ is complete.


a,b are positive integers p,q are rational numbers

1/p+1/q=1  , p>1

prove

a^p/p  +b^q/q  >= ab

Please use $\LaTeX$.

$\frac{1}{p}+\frac{1}{q}=1\qquad p>1$

$\frac{a^{p}}{p}+\frac{b^{q}}{q}\geq ab$

You can either use AM-GM or the fact that $-\log x$ is a convex function.  Then the desired inequality is just the definition of convexity.

[quote="abhishek.nanda"]i had the idea about  AM GM but how exactly do u use it?[/quote]

[hide="Solution"]The condition specifies $\frac{1}{p} + \frac{1}{q} = 1 \implies p + q = pq$.

Then

$\frac{a^p}{p} + \frac{b^q}{q} = \frac{q a^p + p b^q}{pq} = \frac{q a^p + p b^q}{p + q}$

We can interpret this as an arithmetic mean of $q$ different $a^p$ terms and $p$ different $b^q$ terms.  Then

$\frac{q a^p + p b^q}{p + q} \ge \sqrt[p + q]{ a^{pq} b^{pq} } = \sqrt[pq]{a^{pq} b^{pq}} = ab$[/hide]


logarithms
algebra
function
domain
complex numbers
complex analysis
complex analysis unsolved

They say that for $ Re(z_1 z_2) > 0$ and $ Re(z_i) > 0$ with $ i \equal{} 1,2$, the relation $ Log(z_1 z_2) \equal{} Log(z_1) \plus{} Log(z_2)$ holds.

So taking the principal branch of the logarithm $ Log(z)$ we have

$ Log(z_1 z_2) \equal{} log|z_1 z_2| \plus{} i arg(z_1 z_2) \equal{} log|z_1| \plus{} log|z_2| \plus{} i (arg(z_1) \plus{} arg(z_2)) \equal{} Log(z_1) \plus{} Log(z_2)$

not?

Can we loosen the condition $ Re(z_i) > 0$ to $ arg(z_1) \neq arg(z_2)$, or am I missing something obvious?

p.s.

The principal branch of the logarithm is defined (for me) as follows

Take $ G \equal{} C \minus{} \{z: z \leq 0 \}$. For every $ z \in G$ put $ z \equal{} |z|e^{i \theta}$ with $ \minus{} \pi < \theta < \pi$. Then $ Log(z) \equal{} \log |z| \plus{} i \theta$.

To show that the real part of $ z^{1/2}$ is always positive I can argument as follows: let $ Log z$ be the principal branch of the logarithm, then $ z^{1/2} \equal{} exp(1/2 Log(z)) \equal{} exp(log|z|^{1/2} \plus{} i \theta /2) \equal{} |z|^{1/2} exp(i \theta/2)$. Then because $ \minus{} \pi < \theta < \pi$ we have $ \minus{} \pi/2 < \theta/2 < \pi/2$. So that $ Re(z^{ \minus{} 1/2}) > 0$ or not?

because (for some other branch) if I would take $ z \equal{} \minus{} 1$ then $ z^{1/2} \equal{} i$ which has real part equal to zero (assuming positive means strictly positive here)... confusion  :oops:  :blush:

never mind. I read something that every branch of the log is a translate by some $ 2 \pi k i$. But that means of course not a translate of the domain or anything even near.... (/me Stupid)

ofcourse we can't define log for $ \minus{}1$...

Sure we can, just not on the principal branch.  We can define a branch of the logarithm off any ray from the origin, but we have to be very, very careful when comparing branches.[/url]


Let $ x,y,z$ be positive real numbers satisfying $ \sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx}\equal{}1$. Prove that $ (x\plus{}y\plus{}z)(1\minus{}x)(1\minus{}y)(1\minus{}z)\le 8xyz$

[quote="jerry_smith"]Let $ x,y,z$ be positive real numbers satisfying $ \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx} \equal{} 1$. Prove that $ (x \plus{} y \plus{} z)(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)\le 8xyz$[/quote]
Let $ a\equal{}\sqrt{x}$, $ b\equal{}\sqrt{y}$, $ c\equal{}\sqrt{z}$ then inequality is equivalent to:

$ \left(1\plus{}\frac{(a\minus{}b)(a\minus{}c)\plus{}(b\minus{}c)(b\minus{}a)\plus{}(c\minus{}a)(c\minus{}b)}{ab\plus{}bc\plus{}ca}\right)\left(1\minus{}\frac{(a\minus{}b)(a\minus{}c)}{2bc}\right)\left(1\minus{}\frac{(b\minus{}a)(b\minus{}c)}{2ac}\right)\left(1\minus{}\frac{(c\minus{}a)(c\minus{}b)}{2ab}\right) \leq 1$
so if $ a \geq b \geq c$, and $ ca\plus{}cb \geq ab$ then

$ \frac{(a\minus{}b)(a\minus{}c)\plus{}(b\minus{}c)(b\minus{}a)\plus{}(c\minus{}a)(c\minus{}b)}{ab\plus{}bc\plus{}ca} \leq \frac{(a\minus{}b)(a\minus{}c)}{2bc}\plus{}\frac{(b\minus{}a)(b\minus{}c)}{2ac}\plus{}\frac{(c\minus{}a)(c\minus{}b)}{2ab}$ by Vornicu-Schur so

$ (x \plus{} y \plus{} z)(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)\le 8xyz$

[quote="Nak"]
so if $ a \geq b \geq c$, and $ ca \plus{} cb \geq ab$ then

$ \frac {(a \minus{} b)(a \minus{} c) \plus{} (b \minus{} c)(b \minus{} a) \plus{} (c \minus{} a)(c \minus{} b)}{ab \plus{} bc \plus{} ca} \leq \frac {(a \minus{} b)(a \minus{} c)}{2bc} \plus{} \frac {(b \minus{} a)(b \minus{} c)}{2ac} \plus{} \frac {(c \minus{} a)(c \minus{} b)}{2ab}$ by Vornicu-Schur[/quote]
That's wrong! Try to a=3/4, b=4/5, c=8/31

[quote="N.N.Trung"][quote="Nak"]
so if $ a \geq b \geq c$, and $ ca \plus{} cb \geq ab$ then

$ \frac {(a \minus{} b)(a \minus{} c) \plus{} (b \minus{} c)(b \minus{} a) \plus{} (c \minus{} a)(c \minus{} b)}{ab \plus{} bc \plus{} ca} \leq \frac {(a \minus{} b)(a \minus{} c)}{2bc} \plus{} \frac {(b \minus{} a)(b \minus{} c)}{2ac} \plus{} \frac {(c \minus{} a)(c \minus{} b)}{2ab}$ by Vornicu-Schur[/quote]
Why?[/quote]
It is equivalent to:

$ (a \minus{} b)(a \minus{} c) \left( \frac {1}{2bc} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \plus{} (b \minus{} c)(b \minus{} a) \left( \frac {1}{2ac} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \plus{} (c \minus{} a)(c \minus{} b) \left( \frac {1}{2ab} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \geq 0$.

1. $ (c \minus{} a)(c \minus{} b) \geq 0, (a \minus{} b)(a \minus{} c) \geq 0, (b \minus{} c)(b \minus{} a) \leq 0, |(a \minus{} b)(a \minus{} c)| \geq |(b \minus{} c)(b \minus{} a)|$.

2.  $ \frac {1}{2ab} \minus{} \frac {1}{ab \plus{} bc \plus{} ca} \geq 0, \frac {1}{2ac} \minus{} \frac {1}{ab \plus{} bc \plus{} ca} \geq 0, \frac {1}{2bc} \minus{} \frac {1}{ab \plus{} bc \plus{} ca} \geq 0$.

3. $ \left(\frac {1}{2bc} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \geq \left(\frac {1}{2ac} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right)$.

$ 1,2,3 \Longrightarrow (a \minus{} b)(a \minus{} c) \left( \frac {1}{2bc} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \plus{} (b \minus{} c)(b \minus{} a) \left( \frac {1}{2ac} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \plus{} (c \minus{} a)(c \minus{} b) \left( \frac {1}{2ab} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}\right) \geq 0$

[quote="N.N.Trung"]I'm sure that your post doesn't work.
Try to a=3/4, b=4/5, c=8/31[/quote]
I wrote that inequality is true if $ a \geq b \geq c$, and $ ca \plus{} cb \geq ab$


The latest issue of Forbes Magazine has an article about Akamai Technologies.  I understand it will be available on newsstands tomorrow (April 7, 2007).  Many readersof this forum will recognize that the Akamai Foundation affiliated with the company generously supports the American Mathematics Competitions by providing the scholarships awarded to the top three scorers on the USAMO.  The Akamai Foundation also provides the support for the expanded MOSP that allows us to invite about 25 young students to the MOSP each summer.

As an adjunct to the magazine article, acknowledging that the company  is built on mathematical concepts, the on-line version of the magazine has a "quiz" made from some recent problems on the 2006 AMC 8, the 2007 AMC 10 and 2007 AMC 12 and the 2007 AIME Alternate.  You might enjoy reading the article, and clicking thorugh the quiz.  It is at http://www.forbes.com/technology/
The quiz is titled "How Number Nifty are You?"

