Acceleration is defined as change in velocity divided by time.  So $ a=(15-5)/6=10/6=5/3\mbox{ m/s^{2}}$.

Do you know what acceleration is? It is basically a change in velocity. It is measured in $ ms^{\minus{}2}$ (m being metre, the standard). The $ s^{\minus{}2}$ might be confusing, and $ s^{\minus{}1}$ may also be confusing, if you can imagine velocity as a rate, or in other words a fraction, a fraction of speed on time, such like this $ \frac{speed}{time}$, we could ask our self what the units maybe, lets say they are the second and metre; then the rate is $ m/s$ or, in fraction form, $ \frac{m}{s}$.

Now, the second is in the bottom, and so if this were two multiples, then it would be $ m \times \frac{1}{s}$, and with the $ \frac{1}{s}$, as we know from the basic indice law $ a^{\minus{}1} \equal{} \frac{1}{a}$, we know that now $ \frac{1}{s}$ can be written as $ s^{\minus{}1}$. So thats basically it.

With your question, you must know a simple formula, that is $ a \equal{} \frac{v\minus{}u}{t}$. Where

$ a$ is acceleration
$ v$ is the final velocity
$ u$ is the initial velocity
$ t$ is time

Now let me explain this formula. This formula is for average acceleration, the difference between average acceleration and just acceleration is that average acceleration is just simply the average, because, as we know, if the acceleration is not constant, then it is always changing, and this gives us a curve, which we use calculus to deal with. Now we should understand that acceleration is the rate of change of velocity, the chance in velocity is the top part in this equation, that is the change, and the bottom is the other change, as acceleration is in metres per second... per second.

[quote="slowlearner"][size=117]what is the acceleration of a car that moves from a speed of 5.0 m/s to a speed of 15 m/s during a time of 6.0 s? Thanks for any help.[/size][/quote]

In this question, it is obvious that the initial velocity is $ 5.0 ms^{\minus{}1}$, the final velocity is $ 15.0 ms^{\minus{}1}$ and the time is 6.0 seconds. Lets sum this up

$ v \equal{} 15.0 ms^{\minus{}1}$
$ u \equal{} 5.0 ms^{\minus{}1}$
$ t \equal{} 6.0 s$

If we plug this into our equation to find the velocity we get

$ a \equal{} \frac{v\minus{}u}{t}$

$ a \equal{} \frac{15.0 ms^{\minus{}1}\minus{}5.0 ms^{\minus{}1}}{6.0 s}$

$ a \equal{} \frac{10.0ms^{\minus{}1}}{6.0 s}$

$ a \equal{} 1\frac{2}{3}  ms^{\minus{}2}$

This is roughly $ 1.7 ms^{\minus{}2}$

wat r u trying to explain or especially to whom?
 :mad:


A circular clock with centre C has twelve hour marks.  The one o'clock and two o'clock hour marks are O and T respectively.  The circle centre O thorugh T cuts CT again at D.  Where does OD cut the circumference of the clockface? Explain!

[hide][color=indigo]This was hard! After I did the stuff in the problem, I named the point where OD intersected the circle again F. Then I drew OC, CF, OT, and TF.
1. COT is an isosceles triangle (all radii of a circle are equal). :ang: OCT=1/12*360 = 30 :^o: so :ang: COT and :ang: OTC equal 75 :^o:
2. OTD is an isosceles triangle (same reason as #1). :ang: OTD = :ang: OTC = 75 :^o:
3. CT is a straight line so 180 :^o: - 75 :^o: = :ang: CDO. :ang: CDO = 105 :^o:
4. DOT = 30 :^o: , and 75 - 30 = 45, so :ang: COF =  45 :^o:
5. COF is another isosceles triangle (still the same reason).  CFO = COF = 45 :^o: , so OFC = 90.
6. 90 = 1/4*360 so it is three hours from O to F. That means F is at the 4 o'clock mark.  :D [/color][/hide]

This wasnt as easy as I thought but I got the ans too
point F :first:


[b] Let [/b]  $ A \in M_{n} \mathbb{(R)}$ , I is identity nxn matrix ,$ m \in N^{*}$ ,
$ m \geq 2$
Prove that :
  $ Lim_{n \to \infty} \sum_{k = 0}^{2008n} {[(A + \frac {I}{n + k})^{m} - A^{m}]} = m.ln(2009).A^{m - 1}$

$ \sum_{k \equal{} 0}^{2008n} {[(A \plus{} \frac {I}{n \plus{} k})^{m} \minus{} A^{m}]}\equal{}mA^{m\minus{}1}\sum_{k \equal{} 0}^{2008n}\dfrac{1}{n\plus{}k}\plus{}\dfrac{m(m\minus{}1)}{2}A^{m\minus{}2}\sum_{k \equal{} 0}^{2008n}\dfrac{1}{(n\plus{}k)^2}\plus{}\cdots$.
$ \sum_{k \equal{} 0}^{2008n}\dfrac{1}{n\plus{}k}\equal{}ln(2009n)\plus{}\gamma\minus{}ln(n)\minus{}\gamma\plus{}O(1/n)\equal{}ln(2009)\plus{}O(1/n)$.
$ \sum_{k \equal{} 0}^{2008n}\dfrac{1}{(n\plus{}k)^2}<\sum_{k \equal{} n}^{\infty}\dfrac{1}{(k)^2}$ converges to $ 0$ when $ n$ tends to infinity.
The factors of $ A^{m\minus{}3},\cdots$ converge also to $ 0$.                            QED


I am stuck... need help on how to start this, or something... anything...

question:

It is a common experience to hear the sound of a low flying airplane, and look at the wrong place in the sky to see the plane.  Suppose that a plane is traveling directly at you at a speed of 200 mph and an altitude of 3000 feet, and you hear the sound at what seems to be an angle of inclination of 20 degrees.  At what angle should you actually look in order to see the plane?  Let the speed of sound be 1100 ft/sec.

Is this homework? If so, AoPS is not the place to post it.

[hide="Hint"]Calculate the distance the plane flies as the sound comes to you, and it's position relative to where it was before.[/hide]

no its not homework....

 its a challenge problem for a class I am taking... its not graded... just a challenge...  I dont want the answer, just some help getting it started...

[hide="To Start"]Draw a line $ m$ on your paper. Place a point, $ P$, on the line that represents yourself. Draw a line $ d$ that goes $ 20^{\circ}$ (relative to $ m$) either to the left or right. Construct a line $ n$ parallel to $ m$ and at a distance of 3000 ft (scaled down, of course) from it. The intersection of $ n$ and $ d$ (call this $ Q$) is the place at which you looked. Now calculate how long it took for the sound to get to $ P$ from $ Q$, and then find how much the plane would have flown in this time. Finally, mark off a point on $ n$ where the plane is when you hear it's sound coming from $ Q$, and calculate the angle.[/hide]


[quote="GunUltimateID"]how to solve this equation by hand
2x^3-3x^2-12x+12=0[/quote]

As Dr. Graubner pointed out, the solutions are very ugly. To attempt to solve this by hand, you can try memorizing (Although I would [b]not[/b] recommend it) the [url=http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots]Cubic Formula[/url].

[quote="Neil30z"]Instead of using the "cubic formula" we can try to use the rational zero theorem, Descarte's rule of signs. If that doesn't work, you can always try to take a look at the how the sum of the roots, product of the roots, and the sum of the product of the roots taken 2 at a time (in this case), relate to your roots.
(intuition)[/quote]

Yes, but this is clearly not a case for intuition (Look at the solutions).


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

I am looking for someone to solve the question below using steps as you go through the lesson. I would like to learn how to solve similar questions on my own. Ready?

How many points are equidistant from point R(3,1)
and S(3, 6) and also 4 units from the origin?

As people explained in another topic, the locus of points equidistant from two points is the perpendicular bisector of those two points.

The perpendicular bisector of $RS$ is the line $y=3\frac{1}{2}$ (see why?).

The locus of points distance 4 from the origin is a circle.

We need to find the number of points that this line and this circle meet at. The maximum number that any circle and line can meet at is 2- and if we draw the circle (compass point on the origin, circle passing through $(0,4)$) and the line we see that this circle and this line do meet in exactly two places.

There is no need to find the points (!) but if we did want to find them, we could draw a right-angled triangle with one corner at the origin and the other end of the hypotenuse on one of the two points, and do pythagoras.

How did you get y = 3.5?

Where did 3.5 come from?

You stated:

"There is no need to find the points (!) but if we did want to find them, we could draw a right-angled triangle with one corner at the origin and the other end of the hypotenuse on one of the two points, and do pythagoras."

Can you give me an example using the steps in the above statement?

$y=3.5$ is the line that contains the points equidistant from $(3,1)$ and $(3,6).$ To find it, you first find the midpoint of the two points, find their slope, and then use point-slope form to find the perpendicular bisector. The midpoint is at $\frac72$ and the slope is $\text{undefined},$ so the perpendicular bisector is $y=\frac72,$ or $y=3.5.$

To find the points you are looking for, you construct a circle with radius $4$ centered at the origin, and find it's intersections with the line $y=3.5.$

Tim was saying you could construct a triangle with one side being a radius of the circle (from the origin to one of the intersections with the line), the second being a segment on the $x$-axis, and the third being a vertical segment from the point of intersection to the $x$-axis.

$\left(\frac72\right)^{2}+x^{2}=4^{2}\Rightarrow x=\frac{\sqrt{15}}2$

The $x$ coordinates are $\pm\frac{\sqrt{15}}2,$ so the points of intersection are $\left(\frac{\sqrt{15}}2,\frac72\right)$ and $\left(-\frac{\sqrt{15}}2,\frac72\right).$

Strange-looking points but I follow your reasoning.


geometry
perpendicular bisector
angle bisector
geometry proposed

Let a triangle $ ABC$, a point $ P$ inside triangle so that $ \angle{ABP}\equal{}\angle{ACP}$. Denote $ D,E$ are midpoint of $ AP,BC$. Prove that: $ DE$ pass through perpendicular feet of $ P$ on bisector of angle $ \angle{BAC}$.

Circle (E) with center E and diameter AP cuts AB, AC, ED at $ U \in AB,\ V \in AC,\ Q \in ED.$ Since AP is a diameter of (E), the points U, V are perpendicular projections of P on AB, AC and $ PQ \perp AQ.$ By solution of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=174227]Two opposite angles of a quadrilateral equal[/url] also discussed in [url]http://www.mathlinks.ro/viewtopic.php?t=172417[/url], posts 6-7, ED is the perpendicular bisector of UV. This makes the $ \triangle QUV$ is isosceles with QU = QV. From the circle (E), $ \angle QAC \equiv \angle QAV \equal{} \angle QUV \equal{} \angle QVU \equal{} \angle QAU \equiv QAB,$ AQ bisects the $ \angle CAB.$

Let $ M$, $ N$ and $ K$ the feet of perpendicular lines from $ P$ to $ AB$, $ AC$ and the angle bisector of angle $ BAC$, respectively. We can easily to prove that $ DK$ is the perpendicular bisector of the segment $ MN$ $ (1)$.
Now, let $ R$, $ S$ be the mid-points of the segments $ PB$, $ PC$, respectively. We have $ \triangle MRE\equal{}\triangle ESL$ (S.A.S). It implies $ EM\equal{}EN$, i.e. $ E$ lies on the perpendicular bisector of the segment $ MN$ $ (2)$.
From $ (1)$ and $ (2)$, we conclude that $ D$, $ E$ and $ K$ are collinear as desired.


Noah was walking down Elm St. He then made 2 left turns and 3 right turns and is now facing east. Every turn is at a $ 90^\circ$ angle. What direction was Noah originally going on Elm St.?

2 left turns+3 right turns=1 90 degree turn to the right. If you turn 90 degrees right once and face east, you originally faced north.

Answer: Noah was originally going $ \boxed{\text{north}}$ on Elm St.


Given the triangle $ ABC$ such that $ (AB)^2\plus{}(AC)^2\equal{}(BC)^2$ and $ (AB)\equal{}(AC)\equal{}10$, consider the line $ t$ , passing through the midpoint of the larger side and perpendicular to the plane $ \alpha$ of the given triangle. Given the point $ P$, such that $ P\in t$ , calculate the volume of the pyramid $ PABC$ $ (P\cancel{\in}\alpha)$, knowing that the radius of the inscribed sphere is $ \sqrt{2}$ .

Let $ M$ be the midpoint of $ BC$, and $ h: \equal{} MP$.

Since $ MA\equal{}MB\equal{}MC\equal{}5\sqrt{2}$, we get $ PA\equal{}PB\equal{}PC\equal{}\sqrt{h^2\plus{}50}$

If $ V$ is the volume and $ S$ the surface area of the pyramid, then

$ V\equal{}{Sr\over 3}$

where $ r$ is the inradius.

The triangles $ PAB$ and $ PAC$ are isosceles, with the sides $ 10,\sqrt{h^2\plus{}50},\sqrt{h^2\plus{}50}$, so we easily find that their combined area is $ S'\equal{}10\sqrt{h^2\plus{}25}$

The area of $ \triangle PBC$ is $ S''\equal{}{5h\sqrt{2}}$

The area of $ \triangle ABC$ is $ S'''\equal{}50$

So we get $ S\equal{}S'\plus{}S''\plus{}S'''\equal{}50\plus{}5h\sqrt{2}\plus{}10\sqrt{h^2\plus{}25}$

On the other hand, $ V\equal{}{S'''h\over 3}\equal{}{50h\over 3}$

Therefore we get the equation

$ 50h\equal{}(50\plus{}5h\sqrt{2}\plus{}10\sqrt{h^2\plus{}25})\sqrt{2}$

After simplyfing and rearranging, we get

$ 4h\minus{}5\sqrt{2}\equal{}\sqrt{2h^2\plus{}50}\implies h\equal{}{20\sqrt{2}\over 7}$

Now $ V\equal{}{50h\over 3}\equal{}{1000\sqrt{2}\over 21}$


quadratics
geometry
perpendicular bisector
algebra

Let $S_1$ be a semicircle with centre $O$ and diameter $AB$.A circle $C_1$ with centre $P$ is drawn, tangent to $S_1$, and tangent to $AB$ at $O$. A semicircle $S_2$ is drawn, with centre $Q$ on $AB$, tangent to $S_1$ and to $C_1$. A circle $C_2$ with centre $R$ is drawn, internally tangent to $S_1$ and externally tangent to $S_2$ and $C_1$. Prove that $OPRQ$ is a rectangle.

