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$(1)$ Obviously this dont converges........

It's funny when obvious things turn out to be wrong ... near 0, $\sin x$ looks very much like $x$, so for large $n$ that expression looks very much like $\sum \frac{2^{n}}{3^{n}}x$ and so converges.  It does not look to me like the value has a nice form, however, if that's what "solve" means.

[quote="JBL"]It's funny when obvious things turn out to be wrong[/quote]
Not funny at all when this happens in your paper.  :(

I think the question is: for what values of $x$ does the series converge? and the answer is all $x$ for (1) and $x\ne 1$ for (2)... probably.

Yes. That product is close to $\frac1{\sqrt{n}}$ in the second case.

Evaluating the second series:
Let $f(t)=\sum_{n=0}^{\infty}\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}t^{2n}$. Then $f'(t)=\sum_{n=1}^{\infty}\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n-2)}t^{2n-1}$, $\frac1tf'(t)=\sum_{n=1}^{\infty}\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n-2)}t^{2n-2}$, and $\int_{0}^{t}\frac1yf'(y)=\sum_{n=1}^{\infty}\frac{1\cdot 3\cdots (2n-3)}{2\cdot 4\cdots (2n-2)}t^{2n-1}=tf(t)$.

Differentiating both sides, $\frac1tf'(t)=f(t)+tf'(t),$ and $\frac{f'(t)}{f(t)}=\frac{t}{1-t^{2}}.$ This is a separable differential equation, with solutions $\ln f(t)=-\frac12\ln(1-t^{2})+C$ on $(-1,1)$; exponentiating, the solutions are $f(t)=\frac{A}{\sqrt{1-t^{2}}}$. Since $f(0)=1$, $A$ is 1. Substituting $t^{2}=\frac{2x}{1+x^{2}}$, the sum of the series in the post is $f\left(\sqrt{\frac{2x}{1+x^{2}}\right)-1= \frac1{\sqrt{1-\frac{2x}{1+x^{2}}}}-1= \frac{\sqrt{1+x^{2}}}{|x-1|}-1}$

This could also have been found by recognizing the binomial series, but generating function techniques are good to know.

Dear JBL

$\sum_{n = 1}^\infty{2^{n}\sin \frac{x}{{3^{n}}}}$

It's different of

${\sum_{n = 1}^\infty{(\frac{2}{3})^{n}\sin(x)}}$

the question is. if we make a substituition of the Sum for a Integral. the integral dont converges.

If converges just for one value, It's a really strange question.

I Solved the question for mora natural solutions, and not for trivial solutions.

The question is to evaluate, not approximate.

so x not equal to zero.
x reals.
etc.

like division by zero

yagaron: I cannot decipher the meaning of your post. But JBL is absolutely correct, and the series converges for all $x.$ He's not saying that the two series are the same series; rather, he's invoking the limit comparison test for convergence.

Finding an exact value for that sum looks like a very tricky question. If that were a $2^{n}$ in the denominator rather than $3^{n}$, there would be more tricks we could use. There are enough trigonometric identities in the world that I won't swear that an exact evaluation is out of reach, but I don't see the method.

sorry but I do not see how $\sin\left(\frac{x}{3^{n}}\right)\equiv \frac{x}{3^{n}}$ for $x$ close to 0, and how does this help prove that the sum converges?

[quote="maokid7"]how $\sin\left(\frac{x}{3^{n}}\right)\equiv \frac{x}{3^{n}}$ for $x$ close to 0[/quote]
JBL wrote that those two things "looks very much like" each other, not that they are equal. The precise meaning of his statement is $\lim_{n\to\infty}\frac{\sin\left(\frac{x}{3^{n}}\right)}{\frac{x}{3^{n}}}=1$

nobody prooves nothing.

I Test for some values in mathematica and it SAys that "does not converges"

so, how can it converges for all $x$ if dont converges for some x?

Zeno paradox maybe?

That's wrong. A simple proof of convergence:
First, $|\sin(t)|\le |t|$ for all real $t$. Now $|2^{n}\sin\frac{x}{3^{n}}|<2^{n}\cdot |\frac{x}{3^{n}}|=\left(\frac23\right)^{n}\cdot x$.
By the comparison test, the sum converges for any $x$ and the sum is at most the sum $2x$ of the dominating series.

That was redundant- we've already shown convergence for $x\neq 1$ and evaluated that series exactly.

$\sum_{n = 1}^{\infty}{\frac{{1\cdot3\cdot\cdots\cdot(2n-1)}}{{2\cdot4\cdot\cdots\cdot2n}}}\geq\sum_{n = 1}^{\infty}\frac{1}{2\sqrt{n}}=+\infty$.


Hi,

i came across this problem in a programming contest.
Its somewhat similar to the Nim game that i posted a couple of days earlier, in the fact that the approach to solve the problem is through xor operations.

Here is the description of the game:

There are k stone piles with n1, n2, ... , nk stones. One is permitted to take any positive number of stones from one pile for a move, but no more than half of the pile.

The one who can't make his move loses.

Examples:

3 2 1  --------- First wins.

1 3 7 63  ------ Second wins.

Any solution to this problem is much appreciated.

[color=olive][u]Preface[/u]
I solved a large part of the problem. I can find an infinite number of initial situations such that you can win. But there are some other situations that leads you to victory, and I didn't find yet a formula to express them [/color]

Let $L$ be the set of the loosing positions
and $W$ the set of the winning positions

the final situation is $(1,1,...,1)\in L$

We assume that [u]your opponent[/u] goes first. If the initial situation is [b]L[/b], then you are the winner

[color=indigo][size=150][hide="The Equivalence Relation (click)"][/size][/color]
[color=red][b][u]Definition[/u]:[/b]
Two numbers $a,b$ are said to be [u]equivalent[/u] $(a\equiv b)$ iff there is $n\geq 0$ such that either $\boxed{\frac{a+1}{b+1}=2^{n}}$ or $\boxed{\frac{b+1}{a+1}=2^{n}}$[/color]

[color=indigo]Notice that $\boxed{a\equiv a}, \boxed{\ a\equiv b \iff b \equiv a}$, and $\boxed{a\equiv b \wedge b\equiv c \Rightarrow a\equiv c}$

This definition allows to define the classes
Let $c_{0}$ be the smaller element of some class (this will be the [b][u]generator[/u][/b]). Then we can find all the members of the class by using the formula
$c_{n}= 2^{n}(c_{0}+1)-1$ or by the recursive formula $c_{n+1}= 2\cdot c_{n}+1$. Check it!

The class of $1$ is $C[1]=\{1,3,7,15,31,63,127,... \}$
Also
$C[2]=\{2,5,11,23,47,95,191,... \}$
$C[4]=\{4,9,19,39,79,159,319,...\}$
$C[6]=\{6,13,27,55,111,223,447,...\}$
$C[8] =\{8,17,35,71,143,287,575,...\}$
[/color][/hide]$\hline$

Suppose that there is only one pile with $127$ stones. This is an $L$-position. Indeed,
if your opponent remove $1$ stone, you will remove $63$ stones. Result=$63$
if your opponent remove $63$ stones, you will remove $1$ stone. Result=$63$

So, you can walk your opponent in the path $127 \to 63 \to 31 \to 15 \to 7 \to 3 \to 1$
Below, such path is called "transformation". The number $127$ has been transformed to the number $1$.
The transformation reffers only to [b]L[/b]-positions, the positions that the opponent is called to make some move.

[color=brown][size=150]Theorem 1[/size]
Let $c_{0}$ be a generator of some class ($c_{0}=1$ or an even number), and $m\in C[c_{0}]$ any odd number $m>c_{0}$.
Then $m$ can be transformed to $c_{0}$[/color]
[color=blue][size=150]
[hide="Proof"][/size]
The proof is easy, let $n\geq 0$ such that $m=c_{n+1}$. Then $m=2\cdot c_{n}+1$
The opponent removes $s\leq c_{n}$ stones.
Then you remove $c_{n}+1-s$ stones and the result is $2\cdot c_{n}+1-s-(c_{n}+1-s)=c_{n}$

You'll need $n$ steps more to reach $c_{0}$[/color]
[/hide]$\hline$

So we can suppose that in the initial situation $(n_{1},n_{2},...,n_{k})$ all the numbers are either [b]1[/b] or even numbers

[color=green][size=150]Conclusion[/size]
If the initial situations consists only of members of $C[1]$, you will win
That's why "1 3 7 63 ------ Second wins"[/color]

Thus, every pile which has $n\in C[1]$ stones could be ignored

[color=brown][size=150]Theorem 2[/size]
If two piles have the same number of stones, then they can be ignored[/color]

[color=blue][size=150][hide="Proof"][/size]
If there are two piles with $m$ stones each, then you can do what your opponent last did.
If he remove $s$ stones from one pile, then you remove $s$ stones from the other pile. Under this sequence, some time both piles will have [b]1[/b] stone each, so they can be ignored[/color]
[/hide]$\hline$

[color=brown][size=150]Theorem 3[/size]
When we remove some stones from a pile with [b]m[/b] stones, the possible results are pairwisely not equivalent[/color]

[size=150][hide="Proof"][/size]
Let $m=2a+r$, where $r=0\text{ or }1$

We can remove at most $a$ stones. The result is $x=(a+r)+s,$ with $s=0,1,...,a-1$
Total $a$ different results

If $c_{0}$ is the generator of $C[x]$ then $c_{0}\leq x \Rightarrow c_{0}<m$

If $m=2a$ then there are $a$ classes with generator $<m$; these are the classes $C[1],C[2],C[4],...,C[2(a-1)]$

If $m=2a+1$ then there are $a+1$ classes with generator $<m$; these are the classes $C[1],C[2],...,C[2(a-1)], C[2a]$
[i]Notice that, in this case,[/i] [b]C[a][/b] [i]is not possible, so there are again[/i] [b]a[/b] [i]possible classes to contain the result[/i]

We will prove that each of $x=a+r+s$ belongs to a different class.

We suppose conversely that $\exists b,c$ with $0\leq b<c\leq a-1$ such that $a+r+b\equiv a+r+c$. Then

$\frac{a+r+c+1}{a+r+b+1}=2^{n}$, where $n>0$ because $c\neq b$

$\Rightarrow a+r+c+1 = 2^{n}(a+r+b+1) \Rightarrow$
$c+(a+r+1) = 2^{n}(a+r+1)+2^{n}b \Rightarrow$
$c = (2^{n}-1)(a+r+1)+2^{n}b \Rightarrow$
$c = (2^{n}-1)(a+r+1)+(2^{n}-1)b+b \Rightarrow$
$c-b = (2^{n}-1)(a+r+1+b)$

$n\geq 1 \Rightarrow 2^{n}-1\geq 1$, so the right part is $\geq a+r+1+b\geq a+r+1$

But the left part is $c-b\leq a-1$, contradiction[/hide]$\hline$

This theorem says that, depending on your opponents' move, you can take some pile with $m$ and convert it to any other class you want with generator $<m$, such that it will cancel out with some other pile.
Your opponent always breaks the balance. You always fix it.

[color=darkblue][size=150][hide="Examples"][/size]
$\blacksquare$ We can combine [b]L[/b]-positions to find a new [b]L[/b]-position.
For example $(7)\in L$ and $(4,4)\in L$. Then $(7,4,4)\in L$
$(2,5)\ (+)\ (8,8) = (2,5,8,8) \in L$
$(1,3,7)\ (+)\ (8,71)\ (+)\ (13,55) = (1,3,7,8,71,13,55) \in L$

$\blacksquare$ In the start of my post, I wrote that there are some [b]L[/b]-positions that I cannot determine; here is an example:
$(2,4,6)$ is an [b]L[/b]-position. The opponent can remove from [b]2[/b],[b]4[/b] or [b]6[/b]. Always you'll have a strategy

$(2,4,6)\to (1,4,6)$ The pile with [b]1[/b] stone doesn't matter; You remove [b]2[/b] from [b]6[/b] $\Rightarrow (1,4,4)$. Remember that it is a "balanced" situation, hence $L$

$(2,4,6)\to (2,3,6)$ Forget [b]3[/b]. Move $(2,3,6) \to (2,3,5)$, which is $\equiv (2,3,2)$

$(2,4,6)\to (2,2,6)\to (2,2,3)$

$(2,4,6)\to (2,4,5)\to (2,3,5)\equiv (2,3,2)$

$(2,4,6)\to (2,4,4)\to (1,4,4)$

$(2,4,6)\to (2,4,3)\to (2,2,3)$[/color]
[/hide]$\hline$

The above example $(2,4,6)$ consists of three different classes.

Which is the relation between the three classes?
Why doesn't $(2,4,8)$ work?
What happens if we have 4 piles with different number of stones? (different classes, for example 2,4,6,8)

These are questions that I have not answered yet
But I enjoyed this game!  :)

[i]PS
Sorry for the many edits, but they were necessary[/i]

I define the class generator in a different way
Let $A(x)$ be the class generator of $x$.

There are multiple ways to find the class generator of some number $x$.

For example, the class $C[6]=\{6,13,27,55,...\}$ in the binary system is $\{110,1101,11011,110111,...\}$
Notice that, to find the genarator of some element, we should remove all the last 1s.

So, if $n$ is the smallest natural number such that $2^{n}\mid x+1$ then $A(x)=\frac{x+1}{2^{n}}-1$

And $x=(2a)\cdot 2^{n}+2^{n}-1$

- If $x$ is even then $A(x)=x$
- If $x$ consists of only ones, then $A(x)=0$
- $\forall x, \ A(x)\leq x$
- and at any case $A(x)=2a$, which is even

[color=brown][size=150]Theorem 4[/size]
$(n_{1}, n_{2}, ..., n_{k})\in L\iff A(n_{1}) \otimes A(n_{2}) \otimes ... \otimes A(n_{k})=0 \ \ (1)$[/color]

[size=150][hide="Example"][/size]There are [b]10[/b],[b]13[/b], and [b]12[/b] stones.
We write each number in the binary form and seperate it to two parts: the left and the right.
The left part ends with the last zero of the number. So the left part of $x$ is $A(x)$
The right part consists only of 1s. We just [u]don't care[/u] about these ones.

