geometry
calculus
integration
limit
calculus computations

What is the area of the following region?

$ 0 \le y \le e^x$, $ x \le 0$.

You're looking to evaluate this integral:
$ \int_{\minus{}\infty}^{0}e^{x}\ dx$
This is just
$ e^{x}\bigg|_{\minus{}\infty}^{0}\equal{}1\minus{}\lim_{x\rightarrow\minus{}\infty}e^{x}\equal{}1\minus{}0\equal{}1$

Good, but this topic will be left open for other solutions.

BTW, find a formula for this:

$ \int_{a}^{b}e^x$

I'll assume there's supposed to be a $ dx$ following the $ \int_{a}^{b} e^x$. (I don't know how to do it without the $ dx$)
Using a similar format as JRav's solution, it's just $ e^{x}\bigg|_{a}^{b} \equal{} e^b \minus{} e^a$

Without the $ d$_, it's just nonsense. The $ d$_ is part of the integral sign.


geometry
incenter
rectangle
cyclic quadrilateral
geometry proposed

$ABCDE$ is a convex pentagon inscribed in a circle. Prove that the centers of mass of triangles $I_{ABC}I_{ACD}I_{ADE}$ and $I_{BAE}I_{BED}I_{BCD}$ are identical. Here $I_{XYZ}$ means the incenter of triangle $XYZ$.

First, we show that centroids of the triangles $\triangle I_{ABC}I_{ACD}I_{ADE}$ and $\triangle I_{DEA}I_{DAB}I_{DBC}$ are identical. Their vertex $I_{ADE}\equiv I_{DEA}$ is common. ABCD is a cyclic quadrilateral, the remaining 4 incenters form a rectangle $I_{ABC}I_{DBC}I_{ACD}I_{DAB}$ with equal diagonals  $I_{ABC}I_{ACD}$ and $I_{DAB}I_{DBC}$ cutting each other at the common midpoint $M_{ABCD}$ of these 2 segments, hence the triangles $\triangle I_{ABC}I_{ACD}I_{ADE},\ \triangle I_{DEA}I_{DAB}I_{DBC}$ have a common centroid $G_{AD}.$ In exactly the same way, the triangles $\triangle I_{DEA}I_{DAB}I_{DBC}$ and $\triangle I_{BCD}I_{BDE}I_{BEA}$ have a common centroid $G_{DB}\equiv G_{AD}$ and the conclusion follows.


The function $ f$ is not defined for $ x \equal{} 0$, but, for all non-zero real numbers $ x$,
\[ f(x) \plus{} 2f\left(\frac {1}{x} \right) \equal{} 3x.
\]
The equation $ f(x) \equal{} f( \minus{} x)$ is satisfied by

$ \textbf{(A)} \ \text{exactly one number} \qquad \textbf{(B)} \ \text{exactly two real numbers} \qquad \textbf{(C)} \ \text{no real numbers} \\
\textbf{(D)}\ \text{infinitely many, but not all, non \minus{} zero real numbers} \qquad \textbf{(E)} \ \text{all non \minus{} zero real numbers}$

[size=75]EDIT - Fixed the LaTeX in the answer choices.  Thanks, Brut3Forc3.[/size]

[quote="Octahedr0n"]The function $ f$ is not defined for $ x \equal{} 0$, but, for all non-zero real numbers $ x$,
\[ f(x) \plus{} 2f\left(\frac {1}{x} \right) \equal{} 3x.
\]
The equation $ f(x) \equal{} f( \minus{} x)$ is satisfied by

$ \textbf{(A)} \ \text{exactly one number} \qquad \textbf{(B)} \ \text{exactly two real numbers} \qquad \textbf{(C)} \ \text{no real numbers} \\
\textbf{(D)}\ \text{infinitely many, but not all, non \minus{} zero real numbers} \qquad \textbf{(E)} \ \text{all non \minus{} zero real numbers}$[/quote]
LaTeX fix; you need to separate lines, otherwise it's too long...

[hide="Solution"]Plug in $ \frac{1}{x}$ to obtain $ f\left(\frac{1}{x}\right)\plus{}2f(x)\equal{}\frac{3}{x}$. Adding this to the original equation, we have $ f(x)\plus{}f\left(\frac{1}{x}\right)\equal{}x\plus{}\frac{1}{x}$. Now we subtract this from the original equation to get $ f\left(\frac{1}{x}\right)\equal{}2x\minus{}\frac{1}{x}$, or $ f(x)\equal{}\frac{2}{x}\minus{}x$. To have $ f(x)\equal{}f(\minus{}x)$, we must have $ \frac{2}{x}\minus{}x\equal{}\minus{}\frac{2}{x}\plus{}x \iff \frac{4}{x}\equal{}2x \iff 2\equal{}x^2$, which is satisfied for exactly two real numbers, or [b]B[/b].[/hide]

I got exactly two real numbers, but I'll wait for more people to post.

[hide="solution"]

$ f(x) \plus{} 2f\left(\frac {1}{x} \right) \equal{} 3x \qquad(1)$

Let $ x \equal{} \frac {1}{x}$

$ f\left(\frac {1}{x}\right) \plus{} 2f(x) \equal{} \frac {3}{x} \qquad(2)$

Subtracting $ 2(2) \minus{} (1)$ we get

$ f(x) \equal{} \frac {2}{x} \minus{} x$

$ \therefore$ if $ f(x) \equal{} f( \minus{} x) \Longrightarrow \frac {2}{x} \minus{} x \equal{} x \minus{} \frac {2}{x}$

$ \Rightarrow x \equal{} \pm\sqrt {2} \longrightarrow \boxed{B}$[/hide]


hello, i see no inequality here.
Sonnhard.

[quote="arqady"][quote="thanhak5"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$ . Prove that
           $ \dfrac{a}{(ab \plus{} b \plus{} 1)^2} \plus{} \dfrac{b}{(bc \plus{} b \plus{} 1)^2} \plus{} \dfrac{c}{(ca \plus{} c \plus{} 1)^2} \geq \dfrac{1}{a \plus{} b \plus{} c}$[/quote]
Maybe $ \dfrac{a}{(ab \plus{} b \plus{} 1)^2} \plus{} \dfrac{b}{(bc \plus{} c \plus{} 1)^2} \plus{} \dfrac{c}{(ca \plus{} a \plus{} 1)^2} \geq \dfrac{1}{a \plus{} b \plus{} c}$  $ ?$  :wink:
It's true.[/quote]We can let $ a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}$ so that $ \sum\frac{\frac{x}{y}}{\left(\frac{x}{z}\plus{}\frac{y}{z}\plus{}\frac{z}{z}\right)^2}\geq\frac{1}{\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}}$
$ \Longleftrightarrow \left(\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\right)\left(z^2\frac{x}{y}\plus{}x^2\frac{y}{z}\plus{}y^2\frac{z}{x}\right)\geq (x\plus{}y\plus{}z)^2$
But we have $ \left(\frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\right)\left(z^2\frac{x}{y}\plus{}x^2\frac{y}{z}\plus{}y^2\frac{z}{x}\right)\geq \left(\frac{zx}{y}\plus{}\frac{xy}{z}\plus{}\frac{yz}{x}\right)^2$ by Cauchy Schwarz and $ \frac{zx}{y}\plus{}\frac{xy}{z}\plus{}\frac{yz}{x}\geq x\plus{}y\plus{}z$ by AM-GM.

Thank you minsoens , arqady , Math pro very much . But I think your solution is long . We try giving to a solutoin shorter for it


function
geometry
3D geometry
number theory
prime factorization
geometric series

Prove that there are no perfect numbers that are also perfect squares.

[hide]Let n be a perfect square.

Case 1:

n is odd.

n has only odd factors.  Since it is a square, it has an odd number of factors and an even number of proper factors.  The sum of an even number of odds is always even and therefore can not be n.

Case 2:

n is even.