Steve Dunbar
AMC Director

[quote]You got 7 out of 8[/quote]

Ok, I might have made some sort of mistake, but I'm pretty sure that there were only 7 questions on that quiz.  I didn't get any wrong.  :huh:

And there are only 9 pages, with the first being an introduction and the last being the one giving your score,

Yeah, I even tried changing the url to say "score=8", and it just said:

You got 8 out of 8.

undefined.

wheras the 7 out of 8 said "you're a marvel" or something.


Find all integers like $a$ such that $\frac{a^{2000}-1}{a-1}=t^2,t \in Z$).

I think you mean  $\frac{a^{2000}-1}{a-1}=t^2,t \in Z$   :D


Let $x,y,z$ be positive reals such that $x+y+z=1$. Prove that \[\frac{1}{\sqrt{x^{2}+xy+yz}}+\frac{1}{\sqrt{y^{2}+yz+zx}}+\frac{1}{\sqrt{z^{2}+zx+xy}}\geq3\sqrt{3}.\]

[quote="Albanian Eagle"]first note that $\sum_{cyc}\frac{1}{\sqrt{x^{2}+xy+xz}}=\sum_{cyc}\frac{1}{\sqrt{x}}$
from holder we know that $(x+y+z)\left( \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^{2}\geq 27$ which gives the desired inequality. :wink:[/quote]

Isn't it supposed to be $\sum_{cyc}\frac{1}{\sqrt{x^{2}+xy+yz}}$, not $\sum_{cyc}\frac{1}{\sqrt{x^{2}+xy+xz}}$. Or are they somehow equivelent?

But anyway here is my solution:

From Cauchy-Schwarz we have:

$3 = (1+1+1)(\sum_{cyc}x^{2}+xy+yz)\geq(\sum_{cyc}\sqrt{x^{2}+xy+yz})^{2}$

$\sqrt{3}\geq\sum_{cyc}\sqrt{x^{2}+xy+yz}$

Then, we have:

$\sum_{cyc}\frac{1}{\sqrt{x^{2}+xy+yz}}\geq\frac{1}{\sqrt{3}}(\sum_{cyc}\sqrt{x^{2}+xy+yz})(\sum_{cyc}\frac{1}{\sqrt{x^{2}+xy+yz}})\geq\frac{1}{\sqrt{3}}(1+1+1)^{2}=3\sqrt{3}$

let us call the square roots $A,B,C$ then :
$A^{2}+B^{2}+C^{2}=1$ and we have to show that $(\sum \frac{1}{A})^{2}\geq 27$
this is obvious by holder or jensen (they coincide in this case :D )

Indeed, just use Jensen on $\frac{1}{\sqrt{x}}$  :D

But how do you use Hölder on this?

from holder we have $(a^{2}+b^{2}+c^{2})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}\geq 27$
and it gives the same result :) as jensen


Solve
$ \frac{1}{x\plus{}\frac{1}{y\plus{}\frac{1}{z}}} \equal{} \frac{7}{30}$
if
a) x, y, z  is natural numbers
b) x, y, z  is integer numbers

There are infinitely many solutions since there are 3 variables with 1 equation. For example, another one is:

$ \frac{7}{30}\equal{}\frac{1}{\frac{30}{7}}\equal{}\frac{1}{3\plus{}\frac{9}{7}}\equal{}\frac{1}{3\plus{}\frac{1}{\frac{7}{9}}}\equal{}\frac{1}{3\plus{}\frac{1}{0\plus{}\frac{7}{9}}}\equal{}\frac{1}{3\plus{}\frac{1}{0\plus{}\frac{1}{\frac{9}{7}}}}$

So another solution is $ (3, 0, \frac{9}{7})$, but they aren't integers.

@powerofpi, it says that the variables have to be either natural numbers or integers.

[quote="small_unicorn"]Solve
$ \frac {1}{x \plus{} \frac {1}{y \plus{} \frac {1}{z}}} \equal{} \frac {7}{30}$
if
a) x, y, z  is natural numbers
b) x, y, z  is integer numbers[/quote]

Hello! My name is Alfian Rahman, I am from Indonesia, I am in the 6th grade in elementary school now
I always join Math Olympiad in my country

My answer

We have $ \frac{yz\plus{}1}{xyz\plus{}x\plus{}z}\equal{}\frac{7}{30}$

and we have $ yz$ is $ 6$ so $ xyz\plus{}x\plus{}z\equal{}6x\plus{}x\plus{}z\equal{}7x\plus{}z\equal{}30$

and then $ x$ is $ 4$ and $ z$ is $ 2$

so, $ y$ is $ 3$

The answer is $ x,y,z$ is $ 4,3,2$

I would have solved it the way AndrewTom did, but I don't know how to do that with LaTeX.


superior algebra
superior algebra theorems

Define $Z_{4}\mapsto <i>$ by the rule $f(x1) = i^{x}$, where $0\leq x \leq 3$. Prove that f is an equivalence relation.
Note: 1 in f(x1) is a complex number.

I'm not sure I understand this notationally. Is this $f: \mathbb{Z}_{4}\rightarrow <i>$ by $x\mapsto i^{x}$? If this is the case what purpose is the $1$ serving in $f(x1)$? Possibly $Z_{4}$ is something I haven't seen and is not $\mathbb{Z}_{4}$?

I think [b]guile[/b] wanted to emphasize the difference between integers such as $x$ and elements of $\mathbb Z_{4}$ such as $x1$. Here $1$ is the unit element of $\mathbb Z_{4}$.

[quote="mlok"]You mean $f$ is an isomorphism.[/quote]

Yup. Sorry, it was suppose to be isomorphism.

Let $x,y\in\mathbb{Z}_{4}$, then $f(x+y)=i^{x+y}=i^{x}i^{y}=f(x)f(y)$. The identity maps to identity since $f(0)=1$, so we have a homomorphism. Suppose $i^{a}, i^{b}\in <i>$ where $i^{a}=i^{b}$. Then we have $f(a)=f(b)$ and since $0\leq a,b,\leq 3$ it follows that $a=b$, so $f$ is 1-1. Since the elements of $<i>$ are $i,-1,-i, 1$, each of $1,2,3,0$ map them respectively and so $f$ is onto. Therefore $f$ is a bijection and thus an isomorphism.


A sequence is given:
$ x_{n}\equal{}\minus{}1\minus{}n$ for natural $ n>0$.
A sequence $ y_{n}$ is a sequence such that points (coordinates):
$ (x_{n},0), (\minus{}1,1), (0,y_{n})$ lie on a line.
Find the formula of $ y_{n}$

Help me please.

Thank you.

A line is defined by two points.

The point $ Y_n(0,y_n)$ would lie on the line defined by $ X_n(x_n,0)$ and $ P( \minus{} 1,1)$ if and only if the slope of $ Y_nP$ (or $ Y_nX_n$) is the same as the slope of $ X_nP$. This is valid since $ X_nP$ is not vertical.

That is: $ \frac {y_n \minus{} 1}{0 \plus{} 1} \equal{} \frac {1 \minus{} 0}{ \minus{} 1 \minus{} x_n}\Leftrightarrow y_n \minus{} 1 \equal{} \frac {1}{n}\Leftrightarrow y_n \equal{} \frac {1}{n} \plus{} 1$ for all $ n>0$

Verifying this works for $ X_nY_n$ :  $ \frac {\frac {1}{n} \plus{} 1 \minus{} 0}{0 \minus{} x_n} \equal{} \frac {n \plus{} 1}{n(n \plus{} 1)} \equal{} \frac {1}{n}$

So $ y_n \equal{} \frac {1}{n} \plus{} 1$, $ n\ge 1$

 :)


arithmetic sequence
combinatorics unsolved
combinatorics
arithmetic
Arithmetic Progression
China
TST

Find $k \in \mathbb{N}$ such that

[b]a.)[/b] For any $n \in \mathbb{N}$, there does not exist $j \in \mathbb{Z}$ which satisfies the conditions $0 \leq j \leq n - k + 1$ and $\left(
\begin{array}{c}
n\\
j\end{array} \right), \left( \begin{array}{c}
n\\
j + 1\end{array} \right), \ldots, \left( \begin{array}{c}
n\\
j + k - 1\end{array} \right)$ forms an arithmetic progression.

[b]b.)[/b] There exists $n \in \mathbb{N}$ such that there exists $j$ which satisfies $0 \leq j \leq n - k + 2$, and $\left(
\begin{array}{c}
n\\
j\end{array} \right), \left( \begin{array}{c}
n\\
j + 1\end{array} \right), \ldots , \left( \begin{array}{c}
n\\
j + k - 2\end{array} \right)$ forms an arithmetic progression.