$\begin{array}{c|llll|ll}10&1&0&1&0&&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&0&0&0&0&&\\ \end{array}$

Notice that there exists an XOR balance (the bitwise XOR of $A(n_{i})$ is 0).
Assume that the opponent chooses to remove from [b]10[/b]. He should leave [b]5,6,7,8[/b] or [b]9[/b] stones

$\blacksquare$ Opponent's move: $10\to 5$

$\begin{array}{c|llll|ll}5&&&1&0&1&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&1&0&0&0&&\\ \end{array}$

The XOR is [b]1000[/b], so your move should be such that fixes this error. You should remove $x$ stones from the pile-[b]i[/b], such that $A(n_{i}) \otimes A(n_{i}-x)=1000 \Rightarrow A(n_{i})\otimes 1000 = A(n_{i}-x)$

Let's XOR each of $A(5),A(13),A(12)$ with $1000$

$\begin{array}{c|llll|ll}5&1&0&1&0&=10\\ 13&1&1&1&0&=14\\ 12&0&1&0&0&=4\\ \end{array}$
This table tells us how to play. If we change $5$ then we should make it $10$ to fix the balance. But this is not possible, because $5<10$. Also $13<14$. The only solution is $12\to 4$

But $4$ is too small for $12$ (we can't subtract $8$), so we add as many 1s we need to the right of "100". So we have $100\to 1001$

So, the appropriate move is "remove 3 stones from 12"
and the result is
$\begin{array}{c|llll|ll}5&&&1&0&1&\\ 13&&1&1&0&1&\\ 9&&1&0&0&1&\\ \hline \otimes&0&0&0&0&&\\ \end{array}$

Similarly, if the opponent leaves 6,7,8,9 stones we have:

$\blacksquare$ Opponent's move: $10\to 6$

$\begin{array}{c|llll|ll}6&&1&1&0&&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&1&1&0&0&&\\ \end{array}\Rightarrow \left(\begin{array}{c|llll|ll}6&1&0&1&0&=10\\ 13&1&0&1&0&=10\\ 12&0&0&0&0&=0\\ \hline &&&&&\\ \end{array}\right) \Rightarrow \begin{array}{c|llll|ll}6&&1&1&0&&\\ 13&&1&1&0&1&\\ 7&&&&0&111&\\ \hline \otimes&0&0&0&0&&\\ \end{array}$

Notice that we have two alternative choices, $13\to 10$ or $12 \to 7$

$\blacksquare$ Opponent's move: $10\to 7$

$\begin{array}{c|llll|ll}7&&&&0&111&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&1&0&1&0&&\\ \end{array}\Rightarrow \left(\begin{array}{c|llll|ll}7&1&0&1&0&=10\\ 13&1&1&0&0&=12\\ 12&0&1&1&0&=6\\ \hline &&&&&\\ \end{array}\right) \Rightarrow \begin{array}{c|llll|ll}7&&&&0&111&\\ 12&1&1&0&0&&\\ 12&1&1&0&0&&\\ \hline \otimes&0&0&0&0&&\\ \end{array}$

$\blacksquare$ Opponent's move: $10\to 8$

$\begin{array}{c|llll|ll}8&1&0&0&0&&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&0&0&1&0&&\\ \end{array}\Rightarrow \left(\begin{array}{c|llll|ll}8&1&0&1&0&=10\\ 13&0&1&0&0&=4\\ 12&1&1&1&0&=14\\ \hline &&&&&\\ \end{array}\right) \Rightarrow \begin{array}{c|llll|ll}8&1&0&0&0&&\\ 9&&1&0&0&1&\\ 12&1&1&0&0&&\\ \hline &0&0&0&0&&\\ \end{array}$

$\blacksquare$ Opponent's move: $10\to 9$

$\begin{array}{c|llll|ll}9&&1&0&0&1&\\ 13&&1&1&0&1&\\ 12&1&1&0&0&&\\ \hline \otimes&1&1&1&0&&\\ \end{array}\Rightarrow \left(\begin{array}{c|llll|ll}9&1&0&1&0&=10\\ 13&1&0&0&0&=8\\ 12&0&0&1&0&=2\\ \hline &&&&&\\ \end{array}\right) \Rightarrow \begin{array}{c|llll|ll}9&&1&0&0&1&\\ 8&1&0&0&0&&\\ 12&1&1&0&0&&\\ \hline &0&0&0&0&&\\ \end{array}$[/hide]$\hline$
[color=blue][size=150][hide="Proof"][/size]
The final situation $(1,1,...,1)$ is clearly $L$ because $A(1)=0$ and $0\otimes 0\otimes ...\otimes 0=0$

The opponent plays from a situation where $(1)$ holds.
We will prove that you can play such that, after your move, again $(1)$ holds

$\blacksquare$ If the opponent removes from a pile with $2m+1$ number of stones, then you can always remove from the same pile, such that the result is $m$ (see [b]Theorem 1[/b]). But $A(2m+1)=A(m)$, so again  $(1)$ holds ([b]L[/b]-situation)

$\blacksquare$ Suppose that the opponent removes from a pile with an even number of stones, $n_{k}=2a$.
Notice that $A(n_{k})=n_{k}$ because $n_{k}$ is even

Before his playing:$A(n_{1}) \otimes A(n_{2}) \otimes ... \otimes A(n_{k})=0 \Rightarrow$
$A(n_{k})=A(n_{1}) \otimes A(n_{2}) \otimes ... \otimes A(n_{k-1})\Rightarrow$
$2a=A(n_{1}) \otimes A(n_{2}) \otimes ... \otimes A(n_{k-1})$

Let $s$ be the remaining stones. Then $a\leq s\leq 2a-1\Rightarrow A(s)\leq s<2a$

Now, the XOR of all $A(n_{i})$ is
$X=A(s)\otimes A(n_{1}) \otimes A(n_{2}) \otimes ... \otimes A(n_{k-1}) \Rightarrow \boxed{X = A(s) \otimes 2a\neq 0}$, because $A(s)< 2a$

This is a broken balance. And you should remove some stones from a pile such that the result fixes the balance.

From this point, the pile-[b]k[/b] has $s$ stones, so we can say that $n_{k}=s$
$\hline$

[u]How do you decide your move[/u]:

The most significant bit of $X$ was contributed by an odd number of $A(n_{i})$.

Since it makes no difference, we can assume that it's $A(n_{1})$ that has contributed the most significant bit of $X$.

All the numbers $A(n_{i})$ are even, so $X$ is even

Let $A(n_{1})=y_{m}y_{m-1}...y_{r}...y_{2}y_{1}0$ and $X=x_{r}...x_{2}x_{1}0$

We are to prove that $A(n_{1})\otimes X<n_{1}$

$A(n_{1})\otimes X = y_{m}y_{m-1}...y_{r+1}0z_{r-1}...z_{1}0$

$y_{m}y_{m-1}...y_{r+1}0z_{r-1}...z_{1}0<$
$y_{m}y_{m-1}...y_{r+1}1z_{r-1}...z_{1}0<$
$y_{m}y_{m-1}...y_{r+1}1z_{r-1}...z_{1}01...1$

So, $A(n_{1})\otimes X<A(n_{1})\leq n_{1}$

Let $q=A(n_{1})\otimes X$, which is [u]even[/u]. Of course $q<n_{1}$

Let $p_{m}= (q+1)2^{m}-1$ (in other words, $p_{m}$ are all the elements of the class $C[q]$)

Let $m$ be the smallest natural number such that $n_{1}\leq 2p_{m}$
Then of course $n_{1}>2p_{m-1}\Rightarrow n_{1}>(2p_{m-1}+1)-1\Rightarrow n_{1}>p_{m}-1$

$n_{1}\leq 2p_{m}\Rightarrow 2n_{1}\leq 2p_{m}+n_{1}\Rightarrow \boxed{n_{1}-p_{m}\leq \frac{n_{1}}{2}}$
$n_{1}>p_{m}-1 \Rightarrow n_{1}-p_{m}>-1 \Rightarrow \boxed{n_{1}-p_{m}\geq 0}$

So $\boxed{0\leq n_{1}-p_{m}\leq \frac{n_{1}}{2}}$

But $0\neq n_{1}-p_{m}$. Indeed, if $p_{m}=n_{1}$ then $A(p_{m})=A(n_{1}) \Rightarrow q=A(n_{1}) \Rightarrow$
$A(n_{1})\otimes X = A(n_{1}) \Rightarrow X=0$, contradiction.

So $\boxed{1\leq n_{1}-p_{m}\leq \frac{n_{1}}{2}}$

This means that you can remove $n_{1}-p_{m}$ stones from $n_{1}$ and the remaining stones wil be $p_{m}$

The result of your move is
$A(p_{m})\otimes A(n_{2})\otimes...\otimes A(n_{k})=$
$A(p_{m})\otimes A(n_{1})\otimes A(n_{1})\otimes A(n_{2})\otimes...\otimes A(n_{k})=$
$A(p_{m})\otimes A(n_{1})\otimes X=$
$A(p_{m})\otimes q =q\otimes q = 0$, hence $L$ and that completes the proof[/color][/hide]


function
inequalities
algebra proposed
algebra

Find all functions $f: R \to R$ that obey the equation: $f(\frac{x+y+z+t}{4})=\frac{f(x)+f(y)+f(z)+f(t)}{4}$
 ( :maybe: )
[i]Louis Latin and Vicky [/i]

you just find all the functions which is not convex and concave function. Means $f''(x)=0$. For example, $f(x)=kx+C$ satisfies the question.

I don't think you are right,Mathsniper,because f"(x) may do not exist

It like the JenSen  functions , from the same problems :
 Find all functions  that obey the equation :
i) $f(\frac{x+y}{2}) = \frac{f(x)+f(y)}{2}$
ii) $f(\frac{x+y+z}{3}) = \frac{f(x)+f(y)+f(z)}{3}$
But for this problems can't solve
( :maybe: )
[i]Louis Latin and Vicky[/i]

If $f(x)=a+g(x),a=f(0), g(x)=f(x)-f(0).$ If g(x) is additive (g(x+y)=g(x)+g(y)), then f(x)=f(0)+g(x) is solution. It is easy to show, that from $f(\frac{x+y+z+t}{4})=\frac{f(x)+f(y)+f(z)+f(t)}{4}$ we have $g(x)$ is additive. If exist y, suth that $g(x)$ continiosly in y, or exist y, suth that $g(x)>y \ \forall x\in R$ (or $g(x)<y$) then $g(x)=ax$.


Hi, let's have a word z with just 'a' and 'b' symbols. Let count(a) = count(b), let z begins and ends with the same symbol. Then we can split this word into to non-empty parts z = xy and both x and y have same count of 'a' and 'b':

example:

z = aababbabba - count(a) = count(b) = 5

z = xy , x= aababb, y = abba

And I'm looking for a proof, that this works for every word z. Thanks a lot!!!

[sorry for my English]

Could to be a nice problem but your example doesn't work !

Could you check the terms of the problem ?


geometry
inradius
perimeter
area of a triangle
Heron\u0027s formula

I was just wondering, how would you find the area of a circle inscribed in a triangle when you know the sides of the triangle?

This requires that you calculate the inradius of the triangle.

[hide="The general method"] We will calculate the area of the triangle in two ways.  First, given the sides $a, b, c$, define

$s = \frac{a+b+c}{2}$

Then

$\text{Area}= \sqrt{ s(s-a)(s-b)(s-c) }$

The second method of calculating area involves the inradius $r$ of the incircle.  This is given by

$\text{Area}= rs$

Equating the two allows you to calculate $r$ in the general case. [/hide]
The problem is much simpler for the case of a right triangle, but I'm not sure how relevant that is to your question.

The formula rs. Where r stands for inradius, and s stands for semi-perimeter.Lets say we have a triangle with sides 13,14,and 15. The area of the traingle is 84 by herons. Then set up a equation. So the semi-perimeter is 21. So the inradius is 4. So the area of the inscribed circle is 16pi.

For a right triangle it's:

$r = \frac{a+b-c}{2}$

Where a and b are the legs of the triangle.

Draw a circle inside a triangle.Then by the intersecting tangents property, two intersecting tangents to a circle have equal lengths.  Let these lengths be $x,y,z.$ Then the area of the triangle is $\frac{(x+y)r}{2}+\frac{(y+z)}{2}+\frac{(x+z)}{2}=(x+y+z)r=sr=A$, where $s=x+y+z$ is the semiperimeter, and $r$ is the inradius. By Heron's formula, $A=\sqrt{s(s-a)(s-b)(s-c)}=rs\implies r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$.  Using the general area of a circle formula, we have that the area of the incircle is $\frac{\pi(s-a)(s-b)(s-c)}{s}$. $\blacksquare$

Masoud Zargar


function
algebra
polynomial
complex analysis

How would I find the Taylor series of $ z^2 \minus{} 3z\plus{}i$ about $ 2\minus{}i$?

Write $ z\equal{}2\minus{}i\plus{}w$, and do the algebra.

Thanks jmerry.

I just wanted to know why you decided to take $ z$ as that value. Also, how come my answer isn't a Taylor series expression, with summation and all that junk?

$ \sum$ is just notation. The Taylor series of a polynomial is finite, and what I did is really standard technique.


Hi everybody!

There are 10 anagrams of this word, knowing that "B" can't be close: BABABAA

They are:
BABABAA
BABAABA
BABAAAB
BAABABA
BAABAAB
BAAABAB
ABABABA
ABABAAB
ABAABAB
AABABAB

But what are the logical passages to achieve this solution, without making the list one by one?
Thanks!

If the question you're asking is how many words are there that have exactly $ 4$ A's and $ 3$ B's without consecutive B's, then the reason there are $ 10$ is simple.  Notice that each of the three B's must be placed in between two A's, and because there are only $ 5$ places each B can go, we have simply $ {5\choose 3}\equal{}10$ different words.

Ok, now I understand!
Thank you very much!


inequalities
trigonometry
geometry
inequalities unsolved
China
algebra

I think It's proved inmediate by Tchebyseff.

[quote="Lovasz"]I don't think using Chebychev is appropriate.
L.H.S >= (cosA+cosB+cosC)^2/CosA+cosB+cosC+3=Q
But Q<=1/2 !?!?!? :?:[/quote]
yes,that is where the difficulty is.
many students in china tried to use Cauchy ineq to prove it,but failed.
maybe it's the hardest problem in the match!!

[quote="e.lopes"]denote by S the left side.

the function f(x) = cos²x/(cos(x)+1) is convex.
so, by jensen:

S>= 3.f((A+B+C)/3) = 3 . ((1/2)²/1+ 1/2) = 1/2  :)[/quote]
f(x) = cos²x/(cos(x)+1) is convex????????????????
sorry.I think you are wrong. ;)

I remember that I had solve it 1 year ago. I still remember it's quite hard, but not so.

It's $\cos A+\cos B+\cos C +\frac{1}{1+\cos A}+\frac{1}{1+cos B}+\frac{1}{1+cos C} \ge \frac{7}{2}$

Use mixxing variable, try to prove $A=B$.