The odd prime factors of n must be raised to even powers in n's prime factorization because n is a perfect square.  The number of odd factors of n is equal to the number of factors of the product of all of n's odd prime .  The number of factors is the product of (the powers of the prime factors plus one).  Interpret those parentheses mathematically .  Since all the powers of even, the powers plus one are all odd, and therefore the number of factors is odd.  An odd number of odd factors sum to an odd number, which can not be n because n is even.  Note that we did not exclude n from the list of odd factors because n is not an odd factor.

Sorry for being confusing, but it's hard to explain.

QED[/hide]

Here's how i proved it:

[hide]Let there be a function p(n) which tests the "perfection" of n by adding all its factors. I started exploring a bit with a program i made on my calculator and found that for any power of 2, p(2^x) = 2^x - 1. I tried powers of 3 and found that p(3^x) = (3^x - 1)/2. Also,

p(5^x) = (5^x - 1)/4 and

p(7^x) = (7^x - 1)/6. In general, for any prime n, p(n^x) = (n^x - 1)/(n-1).

So if we want to find a perfect square, p(n :^2: ) = (n :^2:  - 1)/(n-1). This will only work if p(n) = n, so...

n = (n :^2:  - 1)/(n-1).

n(n-1) = (n :^2: -1)

n :^2: -n = n :^2: -1.

n = 1

And 1 doesn't even count.

So no perfect squares.

[/hide]

Here's another challenge:

Are there any perfect cubes? 4ths? 5ths? nths?[hide][/hide]

Your proof doesn't work, I'm sorry to tell you.  You found (though you didn't prove) that for n^x, the sum is (n^(x+1)-1)/(n-1), where is n is a prime.  Unfortunately, that doesn't work for all squares, only the squares of primes.  However, for example, 6^2, the sum is not that given by your formula.  Your formula predicts a sum of 43, when the actual sum is 98.  So, you've managed to prove it for prime squares, but not for all squares.

And, just for kicks, a proof of the formula.  Nice and simple.

[hide]The only divisors of p^n, where p is a prime, are p^0, p^1,...,p^(n-1),p^n.

Therefore, their sum is:

p^0+p^1+...+p^(n-1)+p^n

Using the formula for the sum of geometric series, we find that that sum is given by

(p^(n+1)-1)/(p-1)

[/hide]

But if you're interested in the sum of the divisors, look into the number theoretic function S(n).


The domain of f is [-2,10] and its range is (-15,2].

The domain of g is (-5,7) and its range is [-2,3).

What is the domain and range of:

a) [f(x)]^2

b) f(x)+g(x)

c) f(x) * g(x)

Ask yourself what the domain and range are.  The domain is the set where the function is defined, and the range is the set that the function maps to.

To "discover" what the correct answer is, make up some functions for $ f$ and $ g$ (try making a straight line between points so as to guarantee the range is spanned).  See what happens and try to generalize.


algorithm
combinatorics proposed
combinatorics

The digits of a calculator (with the exception of 0) are shown in the form indicated by the figure below, where there is also a button ``+":
[img]6965[/img]
Two players $A$ and $B$ play in the following manner: $A$ turns on the calculator and presses a digit, and then presses the button ``+". $A$ passes the calculator to $B$, which presses a digit in the same row or column with the one pressed by $A$ that is not the same as the last one pressed by $A$; and then presses + and returns the calculator to $A$, repeating the operation in this manner successively. The first player that reaches or exceeds the sum of 31 loses the game. Which of the two players have a winning strategy and what is it?

[quote="i_like_pie"][hide]$B$ has a winning strategy:

Each time $B$ adds a number to the current sum, he should add $10-a$, where $a$ is the number $A$ added on his turn. This way, the sum will be $30$ at the end of the third turn (1 turn = both go once), and $A$ loses.[/hide][/quote]
It seems that this problem isn't so simple. You are not able to do that at any case.
For example, if he add 7 you can only add 1,4,8, or 9

[color=blue]By the way, I have found [u]all[/u] the winning conditions, at each step.
But I can't find a pattern  :(

I mean that, I can't find a formula to express the winning strategy. If I play this game I have to look to the table of the best moves
[/color]

See the table below.
I made this table by a reccursive way, from the end to the beginning.
The winner is the player $A$

[img]6978[/img][img]6979[/img]

The top of each column shows the current sum.
You find the current sum and then you look at that column.
You will find the best moves. If some move is prohibited, then you choose some other move, in the same column.

So, when the player $A$ starts playing, he should press 3 or 9 exclusive.

Say that $A$ presses 3
If then the player $B$ presses the number 9, we go to the sum 12.

The best moves are 2,4,6. But $A$ can only press the number 6 (because the last pressed digit was 9).
And so on.

[hide="How this table was created - The Algorithm"]
We will express the state just with the current sum $S$
Let:
$BM(S)$ be the set of the best moves, when the sum is $S$
$N(a)$ be the set of the prohibited keys starting from the key $a$. So, every key in this set is in the same row or in the same column with $a$ [u]exclusive[/u].

${X(S)=\bigcap_{i\in BM(S)}{N(i)}}$

So, each $k\in X(S)$ is a key not connected to any of the best moves. That is, if we press the number $k$, starting from the state $S-k$, then the opponent won't be able to play some of the winning moves.

Clearly, if the sum is [b]30[/b] then the player will loose. So there are not best moves on the [b]state-30[/b].
That is, $BM(30)=\emptyset$

The procedure is as follows:
$\forall k \in X(S)$ find $S-k$. Then add $k$ to the set $BM(S-k)$

If you lead your opponent to the [b]state-30[/b], you'll win. So, $k$ is a best move on the state $30-k$.
In other notations, $k\in BM(30-k), \ \ \forall k \in \{1,2,...,9\}$

Now in the [b]state-29[/b].
The set $BM(29)$ is already formed; it is $BM(29)=\{1\}$

We have $X(29)=N(1)=\{1,5,6,8,9\}$.
So we repeat the procedure:

$\forall k \in N(1)$ find $S-k$. Then $k\in BM(S-k)$

In our example it is [b]S=29[/b]. Then
$S-1=28 \Rightarrow  1 \in BM(28)$
$S-5=24 \Rightarrow  5 \in BM(24)$
$S-6=23 \Rightarrow  6 \in BM(23)$
$S-8=21 \Rightarrow  8 \in BM(21)$
$S-9=20 \Rightarrow  9 \in BM(20)$

After that, we continue to the [b]state-28[/b], knowing that $BM(28)=\{1,2\}$. Then $X(28) = N(1) \cap N(2)$
And so on...[/hide]

The above algorithm helps us to find all the winning moves. But you can't have this table on your mind while you are playing.
Can somebody see a pattern in this table?
Can we find some useful formula for it??  :huh:

Ahh... While I was solving the problem I forgot about that complication. :(

Let’s say number A starts. He picks number 9. The sum now equals 9. Let’s now assume every B’s movement.

B picks 8. The sum is now 17. Now since we are A, we decide to pick 9 again. The sum is 26 now. B’s possible movements now are {8,7,6,3}, and out of those ones, only the last one can be picked to not overpass the limit. B picks 3. The sum is 29. And A picks 1 now, so the sum is 30 and A has won the game.

B picks 7. The sum is 16. We pick 9. The sum is 25. The possible B’s movements are {8,7,6,3} and only “3” is a pickable choice. B picks 3. The sum is 28. A picks 2. The sum is 30 and A wins again.

B picks 6. The sum is 15. A picks 5. The sum is 20. And we may see, every number B picks, A can chose the opposite option to make it sum 30 and take the game. {2:8,8:2,4:6,6:4}

B picks 3.  The sum is 12. A picks 6. The sum is 18. Now B has 4 new choices:

   B picks either 3 or 9. The sum is either 21 or 27, so A will just have to pick the other one and make it add 30. A wins the game.
   B picks 4. The sum is 22. A picks 6. The sum is 28 and since we are in 6, the only available choices are {9,5,4,3} which all lend to pass the max sum and A wins the game.
   [code][/code]B picks 5. The sum is 23. A picks 6. The sum is 29. And, as in the previous example. A has won the game since there aren't more possible choices.


calculus
integration
logarithms
trigonometry
calculus computations

Compute the intergral $\int^{\dfrac \pi 4}_0 \ln {(1 +\tan x)}$.

