rotation
geometry
3D geometry
geometric transformation
counting
distinguishability
group theory

I believe the official solution miscounted.  I think they miss one of the wys that you can choose a corner and an edge (picture below, my addition/changes in red)

I think you're right......

either way, i phailed.

Your "extra" case (in red) is the same as the one immediately next to it: rotate the 4th one 120 degrees (pivoting around the white corner) to get the 5th one.

(It's hard to describe: if you have a Rubik's cube or something you can actually do it.)

[quote="DPatrick"]
(It's hard to describe: if you have a Rubik's cube or something you can actually do it.)[/quote]
I borrowed a Rubik's cube when I was doing this problem. :D

Another way to tell that there are 4 "corner-edge" configurations: you can rotate the cube so that the white corner lies where it is in 3 ways.  There are 12 edge cubes, but dividing by 3 to account for the rotations (since they all move) gives 12/3 = 4 positions.

(This is what people who refer to "Burnside's Lemma" mean, but we're not going to require you to say anything that technical.)

d*** i got something completely unrelated.

I see it now... ugh...

I thought that this was the easiest problem.. and then I missed it.  I got $ 23$ cases.

(after looking at my solution)

uh... that was not the case I double counted. :-0 ?????????  Somewhere I still got a 23rd case....

found it... I counted 6 ways to have an edge with another edge.... gg

So I suppose now I'm goin to get around 19 or 20 for the round... so much for gold...

Eh, it's just like me to fail this problem... direct counting :(

i got 21. argh, epic fail
and i did the exact same casework as the official solution too
with a rubik's cube
there goes 5 points

I failed. I got 21. I did this problem like last minute and undercounted a case for edge-edge.  :|

[quote="timwu"]I did this problem like last minute and undercounted a case for edge-edge.[/quote]

Edge-edge is the hardest to visualize. Was the one you missed the case of two edges where one must take a minimum of 4 steps (going from one cube to an adjacent one), but is not where the edges are directly opposite?

Yeah. (After fixing an edge first) I did not realize that there were 2 different colorings among the 4 of such edges and only counted 1 (if this sentence makes sense).

lol, I got the answer to the problem, but my solution wasn't ideal.  I just used a bunch of casework with the help of my rubix cube.

[quote="pythag011"]BTW, what did you use to solve 1? I used Burnside's Lemma...

However, it gave the same answer, and the Burnside's Lemma solution was different, so I don't think there's a mistake...[/quote]
whats burnside's lemma?

[quote="gauss1181"]whats burnside's lemma?[/quote]

I'm not trying to be mean or anything, but can't you just look [url=http://en.wikipedia.org/wiki/Burnside_lemma]it[/url] up?

Your one in red is not correct. There are indeed 22 distinguishable faces.

Draw an imaginary line from the white cube in the corner to the black cube on the exact opposite of the cube. Put a white cube where that edge is in the pic. Now, use the imaginary line as an axis of rotation, and turn the cube 60 degrees along the axis. As you can see, it yields the same result.

This question was easy for me since I'm a Rubik's Cube nerd. 5x5's ftw.

I've never studied or seen any group theory, but I did use the idea which DPatrick said was the essence of Burnside's Lemma, to assure myself that I had thought of all the possible cases (the fact that each congruence class that pivots around a corner can only have at most three different cubes, etc.).  I was going to include the argument to prove that I had gotten to all the cases; however, because the argument would have to get longer and more complicated in a few of the cases, and because I had already spent half of my solution talking about congruence classes instead of counting, I decided to just hope they weren't looking for formal proofs.  I mean, the second problem was supposed to be harder, wasn't it?


vector
inequalities proposed
inequalities

Suppose that $a, b, c$ are real numbers such that for all positive numbers
$x_1,x_2,\dots,x_n$ we have

$(\frac{1}{n}\sum_{i=1}^nx_i)^a(\frac{1}{n}\sum_{i=1}^nx_i^2)^b(\frac{1}{n}\sum_{i=1}^nx_i^3)^c\ge 1$

Prove that vector $(a, b, c)$ is a nonnegative linear combination of vectors
$(-2,1,0)$ and $(-1,2,-1)$.

There must be some condition missing. If we put for exemple $(a,b,c)=(-1,-1,1)$, the ineq is equivalent to $(\frac{1}{n}\sum_{i=1}^nx_i^3)\geq (\frac{1}{n} \sum_{i=1}^nx_i^2)(\frac{1}{n}\sum_{i=1}^nx_i)$ which is true by Chebyshev.
But if we add the condition $c\le 0$, then it can be easily shown:
Denote the $j$-th power mean by $M_j$, so the ineq reads $M_1^aM_2^{2b}M_3^{3c}\ge1$. If this is to be valid for all $x_i$, it must be homogenous so $a+2b+3c=0.$ So all feasible vectors are part of a 2-dimensional space.

The vectors $(-2,1,0)$ and $(-1,2,-1)$ are feasible, as for the first one the ineq is equivalent to $M_2^2\ge M_1^2$ which is obvious, and for the second one it is equivalent to $M_2^2\ge M_1M_3$ which follows by the well-known concavity of the power means.
If two vectors are feasible, so is any positive linear combination of them.
With the condition that $c\le0$, it suffices to show that the first one, $(-2,1,0)$, cannot have a negative coefficient. This is equivalent to showing that $a+b+c\le0$, i.e. (substracting that from $a+2b+3c=0$)  $b\ge-2c$. Assume $c\le0$ and let $C: =-c\ge0$. So the ineq is $M_1^{3C-2b}M_2^{2b}\ge M_3^{3C}$.
if $b\leq \frac32C$, we get $LHS \le M_3^{3C-2b}M_3^{2b}=M_3^{3C}=RHS$, so  all $x_i$ are equal, contradiction.
Just missing the case $\frac32C\le b<2C.$ :?:


Let (I) be the inscribed circle of the hexagon ABCDEF
Prove that AD,BE,CF meet at the same point
I think it's well-known but I don't know what its name and how to solve it
So someone please help me !!!!
(Sorry I can't find a better word, I'm not good at English)

It's called Brianchon's Theorem. Try looking for it with Google, now that you know the name :).

[color=red][Moderator edit: Or look at http://cut-the-knot.com/Curriculum/Geometry/Brianchon.shtml .][/color]


how bout we start 2-D?

find the area of the points inside or within 1 unit of a 3x4 rectangle.

Miths' suggestion is a good one.  Another thought -- [hide]think about just 1 point.  What are all the points a distance 1 or less from a given point?  Now what happens as you move your given point around on the box?[/hide]

As you can imagine...

[hide]This is sort of like a "rounded box". Divide it up into parts...specifically one sphere of radius one, three cylinders with radius 1 and heights 3,4, and 5, and the original box itself.[/hide]

Rep123max wrote:I cant really think where to start

masmune's posting is equivalent to the following.  [The attached illustration should be considered a spoiler, but I don't know how to "spoilerize" it.]

[hide]I think about each "dimension" of the box.  Starting with the zero-dimensional points...

The box has 8 vertices.  Think about a point.  What does the set of points that are within 1 unit of a fixed point look like?

The box has 12 edges.  Think about a line.  What does the set of points that are within 1 unit of a line look like?

The box has 6 faces.  Think about a plane.  What does the set of points that are within 1 unit of a plane look like?

The box has a 3-dimensional interior.  What is the set of points within the interior of the box?

Now, it should be fairly straightforward to come up with a general formula, given height (5"), length (4"), width (3"), and zone-distance (1")[/hide]


Find all positive integers $ n$ for which there exist integers $ a$ , $ b$ and $ c$ such that $ a\plus{}b\plus{}c\equal{}0$ and $ a^{n}\plus{}b^{n}\plus{}c^{n}$ is prime.

We can go further by characterizing all solutions.

[hide]$ a^n\plus{}b^n\plus{}c^n \equiv a\plus{}b\plus{}c \mod{2}$, so $ 2|a^n\plus{}b^n\plus{}c^n$.  Hence $ a^n\plus{}b^n\plus{}c^n\equal{}2$.

Let $ a \ge b \ge c$.  Then, assuming the three integers are not all zero, $ a>0$, $ a\equal{}\minus{}(b\plus{}c)$.  Assume that $ 0 \ge b \ge c$ (if this is not the case, we can take the additive inverses of $ a, b,$ and $ c$).  Then if $ n \ge 3$ and $ n$ is odd, $ a^n\plus{}b^n\plus{}c^n$ is equal to $ bc$ times a thing which is less than $ \minus{}1$.  Hence we have $ b\equal{}c\equal{}\minus{}1$, so $ a^n\plus{}b^n\plus{}c^n\equal{}2^n\minus{}2$ which is not prime.  Hence $ 2|n$.

So now we have $ a^n, b^n,$ and $ c^n$ are all nonnegative squares and add to $ 2$.  Therefore, they are $ 0, 1,$ and $ 1$.  So $ (a, b, c)\equal{}(1, 0, \minus{}1)$, and $ n$ is any even number.[/hide]


superior algebra
superior algebra theorems

[quote="hersheykissx3389"]ok. so if A={1,2,3} would 3 be a subset, or would it have to be written {3} to be one ??

alsooo ... if A={1,2,3,{4,5}} how many combinations could it have ? 16 or 32 ??[/quote]

$ 3$ is not a set. You need to write $ \{ 3\}$ be to be a subset of $ \{1,2,3\}$.

[quote="sylow_theory"]$ 3$ is not a set.[/quote]

sure? :roll:

Let me guess $ 3\equal{}\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$. Clearly not the point of the question.  :rotfl:

sorry fot interrupting but is this the right section to post this question?  :huh:


linear algebra
combinatorics unsolved
combinatorics

N contestants answered an exam with n questions. $a_i$ points are awarded for a correct answer to question i and nil for an incorrect answer. After the questions had been marked it was noticed that by a suitable choice of positive numbers $a_i$any desired ranking of the contestants could be achieved. What is the largest possible value of N?

I remember using linear algebra to show that the answer is $n$. It was posted a long time ago by Pascual. Try searching first, man. Most of the problems you've posted these last couple of minutes are already on the forum.


function
calculus
integration
calculus computations

Obtain the general solution of the equation:

f''(x) + 6f'(x) + 9f(x) = exp(2x)

$k^{2}+6k+9=0$ so $k=-3$ is the only solution to charachteristic polynomial.

The solution (general) to the homogenous equation is:
$y=c_{1}e^{3x}+c_{2}xe^{3x}$

Look for a particual solution of the form,
$y_{P}= Ae^{3x}$ for solve for constant $A$.

[quote="sludgethrower"]Actually your complementary function is wrong...the powers of e should be -3x. And the particular integral (not particular solution!) should have the form Ae^(2x).[/quote]

In that case you should seek solutions of the form $y=At^{3}e^{2t}$


Let $ABC$ be an acute triangle. The interior angle bisectors of $\angle ABC$ and $\angle ACB$ meet the opposite sides in $L$ and $M$ respectively. Prove that there is a point $K$ in the interior of the side $BC$ such that the triangle $KLM$ is equilateral if and only if $\angle BAC = 60^\circ$.

$\left( \exists\right) K\in (BC)$ such that the triangle $MLK$ is equilateral$\Longrightarrow m(\widehat {BKM})=m(\widehat {MKL})=m(\widehat {CKL})=60^{\circ}\ a.s.o.$

I used a remarkable geometrical locus:

Let a fixed line $d$, a fixed point $A\not\in d$ and a mobile point $M\in d$. Then the geometrical locus of the point $L$ for which the triangle $AML$ is equilateral is the reunion of the two lines $l_1,\ l_2$ which are symmetrically w.r.t. the line $d',\ A\in d',\ d'\perp d$ and $m\left( \widehat {d,d'}\right)=60^{\circ}.$

If the angles BKM, MKL and LKC are equal, then the triangle would be isosceles (ML is parallel to BC) :maybe:

Let $ I $ be the incenter of $ ABC $.
If part:Assume that $ \angle A=60^o $, then we have $ a^2=b^2-bc+c^2 \implies BM+CL=BC $.If the lines perpendicular to $ BI,CI $ through $ M,L $, respectively, intersects $ N $ , then $ N $ must lies on $ BC $.On the other hand $ AMIL $ is cyclic and $ \angle ILM= \angle IML= 30^o $ and $ \angle BIM= \angle CIL=60^o $.Thus $ NLM $ is equilateral.

Could anyone solve another part of the problem?

i.e. "if the triangle $KLM$ is equilateral ,then $\angle BAC = 60^\circ$.

ΒΛΕΠΕ  ΣΧΗΜΑ:


trigonometry calculus integration derivative

What would happen to the time period of a simple pendulum if $ \sin x$ is approximated as $ x \minus{} \frac {x^{3}}{6}$ instead of x? I think that the time period will increase.

The reason for the simple approximation is so that you can calculate the period exactly. If you don't use that approximation, you might as well use $ \sin$ and the resulting elliptic integral; there's no point in using a complicated approximation, because it's just as intractible as the exact version.