Find all $n$ which satisfies part [b]b.)[/b]

part a)
1)if $ k \equal{} 1$ then $ \left (\begin {array} {c} n \\
j\end{array} \right)$ an arithmetic progression.
2)if $ k \equal{} 2$ then $ \left (\begin {array} {c} n \\
j\end{array} \right), \left (\begin {array} {c} n \\
{j \plus{} 1}\end{array} \right)$ an arithmetic progression.
3)if $ k \geq 3$ then $ \left (\begin {array} {c} n \\
j\end{array} \right), \left (\begin {array} {c} n \\
{j \plus{} 1}\end{array} \right), \left (\begin {array} {c} n \\
{j \plus{} 2}\end{array} \right)$ an arithmetic progression so that:
$ \left (\begin {array} {c} n \\
j\end{array}\right) \plus{} \left (\begin {array} {c} n \\
{j \plus{} 2}\end{array}\right) \equal{} 2\left (\begin {array} {c} n \\
{j \plus{} 1}\end{array}\right)$
so
$ (j \plus{} 2)(j \plus{} 1) \plus{} (n \minus{} j)(n \minus{} j \minus{} 1) \equal{} 2(j \plus{} 2)(n \minus{} j)$
so
$ 4j^2 \plus{} (4n \minus{} 8)j \plus{} (n^2 \minus{} 5n \plus{} 2) \equal{} 0$
so
$ j \equal{} \frac {n \minus{} 2 \plus{} \sqrt {n \plus{} 2}}{2}$ or $ j \equal{} \frac {n \minus{} 2 \minus{} \sqrt {n \plus{} 2}}{2}$
so $ n$ be a form $ t^2 \minus{} 2$ fo $ t \in \mathbb{Z}$ and $ t \geq 2$.
so for any $ n$ and $ k \geq 3$ there is not exist $ j$.

part b)
1)if $ k \equal{}1,2,3$ similary part a case 1,2 for every n there is exist $ j$.
2)if $ k \geq 4$ similary part a case 3,$ n$ be a form $ t^2 \minus{} 2,t \in \mathbb{Z},t \geq 2$.
so if $ k\equal{}1,2,3$ every $ n$, and if $ k \equal{} 4$ $ n$ be a form $ t^2 \minus{} 2$ for some integer $ t$.


$ a)$ Consider $ x,y,z \in \mathbb{R}$ such that $ xy\plus{}yz\plus{}zx\equal{}1$. Prove that:

$ \frac{x}{x^2\plus{}1}\plus{}\frac{y}{y^2\plus{}1}\plus{}\frac{z}{z^2\plus{}1}\equal{}\frac{2}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$

$ b)$ Using $ a)$ prove that if $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}abc$ then:

$ \frac{a}{a^2\plus{}1}\plus{}\frac{b}{b^2\plus{}1}\plus{}\frac{c}{c^2\plus{}1} \le \frac{abc}{4}$

a. Replace 1 with ab+bc+ac, notice that $ a^2 \plus{} ab \plus{} bc \plus{} ac \equal{} (a \plus{} b)(a \plus{} c)$,...
b. Let $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}$, we have to prove that
\[ \sum {\frac {x}{{x^2 \plus{} 1}}} \le \frac {1}{{4xyz}}
\]
, which is obvious true using a)


[quote="BrainMan1"]Can someone help me solve this PDE:

u[size=75]t[/size] = u[size=75]xx[/size] - (B^2)u, 0 < x < L, t > 0
u(0,t) = U1, u(L,t) = U2, u(x,0) = f(x)[/quote]

Let $ u(t,x)\equal{}v(t,x)e^{\minus{}b^2t}$. Then $ v_t\equal{}v_{xx}$.

Looks like a diffusion problem for one-dimensional container with decay. I'm very tired to describe it now but I think you can make it pliable by setting v=ue^(f(x,t)) (for appropriate f), seeing what equation v satisfies, and using some separation of variables.

ah. as above.

[quote="BrainMan1"]What are the conditions for v(t,x) then? I get a PDE with v(t,0)=exp(B^2*t)U1, etc. How do I solve it from here?[/quote]

In russian http://u-pereslavl.botik.ru/~trushkov/pde/pde.pdf pp.79-80


Indeed , we only need $abc \geq 2^9$  ;)


Find $ x ,y \in Z $ such that : $ 2.(x^3-x)=y^3-y $

I think $\{x,y\}\subset\{-1,0,1\}$ or $(x,y)=\pm(4,5)$ are the only solutions - but I'm only guessing right now...

The following is earlier post.
[url]https://artofproblemsolving.com/community/c205226h33527_equation_from_dkarpov_1[/url]


Hi

Let $ s,p\in \mathbb{N}^2$ such that $ s\geq 2p^2$ and $ M_1,\dots,M_s \in \mathbb{Z}^2$ distinct points. Is it true that there's a subset I of $ [[1,s]]$ with contains at least $ s/p^2$ integers such that
\[ \forall i \neq j \in I \qquad \parallel{}M_i \minus{} M_j\parallel{}\geq p\]
where $ \parallel{}(x,y)\parallel{} \equal{} \max(|x|,|y|)$ ?

It's clear if $ p \equal{} 1$.

Thanks

Just look at the intersections of your point set with the $ p^2$ integer lattices with step $ p$ in each direction. Each such intersection satisfies the desired condition, they are $ p^2$, so at least one of them contains $ s/p^2$ points in our original set. ;)


Hello,

Is there any way you can embed a Geogebra image onto a LaTeX document?

If so, how do you do it (i.e., what are the commands)?

Thanks

Export from Geogebra in PNG format then use \includegraphics in your LaTeX document.


can you give me hints to solve $ y\equal{}2x\dfrac{dy}{dx}\plus{}x^2(\dfrac{dy}{dx})^4$

In principle you could just solve the quartic for $ \frac {dy}{dx}$, to get $ \frac {dy}{dx} \equal{} f_i(x,y)$ for $ 1\leq i \leq 4$. As long as these $ f_i(x,y)$ are continuous in some region (and they will be) we have the standard existence and uniqueness results for each of the four branches.  The $ f_i$ are very ugly, however, and though this approach will give you the solution in principle, it's unlikely to yield explicit expressions without first finding some clever substitutions for $ x$ or $ y$ to simplify the form of $ f_i$.

[hide="cheap unsatisfying answer"]In this case, a family of solutions you can 'eyeball' is $ y \equal{} c\sqrt {|x|} \plus{} \left(\frac {c}{2}\right)^4$ for any $ c\in \mathbb{R}$.  One might arrive at this by noticing that the solutions of the ODE $ y \equal{} 2x\frac {dy}{dx}$, $ y \equal{} c\sqrt {|x|}$, satisfy $ x^2\left(\frac {dy}{dx}\right)^4 \equal{} \text{const}$, and therefore look for solutions of the form $ y \equal{} c\sqrt {|x|} \plus{} d$ for the original ODE.[/hide]

[quote="tamal banerjee"]can you give me hints to solve $ y \equal{} 2x\dfrac{dy}{dx} \plus{} x^2(\dfrac{dy}{dx})^4$[/quote]

Let $ y'\equal{}p$. Then $ y\equal{}2xp\plus{}x^2p^4$.
$ pdx\equal{}2xdp\plus{}2pdx\plus{}2xp^4dx\plus{}4x^2p^3dp$.
$ (1\plus{}2xp^3)(pdx\plus{}2xdp)\equal{}0$
et cetera


analytic geometry
graphing lines
slope
algebra open
algebra

Write the equation of the line (in slope-intercept form y = mx + b) that passes through the points (-2, 8) and (-9, -6).

how do i set this up?

hey angel, just for future reference, this isnt the best place to post these problems, these are more suited for http://www.artofproblemsolving.com/Forum/index.php?f=299 .

but to answer your question, you can first find the slope of the line, which is $m$ in '$y=mx+b$.  Then once you have $m$, you can plug it into the equation.  in this case, $m$=$\frac{8-(-6)}{-2-(-9)}$ = $\frac{14}{7}$ = 2.

So now we know your line is $y=2x+b$.  now all that remains is to solve for $b$, and we can do this by using one of the given points.  We know that (-2, 8) is on the line, so we plug in those values to get $8=2(-2)+b$, which means $b$=12.

so your line is $y=2x+12$.

all of these problems are pretty systematic, so once you get the hang of the procedure its just plugging in numbers.  anything that needs clarification feel free to ask.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ f(x)$ and $ g(x)$ be irreducible polynomials over a field F and let $ a$ and $ b$ belong to some extension $ E$ of $ F$. If $ a$ is a zero of $ f(x)$ and $ b$ is a zero of $ g(x)$, show that $ f(x)$ is irreducible over $ F(b)$ if and only if $ g(x)$ is irreducible over $ F(a)$.

We have that $ F(a)\equal{}F[x]/(f(x))$ and $ F(b)\equal{}F[y]/(g(y))$. Now, if $ f(x)$ is irreducible in $ F(b)$, then $ F(b)[x]/f(x)$ is a field. That is, $ F[x,y]/(f(x),g(y))$ is a field. But $ F[x,y]/(f(x),g(y))\equal{}F(a)[y]/(g(y))$ which means that $ g(y)$ must be irreducible over $ F(a)$. The other direction is identical.