Write $ \ln (1+\tan x)= \ln \frac {(\ sin x+\ cos x) } {\cos x} = \ln (\sqrt {2}\sin (x+\frac {\pi} {4}))-\ln \cos x$ and use the following knowledge :
\[
\displaystyle \int^{\frac {\pi} {2}}_0 \ln \sin x \,dx = \frac {\pi} {2} \ln 0,5
\]


calculus
integration
limit
trigonometry
function
real analysis
real analysis unsolved

Prove that \[ \lim_{n\rightarrow \infty}  \frac{1}{n} \int_{1}^{\infty}
\sqrt[n]{t} \, \frac{\sin t}{t} \, dt\equal{}0.\]

Write $ f_{n}(t) \equal{} \frac{1}{n}t^{\frac{1}{n} \minus{} 1} \sin t$.
1. Can you bound the sequences {$ f_{n}$} by some non-negative function which is (Lebesgue) integrable?
2. What is $ \lim_{n \to \infty} f_{n}$?
3. If you have some results for 1 and 2, then you can invoke the Dominated Convergence Theorem.

[quote="akech"]1. Can you bound the sequences {$ f_{n}$} by some non-negative function which is (Lebesgue) integrable?[/quote]
No, I don't think  you can. After all the $ f_n$ are not themselves nonnegative, and $ \frac{\sin t}t$ is not integrable on $ [1,\infty).$

Try integrating by parts first:

$ \int_1^{\infty}\frac1nt^{\frac1n\minus{}1}\sin t\,dt\equal{} \left.\minus{}\frac1nt^{\frac1n\minus{}1}\cos t\right|_1^{\infty}\plus{}\frac1n\left(\frac1n\minus{}1\right)\int_1^{\infty}t^{\frac1n\minus{}2}\cos t\,dt$

Now you can slap absolute values in, replace $ \cos t$ by $ 1,$ and have an estimate you can use.

Thanks, Kent. I should have been careful.


Consider $ n$ functions of the form $ f_n: R\rightarrow R$, $ f_n(x)\equal{}a_nx^2\plus{}b_nx\plus{}c$, $ a_n\ne 0$. Which is the greatest possible number of parts that  the plane $ xOy$ can be divided by the graphs of these $ n$ functions?

Is it intentional that $ c$ is independent of $ n$?

I still see no solution.
Is this problem so hard?


Blah.

OK guys so all our tournaments are going to be in like, an uberness cup. I will tally the results whether you like it or not, so if you don't like it, then you can  :cursing:

Yeah sorry about the outburst. I sense a mod will delete both this line and the above paragraph.

On to business!

[size=200]Welcome to CTC!!!![/size]

Like the acronym?

So yeah, I'll be tallying tourney results, from the tournament after Dojo's on. You get points based on place.

1=50
2=25
3/4=15 (if there is a distinction, like a 3rd place match, then it becomes 20/10)
5-8=5
9-16=2

That is for a 16. It will be adjusted accordingly for more or less.

CURRENT STANDINGS:

You all are tied with a bunch of first graders. How does that make you feel?

Erm, why does the acronym CTC sound so familiar? :|

I don't know. Perhaps you were involved with the original[url=http://cxa.marketwatch.com/TSX/en/Market/quote.aspx?symb=CTC]CTC?[/url]

[quote="mathblitz"]based on the acronym i dont like where this is going but i will remain positive and try not to go cynical. but why do we need this? we already have tournaments we dont need an uber tournament of tournaments thats just wrong and it would take to long until any good info on whos the best would have some importance. And then i press the question:

WHY DO YOU or anyone for that matter... NEED TO KNOW YOUR THE BEST??? why do you need to lower others self esteem to satisfy yourself? the tournaments do enough.
I dont think this is a good idea...

Uh oh cynical side just came out. Sorry.[/quote]

OK if we are going to go all cynical.

1. Proper grammar please. No one likes to see:

omg u suck ctc is a stupid acomrny and u suck and i hatez u cuz u w3ant to win blagh lets not use punctuation or capital l3tt3rz...

2. Also, you don't have to participate.

3. Why is it that you never post with the exception of saying how stupid one of my ideas is?


calculus
integration
LaTeX
algebra
partial fractions
calculus computations

I don't know how to use the symbols yet, so INT is the integration sign

INT 1/(x^9+x^6) dx[/code]

Is this what you want to compute: $\int \frac{1}{x^{9}+x^{6}}\,dx$

Here's the $\text{\LaTeX}$ website:
[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php]Learn LateX[/url]

Anyway, here is what I would do with this problem:
$\int \frac{1}{x^{9}+x^{6}}\,dx$
${=\int \frac{1}{x^{6}(x^{3}+1)}}\,dx$
$=\int \frac{1}{x^{6}}\cdot \frac{1}{(x^{3}+1)}\,dx$

Now you can probably apply integration by parts and then use partial fractions...
OR you could keep factoring:

$=\int \frac{1}{x^{6}}\cdot \frac{1}{(x+1)(x^{2}-x+1)}\,dx$


function
calculus
derivative
complex analysis
complex numbers

I've just started reading a complex analysis book (Ahlfors) and I was wondering if someone could help me with some concepts i don't understand.

After deriving the Cauchy-Riemann equations, $ \frac { \partial u }{\partial x } \equal{} \frac { \partial v }{ \partial y }$, $ \frac {\partial u }{\partial y } \equal{} \minus{} \frac { \partial v }{ \partial u }$, the book then says :
$ \bigtriangleup u \equal{} \frac { \partial ^2 u } {  \partial x^2} \plus{} \frac {  \partial ^2 u } {  \partial y^2 } \equal{} 0$ (and similarly for $ v$). I don't understand how that is obtained from the previous equations.

I'm also not sure how to rigorously prove the following (which seem obvious):
1. If $ g ( w)$ and $ f(z)$ are analytic function, show that $ g(f(z))$ is also analytic. [ Is the normal proof of the Chain Rule applicable here, without modification ? ]
2. Prove rigorously that the functions $ f(z)$ and $ \overline {f ( \overline z )}$ are simultaneously analytic.
3. Prove that the function $ u(z)$ and $ u( \overline z)$ are simultaneously harmonic.

1. Differentiate the CR equations to get harmonicity.
2. Yes, usual chain rule proof is fine.
3. Write down the local power series, and take the conjugate actions you want.
4. Explicitly compute the partial derivative with respect to x and y, and see how they are different for the two functions.

In ahlfors analytic functions are not introduced as having power series locally, so for 3) you have to take a different route than suggested. Almost always, it is easiest to do it through the cauchy-riemann equations. I imagine that there might be a slight problem because Ahlfors only shows that $ f \equal{} u \plus{} iv$ is analytic if the partials of $ u,v$ are continuous and satisfy C-R. However, later it will be shown that if $ f$ is analytic, then the partials are continuous. Or you may want to directly prove that $ f$ is complex differentiable at a point iff it it is differentiable as a map from $ R^2$ to $ R^2$ and the partials satisfy the C-R equations in that point.


polynomial
number theory unsolved
number theory

Prove that all prime of the form $ p = 3k + 1$ can be written as $ p = a^{2} + 3b^{2}$ where $ a,b \in \mathbb{Z}$.

If p is a prime in the form $ 3k+1$ then $ (\frac{-3}{p})=1$,it means that exist an integer t such that $ t^2\equiv -3 (\mod p)$
Let $ m=[\sqrt{p}]$
Consider the set ${ A=\{x+ty},x,y\in \{0,...,m\}$
$ |A|=m^2>p$ therefore exist $ (x_0,y_0),(x_1,y_1)$ such that
$ x_0+ty_0\equiv x_1+ty_1 (\mod p)$
Let :$ a=x_0-x_1,b=y_1-y_0$
Then $ a\equiv tb (\mod p)$
$ \Rightarrow a^2\equiv -3b^2 (\mod p)$
Because $ a,b\in \{0,..,m\}$ so $ a^2+3b^2\in \{p,2p,3p\}$
[u] Case 1 [/u]
 $ a^2+3b^2=2p^2$ then $ a\equiv b \equiv 1 (\mod 2)$
It gives that $ 4|a^2+3b^2$ or $ 2|p$ (contradiction )
[u] Case 3 [/u]
 $ 3p=a^2+3b^2$ then $ a=3a'$ and so $ p=b^2+3a'^2$
[u]Case 3 [/u] $ p=a^2+3b^2$ then problem is true.
So problem claim.