I hope I am not wrong, but it does seem that cauchy work.
Put $cos A = \frac{b^2+c^2-a^2}{2bc}$, then it suffices to prove
$\sum \frac{(b^2+c^2-a^2)^2}{2bc(b+c)^2-2a^2bc} \ge \frac{1}{2}$ or $\sum \frac{(b^2+c^2-a^2)^2}{bc(b+c)^2-a^2bc} \ge 1$
By Cauchy, $L.H.S. \ge \frac{(a^2+b^2+c^2)^2}{\sum bc(b+c)^2 - \sum a^2bc} \ge 1$
$a^4+b^4+c^4+2\sum a^2b^2 \ge \sum_{sym} a^3 b + 2\sum a^2b^2 - \sum a^2bc$
$a^4+b^4+c^4+\sum a^2bc \ge \sum_{sym}a^3 b$
$\sum a^2(a-b)(a-c) \ge 0$
which is schur

[quote="zhaobin"][quote="e.lopes"]ops!  :blush:

sorry.

i have one solution opening the inequality and using that cos²A + cos²B + cos²C + 2.cosA.cosB.cosC = 1.

i will post  :)[/quote]
nice.
and what about your solution?[/quote]

i found one error  :blush:  :(

[quote="lccs"]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$ \sum \frac {\cos^2{A}}{\cos{A} \plus{} 1}\ge\frac {1}{2}$[/quote]
I will post a solution by AM-GM of me :)
We need to prove that for any $ a,b,c \ge 0$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc \equal{}1$, then
\[ \frac{a^2}{a\plus{}1}\plus{}\frac{b^2}{b\plus{}1}\plus{}\frac{c^2}{c\plus{}1} \ge \frac{1}{2}\]
The inequality is equivalent to
\[ \frac{2a^2}{a\plus{}1}\minus{}a^2\plus{}\frac{2b^2}{b\plus{}1}\minus{}b^2\plus{}\frac{2c^2}{c\plus{}1} \minus{}c^2 \ge 1\minus{}a^2\minus{}b^2\minus{}c^2\]
\[ \Leftrightarrow \frac{a^2(1\minus{}a)}{1\plus{}a}\plus{}\frac{b^2(1\minus{}b)}{1\plus{}b}\plus{}\frac{c^2(1\minus{}c)}{1\plus{}c} \ge 2abc\]
By AM-GM Inequality, we have that
\[ LHS \ge 3\sqrt[3]{\frac{a^2b^2c^2(1\minus{}a)(1\minus{}b)(1\minus{}c)}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}}\]
It suffices to prove that
\[ (1\minus{}a)(1\minus{}b)(1\minus{}c) \ge \frac{8}{27}abc(1\plus{}a)(1\plus{}b)(1\plus{}c)\]
\[ \Leftrightarrow \left( \frac{1}{a}\minus{}1\right)\left( \frac{1}{b}\minus{}1\right)\left( \frac{1}{c}\minus{}1\right) \ge \frac{8}{27} (1\plus{}a)(1\plus{}b)(1\plus{}c)\]
By AM-GM Inequality, we have that
\[ (1\plus{}a)(1\plus{}b)(1\plus{}c) \le \left( \frac{a\plus{}b\plus{}c\plus{}3}{3}\right)^3 \le \frac{27}{8}\]
It suffices to prove that
\[ \left( \frac{1}{a}\minus{}1\right)\left( \frac{1}{b}\minus{}1\right)\left( \frac{1}{c}\minus{}1\right) \ge 1\]
which is just a known one. :)
Equality holds in the original inequality when $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$ or $ a\equal{}1,b\equal{}c\equal{}0$. :)

[quote="lccs"]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$[/quote]Sorry to revive this old thread, but I want to mention that this inequality is actually true in [b]any[/b] triangle $ABC$ ... Next I will give the proof of this fact :)
[quote]$\triangle\, ABC\ \ \implies\ \ \boxed{\ \frac {\cos^2A}{1+\cos A}+\frac {\cos^2B}{1+\cos B}+\frac {\cos^2C}{1+\cos C}\, \ge\, \frac 12\ }$[/quote]
[color=darkblue][u][b]Proof.[/b][/u] The left hand side of the inequality can be written as : $\sum\, \frac {\cos^2A}{1+\cos A}=\sum\, -[\frac {1-\cos^2A-1}{1+\cos A}]=$

$=-\sum\, (1-\cos A)+\sum\, \frac 1{1+\cos A}=$ $\sum\, \cos A+\frac {\sum\, (1+\cos B)(1+\cos C)}{\prod\, (1+\cos A)}-3$ . Making use

of the following set of identities: $\{\begin{array}{cccc}
\sum\, \cos A=1+\frac rR \\ \\
\sum\, \cos B\cos C=\frac {s^2+r^2-4R^2}{4R^2} \\ \\
\prod\, \cos A=\frac {s^2-(2R+r)^2}{4R^2}\end{array}$ one obtains (after some computations) that :

$\boxed{\ \sum\, \frac {\cos^2A}{1+\cos A}=\frac {R(4R+r)^2-(3R-2r)s^2}{2Rs^2}\ }$ . Therefore, it remains to prove the following inequality :

$\frac {R(4R+r)^2-(3R-2r)s^2}{2Rs^2}\, \ge\, \frac 12\iff\boxed{\ s^2\, \le\, \frac {R(4R+r)^2}{2(2R-r)}\ }$ , which is true since it follows from the

remarkable distance $\boxed{\ H\Gamma^2=4R^2[1-\frac{2s^2(2R-r)}{R(4R+r)^2}]\ }$ , where $H$ and $\Gamma$ represent the [u][b]orthocenter[/b][/u] and

[u][b]Gergonne[/b]'s point[/u] respectively of $\triangle\, ABC$ . $\square$
[/color]

Also we see that if $f(x)=\frac{\cos^2x}{1+\cos x}$ so $f''(x)\geq0$ for all $x\in\left[\frac{\pi}{3},\pi\right)$.
Hence, by Vasc's RCF Theorem it remains to prove this inequality for $\beta=\alpha$ and $\gamma=\pi-2\alpha$, which gives
$\frac{\cos^2\alpha}{1+\cos\alpha}+\frac{\cos^2 2\alpha}{1-\cos2\alpha}\geq\frac{1}{2}$, which is easy.
Id est, indeed, the inequality is true for all triangle.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3592156&sid=1b067bf2c782557d96a50080bed11b7e#p3592156]2005 China Second Round Olympiad[/url]

[quote="Mateescu Constantin"]
$\triangle\, ABC\ \ \implies\ \ \boxed{\ \frac {\cos^2A}{1+\cos A}+\frac {\cos^2B}{1+\cos B}+\frac {\cos^2C}{1+\cos C}\, \ge\, \frac 12\ }$[/quote]
$$\frac {\cos^2A}{1+\cos ^2A}+\frac {\cos^2B}{1+\cos^2 B}+\frac {\cos^2C}{1+\cos^2 C}\, \ge\, \frac 12 $$
( Thanos Magkos)
P3657

[quote=xzlbq][quote=lccs]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$[/quote]

Let ABC be an any triangle triangle,Prove:
\[\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}\]

del acute...[/quote]
See 14#

[quote=sqing][quote=xzlbq][quote=lccs]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$[/quote]

Let ABC be an any triangle triangle,Prove:
\[\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}\]

del acute...[/quote]
See 24#[/quote]

See 24#
where?

[quote=xzlbq]

See 24#
where?[/quote]
I want to mention that this inequality is actually true in any triangle $ABC$ ... Next I will give the proof of this fact :)

[quote=xzlbq][quote=lccs]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$[/quote]

Let ABC be an any triangle triangle,Prove:
\[\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}\]

del acute...[/quote]

this is a “角代换扩展不等式”.please see:

In triangle,prove
\[\sum{\frac{\cos^2{2A}}{1-\cos{2A}}}\geq \frac{1}{2}\]

equality hold when $a=b=c$ or $a =  b = 1/2\sqrt{2}c$

and

\[\sum{\frac{\cos^2{4A}}{1-\cos{4A}}}\geq \frac{1}{2}\]

扩展级可能是无穷大！

[quote=lccs]Has it been posted?
Let ABC be an acute-angled triangle,Prove:
$\sum \frac{\cos^2{A}}{\cos{A}+1}\ge\frac{1}{2}$[/quote]
https://artofproblemsolving.com/community/c6h505814p2841217
https://artofproblemsolving.com/community/c6h1732694p11237497
\[\frac{1+\cos{B}\cos{C}}{1+\cos{A}}+\frac{1+\cos{C}\cos{A}}{1+\cos{B}}+\frac{1+\cos{A}\cos{B}}{1+\cos{C}}\geq \frac{5}{2}\]

Let $A, B, C$ be the angles of triangle $ABC$. Prove that
$$\frac{cosA}{2+cosB+cosC}+\frac{cosB}{2+cosC+cosA}+\frac{cosC}{2+cosA+cosB}\geq \frac{1}{2}$$
$$\frac{cos^2A}{2+cosB+cosC}+\frac{cos^2B}{2+cosC+cosA}+\frac{cos^2C}{2+cosA+cosB}\geq \frac{1}{4}$$
$$\frac{sin^2A}{2+cosB+cosC}+\frac{sin^2B}{2+cosC+cosA}+\frac{sin^2C}{2+cosA+cosB}\leq \frac{3}{4}$$
Let $A, B, C$ be the angles of triangle $ABC$. For $2\leq n\in N,$ prove that
$$\frac{cos^nA}{1+cosA} + \frac{cos^nB}{1+cosB} + \frac{cos^nC}{1+cosC} \geq \frac{1}{2^{n-1}}.$$
[hide=n=2,3][quote="Mateescu Constantin"]
$$ \frac {\cos^2A}{1+\cos A}+\frac {\cos^2B}{1+\cos B}+\frac {\cos^2C}{1+\cos C}\, \ge\, \frac 12$$[/quote]
Let $A, B, C$ be the angles of triangle $ABC$. [url=https://artofproblemsolving.com/community/c6h1788689p11821927]Prove that:[/url]
$$\frac{cos^3A}{1+cosA} + \frac{cos^3B}{1+cosB} + \frac{cos^3C}{1+cosC} \geq \frac{1}{4}.$$[/hide]


probability
geometry
3D geometry
rotation

[quote="i_like_pie"]A cube is painted on all 6 sides and is then cut into 8 smaller cubes. Each of these smaller cubes is rotated at random and then they are put back together.

What is the probability that...

a) at least half of the cube's surface is painted?

b) at least two faces are completely painted?

c) both?

Express your answers as simplified fractions.[/quote]
[hide="analyzation"]
each smaller cube will have 3 faces painted. then when they are placed back together, each smaller cube can have 3, 2, 2, 2, 1, 1, 1, or 0 colored faces showing. (different orientations kinda count)
[/hide]
[hide="a"]

the cube has 24 faces. 12 will be needed for half to be painted.
the chances for that is one half
[/hide]
[hide="b"]
there's one half chance any given quarter of a face is colored. that's $\frac{1}{16}$ per face for the whole face to be colored.
multiply that by 6
$\frac{3}{8}$
if the other face is adjacent, there is $\frac{4}{25}\frac{1}{4}=\frac{1}{25}$ chance of it being completely colored
if the other face is opposite, there is $\frac{1}{16}$ chance of it being completely colored.
multiply $\frac{1}{25}$ by 4 because there are four adjacent faces, and add $\frac{1}{16}$ (there is one opposite face) to it.
$\frac{4}{25}+\frac{1}{16}=\frac{89}{400}$
multiply that with the original $\frac{3}{8}$ to get $\frac{267}{3200}$
Edit-actually ignore the $\frac{89}{400}$ because i forgot that the non-colored face opposite the colored face adjacent to the side being checked on is not capable of being exposed.
the answer's just $\frac{6\times5}{16^{2}}=\frac{3}{8}\times\frac{5}{16}=\frac{15}{128}$
[/hide]
[hide="c"]
$\frac{15}{128}\times\frac{1}{2}=\frac{15}{256}$
[/hide]


ratio
number theory proposed
number theory

For K as large as possible, produce a K-digit integer M such that for each N=1,2,...,K, the integer given by the first N digits of M is divisible by N.

An example is K=4, M=7084, because 7 is divisible by 1; 70 is divisible by 2; 708 is divisible by 3; and 7084 is divisible by 4.

I think you can find it as large as you want by computer although the ratio of this kind of number tends to zero.


induction
Euler
number theory
relatively prime
number theory unsolved

1. a,b,c,d are integers greater than 1
solve: a^b - c^d = 6

3. solve in integers: a^2+b^3+c^4=d^5

and one that i've posted a while ago but did not get an answer.
6. solve in positive integers:
a^2-5b^2=p^2
a^2+5b^2=q^2

[done]
2. let a set be S={2^2-3,2^3-3,...,2^n-3,...}
prove there is a subset of S with infinite terms such that every pair in the subset is relatively prime

4. find all n such that 1^(2n) + 2^(2n) + ... +(2n-1)^(2n) is divisble by 2n

5. maximize m^2 + n^2 given 1<=m,n<=1981, and (n^2-mn-m^2)^2=1

[/done]

thx grobber, Arne

Regarding the second prb: consider we have defined such a sequence with n terms a1,..,an and we are trying to find an n+1 th term of the form 2^k-3. Let p1, p2, .. , pt all the primes which divide one of the numbers ai with i=1 to n. Then define a(n+1)=2^(p1-1)(p2-1)..(pt-1)-3. By Fermat's thm 2^(p1-1)(p2-1)..(pt-1)-1 is divisible by p1,..,pt, so 2^(p1-1)(p2-1)..(pt-1)-3 isn't divisible by any of them. By this method we can define our infinite sequence recursively. I think this comes from an IMO, I can't remember which one.

[quote="grobber"]I think this comes from an IMO, I can't remember which one.[/quote]

i looked but i couldnt find this question :S anyways that works, thanks grobber

For prb no 4:

I think we can prove by induction that 2^k/n for any natural k, so such an n does not exist.
Let S=1^2n+..+(2n-1)^2n, S1=1^2n+3^2n+..+(2n-1)^2n (only the odd ones) and S2=S-S1 (only the even ones).

First of all, 2/n. Here?s hoe we prove it: 2/S and 2/1^2n+(2n-1)^2n, 2/2^2n+(2n-2)^2n,.., 2/(n-1)^2n+(n+1)^2n, so 2 must divide n^2n, so 2/n. Call this P1 (Pk is the sentence ?2^k/n?). This means that 4/S. Obviously 4/S2, so 4/S1. Each of the terms of S1 has form 4k+1, so S1 has form 4k+n, so 4/n (P2). Let?s assume that we have proved Pk. We are trying to prove P(k+1). It?s clear that 2^(k+1)/S2 and 2^(k+1)/S because 2^k/n and 2n/S, so 2^(k+1)/S1.