Let $ a_n$ be the sum of the first $ n$ terms of $ 1\plus{}2\plus{}4\plus{}4\plus{}8\plus{}8\plus{}8\plus{}....$ and $ b_n$ be he sum of the first $ n$ terms of $ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}...$. Investigate the quotient $ \frac{a_n}{b_n}$ for $ n\to\infty$

if $ \frac {k(k \minus{} 1)}{2} \plus{} 1 < n\le \frac {k(k \plus{} 1)}{2} \plus{} 1$
then $ a_n\in[|a_{(k(k \minus{} 1) \plus{} 2)/2},a_{(k(k \plus{} 1) \plus{} 2)/2}|]$
where $ A_{k \minus{} 1} \equal{} a_{(k(k \minus{} 1) \plus{} 2)/2} \equal{} \sum_{i \equal{} 1}^{k \minus{} 1}i2^{i} \equal{} ...easy$ and $ A_k \equal{} a_{(k(k \plus{} 1) \plus{} 2)/2} \equal{} \sum_{i \equal{} 1}^{k}i2^{i} \equal{} ...easy$

so $ A_{k \minus{} 1} \le \frac {a_n}{b_n}\le A_k$ and $ \lim_{k\to \plus{} infty} A_k \equal{} \lim_{k\to \plus{} infty} A_{k \minus{} 1} \equal{} ... \equal{} \lim_{n\to \plus{} infty} \frac {a_n}{b_n}$


calculus
integration
trigonometry
calculus computations

This integral has been neglected ,yet it is not too hard.Here are the main points:
Denoting the integral by I[m,n], we can evaluate it using the recurrence relation

I[m,n].[$m^2 +n^2$] =m(m-1).I[m-2,n]

To get this recurrence relation we have to  apply integration by parts, twice ,each time taking$e^{-nx}$ as the factor to be integrated.
Finally we have to take into account that I[m.n] will have different forms when m=even positive integer and when m= odd positive integer.


inequalities
geometry
perimeter
LaTeX
inequalities solved

Let $a_1,a_2,a_3,a_4$ be the sides of an arbitrary quadrilateral of perimeter $2s$. Prove that
\[ \sum\limits^4_{i=1} \dfrac 1{a_i+s} \leq \dfrac 29\sum\limits_{1\leq i<j\leq 4} \dfrac 1{ \sqrt { (s-a_i)(s-a_j)}}.  \]
When does the equality hold?

My solution is the following: we have
\[\frac{2}{9}\sum_{1 \leq i < j \leq 4}\frac{1}{\sqrt{(s - a_i)(s - a_j)}} \geq \frac{2}{9}2\sum_{1 \leq i < j \leq 4}\frac{1}{(s - a_i) + (s - a_j)} = \frac{4}{9}\sum_{1 \leq i < j \leq 4}\frac{1}{a_i + a_j}.\]
Let's denote $a_1 = a$, $a_2 = b$, $a_3 = c$ and $a_4 = d$.
We should prove now that
\[\frac{2}{9}\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d} + \frac{1}{b + c} + \frac{1}{b + d} + \frac{1}{c + d}\right) \geq\]\[\frac{1}{3a + b + c + d} + \frac{1}{a + 3b + c + d} + \frac{1}{a + b + 3c + d} + \frac{1}{a + b + c + 3d}.\]
By Cauchy we have:
\[\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d}\right)\left((a + b) + (a + c) + (a + d)\right) \geq 9\]\[\Rightarrow \frac{1}{9}\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{a + d}\right) \geq \frac{1}{3a + b + c + d}.\]
Adding the three analogous inequalities to this one we get the result.

Arne: yeah, your solution is a piece of cake compared to mine, thanks  :)

Stevem: Thanks a lot for pointing out the error in my LaTeX code. Actually, please excuse me but I have NO CLUE about what I am doing when I write LaTeX tags. I have never worked with LaTeX, instead I have always been using Scientific WorkPlace; this is a program which enables you to write TeX files without really knowing TeX, just by a user-friendly WYSIWYG interface. Hence, when I need a formula, I just type it in this interface and copy&paste it into the forum. If I try to edit it manually afterwards, I usually get lots of errors. Sorry again.

I used \Huge in order to have the fractions large enough such that they can be read properly. But I see it's not really necessary  :)

  Darij

No need to apologise. I too got into LaTeX using Scientific Workplace, which is a great program for typing maths since it's so natural and you don't need to use the mouse a lot. But on occasions it uses its own system which isn't recognised elsewhere, which makes it awkward.

So it's really good to learn LaTeX and it will be extremely useful to you in the  future. Some university mathematics departments have special classes in LaTeX to help their students.

Because it's so easy to make mistakes I use Preview a lot, so it looks great when posted - no-one sees the hundreds of errors on the way!


search
algebra unsolved
inequalities unsolved
inequalities

Please, do not post the same thing 3 times and in the wrong forum. You should post in algebra unsolved. But don't post it, it has been discussed. Search on the first page of algebra unsolved and you'll find it.

I think anhhungdatviet is a Vietnamese , not a Syrian  :D  :D  .


A circle is circumscribed about triangle PQR. Take a point, S, on the circle. Construct three lines through S, perpendicular to lines PQ, PR, and QR. Prove that the three points of intersection are collinear.

It is the Simson line. You can prove it by angle-chasing.


This game board is made with 36 unit squares. A game show company puts prizes behind the shaded squares on the board, but the contestants can't see the shading. What is the probability that a contestant choosing a unit square from the board at random selects a square with a prize behind it? Express your answer as a common fraction.
[asy]for(int i = 0; i < 6; ++i)
{
	for(int j = 0; j < 6; ++j)
	{
		draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,linewidth(2));
	}
}

for(int i = 1; i < 5; ++i)
{
	filldraw((i,i)--(i+1,i)--(i+1,i+1)--(i,i+1)--(i,i)--cycle,black);
}

filldraw((1,3)--(2,3)--(2,4)--(3,4)--(3,5)--(1,5)--(1,3)--cycle,black);
filldraw((3,1)--(3,2)--(4,2)--(4,3)--(5,3)--(5,1)--(3,1)--cycle,black);[/asy]

There are 10 shaded squares, so the probability is 10/36 or 5/18.


1. How many points, no three of which are collinear, determine 91 lines?
2. What is the least number of people you could have in a group and still be sure that at least 6 of them have birthdays in the same month?

(Please try to give a solution and not just an answer when posting. Thanks.)

[b]Reply for problem 2.[/b]

[hide]
If you take 5 people from every month and multiply 12 by 5, you get 60.  Now you have to add one person because this enshers a 6th repeat in some month, to get 6 people.  The answer is 61. :stretcher:

[color=blue][size=75]kyyuanmathcount: hide tags please![/size][/color]
[/hide]

[hide]14[/hide]
[hide]the trainglular numbers. 91 is the 14th triangular #[/hide]

[hide="problem 1"]
for $n$ points, no three of which are collinear, the maximum number of lines that can be connected is $l(n)=\frac{n(n-1)}{2}$.
[/hide]

[hide="1st"] Well, any two lines can intersect. So there are $\binom{n}{2}$ total points of intersections. $\binom{n}{2}=91$, which is solved to n=14[/hide]
[hide="2nd"]According to the pigeonhole, we have 12 holes and we need 6 forced pigeons which gives 61 birds.[/hide]
EDIT: 1700

[hide="1"]
Since no three lines can be collinear, you can just draw a polygon with p sides, where p is the number of points.
So, 91=d+e
D is the number of diagonals, and e is the number of edges.
$91=e+(\frac{e(e-1)}{2}-e)$
So, $e=\boxed{14}$[/hide]
p.s. does anyone know the other variation of the diagonals formula?

[hide="2"]
Since there are 12 months in a year, in the worst case, there will be 5 people in each month.
5*12=60
Then, add a random person to any month.
So, the answer is 61.[/hide]


inequalities
calculus
calculus computations

c>2
$ x_{1}\equal{}0$
$ x_{n\plus{}1}\equal{}\sqrt{c\minus{}\sqrt{c\plus{}x_{n}}}$
Prove that $ x_{n}$ define for all n has a limit.Find it.

[hide="My poor solution"]Let $ f(x) \equal{}\sqrt{c\minus{}\sqrt{c\plus{}x}}$. Then $ f'(x) < 0$ and $ f$ is strictly decreasing.

For $ x\in [0, c^{2}\minus{}c]$, $ f(x)\leq f(0)$. We also have $ f(0) <\sqrt{c}< 2\sqrt{c}\minus{}1 < c^{2}\minus{}c$.

So if $ x_{n}\in [0, f(0)]$, then $ x_{n\plus{}1}$ is also in $ [0, f(0)]$. This means that each $ x_{n}$ is well defined for all $ n$ and is bounded.