Just a thought. I know you can prove that $ p$ is the sum of two squares iff $ p \equiv 1 (mod 4)$ using Gaussian Integers. The proof I know is to show that $ p$ is not prime in $ \mathbb{Z}[i]$. Can we similarly prove that for $ p \equiv 1(mod 3)$ it is not prime in $ \mathbb{Z}[\sqrt{\minus{}3}]$ and hence prove the original problem?

P.S. I don't know if I'm using this $ \mathbb{Z} [ BLAH]$ notation right  :P .

Yes.  The corresponding result is to show that a prime congruent to $ 1 \bmod 3$ divides a number of the form $ x^2 \plus{} x \plus{} 1$ and therefore cannot be prime, but the relevant ring of integers to work with is not $ \mathbb{Z}[ \sqrt { \minus{} 3}]$ but $ \mathbb{Z} \left[ \frac {1 \plus{} \sqrt { \minus{} 3}}{2} \right]$, the [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url].


http://xkcd.com/503
Hasn't this bothered you? What do you think the global terminology should be?

"The West" and "The East" are historical designations that make perfect sense when you place "Me" somewhere around Turkey or Greece.  Nowadays, they take their meaning from their historical connotations, not their geographical denotations.

When I travel to India, I go east, so I don't mind.

Also, isn't the US considered part of the West?


How do I make a equal sign with a question mark on top of it?  =?

\stackrel{stuff on top}{stuff on bottom} works well.  For example, P \stackrel{?}{=} NP gives
\[ P \stackrel{?}{ \equal{} } NP
\]
You could also use \overset in much the same way (or \underset, with self-explanatory function), I'm not sure exactly what makes them different internally.

Of course, there are some specialties:
\doteq
$ \doteq$

\equiv
$ \equiv$

\cong
$ \cong$

Go to the AoPS wiki for more.

How do you type stuff both on top and below a right arrow?

You can always do something like this (defining a new command, if you need the symbol more than once, of course)
$ F\begin{matrix}{\scriptstyle ab}\\[\minus{}4pt] \Rightarrow\\[\minus{}6pt]{\scriptstyle c}\end{matrix}G$
but I suspect stevem knows of better options.

Well \overset and \underset can be stacked without a problem so you could do something like
\[ A\overset{\text{top}}{\underset{\text{bottom}}{\Longrightarrow}}B
\]
If it's for a thin right arrow then there's the specialized command \xrightarrow[botom]{top}:
\[ A\xleftarrow[\text{bottom}]{\text{top}}B\xrightarrow[\text{bottom}]{\text{top}}C
\]
not sure if there's an analogous command for other arrows recognized on these fora ('\xRightarrow' and '\Xrightarrow' don't seem to work)

[quote="fedja"]but I suspect stevem knows of better options.[/quote]JoeBlow beat me to it :) Those commands require amsmath in a document which should be loaded anyway.

[quote="JoeBlow"]not sure if there's an analogous command for other arrows recognized on these fora ('\xRightarrow' and '\Xrightarrow' don't seem to work)[/quote]Unfortunately it won't work here but for documents you can get these extra arrows using the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=extarrows]extarrows package[/url] - you can see examples of them [url=http://tug.ctan.org/get/macros/latex/contrib/extarrows/extarrows-test.pdf]here[/url].


geometry
geometric transformation
reflection

Prove that every $ f: \mathbb{N}\to\mathbb{N}$ which is a bijection can be written as the sum of two involutions.
http://en.wikipedia.org/wiki/Involution_%28mathematics%29

I assume that should read "composition of two involutions".

[hide]
Let $ X_1 \equal{} \mathbb{N}$. We define $ X_n$ iteratively as follows: let $ S_n \equal{} \{x: \exists n \in \mathbb{Z}, f^n(x) \equal{} \text{min}(X_n)\}$, and set $ X_{n \plus{} 1} \equal{} X_n \setminus S_n$; thus, $ \bigcup S_n \equal{} \mathbb{N}$. (here $ f^n$ refers to the composition of $ f$, $ n$ times)

Suppose $ |S_n| \equal{} k \in \mathbb{N}$. If $ k \equal{} 1$, then define $ g_n(x) \equal{} h_n(x) \equal{} x$ where $ x \in S_n$. Otherwise, $ S_n \equal{} \{x_1, \ldots, x_{k}\}$ where $ f(x_i) \equal{} x_{i \plus{} 1}, x_{k \plus{} 1} : \equal{} x_1$, define the involutions $ g_n,h_n : S_n \to S_n$ as follows: $ g_n(x_i) \equal{} x_{k \plus{} 2 \minus{} i}, h_n(x_i) \equal{} x_{k \plus{} 3 \minus{} i}$ (they are involutions due to the definition of $ x_{k \plus{} 1}$, though this is shown in more detail in the hidden tag); obviously $ f_n(x_i) \equal{} h_n(g_n(x_i))$.
[hide="More specifically,"]$ g_n(x_1) \equal{} x_1, g_n(x_{i}) \equal{} x_{k \plus{} 2 \minus{} i}$ for $ 2 \le i \le k$; and $ h_n(x_1) \equal{} x_2, h_n(x_2) \equal{} x_1$, and $ h_n(x_{i}) \equal{} x_{k \plus{} 3 \minus{} i}$ for $ 3 \le i \le k$. Observe that
\[ h_n(g_n(x_i)) \equal{} \begin{cases}x_{k \plus{} 3 \minus{} (k \plus{} 2 \minus{} i)} \equal{} x_{i \plus{} 1}, & \quad 2 \le i \le k \minus{} 1 \\
x_2, & \quad i \equal{} 1 \\
x_1, & \quad i \equal{} k\end{cases} \equal{} f_n(x_i).\] Here's an example, for $ k \equal{} 5$ and $ S_n \equal{} \{1,2,3,4,5\}$:
\[ \begin{tabular}{|r|r|r|} \hline x & g(x) & h(g(x)) \\
\hline 1 & 1 & 2 \\
2 & 5 & 3 \\
3 & 4 & 4 \\
4 & 3 & 5 \\
5 & 2 & 1 \\
\hline\end{tabular}\] where $ g(1) \equal{} 1$, and the remaining elements are 'reflected' by $ g$; and all the elements are 'reflected' by $ h$.[/hide]
If $ S_n$ is countably infinite, select an arbitrary element $ x_1 \in S_n$, and let $ S_n \equal{} \{x_1, \ldots\}$ where $ x_{2k \plus{} 1} \equal{} f^k(x_1)$ and $ f^k(x_{2k}) \equal{} x_1$, $ k \in \mathbb{N}$. Then define the involutions $ g_n,h_n : S_n \to S_n$ as follows: $ g_n(x_1) \equal{} x_1, g_n(x_{2k}) \equal{} x_{2k \plus{} 1}, g_n(x_{2k \plus{} 1}) \equal{} x_{2k}$; and $ h_n(x_1) \equal{} x_3, h_n(x_3) \equal{} x_1, h_n(x_{2k}) \equal{} x_{2k \plus{} 3}, h_n(x_{2k \plus{} 3}) \equal{} x_{2k}$. Verify, much like above, that $ f_n(x_i) \equal{} h_n(g_n(x_i))$.

Then, naturally, we have $ f \equal{} h(g(x))$, where $ g(x) \equal{} g_n(x)$ if $ x \in S_n$ and $ h(x) \equal{} h_n(x)$ if $ x \in S_n$.