Consider the semicircles $S_1, S_2$ as full circles. Let $R_1 = \dfrac{AB}{2}$ be the radius of the initial circle $S_1$ and $r_1, R_2, r_2$ radii of the remaining circles $C_1, S_2, C_2$. Obviously, the circle $C_1$ is centered on the perpendicular bisector of the segment AB and its radius is $r_1 = PO = \dfrac{R_1}{2}$. Let U be the tangency point of the circles $S_1, C_1$ and let D be the other intersection of the circle $S_2$ with the line AB different from the point B. Invert the circles $C_1, S_2$ in the circle $S_1$. The circle $C_1$ tangent to the inversion circle and passing through the inversion center O is carried into a line tangent to the inversion circle $S_1$ at the point U, which is parallel to the line AB. The tangency point B of the semicircles $S_1, S_2$ is on the inversion circle and it stays in place. The circle $S_2$ is carried into a circle $S_2'$ tangent to the inversion circle $S_1$ at the point B and centered on the line AB. The other intersection D of the circle $S_2$ with the line AB is carried into the other intersection D' of the inverted circle $S_2'$ with the line AB. Since the circle $S_2'$ is centered on the line AB and tangent to the line parallel to AB through the point U, i.e., at a distance $OU = R_1$, its radius is equal to $R_2' = R_1$. Thus the distance of the inverted point D' from the inversion center is equal to ${OD' = OB + BD' = R_1 + 2R_2' = 3R_1}$. By the basic property of inversion, $OD \cdot OD' = R_1^2$ and consequently,

$OD = \dfrac{R_1^2}{OD'} = \dfrac{R_1^2}{3R_1} = \dfrac{R_1}{3}$

This allows to calculate the radius $R_2$ of the circle $S_2$:

$R_2 = \dfrac{DB}{2} = \dfrac{OB - OD}{2} = \dfrac{R_1 - \dfrac{R_1}{3}}{2} = \dfrac{R_1}{3}$

The pairwise tangent circles $S_1, C_1, S_2, C_2$  form a quadruple of Soddy's circles and their radii $R_1, r_1, R_2, r_2$ are bound by the relation

$\dfrac{1}{(-R_1)^2} + \dfrac{1}{r_1^2} + \dfrac{1}{R_2^2} + \dfrac{1}{r_2^2} = \frac 1 2 \left(\dfrac{1}{-R_1} + \dfrac{1}{r_1} + \dfrac{1}{R_2} + \dfrac{1}{r_2}\right)^2$

The radius of the circle $S_1$ has to be taken with the negative sign, because its tangencies with the remaining 3 circles are internal. (If the 4 circles had all tangencies external, all radii would have positive signs.) This yields an equation for the radius of the circle $C_2$:

$\dfrac{1}{R_1^2} + \dfrac{4}{R_1^2} + \dfrac{9}{R_1^2} + \dfrac{1}{r_2^2} = \frac 1 2 \left(-\dfrac{1}{R_1} + \dfrac{2}{R_1} + \dfrac{3}{R_1} + \dfrac{1}{r_2}\right)^2$

$\dfrac{14}{R_1^2} + \dfrac{1}{r_2^2} = \dfrac{1}{2} \left(\dfrac{4}{R_1} + \dfrac{1}{r_2}\right)^2$

$14 \dfrac{r_2^2}{R_1^2} + 1 = \frac 1 2 \left(4 \dfrac{r_2}{R_1} + 1\right)^2$

Denoting $x = \dfrac{r_2}{R_1}$, we get the following quadratic equation for x:

$2(14x^2 + 1) = (4x + 1)^2$

$28x^2 + 2 = 16x^2 + 8x + 1$

$12x^2 - 8x + 1 = 0$

$x = \dfrac{8 \pm \sqrt{64 - 48}}{24} = \dfrac{8 \pm \sqr{16}}{24} = \dfrac{8 \pm 4}{24} = \dfrac{2 \pm 1}{6} = \frac 1 2\ \text{or}\ \frac 1 6$

The first root obviously corresponds to a circle congruent to the circle $C_1$ symmetrical to it with respect to the line AB. By the problem conditions (the semicircles), this root is not acceptable. Thus $r_2 = \dfrac{R_1}{6}$. Now we calculate the sides of the quadrilateral OPRQ:

$OP = r_1 = \dfrac{R_1}{2}$

$QR = R_2 + r_2 = \dfrac{R_1}{3} + \dfrac{R_1}{6} = \dfrac{R_1}{2}$

$PR = r_1 + r_2 = \dfrac{R_1}{2} + \dfrac{R_1}{6} = \frac 2 3\ R_1$

$OQ = OD + R_2 = \dfrac{R_1}{3} + \dfrac{R_1}{3} = \frac 2 3\ R_1$

Since the opposite sides of the quadrilateral OPRQ are equal, it is a parallelogram. Since one of its internal angles $\angle POQ = 90^\circ$ is right, it is a rectangle.

Hi shobber, nice to meet you again! :clap:

I think that this sangaku-like problem can be solved in a simple way:

Let  $OA=OB$ be $r$. We have $OP = \dfrac{r}{2}$. If $OQ=t$ we have
\[ \sqrt{t^2 + \dfrac{{r^2 }} {4}} = PQ = \dfrac{r} {2} + (r - t) = \dfrac{{3r}} {2} - t.  \]
Hence  $OQ = t = \dfrac{2r}{3}$ and $QB=\dfrac{r}{3}$. We draw
parallel lines  $PR$ and $QR$ to $OQ$ and $OP$, respectively, and
we get the intersection point $R$ such that $RP=QO=\dfrac{2r}{3}$
and $RQ=PO=\dfrac{r}{2}$. Being $\dfrac{{2r}} {3} - \dfrac{r} {2} = \dfrac{r} {6} = \dfrac{r} {2} - \dfrac{r} {3}$, the circle
 $C_2$ with center  $R$ and radius $\dfrac{r}{6}$ is
tangent to $C_1$ y $S_2$. On the other way, the relations \[ r - OR = r - \sqrt{\dfrac{{4r^2 }} {9} + \dfrac{{r^2 }} {4}} = r - \dfrac{{5r}} {6} = \dfrac{r} {6},  \] tell us that the circle $C_2$ is tangent to $S_1$.

So, by construction, $OPRQ$ is a rectangle.  :o


A sequence of integers is obtained by starting with three digits $ d_1$,$ d_2$, and $ d_3$. The fourth term of the sequence is the units digit of $ d_1 \plus{} d_2 \plus{} d_3$. Each succeeding term is the units digit of the sum of the three previous terms. What numbers belong in the first, second and third positions respectively to complete the sequence.

\[ \text{  \_\_ },\text{ \_\_   },\text{ \_\_   },1,1,1\]

[quote="GameBot"]A sequence of integers is obtained by starting with three digits $ d_1$,$ d_2$, and $ d_3$. The fourth term of the sequence is the units digit of $ d_1 \plus{} d_2 \plus{} d_3$. Each succeeding term is the units digit of the sum of the three previous terms. What numbers belong in the first, second and third positions respectively to complete the sequence.
\[ \text{ \_\_ },\text{\_\_ },\text{\_\_ },1,1,1
\]
[/quote]

Let them be a,b,c

it is clear that a+b+c>10

if a+b+c=11:

This means that b+c+1=11 so b+c=10.

If b=4 c=6 which fails
b=3 c=7 fails
b=2 c=8 fails
b=1 c=9 SUCEEDS!

Thus we have 1,1,9.


How can I prove that
in any triangle the following is true

$(cosA+cosB+cosC)^2\leq sin ^2 A+sin^ 2 B+sin ^2 C$  ??

Thanks for your time and assistance!

 :)  :)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 (some of the latest posts).

  darij

[quote="zhaobin"][quote="tmbtw"]i have a ineq (like that) :
prove that:
 (cosA/2)^2+(cosB/2)^2+(cosC/2)^2 = >(sinA/2 +sinB/2 + sinC/2)^2[/quote]
they are equal more and less :P[/quote]

What do you mean??  :?

Probably that the ineq is not valid in any triangle??

try let
$x=\frac{\pi}2 -\frac A 2$
And y,z similar
you will find they are equal :)


function
induction
algebra proposed
algebra

A sequence of real numbers $\{a_n\}_n$ is called a [i]bs[/i] sequence if $a_n = |a_{n+1} - a_{n+2}|$, for all $n\geq 0$. Prove that a bs sequence is bounded if and only if the function $f$ given by $f(n,k)=a_na_k(a_n-a_k)$, for all $n,k\geq 0$  is the null function.

[i]Mihai Baluna - ISL 2004[/i]

The condition simply says that the set $\{a_n|n\in\mathbb N^*\}$ has at most one non-null value. It's clear that if this condition hiolds, then the sequence is bounded, so the other implication is the interesting one.

We assume the sequence is bounded. There are two cases:

[b]Case 1[/b]: the maximum $M$ of $\{a_n\}$ is reached.

We can see that if $a_n=M$, then $a_{n+1}$ is either $M$ or $0$, and the same holds for $a_{n-1}$. Moreover, $a_{n+1}=M\iff a_{n-1}=0$, and we can construct the sequence in two possible ways, in both directions, according to whether $a_{n+1}=M$ or $0$. In both these situations, the sequence contains only two values, in both directions, $0$ and $M$, so the conclusion is proven.

[b]Case 2[/b]: the maximum $M$ is never reached.

Let $\varepsilon>0$ be small enough (small enough that $2(M-\varepsilon)>M$, for instance) so that we can find $a_n=M-\varepsilon$. It's easy to show that we can also find $n$ s.t. $a_n=M-\varepsilon$ and $a_{n+1}=t,\ 0<t<M-\varepsilon$. We thus have $a_{n+2}=M-\varepsilon+t$, and if $a_{n+3}=M-\varepsilon$, then $a_{n+4}=2(M-\varepsilon)+t>M$ , which is a contradiction. This shows that, in fact, $a_{n+3}=M-\varepsilon+2t, a_{n+4}=t$, and we can repeat the procedure with $\varepsilon$ replaced with $\varepsilon-2t$ (which must necessarily be $>0$). We find that $\varepsilon-2nt>0,\ \forall n\ge 0$, which is absurd.

Hope it's Ok.

I hope my memory still works (I try to stay beyond Alzheimer's club of Pierre, but it becomes harder and harder), but I think this problem was already solved on the forum. I think Darij posted it, after a kind of TST of Germany. Probably I'm wrong, but it sounds too familiar to me and I do not have the ISL 2004.  ;)

Yes, I believe that problem asked whether such a sequence could be bounded :?.

yea it was with the conditions given that $x_0$ and $x_1$ and positive and not equal to each other. And the answer is no. Its in the ISL 2004, and came up in the British FST.

Only the main idea: write all $x_i-s$ in terms of  the first 2 distinct  $x_i-s$. By induction one can easily obtain that all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\in \mathbb{N}$ . There  are finitely many $(m,n)-s$ with $m<\frac M{x_1}, n<\frac M{x_2}$, with $M$ being the bound so  $M$ is attained . Let $x_{i+1}$ be the first term that achieves the maximum, so $0<x_i<x_{i+1}$, $x_{i+2}=x_{i+1}-x_i$, since it cannot be their sum (M is max) so $x_{i+3}=2x_{i+1}-x_i>x_{i+1}=M$, false.

There are still some cases to be considered for the first few terms of the sequence and for the case $0$ occurs in the sequence.

[quote="ikap"]all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\in \mathbb{N}$[/quote]

Why is that? ($m,n \in \mathbb{N}$) I can only prove it with $m,n \in \mathbb{Z}$ and that doesn't help very much...


logarithms
limit
real analysis
real analysis unsolved

Prove that the sequence [tex](x_n)_{n\geq 2}[/tex], defined by [tex]x_n=\log_n n![/tex] does not contain arithmetic progressions with infinitely many terms.

First of all, prove that $x_{n+1}-x_n<1,\ \forall n\ge 2$ and $\displaystyle \lim_{n\to\infty}(x_{n+1}-x_n)=1$. Then try to show that a sequence which satisfies these conditions can’t contain infinite arithmetic progressions. It’s not hard at all.


topology
limit
advanced fields
advanced fields unsolved

$f: \mathbb{C}\to \mathbb{C}$ continuous and ${\lim_{z \to \infty}f(z) = 0}$.
Show that $f$ has a fixed point.

[mod edit]: use \infty for $\infty$

This is a version of the Brouwer theorem; $f$ is bounded, so it maps some sufficiently large disk into itself.

To prove this version, look at the winding number of $f(z)-z$ around zero along the circle of radius $r$ centered at zero. For small $r$, either zero is a fixed point or the winding number is zero. For large $r$, the winding number is 1. The winding number is continuous in $r$ as long as $f(z)-z$ is never zero on the circle; since it can't be globally continuous, $f(z)-z=0$ somewhere.


induction
linear algebra
matrix
geometry
rectangle
strong induction

Hello

Prove the following Identity

$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$

[quote="mathgeniuse^ln(x)"]Hello

Prove the following Identity

$(F_{n+1})^{2}+(F_{n-1})^{2}=F_{2n}$[/quote]
[hide="idea"]Induction[/hide]
[hide="uglier idea"]Binet's formula[/hide]

[quote="mathgeniuse^ln(x)"]Oh, I forgot to mention.

Is it possible to do it without the ugly, but useful formula of Binet?