The period does increase as amplitude increases for the exact version, to the point of infinite period when there's enough energy to reach or flip over the top.

This is a question in a book, so can u post a solution? Thanks!!

Let $ E$ be the total energy above the ground state with the pendulum stationary at the bottom $ y\equal{}\minus{}L$. We have $ E\equal{}\frac12mv^2\plus{}mg(y\plus{}L)$, and $ \frac{dy}{dt}\equal{}v\sin\theta\equal{}v\frac{\sqrt{L^2\minus{}y^2}}{L}$
Rewriting everything in terms of $ y$,
$ \frac{E}{m}\minus{}gL\equal{}\frac12\left(\frac{dy}{dt}\cdot \frac{L}{\sqrt{L^2\minus{}y^2}}\right)^2\plus{}gy$
$ K\equal{}\frac12\left(\frac{dy}{dt}\right)^2\cdot\frac{L^2}{L^2\minus{}y^2}\plus{}gy$ (defining a new constant $ K\equal{}\frac{E}{m}\minus{}gL$)
Solve for the derivative:
$ \left|\frac{dy}{dt}\right|\equal{}\sqrt{2\frac{L^2\minus{}y^2}{L^2}(K\minus{}gy)}\equal{}\frac{\sqrt{2g}}{L}\sqrt{(L^2\minus{}y^2)(H\minus{}y)}$, where $ H$ is the maximum height $ \frac{K}{g}\equal{}\frac{E}{mg}\minus{}L$.

We solve that by separation of variables, getting $ \int dt\equal{}\int \frac{L}{\sqrt{2g}}\cdot\frac{1}{\sqrt{(L^2\minus{}y^2)(H\minus{}y)}}\,dy$
$ T\equal{}4\int_{\minus{}L}^H\frac{L}{\sqrt{2g}}\cdot\frac{1}{\sqrt{(L^2\minus{}y^2)(H\minus{}y)}}\,dy$ (each period goes up twice and down twice)

That's the exact period $ T$ when it tops out at height $ H$. Note that at $ H\equal{}L$, the integrand includes a factor of $ (L\minus{}y)^{\minus{}1}$, making the integral diverge to $ \infty$. There's another way to see the increase: $ 4\int_{\minus{}L}^H \frac{L}{\sqrt{2g}}\cdot\frac{1}{\sqrt{(L\plus{}y)(2L)(H\minus{}y)}}\,dy$
$ \equal{}2\int_{\minus{}L}^H \sqrt{\frac{L}{g}}\cdot\frac{1}{\sqrt{(L\plus{}y)(H\minus{}y)}}\,dy$
$ \equal{}2\sqrt{\frac{L}{g}}\int_{\minus{}1}^1\frac1{\sqrt{(1\plus{}s)(1\minus{}s)}}\,ds$ (substituting $ s\equal{}\frac1{H\minus{}L}\left(y\minus{}\frac{L\plus{}H}{2}\right)$; the two factors of $ \frac{L\plus{}H}{2}$ cancel)
$ \equal{}2\sqrt{\frac{L}{g}}\cdot\left[\arcsin s\right]_{\minus{}1}^1\equal{}2\pi\sqrt{\frac{L}{g}}$

The exact version replaces $ \frac1{\sqrt{2L}}$ with $ \frac1{\sqrt{L\minus{}y}}$ inside the integral. As $ H$ increases, that makes a steadily larger impact for an increasing period.


algorithm
superior algebra
superior algebra unsolved

I'm trying to write an analogue of the division algorithm for Z[i]. help...

$ \mathbb Z[i]$ is an euclidean ring for the norm $ N(a \plus{} bi) \equal{} a^2 \plus{} b^2$. Now try to describe the remainder of a division.

no. read the definition of euclidiean rings.

Claim that for any a & b in Z[i], there exist q and r

such that a=bq+r  and the Norm(r)<Norm(b)

this defines a division alg.

I want to show that it works, and I've figured it'd be enough to show that

Norm(r/b)<1

But i need help with this last part

What does the set of "integer" multiples $ bq$ of $ b$ look like in the complex plane?


Find the no. of Solutions of

$ | [x] -2x | = 4 $ where $ [x] $ is the greatest integer fn...

We have: $\left|\left[x\right] - 2x\right| = 4$ so there are two cases

1) $\left[x\right] - 2x \geq 0$ so it is $\left[x\right] - 2x = 4$ or $\left[x\right] = 4 + 2x$.
It shows that $x$ must be of the form $\frac{1}{2}k$, the only way to obtain an integer from $4 + 2x$.
Also $x < \left[x\right] +1$ or, that is the same $4 + 2x > x - 1$ and then $x > -5$.
The equation can be rewrited as $\left[1/2k\right] = 4 + k$. We have $k > -10$ and it's clear that $k$ can't be $\geq 0$.
So we have to prove case $k \in \left\{-9,-8,-7,-6,-5,-4,-3,-2,-1\right\}$. For $k = -9$ and $k = -8$ equality holds.
So solutions are $x = -4 \vee x= -9/2$

2) $\left[x\right] - 2x < 0$ so it is $\left[x\right] = 2x - 4$ or $\left[1/2k\right] = k - 4$ and $x > 3$ or $k > 6$
So $k \in \left[6,\infty\right]$.The only solutions are for $k = 7$ and $k = 8$.
So solutions are $x = 4 \vee x = 7/2$.


Determine all trple of real numbers $ x;y;z$ such that
$ x^6 \plus{} y^6 \plus{} z^6 \minus{} 6(x^4 \plus{} y^4 \plus{} z^4) \plus{} 10(x^2 \plus{} y^2 \plus{} z^2) \minus{} 2(x^3y \plus{} y^3z \plus{} z^3x) \plus{} 6(xy \plus{} yz \plus{} xz) \equal{} 0$


abstract algebra
superior algebra
superior algebra unsolved

Is it true that if the exponent(the minimal $ m$ s.t. $ x^{m}\equal{}e$ for all $ x$) of a group is equal to the order of the group(say $ n$) then there exists en element of order $ n$?

Isn't it true by definition? if $ n$ is the smallest positive integer s.t. $ \forall x \in G$ $ x^n\equal{}e$, it must exists at least one element of order $ n$, cause if all elements of the group had an order smaller than $ n$, than the exponent would be smaller than $ n$...

The result is true only for abelian groups.

Can someone check if this is why the result is true for abelian groups.
Let $ p^{r} \equal{}$ exp G, where p is prime and r is some positive integer.
Let $ x \neq 1$, $ y \neq 1$ be distinct elements of G.
Then we have:
$ (xy)^{|x|} \equal{} x^{|x|}y^{|x|} \equal{} y^{|x|} \iff |x| \equal{} p^{r}$ for some x.

Can someone explain why if x is an element of a group then |x| must divide the exponent of the group (assuming the exponent exist)?
If |x| divides exp(G), doesn't this imply that |x| is smaller than exp(G) and how come this does not contradict the very definition of the exp(G)?

[quote="akech"]Can someone explain why if x is an element of a group then |x| must divide the exponent of the group (assuming the exponent exist)? [/quote]

Presumably, $ |x|$ means the order of $ x$. Let $ x$ be an element of finite order in any group $ G$, of finite exponent or not. Let $ |x|\equal{}k$, so $ x^k\equal{}1$ but $ x^l \neq 1$ for any $ 0 < l < k$. It follows that $ x^n\equal{}1$ iff $ k|n$. Proof: $ x^{dk} \equal{} (x^k)^d \equal{} 1^d \equal{} 1$, so any multiple of $ k$ works. In the other direction, if $ x^n\equal{}1$, write $ n \equal{} qk\plus{}r$ where $ 0 \leq r < k$, then $ 1\equal{}x^n\equal{}x^{qk\plus{}r}\equal{}(x^k)^qx^r \equal{} x^r$ which contradicts the minimality of $ k$ unless $ r \equal{} 0$.

In particular, if $ e$ is the exponent of the group $ G$, then $ x^e\equal{}1$ and we just saw that this implies that $ e$ is a multiple of $ k\equal{}|x|$.

[quote]
If |x| divides exp(G), doesn't this imply that |x| is smaller than exp(G) [/quote]
Not necessarily, since e.g. $ 7$ divides $ 7$. It does imply that $ |x| \leq e$ where $ w$ is the exponent.

[quote] and how come this does not contradict the very definition of the exp(G)?[/quote]

Why would it contradict that? The exponent is the least positive integer $ e$ such that $ x^e\equal{}1$ for all $ x \in G$. It doesn't mean that $ e$ is the smallest exponent for any particular $ x$. If some elements have order $ 2$, and others have order $ 3$, the exponent must be a multiple of $ 6$.


trigonometry
calculus
derivative
function
inequalities
inequalities proposed

Let $ x,y$ be real numbers and $ x,y\in [0,\frac{\pi}{3}]$. Prove that
\[ \cos{x}\plus{}\cos{y}\le 1\plus{}\cos{xy}\]

Have any proof without derivative?

hello, we have $ \cos(x)\plus{}\cos(y)\equal{}2\cos(\frac{x}{2}\plus{}\frac{y}{2})\cos(\frac{x}{2}\minus{}\frac{y}{2})$, i think this will help.
Sonnhard.

[quote="zaizai-hoang"]Let $ x,y$ be real numbers and $ x,y\in [0,\frac {\pi}{3}]$. Prove that
\[ \cos{x} \plus{} \cos{y}\le 1 \plus{} \cos{xy}
\]
Have any proof without derivative?[/quote]
Use Dr Sonnhard Graubner 's hint
$ \cos(x) \plus{} \cos(y) \equal{} 2\cos(\frac {x}{2} \plus{} \frac {y}{2})\cos(\frac {x}{2} \minus{} \frac {y}{2}) \le 2cos (\frac {x \plus{} y}{2})$
Note $ cos(xy)\ge cos(\frac {(x \plus{} y)^2}{2})$(since $ cos x$ is decreasing function here
Now Let$ |\frac {x \plus{} y}{2}| \equal{} t$ the inequality becomes $ 1 \plus{} cos(t^2)\ge 2cos(t) \forall 0\le t\le \sqrt {\frac {\pi}{2}}$.Which is always true by derivative! :lol: .Now I can't still find a nice proof as you require.


trigonometry symmetry quadratics AMC AIME geometry circumcircle

A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by $\overline{AB}$, has length 31. Find the sum of the lengths of the three diagonals that can be drawn from $A$.

[hide="Solution"]
Label the vertices counter-clockwise, $A$, $B$, $C$, $D$, $E$, and $F$. Let the lengths of the diagonals $AC$, $AD$, and $AE$ be $x$, $y$, and $z$. Consider quadrilateral $ABCD$, which is cyclic. By Ptolemy's Theorem, we have $y\cdot BC+CD\cdot AB=BD\cdot x$, or since $BD=AE=z$, \[ 81y+81\cdot 31=xy. \] Similarly, if we apply Ptolemy to quadrilateral $ADEF$, we get \[ 81y+81^2=z^2. \] Finally, in quadrilateral $CADE$, we have \[ 81x+81z=yz. \]

Is there an elegant way to solve this system? :stink:
[/hide]

I have a very cumbersome trig solution.

[hide]Let $O$ be the center of the circle and its radius $r$. If $\angle AOF=2\phi$, then $\angle AOB=2\pi-10\phi$, hence

$81=2r\sin\phi$
$31=2r\sin\frac{2\pi-10\phi}{2}=2r\sin 5\phi$

Now

$AC=2r\sin\frac{2\pi-10\phi+2\phi}{2}=2r\sin 4\phi$
$AD=2r\sin\frac{2\pi-10\phi+4\phi}{2}=2r\sin 3\phi$
$AE=2r\sin\frac{2\pi-10\phi+6\phi}{2}=2r\sin 2\phi$

Also

\begin{eqnarray*}\sin 5\phi &=& \Im\{(\cos\phi+i\sin\phi)^5\}\\ &=& 5\cos^4\phi\sin\phi-10\cos^2\phi\sin^3\phi+\sin^5\phi\\ &=& 5(1-sin^2\phi)^2\sin\phi-10(1-\sin^2\phi)\sin^3\phi+\sin^5\phi\\ &=& 16\sin^5\phi-20\sin^3\phi+5\sin\phi\end{eqnarray*}

Now

${31\over 81}=\frac{\sin 5\phi}{\sin\phi}=16\sin^4\phi-20\sin^2\phi+5$

Substitute $t=\sin\phi$:

$t^4-20t^2+{374\over 81}=0\iff t^2={17\over 18}\lor t^2={11\over 36}$

Since $30^\circ<\phi<36^\circ$, we take $t=\frac{\sqrt{11}}{6}\implies\cos\phi={5\over 6}\land 2r=\frac{81\cdot 6}{\sqrt{11}}$

We'll need $\sin 3\phi=-4\sin^3\phi+3\sin\phi=-4\frac{11\sqrt{11}}{216}+3\frac{\sqrt{11}}{6}=\frac{8\sqrt{11}}{27}$

Now

\begin{eqnarray*}AC+AD+AE &=& 2r(\sin 2\phi+\sin 3\phi+\sin 4\phi)\\ &=& 2r(2\sin 3\phi\cos\phi+\sin 3\phi)\\ &=& 2r\sin 3\phi(1+2\cos\phi)\\ &=& \frac{81\cdot 6}{\sqrt{11}}\cdot \frac{8\sqrt{11}}{27}\cdot {8\over 3}\\ &=& 384\end{eqnarray*}[/hide]

[hide]I think I beat all of you for brute force: I found the height of the hexagon (distance between the 31 side and its opposite), built a bunch of right triangles, used triple angle and double angle trig indenties, as well as Pythagorean Theorem like 10 times, and finally factored around 5 5-digit numbers. In the end, the diagonals are 135, 105, and 144, which add up to 384. The height's $40\sqrt{11}$, $FC=144$, and $\sin{\angle AFE}=\frac{5\sqrt{11}}{18}$ if anyone cares...[/hide]

Congratulations, you clearly won the brute-force prize of 2006.  :10:

tralalala no ones posted a complete normal solution so...