Also, $ |f'(x)| \equal{}\frac{1}{4\sqrt{c\plus{}x}f(x)}\leq\frac{1}{4\sqrt{c}f(2\sqrt{c}\plus{}1)}$ for all $ x\in [0, f(0)]$

Since $ c\left( c\minus{}\sqrt{c\plus{}1}\right)$ is increasing for $ c\geq 2$, we have an inequality $ 4\sqrt{c}f(c) \equal{} 4\sqrt{c\left( c\minus{}\sqrt{c\plus{}1}\right) }\geq 4\sqrt{4\minus{}2\sqrt{3}}\equal{} 4 (\sqrt{3}\minus{}1) > 1$. From this, we deduce that $ f$ is contractive on $ [0, f(0)]$ and $ f$ has a unique fixed point $ \alpha$ so that $ x_{n}$ converges to this point.

Finally, we can compute $ \alpha$ by solving $ \alpha \equal{} f(\alpha)$. The answer is $ \alpha \equal{}\frac{\sqrt{4c\minus{}3}\minus{}1}{2}$.
[/hide]


Prove that from any 25 distinct positive numbers one can choose two whose sum and difference do not coincide with any of the remaining 23.

Suppose not. Let the numbers be $a_1<a_2<\ldots<a_{25}$. Consider the pairs $a_{25}, a_i$ for $i=1, 2, \ldots, 24$. Their sum is $>a_{25}$, so it never coincides with any of the $a_i$. Then the differences $a_{25}-a_1>\ldots>a_{25}-a_{24}$ must coincide with $a_1<\ldots<a_{24}$. Since the mentioned differences are ordered from greatest to smallest and are all distinct, we must have $a_{25}-a_i=a_{25-i}$ for $i=1, 2, \ldots, 24$.

Since $a_{24}+a_1=a_{25}$, we have $a_{24}+a_i>a_{25}$ for $i>1$. So $a_{24}-a_2, \ldots, a_{24}-a_{23}$ must coincide with values among $a_{23}, \ldots, a_1$. Moreover, since $a_{23}=a_{25}-a_2>a_{24}-a_2$, we can discard $a_{23}$. Since we have $22$ ordered numbers on each side, we must have $a_{24}-a_{23}=a_1, \ldots, a_{24}-a_{2}=a_{22}$. However, this implies $a_{24}-a_{12}=a_{12}$, so $a_{24}, a_{12}$ is a good pair.


At camp, do they go over the Admission test, do we get our admission test back?

i wanted to know how i got scored on some of my partial solutions

I got accepted and I didn't get my test back, so I assume that they won't give back your admission tests...

what i meant by that was do we get it back at camp?

I know that theres no specific  cutoff score that they are looking for but does anyone think that someone who got like a 5 or 6 on the admisson test has a chance to get accepted?

yeah definitely.

i know someone who solved 4 last year and got in....

Yay!!

I can't believe I got in.

So how useful is the camp? What I mean is, do you learn a lot that would help you in stuff like the AIME and USAMO  and that sort of competition things?

yeah its competition oriented. its pretty useful. especially if you pay attention in class &  throughout the school year, keep up ur math skills/do problems
and even if you dont... i think your subconscious learns stuff... mine did ;P

Is there any chance I can still apply and get in?

Yes, just wait and apply with test C when it comes out.

Realistically, what does that make my chances?


[quote="kunny"]Rewrite L.H.S. as $\frac{y}{x+y+z}x^n+\frac{z}{x+y+z}y^n+\frac{x}{x+y+z}z^n$[/quote]
then what will we do?

[quote="kunny"]Then remark that  $\frac{y}{x+y+z}+\frac{z}{x+y+z}+\frac{x}{x+y+z}=1.$[/quote]

 What do you attempt to do, Kunny? Jensen? But $f(x)=x^n$ then $f''(x)=n(n-1)x{}^n{}^-{}^2 \geq 0$ so...???????

[quote="M4RI0"]$x,y,z>0$ such as $x+y+z=1$ .

Prove that : $x^ny+y^nz+z^nx \le \frac{{n^n}}{(n+1)^n}$[/quote]

  Let check some:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5Eny&t=76726
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5Eny&t=28680
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5Eny&t=2549


function
integration
algebra
polynomial
real analysis
real analysis unsolved

easy but nice :)
Let $f:[0,1]\rightarrow \mathbb{R}$ be continuous function such that $\int _{0}^{1}x^kf(x)dx=0$ $\forall k=0,1\dots n-1$ and $\int_0^1 x^nf(x)dx=1$.prove that there exists x in unit interval such that $|f(x)|\geq 2^n(n+1)$.

It must have been solved before.
You can have a search.

If $P$ is some monic polynomial of degree $n$, then we have $\max_{[0,1]}|f|\cdot\int_0^1|P|dx\ge\int_0^1|Pf|dx\ge \left|\int_0^1 Pf\right|dx=1$, so $\max |f|\ge\frac 1{\int_0^1|P|dx}$.

Now we can simply take $P(x)=\left(x-\frac 12\right)^n$ to get $\int_0^1|P|dx=\frac 1{2^n(n+1)}$ (I hope; I didn't really check it :)).


The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.

Here is what I have so far:
I will first prove the converse. We will assume KBCD is cyclic. Let KC and BD intersect at P. Then ABCP is also cyclic. By power of a point, $ MA \cdot MC \equal{} MP \cdot MB$. So we need to show $ MA \cdot CD \equal{} MB(MD \minus{} MP) \equal{} MB \cdot PD$.
Or $ \frac {MA}{MB} \equal{} \frac {PD}{CD} \implies \frac {MP}{PC} \equal{} \frac {PK}{KB}$. But by Power of a point, $ PC \cdot PK \equal{} PD \cdot PB$. So we need to show $ \frac {KB}{PB} \equal{} \frac {PD}{PM} \implies \frac {CD}{CP} \equal{} \frac {CD}{CM}$ (angle bisector theorem). But then that would mean CP=CM, where's my mistake?

[quote="dgreenb801"]
Or $ \frac {MA}{MB} \equal{} \frac {PD}{CD} \implies \frac {MP}{PC} \equal{} \frac {PK}{KB}$. [/quote]

Here, I think.

And if you prove the converse, you have to show it implies the original problem.  :)

Note:$ MH.CD\equal{}HD.MC$(1)

$ MH\plus{}HD\equal{}MD \longrightarrow MC.MH\plus{}MC.HD\equal{}MC.MD \longrightarrow$ according to(1),$ MC.MH\plus{}CD.MH\equal{}MC.MD$

multiple the equation at $ \frac{MA}{MH}$,then,$ MA.MC\plus{}MA.CD\equal{}\frac{MA.MC.MD}{MH}$(2)

according to assumption we have: $ \frac{MA.MC.MD}{MH}\equal{}MB.MD \longrightarrow MA.MC\equal{}MB.MH$

$ ABCH$ is cyclic,then $ \angle ACH\equal{}\angle ABH\equal{}\angle HCD \longrightarrow KBCD$ is cyclic and $ \angle BKC\equal{}\angle CDB$

(1) draw a circle with center $ C$ and radius $ CD$
(2) let the intersection of that circle and $ AC$ be $ E$
(3) $ MA\cdot MC$+$ MA\cdot CD$=$ AM\cdot ME$=$ BM\cdot MD$ $ \Longrightarrow$ $ A,B,E,D$ are concyclic
(4) $ \angle ABD$=$ \angle KCD$ $ \Longrightarrow$ $ K,B,C,D$ are concyclic
(5) $ \Longrightarrow\angle BKC$=$ \angle CDB$

[quote="77ant"](1) draw a circle with center $ C$ and radius $ CD$
(2) let the intersection of that circle and $ AC$ be $ E$
(3) $ AM\cdot MC$+$ AC\cdot CD$=$ AM\cdot ME$=$ BM\cdot MD$ $ \Longrightarrow$ $ A,B,E,D$ are concyclic
(4) $ \angle ABD$=$ \angle KCD$ $ \Longrightarrow$ $ K,B,C,D$ are concyclic
(5) $ \Longrightarrow\angle BKC$=$ \angle CDB$[/quote]

How did you get (3) $ AM\cdot MC$+$ AC\cdot CD$=$ AM\cdot ME$.

AM*MC+AC*CD=AM*ME=AM(MC+CE)=AM*MC+AM*CE=AM*MC+AM*CD ----> AC=AM which I think is impossible.

My solution(77ant`s idea):

Draw a circle with center C and radius CD. Let the intersection of that circle and AC be E(E lies on the extension of AC).
MA*MC+MA*CD=MA(MC+CD)=MA(MC+CE)=MA*ME
MA*ME=MB*MD so A,B,E,D are concyclic.
<CDE=<CED=<ACD/2=x
<AED=<ABD=<KCD=x, so K,B,C,D are concyclic.

--->   <BKC=<CDB. :)

my mistake in typing  :blush:  I have corrected it right now.

Very interesting question

Note that showing <BKC = <CDB is the same as showing KBCD cyclic.

Extend AC and let a point D' be on AC with CD = CD'. Now the metric condition given becomes MA(MC+CD') = MB(MD) so MA(MD') = MB(MD). This implies that quadrilateral ABD'D is cyclic.