Let?s prove this lemma first:
Lemma  If a is odd a^(2^k)=1 (mod 2^(k+1)). This is true from Euler?s thm, since a and 2^(k+1) are coprime and phi(2^(k+1))=2^k.

Back to the prb: Since 2^k/n every term of S1 = 1 (mod 2^(k+1)) from the Lemma. Then S1 = n (mod 2^(k+1)), so 2^(k+1)/n. This means that induction works and 2^k/n for any k, which means that there are no such n.

[quote="grobber"]Regarding prb no 5

I saw this problem in a book. It was solved and I couldn?t help looking at the soln:
Let?s observe that (1,1) satisfies the equality. We can also see that if (n^2-mn-m^2)^2=1 and 0<m<n then m<n<2m. We get (n^2-mn-m^2)^2=               ((n-m)^2+mn-m^2)^2= ((n-m)^2+m(n-m)-m^2)^2= (m^2-m(n-m)-              (n-m)^2)^2. This means that from the pair (n, m) we can get the pair (m, n-m) where 0<n-m<m. If we keep applying this transformation we will eventually get to the pair (1,1). This means that all solutions are obtained from the pair (1,1) by the transformation (x, y)->(x+y, x). So the pairs are (1, 1); (2, 1);..;(F_(n+1), F_n) where F_n are the terms of the Fibonacci sequence. The greatest Fib term <=1981 is F_16=1597, so the max of m^2+n^2=F_16^2+F_15^2=3514578.

I would have loved to solve this one myself, but that?s life..
 :D[/quote]
seems to me it's imo 81

[quote="bugzpodder"]now that you've mentioned, i went and found #5 to be 1981 IMO question as well.[/quote]

yeah, i found that one on kalva.  i did not know before i posted

gonna try something on this one:

6. solve in positive integers:
a^2-5b^2=p^2  [1]
a^2+5b^2=q^2 [2]

I think that (I don't know for sure.. hope it's good) it only has trivial solutions!
lets consider (a,b,p,q)=1 (otherwise we can divide it by that number)

from [1]+[2] we get that 2a^2=p^2+q^2 [3]
we know that the mod on 2 are {2,4,6 and 8}
so in [3] we need to have p=2p0 and q=2q0
so we get that a^2=2p0^2+2q0^2 [4]

with [1]-[2] we do the same for b^2 and so we get that b is also divizible with 2 [5]

from [4] and [5] we get contradiction!

hope it's okay! cheers! :D :D :D

not exactly sure what you mean by trivial, but the one (only) solution i've found using my compy is (a,b,p,q)=(41,12,31,49)
according to some, Fibonnacci also found this solution.

Guess I am wrong!
I found a problem:

solve in pozitive integers :

x^2+6y^2=z^2
6x^2+y^2=t^2

which is done the way I tried to solve your problem (only you do mod 7).

I thought that the same could be done in this case!

my fault..

BTW : trivial = (a,b,p,q)=(0,0,0,0).

cheers! :D :D

Here's where I think you went wrong: from 2a^2=m^2+n^2 you don't get m and n even. Bye! :D

[hide=Hint for P1]$a^b$ and $c^d$ need to be the same parity. If $a, b, c, $ and $d$ are greater than $1$, then $a^b$ and $c^d$ cannot be even because any even number to the power of a number greater than $1$ will be a multiple of $4$. $4$ is not a multiple of $6$, so $a^b$ and $c^d$ cannot be even. Therefore, they need to be odd, so we need two odd perfect powers that have a difference of $6$.[/hide]

Lol why are you hinting a $2003$ post. But I like this problem

Find all $(a,b,p,q)\in\mathbb N^4$ such that
$$\begin{cases} a^2+5b^2&=p^2\\ a^2-5b^2&=q^2\end{cases}$$

[hide=Ideas]
Suppose there exists a minimal solution. So $\gcd(p,q)=1$ why? Suppose is a prime $r$ dividing $p,q$ so $r\mid 2a^2$ but if $r=2$ implies $p,q$ are even and also $a,b$ so we get a solution $\frac{a}{4},\frac{b}{4},\frac{p}{4},\frac{q}{4}$ contradiction minimality, now if $r\mid a^2$ we get $r\mid 5b^2$ but similar to above we get contradiction.

Summing both equations we have
$$2a^2=p^2+q^2\iff (2a)^2=(p-q)^2+(p+q)^2$$ so it can be parameterized by using Pythagorean Triple. Now subtracting we get
$$10b^2=p^2-q^2=(p-q)(p+q)$$
And $\gcd(p-q,p+q)=\gcd(2p,p+q)=\gcd(2,p+q)$ which since $p,q$ must be same parity we get it's $2$ so $p+q=2x,~p-q=2y$ with $\gcd(x,y)=1$ so
$$10\left(\frac{b}{2}\right)^2=xy$$
i.e $p+q=Au^2$ and $p-q=Bv^2$ where $AB=10$.

So now we have $(2c)^2=(Au^2)^2+(Bv^2)^2$ which seems nice. By Pythagorean Triples we get $Au^2=2mn,~Bv^2=m^2-n^2$ or viceversa. Now plug values of $A,B$ and try find a contradiction of minimality


geometry
trapezoid
trigonometry
trig identities
Law of Sines

Suppose that $ ABCD$ is a trapezoid in which line $ AB$ is parallel to line $ BC$. Given that line $ AC$ is perpendicular to line $ CD$, line $ AC$ bisects $ \angle BAD$, and $ area(ABCD)\equal{}42$, then compute $ area(ACD)$

[quote="erdogankerem123"]Suppose that $ ABCD$ is a trapezoid in which line $ AB$ is parallel to line $ BC$. Given that line $ AC$ is perpendicular to line $ CD$, line $ AC$ bisects $ \angle BAD$, and $ area(ABCD) \equal{} 42$, then compute $ area(ACD)$[/quote]
SUPER EDIT:
WOAH, COMPLETELY MESSED UP...
[hide="Solution"] Okay, so we call $ \angle CAD \equal{} \theta \equal{} \angle ACB \equal{} \angle BAC$.  Thus, we have that, by the Law of Sines, $ \frac {AC}{AB} \equal{} \frac {\sin(180 \minus{} \theta)}{\sin\theta} \equal{} 2\cos\theta$.  So, we have that $ AC \equal{} 2AB\cos\theta$.  Drop perpindicular BX to AC.  We have that $ BX \equal{} AB\cos\theta$.  So, $ [ABC] \equal{} \frac {(BX)(AC)}{2} \equal{} AB^2\cos\theta\sin\theta$.  We also have that $ \frac {CD}{AC} \equal{} \frac {\sin\theta}{\cos\theta}$.  So, we have that $ \frac {CD}{2AB\cos\theta} \equal{} \frac {\sin\theta}{\cos\theta}$.  Thus, $ CD \equal{} 2AB\sin\theta$.  So, we have that $ [ACD] \equal{} \frac {(CD)(AC)}{2} \equal{} 2AB^2\sin\theta\cos\theta \equal{} 2[ABC]$.  So, $ 42 \equal{} [ABCD] \equal{} [ACD] \plus{} [ABC] \equal{} 1.5[ACD]$.  So, $ \boxed{[ACD] \equal{} 28}$. [/hide]
EDIT:
[hide="Alternate Solution"] Again, we drop perpindicular BX and notice that ABC is isoceles.  Thus, we have that $ \frac {AX}{AC} \equal{} \frac {1}{2}$.  Okay, so $ \angle BAX \equal{} \angle CAD$ and $ \angle AXB \equal{} 90 \equal{} \angle ACD$.  Thus, we have that $ \triangle ABX\sim \triangle ADC$.  So, we have that $ \frac {[ACD]}{[ABX]} \equal{} (\frac{AC}{AX})^2 \equal{} 4$.  Now, we have that $ \frac {[ABC]}{2} \equal{} [ABX]$.  So, $ \frac {[ACD]}{\frac {[ABC]}{2}} \equal{} 4$ and thus $ \frac {[ACD]}{[ABC]} \equal{} 2$.  So, $ [ACD] \equal{} 2[ABC]$ and we do the same thing as in the previous solution.  [/hide]
EDIT: Wow... 7 edits... don't worry... they are all more or less from my making typos...

Little problem with your code there: an illegal expression turned into a huge empty space.

[quote="i_like_pie"]How can $ \overline{AB}$ be parallel to $ \overline{BC}$?[/quote]

oh, sorry, a rather significant typo there, it was actually supposed to be  $ \overline{AD}$ parallel to $ \overline{BC}$

[quote="The QuattoMaster 6000"][quote="erdogankerem123"]Suppose that $ ABCD$ is a trapezoid in which line $ AB$ is parallel to line $ BC$. Given that line $ AC$ is perpendicular to line $ CD$, line $ AC$ bisects $ \angle BAD$, and $ area(ABCD) = 42$, then compute $ area(ACD)$[/quote]
SUPER EDIT:
WOAH, COMPLETELY MESSED UP...
[hide="Solution"] Okay, so we call $ \angle CAD = \theta = \angle ACB = \angle BAC$.  Thus, we have that, by the Law of Sines, $ \frac {AC}{AB} = \frac {\sin(180 - \theta)}{\sin\theta} = 2\cos\theta$.  So, we have that $ AC = 2AB\cos\theta$.  Drop perpindicular BX to AC.  We have that $ BX = AB\cos\theta$.  So, $ [ABC] = \frac {(BX)(AC)}{2} = AB^2\cos\theta\sin\theta$.  We also have that $ \frac {CD}{AC} = \frac {\sin\theta}{\cos\theta}$.  So, we have that $ \frac {CD}{2AB\cos\theta} = \frac {\sin\theta}{\cos\theta}$.  Thus, $ CD = 2AB\sin\theta$.  So, we have that $ [ACD] = \frac {(CD)(AC)}{2} = 2AB^2\sin\theta\cos\theta = 2[ABC]$.  So, $ 42 = [ABCD] = [ACD] + [ABC] = 1.5[ACD]$.  So, $ \boxed{[ACD] = 28}$. [/hide]
EDIT:
[hide="Alternate Solution"] Again, we drop perpindicular BX and notice that ABC is isoceles.  Thus, we have that $ \frac {AX}{AC} = \frac {1}{2}$.  Okay, so $ \angle BAX = \angle CAD$ and $ \angle AXB = 90 = \angle ACD$.  Thus, we have that $ \triangle ABX\sim \triangle ADC$.  So, we have that $ \frac {[ACD]}{[ABX]} = (\frac {AC}{AX})^2 = 4$.  Now, we have that $ \frac {[ABC]}{2} = [ABX]$.  So, $ \frac {[ACD]}{\frac {[ABC]}{2}} = 4$ and thus $ \frac {[ACD]}{[ABC]} = 2$.  So, $ [ACD] = 2[ABC]$ and we do the same thing as in the previous solution.  [/hide]
EDIT: Wow... 7 edits... don't worry... they are all more or less from my making typos...[/quote]

Wow the first one is so much lengthier than the second one. But yea i did it the first way to. Notice that $ \triangle{ABX}\cong\triangle{BCX}$ and that they are each a fourth of $ \triangle{ACD}$s area.


[hide="Hint for #2"]Just split the fractions and regroup[/hide]

For number 1, equality holds when x=1, and for number 2, equality holds when a=b=c.

[hide]$(x-1)^{2}=x^{2}-2x+1\ge 0$
$\implies x^{2}+1\ge2x$
$\implies \frac{x^{2}+1}{x}=x+\frac{1}{x}\ge2$.

$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=(\frac{a}{c}+\frac{c}{a})+(\frac{b}{a}+\frac{a}{b})+(\frac{c}{b}+\frac{b}{c})\ge 6$, by application of the first result 3 times.[/hide]

add 1 to each fraction, then it becomes (a+b+c)(1/a+1/b+1/c)>= 9, which is true by am-hm

Or you could simply use AM-GM and get \[\frac{a}b+\frac{a}c+\frac{b}a+\frac{b}c+\frac{c}a+\frac{c}b\geq 6\sqrt[6]{\frac{a^{2}b^{2}c^{2}}{a^{2}b^{2}c^{2}}}=6\]

[quote="alex__ib"]Prove that $x+\frac{1}{x}\geq2$, then prove that
$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq6$

All the numbers are positive and real[/quote]

[hide="the first one"]
$\\ (x-1)^{2}\ge0\\ x^{2}-2x+1\ge0\\ x^{2}+1\ge2x\\ \therefore x+\frac{1}{x}\gex$
[/hide]


geometry
circumcircle
trigonometry
inequalities
function
angle bisector
trig identities

Here are two problems(actually):

1) A triangle ABC is given. The bisectors intersect the circumcircle at A',B',C' respectively. Prove that:
area[A'BC]+area[AB'C]+area[ABC']>=area[ABC]

2)Let AO,BO,CO be the foot of the altitudes of an acute-angled triangle ABC.
Let M be a point from the circumcircle of ABC , A',B',C' the projections of M onto BC,CA,AB and A",B",C" the projections of M onto the tangents from A,B,C at the circumcircle.
Prove that:

a) MA'/MA" * MB'/MB" * MC'/MC"=1
b)MA"*B0C0+MB"*C0A0+MC"*A0B0=2area[ABC]

I haven't tried these, but these look interesting.

[color=cyan]Okay, I have number 1 up until a point where I can hand-wave it, sort of :).

Just consider A, B, A', C.  Let AA' intersect BC at D.  Let AB = c, BC = a, CA = b.  By the angle bisector theorem, BD = ca/(b+c) and
CD = ba/(b+c).  Let AD = ta, A'D = xa.  By Power of a Point,
xa*ta = BD*CD.
[A'BC]/[ABC] = xa/ta.
(1)  [A'BC]/[ABC] = BD*CD / ta^2

Now, by the otehr Angle Bisector Theorem, ta^2 = bc - BD*CD.

Now make substitutions into (1) and simplify:
[A'BC]/[ABC] = a^2/((a+b+c)(b+c-a))

So, lets add this up:
([A'BC] + [AB'C] + [ABC'])/[ABC] = X = a^2/((a+b+c)(b+c-a)) +
b^2/((a+b+c)(c+a-b)) + c^2/((a+b+c)(a+b-c))

Apply AM-GM to the right side of this:
X/3 = Cuberoot[(abc)^2/((a+b-c)(b+c-a)(c+a-b)(a+b+c)^3)]
 = Cuberoot[(16R^2)/(16(a+b+c)^2)]  <-- there's an area squared in the top and bottom that cancel out, using abc/(4R) and Heron's
 = (R/(a+b+c))^(2/3).