[hide="solution..."]Given:$AB=31$, All other sides $=81$

Labeled $A,B,C...$ in clockwise order. We're trying to find diagonals $AC, AD, AE$. (a diagram would be useful)

Due to symmetry, we have $AC=BF=x, AD=BE=FC=y, AE=BD=CE=DF=z$

Using Ptolemy's on Quadralateral $ABDE$ we have
$31 \cdot 81+z^{2}=y^{2}$

Using Ptolemy's on Quadralateral $CDEF$ we have
$81^{2}+81y=z^{2}$

Plug in $81^{2}+81y$ from the second into $z^{2}$ from the first equation to get $31 \cdot 81+81^{2}+81y=y^{2}$

Expanding and grouping like terms gives
$y^{2}-81y-9072$

Using the Quadratic Formula gives us $y=144$

Once we have the length of one of the sides, finding the other 2 are easy.

Using Ptolemy on Quadralateral $ABCF$ we have
$81^{2}+31 \cdot 144=x^{2}\Longrightarrow x=105$

Using Ptolemy on Quadralateral $CDEF$ again, we have
$81^{2}+81 \cdot 144=z^{2}\Longrightarrow z=135$

$x+y+z=144+105+135=\fbox{384}$ as desired.[/hide][/hide]

[hide="props to proposer for nice numbers :)"]
I call the three diags 1,2,3.

Then, $ 81^2\plus{}2\cdot 81\equal{}1^2$ and $ 1^2\plus{}31\cdot 81\equal{}2^2$
Thus $ 2^2\equal{}2\cdot 81\plus{}152\cdot 81$, so \[ 2 \equal{} \frac{81\plus{}\sqrt{81^2\plus{}81\cdot 112\cdot 4}}{2}\equal{}\frac{81\plus{}\sqrt{81}\sqrt{523}}{2}\equal{}144\]
Thus $ 1^2\equal{}2^2\minus{}31\cdot 81$, or trickily, $ (144\minus{}1)(144\plus{}1)\equal{}31\cdot 81\equal{}9\cdot 279$ which add up to 288 thankfully, so $ 1\equal{}135$.
Then we compute $ 3\cdot 1\equal{}31\cdot 81\plus{}2\cdot 81$ or $ 3\cdot 135\equal{}175\cdot 81$ so $ 3\equal{}105$

Happy numbers :)

[/hide]

me@home: as an idea: don't label your diagrams with integers? ;) Put a box around them or something to distinguish diagram numbers from integers.

Sorry for the revive, but I felt that I have to summarize the methods that were used effectively in this difficult problem. And plus, the aime is coming up, so...

1st approach: We are given the side lengths of a hexagon inscribed in a circle. Perhaps this will point us towards a solution with Ptolemy's theorem. What really tells us to use Ptolemy's is the fact that many of the diagonals have the same length, so we might guess to write a few equations using Ptolmy's theorem and then solve them by cancelling out the common diagonals.

2nd approach: We can start by noticing that we are given all the side lengths of a hexagon. So maybe we can use the Law of Cosines... But that gets really messy.

      But then we realize that the hexagon is inscribed in a circle, and if we want to keep the algebra (relatively) easy, then maybe we should use the extended law of sines. This is advantageous because we get to relate the angle, lengths and the circumradius all in one equation. And since we have a cyclic hexagon, then many of the angles will be the same.

       So we start by writing an equation that only involves the angle and five times that angle. We soon get a fourth-degree polynomial that can be simplified to a quadratic. Courageously, we press on, knowing that on an aime, the problem proposers will almost always make the discriminant a perfect square. So we now know the sine and the cosine of one (and thus all) of the angles.

      We can use the extended law of sines to get the length of the circumradius, since we know the side length and know know the sine of the angle. Then, we bash out the algebra...


inequalities
Duke
college
geometry
geometric transformation
reflection
Support

So how awesome is this?  The international governing body for gymnastics essentially used majorization to break the tie in the uneven bars final between Nastia Liukin (USA) and He Kexin (China).

Each contestant had the same A-score (difficulty) of 7.7, and the same B-score (execution) of 9.025.

To break the tie, they looked at the breakdown of the B-score by the individual judges.  Nastia had (9.1,9.0,9.0,9.0).  Kexin had (9.1,9.1,9.0,8.9).

So Nastia got majorized!  (and lost the gold medal... to china. darn)

if two contestants had scores (9.5, 9.0, 9.0, 9.0) and (9.3, 9.2, 9.1, 8.9), the latter would win despite no majorizations in either direction ?

what are you just saying that if one contestant majorizes another, she wins (modulo the fact that they already throw out the highest and lowest scores)

then yes that is true okay

but the point is the tie-breaking procedure is much less complicated than majorization

nastia should so totally have won though

Go China!
hahaha in your face  :lol:
I think that they should have done the proper tiebreaking procedure though.. China would've won without controversy xD

HAHA majorization in the olympics. Very cool.

DUDE, NO

it is pretty annoying when mathematicians try to impose math onto everything

like, if it actually works, then yes, good job

if it doesn't it just looks lame like "that dude is such a nerd he thinks about everything in terms of math even when he is wrong"

this is not majorization, mmkay

[quote="MysticTerminator"]they did do the proper tiebreaking procedure. the controversy is that (1) the proper tiebreaking procedure is STUPID and (2) nastia should totally have won (without any ties and whatnot)

ALSO IF CHINA IS SO GOOD GO LIVE IN CHINA[/quote]
Dude stop pulling the "IF CHINA IS SO GOOD GO LIVE IN CHINA" because it doesn't mean anything. It's okay to be proud of your own culture/people. And besides, I do plan to spend a lot of time in China when I'm older.

uh no, I actually believe in it and refuse to stop pulling it. if you do plan to be tied to china as your home country, then good. I applaud your choice and your patriotism. however most people have clearly made their choice to remain american (okay so normally when I talk to people it's like my roommates and stuff who can actually make their own choices as to where to go to college, for example, rather than you young dudes who can't. still, I bet most of y'all will remain in america for the majority of your life).

thanks for pointing out how it doesn't mean anything without explaining why the obvious meaning which it does have somehow fails to be applicable

yes, okay, be proud of your culture. that's awesome. you can root for them. that too is awesome. what I don't understand is openly rooting for them against and in direct comparison with america.

Heh...
American life is better than life in China. That's for sure. Somehow though, people (like me) can feel more attachment to our homeland than the place where we live. I remember reading in Time yesterday about a person who was in prison for supporting the Tiananmen Square protests, and she still wants to return to China after spending many years abroad.
When my family watches the Olympics, the order of preference (for countries) goes China then America.
Things get strange when you pull some poor Chinese kids who were lucky enough to go to college after the Cultural Revolution and send them overseas to study.

Serial said a lot of things I wanted to say.  I root for China over the U.S. b/c I feel a bigger connection to China than I do to the U.S.
lots of Chinese kids on AoPS have parent's who were young during the cultural revolution, and then were lucky to be among the 1% of ppl who made it to college right after the end of the revolution, and came out to the U.S. (in my parent's case, to pursue further education).  Though born in the U.S., lots of us have been back to China multiple times, and have most of our family there, and still feel a strong attachment to China.

[quote="serialk11r"] Somehow though, people (like me) can feel more attachment to our homeland than the place where we live.[/quote]

yes

this is what I don't understand

if you feel more attachment to china than america, why wouldn't you rather be there

Not necessarily?
Being in America has it's advantages too. That's why all these immigrants are here. I'd like to finish my education here, because there's more room to do things I want to do, and there's a lot less competition (=better school without the work). Working in America also pays a good deal more.
However if I do get a good deal of money or something I'd really want to move back to China. I've already spent 1/3 of my short 15 year life there. I just think it's more interesting there, and I feel more at home because I'm not the minority ;)

Rooting for China in a situation like this is an act of being proud of your own people, your own culture, but it does not mean that you believe China is a better place to live overall.

okay so first, let me clarify what I am saying. I am not saying to spurn your country and your people and all that other wonderful stuff. be as proud of them as you want to be. what I find disconcerting is the open way in which chinese immigrants root for their home country in direct comparison with the united states. breaking it down, the argument seems to run as -- okay we will reap the full benefit of the advantages conferred by living in american society yet still openly root against it ("china is pwning US in gold medals", etc). I find this ungrateful and tacky and would appreciate any elucidation on how this is not the case.

what annoys me is how people come to the US, reap the full benefits of the education system, then just leave without contributing. Its like they're using the country...I understand, i root for my parents country too sometimes,  but for me i am essentially an american b/c thats where i live and i am a citizen.

Also what about the morally indefensible actions of the chinese government in Tibet, dont you have qualms about that? Your going to use the iraq war as a counter example against the us but the fact is most ppl are against it whereas in china nobody cares...

By "is uplifting," I mean something like "brings optimism," and it is natural to think that optimism will be to the benefit of a society and will help the society improve. I mean, I'm not sure that this is in fact the case, but it seems inappropriate to simply assume one way or the other. While there must be some kind of attachment to some essential aspect of the country in question, it may or may not be to the "system" (which isn't particularly essential--countries change dramatically over centuries but still identify with their past selves).

Regarding the mathematical/emotional issue, the hope that one team wins, when carried out to its logical conclusion, is also the hope that other teams lose. Since you were arguing that people's logics were flawed, not that they were emotionally messed up, I assumed that either you intended to show that the hope for a team to win is inextricably tied to support of that country's "system" (in which case we might have some compatibility issues with expressions of "cultural joy") or there is some way to infer from specific statements more information than simply which team a person hopes will win or hopes will lose (in which case we would want to investigate exactly how one can distinguish such cases from simple statements of "I want ___ to win"). I actually thought something along the gratefulness vein would be closer to your fundamental point.

Back on topic, one could argue it's better for them to take the person who minorized the other as then that person has higher consistency.

Personally, I'm going to say they should take the person who's taller, as it's harder to do gymnastics when you're tall :D

Wait, but when China is "OWNING" diving, for instance, doesn't that mean that the American team is losing because they can't win anymore?
To some degree, people have the right to live where they choose because of the benefits they see in living and working there. It may be for the money, the environment, or because their life is already established in place X. However, that doesn't mean they have to support everything in that place. What if you were say, British, and you moved to China because you want to start a business there? Does that mean you shouldn't cheer for the British team for whatever?

[quote="MysticTerminator"]They should certainly support the British team. Did I, at any point, say Chinese-Americans should not root for China? No. I have stated the opposite many, many times and am almost starting to get slightly annoyed at having to restate this continually. I would find it very strange if, however, the Britisher in this case who has rationally chosen to affiliate himself with China would then go about gloating about British dominance over China in whatever event may be relevant.

alex I will think about your point and respond later right now I am trying to remember vector calculus okay[/quote]
When I say rooting for a team, I mean that you are soley supporting that team and implying that you do not want any other team to win.

See this is where I'm losing you, why do you think that this "affiliation" with some entity necessarily means that you have to support their sports team? What if you only went to China because you thought there was a good business opportunity, and you thought it would be interesting to be able to live there and see China, but you still felt like you were a Britishman? What's wrong with that?

okay, this is the main point I think people are getting hung up on:

there is a difference between rooting for some given team A and rooting against some other team B.

observe the quotation examples I gave a couple posts ago. do you agree that there is a fundamental difference between the first and second quotation in each pair?

The main differences are that in each pair, the second is rude whereas the first is not, and the second conveys stronger feelings than the first. If your point is that people should not be rude, then umm.. ok then let's not be rude? As for whether you can conclude that anything is illogical, see my previous post.

LOL I'm surprised that the discussion has gone this far but ok so Mystic:

Children of chinese immigrants in the US are still brought up to respect and retain cultural values by their parents, and they usually visit their country often or occassionally, and when they do, in many aspects they feel more of a connection with China than United States. What's wrong to root for the country that you have more connection with?