Now let <CDD' = <CD'D = <AD'D = <ABD (by cyclic quads) = <1. Since <DCA is an exterior angle, it is the sum of <CDD' and <CD'D, so <DCA = 2<1.
Since CK is the angle bisector of <DCA, <DCK = <KCA = <1. Now let X be the intersection of CK and BD. Since <XCA = <XBA, quadrilateral BAXC is cyclic as well.

Now this question is just a simple angle chase. Let <CAB = <CXB = <2. Then <CKA = <CAB - <KCA (by exterior angles) = <2-<1.
<CDM = <CXM - <DCX (by exterior angles again) = <2-<1. Thus, <CKA = <CDM, so quadrilateral KBCD is cyclic.

Finally, this implies that <BKC = <CDB. QED

(Sorry for no diagram or LaTeX, I don't know how to do either :()

[hide=Solution]Let $P$ be the point where the circumcircle of $\triangle ABD$ intersects line $AM$ at a point other than $A$. From Power of a Point, $$MA\cdot MC + MA\cdot CD = MD\cdot MD = MA \cdot MP.$$ It follows that $MC + CD = MP$ which means that $CP = CD$. From this, we can conclude that $\angle CDP = \angle CPD$. We can easily notice that these two angles also equal $\angle DCK =\angle KCA$ by properties of exterior angles and the fact that $KC$ is an angle bisector of $\angle DCA$. Thus, $$\angle DCK = \angle DPC = \angle DPA = \angle DBA =\angle DBK$$, which implies that quadrilateral $DCBK$ is cyclic. The result follows immediately.[/hide]

[hide="solution"]Let $D'$ be the point on ray $AC$ such that $CD'=CD$ and let $A'$ be the point on ray $DC$ such that $AC=A'C$. The condition tells us by PoP that $ABD'D$ is cyclic, and by PoP, $A'$ is also on the circle. Thus, $\angle ACD = \angle A'CD' = 2\cdot \angle AD'D$. So $$\angle ABD = \angle AD'D = \dfrac{1}{2}\angle ACD = \angle KCD$$ so $BCDK$ is cyclic. Thus, $\angle BKC = \angle CDB$ as desired.[/hide]

Solution:

Let $BD$ meet $KC$ at $N$.
$ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$ implies that $\frac{MB}{MA}$ = $\frac{MC+CD}{MD}$.
Applying angle bisector theorem, we get
$\frac{MC}{MN}$ = $\frac{CD}{DN}$,

or, $\frac{MC\cdot ND}{MN\cdot MD}$ = $\frac{CD}{MD}$,

or, $\frac{MC}{MN}$ = $\frac{MC+CD}{MD}$.

Thus, $\frac{MC}{MN}$ = $\frac{MB}{MA}$,

or, $MC\cdot MA$ = $MB\cdot MN$.
Hence, by the converse of angle bisector theorem, $ABCN$ is cyclic.
This implies that $\angle ABD$ = $\angle ACN$,
or, $\angle KBD$ = $\angle KCD$,
or, $KBCD$ is cyclic.
So, $\angle BKC$ = $\angle CDB$.

[hide=Solution]
Let $CK$ intersect $BD$ at $E$, and $AC$ intersect the circumcircle of $\triangle BCD$ at $F \neq C$. Then
\[MF \cdot MC = MB \cdot MD = MA \cdot MC + MA \cdot CD\]
implies
\[MA \cdot CD = MC \cdot (MF - MA) = MC \cdot AF\]
By this and the angle bisector theorem,
\[\frac{AF}{AM} = \frac{CD}{CM} = \frac{ED}{EM}\]
so $AE \parallel DF$. Hence
\[\angle AEB = \angle FDB = \angle ACB\]
so $A$, $B$, $C$, and $E$ are concylic. Finally,
\[\angle BKC = \angle BEC - \angle KBE = \angle BEC - \angle MCE = \angle BEC - \angle DCE = \angle CDB\]
[/hide]

Nice one.
Let $\omega$ be the circumcircle of $BCD$. Let $E=AC\cap \omega$ and $K'$ be the point on $\omega$, that lies on angle bisector of $\angle ACD$. We ultimately want to show that $A,B,K'$ are collinear.
We have that $$ MA \cdot MC +MA \cdot CD = MB \cdot MD, \quad (1)$$ but by Power of a Point we have that
$$MB\cdot MD=MC\cdot ME.$$
Thus, $(1)$ takes the following form:
$$MA \cdot MC +MA \cdot CD =MC\cdot ME \Longleftrightarrow MA\cdot CD=AE\cdot MC\Longleftrightarrow
 \frac {MA}{AE} = \frac {MC}{CD} \quad (2)$$
We also have that $\triangle MCD\sim \triangle MBE$, since $\angle ECD=\angle EBD$ and $\angle EMB=\angle DMC$, thus $$\frac {MC}{CD} = \frac {MB}{BE} \quad (3)$$
By $(2)$ and $(3)$, we conclude that $\frac {MA}{AE} = \frac {MB}{BE}$, which means that $AB$ is the angle bisector of $\angle DBE$. Also, since $K'$ is the defined as intersection of $\omega$ and angle bisector of $\angle ACD$, we have that $\triangle K'DE$ is isosceles, from here we have that $K'B$ is the angle bisector of $\angle DBE$. We conclude that $A,B,K'$ are collinear.

since it came back in front page due to the last solution,
let's say the origin of the problem: [b] 2000 Belarusian MO B 10.3[/b]

also posted and solved [url=https://artofproblemsolving.com/community/c6h296380p1605623]here[/url], [url=https://artofproblemsolving.com/community/c6h213009p1175536]here [/url] and [url=https://artofproblemsolving.com/community/c6h1920219p13167817]here[/url]

Let $D'$ be on $AC$ such that $C$ lies between $M,D'$ and $CD' = CD$.
$MA.MC + MA.CD = MB.MD \implies MA.MD' = MB.MD \implies ABD'D$ is cyclic.
$\angle KBD = \angle ABD = \angle AD'D = \frac{\angle ACD}{2} = \angle KCD \implies KBCD$ is cyclic so $\angle BKC = \angle BDC$.

Let $N= BD\cap KC$. By the angle bisector theorem, we know that

$\frac{NM}{ND}= \frac{MC}{CD}$.(1)

As $MA(MC+CD)= MB\cdot MD$, by (1), we get $MA(MC+\frac{MC\cdot ND}{NM})= MB\cdot MD$
$\implies MA\cdot MC(1+\frac{ND}{NM})= MB\cdot MD$

$\implies MA\cdot MC\cdot\frac{(ND+NM)}{NM}= MB\cdot MD\implies MA\cdot MC\cdot\frac{MD}{NM}= MB\cdot MD$

$\implies MA\cdot MC= NM\cdot MB$.
So, $ABCN$ is ciclic $\implies \angle DCN=\angle NCM= \angle NBA\implies KBCD$  is ciclic, therefore $\angle BKC=\angle BDC$, QED

Let the angle bisector of $\angle{ACD}$ intersect $BD$ at $X$.

By given condition, we know that $MA=\frac{MB \cdot MD}{MC+CD}$. By angle bisector theorem, we get that $XM=\frac{MC \cdot MD}{MC+CD}$.

Then $XM \cdot MB=\frac{MB \cdot MC \cdot MD}{MC+CD}=AM \cdot MC$, implying that $ABCX$ is cyclic by converse of PoP.

Therefore $\angle{DCX}=\angle{MCX}=\angle{ABD}$, implying that $BCDK$ is cyclic, and $\angle{BKC}=\angle{BDC}$ $\blacksquare$


Prove that for all positive reals [tex]a, b, c[/tex], we have

[tex]
\displaystyle{
\frac{a}{\sqrt[3]{b^{3} + c^{3} + 6abc}} + \frac{b}{\sqrt[3]{c^{3} + a^{3} + 6abc}} + \frac{c}{\sqrt[3]{a^{3} + b^{3} + 6abc}} \ge \frac{3}{2}
}[/tex]

I've compared

[tex]
\displaystyle{
\frac{a}{\sqrt[3]{b^{3} + c^{3} + 6abc}}
}[/tex]

with forms such as

[tex]
\displaystyle{
\frac{a^{r}}{\omega_{1}a^{r} + \omega_{2}b^{r} + \omega_{2}c^{r}}
}[/tex]

and come up empty.  AM-GM and Power Mean make this inequality false rather quickly and the radicals seem to prevent any Jensen based smoothing argument, so I am stuck. . .  ideas anyone?  (I am fairly sure that my conjecture is true.)

WLOG a + b + c = 1.

Jensen applied on the convex function $f(x) = x^{-1/3}$ gives

$\sum af(b^3 + c^3 + 6abc) \geq f\left(\sum \left(ab^3 + ac^3 + 6a^2bc\right)\right)$

so we have to prove $f\left(\sum \left(ab^3 + ac^3 + 6a^2bc\right)\right) \geq 3/2$

or $\sum\left(ab^3 + ac^3 + 6a^2bc\right) \leq 8/27$.