Note in essence that the involutions defined are similar to slightly shifted reflections; will post a more informal explanation.
[/hide]


A triangle $ ABC$ has an area of 12. We extend side $ AC$ to obtain a point $ M$ such that $ A$ is between $ M$ and $ C$ and $ AM\equal{}AC$. We extend side $ AB$ to obtain a point $ N$ such that $ B$ is between $ N$ and $ A$ and $ AB\equal{}BN$. We extend the side $ BC$ to obtani a point $ P$ such that $ C$ is between $ P$ and $ B$ and $ BC\equal{}CP$. How many square centimeters are in the area of triangle $ MNP$?

[hide]it's hard to explain without a picture but....
you can see that triangle ABC's area is equal to the area of triangle BCN,ACP, and ABM.
Triangle BMP's area is equal to the area of triangle ACP.
Triangle BMN's area is equal to area of Triangle ABM.
Triangle CPN is equal to the area of Triangle BCN
Area of triangle MNP=Triangle BMP+Triangle ABM+Triangle BMN+Triangle ABM+Triangle CPN+Triangle BCN+Triangle ABC.
This is equal to Triangle ABC's area * 7= 84 Yay!!!!![/hide]

I think you mixed up the points :wink:

[hide="Picture"][img]http://mathcounts.saab.org/mcgif/p215.gif[/img][/hide]

[hide="hint"]Draw lines and make triangles, then use the ratio of bases and stuff.[/hide]

EDIT:WTH, where did I post "similar" from?

[hide="to be more specific"]
draw segments BM, CN, and AP
[/hide]


What is the least possible value of the expression below if $ a$ and $ b$ are positive integers.

$ \frac{9a^2}{b} \plus{} \frac{243}{ab} \plus{} \frac{9b^2}{a}$

[quote="aadil"]by am gm  the answer is $ 3(9^2 *243)^(1/3)$[/quote]

could you show me your work please?

[quote="kurt.math"]What is the least possible value of the expression below if $ a$ and $ b$ are positive integers.

$ \frac {9a^2}{b} \plus{} \frac {243}{ab} \plus{} \frac {9b^2}{a}$[/quote]

$ \frac {9a^2}{b} \plus{} \frac {243}{ab} \plus{} \frac {9b^2}{a} > \equal{} 3 ({\frac {9a^2}{b} * \frac {243}{ab} * \frac {9b^2}{a}})^{\frac {1}{3}}$

=81

The solution is still not complete yet... Remember that $ a,b$ are positive integers. What aadil found is that the expression attains the minimum for some positive [b]real[/b] numbers $ a,b$ when you apply AM-GM. You must also show that 1) the terms are equal; 2) when they are equal, the values of $ a,b$ are positive integers. If you solve for $ a,b$, you get $ a\equal{}b\equal{}3$. Then [u]now[/u] you can claim that the minimum is 81.

hello, the minimum is $ 81$ and will be attained for $ a\equal{}3$ and $ b\equal{}3$.
Sonnhard.

[quote="kurt.math"][quote="10000th User"] the terms are equal[/quote]

Why must the terms be equal?[/quote]

That is the equality condition of AM-GM. Equality occurs (and thus the minimum occurs) when all the terms are the equal.

i,m a little confused.why must we apply only am gm.what about am hm ?

Well, we want the value of that specific sum on the LHS of an inequality($ \geq$). Also, you should notice that in the GM, the a's and b's cancel. Any other way would [b]not[/b] work.


Given 94 bricks, each 4 by 10 by 19, how many different heights of tower can be built assuming each brick adds 4, 10, or 19 to the height.

The equation x^10 +(13x-1)^10 =0 has 5 pairs of complex roots a1,b1, a2,b2, a3,b3,a4,b4,a5,b5.
Each pair ai bi are complex conjugates find E 1/aibi (the E is supposed to be a lame surrogate for the summation thingy).

1) [hide]Look the following 2 equations:

a+b+c=94---(1)

4a+10b+19c=k---(2)

where a, b, c are non-negative integers.

So we want to find number of possible values of k satisfies the equations

(2)-4(1) we get

6b+15c=k-376

3(2b+5c)=k-376.

k-94 must be divisible by 3 so k(mod 3)=1(mod 3)

Also the maximum value of k is 94*19.

so possible values of k are numbers between 94(19) and 376 and 1(mod 3)

there are 471 of these numbers.

but 2b+5c can never be 1 or 3. so k-376 can never be 3 or 9.

so 471-2=469. But I am done yet. there are still numbers 2b+5c can't be when b+c :le: 94. Let b+c=n, we have 2b+5c=5n-3b, n :le: 94.

Since 5(94), 5(93), 5(92) each :equiv: 2, 0, 1(mod3), so 2b+5c can be any number  :le: 5(92). There are 4 numbers, 469, 468, 466, 463, that can't be value of 2b+5c. So there are 469-4=465 numbers[/hide]

2)

[hide]Rearrange the equation to get (1-13x)^10 = -(x)^10, i.e. (1/x-13)^10 = -1. The 10th roots or -1 are easy to find, and they are [tex]\cos\theta+i \sin\theta[/tex] where [tex]\theta [/tex] is equal to [tex]\frac{\pi+2\pi n}{10}[/tex] and n ranges from 0 to 9. We now find that the reciprocals of the roots of the equation can be written in their conjugate pairs as

[tex](13+\cos\theta)\pm i\sin\theta[/tex]

for [tex]\theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}[/tex]. Now, the expression 1/aibi equals

[tex]\left((13+\cos\theta)+i\sin\theta\right)\left((13+\cos\theta)-i\sin\theta\right) = 170+26\cos\theta[/tex], so that the desired sum equals

[tex]850+26\sum\cos\theta[/tex]

where the sum ranges over the 5 values of theta. But since cos(pi/10)+cos(9pi/10) = cos(3pi/10)+cos(7pi/10) = cos(5pi/10)=0, the sum on the right equals 0 so the answer is 850.[/hide]


[quote="larry09"]Let $a,b,c,d$ be positive numbers so that $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove that
\[3(a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2})-2(abc+bcd+cda+dab)\leq 8. \]
[/quote]
Nice, using this one, we can prove the following
\[\frac{1}{3-bcd}+\frac{1}{3-cda}+\frac{1}{3-dab}+\frac{1}{3-abc}\le 2 \]
where $a,b,c,d$ be nonnegative real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4.$ :)


inequalities
function
calculus
derivative
real analysis
real analysis unsolved

Let $ f: \mathbb{R} \to \mathbb{R}$ be a twice differentiable function with the second derivative nonnegative :
 Prove that : $ f(x\plus{}f^{(1)}(x)) \geq f(x)$ , $ \forall x \in \mathbb{R}$

Let $ x$ be such that $ f'(x) >0$ then by the mean value theorem we have $ f(x\plus{}f'(x))\minus{}f(x)\equal{}f'(x)f"(h) >0$ where $ h \in (x,x\plus{}f'(x))$.Similarly if $ f'(x) <0$

[quote="pardesi"]Let $ x$ be such that $ f'(x) > 0$ then by the mean value theorem we have $ f(x \plus{} f'(x)) \minus{} f(x) \equal{} f'(x)f"(h) > 0$ where $ h \in (x,x \plus{} f'(x))$.Similarly if $ f'(x) < 0$[/quote]

Not sure what mean value theorem you have in mind, but the above doesn't look right to me. I would use Taylor's theorem: $ f(x\plus{}h)\minus{}f(x) \equal{} f'(x)h \plus{} f''(c)h^2/2$ for some $ c$ between $ x$ and $ x\plus{}h$. If $ h \equal{} f'(x)$, we have nonnegativity as desired.

[quote="pardesi"][quote="WWW"]
Not sure what mean value theorem you have in mind, but the above doesn't look right to me. I would use Taylor's theorem: $ f(x \plus{} h) \minus{} f(x) \equal{} f'(x)h \plus{} f''(c)h^2/2$ for some $ c$ between $ x$ and $ x \plus{} h$. If $ h \equal{} f'(x)$, we have non-negativity as desired.[/quote]
I used the [url=http://www.artofproblemsolving.com/Wiki/index.php/Lagrange%27s_Mean_Value_Theorem]Lagrange's mean value theorem[/url].Your taylor expansion is incomplete without the remainder term and the second term should have $ c$ not $ x$[/quote]

OK, just the ordinary mean value theorem then. In this case you want $ f'(h)$, not $ f''(h)$, i.e., $ f(x \plus{} f'(x)) \minus{} f(x) \equal{} f'(x)f'(h)$ where $ h \in (x,x \plus{} f'(x))$.