So, I claim a+b+c <= 3R*sqrt3

Now, we use extended law of sines to convert this to

SinA + SinB + SinC <= 3/2*sqrt3.

Now, the only simple way I know to prove this is to appeal directly to (I think it's called) Jensen's Inequality for concave-down functions, since we know A + B + C.  So then we're done.[/color]

[color=cyan]By the way, I would love some feedback on these.  As for the second one, the best I've done so far is to make a pretty picture using Geometer's Sketchpad and verify that it does in fact appear to work.  I'm sure someone has a better idea than that.  There are too many possibilities, though -- not friendly.  Cyclic quads, angles, drawing new pieces of the diagram, lengths and angles, areas, I just don't know where to begin.[/color]

For pr. 2) a):
$\triangle MB"C'\sim\triangle MC'A"\implies MC'^2=MA"\cdot MC"$ and same way we get other 2 similar relations which, multiplied side by side will give $MA'^2\cdot MB'^2\cdot MC'^2=MA"^2\cdot MB"^2\cdot MC"^2$, done.

Best regards,
sunken rock


Suppose that the integers $a_{1}$, $a_{2}$, $\cdots$, $a_{n}$ are distinct. Show that \[(x-a_{1})(x-a_{2}) \cdots (x-a_{n})-1\] cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Let call our polynomial $f$ and suppose on the contrary that:
$f(x) = p(x)q(x)$
for some $p, q \in \mathbb{Z}[x]$ non constant. Take $w(x) = p(x)+q(x)$. For $x=a_1, a_2, ..., a_n$ exactly one of $p(x)$ and $q(x)$ is $1$ exactly one is $-1$ so $w(a_1)=w(a_2)=...=w(a_n)=0$. This is a contradiction since degree of $w$ is less then $n$.

For the sake of completeness: you need to exclude that $p(x)= - q(x)$.

But if this would happen, we would get $f(x) = - p(x)^2$, contradicting that the highest nonzero coefficient is $>0$.


I am searching for an easy math site for "dummies" that explains how to solve basic vector algebra operations such as adding, subtracting, multiplying, etc.

Look around [url=http://mathworld.wolfram.com/]Wolfram Mathworld[/url].

I don't know if I would call Mathworld a site for dummies; it probably gives enough information to satisfy professionals.

Some sites I googled:
http://www.physics.uoguelph.ca/tutorials/vectors/vectors.html
http://mathforum.org/~klotz/Vectors/vectors.html
http://mathforum.org/library/drmath/sets/select/dm_vectors.html


A month before, i post this inequality, which is own.

If $ a,b,c > 1$ and $ 0 < a,b,c < 1$, then

$ \left( \frac {1 \minus{} a^3}{1 \minus{} a} \right) a \plus{} \left( \frac {1 \minus{} b^3}{1 \minus{} b} \right) b \plus{} \left( \frac {1 \minus{} c^3}{1 \minus{} c} \right) c \geq \left(\frac {1 \minus{} abc}{1 \minus{} \sqrt [3]{abc}} \right) (a \plus{} b \plus{} c)$.

peine had proven it.

Now, I post another new inequality:

Let $ a,b,c > 1$ and $ 0 < a,b,c < 1$. Prove that

$ \left(\frac {1 \minus{} abc}{1 \minus{} \sqrt [3]{abc}}\right) \left(\frac {a^4 \plus{} b^4 \plus{} c^4}{abc} \right) \geq \left( \frac {1 \minus{} a^3}{1 \minus{} a} \right) a \plus{} \left( \frac {1 \minus{} b^3}{1 \minus{} b} \right) b \plus{} \left( \frac {1 \minus{} c^3}{1 \minus{} c} \right) c$.

Futhermore, i want to ask for proving the following inequality,

Let $ a,b,c > 0$. Prove that

$ \frac {1}{(1 \plus{} \sqrt {ab})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {bc})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {ca})^2} \geq \frac {3 \plus{} a \plus{} b \plus{} c}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}$.

Let $ a,b,c >0$. Prove that

$ \left(a\plus{}\frac{1}{a} \right) \left(b\plus{}\frac{1}{b} \right) \left(c\plus{}\frac{1}{c} \right) \geq \frac{4}{3} \left(a\plus{}b\plus{}c\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \right)$.

[quote="4865550150"]Let $ a,b,c > 0$. Prove that

$ \left(a \plus{} \frac {1}{a} \right) \left(b \plus{} \frac {1}{b} \right) \left(c \plus{} \frac {1}{c} \right) \geq \frac {4}{3} \left(a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right)$.[/quote]

Its easy
$ (a \plus{} \frac {1}{a} ) \left(b \plus{} \frac {1}{b} \right) \left(c \plus{} \frac {1}{c} \right) \ge 4(c\plus{}\frac{1}{c})$
$ (a \plus{} \frac {1}{a} ) \left(b \plus{} \frac {1}{b} \right) \left(c \plus{} \frac {1}{c} \right) \ge 4(b\plus{}\frac{1}{b})$
$ (a \plus{} \frac {1}{a} ) \left(b \plus{} \frac {1}{b} \right) \left(c \plus{} \frac {1}{c} \right) \ge 4(a\plus{}\frac{1}{a})$
After summing we will get:
$ \left(a \plus{} \frac {1}{a} \right) \left(b \plus{} \frac {1}{b} \right) \left(c \plus{} \frac {1}{c} \right) \geq \frac {4}{3} \left(a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right)$


What is the greatest three-digit number which is a multiple of
13?

I just used guess and check. Any better way?

We know $ 1300$ and $ 130$ are multiples of $ 13$, so we can subtract $ 130\plus{}130$ from $ 1300$ to obtain $ 1040$.  Now subtracting $ 13$ multiple times yields $ 1027,1014,1001,\boxed{988}$.

Or, note that $ 1001\equal{}1000\plus{}1\equal{}10^3\plus{}1^3\equal{}(10\plus{}1)(10^2\minus{}10\plus{}1)\equal{}11 \times 91\equal{}7 \times 11 \times 13$, so our answer must be $ 1001\minus{}13\equal{}\boxed{988}$.

Yes, knowing that 1001=7*11*13 should be known. Then subtract 13.


The percent that $M$ is greater than $N$, is:

(A) $\frac{100(M-N)}{M}$

(B) $\frac{100(M-N)}{N}$

(C) $\frac{M-N}{N}$

(D) $\frac{M-N}{M}$

(E) $\frac{100(M+N)}{N}$

I got A because we're finding the percentage here, so we have to multiply by 100. Next, we're finding the (m-n)/m because m is greater than n, and we're also finding the difference of m and n.

[quote="math92"][hide]I got C.  I plugged in 3 for M and 2 for N and i think the answer comes out to be 1/2.[/hide][/quote]
math92, you had the right answer, but the question asked for the PERCENT  ;) .
[hide]
$B$, because you can plug 10 and 20, M is 20, N is 10.
A doesn't work because it would be 1000/20=500, but it increaases by 100%.
B works because 1000/10=100%[/hide]

i am weird... you guys are right.  I need to be more careful. :)


function
integration
logarithms
search
real analysis
real analysis solved

Let $f$, a strictly positive function continues in $[a,b]$ and either \[F(x) = \int_a^b {[f(t)]}^xdt, \ \ x \in \mathbb{R}.\] Prove that the function $x \mapsto \ln{F(x)}$ is a convex function on $\mathbb{R}$.

[quote="Myth"]Indeed, $(\ln F(x))''=\int_{a}^b[f(t)]^x\ln^2f(t)dt\geq 0$. Thus $\ln F(x)$ is a convex function.[/quote]
I break my head to search $F'(x)$  :rotfl: ! Thank you Myth.

[quote="Myth"]Indeed, $(\ln F(x))''=\int_{a}^b[f(t)]^x\ln^2f(t)dt\geq 0$. Thus $\ln F(x)$ is a convex function.[/quote]
Actually, it's $F''(x)=\int_{a}^b[f(t)]^x\ln^2f(t)dt\geq 0$.

:blush:  :blush:  :blush:
I don't know how could it happen...

\[\displaystyle (\ln F(x))''=\frac{\int_a^b[f(t)]^x\ln^2f(t)dt\,\int_a^b[f(t)]^xdt-\left(\int_a^b[f(t)]^x\ln f(t)dt\right)^2}{\int_a^b[f(t)]^xdt}\geq 0\]
due to Cauchy-Bunyakowski inequality.


Eve and Odette play a game on a $ 3 \times 3$ checkerboard with black and white checkers. The rules are as follows:

$ (1)$ They play alternately.
$ (2)$ A turn consists of placing a checker on an unoccupied square of the board.
$ (3)$ A player in turn may select a checker of any color.
$ (4)$ When the board is full, Eve obtains one point for every row, column or diagonal with an even number of black checkers, and Odette obtains one point for every row, column or diagonal with an odd number of black checkers.
$ (5)$ The player obtaining at least five of the eight points wins.

$ (a)$ Is a $ 4 : 4$ tie possible?
$ (b)$ Describe a winning strategy for the girl who plays first.

posted before. it's an old problem from canada.


let <AOB is an angle that $ A_1,...,A_4$ are 4point on AO and $ B_1,...,B_4$ are 4point on BO.
$ (A_1B_1,A_2B_2)\equal{}N$ and $ (A_4B_4,A_3B_3)\equal{}M$.prove that:O,N,M are collinear if and only if:
\[ \frac{OB_1}{OB_3}. \frac{OB_2}{OB_4}.\frac{B_3B_4}{B_1B_2}\equal{}\frac{OA_1}{OA_3}.\frac{OA_2}{OA_4}.\frac{A_3A_4}{A_1A_2}\]

Let $ P$ and $ Q$ be the intersections of the ray passing through $ O$ and   $ A_1B_2 \cap A_2B_2$ with $ A_2B_2$ and $ A_4B_4$ respectively. From the harmonic pencil $ O(A_2,B_2,P,N),$ we get

$ O,M,N$ are collinear $\Longleftrightarrow  (A_4,B_4,Q,M) \equal{} \minus{} 1$ $ \Longleftrightarrow  \frac {QA_4}{QB_4} \equal{} \frac {MA_4}{MB_4} \ (\star)$

By Menelaus' theorem for $ \triangle OA_4B_4$ cut by the tranversal $ \overline{A_3B_3M}$ we get

$ \frac {MA_4}{MB_4} \equal{} \frac {A_4A_3}{OA_3} \cdot \frac {OB_3}{B_3B_4} \ \ (1)$

On the other hand, by  Menelaus' theorem for $ \triangle OA_2B_2$ cut by the tranversal $ \overline{A_1B_1N},$ we deduce that

$ \frac {QA_4}{QB_4} \equal{} \frac {OA_4}{OB_4} \cdot \frac {\sin \widehat{AOQ}}{\sin \widehat{BOQ}} \equal{} \frac {OA_4}{OB_4} \cdot \frac {PA_2}{PB_2} \cdot \frac {OB_2}{OA_2}$

$\Longrightarrow \ \frac {QA_4}{QB_4}\equal{} \frac {OA_4}{OB_4} \cdot \frac {OB_2}{OA_2} \cdot \frac {A_1A_2}{OA_1} \cdot \frac {OB_1}{B_1B_2} \ \ (2)$

Substituting $ (1)$ and $ (2)$ into $ (\star)$ gives

$ \frac {OA_4}{OB_4} \cdot \frac {OB_2}{OA_2} \cdot \frac{A_1A_2}{OA_1} \cdot \frac {OB_1}{B_1B_2} \equal{} \frac {A_4A_3}{OA_3} \cdot \frac {OB_3}{B_3B_4}$

$ \Longrightarrow \ \frac {OB_1}{OB_3} \cdot \frac {OB_2}{OB_4} \cdot \frac {B_3B_4}{B_1B_2}$ $ \equal{} \frac {OA_1}{OA_3} \cdot \frac {OA_2}{OA_4} \cdot \frac {A_3A_4}{A_1A_2}$


So where do you think they'll bring us for our days off?
I reckon we might go to the Pontonjia caves, and maybe to the beach, any other suggestions? :?:

More importantly... where do you reckon they'll keep us in the time between? Any info on accommodation yet?

I find the IMO 2006 website a little less informative than last year's site, although I like a few aspects about it too.

But, yeah, I hope we'll go see the caves. I'd say going to some place by the shoreline would be rather likely, although I couldn't say about the beach. I don't think there are that many beaches in Slovenia.

[quote="einaraxel"]I find the IMO 2006 website a little less informative than last year's site, although I like a few aspects about it too.[/quote]

BTW: "[b]Which[/b]" last year's site :huh: ?

[quote="einaraxel"]I find the IMO 2006 website a little less informative than last year's site, although I like a few aspects about it too.[/quote]You must be confusing websites. This year's site is much better than last year's site.

I agree with Valentin. The 2006 site is really good. It is well-organized and has information on basically everything, and even has pictures of each day. I hope the furute websites will be as good as this one. :)

[size=67]btw post number 500[/size]

Well, ok, it is now, seeing how fast it's updated, but on last year's website you could find relevant info much earlier than you could this year. In particular accomodation, where we were going on the excursions and so on.


Someone can but I can't. Sorry, don't know enough number theory yet. Still waiting on a reply for James as well. Good luck with your proof though. Very hardcore.

what means "one solution [b]coprimely[/b]to the equation x^3+y^3+z^3=6xyz" ?
a solution $(x,y,z)$ such that $x,y,z$ are coprime ?

That is good question. I am not completely sure what he is trying to say either.

[quote="bomb"]hey.

Can anyone out there do...

Prove there exists one solution coprmiely to trhe equation x^3+y^3+z^3=6xyz in the positive integers.

Bomb  :no:[/quote]

$x=1$, $y=2$ and $z=3$. That's all.

Oh. Alright. That wasn't much but if that's all then that's all.

hehe.  no.  *prove* (or disprove; i don't know if it's true yet) that's the only solution.


When five brothers gathered to trade cards, only Jeb had some cards to trade.  Jeb gave $ \frac{1}{2}$ of his cards to Karl; then, Jeb gave Karl 2 more cards.  Karl then gave $ \frac{1}{2}$ plus 2 of the cards he received to Doug.  Likewise, Doug gave $ \frac{1}{2}$ plus 2 of the cards he received to Chris.  Chris received 10 cards from Greg.  How many cards did Jeb have before he gave any to Karl?

umm..in our game all it said was 1/2....lol

If I am not mistaken, then I think that the 2's are 1/2.  :D


inequalities
arithmetic sequence
number theory solved
number theory

$R=(R_{n},R_{n-1},...,R_{1})$ is the binary show of integer $R$ and the $k_{th}$ number of $R$ mean the
$k_{th}$ number of $R$ in the binary show or $R_{k}$.
Let $P,Q$ two primes .
prove that there exist a prime like $\mathcal H$ s.h. the $\mathcal H_P=1$ and $\mathcal H_Q=0$

I think it is quite easy or did I misunderstood the problem?