I don't know whether the last inequality is true. (Didn't think about it yet.)

The latter inequality reduces to \[4\sum_{sym}  a^4 + 5\sum_{sym}a^3b + 24\sum_{sym}a^2b^2 \geq 33\sum_{sym} a^2bc\] and is a direct consequence of Muirhead.
QED

Very nice.  I'm stupefied - That has to be the most elegant use of weighted Jensen that I've ever seen!

[quote="Mildorf"]Prove that for all positive reals $a, b, c$, we have

\[
\displaystyle{
\frac{a}{\sqrt[3]{b^{3} + c^{3} + 6abc}} + \frac{b}{\sqrt[3]{c^{3} + a^{3} + 6abc}} + \frac{c}{\sqrt[3]{a^{3} + b^{3} + 6abc}} \ge \frac{3}{2}
}\]
[/quote]
By Holder $\left(\sum_{cyc}\frac{a}{\sqrt[3]{b^{3} + c^{3} + 6abc}} \right)^3\sum_{cyc}a(b^{3} + c^{3} + 6abc)\geq(a+b+c)^4$.
Hence, it remains to prove that $8(a+b+c)^4\geq27\sum_{cyc}(a^3b+a^3c+6a^2bc)$, which is obvious.


Use a 36 degree - 72 degree - 72 degree triangle to prove that sin 18 = (root (5) - 1)/4 and cos 36 = (root (5) + 1)/4

Use the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1335229#1335229]Golden Triangle[/url]. Or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1388778#1388778]bash it[/url].


$ (a_n)$ - bounded sequence of integers, and infinitely many $ a_n\neq 0$.

is it true that $ \sum\frac{a_n}{n!}$ is irrational ?

Consider $ a_n\equal{}n\minus{}1$. It is bounded below, yet the sum
$ \sum^\infty_{n\equal{}0}\frac{n\minus{}1}{n!}\equal{}0$ is not irrational.

the sum is 0, but the sequence is unbounded :D

lol its bounded below since $ a_n\ge \minus{}1, \forall n\in\mathbb{N}_0$, just not bounded above; I didn't know you meant "bounded above" by saying "bounded".

bounded = bounded from below and from above;

$ (a_n)$ is bounded iff there is $ M>0$ s.t. $ |a_n|<M$ for all $ N$ :)

It's the standard proof actually.

If $ |a_n|\le M$ for all $ n$ and $ \sum_{n\ge 0}{a_n\over n!} = {p\over q}$ with $ p\in\mathbb{Z},q\in\mathbb{N}$,
then for $ N > q,N > M$ and $ a_N\not = 0$ we have $ \sum_{n\le N}{a_n\over n!}\not = {p\over q}$,
or otherwise $ 0\not = {a_N\over N} = (N - 1)!\left({p\over q} - \sum_{n < N}{a_n\over n!}\right)\in\mathbb{Z}\Rightarrow 1\le \left|{a_N\over N}\right|\le {M\over N}$.
Contradiction!

So $ 0\not = N!\left({p\over q} - \sum_{n\le N}{a_n\over n!}\right)\in\mathbb{Z}$
and therefore
\begin{align*}1 & \le \left|N!\left({p\over q} - \sum_{n\le N}{a_n\over n!}\right)\right| = \left|N!\sum_{n > N}{a_n\over n!}\right| = \left|\sum_{n > N}{a_n\over (N + 1)\cdots (n - 1)n}\right| \\
& \le \sum_{k\ge 1}{|a_n|\over (N + 1)\cdots (N + k)}\le \sum_{k\ge 1}{M\over (N + 1)^k} = {M\over N}.\end{align*}
Contradiction!

thanks very much !

one more question: $ e$ is known to be transcendental.
Is every sum of this form transcendental too ?

I don't ask for a proof here, maybe there is a some kind of general theorem about it ?


I am a high school geometry teacher. We use Glencoe Geometry Integration and Application. I supplement (a lot) and my main supplementary materials are:

Introduction to Geometry from AoPS (as well as some geometry material from Vol 1 and Vol 2)
Geometry for Enjoyment and Challenge
Geometry: Plane and Solid by Jeffrey Weeks
Challenging Problems in Geometry by Posamentier

Some other textbooks are alright such as Geometry by Larson. Does anyone else know of any good geometry books in the context of the list above?

HOw about the McDougal Littell Geometry book by Larson, Kenton, Stiff, and those guys?

I used Geometry For Enjoyment and Challenge as well.


In circle O, Chord AB has length 2x. Chord CD bisects AB and intersects it at point E. Given that CE = 1, prove that CD > AB.

[hide]
$ CE\cdot ED=AE\cdot EB\Rightarrow ED=x^{2}$.
Then $ CD=CE+ED=1+x^{2}>2x=AB$.
[/hide]


Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt{\frac{x^2y}{y\plus{}z}}\plus{}\sqrt{\frac{y^2z}{z\plus{}x}}\plus{}\sqrt{\frac{z^2x}{x\plus{}y}} \le\sqrt{\frac 32(x^2\plus{}y^2\plus{}z^2)}\]

[quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt {\frac {x^2y}{y \plus{} z}} \plus{} \sqrt {\frac {y^2z}{z \plus{} x}} \plus{} \sqrt {\frac {z^2x}{x \plus{} y}} \le\sqrt {\frac 32(x^2 \plus{} y^2 \plus{} z^2)}
\]
[/quote]
I have solved it during the lesson. :D
Let $ x \plus{} y \plus{} z \equal{} 1.$ Then $ \sum_{cyc}\sqrt {\frac {x^2y}{y \plus{} z}}\leq\sqrt {\sum_{cyc}\frac {xy}{y \plus{} z}}.$
Id est, it remains to prove that $ 2\sum_{cyc}(x \plus{} y \plus{} z)xy(x \plus{} y)(x \plus{} z)\leq3(x^2 \plus{} y^2 \plus{} z^2)(x \plus{} y)(x \plus{} z)(y \plus{} z),$
which equivalent to the following
$ \sum_{cyc}(x^4y \plus{} 3x^4z \minus{} x^3y^2 \plus{} x^3z^2 \minus{} 4x^2y^2z)\geq0,$ which obviously true.

[quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt {\frac {x^2y}{y \plus{} z}} \plus{} \sqrt {\frac {y^2z}{z \plus{} x}} \plus{} \sqrt {\frac {z^2x}{x \plus{} y}} \le\sqrt {\frac 32(x^2 \plus{} y^2 \plus{} z^2)}
\]
[/quote]
By Cauchy-Schwarz inequality, we have:
$ (\sqrt {\frac {x^2y}{y \plus{} z}} \plus{} \sqrt {\frac {y^2z}{z \plus{} x}} \plus{} \sqrt {\frac {z^2x}{x \plus{} y}})^2 \le (xy\plus{}yz\plus{}zx)(\sum \frac{x}{y\plus{}z})$
So we need to prove:
$ (xy\plus{}yz\plus{}zx)(\sum \frac{x}{y\plus{}z}) \le \frac 32(x^2 \plus{} y^2 \plus{} z^2)$
$ \Leftrightarrow (xy\plus{}yz\plus{}zx)(\sum \frac{(x\minus{}y)^2}{2(x\plus{}z)(y\plus{}z)}) \le \frac{3}{4}((x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2)$
$ \Leftrightarrow \sum S_x(y\minus{}z)^2 \ge 0$
$ S_x\equal{}\frac{3}{2}\minus{}\frac{xy\plus{}yz\plus{}zx}{(y\plus{}x)(z\plus{}x)} >0$
 $ S_y\equal{}\frac{3}{2}\minus{}\frac{xy\plus{}yz\plus{}zx}{(x\plus{}y)(x\plus{}z)} >0$
$ S_z\equal{}\frac{3}{2}\minus{}\frac{xy\plus{}yz\plus{}zx}{(y\plus{}z)(x\plus{}z)} >0$
So we are done! :)

[quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt {\frac {x^2y}{y \plus{} z}} \plus{} \sqrt {\frac {y^2z}{z \plus{} x}} \plus{} \sqrt {\frac {z^2x}{x \plus{} y}} \le\sqrt {\frac 32(x^2 \plus{} y^2 \plus{} z^2)}
\]
[/quote]
By CBS we have
$ LHS\le \sqrt{(x^2\plus{}y^2\plus{}z^2)(\sum_{cyc}\frac y{y\plus{}z})}\le RHS$ because $ \sum_{cyc}\frac y{y\plus{}z}\le \sqrt{\frac3{2}}$   :D

[quote="NguyenDungTN"][quote="dduclam"][quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \sqrt {\frac {x^2y}{y \plus{} z}} \plus{} \sqrt {\frac {y^2z}{z \plus{} x}} \plus{} \sqrt {\frac {z^2x}{x \plus{} y}} \le\sqrt {\frac 32(x^2 \plus{} y^2 \plus{} z^2)}
\]
[/quote]
By CBS we have
$ LHS\le \sqrt {(x^2 \plus{} y^2 \plus{} z^2)(\sum_{cyc}\frac y{y \plus{} z})}\le RHS$ because $ \sum_{cyc}\frac y{y \plus{} z}\le \sqrt {\frac3{2}}$   :D[/quote]
It's wrong! Try x=y=z![/quote]

Sorry,it's
$ LHS\le \sqrt {(x^2 \plus{} y^2 \plus{} z^2)(\sum_{cyc}\frac y{y \plus{} z})}\le RHS$ because $ \sum_{cyc}\frac y{y \plus{} z}\le \frac3{2}$
but this is not true,too  :(


I understand how to use mass points in a triangle with maybe to cevians, but when we get more than that, I've been getting confused.  The problem I'm working on involves a median to one side, and then 3 lines splitting another side into 4 congruent segments.