As for my Taylor expansion, it is just as I said. Review the Lagrange form of the remainder.


Let x, y be all reals but x ≠ ±y, prove that

$ \left | \frac {x \plus{} y}{x \minus{} y} \right |^{xy} \geq 1$

[quote="whomer"]Let x, y be all reals but x ≠ y, prove that

$ \left | \frac {x \plus{} y}{x \minus{} y} \right |^{xy} \geq 1$[/quote]

$ x\equal{}1,y\equal{}\minus{}1$ if x+y≠0 Than we can asume $ x>y$ and $ x\plus{}y>0$ (so $ x\ge 0$)so
$ \frac {x \plus{} y}{x \minus{} y}\ge 1$
$ x\plus{}y\ge x\minus{}y$
$ 2y\ge 0$

if $ y\ge 0$  then $ \frac {x \plus{} y}{x \minus{} y}\ge 1$ and $ xy\ge 0$
if $ y\le 0$ then $ \frac {x \plus{} y}{x \minus{} y}\le 1$ and $ xy\le 0$ and we are done :)


number theory
least common multiple
geometry
inequalities
calculus
circumcircle
trigonometry

I think I got 5/5/5/4/3.

My explanation for #4 part b. was a bit unclear.
I didn't completely solve the problem in number 5.

What about problem difficulty? Post your opinions here.

They are...

it's wednesday

I think 5/5/5/0/5

I don't see solutions up on the website...

Well the solutions should be up sometime today, so I'd just wait it out until this evening maybe.

The solutions are up.

For #4, I didn't realize that 2 mod 4 cannot be a square and instead argued the denominator (LCM) is at most divisible by 2. For #5, I used a different approach: first use area to prove $ PQ \equal{} BC \cdot \frac{BQ\cdot CQ}{BQ\cdot AB \plus{} CQ\cdot AC}$, then use Cauchy to prove that is at most $ \frac{125}{49}$ (unfortunately I typoed some inequality signs in the middle, hopefully it's easily recognizable.) Just for fun, I calculated the absolute maximum to $ \frac{25}{\sqrt{(\sqrt[3]{9}\plus{}\sqrt[3]{16})^3}}$.

For 5, I used power of a point and Stewart's Theorem, and then a little (gasp) calculus... I couldn't think of another way to do it.
Its okay to use calculus, right? :/

I think 5/5/5/0/5 for me, which coincidentally is the same distribution as last time  :P
I'd say that #5 was the hardest and #3 was the easiest.  I spent all my free time on Sunday working #5, so I didn't get a chance to even try #4  :(

You can use calculus as long as the proof is solid

Hm I think I did ok. 1,2,3 were same as intended solutions.

#4 i overkilled a bit; I showed that a frog could achieve a move of 1/c for any c, but the concept was still there.

#5: lol. I put A=(0,0), B=(0,3), C=(4,0) and bashed with line AP being y=kx. Then I get the length of PQ is $ \frac{25k}{(4k\plus{}3)\sqrt{k^2\plus{}1}}$ and then a simple application of the product rule for derivative shows its maximixed when k=cbrt(.75), which is less than 25/4rt6.

5/5/5/4/5

First 3 are fine.  Number 4 I wrote a completely different proof than the one in the solutions but I think it's right...  Probably a 4ish since my explanation might not have been that good.
I shamelessly bashed #5 with calculus :) I think the proof is solid though

#1: I thought this was pretty easy. I did it in about 20 minutes.
#2: Probably easier than #1.
#3: This one was not easy. I spent around 4-5 hours working on it.
#4: This one was hard. I didn't really have a clear solution for part b. I think I lost a point (maybe two.)
#5: Wow. Nice solution! I didn't think of that. I used the chord power theorem (power of a point) and I checked from 2.5 to 3.2 what PQ would be. I checked every 0.000001 interval using a TB (True Basic) program. Then, I said that the value of the function would not "go crazy" and jump really fast in an interval of 0.000001, except when the denominator of my function was around 0, which it couldn't have been close to. I think I lost 2 points on this one.

Number 3 I wasted a lot of time factoring the $ a^2 \plus{} b^2 \equal{} k * 7^{2009}$ into something like $ 7^2(a \plus{} b)^2 \minus{} k * 7^{2009} \equal{} 2 * a * b$ and trying to prove the left side was divisible by $ 7^{2010}$... meh

I'm hoping for 5/5/5/5/5.  1, 2, and 4 weren't hard, and 3 I had seen before with 7 replaced by 31 and 2009 replaced by 1995. 5 took me a long time though, and my solution was a lot uglier than the official 1.  (I think they were equivalent, but in the end I law-of-sines bashed to turn it into a coordinate problem, which then turned it into an AM-GM problem.)

What's wrong with this method for #5?
By power of a point, AP * PQ = CP * PB
Say the length of CP is x, then the length of PB is 5-x (CP + PB is hypotenuse of length 5)
So PQ = (x)(5-x) / AP
and PQ <= maximum value of (x)(5-x)/AP
to maximize this, we find the maximum numerator value (2.5^2) and minimum denominator value (3 (also length of AB) since AP must lie on BC)
Hence, PQ <= (2.5)^2/3 = 25/12 < 25 / 4*root6

Seems right but seems too easy...let me know when/if you find my error.

(Please excuse my lack of Latex knowledge)

The minimum value of AP isn't 3, it's 12/5 (when AP is perpendicular to BC).

[quote="tenniskidperson3"][url=http://usamts.org/Papers/Year21/Round2/9877-1257293445-solutions.pdf]Here's mine.[/url]

I'm not so sure about number 5, but it is a solid proof with no calculus.  I would've used calc, but then I realized that there are always noncalc solutions.  All the others were "trivial", I might say.[/quote]

You could've taken your polar equation for the length of PQ and taken the reciprocal, and then just AM-GM. haha

Dang, you're right.  That would have eliminated a full page.  The thing is, I tried going down that alley, but i didn't get any immediate results so I stopped.

hmmm... I did not calc bash at all...

hopefully it's right...  :maybe:

[quote="123456789"]Do you think me saying "WLOG" instead of saying I used directed segments will merit points off?[/quote]

To me, WLOG implies that you are assuming the case where $ BP<PC$ is the same as the case where $ BP>PC$. This is my first year of USAMTS, so I'm really not sure how strict the grading is, but my guess is that you might get 1-3 points off.

[quote="MathWise"][quote="123456789"]Do you think me saying "WLOG" instead of saying I used directed segments will merit points off?[/quote]

To me, WLOG implies that you are assuming the case where $ BP < PC$ is the same as the case where $ BP > PC$. This is my first year of USAMTS, so I'm really not sure how strict the grading is, but my guess is that you might get 1-3 points off.[/quote]

Darn there goes perfect year.

[quote="winternights1"]I think 5/5/5/0/5 for me, which coincidentally is the same distribution as last time  :P
I'd say that #5 was the hardest and #3 was the easiest.  I spent all my free time on Sunday working #5, so I didn't get a chance to even try #4  :( [/quote]

That's unfortunate, for this round you could have at least done the first half of #4 which is simple arithmetic and scraped 1 or 2 points. Last round's #4 was ridiculous...

Here are my solutions.

As you can see, I fail at explaining. How many times did I screw up?   :lol:

I thought I did really well until I saw the solutions. Oh dear, here's my guesses...

1: 4 - my proof strategy was intuitive to me, but writing it down proved to be a monster, and I probably made mistakes. Turns out the actual solution is much shorter.
2: 3 - i can't read, but correct concept (long story)
3: 5 - hope this one is okay
4: 5 - yep although I overdid the first one
5: 4 - graphing calculator anyone?