Thanks[/quote]
This seems definitely like an induction problem.

I have not thought about this problem (excluding induction and Binet's formula) yet, but the idea that first came to my mind was to simplify the problem by subtracting $2F_{n-1}F_{n+1}$ and using $\left( \begin{matrix}1&1\\1&0\end{matrix}\right)^{n}=\left(\begin{matrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{matrix}\right)$ along with $\det \left(\prod A_{i}\right)=\prod\det (A_{i})$.

[quote="iamagenius"]hey good idea!

 :P[/quote]

That was totally retarded, annoying, unnecessary, and spammy.

A well known Fibonacci formula is:

$F_{2n+1}=F_{n}^{2}+F_{n+1}^{2}$

I don't know if that's what you were going for. I don't think your formula works.

Edit: I lied, I think rnwang is right.

For induction (this is for the identity I had in my previous post), it looks like you'll have to prove

$F_{n+1}\cdot (F_{n+2}+F_{n})=F_{2n+2}$.

Use the definition and you should get a more general formula:

\[F_{2n}=F_{j+1}F_{2n-j}+F_{j}F_{2n-(j+1)}\]

So for $j=n$, we have

\[\begin{aligned}F_{2n}&=F_{n+1}F_{n}+F_{n}F_{n-1}\\ &=(F_{n+1})^{2}-(F_{n-1})^{2}\end{aligned}\]

[quote="rnwang2"][quote="tjhance"]... Am I being totally stupid or wut

$n=3$
$(F_{4})^{2}+(F_{2})^{2}=F_{6}$
$3^{2}+1^{2}=8$
$9+1=8$
$10=8$

??????????[/quote]
It seems that the identity should be

$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$.[/quote]
No wonder, I tried proving it by induction and I went in circles...

[quote="rnwang2"][quote="tjhance"]... Am I being totally stupid or wut

$n=3$
$(F_{4})^{2}+(F_{2})^{2}=F_{6}$
$3^{2}+1^{2}=8$
$9+1=8$
$10=8$

??????????[/quote]
It seems that the identity should be

$(F_{n+1})^{2}-(F_{n-1})^{2}=F_{2n}$.[/quote]

Yeah, my mistake.

But I have a question, is there any way that might be possible to prove without any formulae.

[quote="mathgeniuse^ln(x)"]But I have a question, is there any way that might be possible to prove without any formulae.[/quote]

You might be able to use the fact that $F_{n}$ represents the number of ways to tile a board of length $n$ with $1\times 1$ squares and $2\times 1$ rectangles

whats wrong with binet's formula...it is not messy computation at all

[quote="paladin8"][quote="boxedexe"]The so-called "formula" in your post is incorrect (try n=2).[/quote]

Sorry, that formula has the indices shifted (i.e. $F_{0}= 1$, $F_{1}= 1$, etc.).[/quote]But, if you shift the index, you will make this statement false: $F_{5n}$ is divisible by 5 for all $n\in\mathbb{N}$ :D

[quote="mathgeniuse^ln(x)"][quote="paladin8"][hide]First, rewrite the equality as

$F_{2n}= F_{n+1}^{2}-(F_{n+1}-F_{n})^{2}= F_{n}(2F_{n+1}-F_{n})$.

The base case is easy, since we have $F_{4}= 3 = 2^{2}-1^{2}= F_{3}^{2}-F_{1}^{2}$. Now we will show that it is true for $n = k+1$ assuming it is true for $n \le k$. We have

$F_{2k+2}= F_{2k+1}+F_{2k}= 3F_{2k}-F_{2k-2}$.

Now we apply the inductive hypothesis to get

$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-F_{k-1}(2F_{k}-F_{k-1})$,

which we can simplify with the substitution $F_{k-1}= F_{k+1}-F_{k}$ to get

$F_{2k+2}= 3F_{k}(2F_{k+1}-F_{k})-(F_{k+1}-F_{k})(3F_{k}-F_{k+1}) = 2F_{k}F_{k+1}+F_{k+1}^{2}$

and finally from $F_{k}= F_{k+2}-F_{k+1}$ we get

$F_{2k+2}= F_{k+1}(2F_{k+2}-F_{k})$

as desired.[/hide]

The interpretation of the Fibonnaci sequence that renatzu mentioned leads to a simple proof that $F_{2n}= F_{n}^{2}+F_{n-1}^{2}$, but I'm not quite sure how to apply it to this problem.[/quote]

Okay that is a good proof.

Are we allowed to assume the base cases are true and then prove that for $n=k+1$, we can use the cases for $n=k-1$ and $n=k$ and prove our statement?

Is that okay?

I didn't know that, that is why I didn't approach it like that.[/quote]

Yes, though my proof is incomplete because I need two base cases :oops:. That's ok, though, you can fill that part in. Formally, it's called strong induction (assuming it is true for $n \le k$ and then proving it for $n = k+1$). One flawed "proof" using strong induction is this.

Prove: $a^{n}= 1$ for all positive reals $a$ and integers $n \ge 0$.

Base Case: $a^{0}= 1$ trivially.

Induction Step: $a^{k+1}= \frac{a^{k}\cdot a^{k}}{a^{k-1}}= \frac{1 \cdot 1}{1}= 1$ by the induction hypothesis that $a^{n}= 1$ for all $n \le k$.

Were teh two base cases n=2 and n=k?

Thanks

And paladin, I think since we know that my formula is true, we can go to the next case and subtract the two to get it, I don't konw.

[quote="mathgeniuse^ln(x)"]Were teh two base cases n=2 and n=k?[/quote]

The base case is n=1, since he assumes it holds true for k and k+1..


Find the number of distinct integer-sided triangles with the perimeter of 2007.

[hide="hint"]the length of each of the sides is less than 1004[/hide]


Έχετε να προτείνετε ένα καλό θέμα με διανύσματα , να περιέχει (εσωτερικό γινόμενο και συντεταγμένες) :lol:

Parathetw me tin eukairia mia wraia askisi pou eixa dei pro kairou.
Dinontai dio statheres eutheies sto epipedo oi opoies temnontai sto simeio O. Apo to O ksekinoun stis tesseris kateuthinseis twn aksonwn ta kinita A,B,C,D me diaforetikes taxitites. Na apodeiksete oti i posotita

$\frac{AB^{2}-BC^{2}+CD^{2}-DA^{2}}{AC\cdot BD}$, paramenei statheri.

Alexandros

Check μερικές ασκήσεις εδώ:

http://www.telemath.gr/mathematical_lycee/mathematical_lycee_second/pos_tech/index.htm

Νομίζω ότι βγαίνει με θεώρημα τεμνουσών.

Θύμισέ μου ποιό είναι το θεώρημα τεμνουσών ?? τί λέει ακριβώς???

Απο εξωτερικό σημείο Μ του κύκλου φέρνουμε τις τέμνουσες που τεμνουν τον κύκλο στα σημεία Α, Β , Γ ,Δ
τότε ΜΑ ΜΒ =ΜΓ ΜΔ

σε ευχαριστώ πολύ αλλά αυτό το θεώρημα των τεμνουσών να ξέρεις ότι το συμπληρώνουμε με την έκφραση του κύκλου ΘΕΩΡΗΜΑ ΤΕΜΝΟΥΣΩΝ ΤΟΥ ΚΥΚΛΟΥ για να το ξέχωρίζουμε με άλλα θεωρήματα τεμνουσών οκ??
πχ Μενέλαος

[b][size=109]giannis,[/size][/b] καλωσόρισες στην παρέα μας και καλή συνέχεια.

Κώστας Βήττας.


$f(x)=a\log_{b}(x)$, where $a>0$, $b>0$, and $b\not=1$.
If $f \left(\frac{16}{81}\right)=-2$ and $a=\log_{k}b$, determine the positve value of $k$.

[quote="tiredepartment"]$ f(x) \equal{} a\log_{b}(x)$, where $ a > 0$, $ b > 0$, and $ b\not \equal{} 1$.
If $ f \left(\frac {16}{81}\right) \equal{} \minus{} 2$ and $ a \equal{} \log_{k}b$, determine the positve value of $ k$.[/quote]
[hide]
$ \minus{} 2 \equal{} a\frac {\log(16) \minus{} log(81)}{\log b}$
$ a \equal{} \frac {\log b }{\log k}$
$ \minus{} 1 \equal{} \log_k (\frac {4}{9})$
$ k \equal{} \left( \frac {4}{9} \right)^{ \minus{} 1}$
$ k \equal{} \frac {9}{4}$
[/hide]


We just got the graded papers back from the NSA today.  It'll take us a couple of days to get the scores and comments scanned in, so we hope to have the Round 3 scores and feedback released on Thursday or Friday of this week.

Round 3 scores and feedback are now available on the USAMTS website.

We will make an announcement about AIME qualification during the week of Feb 23-27.  We will develop a procedure so that students who did not resubmit Round 2 by mail will still be eligible to qualify for the AIME.  [b]Please do not inquire about this prior to our announcement[/b].


[quote="woohooxd"]Can someone tell me why the answer is E?!?
Peterson's online test told me it was E, however I dont see how that is possible.

If –2 < a < 3 and 5 < b < 10, which of the following represents the range of possible values for ab ?

a. 2 < ab < 13

b. –2 < ab < 11

c. –10 < ab < 30

d. –20 < ab < 15

e. –20 < ab < 30[/quote]

[hide]To find the range of possible values, try to find the smallest possible value for $ab$, and then the largest possible value.

Pretending that the $>$ and $<$ signs were $\ge$ and $\le$ instead, the smallest possible value of $ab$ would be when $a=-2$ and $b=10$, when $ab=-20$.

The largest possible value of $ab$ would be when $a=3$ and $b=10$, when $ab=30$

So the range RIGHT NOW would be $-20\le{ab}\le30$

Changing back the signs wouldn't do us any different, so the answer must be E. [/hide]

[hide]We want the smaller number ab to be negative. The more neagative it is, the better, so pick -2 for a (to make it negative) and 10 from b to make it -20.
If you picked -2 and 5, you would've gotten -10, not smaller than -20.

The biggest number ab should be positive and the biggest numbers, 3 and 10, which make 30.

so -20<ab<30

[/hide]

but one thing doesnt make sense....
if -2<a<3...it doesnt mean -2 is one of the choices...because it isnt saying that a is greater than or equal to. IF -2 was greater than equal to a and less than equal to 3, then we can use -2.
But from that equation only possible values are -1,0,1,2
Same goes for 5<b<10. We cant use 10 because it isnt less than or equal to b. only numbers possible would be 6,7,8,9
right? am i wrong?

That's a good point, but the problem never said that we had to use integer. So the number can be -1.99999999999..... and still work or 2.999999999.....


limit
function
calculus
derivative
logarithms
algebra
rational function

[b]1. The problem statement, all variables and given/known data[/b]
Determine whether the series $ \sum_{n = 2}^{\infty}a_n$ is absolutely,conditionally convergent or divergent
${ a_n = \frac {( - 1)^n}{\sqrt {n}}(\frac {2n}{n + 1})^\pi}$

[b]3. The attempt at a solution[/b]
from Abel's test.$ c_n = \frac {( - 1)^n}{\sqrt {n}}$is convergent.and

${ b_n = (\frac {2n}{n + 1})^\pi} = \frac {2^{\pi}}{(1 + \frac {1}{n})^{\pi}} = \frac {2^{\pi}}{1 + \frac {\pi}{n} + o(\frac {1}{n^2})}$.Which has limit $ 2^{\pi}$.So a_n is convergent.

$ |a_n| = \frac {2^{\pi}}{\sqrt {n}(1 + \frac {1}{n})^{\pi}} = \frac {2^{\pi}}{\sqrt {n} + \frac {\pi}{\sqrt {n}} + O(\frac {1}{\sqrt {n}n})}$

I don't know exactly but it seems to me that the last equation is divergent.So a_n is conditionally convergent.

absolutely divergent because  equivalent to $ a_{n}$ is what you have already post $ c_{n}$
and conditionally convergent because $ a_{n}\minus{}>0$ and changed sign

Since $ \left(\frac {2n}{n \plus{} 1}\right)^{\pi}$ tends to a number other than $ 0$ or $ \infty,$ then we should just start looking at it as if it were that constant. We need to "de-clutter" and just remove it from our sight.

The workhorse test for "de-cluttering" is the limit comparison test: if $ a_n$ and $ b_n$ are both eventually positive and if $ \lim_{n\to\infty}\frac {a_n}{b_n} \equal{} L,$ then $ \sum a_n$ and $ \sum b_n$ either both converge or both diverge.

Hence in this case, $ \sum_{n \equal{} 2}^{\infty}|a_n|$ is limit comparable to $ \sum \frac1{\sqrt {n}},$ which is known to diverge.

So the original series does not converge absolutely. Since the alternating series test does apply to it, then the series converges conditionally.

As part of checking that the alternating series test holds, we do have to show that $ |a_n|$ is an eventually decreasing function of $ n.$ In this case, that's a little annoying. For completeness, let me accept the annoyance and show that:

Let $ f(x) \equal{} \frac1{\sqrt {x}}\left(\frac {2x}{x \plus{} 1}\right)^{\pi} \equal{} x^{ \minus{} 1/2}\left(2 \minus{} \frac2{x \plus{} 1}\right)^{\pi}.$

Then $ f'(x) \equal{} \minus{} \frac12x^{ \minus{} 3/2}\left(2 \minus{} \frac2{x \plus{} 1}\right)^{\pi} \plus{} x^{ \minus{} 1/2}\pi\left(2 \minus{} \frac2{x \plus{} 1}\right)^{\pi \minus{} 1}\left(\frac2{(x \plus{} 1)^2}\right)$

$ \equal{} x^{ \minus{} 1/2}\left(2 \minus{} \frac2{x \plus{} 1}\right)^{\pi \minus{} 1} \left[ \minus{} \frac1{2x}\left(2 \minus{} \frac2{x \plus{} 1}\right) \plus{} \frac {2\pi}{(x \plus{} 1)^2}\right].$

Examination of the quantity in the brackets to find the dominant terms (namely $ \minus{} \frac1x$ ) shows that this derivative is negative for all sufficiently large $ x.$ Hence $ f(x)$ is an eventually decreasing function and thus $ |a_n| \equal{} f(n)$ is an eventually decreasing sequence.