If $p>q$ let $a=2^p$ and $b=2^{p-1}+1$.
Now since $a$ and $b$ are coprime, and using Dirichlet's theorem, the arithmetic progression $\{an+b \}$ contains a prime $H$. Clearly the binary expansion of $H$ starts with $H_1=1=H_p$ and $H_i = 0$ for $i=2, \cdots , p-1$ (in particular $H_q=0$). The choice of $n$ does not affect it since it is multiplied by $2^p.$

If $p<q$ we can do as above with $\{2^qn + 2^{p-1}+1 \}.$

Pierre.

wow ,Perfect sir,Its completely correct,i was trying to find the $H$ by chebyshev thoerem Never thought of  Dirichlet's  :(

Anyway the problem i gave above is the equal to the first part of the Iran final exam of summer camp.
There is the complete problem :
We denote the relation $\leftrightarrow$ between the two numbers $m,n$ and we write $m \leftrightarrow n$ if and only if the $m_{th}$ number of $n$ in the binary show be one OR the $n_{th}$ number $m$ in the binary show be one .

we call a set like $A$ golden if
1.$A \subset \mathbb{N}$
2.for every two numbers like $x,y \in A$ there exist  a number like $r \in A$ such that $r \leftrightarrow x$ and $
r \not \leftrightarrow y$.

1.Prove that the set of the primes is a golden set.(which is solved by Mr.pbornsztein,in above,Nice solution indeed)
2.If we devide the set of primes in to sets like $P_1,...,P_n (P_i \cap P_j =\emptyset)(\bigcup P_i=\text{Set of primes})$
prove that one of the $P_i$s is a golden sets.


geometry
rectangle
analytic geometry
combinatorics unsolved
combinatorics

Every natural number is coloured in one of the $ k$ colors. Prove that there exist four distinct natural numbers $ a, b, c, d$, all coloured in the same colour, such that $ ad \equal{} bc$, $ \displaystyle \frac b a$ is power of 2 and $ \displaystyle \frac c a$ is power of 3.

No body interest this problem  ?   I think it's very nice

I agree with you but I can't give you any idea because I spent about an hour thinking but with no result. :(

I'll give you the hint: consider only the numbers in the form $ 2^a \cdot 3^b$. You can arrange them in the 2D table...

Indeed a very nice problem. Let a be any positive integer. Fix a and label the lattice points of the first quadrant of the Cartesian plane as follows: (x,y) is labelled with $ 2^x\cdot 3^y \cdot a$, for x,y nonnegative integers. Colour each of these points with the colour of the number it represents. Then we simply need a monochromatic rectangle with sides parallel to the x and y axes. Define a segment starting at (x,y) to be the collection of points (x,y+i) for i=0,1,2,...,k. Since there are k+1 points every segment contains two points of the same colour. Furthermore for n sufficiently large some two of the segments starting at (0,0), (0,1), (0,2),...,(0,n) are coloured in exactly the same way, i.e. each pair of corresponding points have the same colour. Now these two segments contain a rectangle because each segment contains two points of the same colour, and the two segments are coloured in exactly the same way.

I don't understand the question. What does it mean when it discusses the colors? I can find $4$ natural numbers that satisfy the given conditions ($1,2,3,$ and $6,$ for example) but I know it must be more difficult than that. The colors are a bit confusing....

We prove the stronger result that $a,b,c,d$ exist where $\{a,b,c,d\}\subset\mathbb{S}= \{2^x3^y|(x,y)\in\mathbb{N}_0^2\}$. Coordinatize the plane such that the point in the first quadrant with coordinates $(x,y)$ corresponds to the number $2^x3^y$. Then the condition $ad=bc$ is equivalent to $a,b,c,d$ corresponding to a rectangle with $a,d$ at opposite corners. Given a monochromatic rectangle in the coordinate plane, we can easily construct $a,b,c,d$ as follows.
[asy]
import cse5; import olympiad;
pathpen = black; pointpen = black;
size(14cm);

pair O=origin, D=dir(30), A=dir(210), B=dir(150), C=dir(-30);
DPA(A--B--D--C--cycle, red);
D("a",A,A);
D("c",B,B);
D("b",C,C);
D("d",D,D);
[/asy]
So it suffices to prove the existence of a monochromatic rectangle if we color the lattice points in the first quadrant with $k$ colors.

Consider the $k+1\times k^{k+1}+1$ grid with lower left corner $(0,0)$ and upper right corner $(k,k^{k+1})$. There are $k^{k+1}$ different colorings of a row consisting of $k+1$ points, so by pigeonhole principle two rows in this grid are identical. Consider one of these rows. There are $k$ possible colors in a row, so there are two points in this row that are the same color. But these points and the corresponding points in the identical row give a monochromatic rectangle so we are done. This also gives the trivial bound that there exist $a,b,c,d\le k^{k+1}$.

The condition $ad=bc$, implies that $a=xy$,$d=zw$,$b=xz$ and $c=yw$, for some $x,y,z,w \in \mathbb{N}$.
The other conditions imply that $x=3^hw$ and $y=2^tz$, where $t$ and $h$ are some natural numbers.
Thus giving us $a=3^h2^tzw$, $b=3^hzw$, $c=2^tzw$ and $d=zw$.

So now obviously we must have at least one coloring that contains $\infty$ composite numbers.
Because of this fact we can set $z,w,t,h$ so that $a,b,c,d$ are of the same color.
(Alternately we could have went on case by case on the specific situations and forms of $a,b,c,d$ in terms of coloring).

Place each number of the form $2^x3^y$ in the lattice point $(x,y)$ of the Cartesian coordinate plane. Then, it is sufficient to find a monochromatic rectangle. Now, there are two ways to show one exists:
1) By infinite pigeonhole (twice) there are an infinite number of columns with the same color an infinite number of times. Now, if no monochromatic rectangle is present, that means that of the remaining numbers in these columns, painted of the other $k-1$ colors, there are an infinite number of columns with the same color an infinite number of times. Continuing with this logic (with an induction argument), we eventually reach inifinite number of columns of the same color, which obviously contain a rectangle.
2) Consider the first $k+1$ numbers in a row. There is a finite amount of ways to color them. Hence, if we look at sufficiently many rows, there are two colored in the exact same way. Since there are $k+1$ numbers, at least $2$ are of the same color, and this gives the desired rectangle. $\blacksquare$


For all m \in  N , let f_m(x)\in  Q[x]:
   1^(2m+1) + 2^(2m+1) + ...+n^(2m+1)=f_m(x) , for all n \in  N*
prove that f_m(x) is divisible by x^2

??? where is x in the formula of f_m(x)?

Pierre.

1^(2m+1) + 2^(2m+1) + ...+n^(2m+1)=f_m(x)

Where does x appear on the left ? Did you mean f_m(n)=...?

Oh , Sorry!
Correct this problem to :
1^(2m+1) + 2^(2m+1) + ...+n^(2m+1)=f_m(n)

I don’t understand this problem!!   Manifestly, for m=1 and n even , f_m(n) is not divisible by n^2 ...

You must note that f_m(x) is a polynomial .
Ex :f_m(x)=x^2.q(x) , q(x) \in Q[x]
    then f_m(x) is divisible by x^2

I have something terrible on it.

Each odd polynomial $P(x)$ of degree $2m+1$ can be represented as \[P(x)=\sum_{k=0}^{m}c_{k}\binom{x+k}{2k+1}.\] So let us do it with monom $x^{2m+1}$ and $c_{k}$ are required coefficients.

Note that $\binom ab=\binom{a+1}{b+1}-\binom{a}{b+1}$.

It means that we have the following representation for $f_{m}(x)$: \[f_{m}(x)=C+\sum_{k=0}^{m}c_{k}\binom{x+k+1}{2k+2}.\] But it is easy to see that $C$ is zero, since $f_{m}(0)=0$.

We have $f_{m}(0)=0$, therefore $x\mid f_{m}(x)$. It is sufficies now to show $f'_{m}(0)=0$. Really, \[f'_{m}(0)=\sum_{k=0}^{m}c_{k}\frac{(-1)^{k}(k+1)!k!}{(2k+2)!}= \frac{1}{2}\sum_{k=0}^{m}c_{k}\frac{(-1)^{k}k!k!}{(2k+1)!}=\frac{1}{2}(x^{2m+1})'|_{0}=0.\]


AMC
AIME
quadratics
Support
search
algebra
system of equations

$ x^3\equal{}13x\plus{}3y$
$ y^3\equal{}3x\plus{}13y$
Compute $ (x^2\minus{}y^2)^2$

[hide="My solution"]We can quickly see relations: recall the cubic factors.
1) $ x^3\minus{}y^3\equal{}(x\minus{}y)(x^2\plus{}xy\plus{}y^2)$
2) $ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)$
3) As for the solution itself, we want $ (x^2\minus{}y^2)^2$ and $ x^4\minus{}y^4\equal{}(x^2\plus{}y^2)(x^2\minus{}y^2)$ can help us get that.

Thus,
$ x^3\minus{}y^3\equal{}10(x\minus{}y)\equal{}(x\minus{}y)(x^2\plus{}xy\plus{}y^2)\Rightarrow x^2\plus{}xy\plus{}y^2\equal{}10$
$ x^3\plus{}y^3\equal{}16(x\plus{}y)\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)\Rightarrow x^2\minus{}xy\plus{}y^2\equal{}16$
Summing our resultants we obtain $ x^2\plus{}y^2\equal{}13$ and it follows directly that $ xy\equal{}\minus{}3$

Now we begin to work towards our goal $ (x^2\minus{}y^2)^2$:
$ x^4\equal{}13x^2\plus{}3xy\Rightarrow x^4\minus{}13x^2\plus{}9\equal{}0$
$ y^4\equal{}3xy\plus{}13y^2\Rightarrow y^4\minus{}13y^2\plus{}9\equal{}0$
Although using $ x^4\minus{}y^4$ seemed good at first, this seems way better.
$ k^4\minus{}13k^2\equal{}\minus{}9\Leftrightarrow \displaystyle\left(k^4\minus{}13k^2\plus{}\frac{169}{4}\right)\equal{}\frac{169}{4}\minus{}9\Leftrightarrow \left(k^2\minus{}\frac{13}{2}\right)\equal{}\frac{133}{4}\Leftrightarrow$
$ (2k^2\minus{}13)^2\equal{}133$

Now just a quick substitution, we allow $ k\equal{}x$ and $ 13\equal{}x^2\plus{}y^2$ yielding $ \boxed{(x^2\minus{}y^2)^2\equal{}133}$

[/hide]

[hide]It's not difficult to see that \[ x^3\minus{}y^3\equal{}10(x\minus{}y)\\ x^3\plus{}y^3\equal{}16(x\plus{}y)\]
If $ x\equal{}\pm y$ then $ (x^2\minus{}y^2)^2\equal{}0$. Otherwise we can divide by $ x\plus{}y$ and $ x\minus{}y$ and we eventually have \[ x^2\plus{}xy\plus{}y^2\equal{}10\\ x^2\minus{}xy\plus{}y^2\equal{}16\]
Adding these two we get $ x^2\plus{}y^2\equal{}13$. On the other hand we get $ 2xy\equal{}\minus{}6$ by subtracting. Thus
\[ (x^2\minus{}y^2)^2\equal{}(x^2\plus{}y^2)^2\minus{}(2xy)^2\equal{}13^2\minus{}6^2\equal{}169\minus{}36\equal{}\boxed{133}\][/hide]

Hmmm. My solution also states that $ (y^2\minus{}x^2)^2\equal{}133$, which isn't surprising: $ (x\minus{}y)^2\equal{}(y\minus{}x)^2$

I wonder though, if we can take it one step further and solve it?

[quote="igiul"]I wonder though, if we can take it one step further and solve it?[/quote]

[hide]We have \[ x^2\plus{}y^2\equal{}13\\ x^2\minus{}y^2\equal{}\pm\sqrt{133}\]
Addition and subtraction yields
\[ x^2\equal{}\frac{13\pm\sqrt{133}}{2}, y^2\equal{}\frac{13\mp\sqrt{133}}{2}\]
Thus \[ (x,y)\equal{}\left(\sqrt{\frac{13\pm\sqrt{133}}{2}}, \sqrt{\frac{13\mp\sqrt{133}}{2}}\right)\][/hide]

Why can't it be $ 0$?

[hide="To solve the system"]
If $ (x \plus{} y)\neq 0, (x \minus{} y)\neq 0$, citing the above results by @nayel: $ x^2 \plus{} y^2 \equal{} 13, xy \equal{} \minus{} 4$. Some simple addition gives us
$ (x \plus{} y)^2 \equal{} 7 \Leftrightarrow x \plus{} y \equal{} \pm \sqrt {7}$,
so $ x,y$ are the two roots of quadratic equation
$ t^2\pm \sqrt {7}t \minus{} 3 \equal{} 0$
Solving it
$ t_{1,2} \equal{} \frac {\pm \sqrt {7}\pm \sqrt {19}}{2}$.
Therefore, $ \{x,y\} \equal{} \frac {\sqrt {7}\pm \sqrt {19}}{2} \text{ or } \frac { \minus{} \sqrt {7}\pm \sqrt {19}}{2}$.

If $ x \plus{} y \equal{} 0$ or $ x \minus{} y \equal{} 0$, we have these roots:
$ (0,0), (\sqrt{10}, \minus{}\sqrt{10}), (\minus{} \sqrt{10}, sqrt{10}),(4, 4), (\minus{}4,\minus{}4)$.
[/hide]

[quote="nayel"]
$ x^2 \plus{} y^2 \equal{} 13$
$ x^2 \minus{} y^2 \equal{} \pm\sqrt {133}$[/quote]

Nice solution Nayel.  I'll just wrap up what I started.

[hide]$ (2x^2\minus{}13)^2\equal{}133\Rightarrow x\equal{}\sqrt{\frac{13\pm\sqrt{133}}{2}}$
Now well just do $ \frac{13\plus{}\sqrt{133}}{2}\plus{}y^2\equal{}13\Rightarrow y\equal{}\sqrt{\frac{13\mp\sqrt{133}}{2}}$[/hide]
This supports Nayel's response.  :)

I believe $ 0,0$ is trivial solution. I was referring specifically to the pair $ (x,y)\Leftarrow (x^2\minus{}y^2)^2\equal{}133$

please forgive my peskiness but what number is this; i would really rather not enter a system of equations into the search field.