Final: 21. Okay that isn't bad at all. Something just tells me that my solutions are not up to par with such a high score but we'll see.

Hopefully I didn't lose points on #5. Not worried about anything else.

#5 = calculus bash + Mathematica/WolframAlpha = win. (they should make non-bashable problems)


geometry
3D geometry
geometric transformation
rotation

I think the answer is too obvious but in case it's bit hard, I post here.

Each face of a cube is colored either red or blue.  How many distinct colorings of the cube are possible independent of orientation? (For example, there is only one way to color the cube such that one face is red and the remaining faces are blue).

[hide="Answer"]

This was #34 on MMPC.. I get 10..

No red face ad 9 blue
1 red face..
2 red face...
.... all red face and no blue.. That's 10...[/hide]

Is this right? ;)

[hide]There is 1 cube with no faces red.
There is 1 cube with one face red.
There are 2 cubes with two faces red (opposite or adjacent).
There are 2 cubes with three faces red (chain or all adjacent).
If four faces are red, then two faces are blue, and there are 2.
If five faces are red, then one face is blue, and there is 1.
If six faces are red, then no faces are blue, and there is 1.

Therefore, the answer is 1+1+2+2+2+1+1=10.[/hide]

I think the answer is 8.  I did it by pure enumeration.


For my blog, I am labeling a triangle's vertices and it's side lengths with the variables $ A,B,C,a,b,c,a \minus{} d,b \minus{} e,d \minus{} f,d,e,f,h_1,h_2,h_3$. However, I can only use the three of them that are not single digit. If I put in a single letter in the 'Insert Text' box, the result is a random decimal number? I also need to know how to place a subscript such as $ h_1,h_2,h_3$. Asymptote is too hard for me.

Any help will be appreciated.

I'm not really sure what picture you want, so at this point I can only say that a-b is an expression (the difference of a and b), not a valid variable name. Thus, unless you put it in a text box as a string, it won't be allowed as a label. As to $ h_1$, just type h_1 as you would do in LaTeX. This is an attempt to guess what you had in mind (but it is just a wild guess): [geogebra]b82747f87ee6a5df66abe68ca67bca5df83d88f3[/geogebra]
(I left the algebra window, so you can see most of the variables except the text boxes).


We should be able to change our screen name.  See I want my screen name to be JoBro but I used jobro_gold because that was my Yahoo.com name and I was really into Golden Sun.

Press the print-screen key to capture a screenshot. (Alt-print-screen to capture a shot of the currently active window.)

I then loaded it into photoshop, circled the row, cropped it, and saved it as a GIF.

If you want to use paint,

1. open paint
2. edit -> paste; this will copy the screenshot into the paint program
3. do what you want, and
4. save. (preferably as a GIF.)

[quote="Osiris"]Press the print-screen key to capture a screenshot. (Alt-print-screen to capture a shot of the currently active window.)

[/quote]

Am I just being incredibly stupid?...

I can't seem to find this "print-screen" option...unless...do u mean "print preview"?...i doubt it...

[quote="zscool"]it should be a key on your keyboard[/quote]

ah...how stupid of me...I was searching thru the menus and stuff...

however, the key doesn't work!...argh...

[quote="Osiris"]Can't you change it here? (This is the PROFILE page)[/quote]

I don't think you can.  I tried it as an experiment, and it didn't work.  But that was several months ago...

[quote="TripleM"]After pushing the key you then have to go and paste it somewhere :) (ctrl+v)[/quote]

ah...yes...that would be the problem! I was just kinda expecting my printer to jump to life there...lol...thanks. (you can tell i've never used this button before)

For better control. Like specific command buttons or like a specific area. I have Fullshot. My schools have Snagit which does the same thing.

It doesn't work. (something to do with phpBB or something)

[quote="h_s_potter2002"]Is it posible to get a screenshot using a mac, becuase I can't find it anywhere on my mac computer.[/quote]
Click the link that says "Profile" right next to "Private Messages" up at the top. It's a little blue box.

[quote="confuted"][quote="h_s_potter2002"]Is it posible to get a [b]screenshot[/b] using a mac, becuase I can't find it anywhere on my mac computer.[/quote]
Click the link that says [b]"Profile"[/b] right next to "Private Messages" up at the top. It's a little blue box.[/quote]

Hilarious :)

the button could be named:  PrtSc or something

[quote="h_s_potter2002"]Is it posible to get a screenshot using a mac, becuase I can't find it anywhere on my mac's keyboard.[/quote]

There is no PrintScreen button on a mac, but there are tons of ways to take screenshots.

1.  A million different shortcut keys.

Capture screen to file - Command (apple key) - Shift -3
Capture screen to Clipboard - Command - Control - Shift - 3
Capture Selection to File - Command - Shift - 4
Capture Selection to Clipboard - Command - Control - Shift-4

If you ever forget, go to system preferences -> Keyboard & Mouse -> Keyboard Shortcuts

2.  Go to Finder, click "Applications", go to the folder "Utilities", and open Grab.  This program operates on menus and should be easy to figure out.


If $ a,b,c$ are positive Real numbers then
$ \frac {1}{a^{3} \plus{} b^{3} \plus{} abc } \plus{} \frac {1}{b^{3} \plus{} c^{3} \plus{} abc} \plus{} \frac {1}{c^{3} \plus{} a^{3} \plus{} abc} \leq \frac {1}{abc}$

[hide]
$ x^3 \plus{} y^3 \ge xy(x \plus{} y)$

Therefore

$ \sum \frac {1}{a^3 \plus{} b^3 \plus{} abc} \leq \sum \frac {1}{ab(a \plus{} b \plus{} c)} \equal{}  \sum \frac {c}{abc(a \plus{} b \plus{} c)} \equal{} \frac {1}{abc}$
[/hide]


Does anybody know the qualifying scores to get to GPML regionals?  Is it half right just like the state qualifying scores?

Please respond ASAP.  State is in two days!  :)

I think it's placing, or having your team place, or both.  There might be some other way, too, though.  I don't really remember...  I assume (since you say you are from Iowa) that I will see you at West tomorrow!   :D

Thanks for the info.  I qualified!  Did you?

Yup, I think 3 times...  Sprint + Target + Sweepstakes :D


The mapping $ f: \mathbb N \to \mathbb N$ is defined as ''To move its top digit into one's digit'', for instance, $ f(12345)\equal{}23451, f(901234)\equal{}12349$.

Determine the least $ n$ such that $ f(n)\equal{}\frac{8}{5}n$.

Let $ n\equal{}A_1A_2...A_m$. Where $ A_i$ are digits.

Then:

$ A_2A_3...A_m A_1\equal{} \frac {8}{5}A_1A_2...A_m$

$ 5.A_2A_3...A_m A_1\equal{}8.A_1A_2...A_m$

$ 42.A_2A_3...A_m\equal{}(8.10^{m\minus{}1}\minus{}5)A_1$

Then $ A_1$ is even, and because that is not a multiple of 7.
Then $ (8.10^{m\minus{}1}\minus{}5)$ is a multiple of 7.

Then the minimum number $ m$ possible is 6.

So $ 42.A_2A_3...A_6\equal{}799995.A_1$
$ \implies 2.A_2A_3...A_6\equal{}38095.A_1$

Like $ A_1$ is not 0, the minimum of $ A_1$ is 2. Then The minimum $ n$ is:
$ n\equal{}A_1A_2...A_6\equal{}238095$


$ a,b,c > \equal{} 0$,$ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} > \equal{} \frac {5}{2}$

Can anyone post some inequalities similar to this problem? The solution is not "a=b=c".
And please post the [color=red]solution to your problem[/color], thank you!