But then you mentioned that they reap benefits from the America, sounding as if it's contradictory. But really, the benefits they reap from the country China is much much more. First of all, if the native people from that country and the country itself did not exist, Chinese people who live in America wouldn't exist either. How can you objectively argue that they've reaped more benefits from America?

Or you say that they live the American lifestyle and hence should root for America. No, in fact, the average chinese/asian dude lives a life different than Americans of other ethnicity.

And if they root for China, it's quite normal for them to say kinda like "IN YOUR FACE" to any country, including the US

All at the end, there's not even a point arguing about this at all, because rooting for a country that you want is not wrong under any circumstances

Another problem with your logic is, you're trying to say that rooting against the country you live in is wrong, and rooting for some country is different from rooting against another. The only problem is, that forces you to root for the country you live in instead of another or else you would be doing something "wrong".

okay, so serialkiller has apparently completely failed to think through the basic logical principles here. I don't really feel like responding to this lack of comprehension anymore on here since it simply distracts from the main points. IM me again if you want.

the first four paragraphs of jimhu's post are mostly irrelevant

the last two are wrong, or at least this is the position I am taking, which I have sketched out pretty fully. you kind of make the claim that they are not wrong without elaborating any further. this doesn't really merit a response.

for alex's second post, yes, to some degree rooting against teams is considered rude to that team. if that's the scale you want to measure things by, then all right, that's fine. now our positions differ in that you say "let's not be rude" in general whereas I am only making the specific point "let's not be rude to the country which provides you with your subsistence". however, I should note that the change of variables from "rooting for & rooting against" to "being polite & being rude" is not necessarily as clear-cut as it appears, since to a large extent in sports rivalries, what would thence be classified as "being rude" is perfectly natural and acceptable.

regarding alex's general point about logic -- I am using logical arguments because that is the underlying system of axioms which I use to formulate my points. however, I am in the present case discussing sports rivalry, which is far more relevant to the emotional states than the logical trains of thought in consideration. in other words, the logic which I am referring to is the one that I am manually introducing in a second layer over the actual substance of consideration, which happens to be largely emotional. for example, one of my points may be phrased as follows: "it is [i]logically[/i] inconsistent to [i]emotionally[/i] root against the country with which one has chosen to affiliate oneself" -- I hope this clarifies at which specific layer the logic comes into play. hence, I stand by my main postulate, namely that there is a large (emotional) difference between rooting for a team and against another.

[quote="MysticTerminator"]
 now our positions differ in that you say "let's not be rude" in general whereas I am only making the specific point "let's not be rude to the country which provides you with your subsistence". [/quote]
Whoa, whoa, whoa hold the phone.  Why should we be rude to any country in general?  That looks like a total disregard of Olympic morals to me.  Isn't the whole idea of the Olympics for the watchers and the athletes to have a great time and to build International bonds?  I don't see any difference between hating on our country or hating on China... they are both completely and totally morally repugnant...

yeah, see the sentence [i]right after that[/i].

[quote="MysticTerminator"]okay, so serialkiller has apparently completely failed to think through the basic logical principles here. I don't really feel like responding to this lack of comprehension anymore on here since it simply distracts from the main points. IM me again if you want.[/quote]

Arnav, if you'd like to engage in personal attacks, start your own thread.  I never intended for a light-hearted discussion about the rules of gymnastics to degenerate into this nonsense.

Indeed, serialkiller's claim does not follow from your postulates, but perhaps your arguments would be easier to understand if they were conveyed in complete sentences, according to the conventions of standard English grammar, rather than in an in-your-face style of writing so incoherent it makes stream-of-consciousness seem like Strunk and White in comparison.

The fact that he took the trouble to IM should demonstrate to you that, however inadequately his arguments are expressed, he is at least interested in engaging in a constructive dialogue.  Perhaps you shouldn't be so quick to presume it beneath you to reciprocate.  Perhaps you shouldn't be so quick to demonize those who, unlike you, make the effort to seek out and understand the opposing point of view.

If you were a true champion of this nation and its democratic ideals--the nation that welcomes newcomers from the distant shores of every corner of the globe and with pride grants them participatory rights in its society--you'd know that listening to the voices of passionate Americans is beneath no one.

Just a thought.

--DBR

hahaha, you're DBR? that seems pretty sensible, as just about everything in your post is wildly inaccurate except for the fact that I do use nonstandardized grammar. I think the main difference lies with my capitalization tendencies, however, and this really shouldn't be enough to trip people up. anyway, you'll note that at this point I have ceased actually responding to all the illogical attacks on me directly in this thread as they do get rather tiring (in just about every other thread in the Round Table, I continue to explain because usually after a while, people get it, but here it doesn't seem to be happening for some reason. I don't think I've been impatient in this regard -- it's just that if I continue to simply engage in this gentle clarification, it doesn't really seem like we would get anywhere). If you'd like clarification on any point, you too can feel free to IM me (and yes, I will respond constructively unless I am afk). If for some reason that doesn't suit you, you can make another thread for me to explain whether or not I am actually making personal attacks or pointing out basic flaws in reasoning, et cetera -- however, as stated, I am not going to do that here any further as it distracts from the discussion proper.


integration
trigonometry
calculus
function
real analysis
real analysis unsolved

Find
\[ \int\limits_0^\pi  {{{\sin }^4}(x \plus{} \sin 3x)dx}\]
.

with $ 8sin^4(t)\equal{}cos(4t)\minus{}4cos(2t)\plus{}3$

Denote the initial integral $ A$

$ 8A\equal{}B\minus{}4C\plus{}3\pi$  with

$ B\equal{}\int_{0}^{\pi}\cos(4x\plus{}4\sin(3x))dx$

$ C\equal{}\int_{0}^{\pi}\cos(2x\plus{}2\sin(3x))dx$

look like a kind of Bessel function see formulae (69)

http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html


calculus
derivative
function
real analysis
real analysis solved

Hi.

I need help with the following.

How do I show that

$f(x) = \frac{x}{1+x^2}$

is uniformely continuous on $\mathbb{R}$?

Thanks!

Mike

If $\left\vert xa\right\vert \leq1$ then
\[
\left\vert \frac{x}{1+x^{2}}-\frac{a}{1+a^{2}}\right\vert =\left\vert
\frac{\left(  x-a\right)  +xa\left(  a-x\right)  }{\left(  1+x^{2}\right)
\left(  1+a^{2}\right)  }\right\vert \leq\left\vert x-a\right\vert +\left\vert
xa\left(  a-x\right)  \right\vert \leq2\left\vert x-a\right\vert
\]
If $\left\vert xa\right\vert \geq1,$ then $\left\vert xa\right\vert ^{2}%
\geq\left\vert xa\right\vert ,$ then
\[
\left\vert \frac{x}{1+x^{2}}-\frac{a}{1+a^{2}}\right\vert =\left\vert
\frac{\left(  x-a\right)  +xa\left(  a-x\right)  }{\left(  1+x^{2}\right)
\left(  1+a^{2}\right)  }\right\vert \leq\left\vert x-a\right\vert +\left\vert
\frac{xa\left(  a-x\right)  }{\left(  xa\right)  ^{2}}\right\vert
\leq2\left\vert x-a\right\vert
\]

So for given $\varepsilon>0$ take $\delta=\varepsilon/2$ . If $\left\vert
x-a\right\vert <\delta=\varepsilon/2,$ then
\[
\left\vert f\left(  x\right)  -f\left(  a\right)  \right\vert \leq2\left\vert
x-a\right\vert <\varepsilon
\]

I hope it is correct.

it must be uniformly continuous cause it's bounded and it's monotonic in a finite number of sets

Two more reasons why $f$ is uniformly continuous.

1. Its derivative can be explicitly computed: $f'(x)=\frac{1-x^2}{(1+x^2)^2}.$ This is clearly a bounded function.

If $f$ has a bounded derivitave, then it is clearly Lipschitz. In particular, if $|f'(x)|\le M$ then by the Mean Value Theorem, $|f(x)-f(y)|\le M|x-y|$ for all $x,y.$

If $f$ is Lipschitz, then it is uniformly continuous. This is a direct application of the definition of uniform continuity.

2. $f\in C_0(\mathbb{R}),$ where $C_0(\mathbb{R})$ is the set of continuous functions that tend to zero at infinity. Every member of $C_0(\mathbb{R})$ is uniformly continuous on $\mathbb{R}.$

I won't write out the whole proof, but the basic ideas look like this: outside some compact set, $f$ is small, and can't vary much from point to point just because it's small. Inside that compact, set, $f$ is uniformly continuous becuase it's a continous function on a compact set. Putting this together (which involves choosing the compact set after we already have our $\epsilon,)$ we have that $f$ is uniformly continuous everywhere.


function
integration
logarithms
real analysis
real analysis unsolved

Find $ B_n$, where $ B_n\equal{}\sum_{i\equal{}0}^{n\minus{}1}B_{n\minus{}i\minus{}1}{n\minus{}1\choose i}$ and $ B_0\equal{}1$, using generating fuction.

Perhaps use the "exponential" type generating function.

Make the RHS look like a convolution.

First rewrite the recursion as:

$ (n+1)\frac{B_{n+1}}{(n+1)!} = \sum _{i=0} ^n \frac{B_{n-i}}{(n-i)!}\frac{1}{i!}$

Now, use the generating functions defining $ f(x) = \sum ^\infty_{n = 0} \frac{B_n}{n!} x^n$:

$ \sum ^\infty_{n = 0}  x^n  (n+1)\frac{B_{n+1}}{(n+1)!} = \sum ^\infty_{n = 0}  x^n \sum _{i=0} ^n \frac{B_{n-i}}{(n-i)!}\frac{1}{i!}$

For the RHS we can use:

$ (\sum ^\infty_{i=0} a_i x^i ) \cdot (\sum ^\infty_{j=0} b_j x^j ) = \sum ^\infty_{k=0} x^k \sum^k _{l = 0} a_l b_{k-l}$

This gives us:

$ \sum ^\infty_{n = 0}  x^n  (n+1)\frac{B_{n+1}}{(n+1)!} =( \sum ^\infty_{n = 0}  x^n  \frac{B_{n}}{n!}) \cdot (\sum ^\infty_{n = 0}\frac{x^n}{n!} )$

$ \sum ^\infty_{n = 0}  x^n  (n+1)\frac{B_{n+1}}{(n+1)!} =e^x \sum ^\infty_{n = 0}  x^n  \frac{B_{n}}{n!}$

For the LHS use the fact that:

$ f'(x) = \sum ^\infty_{n = 0} n \frac{B_n}{n!} x^{n-1} = \sum ^\infty_{n = 0} (n+1) \frac{B_{n+1}}{(n+1)!} x^n$

This gives us:

$ f'(x) = e^x f(x)$

Solve this by separation of variables:

$ \frac{\text{d} f}{f} = e^x \text{d}x \, / \int$

$ \ln f = e^x + C$

$ f(x) = C e^{e^x}$

From $ B_0 = f(0) = C e = 1$ we get $ C = \frac{1}{e}$.