If this question were to appear on one of my tests, I would not expect my students to go to that much trouble to show the monotonicity.

I think there's an old theorem (Hardy?) that says that any rational function of finitely many terms like $ x^p$, $ \log x$, and $ a^x$ is monotone for sufficiently large $ x$. Not particularly interesting or surprising, but would be helpful here.
Of course, the monotonicity part was needed for AST, not for LCT.

[color=green][Oops. Thanks for pointing that out. I've edited. - K.M.][/color]

Thanks a lot.I haven't known about de-cluttering method.


The hour hand of a broken clock points to $ 5$ and the minute hand to $ 12$. To the nearest percent what part of the clock face is shaded?

[asy]draw(circle((0,0),1));
filldraw(arc((0,0),1,-75,90)--(0,1)--(0,0)--(0.25,-0.96)--cycle,gray);
draw((0,0)--(0,1),Arrow);
draw((0,0)--(0.25,-0.96),Arrow);[/asy]

The shaded portion represents $ \frac{5}{12}$ of the whole area. $ \frac{1}{12} \equal{} 8.\overline{3}\%$. Multiplying this by $ 5$ gives $ 41.\overline{6}\% \approx \boxed{42\%}$


modular arithmetic
abstract algebra
number theory
relatively prime
group theory

Find the remainder when the integer 1x3x5x7x....x2003x2005 is divided by 1000

obviously , the product include the number 125 so we only need to find the remainder of $ 1x3x...x2005$ when we divided it by 8
notice that every numbers in the product can be rewritten as $ 2k\plus{}1$ where k from 0 to 1002
if k is divided by 4 we found $ 2k\plus{}1 \equiv 1 (mod 8)$
if k \equiv 1 (mod 4) we found $ 2k\plus{}1 \equiv 3(mod 8)$
if k \equiv 2(mod 4) we found $ 2k\plus{}1 \equiv 5 (mod 8)$
if k \equiv 3 (mod 4) we found $ 2k\plus{}1 \equiv 7 (mod 8)$
take a look at the product $ 1x3x5x7....2005 \equal{} (1x3x5x7)x..x(1993x1995x1997x1999)$$ 2001x2003x2005 \equiv (1x3x5x7)^{250}x2001x2003x2005 \equiv 7 (mod 8)$
and now you got the answer ...

The factor or $ 125$ is a very good observation! However, your answer is incorrect. We can fix that by cleaning up the rest of the proof a bit.

[hide]Now, note that every odd number is $ \pm 1, \pm 3$ in mod 8.

For ease of counting, let us consider all odds less than 2008 first. There are 2008/8 = 251 sets of $ \pm1, \pm 3$ factors in mod 8 within this product.

However:
- one factor of $ \minus{} 1 \equiv 2007$ was artificially added for the above calculation; we will remove this now
- one factor of  $ \minus{} 3 \equiv 125$ will be removed for now; we will multiply in the 125 later
Thus, these two will only have 250 factors. So, we have, in $ \pmod{8}$:

$ \frac {\Pi}{125}$ $ \equiv ( \minus{} 1)^{250} (1)^{251} ( \minus{} 3)^{250} (3)^{251} \equiv 3^{250} 3^{251} \equiv 9^{250}(3) \equiv 1^{250} (3) \equiv 3$ $ \pmod{8}$

Hence, our product is of the form $ \prod \equal{} 125 (8k \plus{} 3) \equiv 125 \cdot 3 \equiv 375 \pmod {1000}$

[b]@HTA[/b] You cannot count the 125 towards the mod 8 calculation, since it is already being set aside to facilitate the final multiplication.
[/hide]

[quote="t0rajir0u"][b]Follow-up:[/b]  Find the remainder when the integer $ 1 \times 3 \times 7 \times ... \times 2003 \times 2007$ is divided by $ 1000$, where the product is taken over all integers from $ 1$ to $ 2007$ not divisible by $ 2$ or by $ 5$.[/quote]

[hide]  Hmm... if we take this as 2001 x 2003 x 2007 x ( 1 x 3 x...999) x (1001 x 1003 x 1999)

note that the last two terms are the same mod 1000, and each congruent to the product of all of the number relatively prime to 1000.  And note that since the numbers relatively prime to 1000 form an abelian group under multiplication (mod 1000) we can just pair each number up with its inverse.  Either a number has an inverse other than itself, in which case we can match it up with the number in its term (ie match 3 with 667 and 1003 with 1667) or if it is its own inverse we can match it up with itself from the other term (ie 999 and 1999).  From there we have 2001 x 2003 x 2007 x (a bunch of 1's mod 1000) which comes out to be 21 mod 1000
[/hide]

[quote="Farenhajt"][quote="Hamster1800"]Am I missing something or is $ 375\equiv7\pmod8$, hence HTA's solution is fine[/quote]

He found the remainders modulo 125, which is zero, and modulo 8, which is 7 - but what's asked for is COMBINED remainder modulo 1000, so we must find a number which is divisible by 125, but gives remainder 7 when divided by 8, and that's 375.[/quote]

As far as I've seen on these forums, finishing a solution with the Chinese Remainder Theorem is often left out of these types of problems. Therefore, I felt like there was nothing lacking in his solution.

If that's the case, that habit should be fought against, since abandoning a problem without giving a final answer is worth almost nothing.

It's not abandoning so much as just knowing that you just have to check $ 0, 125, 250, 375, 500, 625, 750, 875$ and check which one gives you the correct remainder upon division by $ 8$. Alternatively, you could algorithmically arrive at the answer, but that is less common.

Also, why do you say it has almost no value? The key insights were made, all that was left out was some computation. Clearly, on a competition you would finish the problem, but when it's practice, I see nothing wrong with leaving out computation.

Sketch of a solution is one thing, but going all the way through, and then abandoning it at the very end out of sheer laziness (no other reason possible) really looks bad.

I'm sorry, I misunderstood HTA's intention. Personally, I think that even if a proof is not fully completed, the full intent of the solution sketch should not have to be implied. A better choice of words and neater typesetting on his part (or more time spent on my part) would have made the intent more obvious.

i'm sorry , i didn't give the final answer because i was running out of time , i had to go to school , so here is the rest of the solution
as Hamster1800 said , we have to check 0;125;250;375;500;625;750;875 to see which one gives the correct remainder , but notice that it gives the remainder 7 upon division by 8 so it must be an odd number,
$ 125 \equiv 5 (mod 8)$
$ 375 \equiv 7 (mod 8)$
so the answer is 375

p/s it's no need to check all of them , stop when you find the desire answer


1)A piece of copper of density 8.9g/cm^3 floats on mercury of density 13.6g/cm^3.What fraction of its volume will be above the surface?

2)A helium ballom is tied to  the back seat by a massless string of a car and at rest,the bllon floats upwards.Now,the car is closed.If the car is suddenly accelerated,where will the helium ballon go?(Front,back,sideways?)

[quote="shadysaysurspammed"]
2)A helium ballom is tied to  the back seat by a massless string of a car and at rest,the bllon floats upwards.Now,the car is closed.If the car is suddenly accelerated,where will the helium ballon go?(Front,back,sideways?)[/quote]
[color=blue]The answer is [b]FRONT[/b][/color]

1)  I would do this one by assuming a mass of copper, calculating it's volume, than the volume of an equal weight of mercury and the percentage of copper's volume that is the mercury's volume is the percentage that is submerged in mercury when the copper if floating in mercury.

2)  The balloon will move back.  An object at rest remains at rest until a force acts on it.  When the car starts accelerating, no force has had an opporutnity to act on the balloon, so it will intially seem to move back, as the car moves forward.  (I've actually done this one, so they theoritical answer is confirmed by experiement.)

I'm sorry i have less time to thiunk on the 2nd one bcos i have xams right now,but answer is front(i think) bcos of archimedes.There is some air in the car that will go back and thus push the ballon forward.

Let me explain the second one:-

There is some air in the car as the question says that the balloon is floating in the car.Now we close the doors.

When we acclelrate,the air moves to the back of the car.Since the helium balloon is lighter than air and has a massless string(so no inertia,same for rubber skin),the helium balloon will move forward by buoyant force (Consider it like a cork on water).

[quote="perfect_radio"]1) We must have $F_A = G$.

$\Longleftrightarrow \rho_{\textrm{liquid}} V_{\textrm{submerged}} g = \rho_{\textrm{body}} V_{\textrm{total}}g$;

$\Longleftrightarrow \, \frac{V_{\textrm{submerged}}}{V_{\textrm{total}}} = \frac{\rho_{\textrm{body}}}{\rho_{\textrm{liquid}}}$. Right, guys?

2) I don't really understand what force pushes the balloon to the front  :blush:  I didn't learn physics when I should have.[/quote]

You are right about the first one.

In the second one,[b]buoyant force[/b] pushes the balloon
forward

Like a cork floats up on water due to buoyant force,you know about this right?So in the same way,the air pushes the balloon forward by buoyant force.

Why does a balloon float in air in the first place?Due o buyant force exerted by the air around it.If you take a balloon filled with air to the moon(where there's no atmosphere),it won't float up since there's no atmosphere.

Concerning the second problem about a helium balloon in an accelerating vehicle, you can find a useful discussion at this site:

http://www.amherst.edu/physicsqanda/Helans.htm

The link is NOW CORRECT .I am sorry about this mistake.

Also a Google search under, inertial frames helium balloon,will turn up a lot of other  similar sites.

Another dumb question, related to the link provided above: when an elevator is going upwards, you feel heavier because of what? The ??? force, or what?

Sorry for the stupid question
 :D

If you stand on a weigjhing machine,your wieght will be more than usaual if the elevator is accelerating upwards because now the downward force would be $m(g+a)$ instead of $mg$ where $a$ is the $net$ acceleration with which the elevator is going upwards.The elevator is actually going up with acceleration $g+a$,so by newton's third law,you should get weight=$m(g+a)$ :)

I searched on the net about this subject and I found that weight still stays equal to $mg$. You feel heavier because of the normal force. Kind of logical, if you think about it.

Yes,that is what i said,the weiht on the scale will be more but you won't get fatter or anything.Didn't i say newton's third law?That is basically normal force.


Let $T$ be a point inside a square $ABCD$. The lines $TA, TB, TC, TD$ meet the circumcircle of $ABCD$ again at $A',B',C',D'$, respectively
. Prove that
$A'B'.C'D'=A'D'.B'C'$.

We have:

$\triangle TA'B'\sim \triangle TBA \Longrightarrow \frac{A'B'}{a}=\frac{TB'}{TA}$,
$\triangle TC'D'\sim \triangle TDC \Longrightarrow \frac{C'D'}{a}=\frac{TC'}{TD}$,

$\triangle TA'D' \sim \triangle TDA \Longrightarrow \frac{A'D'}{a}=\frac{TA'}{TD}$,
$\triangle TB'C' \sim \triangle TCB \Longrightarrow \frac{B'C'}{a}=\frac{TB'}{TC}$,

where $AB=a$.

It follows that $A'B' \cdot C'D'=a^{2}\frac{TB' \cdot TC'}{TA \cdot TD}$, and $B'C' \cdot A'D'=a^{2}\frac{TA' \cdot TB'}{TC \cdot TD}$. Now it's enogh to show $\frac{TC'}{TA}=\frac{TA'}{TC}$, which is trivial w.r.t. power of the point theorem.


Consider all the ways of writing exactly ten times each of the numbers $0, 1, 2, \ldots , 9$ in the squares of a $10 \times 10$ board.
Find the greatest integer $n$ with the property that there is always a row or a column with $n$ different numbers.

$4$ acording to Iran's problems, and beautiful Fiachra solution:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16445[/url]


Anyone know how to solve the first question...


calculus
integration
trigonometry
calculus computations

hey,
Here are a few problems I'v tried to solve but with no success.

Could anyone help?

integrate (x^5)cos(x^3) dx (First make a substitution then use integration by parts)

integrate within the limits of 9 and 5 the expession (sin(5x))^2 (cos(5x))^2 dx  (Hint: you need to use some tig identities such as the double angle formulas for sine and cos.)

Integrate within the limits of pi/2 and 0 the expression (sinx)^5 (cosx)^20 dx

Integrate within the limits of pi/15 to 0 the expression ((sec(5x))^12)/(cot(5x)) dx

Thanks a lot

[quote="someoneELSE"]

integrate (x^5)cos(x^3) dx [/quote]
$\int x^{5}\cos{x^{3}}dx=\frac{1}{3}\int x^{3}\cos{x^{3}}d(x^{3})$. Now, $\int t\cos{t}dt=\int td\sin{t}=t\sin{t}-\int\sin{t}dt=...$


[quote="chantown"]Rusty ate an average of two hamburgers a day in January. What is the largest possible number of hamburgers that he could have eaten on January 15?[/quote]
[hide] There are 31 days in January.  So, the total number of hamburgers he at must me (average)(number of days)=2(31)=62.  So, he ate 62 hamburgers.  Now, if he doesn't have to eat any hamburgers in a day, then he could have simply ate all 62 on January 15th.  If he must eat at least once a day, then say that he ate 1 each day for the other 30 days.  So, there are 62-30(1)=32 hamburgers he ate on January 15th.
[/hide]


CGMO好像是在8月10日至15日舉行... 請問有沒有人知道題目呢？

请问有没有2004年女子数学奥林匹克(南昌)的前三名的获奖名单?

順便問一下有沒有05年CGMO的獲獎名單？

这些印题的长春人，竟然在卷纸上写华罗庚的名言？脑子长了虫子。多影响选手心情。[至少是我不爽，这不着四六的……]
不过感觉今年比去年难？

呵呵. 华老的话到没那么倒胃口.