I guess if you could explain like if you have a line divided into 4 segments and you're given the ratios of them (a:b:c:d) what would the use of mass points look like?

And is there a chapter in AoPS that talks about mass points?

Generally speaking, the way to approach such a problem, where you have multiple points of intersection of cevians within the triangle, is to apply mass points about each intersection point.  That is, take all cevians that intersect at one of the intersection points, apply mass points to find the ratios of those pieces, then repeat.  After doing this, you will probably be able to combine the various ratios you've found to get the ratios you want.


The figure below is a $ 4 \times 4$ grid of points.
[asy]unitsize(15);
  for ( int x = 1; x <= 4; ++x )
    for ( int y = 1; y <= 4; ++y )
      dot((x, y));[/asy]Each pair of horizontally adjacent or vertically adjacent points are distance 1 apart.  In the plane of this grid, how many circles of radius 1 pass through exactly two of these grid points?

For a unit circle to pass through two grid points, the distance between those two points A and B must be at most 2. There are three ways for this to occur: A and B are adjacent horizontally or vertically, A and B are adjacent diagonally, or A and B are a distance 2 apart either horizontally or vertically.

Consider two horizontally adjacent points A and B. There are exactly two unit circles that pass through A and B, with centers O and O', and triangles OAB and O'AB are both equilateral triangles with side length 1. Points O and O' are not a unit length from any other points of the 4x4 grid (their distance to two other lattice points is less than one, and greater than one to other lattice points).

There are 12 pairs of horizontally adjacent points in the grid, each pair is on exactly two unit circles that do not coincide with any other points from the grid. This accounts for 24 such circles. Likewise, there are 24 unit circles, each contains two vertically adjacent points from the grid.

The center of any unit circle passing through two diagonally adjacent grid points must be a point of the grid.

If the center is one of the 4 corner points, then the unit circle does not pass through any other grid point.

Otherwise, the circle passes through other grid points.

A unit circle that passes through two points at distance 2 apart must pass through at least one other grid point.

Thus the total number of circles to be counted is 24 + 24 + 4 = 52.

I have counted 3 times but every time I found only 24 circles. Where I made the mistake I am not getting.

Can you explain your method?  Then we can see where you went astray.  You can also study Albert's solution above.  You can also look over the "official" solutions here:
[url]https://mathprize.atfoundation.org/resources[/url].

[hide=sol]
Note that two points in the grid can only be $1, \sqrt{2}, \sqrt{3}, 2$ apart(or else they cannot both be points of a circle of radius $1$). There are $48$ circles of radius $1$ with distance of $1$, and $4$ of distance $\sqrt{2}$. The other two cases do not work. Thus the answer is $52$

[/hide]


trigonometry
vector
complex numbers
geometry proposed
geometry

[b]See[/b] the [u]Armo's Lemma[/u] posted at the [u]Tiks's proposed problem[/u] from
http://www.mathlinks.ro/Forum/viewtopic.php?t=38041

[b]A generalization.[/b] On the sides of the triangle $ABC$ externally we construct
the triangles $ABF,\ ACE,\ BCD$ such that $DB=DC$, $m(\widehat {BCD})=m(\widehat {CBD})=z$, ${m(\widehat {BAF})=m(\widehat CAE})=x$, $m(\widehat {ABF})=m(\widehat {ACE})=y$ and $x+y+z=90^{\circ}$.
Prove that $DE=DF$ and $m(\widehat {EDF})=2y$.

[b]Remark.[/b] For $z=0\ (x+y=90^{\circ})$, i.e. $D\in (BC),\ DB=DC;$
$FA\perp FB,\ EA\perp EC;\ \widehat {BAF}\equiv \widehat {CAE}$ we obtain the [b]Armo's Lemma[/b].

$\blacksquare \ \Longrightarrow$ Now [i]Sir Tiks[/i] can offer a generalization of own proposed problem.
I will make the same thing. Come in competition, Tiks ! Who make more quickly ?

Ok! Ph-An ;)
And don't says my "sir" :D ,pleas.

I think that it should be
<DEF = <DFE = 90-y

Let$B_1\in{AF},BB_1\perp{AF}$

$C_1\in{AE},CC_1\perp{AE}$

$M\in{BC},BM=CM$

$D_1\in {DM},D_1M=DM$

$F_1\in{FD_1},FF_1=DF_1$

$E_1\in{ED_1},D_1E_1=EE_1$

Armo’s lemme=>$F_1B_1=FM$,$\angle{F_1MB_1}=z$and

$E_1M=E_1C_1$,$\angle{E_1MC_1}=z$and

$MB_1=MC_1,\angle{B_1MC_1}=2(y+z)$=>$\triangle{F_1B_1M}=\triangle{E_1C_1M}$=>

$F_1M=E_1M$=>$FD=2F_1M=2E_1M=DE$ and $\angle{FDE}=\angle{F_1ME_1}=2(y+z)-(z+z)=2y$

[color=red]REMARK:[/color]Armos’ lemme also the  problem of shortlist[url=http://www.kalva.demon.co.uk/short/sh82.html]www.kalva.demon.co.uk/short/sh82.html[/url]problem7 ;)  ;)

Hi ! Sorry, [i]Jpe[/i], I forgot the digit $\underline {\overline {\left| 2\right|}}$ ! Thanks. I corrected above.

Here is [u]my (metrical) solution[/u] (independent of the [i]Armo's lemma[/i]).

[i]The Sinus' theorem[/i] applied in the triangles $ABF$, $ACE$, $BCD$ respectively $\Longrightarrow$

$\frac{FA}{\sin y}=\frac{FB}{\sin x}=\frac{c}{\cos z}$, $\frac{EA}{\sin y}=\frac{EC}{\sin x}=\frac{b}{\cos z}$, $DB=DC=\frac{a}{2\cos z}.$

The [i]generalized Pitagora's theorem[/i] applied to the sides $DF$, $DE$, $EF$ of the triangles

$DBC$, $DCE$, $AEF$ respectively $\Longrightarrow$ $DE=DF=\frac{EF}{2\sin y}=\frac{\lambda}{2\cos z}$,

where $\lambda ^2=a^2\cos 2x +2(b^2+c^2)\sin ^2 x+4S\sin 2x\ (\lambda\le b+c)$. Finally,

$\phi=m(\widehat {EDF}),\ \sin \frac{\phi}{2}=\frac{FE}{2\cdot DF}=\sin y\Longrightarrow \phi =2y\Longrightarrow m(\widehat {EDF})=2y.$

[u]Remark.[/u] There are a [u]syntetical solution[/u] (see the [u]Ph-An's solution[/u] , independently of [i]A.'s lemma[/i] and the [u]Tiks' solution[/u], but dependently of the [i]A.'s lemma[/i]), a [u]solution with complex numbers[/u], a [u]vectorial solution[/u] and a [u]analytical solution[/u]. In the my opinion, the best from these is a [u]Tiks' solution[/u] (with the defect of the dependence upon [i]Armo's lemma[/i]). I came to the conclusion that Tiks will must find a [b]independent[/b] (of the [i]A.'s lemma[/i]) solution. Tiks, I wish you every success !

Ph-An I have solution with complex nambers and solution with vectors,but I think it isn't interesting :(  :)

I forgot and the vectorial solution. I added it above. At the first you find a syntetical solution for the geometrical locus from http://www.mathlinks.ro/Forum/viewtopic.php?t=61911. I think that I didn't understood: wish you an another problem for a competiton between I and you ? On no account ! I joke with you. I respect you as a friend, and no as a disputant ! [b]The love is not to be trifled with ![/b]

Ok,I see ;) ,but if you can solve this problem[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61826]www.mathlinks.ro/Forum/viewtopic.php?t=61826[/url][color=darkblue]PLEASE[/color]!!!!


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complex analysis unsolved

Let $ A\subseteq\mathbb{C}$ be an open set and $ (g_{n}) ,n\equal{}1,2,3..$ a [b]normal[/b] sequence of analytic functions $ g_{n}: A\rightarrow\mathbb{C}$
Suppose $ Reg_n(z)\rightarrow Reg(z)$ for all $ z\in A$.

Prove that $ g'_{n}\rightarrow g'$ uniformly on the compact subsets of $ A$