Now, all that we have to do is calculate $ f^{(n)}(0) = B_n$:

$ f(x) = \frac{1}{e} \sum ^ \infty _{n = 0} \frac{e^{x n}}{n!} = \frac{1}{e} \sum ^ \infty _{n = 0} \frac{1}{n!} \sum ^\infty _{i = 0} \frac{x^i n^i}{i!} = \frac{1}{e} \sum ^\infty _{i = 0} \frac{x^i}{i!} \sum ^ \infty _{n = 0} \frac{n^i}{n!}$

From this we can see that:

$ \boxed{\boxed{B_n = \frac{1}{e} \sum ^ \infty _ {i = 0} \frac{i^n}{i!}}}$

(The numbers $ B_n$ are called Bell numbers - [url]http://mathworld.wolfram.com/BellNumber.html[/url])


[quote="hardsoul"][quote="zhaobin"]:) .
it is a strang problem.who can help me to prove of disprove:
if $a,b,c>0$ can we prove
$a^8+b^8+c^8+4a^2b^2c^2(ab+bc+ca) \geq a^4b^4+b^4c^4+c^4a^4+4a^2b^2c^2(a^2+b^2+c^2)$
I can't slove it by $\sum S_a(b-c)^2 \geq 0$
thanks.[/quote]
I think this inequality can be proved by method of mixing variables:
Define $f(a,b,c)=RHS-LHS$ and $a\ge b\ge c$
Then $f(\sqrt{ab},\sqrt{ab},c)-f(a,b,c)=(a^4-b^4)^2-4a^2b^2c^3(a+b-2\sqrt{ab})=(\sqrt a-\sqrt b)^2\left((\sqrt a+\sqrt b)^2(a+b)^2(a^2+b^2)^2-4a^2b^2c^3\right)\ge 0$
Now $f(x,x,y)=2x^8+y^8+4x^4y^2(x^2+2xy)-x^8-2x^4y^4-4x^4y^2(2x^2+y^2)=(x^4-y^4)^2-4x^4y^2(x-y)^2=(x-y)^2((x+y)^2(x^2+y^2)^2-4x^4y^2)\ge 0$[/quote]

thanks.nice solution. :)

[quote="nttu"]SOS is an useful method and I think we can use it to solve this problem .
 Note that : $\sum A(b-c)^2 = \sum\frac{A}{(b+c)^2}(b^2-c^2)^2 =\sum\frac{A}{a^2}(ac-ab)^2=\sum b^2c^2A(\frac{1}{b}-\frac{1}{c})^2$
 Here we use $\sum A(b-c)^2 = \sum\frac{A}{(b+c)^2}(b^2-c^2)^2$ , where $A =(b+c)^2(b^2+c^2)^2-4a^2b^2c^2$
 The ineq is equivalent to :
 $\sum[(b^2+c^2)^2-\frac{4a^2b^2c^2}{(b+c)^2}](b^2-c^2)^2 \geq 0 \iff \sum A'(b^2-c^2)^2 \geq 0$
 Suppose $a \geq b \geq c$
 Then $B' \geq 0$
 We have : $(b^2+c^2)^2 \geq 4b^2c^2$ and $(c^2+a^2)^2 \geq 4c^2a^2$
  So $B'+C' \geq 0$
 Similarly , $A'+C' \geq 0$ , $A'+B' \geq 0$
  Then the conclusion is followed[/quote]
you  are right.thanks

I have another method that proves a more general case:
Write the inequality in the form $\frac{\sum(a^4-b^4)^2}{\sum(a-b)^2}\geq 4(abc)^2$.
This looks as if it wants to be generalized, doesn't it? :)… So let’s assume WLOG $abc=1$ (i.e. divide $a,b,c$ by $\sqrt[3]{abc}$) and claim $\frac{\sum(a^n-b^n)^2}{\sum(a-b)^2}\geq n$. This is in fact equivalent to the ineq of
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63954]www.artofproblemsolving.com/Forum/viewtopic.php?t=63954[/url]
Note that we can also formulate it as follows
$s_n: = \frac1n\sum_{cyc}(a^n-b^n)^2$ for real $n>0$ is monotonously increasing. (*)
So the original ineq is equivalent to $s_n\geq s_m$ for $(n,m)=(4,1)$. It can be written as
$[8,0,0]+4[3,3,2]\geq [4,4,0]+4[4,2,2]$.
Note that (*) yields a bunch of symmetric inequalities of similar form. Replacing (where useful) $a$ by $a^3$ etc, we put e.g.
$(n,m)=(5,2)$ and obtain $[10,0,0]+\frac52 [4,4,2]\geq [5,5,0]+\frac52 [6,2,2]$
$(n,m)=(2,1)$ and obtain $[12,0,0]+2[5,5,2]\geq [6,6,0]+2[8,2,2]$
$(n,m)=(7,4)$ and obtain $[14,0,0]+\frac74 [6,6,2]\geq [7,7,0]+\frac74 [10,2,2]$
$(n,m)=(3,2)$ and obtain $[18,0,0]+\frac32 [8,8,2]\geq [9,9,0]+\frac32 [14,2,2]$
$(n,m)=(3,1)$ and obtain $[18,0,0]+3[7,7,4]\geq [9,9,0]+3[10,4,4]$

etc. :)

[quote="spanferkel"] $\frac{\sum(a^n-b^n)^2}{\sum(a-b)^2}\geq n$. This is in fact equivalent to the ineq of
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63954]www.artofproblemsolving.com/Forum/viewtopic.php?t=63954[/url]

etc. :)[/quote]
sorry,I can't see :(

Zhaobin,
This is because of $abc=1$:
$\frac{\sum(a^n-b^n)^2}{\sum(a-b)^2}\geq n$
$\Leftrightarrow \sum(a^{2n}+b^{2n})-2\sum a^nb^n \geq n \sum(a^{2}+b^{2})-2n\sum ab$
$\Leftrightarrow 2\sum(a^{2n})-2\sum \frac1{c^n} \geq 2n \sum(a^{2})-2n\sum \frac1c$
$\Leftrightarrow 2\sum(a^{2n}- \frac1{a^n}) \geq 2n \sum(a^{2}- \frac1a)$
$\Leftrightarrow\sum\frac{a^{3n}-1}{a^n}\geq n\sum\frac{a^3-1}a.$

(And this is how I came up with the other thread :) :) )

ok,thanks.spanferkel,
But can you proved $s_n$ is monotonously increasing？

[quote="spanferkel"]Do you really still fear this one?? Or rather joking? So easy:
Take two reals $n\geq m>0$, so $n=mr$ with $r\geq1.$ So we can apply the claim $\frac{\sum(A^r-B^r)^2}{\sum(A-B)^2}\geq r$ to the variables $A=a^m,B=b^m, C=c^m$  (note that still $ABC=1$). This yields $\frac{ns_n}{ms_m}=\frac{\sum(a^{mr}-b^{mr})^2}{\sum(a^m-b^m)^2} \geq r \Leftrightarrow s_n\geq s_m$.
We can even observe that with putting $s_0: =0$, the $s_n$ are monotonously increasing  for [i]all [/i]real $n$. :roll:  :P[/quote]
ok,thanks very much.
And I think you use the fact $\sum\frac{a^{3n}-1}{a^n}\geq n\sum\frac{a^3-1}a.$,$n \in R,n \ge 1$(at least n is rational number) when you prove  $s_n$ is monotonously increasing.
else:we may use hardsoul's or nttu's method to prove $\frac{\sum(a^n-b^n)^2}{\sum(a-b)^2}\geq n$
that means you inequality has another proof :) without RCF :P

[quote="zhaobin"]
else:we may use hardsoul's or nttu's method to prove $\frac{\sum(a^n-b^n)^2}{\sum(a-b)^2}\geq n$
that means you inequality has another proof :) without RCF :P[/quote]

May you please give links for these two methods? would be fun trying them :rotfl:
Thanks

did you kidding me???
there are just in #2 and #4 :P


3x + y = 23
x = 5y = 31

93+y=23
y should =-70
but the answer isn't right...
are you sure you typed it in right?
isn't is suppose to be
x+5y=31?
then its

3x+15y=93
3x+y=23
14y=70
y=5
x=6

[quote="learningfish"]3x + y = 23
x = 5y = 31

Do i multiply the second line by 3 to cancel out x?
This doesn't seem to give the correct answer[/quote]
[hide]Yea. I think learningfish meant x+5y=31. So using systems you get that y=5. Plugging that asnwer you get that x=6.[/hide]

[quote="learningfish"]Yes it should be

 x + 5y = 31

Are we multiplying by -3?
Its been ages since I've had to do this stuff and I forget the rules of operation[/quote]
So we can get rid of one variable to just solve for one variable.

[quote="learningfish"]3x + y = 23
x = 5y = 31

Do i multiply the second line by 3 to cancel out x?
This doesn't seem to give the correct answer[/quote]

Learningfish, your system is inconsistent.

If it was
$3x+y=23$

$x+5y=31$,

then

$3(x+5y)=3(31)$

$3x+15y=93$

$14y = 70$

$y = 5$

$x = 6$

$(x,y) = (6,5)$

Assuming you meant $x+5y=31$...

[hide]$3x+y=23;\;x+5y=31$

Multiply the second equation by $3$ and subtract the first equation from it to get $14y=70$, or $y=5$.

Then, plug in $y$ and find $x=6$.

$\boxed{(6,5)}$[/hide]

[quote="learningfish"]3x + y = 23
x = 5y = 31

Do i multiply the second line by 3 to cancel out x?
This doesn't seem to give the correct answer[/quote]

well to cancel the x you must have a negative and a positive...so therefore multiply by a -3 and then add...it seems to me it's easier to add the equations rather than subtracting

wen u do this u can either add 2 opposites (like 3x + -3x) or minus the same numbers (3x - 3x)

[quote="jackdillon"][hide]Using substitution, you find that x=6, and y=5.[/hide]

This should be in HSB. you learn solving with 2 unknowns in grade 10(at least in canada.)[/quote]
Yea I learned this in 6th grade. This is in the right place.

ya, we covered this in math class in 7 or 6th grade (i can't remember), also. we finished working with 3+ variables with the corresponding # of equations now (8th grade).

[quote="learningfish"]Yes it should be

 x + 5y = 31

Are we multiplying by -3?
Its been ages since I've had to do this stuff and I forget the rules of operation[/quote] We can multiply by -3 but we could also multiply by 3 and subtract the two equations. I would advise solving for x (x = 31 - 5y subtract 5y from either side) then substitute tht into the first equation.

One wrong number and things can change so much.


Suppose that $ x$ and $ y$ are positive integers satisfying $ 2x^{2}-y^{2}=1$. Show that if x terminates in the digit $ 5$ (in decimal representation), then $ y$ is divisible by $ 7$

Answer (Partial)

$ x$ takes the from $ 10k+5$

$ \Rightarrow x \equiv 10k+5 \equiv 3k+5\,(mod\,7)$ where $ k = 0,1,2,3,4,5,6$

$ \Rightarrow x^{2}\equiv 9k^{2}+30k+25 \equiv 2k^{2}+2k+4\,(mod\,7)$

$ \Rightarrow 2x^{2}\equiv 4k^{2}+4k+1\,(mod\,7)$

$ \Rightarrow y^{2}+1 \equiv 4k^{2}+4k+1\,(mod\,7)$

$ \Rightarrow y^{2}\equiv 4k^{2}+4k\,(mod\,7)$

$ \Rightarrow y^{2}\equiv 0, 1, 3, 6 \,(mod\,7)$

Any integer can be expressed as one of:

$ 0,1,2,3,4,5,6 \,(mod\,7)$

The respective squares can be represented as:

$ 0,1,4,2,2,4,1 \,(mod\,7)$

Hence $ y^{2}\equiv 0$ or $ 1\,(mod\,7)$

This is as far as I got. I am having touble showing that $ y^{2}\not \equiv 1\,(mod\,7)$.


vector
invariant
geometry
abstract algebra
linear algebra
matrix
linear algebra unsolved

Let $E$ be a Complex vector space with finite dimension $n\geq2$. Assume that $A$ is a sub Algebra of $L(E)$, which is [b]unitaire[/b] ( $Id_E\in A$) and [b]transitive[/b] ( i.e if $V$ is a subspace of $E$ that is invariant (or stable) under all element of $A$, then $V$ is either {0} or $E$).
a) Let $x\in E, x\not =0$. Prove that $\{u(x),u\in\ A\}=E$.
b) Let $u\in A,ranku\geq2$. Prove that there exists $v\in A$ such that $uvu$ and $u$ formed an independent system.
c) Show that there are $\lambda\in C, z\in imu, z\not = 0$ such that $uv(z)=\lambda z$.
d) Show that $A$ contains at least one rank 1 element.
e) Find out $A$.

[quote="harazi"]First, please don't post it in the solved questions area. Secondly, may I ask how you have Gourdon's book in Vietnam? That's really interesting.[/quote]
It is not my book, but my teacher's. Why did you ask me this question?? And much sorry about the wrong area post.

It was just curiosity, since from France to Paris there is a quite long way. Anyway, please, a moderator should move this into the proposed section area. [color=red][Done.][/color] But something tells me that sam-n already posted a proof in the topic products of powers or something like this (I posted it some time ago, Molien's theorem generalized by Burnside).

even without the assumption $I \in A$ you can prove that $A=L(E)$ or $A=\{0\}$.

Here's a an approach different from the one in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=burnside&t=48094]other topic[/url], using the theory of semisimple rings. All rings and algebras are assumed to be unital here, but I'm sure slight modifications can be made to prove the general case alekk is referring to; that's just not something I want to worry about :).

I'll restate the problem:

Let $k$ be an algebraically closed field, $n$ a positive integer, and $A$ a transitive subalgebra of $\mathcal M_{n}(k)$. Then $A=\mathcal M_{n}(k)$.

(here, a subalgebra $A$ of $\mathcal M_{n}(k)$ is said to be transitive if $k^{n}$, when regarded as a left $A$-module in the natural way, has no non-trivial proper submodules)

The ring $A$ is, on the one hand, left Artinian: every left ideal is, in particular, a finite-dimensional $k$-vector space, so there can be no infinite strictly descending chains of left ideals. On the other hand, its Jacobson radical is zero: indeed, the Jacobson radical is the intersection of the annihilators of all the left simple $A$-modules. By hypothesis, $k^{n}$ is a simple left $A$-module, and its annihilator is clearly trivial (no non-zero matrix in $\mathcal M_{n}(k)$ can annihilate every vector in $k^{n}$). Putting these two facts together, we find that $A$ is a semisimple ring.