BTW: 去年的题能不能发上来看看?

[quote="Soarer"][quote="ifai"]我們澳門也取得一銅獎 :winner_third: , 雖不算太好, 但亦不錯吧!  :D[/quote]
應該是去imo 那個女孩吧[/quote]
是的. 她在imo亦得銅獎.


Who is Kevin Chen? Everybody seems to be talking about him. What is up???? :?:

Kevin Chen was the 2007 Mathcounts national champion.


linear algebra
matrix
algebra
polynomial
quadratics

How would I solve the following:

$\vec{x'}= \begin{bmatrix}\; \; \; 0 & \; \; \; 1 & \; \; \; 2 \\-5 &-3 &-7 \\ \; \; \; 1 & \; \; \; 0 & \; \; \; 0 \end{bmatrix}\vec{x}$?

I found 3 eigenvalues using Maple.

I got it down to $\lambda^{3}+3\lambda^{2}+7\lambda-1 = 0$

How would I factor this?

I don't see where you got the polynomial from. The matrix has eigenvalues $-1,-1,-1$. (Which means that my link above is not so helpful; it assumes distinct eigenvalues.) Anyway, the general solution is $\vec x=C\exp(tA)$, and the [url=http://en.wikipedia.org/wiki/Ordinary_differential_equation]Wikipedia page[/url] explains how to use this formula.

How did you get those eigenvalues? Did you use $|A-\lambda I| = 0$ and find the determinant by expansion by minors? When I did expansion by minors, I got:

$-\lambda \left|\begin{matrix}-3-\lambda &-7 \\ 0 &-\lambda \end{matrix}\right|-\left|\begin{matrix}-5 &-7 \\ 1 &-\lambda \end{matrix}\right|+2 \left| \begin{matrix}-5 &-3-\lambda \\ 1 & 0 \end{matrix}\right|$

I ended up getting:

$\vec{x}(t) = e^{-t}\begin{bmatrix}1 \\ 1 \\-1 \end{bmatrix}+\left(t\begin{bmatrix}1 \\ 1 \\-1 \end{bmatrix}+\begin{bmatrix}-3 \\ 0 \\ 2 \end{bmatrix}\right)e^{-t}+\left(\frac{1}{2}t^{2}\begin{bmatrix}1 \\ 1 \\-1 \end{bmatrix}+t\begin{bmatrix}-3 \\ 0 \\ 2 \end{bmatrix}+\begin{bmatrix}7 \\ 0 \\-5 \end{bmatrix}\right)e^{-t}$

Does this look right?

Looks about right: the solution should have components $p_{i}(t)e^{-t}$, where $p_{i}$, $i=1,2,3$, is a quadratic polynomial.


Three pencils and two erasers cost $ \$0.60$. Two pencils and three erasers cost $ \$0.55$. How much will it cost to buy seven pencils and seven erasers?

We can set up two equations:

1) 3p + 2e = 0.60
2) 2p + 3e = 0.55

If we add them together, we get 5p + 5e = 1.15. Dividing both sides by 5, we get p + e = 0.21. Since we want 7p + 7e, we have 0.21 times 7 = 1.47.

If we add the equations together we get $ 5p \plus{} 5e \equal{} \$1.15$ Now we get $ p \plus{} e \equal{} \$.23$. So $ 7p \plus{} 7e \equal{} \$1.61$.

Edit: beaten..  :( And I think your solution is incorrect izzy.


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ diving $n$. Prove that $G$ is not simple.

It's been shown here before that if $p|n$ is the minimal prime factor of $n$, then a subgroup of index $p$ in $G$ must be normal. Hence the result.

A proof of the result grobber quoted:

Suppose $H$ is a subgroup of $G$ of index $p$ and order $n$, with every prime factor of $n$ at least $p$. Let $C$ be the set of all conjugates of $H$ in $G$; $G$ acts transitively on $C$ by conjugation. $C$ has at most $p$ elements since $aHa^{-1}=(ah)H(ah)^{-1}$ whenever $h\in H$.
This action of $G$ gives a homomorphism $\phi$ from $G$ to $S_{|C|}$. If $|C|<p$, $\phi$ must be trivial since the two groups have relatively prime orders, and $H$ is normal. If $|C|=p$, the image of $\phi$ must have order dividing $p$, and must be cyclic of order $p$. In this case, since $\phi(h)$ fixes $H$ for all $h\in H$, the kernel of $\phi$ must be $H$. This makes $H$ normal, and the whole thing is impossible anyway.


Euler
geometry
circumcircle
trapezoid
geometric transformation
reflection
incenter

Let $ ABC$ be a triangle and $ \omega$ be its circumcircle, $ H$ be its orthocentre. Let $ A'$ be the antipodal point of $ A$ w.r.t. $ \omega$. Let $ A_1$ be the intersection of $ A'H$ with $ \omega$, and $ A_2$ be the intersection of $ AH$ with $ \omega$. Points $ B_1$, $ B_2$, $ C_1$, $ C_2$ are defined similarly. Prove that $ A_1A_2$, $ B_1B_2$ and $ C_1C_2$ are concurrent at a point lying on the Euler line of triangle $ ABC$.

Because of $ H\equiv BB_{2}\cap CC_{2}\cap B'B_{1}\cap C'C_{1},$ we conclude that the quadrilaterals $ BCB'C'$ and $ B_{2}C_{2}B_{1}C_{1},$ are perspective with respect to the point $ H$ and so, it is easy to show, by [b][size=100]Desarques’s theorem[/size][/b], that the points $ O\equiv BB'\cap CC'$ and $ S\equiv B_{1}B_{2}\cap C_{1}C_{2},$ as the intersection points of their diagonals respectively, are also perspective with respect to the point $ H$ $ ($ the line connecting the points $ O,$ $ S,$ passes through the point $ H$ $ ).$

That is, the point $ S\equiv B_{1}B_{2}\cap C_{1}C_{2},$ lies on the line segment $ OH,$ as the [b][size=100]Euler’s line[/size][/b] of the given triangle $ \bigtriangleup ABC.$

By the same way, we can prove that the intersection point of the line segments $ A_{1}A_{2},$ $ B_{1}B_{2}$ $ ($ or $ A_{1}A_{2},$ $ C_{1}C_{2}$ $ ),$ lies also on $ OH$ and the proof is completed.

$ \bullet$ I see by the drawing that the point $ S$ here, is coincided with the concurrency point in the probem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=174452]Three concurrent lines at apoint on the Euler line (1)[/url], but I have not in mind any proof about it.

Kostas Vittas.

[b][size=100]EDIT:[/size][/b] [color=DarkBlue]My apologies dear all my friends, because of the above solution I presented, is absolutely wrong.

I am very sorry for a such that my stupid mistake.[/color]

Kostas Vittas.

$ AA',\  AA_2$ are isogonals of AB, AC through O, H $ \Longrightarrow$ $ BA'A_2C$ is a cyclic trapezoid with $ A'A_2 \parallel BC,$ isosceles. H is a reflection of $ A_2$ $ \Longrightarrow$ $ BA'CH$ is a parellelogram, its diagonals A'H, BC cutting each other in half at $ M_a \in A'A_1.$ Using $ A'A_1 \perp AA_1$ and $ AA_2 \perp BC,$ we have

$ \angle A_1OA_2 \equal{} 2 \angle A_1AA_2 \equal{} 2 \angle A'M_aB \equal{} 2 \angle A_2M_aC \equal{} \angle A_2M_aH \equiv A_2M_aA_1$

$ \Longrightarrow A_1OM_aA_2$ is cyclic. Inversion of its circumcircle in the circle (O) is the line $ A_1A_2$ passing through the inversion $ T_a$ of $ M_a$ is (O), which is the intersection of tangents to (O) at B, C. The circumcevian $ \triangle A_2B_2C_2$ of the $ \triangle ABC$ WRT the orthocenter H is centrally similar to the tangential $ \triangle T_aT_bT_c$ of the $ \triangle ABC,$ because they have parallel sides (perpendicular to the circumcircle radii OA, OB, OC). $ T_aA_2 \equiv A_1A_2,\ T_bB_2 \equiv B_1B_2,\ T_cC_2 \equiv C_1C_2$ are concurent in the similarity center S of these two centrally similar triangles. The line OH connecting their incenters O, H (or their corresponding excenters) passes through their similarity center S.

If $ O_a,\ O_b,\ O_c$ are circumcenters of the $ \triangle OBC,\ \triangle OCA,\ \triangle OAB$ and $ H_a,\ H_b,\ H_c$ feet of the A-, B-, C-altitudes of the original $ \triangle ABC,$ then the $ \triangle T_aT_bT_c \sim \triangle O_aO_bO_c$ are centrally similar with similarity center O and coefficient 2, while the $ \triangle A_2B_2C_2 \sim \triangle H_aH_bH_c$ are centrally similar with with similarity center H and coefficient 2. Thus S is similarity center of the figures $ HH_aH_bH_cA_2B_2C_2 \sim OO_aO_bO_cT_aT_bT_c.$

This problem appeared on Iran, 2005:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49901[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138610[/url]

Dear all my friends.

I would like to present another approach, by [b][size=100]Pascal’s theorem[/size][/b] and I hope that I am not mistaken (again).

[b][size=100][color=DarkBlue]GENERAL  PROBLEM. – A triangle $ \bigtriangleup ABC$ is given, with its circumcircle $ (\omega)$ and let $ H,$ $ O$ be, two arbitrary points inwardly to it. We denote as $ A',$ $ A_{1},$ $ A_{2},$ the intersection points of $ (\omega),$ from the line segments $ AO,$ $ A'H,$ $ AH$ respectively and similarly we define the points $ B',$ $ B_{1},$ $ B_{2}$ and $ C',$ $ C_{1},$ $ C_{2}.$ Prove that the line segments $ A_{1}A_{2},$ $ B_{1}B_{2},$ $ C_{1}C_{2},$ are concurrent at one point lying on the line segment $ OH$[/color][/size][/b].

[b][size=100]PROOF[/size][/b]. – We consider the cyclic non convex hexagon $ A_{1}A_{2}CC_{2}C_{1}A'$ and so, by [b][size=100]Pascal’s theorem[/size][/b], we have that the points $ S\equiv A_{1}A_{2}\cap C_{2}C_{1}$ and $ H\equiv CC_{2}\cap A'A_{1}$ and $ P\equiv A_{2}C\cap C_{1}A',$ are collinear.

Similarly, from the cyclic non convex hexagon $ AA_{2}CC'C_{1}A',$ we have that the points $ H\equiv AA_{2}\cap C'C_{1}$ and $ O\equiv CC'\cap A'A$ and $ P\equiv A_{2}C\cap C_{1}A',$ are collinear.

Hence, we conclude that the line segments $ SHP$ and $ HOP$ are coincided, because of they have two common points.

That is, the point $ S\equiv A_{1}A_{2}\cap C_{1}C_{2},$ lies on the line segment $ OH.$

By the same way, we can prove that the intersection point of $ A_{1}A_{2},$ $ B_{1}B_{2},$ lies also on $ OH$ $ ($ considering the cyclic non convex hexagons $ A_{1}A_{2}B'B_{1}B_{2}A'$ and $ AA_{2}B'BB_{2}A'$ $ )$ and the proof is completed.

Kostas Vittas.

This problem is a simple and nice example of poles and polars (more poles than polars),
 yet rather elementary, so that D.Grinberg probably will not include it ( out of inflationary
 considerations, I guess) in his chapter  on the subject in the book that has already
 become a  bibliographic rarity (even prepayments are not accepted).

  M.T.


[quote="Mathematicus"]Could someone explain how this works? Thanks

tan(Arctan(x/2) + Arctan(2x/3)) = ((x/2) + (2x/3))/(1-(x/2)(2x/3))[/quote]
Use formula $\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}$.  :D

[hide]here you go...
Assume that $Arctan\frac{x}{2}=\alpha$ and $Arctan\frac{2x}{3}=\beta$ thus $tan\alpha=\frac{x}{2}$ and $tan\beta=\frac{2x}{3}$, and we have :
$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$
and this is exactly what you wanted....[/hide]

[quote="Mathematicus"]Could you show/refer me to proof of this? I hate memorizing formulas that I can't explain.[/quote]
You can do it brutally by repalcing $\tan t$ by $\frac{\sin t}{\cos t}$ and multiplying by both denominators.

Or like this:

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\cos\alpha\cos\beta(\tan\alpha+\tan\beta)$
$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos\alpha\cos\beta(1-\tan\alpha\tan\beta)$

Then just divide the two equalities.


Let $f(x)$ be a monic polynomial of degree $n\ge{1}$ with complex coefficients. Let $x_{1},x_{2},\dots,x_{n}$ be ther $n$ comple roots of $f(x)$. The discriminant $D(f)$ of $f(x)$ is the complex number
$D(f)=\prod_{1\le{i}<j\le{n}}(x_{i}-x_{j})^{2}$.
Express the discriminant of $f(x^{2})$ in terms of $D(f)$.

[hide="so far i have"]
$D(f(x^{2}))=2^{2n}D(f)^{2}(x_{1}x_{2}\dots x_{n})$
i think this is the answer since it seems to work for $n=1,2,3$, so i am thinking i should try induction, but don't know how to proceed. also i am not sure that is an acceptable answer since the last part is not in terms of $D(f)$
[/hide]

f(x^2) is not that different...it is just the coefficients "spaced-out" so to speak...that summation is not that hard to evaluate...there are 2n coefficients and x_k =0 if k is odd

Sorry but I do not follow what  you are trying to say.


it is known, that a1 + a2 +.... + an >= 2^n - 1 and every ai > 0
prove that 1/a1 + 1/a2 +... + 1/an < 2

are you sure it's true, J
i can prove it with the condition
x_i {i=1->n} > 0
x_1+x_2+..+x_i \geq 2^i - 1

yes i apologize, i actually forgot to add that every ai > 0

what about the condition x_1+x_2+...+x_i \geq 2^i - 1


Let $x, y>0$ such that $x\neq e$ and $x^{y}=y^{x}$. Prove that $x^{y} >e^{e}.$ ;)

well, seems like your counterexample works here. This is the form appeared in G.M.A, 3/2005. :oops:


Let P be a polinomyal with integer coefficients such that P(n)>n. find all polinomyal P with that condition such that the list 1,P(1),PP(1),PPP(1)........contains a multiple of m for every m.