A start: Notice that $ 3x\plus{}3y\equal{}x^3\minus{}10x\equal{}y^3\minus{}10y$. So $ (x\minus{}y)((x\plus{}y)^2\minus{}xy\minus{}10)\equal{}0$.

If we mulitply the two relations, $ (xy)^3\equal{}39(x\plus{}y)^2\plus{}100xy$. From here, we can see that $ t\equal{}xy$ gives a cubic

$ t^3\equal{}39t\plus{}390\plus{}100t$, so $ t\equal{}\minus{}10,\minus{}3,13$. Then we can solve for $ x\plus{}y$, etc.


LaTeX
calculus
derivative
inequalities
inequalities unsolved

a,b,c,d are positive real number such that a^2+b^2+c^2+d^2=2005 fidnf MIN of
P=  :Sigma: a/((2005 :sqrt: 2005+bcd)

OK let's $\LaTeX$ it :
$a,b,c,d$ are positive real numbers such that $a^2+b^2+c^2+d^2=2005$ Find the minimum of :
$P=\sum \frac{a}{2005\sqrt{2005}+bcd}$
P.S. : Nice problem  ;)

i'm sorry , ERDOS please corrected it find MIN not MAX (but can you find max)

min maybe easy, just derrivative :D

max cost me alot of time with null result :blush:

to ctlhp:

you are 10 or11,12 class ;)

mr Can told me that 30/4 /2005 olympiad have a most difficult problem ;)

- Would you mind showing me your solution using derivative, cuong?-I don't think Jensen inequality is a good choice for this problem.
- And by the way, are you a student of Mr.Căn? He is a cool man with the famous words:" Even a cow can solve this problem if it  has learned caculus."
 :D

i don't think the min value can be found easily , (the equality doesn;t occurs when x=y=z)

this one is interesting,I'll try to solve it
chúng ta đều là đệ tử của thầy cả, làm quyen đi thôi,qua PMs :D  :D  :D

$ a^2 + b^c + c^2 + d^2 = 1$

Find the minimal of: $\sum \frac{a^2}{a+ abcd}$

just a try:

$\frac{a^2}{a+abcd} \ge a^2$

equivalent to:

$1 \ge a + abcd$

.........


function
symmetry
calculus
calculus computations

Prove or disprove (with an example) the existence of a function $ f$ with the following properties:

1) $ f$ is a "nice" function (continuous, differentiable and so on..)

2) $ f : \mathbb{R}^2 \to S\,,\quad S \subseteq \mathbb{R}$

3) $ f(0,0) \equal{} 0$

4) $ \exists (x, y) \in \mathbb{R} ^2 \,\left(f(x,y) \equal{} 0 \,\wedge\, x^2 \plus{} y^2 \equal{}1 \,\wedge\, x \frac{\partial f}{\partial y} \equal{} y \frac{\partial f}{\partial x}\right)$

Of course, the trivial solution $ f(x, y) \equal{} 0$ is excluded..

Thank you very much!

Take $ f(x,y) \equal{} x^2 \plus{} y^2 \minus{} \frac{1}{\sqrt{2}}(x\plus{}y)$.

Then $ f(0,0) \equal{} 0$, $ f$ is $ C^\infty$ (as nice as you are going to get), and
property 4 is satisfied at the point $ x \equal{} y \equal{} \frac{1}{\sqrt{2}}$. The symmetry
property is immediate since $ f$ is symmetric and we're on the line $ y\equal{}x$.

I just thought of something.. How about a function with the same properties, but:

4)  $ \forall R \in \mathbb{R}\,\exists (x, y)\in\mathbb{R}^{2}\,\left(f(x,y) \equal{} 0\,\wedge\, x^{2}\plus{}y^{2}\equal{}R^2\,\wedge\, x\frac{\partial f}{\partial y}\equal{} y\frac{\partial f}{\partial x}\right)$

Is there such a function!?

The function $ f(x, y) \equal{} (x \minus{} y)^2$ satisfies $ f \equal{} \frac{\partial f}{\partial x} \equal{} \frac{\partial f}{\partial y}\equal{}0$ along the entire line $ y \equal{} x$.

Thank you (again).. I just have a final question, I hope you will find a bit of effort to help me.. How about if the final condition is that the function mustn't be definite or semi definite (positive or negative). In other words, it must be positive and negative for some values of $ (x,y)$..

Found it...

$ f(x,y) \equal{} (x\minus{}y)^3$ works just fine.. Thanks anyways..


Let $ m,n \in \mathbb{N}^*, m > n$, fixed numbers. Find all $ z \in \mathbb{C}$, such that:
\[ \left\{ \begin{array}{c}{1} | z^m + 1 | = | z^n - 1 | , \
 | z^m - 1 | = | z^n + 1 | \ \end{array} \right.  .\]

If $ |z^m \minus{} 1| \equal{} |z^n \plus{} 1|$ and $ |z^m \plus{} 1| \equal{} |z^n \minus{} 1|$, then triangle $ 0, z^m \minus{} 1,z^m \plus{} 1$ is equate triangle $ 0,z^n \plus{} 1,z^n \minus{} 1$. Ten $ |z|^m \equal{} |z|^n$ and (because m>n) $ |z| \equal{} 1$ or $ z \equal{} 0$. If $ z \equal{} 0$ then all done. If $ |z| \equal{} 1$ then $ z \equal{} e^{ia}$. And it is obvious geometry that $ z^n \equal{} \minus{} z^m$ or $ (z^n \minus{} z^m) \in \mathbb{R}$. Then $ z^{(m \minus{} n)} \equal{} \minus{} 1$ or $ z^{(m \plus{} n)} \equal{} \minus{} 1$. Finally answer is $ 0$, and roots from $ \minus{} 1$ by power $ m \minus{} n$ and $ m \plus{} n$.

Fuf, i really don't know enough english words  :(
Hmm... why i say that $ z \equal{} e^{ia}$? Who knows...


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_0^{\frac{\pi}{4}} \frac{1\plus{}2\sin x\minus{}\sin ^ 2 x}{(1\plus{}x)\cos ^ 4 x}\ dx$.

The problem was wrong. I'm very Sorry for you to take time in solving.

kunny

I would reedit the problem as follows.

Evaluate $ \int_0^{\frac {\pi}{4}} \frac {2(1 \plus{} x)\ln (1 \plus{} x)\cdot \sin x \plus{} \cos x}{(1 \plus{} x)\cos ^ 3 x}\ dx$.


You are in a 50 foot by 50 foot by 50 foot room that is levitating 100 feet above the ground. There is an open window near the floor in one corner of the room. Beside the window in the floor is a hook. Two 50-foot long ropes, about one foot apart, are cemented into the ceiling in the center of the room. The ropes are just strong enough to hold your weight.

If you could get the ropes down, you could tie them together and then tie them to the hook by the window so that you could lower yourself to the ground.

You are an expert rope climber and you can tie wonderful knots. You also have a very sharp knife. You have nothing else to use in your escape. How can you safely make it to the ground without falling any distance?

cut the ropes as far as you can near the ceiling then cut them in half and splice them together.

[hide="solution... i think"]Cut both ropes partially, then climb down. Then jump on one and it should snap. Jump on the second. You have two 50 foot ropes. Tie, tie again, hook, climb down, and brag to your friends.[/hide]

I believe the major point of the riddle is that once one rope is cut down, you can't climb and cut the other one and cut it, otherwise you'll fall down onto the floor to hard and die. Am I correct?

In that case, 1=2's solution wouldn't work, because if jumping on a rope would be enough for the rope to snap, then you wouldn't be able to partially cut the second rope without snapping either it or the first one and falling to your doom.

This is a problem from ACoPS, right? I think there was another condition that you could survive a fall of no more than 10 ft. Otherwise, my solution doesn't work.

[hide="Solution"] Climb up one rope and cut the other one, let's say, 5 ft from the top. Using the 5 ft cemented into the ceiling, tie a knot so that it forms a loop. Then, thread the rope that you cut off through the loop, creating a wheel-less pulley, and transfer your weight onto it. Finally, cut off the other rope on the ceiling, and tie it to your pulley-rope. Then, climb down the 95 ft rope, and from there, lower yourself to the ground.[/hide]

on the topic of ACoPS, are there solutions to the problems in the book? or is there a solution manual?

Art and Craft of Problem Solving by Paul Zeitz.

And there's no solutions or solution's manual except for the Teacher's Manual, and you need to be a teacher to obtain it from wiley.com

[quote="r15s11z55y89w21"]Can't you just make increments smaller to make them within the 10ft limit?[/quote]

Yes, you're right; you can make it infinitesimally small (depending on sharpness of knife). 5 ft was just an example :P


Just wondering cause I thought 4 would be last but 5 came.

There are only 5. We saw the last one since we won't have time to do anymore. We only covered half the book, so you will do the rest of the book when you sign up for Algebra 2

Ooooh I see. Thanks! By the way, want to do Countdown in FTW?

Sorry, nowhere near done with the Challenge Set 4.
If you look at our Algebra 1 homepage, we have a subject called Epic FTW Tournament by maybach.
You should join. :lol:

Are the number of challenge sets of the Problem Series classes just one per week?

There are no challenge sets for Problem Series Classes.  :wink:

In addition, I think there are no Challenge Sets for Intermediate NT.


algebra
polynomial
number theory
relatively prime

prove that if there exists two integers a1 and a2 and a polynomial p(x) such that p(a1) and p(a2) are relatively prime, there exists an infinite set of integers a1, a2, a3,.... such that p(a1), p(a2)... share no factors

Hint:
[hide]
What happens if you define $a_{n + 1} = p\left(a_n\right)$?[/hide]


The side length of the regular hexagon is 10cm. What is the number of square centimeters in the area of the shaded region? Express your answer in simplest radical form.

[asy]pair A,B,C,D,E,F;
A = dir(0); B = dir(60); C = dir(120); D = dir(180); E = dir(240); F = dir(300); draw(A--B--C--D--E--F--A);
fill(B--C--E--F--cycle,grey);[/asy]

Or you can notice that a triangle can be made:
[asy]
pair A,B,C,D,E,F;
A = dir(0); B = dir(60); C = dir(120); D = dir(180); E = dir(240); F = dir(300); draw(A--B--C--D--E--F--A);
fill(B--C--E--F--cycle,grey);
draw(B--E);
[/asy]
You should be able to see that the two triangles I made are congruent, and also that they are 30-60-90 triangles.

30-60-90 triangles have side length-ratios of $1:\sqrt{3}:2$. Since we have side length $10$, we know that the height is $10\sqrt{3}$.

Multiply the two lengths to get the area of the square, $\boxed{100\sqrt{3}}$.


this may not be the forum for this question... but SAT IIs are tomorrow, and I am taking IIC, Chemistry, and US History..

I am not ready at all for Chemistry and History, even with being enrolled in the classes.. I wouldn't be surprised if I didn't break 600 on either.

Someone told me that I can go in and take just Math IIC if I want, and that I don't have to take all the tests I signed up for. I don't think this is true. On the CB website, it says cancelling one test requires cancelling all of the tests.. Someone enlighten me..

If this isn't the case, I'm going to have to cancel all of them to prevent colleges from seeing the terrible scores.

Any suggestions or anything?

You don't have to take all the tests you signed up for.  Not taking a test is different from cancelling a test score.  You can take any subset of the tests you signed up for.  You can even take SAT II tests you didn't sign up for (with a few exceptions, like language listening tests).  You can even take more tests than you signed up for (they will send you a bill for the extra).

So, just go in and take the test(s) you want to.

(I am a tutor; that's how I know.)

Wow, thank you.. that relieves me a whole lot. I'll be taking just one test then.

Thanks for the info.


Prove that:
If   $n \in \mathbb{N},n\geq 2,z \in \mathbb{C}$   and   $2^{n - 1} \left( {z^n  + 1} \right) = \left( {z + 1} \right)^n $   then    $\left| z \right| = 1$

Hey $\Pi\Delta\Phi$,  thanks for your help !
[url]http://www.sosmath.com/CBB/viewtopic.php?t=10625[/url]


Let $ P$ be a plane. Prove that there exists no function $ f: P\rightarrow P$ such that for every convex quadrilateral $ ABCD$, the points $ f(A),f(B),f(C),f(D)$ are the vertices of a concave quadrilateral.

[i]Dinu Şerbănescu[/i]

[hide="Is this correct?"]
Consider the image of a circle in $ P$; then any four of the resulting points must form a concave quadrilateral. Thus the image is an infinite number of points such that no four are collinear [*]. The convex hull of the resulting points must be a triangle (otherwise, four of the vertices of the hull form a convex quadrilateral). Let the triangle be $ \triangle ABC$, and consider a point $ D$ in the triangle. The lines $ AD, BD, CD$ partition the triangle into six smaller triangles. Consider a point $ E$ in a region bounded by (wlog) the lines $ AD$ and $ BD$ (existence guaranteed by [*]); all of the angles of $ ABDE$ are less than $ 180^{\circ}$, which is fairly simply to verify, and so is convex.
[/hide]

[hide="Slight variation"]
My solution is similar but perhaps simpler since it doesn't deal with an infinite number of points: just prove that among any $ 5$ points in the plane in general position (no three on a line) there are $ 4$ that form a convex polygon. The reason is that the convex hull must be a triangle (otherwise it's obvious), and the line passing through the two points inside the triangle separates the plane into two pieces, one of which must contain two vertices of the triangle. These two vertices and the two inner points form a convex polygon.

Now consider a convex pentagon in the plane: any $ 4$ points form a convex polygon, and the image under $ f$ also contains $ 4$ convex points by what we just proved, contradiction.
[/hide]

So there are no such functions...

You could also flat out take a pentagon and do casework (there are three cases)...

I suppose that the crossed quadrilaterals are not considered, could someone confirm that for me?


consider $n$ a positive integer $n>1$ and $p$ a prime number such that $p-1$ is divisible by $n$ and $n^3-1$ is divisible by $p$.
Prove that $4p-3$ is a perfect square.

http://www.mathlinks.ro/Forum/viewtopic.php?t=5863

  darij


limit
logarithms
calculus
calculus computations

Because:$\lim_{n \to \infty}S_n=\ln 2$ and  $S_n$ is increasing by n.
So we can Stolz Theorem:
\[ \lim_{n\to\infty} n(\ln 2-S_n)=\lim_{n\to\infty} \frac{\ln 2-S_n}{1/n}=\lim_{n\to\infty} \frac{S_n-S_{n+1}}{1/(n+1)-1/n}=\lim_{n\to\infty} \frac{1/(2n+1)-1/(2n+2)}{1/n-1/(n+1)}=\frac 1 4 \]

I can't use Stolz theorem, anyway your answer is correct. :)

I  find a link for it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15402]www.mathlinks.ro/Forum/viewtopic.php?t=15402[/url].
But I think the proof there is still not elemtary.  :(


calculus
integration
function
geometry
logarithms
calculus computations

[b][b][b](1) Is there is any shorter solution for (e^y - e-^y)/2 = x, Then prove that
dy/dx = 1 / (1+x^2)^1/2[/b]

(2) Integration of |(x-1)*(x-2)*(x-3)|dx
here Lower Limit =0 ,Upper Limit = 4 ( shorter solution).[/b][/b]

Hi jagdish.