By the Wedderburn-Artin Theorem (the structure theorem for semisimple rings), $A$ is of the form $\prod_{i=1}^{k}\mathcal M_{n_{i}}(D_{i})\ (*)$, where $D_{i}$ are finite-dimensional division algebras over $k$. Since $k$ is algebraically closed, all $D_{i}$ must in fact be equal to $k$ here. Also, the general theory says that the simple left modules over $(*)$ are precisely $D_{i}^{n_{i}},\ i=\overline{1,k}$, on which $\mathcal M_{n_{i}}(D_{i})$ acts by multiplication to the left in the obvious way. In our case, by hypothesis, $A$ has a simple left module which is $n$-dimensional over $k$, so one of the $n_{i}$ must be equal to $n$. In turn, this implies that the dimension of $A$ over $k$ is at least $n^{2}$, and since $A$ is contained in $\mathcal M_{n}(k)$, the desired result $A=\mathcal M_{n}(k)$ follows.


ARML
MATHCOUNTS
mathleague.org
AoPS Books

Greetings and welcome to the Great Plains Math League forum.  GPML is a contest with format similar to those of ARML and MathCounts and currently operates the state high school championships in Missouri, Iowa, and Kansas and also distributes its contests around the world.

Check out GPML's website at [url]http://mathleague.org[/url] for information on:
* Contest format (both individual and team events)
* How to participate in local GPML events (state meets in MO, IA, KS are in April)
* How to host a tournament in your location (GPML is looking to expand to other states)
* How to order practice tests (past test series are available dating back to 1997)

[quote="Euler1001"]Does anyone who belongs to this site also compete in the GPML besides me?[/quote]

I know a few do.  There will probably be more given that it's a similar curriculum to ARML.  farmertim was the president of ARML last term and likes to angle students at a similar curriculum of problem solving.

Is the competition restricted only to those who live in Missouri, Iowa and Kansas?

Not at all!  Currently most of our contests are located in MO, IA, and KS, but GPML is always looking for additional host sites around the country.  Anyone who is interested in getting something started locally is encouraged to contact me so we can discuss this further..

Hey farmertim, do you know of any good way to prepare for the GPML state competition?  I have a pretty poor teacher/coach and I have gone through the old tests this year but I am not sure how to prepare best.  Any thoughts?

i know im not asked to answer that question, but:
i think if you can understand 60+% of the material in the AoPS books (both volumes), then you should be well prepared.
And yeah, even if you don't have the books, going through the old contests helps a great deal as I have noticed that the types of problems given are repeated often.


function
trigonometry
topology
real analysis
real analysis unsolved

Here's another cute one (although I'm sure it's extremely easy :)) from the web page in the source field above:

Show a connected compact subset of $\mathbb R^2$ that is not the range of any continuous function from $\mathbb R$.

Hmm... I know that all problems are easy for you, but they are not easy for many other people ;)

I am thinking abouth the following construction.
Consider a Cantor's set on $[0,1]$ on $Ox$ and draw vertical segments of length 1 for each point in this set. Then consider a Cantor's set on $[0,1]$ on $Oy$ and do the same thing with horizontal segments. We obtain a connected compact set.

Yes. The image of a continuous function is path-connected; we just need to find a compact and connected set which is not path-connected.

My example: $(x,\sin\frac1x)$ for $0<x\le1$ and $(0,y)$ for $-1\le y\le1.$

I considered the exact same example as jmerry. Thank you both, but Myth, is there any obvious reason, that I am missing, for which your set is not the range of a real continuous function? :? Sorry for what I am sure is a very stupid mistake..

Actually, Myth, I think your example works as well (even though I bet nobody is interested inthis anymore :)).

Yup my example is same as that of jmerry's. :lol:
But i didnt quite understand Myth's example.Maybe the construction went above my head. :(

Myth's example is more interesting because it is path connected. It can be simplified, though: take a Cantor set $K$ on the line and some point $O$ outside the line and connect $O$ by a straight segment with every point $x\in K$. The reason that this set cannot be a range of any continuous function $f$ is that the pre-image of every open segment $(Ox)$ must be  a non-empty open set in $\mathbb R$ (you cannot jump from one segment to another without passing through $O$) and you can fit only countably many disjoint open sets on a line.

Now here is a real challenge: is it possible to find such an example that is locally path connected?  :) (locally path connected means that for any point $x$ and any neighborhood $U$ of $x$ there exists a neighborhood $V$ of $x$ such that $x$ can be connected with any point $y\in V$ by a path contained in $U$)

I think local connectedness instead of local path connectedness suffices for a negative answer to that question. It follows from the Hahn-Mazurkievicz theoirem: a Hausdorff topological space is the continuous image of $I=[0,1]$ iff it's connected, locally connected, compact and metrizable. Since our examples are subspaces of $\mathbb R^2$, they are metrizable, and the extra hypotheses assert that this example should be compact, connected, and locally path connected and thus locally connected.

In this context I want to ask 1 question how can I show that in a Hausdorff space arc connection means path connection,what I mean to say that though the definitions of the above terms are different but in Hausdorff space the 2 concepts become the same.


calculus
derivative
inequalities unsolved
inequalities

[quote="LoveStorm"]Let $x,y,\alpha\in[0,1]$. Prove that \[(1-x)^{\alpha}+(1-y)^{\alpha}\geqslant (1-xy)^{\alpha}\] [/quote]

$a=1-x, b=1-y$ then $\iff a^{\alpha}+b^{\alpha}\geq (a+b-ab)^{\alpha}$. This is done by $a^{\alpha}+b^{\alpha}\geq (a+b)^{\alpha}$ from $(x^{\alpha})'' \leq 0$ as $\alpha \leq 1$, and $a+b \geq a+b-ab$.

$a^{\alpha}+b^{\alpha}\geq (a+b)^{\alpha}$
$\iff x^{\alpha}+y^{\alpha}\geq 1$ for $x=\frac{a}{a+b}, y=\frac{b}{a+b}$
$\iff x^{\alpha}+y^{\alpha}\geq x+y$, which is trivial as $x, y, \alpha \leq 1$.


As many of you will know, Waterloo is not so enthusiastic on running contests..
(As far as I know, anyways..  :D )

I was told that the Tournament of Towns can be as hard as IMO, but apparently your town has to participate..  :(

Does anybody know how a town can start participating? I couldn't find any info on the internet (but then I probably didn't use enough time on searching..) If anyone knows, could i please get the info, so that I can convince my math teacher of profs from UW??

Thanks,

Jenn Park

U of Toronto maintains a pretty big site on T of Towns: http://www.math.toronto.edu/oz/turgor/index.php

You could always come to Toronto to do it  :D. I don't know if it would be worth it though.

Apparantly, Towns is only held in four cities in Canada: Toronto, Calgary, Edmonton and Vancouver. It may help to contact these [url=http://www.math.toronto.edu/oz/turgor/canada.php]organizers[/url], or [url=http://www.math.toronto.edu/oz/turgor/about1.php]Dr. Ivrii[/url] for further information.

Thanks, i'll look into it.

so yufei, in your expert opinion, how hard is the tournament of towns?

[quote="jpark"]so yufei, in your expert opinion, how hard is the tournament of towns?[/quote]

"Expert opinion" lol :lol:

Tournament of towns is not necessarily more difficult than an average Olympiad, but it is certainly different. Its test format and style of problems can really put one in a Russian environment (not that I know how [i]that [/i]feels).

In the beginning, I used to hate tournament of towns exams, perhaps because they were so full of combinatorics, and they don't really emphasize on the other mainstream Olympiad topics. But now I guess I'm more used and adapt to it.

On the other hand, the solutions for the problems (at least the ones they provide) are amazingly short!

So basically, towns may seem difficult at first because one is not used to it. Then again, diversity is important to the survival of us Olympians (wow that sounds like bio).

Hmm.. it doesn't sound too bad.. Although it would depend on the difficulty of the problems.

Even if I don't get to write it this spring, UW could probably organize it by next fall.. (I hope.. they probably could if they tried..)

Thanks for your resourcefulness, Yufei!  :lol:

In Calgary, we call it the Torture of the Towns.  Lol... we get pizza everytime we write it though!

That is odd, I mean, we have Dr. Sands...and Dr. Sands....hmm....I'm missing one... oh yeah!  Dr. Sands!!  Bug Ian into it, lol.  Maybe he's scared of staying 5 hours in one room with you around though, you never know   :P

Ack, enough of a break.  Back to Bio final studying!  AH!

Hey, that's actually a good idea..  :lol:

That's like a two-in-one deal, isn't it? I get to write a contest + I get to bug Ian...
The only drawback is that I don't see him anymore.. In fact, I hardly ever go to the university now..

But then, it'd be worth it to go there for this, right?  :D

[quote="jpark"]lol, you don't know how lucky you are..

I don't get why none of the profs want to organize it in Waterloo.. I mean, there's Ian, Dr. Rice, Dr. Crippin, and countless others involved with CMC..[/quote]

Hi Jenn, same situation occured to me two years ago in my town, I cried and beg the local proffessors to organize the tourn. of towns, and it didn't get me anywhere! :(   It seems that because of the lack of demand and the cost is an issue so I had no chance of writing...and the nearest place to write it is like thousands of kilometers away!!!  Fortunetly for you, Toronto is half and hour away  ;) , if it doesn't go well with Waterloo, Toronto is near by!  Good luck.

Argh!!!

WHO ARE YOU?????

That means I have to go through my stacks of papers to find that roommate-sheet!!!!
(which I'm not even sure if I have it..)

I don't even remember 'the flute guy'!!!

Let me place a few guesses:
wait, I've forgotten people's names, I just remember faces.. Are you the one with curly hair + glasses?  :blush:
Or... Argh, I give up..

[quote="jpark"]Wait, are you Robert Embree?? (one of the few people I remember clearly..  :lol: )[/quote]

 :)  Ahh Robert Embree, a GOOD FRIEND of mine!  I think he is at queens now.
ha ha ha...  :lol: Keep Trying Jenn!

If the flute guy is the one who sat on the same table with me for one of the banquets, I know who you are, mak.

And if you're who I think you are, I know where your username comes from too.

Am I correct?

I've become curious too.  Do I know mak?  lol, the mystery man...

Well, Sarah,

As far as I remember,  you were my roommate from 2003 Seminar, meaning that you must have sat at the same table as me.
And mak has specifically said that he was roommate to Andrei Borac, the "flute guy," who also sat at the same table as me.
That deduces to the fact that mak must have been with Andrei, so yes, you do know mak..

Hmm.. confusing..  ;)

Ack, I can picture Andrei perfectly, but I can't remember his roommate! and I couldn't find the roommate list from two years ago...(sooo long ago!)

I have a list of all the names though....that doesn't help, I could list out all 50 or so, lol

Keep trying..

I like the fact that I know the 'mysterious' guy mak!!  :D
Makes me feel superior over those who don't know mak..

Ack, i hate being curious.  Gimme a hint?  Pretty pretty please?

Bah...being backstabbed my an evil twin.  Clever...yet low. lol.  Never trust Jenn again!!!!!  BTW, when's Fermat?  I forgot.

[quote="Sunny"] Clever...yet low.[/quote]

Hey!! Just the fact that I'm living in the same city as Ian justifies all that! Consider that as my survival skill..

Fermat is on the February 23rd, two days before my birthday.

The Tournament of Towns apparently requires U.S. dollars, right? I found some information on the website, but most of the info is new to me, so I don't understand everything. According to the website, anyone can participate but to every grade a special coefficient is assigned. In everyday terms, what exactly are they stating? Is the competition written at the University of Toronto (for Toronto students)? What awards are given to individuals; and for teams, what other types of awards? Do you end up competiting against other towns in the world? If so, is there a cut-off mark to get into a town-competiting team? Does this contest have any effect into consideration for the Canadian IMO team? I am awfully confused, it would be greatly appreciated if someone could give me some insight about the test.

Hi d.lam.86

This competition is open to anyone from anywhere, however you must travel to cities that organize and run the competition.

The biggest awards you get in this competition, is an invitation to a research seminar in Russia.

Tornument of towns competition has questions that are very different from the standard olympiad problems (including IMO); I don't think it has any effect into consideration for the Canadian IMO team.  However, writing this contest will give you more experience in math contests!

The competition is for fun, helps you gain more insight in various math topics and it allows you to experience the "Russain Mathematics"  ;)

Thanks, mak, for giving me some info. I still don't know whether you have to pay US dollars and where the Tournament of Towns is written. Help?

What US dollars?

Since you live in Toronto, you just need to register on the [url=http://www.math.toronto.edu/oz/turgor/index.php]Toronto ToT website[/url] (when the fall comes), and show up accordingly. There is no charge.

As well, ToT has no effect whatsoever on the IMO. However, the students in the top group get $\$50.

Okay that's good to know. I haven't seen any middle school students in the Honour Roll. Does that mean that middle school students don't tend to participate or are unable to score high? I want to register as I begin grade 9 but I don't really want to participate at this moment. Should I register anyway?

if all you care about it getting some sort of prize or reward in the long run, forget it.  It is clearly a waste of your time.