It is Iran TST 2004
I had sent it 2 months ago

but do you have a solution? i think its to easy for being iran tst 2004, what do you think?

I think I solved it when Omid posted it. Let's denote those nasty compositions of $P$ by $a_n=P^{(n)}(1)$. Let $k=P(1)-1>0$. There is some $n$ s.t. $k|a_n$. However, from the fact that, in general, $m-n|P(m)-P(n)$, we get $k|a_2-a_1|a_3-a_2|\ldots|a_n-a_{n-1}$, so $k|a_{n-1}$. We repeat the process until we get $k|a_1=P(1)\Rightarrow P(1)-1|P(1)\iff P(1)=2$. As you can see, the method works perfectly fine if we take $k=a_{t+1}-a_t$ for any $t\ge 1$. In the end we get $P(n)=n+1$.

Grobber: Indeed your solution is nicer than mine. Thanks.

Of course this problem had the least mark form the other questions.

I know this problem for ages, and it was proposed for 9th grade here in Russia.


Let the series (a_n)n from N be difined like so: a_(n+2)=(a_(n+1))^(1/a_n) for any n natural  with a_0 , a_1 from (0, oo). Prove that the series (a_n) converges and compute lim n-->oo (a_n).

bye

I don't think it's too hard: just consider the cases a_1<1, a_1=1, a_1>1. If a_1<1 then a_n<1 for any n>=1. Because of this the sequence will become strictly decreasing for n>=2, so the sequence converges to L. Because 0<=L<1 and L=L^(1/L) we find L=0. If a_1=1 then a_n=1 for any n>=1. The sequence converges to L=1. If a_1>1 a_n>1 for any n>=1. The sequence becomes strictly decreasing for n>=2 so the sequence converges to L>=1. Because L=L^(1/L) we get L=1. Think it's correct?


AMC USA(J)MO USAMO geometry Harvard college MIT

[quote="tAsDoOm"]I live in the same region that you do, ComplexZeta, and I've had enough...[/quote]

Come on over to Minnesota. Heck, I could probably drum up a job for you coaching the young mathletes of my acquaintance (unless the local guru, Sly Si, locks it up first).

After I returned from my first year of MOP (this was the year with 10 qualifiers from SFBA), Upstate NY seemed like the middle of nowhere math-people-wise.  I didn't have any local math clubs or teams for high school students, and the nearest USAMO qualifier I knew of lived 3 hours away.  Things have improved since, but at the time I would have killed to live in the Bay Area.

[quote="ComplexZeta"]Is it really worth it? Sure, you'll qualify for USAMO. But that's about it. You won't have any other mathematicians your age within a few hundred miles. I think most of the time people do better when they have competition around them. I certainly benefited tremendously by living in one of the acknowledged mathematical centers of the country. The alternative would be to work a bit harder over the next year and qualify for USAMO the way most people do -- by getting a high USAMO index. Then you get a whole lot more -- you'll be better at math, you'll have mathematical friends, and you'll still make the USAMO. Isn't that better?[/quote]

I'll agree.  Nevada is no fun if you're interested in maths - no one else likes math, you get made fun of (sometimes), you lose motivation, etc.

I'll admit, I was quite silly at the time.  (And I had ulterior motives.)  I was highly deprived as a freshman -- there was no AoPS community then :-P  But having friendly smart math people to work with in person has made a big difference for me, when I've had the chance to do so.

[quote="tAsDoOm"]Exactly. After all, when you have the AoPS books and online community, what else do you need[/quote]

Just curious, if you are an US citizen living overseas who happen to qualify for USAMO, which region do you compare yourself with?

Ditto.  What do you mean by region Oasis?  Well if you wanted to know, it's APO AP or FPO AP.  Something of that sort.  I'm not sure though.

What would you do if some other good mathematicians moved to Wyoming?  Now you're stuck in a hellhole.

Ha!  A hellhole full of cheese!  At least the Great Lakes are nearby so you could douse yourself if you were on fire.

I think you're confusing Wyoming with Wisconsin ... Wisconsin isn't a bad place.  Wyoming has more cows than people.

hey! don't make fun of the USACO! the cows will have their revenge!

California does have the advantage of having more than 5 times the population of Massachusetts.

haha i'm from MA so i'll just chime in and say that massachusetts clearly is better at math.

btw the 2nd level massachusetts olympiad was today.. here was one of the problems....

Prove that among any 14 consecutive positive integers there is at least 1 not divisible by any of {2,3,5,7,11}.

I dont have any formal education in number theory.  I was able to write a proof without any BS, but it was over a page long, does anyone know a quick way to do it?

Can you just do it by inclusion-exclusion?  There are 7 divisible by 2, 4 by 3, and so on, then do number divisible by 2 of them, then by 3 of them, and so on.  You'd have to be very careful about what the best-case scenarios are, but they're definitely provably best-case.   Was this your method?

[quote="beta"]Which state is generally better at math?
California or Massachussetts
They both seem pretty good although I feel California is better.[/quote]

California is by far the best math state.

namstap wrote:Prove that among any 14 consecutive positive integers there is at least 1 not divisible by any of {2,3,5,7,11}.

Hint: [hide]Prove that among 7 consecutive odd integers, there are at least three not divisible by 3 or 5.[/hide]

[quote]Which state is generally better at math?
California or Massachussetts
They both seem pretty good although I feel California is better.[/quote]

Are we talking about high school math competitions or math in general.

For math in general, I would definitely say MA is preferable, with all of the math activities present at HArvard and MIT in Cambridge.

For high school math competitions, it is hard to say. Going by ARML stats, SF Bay beat MA by a few places in 2003, MA beat SF Bay by a few places in 2002, and the two were right next to each other in the standings in 2001.

[quote="tAsDoOm"]Exactly. After all, when you have the AoPS books and online community, what else do you need? :D The Calif. Bay Area is, in my personal opinion, not at all what it's often hyped up to be. The area is plagued with rich, intelligent snobs, a lot of competition, and a load of other problems that I don't even want to get into...[/quote]

is not.  i live here too.  what are you talking about?  i realize some people are incredibly arrogant, but yanno, then again, people elsewhere are too.  what are you talking about?  who is this?

[quote="beta"]Which state is generally better at math?
California or Massachussetts
They both seem pretty good although I feel California is better.[/quote]

definitely Cali.  especially SFBA.  we hardly train for arml, and we almost beat TJ last year.  what can I say?  and the weather is nicer.  San Diego's pretty good, too...  especially now that Richard is training the soucal people...

yea, i would have to say cali too... of course, i don't have much MA experience, but the Bay Area is pretty nice and relaxed.

Of course, losing 6 or 7 seniors is going to hurt our ARML team, so we may or may not place higher than MA this year...

Wow!  I was completely tricked... When someone said a hellhole full of cheese, I automatically thought of Wisconsin instead of Wyoming!  lol

Hey hey hey, y'all better watch out, SoCal shall bloom into one of the foremost mathematical centres in the coming years. Just give us a few years.

Ah...here in the great state of Texas there's loads of good maths people (sorta), lots of space (if u live outside of houston and dallas/ft worth), and great food (if u live inside of houston and dallas/ft worth and austin and san antonio).

However, I think it would benefit us greatly (as a mathematical society) if Mr. Ruzczyk and Mr. Crawford moved down here :-)

[quote="ComplexZeta"]It's scaring me. SFBA keeps getting weaker, and other regions keep getting stronger. That means that if you are from SFBA and you are reading this, do more math, and convince everyone else within a 50-mile radius to do more also, especially if I usually tell you to do more math anyway.[/quote]

scares me too.  i'm going to hold faux arml practices at my school.  what years should i use?

[quote]For high school math competitions, it is hard to say. Going by ARML stats, SF Bay beat MA by a few places in 2003, MA beat SF Bay by a few places in 2002, and the two were right next to each other in the standings in 2001.[/quote]

Um, I'd like to note that the SFBay area is a pretty small portion of California as a whole.  I would guess the population of that area is smaller than MA, as well.  So, I'd argue that if MA = Bay Area, then MA << CA in terms of math.

As for MA having MIT and Cambridge, do keep in mind that CA has Caltech, Stanford, and Berkeley.  The Bay Area also has many of the top Olympiad educators (Zeitz, Davis, Vakil, Poonen, women at San Jose State and Berkeley, etc).

Regardless, I don't think you can go wrong as a student in either the Bay Area or Boston.

What?  More California / Massachusetts rivalry?  I thought the question was settled nearly three years ago with the creation of the grand state of Calimass.

MOPperwise, Massachusetts has really gone down over the past few years.  (Then again, CA isn't what it was in 2000 either.)


Can you prove

$ \sum_{n\equal{}0}^\infty \left(\sqrt{n\plus{}1}\minus{}\sqrt{n}\right)^3\equal{}\frac{3}{2\pi} \, \zeta\left(\frac32\right)$

$ \sum_{n\equal{}0}^\infty \left(\sqrt{n\plus{}1}\minus{}\sqrt{n}\right)^5\equal{}\frac{15}{2\pi^2} \, \zeta\left(\frac52\right)$

Now I know how to show it.

$ (\sqrt{k}\minus{}\sqrt{k\minus{}1})^3\equal{}k^\frac32\minus{}3k(k\minus{}1)^\frac12\plus{}3 k^\frac12 (k\minus{}1)\minus{}(k\minus{}1)^\frac32$

$ \Rightarrow \sum_{k\equal{}1}^n \left(\sqrt{k}\minus{}\sqrt{k\minus{}1}\right)^3\equal{}\sum_{k\equal{}1}^n k^\frac32\minus{}\sum_{k\equal{}1}^n (k\minus{}1)^\frac32\plus{}3\sum_{k\equal{}1}^n k^\frac12 (k\minus{}1)\minus{}3\sum_{k\equal{}2}^n k(k\minus{}1)^\frac12$

$ \equal{}n^\frac32\plus{}3(n\plus{}1)n^\frac12\minus{}6\sum_{k\equal{}1}^n k^\frac12\equal{}\minus{}6\left(\sum_{k\equal{}1}^n k^\frac12\minus{}\frac23 n^\frac32\minus{}\frac12 n^\frac12\right)\to \minus{}6\,\zeta\left(\minus{}\frac12 \right)$

and this is $ \frac{3}{2\pi}\zeta\left(\frac32\right)$ after Riemann's functional equation.

The next series goes similar.

More general one can show

$ \sum_{k\equal{}0}^\infty \left(\sqrt{k\plus{}1}\minus{}\sqrt{k}\right)^{2m\plus{}1}$

$ \equal{}\sum_{k\equal{}0}^m \sum_{j\equal{}k}^m \Big((\minus{}1)^j\minus{}1\Big)\, {j\choose k} {2m\plus{}1\choose 2j} \zeta\left(\minus{}\left(m\minus{}k\plus{}\frac12\right)\right)$.


Suppose $ a,b,c,d$ are positive real numbers.

Prove that

$ \frac{ab}{a\plus{}b}\plus{}\frac{cd}{c\plus{}d}\le\frac{(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}$

 :P


Ring Theory
superior algebra
superior algebra solved

How to prove directly from the definitions of maxiaml and prime ideals that every maximal ideal of a commutative ring R with unity is a prime ideal?

[quote="fredbel6"]suppose M is maximal but not prime
then there a and b not in M such that $a b \in M$

a is not in M and M is maximal : we can write :$a r_{1}+m_{1}=1$ and $b r_{2}+m_{2}=1$ for $r_1,r_2 \in R$ and $m_1,m_2 \in M$

multiplying both equations gives you that $ab r_{1} r_{2}-1$ is in M
but we assumed $a b \in M$ , so $1\in M$

contradiction : a maximal ideal cannot contain unity[/quote]

Thanks a lot! :)

Here is another proof.
Let $A$ be a commutative rings and $M$ a maximal ideal of $A$.
Let $x, y \in A$ be elements of $A$ such that $xy \in M$.
We will prove that $x \not \in M \Longrightarrow y \in M$ (which is equivalent to saying that $M$ is a prime ideal).

Suppose that $x \not \in M.$
[b]Note that $M$ is proprely contained in $M + Ax$[/b]
Since $M$ is maximal, any proper ``overset`` of $M$ is equal to $A$.
Therefore $M + Ax = A$.
Hence there is some $u \in M$ and $a \in A$ such that
\[ 1 = u + ax  \]
Multiply both sides by $y$ to obtain
\[ y = yu + yax  \]
\[ y = yu + axy  \]
(this is were commutativity comes into play).
But $yu \in M$ since $u \in M$ and $axy \in M$ since $xy \in M$ (just recall the definition of an ideal).
Moreoever $y = yu + axy$ and both summands $yu$, and $axy$belongs to $M$. Hence $y \in M$.


symmetry
analytic geometry
function
calculus

Can someone please explain how to derive "-b/a" as the x value for the (x,y) that is the origin of symmetry in a cubic equation?

What is $a$ and what is $b$?
So you want to find k where $-f(x+k) = f(k-x)$, or am I confused?

Sorry, I meant 	$-b/3a$ as the $x$ value of the $(x,y)$ that is the coordinate of the origin of symmetry for a cubic function of the form $ax^3 + bx^2 + cx + d$

[hide]If $f(x)=ax^3+kx$, then $f(-x)=-f(x)$, so symmetric about origin. Then $a(x-p)^3+k(x-p)+q$ is symmetric about $(p,q)$. Set equal to $ax^3+bx^2+cx+d$ and set $x^2$ coefficients equal, get $-3ap=b$, so $p=-\frac{b}{3a}$.[/hide]

Why does $a(x-p)+kx+q$ have its point of symmetry at $(p, q)$?