For your first question, I have

$ \frac{(e^{y} \minus{} e^{\minus{}y})}{2}$ gives $ x\equal{}\sinh y$.  Differentiating, $ \frac{dx}{dy} \equal{} \cosh y \frac{dy}{dx} \equal{} \frac{1}{\cosh y} \equal{} \frac{1}{\sqrt{1\plus{}x^{2}}}$.

For the second,, I think the only way to do it is to multiply out the integrand and integrate term by term. I get $ x^{4} \minus{}2x^{3} \plus{}\frac{11x^{2}}{2} \minus{}6x$.

Putting in the limits, I get $ 0$.

AndrewTom, your answer to the second question is incorrect. It cannot possibly be 0, because I believe he is integrating a function that takes nonnegative values on the interval.

The trick to the second question is noticing the symmetry. Considering $ f(x) = (x - 1)(x - 2)(x - 3)$, we note that $ f(x) = - f(4 - x)$, and so $ |f(x)| = |f(4 - x)|$. Thus $ \int_0^2 \!|f(x)|\, dx = \int_2^4 \!|f(x)|\, dx$, so we need only work on one of the two intervals $ [0,2]$ and $ [2,4]$. We'll pick $ [0,2]$. Noting that $ f(x)$ is negative on $ (0,1)$ and positive on $ (1,2)$ we finally get

$ \int_0^2 \!|f(x)|\, dx = \int_1^0 \!f(x)\, dx + \int_1^2 \!f(x)\, dx$

Then, consider $ g = (x - 1)(x - 3), dg = 2(x - 2) dx$, and we can make the needed substitution to get

$ \int_1^0 \!f(x)\, dx + \int_1^2 \!f(x)\, dx = \frac {1}{2}(\int_0^3 \!g\, dg + \int_0^{ - 1} \g\, dg)$ which is a lot easier to integrate; using the antiderivative $ F(g) = \frac {g^2}{2}$ gives us $ \frac {5}{2}$. But, that's only half the area (we only took the integral over half the interval), and so we multiply by $ 2$ to get $ 5$.

ya thats correct .

since (x-1) (x-2) (x-3)  is positive in the interval $ [ 1, 2 ]$ and  $ [ 3 , \infty  ]$ . the integral can be written as

$ \int_{0}^{4} | (x-1) (x-2) (x-3) | dx  =  -\int_{0}^{1}  (x-1) (x-2) (x-3)  dx + \int_{1}^{2}  (x-1)
(x-2) (x-3)  dx   - \int_{2}^{3}  (x-1) (x-2) (x-3)  dx   +\int_{3}^{4}  (x-1) (x-2) (x-3)  dx$

 to  evaluate each term put $ x-2 = t$  and the integral becomes :

$ -\int_{-2}^{-1} t^3 - t  dt  + \int_{-1}^{0} t^3 - t  dt  - \int_{0}^{1} t^3 - t  dt  +
\int_{1}^{2} t^3 - t  dt$

on evaluating this integral we get $ 5$ .

Well, -Elixir- and vijaymenon, I'm afraid I have to disagree with both of you:

The definite integral is not the same as the area, as the curve $ y\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)$ is below the $ x$-axis from $ x\equal{}0$ to $ x\equal{}1$, above it from $ x\equal{}1$ to $ x\equal{}2$, below it from $ x\equal{}2$ to $ x\equal{}3$ and above it from $ x\equal{}3$ onwards.

$ \int_{0}^{1} \equal{} \frac{\minus{}9}{4}$

$ \int_{1}^{2} \equal{} \frac {1}{4}$

$ \int_{2}^{3} \equal{} \frac{\minus{}1}{4}$

$ \int_{3}^{4} \equal{} \frac{9}{4}$

So the area is $ \mid \frac{\minus{}9}{4} \mid \plus{} \mid \frac{1}{4} \mid \plus{} \mid \frac{\minus{}1}{4} \mid \plus{} \mid \frac{9}{4} \mid \equal{} 5$, whereas the define integral has the value $ \frac{\minus{}9}{4} \plus{} \frac{1}{4} \plus{} \frac{\minus{}1}{4} \plus{} \frac{9}{4} \equal{} 0$.

AndrewTom: Reread his post. There's an absolute value sign there. ^__^

Oops, thanks _Elixir-, I didn't spot the absolute value sign.

#1. For $ e^y \minus{} e^{ \minus{} y} \equal{} 2x$, let $ e^y \plus{} e^{ \minus{} y} \equal{} 2z$ we have $ 4(z^2 \minus{} x^2) \equal{} 4\Longleftarrow {\boxed{(A \plus{} B)^2 \minus{} (A \minus{} B)^2 \equal{} 4AB}}$, yielding $ z \equal{} \sqrt {1 \plus{} x^2}\ \because z\geq 1\ by\ A.M. \minus{} G.M.$ $ \therefore e^y \equal{} z \plus{} x\Longleftrightarrow y \equal{} \ln (x \plus{} \sqrt {1 \plus{} x^2})$, yielding $ \frac {dy}{dx} \equal{} \frac {1}{\sqrt {1 \plus{} x^2}}$

#2.

$ \int_0^4 |(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)|\ dx \equal{} 2\int_0^2 |t^3 \minus{} t|\ dt$

$ \equal{} 2\left\{ \minus{} \int_0^1 (t \minus{} t^3)\ dt \plus{} \int_1^2 (t^3 \minus{} t)\ dt\right\} \equal{} \boxed{5}$.


Hey Guys! It's been a while...

anyway, so I'm trying to develop a proof for Planck's BackBody Radiation Law... which, for those of you who don't know and want to know it's:

$u(\lambda) = \frac{8\pi h c \lambda^{-5}}{e^{hc/\lambda kT}-1}$.

The part I need is:

$\overline{E} = \frac{hc/\lambda}{e^{hc/\lambda kT}-1}$

where $\overline{E}$ is the average energy of infinite electromagnetic oscillators.

The problem is, I need to solve the summation $\sum_{n=0}^\infty \frac{n}{a^n}$. For some reason, I can't seem to do it. I know the result but I need the method. Some help would be appreciated.

Thanks!

So basically you just need
$\sum_{n=0}^\infty{\frac{n}{a^n}}=0+\frac{1}{a}+\frac{2}{a^2}+\cdots=S$

You ll see why I put =S at the end but it is basically just assigning variables.  Well if you multiply by 1/a you get:

$\frac{1}{a}\sum_{n=0}^\infty{\frac{n}{a^n}}=0+\frac{1}{a^2}+\frac{1}{a^3}+\cdots=\frac Sa$

Now we note that $S-\frac 1a S=\frac 1a+ \frac{1}{a^2}+\cdots$ which is geometric.  Evaluating it such yields:

$\frac{a-1}{a}S=\frac{\frac{1}{a}}{1-\frac 1a } \Rightarrow S=\frac{a}{(a-1)^2}$

[quote="Spoon"]Thank you.[/quote]

Here's a similar idea, which I first saw a long time ago on an old AHSME problem.

Sum(n/a^n) = 0 + 1/a + 2/a^2 + ..... + n/a^n
= 0 + 1/a + 1/a^2 + .....
+ 0 + 0 + 1/a^2 + ......
+ 0 + 0 + 0 + 1/a^3

So we have essentially a triangular array of infinite geometric series.

However, what we end up getting is a geometric series of geometric series, with first term 1/(a-1) and common ratio 1/a

So we get 1/(a-1)*(1-1/a) = a/(a-1)^2 which is again what joml posted.

And of course, for any of these summations to be valid, we must have |a| < 1.


[quote="STARS"]Tangents $ l_1$ and $ l_2$ common to circles $ c_1$ and $ c_2$ intersect at point $ P$, whereby tangent points remain to different sides from $ P$ on both tangent lines. Through some point $ T$, tangents $ p_1$ and $ p_2$ to circle $ c_1$ and tangents $ p_3$ and $ p_4$ to circle $ c_2$ are drawn. The intersection points of $ l_1$ with lines $ p_1, p_2, p_3, p_4$ are $ A_1, B_1, C_1, D_1$, respectively, whereby the order of points on $ l_1$ is: $ A_1, B_1, P, C_1, D_1$. Analogously, the intersection points of $ l_2$ with lines $ p_1, p_2, p_3, p_4$ are $ A_2, B_2, C_2, D_2$, respectively. Prove that if both quadrangles $ A_1A_2D_1D_2$ and $ B_1B_2C_1C_2$ are cyclic then radii of $ c_1$ and $ c_2$ are equal.[/quote]
[hide="Solution"]
Let the incircle of $ \triangle A_2PA_1$ have radius $ r$ and the incircle of $ \triangle D_1PD_2$ have radius $ R$.  Notice that since $ B_2C_1C_2B_1$ and $ A_1A_2D_1D_2$ are cyclic, we have that $ \angle TA_1B_1 = \angle TD_2C_2$ and $ \angle TB_1P = \angle TC_2P$.  Thus, $ \angle TB_1P - \angle TA_1B_1 = \angle TC_2P - \angle TD_2C_2$.  It follows that $ \angle A_1TB_1 = \angle C_2TD_2$, so $ \triangle TB_1A_1\sim \triangle TC_2D_2$.  Furthermore, $ \triangle A_2PA_1\sim \triangle D_1PD_2$.  Thus, their incircles are in porportion, so $ \frac {A_1P}{D_2P} = \frac {r}{R}$.  Yet, $ \triangle B_2PB_1\sim \triangle C_1PC_2$, so their excircles are in porportion, so $ \frac {PB_1}{PC_2} = \frac {r}{R} = \frac {A_1P}{D_2P}$.  This, combined with the fact that $ \triangle TB_1A_1\sim \triangle TC_2D_2$ implies that $ \triangle TPA_1\sim \triangle TPD_2$, so $ \angle A_2TP = \angle D_1TP$.  Also, \[ \angle TPA_2 = \angle TPA_1 - \angle A_2PA_1 = \angle TPD_2 - \angle D_1PD_2 = \angle TPD_1\]  Furthermore, $ TP = TP$, so $ \triangle TPA_2\cong \triangle TPD_1$.  It follows that $ TA_2 = TD_1$.  Since $ A_1A_2D_1D_2$ is cyclic, we conclude that $ A_1A_2 = D_1D_2$, so $ \triangle A_1A_2P\cong \triangle D_2D_1P$.  Thus, $ r = R$, as desired. [/hide]


HMMT Harvard college MIT email geometry AMC

Hey everyone,

This forum is for discussion of the Harvard-MIT Mathematics Tournament (HMMT).  For official information about the contest please visit the [url=http://web.mit.edu/hmmt/www/]HMMT Website[/url].

[quote="bpms"]Do you have to do it through your school?[/quote]

uh no. see: Quagga.

does a team have to have a coach, or can the team manage the registration stuff by itself?

Not necessarily an adult, though having an adult go with you/do registration is always much easier and organized. They do ask for a sponsor to help with grading and helping out.

A team can be fully student run and the students do not have to go to the same school.

if we do have people from other schools, should we put "Various" for the "school" account, and just choose 1 school's address to put for the account?

ok hopefully my last question.

when we send our payment by mail, we just address it to MIT's mailing address, right?

also, do we include just our school name in the mail, or should we also add school address and team name??

You aren't supposed to pay until you get an email confirmation that you can come.  Also, please don't just send it to MIT, I'm pretty sure there's another address for you to send it to.  I don't what address they'll want you to send it to though.

Is it true that winning the HMMT guarantees admission to Harvard and MIT?

[quote="joml88"]lol, no.  Where did you hear that?[/quote]
I didn't mean like, once you win you join instantly, I meant that like if you win HMMT, then when you apply they let you in for sure. I heard it from a coach of mine...

Yes, I know what you meant.  I don't think there is any policy like that.  The only way to get in is through the application and it's very unlikely they'll let you in based on one thing.  I think your coach was being a little facetious.  I mean, chances are good that if you win HMMT you'll also have the additional credentials to get into both schools.  But as far as a guarantee that you'll get in, none that I know of exist.

Do any middleschoolers participate or they have no chance?
Like Geometry class 8-graders? Geometry is officially a high school subject.
Or the math level is not enough, they need years of math competition preparation (which they do not have in 8th grade) to be ready?

Does the payment check have to arrive by February 1st or does it just have to be postmarked by then?

is it only for high schoolers or can talented 7th graders go?

@ilikecake: I don't see any rules prohibiting those not yet in high school. However, a word of caution: HMMT is one of the harder college tournaments, and when writing questions they keep the high-school demographic in mind; this means that the problems might be quite hard. But if you think you can handle it, all the more power to you.

@ilikecake: I've participated since 6th grade. Anyone in high school or below can participate, but obviously it will more challenging for those below high school (in general).

So, uh, if I'm the only kid interested in math competitions at my school, and just about the only one with contest experience in my county, do I have to find five people from another county in my state (say, approximately 97 miles away) in order to participate in this as a team?

Also, so I see there's no number theory category for the individual tests. Bummer. I'd go with combo+alg then, probably.

Yeah, it seems so. Also note that on the rules page [url]http://mit.edu/hmmt/www/faq.shtml#registration[/url] it says that you must live within 100 miles of those other people. (Though it also says that exceptions can be made.)

More pertaining to your specific case, we sent an NC team to HMMT last year, and got 3rd place. There seems to be no reason why we won't do the same again this year (and we might even send two teams?). If so, its pretty likely (almost guaranteed) that you'd be offered a spot on one of the teams. Just keep an eye out in the NC forum for any contest signups (for this and PUMAC).

When will we be able to sign up for the HMMT?

We are going to try to open registration within the next week. We will make sure that word gets out.

Can anyone say what the date for the 2012 HMNT will be?

I have a question:

I am an international student (from Guatemala) and I don't have a team but I really want to participate on HMMT or HMNT, what can I do?

[quote="JSGandora"]I think these questions would better be directed at the HMMT officers themselves: [url]http://web.mit.edu/hmmt/www/contact.shtml[/url][/quote]

thank you, I already tried that more than a month ago, maybe 2 but never heard back...

I believe that HMMT is only open to US students.  You could also pm one of the officers on AoPS.

Hi! It's a little late to register for the 2013 Tournament, but we do actually allow international students to compete at HMMT February.