Otherwise, if you are interested in looking at some cool problems and writing up the solutions, good.

I personally feel that's what olympiads are all about...there are way too many people who only care about making the IMO team just because, not because they like the math or appreciate its beauty.  But then again, most IMO people come from the "Math is cool" group, not the "Must make IMO group"

[quote="billzhao"]However, the students in the top group get $\$50.[/quote]
?#%^*@)(@)?!^%(?
Really? I didn't know that. How come I never got anything?(Maybe I wasn't in the top group :( )

Anyway, Tournament of Towns is [b]THE BEST[/b] competition [b]IF[/b](note that if is bold) you like beautiful problems. I love the problems. They are so insightful.
[quote="d.lam.86"]middle school students don't tend to participate or are unable to score high?[/quote]
I really don't think it matters. The problems are for all grades, even smart elementary students can understand and solve the problem with tiny knowledge of math(maybe I exaggerated little bit :blush: ). That's why the problems are so good.

In conclusion, if you like math, you [b]HAVE TO [/b]write it. It is shame not to solve such beautiful problems.

Only Toronto students get the $\$50. It varies from town to town.


Prove that there are infinitely many primes of the form $ 10k \plus{} 9$ where $ k$ is an integer.
Can this be proved without citing the Dirichlets theorem

Suppose there are finitely many $ p_1,p_2,...,p_k$ with $ P\equal{}p_1p_2\cdots p_k$ and consider the prime divisors of $ 20P^2\minus{}1$  :wink:


Solve in natural the  equation $ x^{3}\plus{}y^{3}\equal{}(x\plus{}y)^{2}$   :?:

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :ninja:

I can see that $ (x, y) \equal{} (n, \minus{} n)$ for natural numbers $ n$ will work and also $ (1, 2), (2, 1)$ and $ (2, 2)$.

(Otherwise, I suspect that $ x^{3} \plus{} y^{3} > (x \plus{} y)^{2}$. If $ x\equal{}y$, then we have $ 2x^{3} > 4x^{2}$, which is true for $ n>2$.)

So I think that the only solutions for natural numbers are $ (1, 2), (2, 1)$ and $ (2, 2)$.

Since $ x,y > 0$, we have that $ x\neq \minus{} y$.
So, the equation becomes $ x^2 \minus{} xy \plus{} y^2 \equal{} x \plus{} y$
$ x^2 \minus{} xy \plus{} y^2 \minus{} x \minus{} y \equal{} 0 \rightarrow 2x^2 \minus{} 2xy \plus{} 2y^2 \minus{} 2x \minus{} 2y \equal{} 0$
Which is:
$ (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \plus{} (x \minus{} y)^2 \equal{} 2$

So, the solutions are:
$ (1,2),(2,2),(2,1)$

[quote="Gabuntu94"]Since $ x,y > 0$, we have that $ x\neq \minus{} y$.
So, the equation becomes $ x^2 \minus{} xy \plus{} y^2 \equal{} x \plus{} y$ . . . (*)

[/quote]

Alternatively:

Also, solving the equation (*), which is in the quote box, for y, you get:

$ y \equal{} \frac {(x \plus{} 1) \pm \sqrt { \minus{} 3x^2 \plus{} 6x \plus{} 1}}{2}$

This can become:

$ y \equal{} \frac {(x \plus{} 1) \pm \sqrt {4 \minus{} 3(x \minus{} 1)^2}}{2}$ . . . (**)

It is necessary for $ 4 \minus{} 3(x \minus{} 1)^2$ to be a perfect square so that $ y$ is an integer,
but it is not sufficient.

So $ 3(x \minus{} 1)^2$ must be equal to $ 0, 3$, or $ 4$, potentially.

If $ 3(x \minus{} 1)^2 \equal{} 0$, then $ x \equal{} 1$.
If $ 3(x \minus{} 1)^2 \equal{} 3$, then $ x \equal{} 0$ or $ x \equal{} 2$.
If $ 3(x \minus{} 1)^2 \equal{} 4$, then there are no integer x-values.

Ignore $ x \equal{} 0$ for positive values/solutions for $ x$ and $ y$.

In equation (**),  if $ x \equal{} 1$, then $ y \equal{} 0$ or $ 2$.

Ignore $ (1, 0)$ because they are not both positive integers.

In equation (**),  if $ x \equal{} 2$, then $ y \equal{} 1$ or $ 2$.

Then the solutions in positive integers are:

$ (1, 2)$
$ (2, 1)$
$ (2, 2)$


It's finally coming out everyone! I've read it and I really want to see the movie, but I'm sad to say there's no way the movie is going to live up to the book. Douglass Adams was such a genious. Are you anticipating it?

I'm excited to see it, but I'm worried about the fact that it appears to be more of an action movie than a comedy, and the lack of Adams' narration, which is half of what makes the books funny.

On the flip side of my first issue, though, if they ever decide to make movies out of the sequels, those will have to belong more to the "serious movie" genre, as opposed to being comedies.

I've been wanting to read the book...

So I presume from the responses that the book is good and fun to read?

ankur: Yes, the book is extremely funny, and I highly recommend it.  I would also recommend the other four books in the "trilogy" (Restaurant at the End of the Universe; Life, the Universe, and Everything; So Long, and Thanks for All the Fish; and Mostly Harmless).  However, I don't recommend them as strongly as the first one; each one is more serious and less funny than the last (although equally good reading).

I read the first three; they were all pretty good. But the first one was definitely one to read.

As for the movie, it looks pretty cool I guess, but it has a reputation to live up to, so we'll see.

[quote="Danbert"]The first several books were great; the last two weren't that good, in my opinion.  When I saw a commercial for the movie, I winced--there's absolutely no way a movie could live up to the book; they'll probably end up butchering the storyline as well.  I think that, with a book like that, it's not possible for a movie to match the imagination.[/quote]

There's already a review.  They cut almost all the jokes out:

http://slashdot.org/article.pl?sid=05/04/10/1447214&tid=97&tid=214

Ouch. That's quite a beating on the movie. I'm quite curious to see it now :P.

[quote="Hexaditidom"][quote="nr1337"]This movie is from Disney. That's definitely... something to be concerned about.[/quote]

DISNEY???

Where did you find that out?[/quote]

Lots of places.  Just let me find them...

EDIT: Looky here:
http://www.allmoviephoto.com/photo/2005_The_Hitchhiker's_Guide_to_the_Galaxy_photo.html

QUOTE: "Martin Freeman as Arthur Dent, Sam Rockwell as Zaphod Beeblebrox and Mos Def as Ford Prefect in Disney Pictures' adventure The Hitchhiker's Guide to the Galaxy, directed by Garth Jennings."

Dammit, I just can't believe they are going to screw up Hitchhiker too. Movie studios did it with James and the Giant Peach, Tuck Everlasting, those Unfortunate Events books (although those were going down the drain anyway), and whose bright idea was it to make a two hour version of the Count of Monte Cristo? (Actually, despite the obvious, that was quite good.) And don't even get me started about Harry Potter, which was losing its appeal anyway when everybody [i]else[/i] started to like it. QUIT RUINING PERFECTLY GOOD BOOKS!


algorithm
combinatorics unsolved
combinatorics

A rectangular array has 9 rows and 2004 columns. In the 9 * 2004 cells of the table we place the numbers from 1 to 2004, each 9 times. And we do this in such a way that two numbers, which stand in exactly the same column in and differ around at most by 3. Find the smallest possible sum of all numbers in the first row.

I think the following algorithm gives the best result:
At each moment, place in the rightmost column the 8 biggest numbers remaining and place in first row the smallest possible number to satisfy the conditions of the problem.
I have a proof, but it's not very rigorous and I'm not sure..
Here is an example for n=6
1 1 1 2 3 3
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
2 3 4 5 5 6
2 3 4 5 6 6

I dont know why nobody post a solution yet!
lets solve this problem for N
in this problem N=2004
the solution is = (n-3)(n-2)/2 + n-1
use strong induction on (N)and show that if
a1<=a2<=....<=an are the numbers in first row we have :
a(n)>=n-3
a(n-1)>=n-3
a(n-2)>=n-4
a(n-3)>=n-5
......
a4>=2
a3>=1
a2>=1
a1>=1
and we have 1+1+1+2+3+....(n-5)+(n-4)+(n-3)+(n-3)=(n-3)(n-2)/2 + n-1
create an example for holding quality isnt hard.


In right triangle hypothenus = m and incircle's radius =r. Find other sides of triangle. In what relationship between m and r this problem has solution.

hello, we have $ m\plus{}2r\equal{}a\plus{}b$ and from here by squaring both sides $ ab\equal{}2rm\plus{}2r^2$, solving this system for
$ a$ and $ b$ we get
$ a\equal{}\frac{1}{2}m\plus{}r\minus{}\frac{1}{2}\sqrt{m^2\minus{}4rm\minus{}4r^2}$
$ b\equal{}\frac{1}{2}m\plus{}r\plus{}\frac{1}{2}\sqrt{m^2\minus{}4rm\minus{}4r^2}$
Further we get $ m^2\minus{}4rm\minus{}4r^2\geq0$ this is equivalent with $ |m\minus{}2r|\geq2\sqrt{2}r$, in the case of
$ m\geq2r$ we get the condition $ m\geq2r(\sqrt{2}\plus{}1)$.
Sonnhard.


Hy i  am having problems understanding this type of problems, because i am not really sure of the math that i am doing regarding signs and inequations:

[tex]\|{x/(x+2)}\|\leq{2}[/tex]

I do these:

[tex]{x/(x+2)}\leq{2}[/tex]

and

[tex]-{x/(x+2)}\leq{2}[/tex]

That would be correct, but the common way to express it would be:

$\frac{x}{x+2}\leq{2}$ and $\frac{x}{x+2}\geq{-2}$

Thanks 4everwise!
I have another question:

I have this problem "seeing" what this is. I have this inequation: [tex]\|{(2x-1)/x}\|>2[/tex] So i draw the graph also on derive: http://img339.imageshack.us/img339/3338/modulo8gr.jpg

Now the problem is that for me what this inequation with absolute value expresses is that when you take a number in the Dominium this inequation tells me that the only possibility to get a result > 2 is from (-inf, x=1/4), this of course in the Range. I mean, this is not the same as saying that it expresses a distance > 2 of two points on the x axis? I understand this as using the same principle about distance (absolute value) and interval (what inequations express).
So in conclusion, i see this as the distance from a value in the Dominium of the function to that value in the Range of the function, which by the way is a homographic. I am wrong? Thanks a lot for any help!


i really like math..and im pretty sure everyone here does as well...and im wondering what can u really do with a math degree....like most jobs out there like engineering has math in it but also stuff like phyics or chemistry..so what do people who really like math especially pure mathematics do as a job...im planning on getting an education in math, pure math and im not really interested in any other subject and to a small extent i dont even really like applied mathematics..so what is out there in the world that a mathematician can get into except research and teaching?

There's plenty of stuff you can do with a pure math degree. Don't worry about it, and enjoy math.

Any examples?

I'm in the same boat as SLeon. Exact same boat.

I'm thinking about doubling in Engineering and Math just so I can get a job thats not research/teaching.

What kind of jobs use math and are exciting at the same time?

You can really do anything with a math degree. The typical jobs are professor and working for the NSA. Of course, there are others that, truthfully, I haven't researched. (I'd much rather make 60K a year as a pure mathematician than 1M a year as an engineer.) Check out the websites and magazines of the AMS and MAA. I'm sure you can find a lot of articles on the subject.

You might want to check out the Mathematical Sciences Career Information page on non-academic employment on the American Mathematical Society's web site:

[url]http://www.ams.org/careers/[/url]

[quote="ComplexZeta"]You can really do anything with a math degree. .[/quote]

I'll add a caveat to that: [i]if you learn how to program a computer[/i].

Very true. I still cant decide if what Im more interested in besides math, cognitive psychology or comp sci.

[quote="rrusczyk"]I'll add a caveat to that: [i]if you learn how to program a computer[/i].[/quote]

Yeah, but you can pick that up whenever you want in a month or so.

The trouble is convincing a prospective employer you have done so.

you guys are really funny..and ive been looking at those websites and im really beginning to think that i better start likeing computers and science... becuase it does seem like all the jobs out there with math in it is more computer science or just science...I'm guessing it's sort of part of the job?...like jobs out in the world are jobs where u have to learn something with it...now im getting pretty scared about getting just a math degree...

I know how you feel, SLeon.

I want to major in pure math, but I also want to make a decent living. I might end up majoring in computer science, although I don't think I'll enjoy doing it for a living...

If only there were cool jobs for mathematicians that didn't involve teaching, reserach, and financial planning.

If you want to do pure math but not teaching or research, then what exactly do you want to do?

You can always try to become an olympiad trainer :D

Problem solving, which is one of the most important concepts in math competitions, is a skill that can be applied to other fields. Several other invaluable skills can be gained through mathematics as well, such as persistence, (I SWEAR I WILL GET THIS PROBLEM DONE OR ELSE!!!!!!), circumspection, (as in, being careful enough not to make stupid mistakes) and dealing with pressure (deadlines for math competitions, and the mere act of competing).