Is point of symmetry when ever $f(x) = f(-x)$ in a function where $-f(x) = f(-x)$?

If so, then I understand.

A point of symmetry is a point $(p,q)$ such that $f(p-x)-q=q-f(p+x)$ for all $x$. When $p=q=0$ this just becomes $f(-x)=-f(x)$, and if $f(x)$ is symmetric about the origin, $g(x)=f(x-p)+q$ will be symmetric about $(p,q)$ since it's translating the graph.

o wait that should read $a(x-p)^3+k(x-p)+q$. But this only changes the constant, which is not used anyway.

you could also prove this with some basic calculus. but i like the proof that was provided more than the calculus one.

[quote="breez"]Nice explanation scorpius, I understand this now.

However, can you please elaborate on why $f(p-x)-q=q-f(x-p)$ is true about any point of symmetry $(p,q)$?[/quote]

Ouch, I typed the wrong thing again! It now reads $f(p+x)$ instead of $f(x-p)$. Now this is accurate. If you start at $(p,q)$ and move over $x$ to the left, you will move $f(p-x)-q$ up, while if you move $x$ to the right, you will move $f(p+x)-q$ up, or $q-f(p+x)$ down. The distance moved up when traveling $x$ to the left will be the same as the distance moved down when traveling $x$ to the right (by symmetry of $(p,q)$).

Yeah, I was wondering about that, cause like $f(x)$ and $f(-x)$ wouldn't be related as an ordered pair unless an odd function.

Thanks for the explanations anyhow, they were very helpful.


Find all positive integers $ a$ such that

$ \frac{a^2\plus{}2a\minus{}2}{a^3\minus{}5a\plus{}3}$

is a fraction in reduced form. Prove your answer.

[hide="Hint"] Euclidean Algorithm.  :) [/hide]


Let $ a,b,c,d > 0$ and $ a\plus{}b\plus{}c\plus{}d \equal{} 1$. Prove that $ abc\plus{}bcd\plus{}cda\plus{}dab\leq \frac{1}{27}\plus{}\frac{176}{27}abcd$

[quote="Erken"]ISL 1993 :wink:[/quote]
Yes , you are true http://www.mathlinks.ro/Forum/viewtopic.php?p=463192#463192  :wink:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let P be a p-Sylow subgroup of $G$. Show that $N_{G}[N_{G}(P)]=N_{G}(P)$

I also need help with this one
Suppose H and K are subgroups of G and that $K\triangleleft H$. Show that $H\leq N_{G}(K)$

$N_{G}(K)$ is the normalizer.

$P$ is a normal $p$-sylow subgroup of $N(P)$. Hence, it's the unique one. Now suppose there exists $x \in N(N(P))$ s.t. $x \notin N(P)$. Then $x P x^{-1}\neq P$, and $x P x^{-1}\subseteq x N(P) x^{-1}= N(P)$. So, $x P x^{-1}$ is another $p$-sylow subgroup of $N(P)$, a contradiction.

The second exercise is a trivial calculation ... this should not make any problems.

the second one is straight from the definition of the normalizer
the first :
suppose that $x \notin N_{G}(P)$ but $x \in N_{G}[N_{G}(P)]$ thus
$xN_{G}(P)x^{-1}=N_{G}(P)$ so $\phi_{x}(g)=xgx^{-1}$ is an automorphism
on $N_{G}(P)$ so $\phi_{x}(P)=P$ since $P$ is the only Sylow $p$-subgroup in $N_{G}(P)$ (since it's normal in $N_{G}(H)$), thus $xPx^{-1}=P$ - contradiction with our assumption.

Thanks, got the first one!

Hmmm, regarding the second one, I had $<$ floating around in my mind when I was thinking of how to solve it and not $\leq$. Well, it makes sense now!


Unweighted Grade point average is calculated as follows $\frac{4x+3y+2z+1t+0f}{x+y+z+t+f}$

Where...
$x=$number of A's
$y=$number of B's
$z=$number of C's
etc.

What are the minimum number of total grades required so that every tenths of a GPA can be represented [no rounding] (any combination of grades totaling the amount is allowed; from the range of $0 \le g \le 4$ where $g=$grade point average)?

One needs ten classes because if the denominator is 10, one can guarantee every tenths place will be picked.
Any smaller number would skip at least one number in the denominator simply because there's not enough decimal representations of fractions to cover every tenth.

[quote="JBL"]It is not finished unless you show that each possible point can in fact be achieved using only ten classes.[/quote]
Well, $4x+3y+2z+1t+0f$ must be all the numbers between 0 and 40. Starting at 40, $x=10$ and $y=z=t=f=0$. Then going to 39, we can move one of the grades down to a B with $x=9$, $y=1$, $z=t=f=0$. Therefore, we can keep moving grades down one letter grade to get all the numbers between 0 and 40.

As you can see, I'm horrible at writing proofs. I think this is call induction or something.


Consider the six by six grid of unit squares below. How many
rectangles of area 3 square units can be formed using only the
line segments of the grid as the sides of the rectangles?

[asy]unitsize(0.1inch);
draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);
draw((1,0)--(1,6));
draw((2,0)--(2,6));
draw((3,0)--(3,6));
draw((4,0)--(4,6));
draw((5,0)--(5,6));
draw((0,1)--(6,1));
draw((0,2)--(6,2));
draw((0,3)--(6,3));
draw((0,4)--(6,4));
draw((0,5)--(6,5));[/asy]

[hide=Solution]4 rectangles on each row
4*6=24 on the horizontal rows
But we also need to count the Vertical rows
So: $(4\times 6)\times 2 = 48$[/hide]

[quote=venusalien][quote="Maxima"]Where in the world does this come from?[/quote]
i know, right? How did you get this[/quote]

Source: MathCounts 2003 State Team

[quote=arnolde1234][quote=venusalien][quote="Maxima"]Where in the world does this come from?[/quote]
i know, right? How did you get this[/quote]
[/quote]
Source: MathCounts 2003 State Team

EDIT: Sorry for revive. just thought u still wanted to know :blush:

[quote=Ridvik]they are alcumas problems[/quote]

Yeah, it probably is, but it's not the problem stated above.


For which primes x,y,z is (x + y + z)xyz a square ?

What a lousy answer, no proof. :D  :D  :D

if x|(x+y+z),y|(x+y+z),z|(x+y+z),so x|(y+z),y|(x+z),z|(x+y)
assume that x+z=sy x+y=tz
so y=(1+t)x/(ts - 1 ), z = (s+1)x/(ts-1)
(y+z)/x = (s+t+2)/(ts-1)
s+t+2 >= st - 1
so we can know that (t-1)(s-1) <= 4
if t=1 because (s+t+2)/(ts-1)=(s+3)/(s-1) is a integer,(s-1)|4
 if s = 2,y=2x,not a prime
 if s = 3,y=x,z=2x,not a prime
 if s = 5,y=x/2,y=3x/2,so x = 2,y=1,not a pime
else if t=2,because (s+t+2)/(ts-1)=(s+4)/(2s-1) is a integer,s+4>=2s-1
 if s = 2,y=x,z=x,so (x+y+z)xyz=3*x^4,not a square
 if s = 3,4,(s+4)/(2s-1) not a integer
 if s = 5,y=x/3,z=2x/3,so x = 3,y=1,not a prime
else if t = 3,because (s+t+2)/(ts-1)=(s+5)/(3s-1) is a integer,s+5>=3s-1
 or s <= 3,that s = 3,y=x/2,not a prime
else if t = 4,5,the same we can know there don't exit the (x,y,z)

so there is x,y,x=y
(x+y+z)xyz=(2x+z)z*x^2
then (2x+z)*z must be a square
we know that z|(2x+z),
so z = 2,and (x+y+z)xyz = 4*(x+1)*x^2
then (x+1) must be a square,assume x+ 1 = n^2
x = (n-1)*(n+1)
so n - 1 = 1,x=3

Above prove that (3,3,2),(3,2,3)and so on is the all(x,y,z)

Wow that's a long proof :)

I got this question from arne long time ago, I think i solved in 4 lines? I'll see if i can find it back, and post it.

If we have x=y then z(z+2x) is a square and z|z+2x, so z=2.
Then 4x+4 is a square and x=3.
If x,y,z are distinct, WLOG let x<y<z, then we have xyz|x+y+z which is impossible since xyz>x 2 z>3z>x+y+z.

Q. E. D.  (Exactly 4 lines :D )


On three tests, Molly scored 84, 83, and 91. On the same three tests, Milly scored 86, 87 and 88. What is the positive difference between their average scores?

Molly's average is $ \frac{84\plus{}83\plus{}91}{3}\equal{}\frac{258}{3}\equal{}86$. Milly's average, by inspection, is $ 87$. The difference is $ 87\minus{}86\equal{}\boxed{1}$.

Faster way.
Compute the difference of each element then divide by 3
$\frac{(84-86)+(83-87)+(91-88)}{3}=-1$
Positive is 1

how is that faster the first one is faster

Note that now you only have to compute small numbers (-2, -4, 3) rather than adding three large ones together.


We can use
$ \sqrt{1\plus{}\frac{1}{n^2}\plus{}\frac{1}{(n\plus{}1)^2}}\equal{}1\plus{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$


[quote="mathwizarddude"]1. Prove that $ \frac {1*3*5*7*...*9997*9999}{2*4*6*8*...*9998*10000} < 1/100.$

2. Prove that: $ 1/1^2 \plus{} 1/2^2 \plus{} 1/3^2 \plus{} 1/4^2 \plus{} .... \plus{} 1/9999^2 \plus{} 1/10000^2 < 2 \minus{} 1/10001.$[/quote]

mathwizarddude,

I amended your subject description; I left
it blank.  You can just let other readers decide
how THEY feel about the problems' difficulty
level, and you don't need to set out a lure
for the problems.  Let the problems speak
for themselves.


trigonometry
LaTeX
analytic geometry
algebra
polynomial
quadratics
quadratic formula

This problem is in volume 2, page29, #27.

  If 0<x<180 and cosx + sinx =1/2 then find (p,q) such that tanx= - (P +sqrt(q))/3. I squared the given equation to find sin2x and found it to be -3/4. Can I then divide -3/4 by 2  to find sinx and then find cosx and tanx? But even if I do that how can I manipulate tan x to find - (p + sqrt(q))/3? What do you usually do in situations like this?

 Thank you.

[quote="Archimedes27"]
... I squared the given equation to find sin2x and found it to be -3/4. Can I then divide -3/4 by 2  to find sinx ?[/quote]

No, $\displaystyle \frac{sin 2x}{2} \neq \sin x$. The $2x$ goes [i]into[/i] the $\sin$ function. The $2x$ and the $\sin$ are not multiplied, so division cannot separate them.

[hide]You're going to have to use one of your pythagorean identities here.[/hide]

Please use Latex, I have trouble understanding what you're saying. If you don't know how, use the testing forum to practice.

no you cannot divide $\sin 2x$ by 2 to find $\sin x$ because $\sin 2x\neq 2\sin x$

sorry i have to change the variable 'x' to $\theta$ because it is confusing if you put it into the $xy$-coordinate plane, whether $x$ is a $x$-coordinate or an angle.

first remember that $0<\theta<180$, believe it or not, this is very important than anything else when you start. if $0<\theta<180$ then $\sin\theta$ is always positive. why? because $\sin\theta$ is always the $y$-coordinate of the upper part of the circle going counterclockwise from positive $x$-axis to negative $x$-axis.

now let's start solving the equation $\cos\theta+\sin\theta=\frac12$ and find $\tan\theta$. from your previous calculation, you found
\[2\sin\theta\cos\theta=-\frac34\]
now solve for $\cos\theta$ and get
\[\cos\theta=-\frac{3}{8\sin\theta}\]
use this new expression and plug it back into the equation you started with
\[\cos\theta+\sin\theta=-\frac{3}{8\sin\theta}+\sin\theta=\frac12\]
simplify and write it into a polynomial form
\[8\sin^2\theta-4\sin\theta-3=0\]
use quadratic formula and you should get
\[\sin\theta=\frac14\pm\frac{\sqrt7}4\]
but remember i said earlier that $\sin\theta$ is always positive, so we choose the positive value (i hope you know how to choose which one is positive)
\[\sin\theta=\frac14+\frac{\sqrt7}4\]
then you can find $\cos\theta$ too from the equation you started with
\[\cos\theta=\frac14-\frac{\sqrt7}4\]
and you are finally at the last step! lets finish up with $\tan\theta$
\[\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac14(1+\sqrt7)}{\frac14(1-\sqrt7)}=-\frac{\sqrt7+1}{\sqrt7-1}\frac{\sqrt7+1}{\sqrt7+1}=-\frac{4+\sqrt7}3=-\frac{p+\sqrt{q}}3\]
at long last you can find what $p$ and $q$ is :)

ok finished! :pilot:


After taking the amc10a , what do you think you got?

I think I got [hide][hide]130[/hide][/hide]
That is, considering I bubbled in correctly and stuff,but I still think I could have done better.

I think I didn't answer 4 and kinda guessed on one of them.
So maybe 130

Assuming I bubbled correctly, I got 21 correct, 2 blank = 131.

If I had bubbled correctly and I didn't put 2 answers on 1 question.........I would have gotten a 129........yup......ugly number......The funny thing is....I believe I did better on this test than the AMC8......

[quote="kyyuanmathcount"]I have a 137 assuming the machines didn't screw up on me.  They have the answers out :).[/quote]where can you find the answers?

[quote="Ubemaya"]I got a 127.5... I did horribly. What grades are you guys in?[/quote]

I got a 126.5 and I'm in 8th grade. You?

I'm in 7th grade. Are all of you guys in MathCounts?