Many people who major in physics, chemistry, or mathematics ultimately move onto other fields because they cannot find a job in the fields they majored in. At least they've gained the skill of problem solving.

But only do it if it's fun. Since if it isn't, you could be better off doing something else.


find all value b such that : sequence$(x_{n})$ sastisfying
$x_{1}=b$;$x_{n+1}=x_{n}^2 +(1-2a)x_{n} +a^2$
   sequence $x_{n}$ have limit  when $n$ to  infly .( $a$ is const )

$x_{n+1}-x_n=(x_n-a)^2\geq 0$ so $x_n$ is increasing and the hypothesis tells us that it is convergent so $\lim_{n\to\infty}x_n\geq x_1=b$. When calculating it`s limit we get that  $l-l=(l-a)^2\rightarrow l=a$ so $a\geq b$.

and I think you should also prove $x_n \le a$ when $a \ge b$
but that's easy

[quote="spix"]well $x_1=b\leq a$ and since $x_n$ is increasing and has limit $a$ then obviously $x_n\leq a$.[/quote]
sorry,I don't think you can prove by this way.
because we must prove $x_n \le a$ first,then prove $x_n$ has limit.... :)

if $x_n>a$ then the sequence is unbounded therefor $x_n\to\infty$


geometry
geometric transformation
rotation
reflection

$8$ boys arrive at a seaside resort, having $7$ suites with $2$ beds each. They want to stay for $7$
day in such a way that

$1)$ Each day, two boys occupy a suite, leaving three suites vacant
$2)$ Each boy occupies a different suite each day
$3)$ Each boy has a different partner each day.

Show how this may be arranged.

There is one basic solution, which after rotations,reflections and transpositions of columns and rows leads to all others.


Rather than visualizing folding the net in your head, there is a simple and concrete way of determining which edges are identified with each other.  Choose a vertex of the net that has an excess of incident edges, and label the two edges on the boundary as the same identified edge.  For example, the exposed edges on 2 and 7 are joined to each other since the incident vertex contains 4 faces but 5 edges as shown by the net.  We do the same for exposed edges on 1 and 4, and 5 and 6, respectively.  Then traverse the remaining edges on the boundary, and it is simple to see how they match up.  Going counterclockwise starting from the free vertex of the "1" face, we would label the edges like this:  abbacddcee.


Find the value of $ x$ that satisfies the equation
\[ 25^{\minus{}2}\equal{}\frac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}.
\]$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 9$

[quote="worthawholebean"]Find the value of $ x$ that satisfies the equation
\[ 25^{ \minus{} 2} \equal{} \frac {5^{48/x}}{5^{26/x}\cdot25^{17/x}}.
\]
$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$[/quote]

We have
$ 25^{ \minus{} 2} \equal{} \frac {5^{48/x}}{5^{26/x}\cdot25^{17/x}}\equal{}25^{\frac{\minus{}6}{x}}$
therefore
$ \minus{}2\equal{}\frac{\minus{}6}{x}$
so
$ x\equal{}3$

To check: a quick way to change bases of an exponent is (without going into logs rules) is that what root you put on the base (like a square of cube root) you would multiply the power by that root number (like 2,3 respectively from the previous example)

[hide="Easier?"]Take the base-5 logarithm of both sides:
$ \minus{}4\equal{}\frac{48}{x}\minus{}\left(\frac{26}{x}\plus{}\frac{34}{x}\right)\implies \minus{}4x\equal{}48\minus{}26\minus{}34\equal{}\minus{}12\implies x\equal{}\boxed{\textbf{(B)}\ 3}$.[/hide]

[hide="Click for solution"]
Write $ 25$ as $ 5^2$.  Raising both sides to the $ x$th power yields

$ 5^{\minus{}4x}\equal{}\frac{5^{48}}{5^{26}5^{34}}\equal{}5^{48\minus{}26\minus{}34}\equal{}5^{\minus{}12}$.

Thus, $ \minus{}4x\equal{}\minus{}12 \implies x\equal{}3$, so the answer is $ \boxed{\textbf{(B)}}$.
[/hide]

It's much easier to work with integers than fractions.


There exist four numbers such that the sum of the factorials of the digits of the number equal the number.
Three are 1, 2, and 145.
Find the fourth.

[hide="Zip code of Lexington KY"]Let us try: 40585   :)[/hide]

:rotfl:  Was that a guess, or did you actually do something to solve it?

Discussion about a bound on the number of digits:

Say that our number has $n$ digits.

Then we have the maximum of the number is $n*9!$ since the largest possible digit is $9$.

Hence, we have that $10^{n-1}\le 9!n<10^{n}\implies n-1< 12.8\log n<n$ since $\log x$ is an increasing function.

Note that $f(x)=\frac{x-1}{\log x}$ is also an increasing function for sufficiently large $x$...a weak bound is $x\ge 2$ [use calculus].

Anyway, we have that $\frac{n-1}{\log n}< 12.8$ which implies that $n\le 16$

Obviously, this is a horrible bound. But it shows us that there are finitely many solutions.

Yea...I did natural logarithm by mistake...

This is almost like problem 34 on http://www.projecteuler.net.
The result is $40585$.
These numbers are called [url=http://mathworld.wolfram.com/Factorion.html][i]factorions[/i].[/url]

[quote="Gyan"][quote="Altheman"]Discussion about a bound on the number of digits:

Say that our number has $n$ digits.

Then we have the maximum of the number is $n*9!$ since the largest possible digit is $9$.

Hence, we have that $10^{n-1}\le 9!n<10^{n}\implies n-1< 12.8\log n<n$ since $\log x$ is an increasing function.

 [/quote]

There may be a simple error/typo ... log of 9! base 10 (not e)  is about 5.6   and you  will get something like (5.6+log n)  in the middle ...
Actually limit is much lower (6-7 digits)... small enough (few seconds of cpu)  even for those who  are not lucky enough to simply   live in KY and guess :)[/quote]
How would one do this without a computer?
This problem is from intermediate number theory...


Two cylinders $ A$ and $ B$ are holed at different places in the inside, such that the moment of inertia of $ A$ is more than $ B$. The two cylinders are let to roll down a slope (with friction), starting from rest. Is it possible that all the time the cylinders are rolling without slipping? If so, which one reaches the ground first?

Once it reaches its final velocity it won't be slipping anymore so $ v\equal{}\omega r$.  But this equation doesn't hold if there is slipping.

[quote="JRav"]Once it reaches its final velocity it won't be slipping anymore so $ v \equal{} \omega r$.  But this equation doesn't hold if there is slipping.[/quote]

So is it possible that $ v \equal{} \omega r$ holds all the way right from start?  :maybe:

If there is friction, then there is no slipping - as is the usual supposition made in problems like this. So we can use $ v \equal{} \omega r$.The purpose here is simply to relate the final velocity to the moment of inertia. However, note that there is conservation of energy, and so part of the initial potential energy is converted not only is translational kinetic energy but also in rotational kinetic energy. As this last one is directly proportional to the moment of inertia, then the cylinder with greater moment of inertia will have less translational kinetic energy, and so it will reach the ground last.


floor function
number theory unsolved
number theory

Prove that $ n!$ is divisible by $ 2^{n\minus{}1}$ if and only if $ n\equal{}2^{k\minus{}1}$ for some positive integers $ k$.

Let m be the greatest integer such that $ 2^m \le n$.
$ \lfloor \frac {n}{2} \rfloor \plus{} \lfloor \frac {n}{4} \rfloor \plus{} \lfloor \frac {n}{8} \rfloor \plus{} ...\lfloor \frac {n}{2^m} \rfloor$ is the number of factors of 2 dividing $ n!$, which must be at least $ n \minus{} 1$ since $ 2^{n \minus{} 1}$ divides n.
We must have
$ n \minus{} 1 \ge \frac {n}{2} \plus{} \frac {n}{4} \plus{} \frac {n}{8} \plus{} ...\frac {n}{2^m} \ge \lfloor \frac {n}{2} \rfloor \plus{} \lfloor \frac {n}{4} \rfloor \plus{} \lfloor \frac {n}{8} \rfloor \plus{} ...\lfloor \frac {n}{2^m} \rfloor \ge n \minus{} 1$.
Equality holds when n is a power of 2.


Find the minimum possible value of $ x^{2}\plus{}y^{2}$ given that $ x$ and $ y$ are real numbers satisfying

$ xy(x^{2}\minus{}y^{2}) \equal{} x^{2}\plus{}y^{2}$ and $ x \neq 0$.

the minimum value of $ A\equal{}x^2\plus{}y^2$ is 4

hello, you can use the method of Lagrange multipliers. I think there is no maximum.
Sonnhard.

Thanks Sonnhard. I guess it's easier (and quicker) to let $ x \equal{} r \cos \theta$ and $ y\equal{}r \sin \theta$, that is change to polar coordinates.


For any positive integer $ n$ prove that there exists a polynomial $ P(x)$ of degree atleast $ 8n$ such that
$ \sum_{k\equal{}1}^{(2n\plus{}1)^2}  |P(k)| \le |P(0)|$

Take $ P(x) \equal{} L(x) \plus{} Q(x)\prod_{k\equal{}0}^{N} (x\minus{}k)$, where $ L(x)$ is the Lagrange interpolation polynomial (of degree at most $ N$) with $ L(0) \equal{} 1$ and $ L(k) \equal{} 0$ for $ 1 \leq k \leq N$.
Then $ \sum_{k\equal{}1}^N | P(k) | \equal{} 0 < 1 \equal{} | P(0) |$, while $ \deg P \equal{} (N\plus{}1)\deg Q$ may be made as large as wanted.


How many pairs of positive integers (a, b) with a<=b satisfy 1/a+1/b=6

I'm assuming you mean $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{6}$, because otherwise no positive integers will work.

[hide="If so"]
$ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{6}$

$ b \plus{} a \equal{} \frac {ab}{6}$
$ 6b \plus{} 6a \equal{} ab$
$ 6b \plus{} 6a \minus{} ab \equal{} 0$
$ \minus{} 6b \minus{} 6a \plus{} ab \equal{} 0$
$ \minus{} 6b \minus{} 6a \plus{} ab \plus{} 36 \equal{} 36$
$ (a \minus{} 6)(b \minus{} 6) \equal{} 36$

Since we want $ a \le b$, we could have pairs $ (b,a)$ of $ (42,7)$, $ (24,8)$, $ (18,9)$, $ (15,10)$, and $ (12, 12)$.
[/hide]


I cant get this problem because their should be at least three numbers to find [i]N[/i] but it only gives you 2?

                                                  35% of 40 is[color=red] N[/color]

                                                  [u] 35%    N[/u]
                                                        N   =   40                     :what?: [/i]

So, 35% of 40 is N. We know 35% is equivalent to 7/20. So, 7/20 * 40 = 14 = N. You just have to understand that it's asking for a fraction or percentage of a given number, so you can indeed find N.

35 * 40 / 100

that's what you should do to get the N

and I didn't understand what you were talking about...

this part..

[quote]35% N
N = 40  [/i][/quote]

I don't see how that is so hard...
I would just do .35*40=n and just multiply .35 and 40.
n=14

[quote="Fowi Hunter"]I don't see how that is so hard...[/quote]

We have forums for users of various levels of math experience.  The purpose of this forum is for discussion of middle school curriculum.  If you are looking for something to challenge you, keep moving up a level in forums until you find it.


calculus integration geometry perimeter inequalities

The number of scalene triangles having all sides of integral lengths, and perimeter less than $ 13$ is:

$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$

just do casework with $ a\plus{}b\plus{}c \equal{} 13$, with $ a<b<c$ and $ c<a\plus{}b$ (combining the inequalities to $ a<b<c\minus{}a$ may help)

[hide="only triangles"]2,3,4
2,4,5
3,4,5[/hide]

It's easy to prove that they are the only ones by the Triangle Inequality.


ABC triangle.  a+b+c=2
Prove that:
a 2 +b 2 +c 2 + 2abc<2

This is a British MO - question, early 90's I think (1992 ?).
Note that $a, b, c < 1$. Hence $2(1 - a)(1 - b)(1 - c) > 0$, so $2 - 2(ab + bc + ca) + 2abc < 0$.
Now add $(a + b + c)^2 = 4$ to both sides to get the result.


[hide]
$\frac{100}{4}-6 = 19 \text{feet, the diameter}$
[/hide]

[quote="sharkman"]A pond is enclosed by a wooden deck that is 3 feet wide.  The fence surrounding the deck is 100 feet long.  If the pond is circular, what is the diameter of the pond?[/quote]
[hide]
The radius of th pond and deck is $r+3$ so $2\pi(r+3)=100$ so $r=\frac{50}{\pi}-3$

Therefore the diameter is $\frac{100}{\pi}-6$.[/hide]