8th...This is my first year in Mathcounts....... :(  :blush:

I got either a 132 or a 138, if I bubbled everything in as I meant to. Not sure if I got one of the questions right, that's all... I know I missed 2 (stupidity) and answered everything.

Where are the answers posted?

I got a 130. 20 correct, 1 wrong, and 4 blank.

On question 10 did anyone else forget 0?

now that I know i got the linked rings question wrong  :blush:
i also got the last one wrong
so maximum of 118 :blush:

At most a tiny bit over 120, but not likely.  :(

131, assuming machine reads everything correctly. Ironically I got 21 right on this test and only 19 right on the AMC8

I completely messed the AMC 8 so I had the same, but the fact that I omitted two because you get so many points for that changes a little.

I think I failed so maybe 0?  :rotfl:  :rotfl:


With a little help from analysis and a  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13575]polynomial exercise[/url] I formulated in another thread. The [b]Fundamental Theorem of Algebra[/b] can be proved in quite elementary manner which can be understood by high school students.

So I am going to state a fact from analysis and ask you to prove the [b]Fundamental Theorem of Algebra[/b].

[b]Fact[/b]. Let [tex]g(z) = a_n z^n + \cdots + a_1 z + a_0[/tex] be a polynomial with complex coefficients, then [tex]|g(z)|[/tex] attains minimum value on any closed disk. (A closed disk is a region in the complex plane which is enclosed by a circle including the boundary. The Fact basically says, any continuous function attains minimum value on any compact subset).

Now prove this:

[b]Fundamental Theorem of Algebra[/b]. Let [tex]f(z) = a_n z^n + \cdots + a_1 z + a_0[/tex] be a polynomial of degree > 0 with complex coefficients. Show that there exists a complex number [tex]z_0[/tex] such that [tex]f(z_0)=0[/tex].

Yep, I'd sure like to see the steps of an elementary proof of the Fundamental Theorem of Algebra (posted in spoiler, of course).

[quote="josejacobi"]I think the same proof appears in Michael Spivak's CALCULUS text book
(chapter 25).[/quote]

I haven't seen Spivak's  proof. Could you post a sketch in spoiler?

Hint:

Step 1. [hide]based on the fact in analysis I stated above, show that |f(z)| attains minimum value on the entire complex plane C.[/hide]

Step 2. [hide]Assume that the Fundamental Theorem of Algebra is false, let m = |f(z_0)| = min |f(z)| on C, then m>0. Define g(z) based on f(z) such that g(0)=0 and |g(z)-1| :ge: 1 for any z in C.[/hide]

Here's a proof using the function g.
I still don't  know TeX. I hope no one has trouble
opening the attached file.

If there's interest in the proof I might
write a new version using TeX.
(I guess that would be good exercise in TeX)


MATHCOUNTS
limit
geometry
rectangle
perimeter
special factorizations

How many pairs $(m,n)$ of integers satisfy the equation $m+n=m \cdot n$ ?

This can't be a Mathcounts problem, but this is how I solved it.

Let $m$ be $x$ and $n$ by $y$.
$x+y=xy$
$x=xy-y$
$x=y(x-1)$
$y=\frac{x}{x-1}$
Now you can see that $\lim\limits_{x\rightarrow\infty}\left(\frac{x}{x-1}\right)=1$ and the same with $x\rightarrow -\infty$, which means that $y$ cannot be less than 1 while $x$ increases and $y$ cannot be greater than 1 while $x$ decreases. So using this knowlage and a graphing calculator, the pairs are (0,0), (2,2), and eventually, $(\infty,1)$ and $(-\infty,1)$, so that's 4 pairs.

I think it can be a Mathcounts problem as 0,0 and 2,2 are common sense solutions, though I'm not sure if it can be proved with Mathcounts level material.  To ln(dx/dy): you can't really treat infinity as a number, it's not an actual number, much less an integer, just a concept.

There is the "complete the rectangle" factoring trick.

$mn=m+n$

$mn-m-n=0$

$mn-m-n+1=1$

$(m-1)(n-1)=1$.

Now you're looking for pairs of integers whose product is 1. I'll let others finish it from here.

Here's a related question: Find all rectangles with integer sides such that the perimeter has the same numerical value as the area.

The same factoring idea works for this one.

Certainly looks like MATHCOUNTS level. By Kent Merryfield's advise, you need only know what makes up an integer and the possible ways to obtain that product, i.e. that the product of two integers is equal to 1.

Sorry to say this but infinity is not a number, and this problem can be solved [b]without[/b] using limits.

[size=75][Edited by rcv to remove sarcasm and excessive quoting][/size]

[quote="3X.lich"]
Sorry to say this but infinity is not a number, and this problem can be solved [b]without[/b] using limits.[/quote]
Even if we're working in the extended real numbers, it's certainly a stretch to say that its an integer.

A standard trick (or tool in your tool box as  my son used to call these   tricks).

[hide]we have  $mn -m -n = 0 $
rewrite it as:

$ (m-1) (n-1) - 1 = 0 $
or $(m-1)(n-1) = 1$

This reduces  into finding all the factors of 1 , which , obviously,   are   (+1, +1) and (-1, -1)

and  $(m-1)=(n-1) = +1 $  gives  $  m=n=2 $
whiel the other  $m-1 = n-1 = -1 $ gives  $m = n = 0 $

The t method can be applied to  the type of equation: $mn + $ (somthing)*$m$  (anothersomething)*$n$  $etc$

Hope this helps.   (There are many such  problems which reduces to this type - so knowing this method is helpful)[/hide]

[quote]The method can be applied to the type of equation:
$ mn +  \mathrm{(something)}*m +\mathrm{(anothersomething)}*n$ etc [/quote]
By analogy to the algebraic technique of completing the square, I like to refer to this as "completing the rectangle."

You will also see the same factoring trick refered to around this site as "Simon's Favorite Factoring Trick."


Just curious how hard this is. Any comments or anything are appreciated.

So, my honest opinion: I would describe it as Intermediate, but it feels very contrived.  That is, it seems like you introduced a lot of stuff just for the sake of it, and to some extent it just serves to make the problem "not pretty."  I don't know if you did your diagram to scale, but it was quite messy, and ideally you would spread things out more so that they were more visible.  I wouldn't say it was a bad problem -- just that it didn't feel natural.  Maybe I would say it seems like you constructed the problem in the wrong way -- instead of starting with a diagram and trying to find things out about it, it seems you started with the answer to the problem and tried to make a question for it, a messy prospect.

I attached an alternate diagram (without labels), which I thought looked clearer.

Note: diagram edited due to eariler error.

JBL in your picture it looks like you made  :ang: D automatically a right angle but the original problem isn't marked that way.

Sorry, you're right.  That was one thing I was going to say about the diagram, actually -- that angle looked right -- but instead I convinced myself it really was right :)  My mistake.


[color=red]Given the convex quadrilateral  ABCD, inscribe another quadrilateral with minimal perimiter.[/color]

The following is what I tried.

(1) The problem must have a solution: ABCD is compact and the "perimiter mapping" is continuous.

(2) Let XYZT an arbitrary quadrilateral inscribed in ABCD. $X\in [DA], Y\in [CD], Z\in [BC]$
Let X' be the symmetric of X with respect of CD. Suppose ZX' meets SEGMENT [CD] in U.
Then perimiter(XYZU)<=perimiter(XYZT).

(3) Consider the former case (2) and see what happens if $ZX'\notin [CD]$. Then perimiter(XYZU))<=perimiter(XYZT)
where U=D.

(4) We now apply the same trick used in (2) and (3) to the new inscribed quadrilateral XYZU.
We now consider two cases: (a) the trick fails because of the reason explained in (2) and (3). (b) the trick does not fail.

(a) Apply (3) again.
(b)  Perimiter is strictly decreasing in general unless we could find an inscribed quadrilateral XYZT such that $\widehat{AXY}=\widehat{DXT}$, $\widehat{DTX}=\widehat{CTZ}$ etc.

If that happens I proved that ABCD must be cyclic. Then I think there are infinitely many solutions.

But what happens if ABCD is NOT cyclic? Recall we know there is solution so I think this implies that if ABCD is  not cyclic then "the trick" fails, ie, we must consider the case explained in (3).
I think that my argument is "weird" if it is correct...

wait, i'm trying the problem, but i don't understand the part "Let U be the symmetric of Z with respect to the side [D,A]." please explain.

drunner2007

I edited my first message  :oops: but now I explained my reasoning :)

Maybe this problem is not for gettin started, sorry

[quote="Iron Fist"]Connecting midpoints intuitively seems right...[/quote]

No... Thats not true in general. Try it. It works for a rectangle as I proved but it isnt the only solution (in fact, there are infintely many others).
Even if ABCD is cyclic the Varignon's quadrilateral is not a solution.


문서를 어떻게 하면 지울수 있죠? -_-

Hello.

How can i delete my postings?

-_-

only edit??

In most cases, you can't delete your posts, after next reply is posted or many hours have passed. Contact a moderator or an administrator.

대부분의 경우에는 자신의 글은 지울 수 없으며 게시판 관리자 혹은 사이트 운영자에게 연락하여야 합니다.

감사합니다. 레이텍 시험을 해볼려고 했는데, 지워지지 않아서 수정밖에 할수가 없었네요.

레이텍보다는 한글의 수식편집기가 익숙해서 변환방법을 찾다가, 결국은 레이텍을 이용할수밖에

없다는 결론을 내렸습니다.

레이텍... 정말 대학교 시절에 써보고 오랜만에 다시 써보네요. (전 93학번. -_-)


limit
logarithms
function
inequalities
real analysis
real analysis unsolved

x>0

compute

$\lim _{n\rightarrow + \infty}n\ln(1+\ln(1+(...+\ln(1+x/n))...)$

$\ln $appear n-times

Choose $x\in\mathbb{R}^+$ and let $f(t)=\log(1+t)$; now, fix $\epsilon>0$  and take $\bar{n}$ such that $x/\bar{n}<\epsilon$. Obviously, $f(t)<t$, so $f^{(n)}(t)<t<\epsilon$ with $t=x/n$ and $n>\bar{n}$.
We have that
\[f^{(n+1)}(x/(n+1))<x/(n+1)<\epsilon\]
So $f^{(n)}(x/n)\to 0$ as $n\to\infty$. As $\log(1+x)\sim x$ if $x\to\infty$, we have
\[f^{(n+1)}(x/(n+1))\sim f^{(n)}(x/(n+1)\sim\ldots\asymp x/(n+1)\quad \textrm{ with }n\to\infty\]
Hence, the limit is $x$.

You can't do that; the effect of $n$ iterations should matter. You can't just throw out the $\frac{x^2}{n^2}$ terms, since we have $n$ of them and we multiply by $n$ at the end.

We have $\ln(1+t)=t-\frac{t^2}{2}+O(t^3)$. The $O(t^3)$ terms will be insignificant here. Thus $f^n(t)\approx \frac{2t}{nt+2}$. In context, the overall answer should be $\frac{2x}{x+2}$. Based on experimental evidence, I'm pretty sure of this, and $x$ is just wrong.

Yes, jmerry's answer is correct.  Here is a rigorous argument.  Define the function
\[
  g(t) = \frac{1}{f(t)} - \frac{1}{t} \, .
\]
Note that $g$ is decreasing on $(0, \infty)$, and $\lim_{t \rightarrow 0} g(t) = \frac{1}{2}$.

First, we will aim for a lower bound on the limit.  By the above facts, we have $g(t) < \frac{1}{2}$ for $t > 0$.  Rearranging that inequality gives
\[
  f(t) > \frac{1}{1/t + 1/2}
\]
for $t > 0$.  Iterating, we get
\[
  f^{(n)}(t) \ge \frac{1}{1/t + n/2}
\]
for $t > 0$.  That gives us a lower bound matching jmerry's answer.

Next, we will aim for an upper bound.  Given $\epsilon > 0$, we have $g(t) \ge g(\epsilon)$ for $0 < t \le \epsilon$.  Rearranging that inequality gives
\[
  f(t) \le \frac{1}{1/t + g(\epsilon)}
\]
for $0 < t \le \epsilon$.  Iterating, we get
\[
  f^{(n)}(t) \le \frac{1}{1/t + g(\epsilon) n}
\]
for $0 < t \le \epsilon$.  That gives us an upper bound on the limit (superior)
of
\[
  \frac{x}{1 + g(\epsilon) x} \, .
\]
As $\epsilon$ approaches $0$, that bound approaches jmerry's answer.


How many cubic numbers are also perfect squares under 1000?

[hide]Perfect squares that are also perfect cubes are powers of six because lcm(2,3) = 6.

Clearly, $1000 = 10^{3}$.

$1000 = 10^{3}= (\sqrt{10}){}^{2}{}^{3}= (\sqrt{10})^{6}$

$\sqrt{10}\approx 3.16$

Thus, where $n$ is an integer, $\forall n \text{ s.t. }0<n<3.16$, $n^{6}< 1000$.

Ergo, $n$ can be 1, 2, or 3, so there are $\boxed{3}$ powers of 6 less than 1000.[/hide]


superior algebra
superior algebra unsolved

prove that a finite dimensional real and commutative division algebra withh identity is isomorphic to $\mathbb{R}$ or $\mathbb{C}$

Commutative division algebras are simply fields. The result we want to prove is almost clear, since any algebraic extension of $\mathbb R$ which is different from $\mathbb R$ itself must have a root of a second degree equation with real coefficients and negative discriminant, i.e. basically, it must contain $i$, so $\mathbb C$. On the other hand, there are no finite dimensional extension fields larger than $\mathbb C$, since $\mathbb C$ is algebraically closed.

A more difficult result would be Frobenius' Theorem: if you remove the commutativity condition, you can also get the quaternions, and these are all of them (I think it's been posted around here somewhere).

the problem is why there is no division algebra of dimension greater than 3 over $\mathbb{R}$.if i am not wrong this classical due to Hopf.