Canada 1999 -
If $ x, y, z \geq 0$ and $ x \plus{} y \plus{} z \equal{} 1$
Prove: $ x^{2}y \plus{} y^{2}z \plus{} z^{2}x \leq \frac {4}{27}$
What is solution to your problem?

to adjust
$ x>\equal{}y>\equal{}z$,$ u\equal{}\frac{x\plus{}y}{2}>\equal{}z$
so $ u^{2}>\equal{}\frac{1}{3}$
When $ u>\equal{}1$,$ f>\equal{}\frac{1}{x\plus{}y}\plus{}x\plus{}y\equal{}2u\plus{}\frac{1}{2u}>\equal{}\frac{5}{2}$
When $ u<1$,$ z\equal{}\frac{1\minus{}xy}{x\plus{}y}>\equal{}\frac{1\minus{}u^{2}}{2u}\equal{}z'>0$
we have to prove $ f(x,y,z)>\equal{}f(u,u,z')>\equal{}\frac{5}{2}$
What is solution to your problem, borislav_mirchev?

[quote="zaya_yc"]$ x,y,x\geq 0$ and x+y+z=1 . n , m different; prove:
$ x^{n}y^{m} \plus{} y^{n}z^{m} \plus{} z^{n}x^{m}\leq\frac {n^{n}m^{m}}{(n \plus{} m)^{n \plus{} m}}$[/quote].
This is also the generalization of the well known canada problem.

Here are written the solutions.
http://www.math.ca/Competitions/CMO/
Your problem is very interesting.

P.S. I've solutions to all problems of CMO - since founding to now.

As the answer to Twin's question, there are even a lot of symmetric  and beautiful inequalities that equalities hold for $ a\not \equal{} b\not \equal{} c$. I called them "unusual inequalities". For above problem, the equality still holds for $ a \equal{} b \equal{} 1$. For the second above problem, the inequality is not symmetric. I mean here is that symmetric inequalities that equalities hold for a,b,c are totally different numbers.

For instance, here is an simple example
\[ \frac {a^2}{b^2 \plus{} c^2} \plus{} \frac {b^2}{c^2 \plus{} a^2} \plus{} \frac {c^2}{a^2 \plus{} b^2} \plus{} \frac {27(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \ge 52.
\]
Serects In Inequalities dedicates one section (about 12-20 pages) to discuss about this type of inequalities (reading it, you can realize that creating this kind of inequalities is very natural).

[quote="Twins"]$ a,b,c > \equal{} 0$,$ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} > \equal{} \frac {5}{2}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31377
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476

What about this problem?
$ x,y,z>\equal{}0$,$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}1$,prove that $ f(x,y,z)\equal{}\frac{x}{1\plus{}yz}\plus{}\frac{y}{1\plus{}zx}\plus{}\frac{z}{1\plus{}xy}<\equal{}\sqrt{2}$
Can anyone teach me how to ajust? I always don't know $ f(x,y,z)<\equal{}f(0,?,?)$

Some beautiful Inequalities
http://www.mathlinks.ro/Forum/viewtopic.php?t=144612
http://www.mathlinks.ro/Forum/viewtopic.php?t=144697
 :wink:


Megan has already saved $ \$18$, and she earns $ \$4$ per hour working four hours on each Saturday and three hours on each Sunday. Rhoni has already saved $ \$22$, and she earns $ \$3$ per hour and works six hours each week. If they combine their money, in how many weeks will they be able to afford a $ \$316$ stereo?

Would this equation work for this?

50 + 46x = 316

I got that by finding how much the two people make all together in one week.

I'm not sure where you got the numbers.

The equation would look like this:

$ 40 \plus{} (4 \cdot 4 \plus{} 4 \cdot 3 \plus{} 3 \cdot 6)x \equal{} 316$

where $ x$ is the amount of weeks it will take for them to afford the stereo. Solving yields $ \boxed{6}$.

If you don't understand how I got the equation:

Rhondi and Megan already saved up 18+22=40 dollars. Together, they earn $ 4 \cdot 4 \plus{} 4 \cdot 3 \plus{} 3 \cdot 6$ dollars each week.

Aylsero your equation is correct except the 50 at the beginning should be 40.


i guess
im searching for [some introductory or just general]  robotics camps out there, preferably residential program that's longer than 2-3 weekks during summer.

if anyone knows any robotics camps out there, could you please post here? :]

You will love to see [url=http://www.searchfreak.blogspot.com/2008/12/robo-cup-soccer-challenge-in-india-at.html]this[/url]. Sadly, two other robotics camps at the IITs ended recently. One more [url=http://robocup.in/]link (official)[/url].

[quote="sooozyy"]oh wow robocup
:P
wowww
thats all i can say[/quote]

What does that mean?  :huh:

Maybe you feel that is stupid, but the competetion is tough. Probably you are underestimating the IITs. :wink:

hahah noo
im not underestimating at all
i was just surprised that there is a thing called robocup out there
i was just surprised. simply that :]

[quote="sooozyy"]i was just surprised that there is a thing called robocup out there [/quote]

Yeah..the name sounds stupid but it is really exciting.

If you are still looking for a great Robotics summer camp, you should definitely check out iD Tech Camps ([url]http://www.internalDrive.com[/url]). Students who enjoy robotics have a great time at iD Tech Camps- America's #1 Tech Camp.  iD Tech Camps offers weeklong, day and overnight, technology summer programs for ages 7-17 at over 60 prestigious universities nationwide including UC Irvine, UCLA, Stanford University, MIT, UT Austin, Columbia, Emory University, Northwestern and many  more.  During the week, students build robots, learn programming in C++ or Java, create video games, design websites with Flash, produce digital movies and more.

They have a great RoboContenders Course for beginner through advanced students.  Here is some info on the course, you can find out more by going to their website, [url]http://www.internalDrive.com[/url] or by calling 1-888-709-TECH (8324) option 1.

[b]Course Description[/b]:
Working in small teams, students in this summer robotics camp will use the VEX® Robotics Design System to compete in fun real-world challenges like robo-soccer, obstacle courses, capture-the-flag, tag, sumo wrestling, maze solving and much more! Each day will bring a new challenge, and your team's robot will have to morph to better adapt and compete. Advanced students will have the opportunity to learn about building complex robotic sensors, advanced robot arms, drive trains and programming "smarter" robots.

This season we have new kits and a more advanced parts inventory, including treads, sensors and accessories to help you take your skills to the next level. Robotics is the main portion of this summer camp, but each student will be given a brief introduction to the Easy C™ programming language. Finally, students will take home and upload video of the week's battles, competitions and collaborative teambuilding events. Welcome to iD! Click to the left for more details, and scroll down for promotions and multi-week packages.

Visit http://www.internalDrive.com or call 1-888-709-TECH (8324) for details.

Adding to the question, are there any good robotic camps(introductory) in California over the summer that any of you know of?

-rts2007

I would say I WENT TO ONE THIS SUMMER! but I live in Alaska.


A circle is inscribed in a triangle with sides 6, 8, and 10. What is the area of the region of the triangle [b]not[/b] occupied by the circle?

Here's how I did it:

[hide]You know that this triangle is a right triangle (3,4,5) and so the area of this triangle is (6x8)/2=24. Then to find the radius of a circle inscribed in it, you use the formula rs, so rs=24 or rx12=24, so r=2, then you know that the area of the circle is 4pi, thus the area of the region not occupied by the circle is [b]24-4pi[/b]. [/hide]

Anybody get any other solutions or ways how they would do it?

[quote="tarquin"]Uh find the inradius and subtract the area of the circle from that of the triangle.[/quote]

What was your answer then?  :?:

Applying $\frac {a + b - c}{2} = r$ we have $r = 2$ so the answer is $24 - 4 \pi$

wow. sure would have helped it i remember that $rs=[ABC]$
anywho, there's my solution:
[hide="warning very ugly and weird"]by extending the radii to the sides, you create 3 pairs of equal sides. by messing around with the numbers i found the only way that works. 6 is broken down into 2+4, 8 is broken down into 2+6 and 10 is broken down into 4+6. the radius is thus 2 and the area desired is $24-4\pi$[/hide]
btw, what do you use to create pictures. a pic would make my solution a lot clearer.


calculus
integration
trigonometry
complex analysis
calculus computations

Evaluate the following
$ \int_{ \minus{} \pi}^{\pi}\frac {cosx}{1 \minus{} a\cdot cosx}\,dx$
(without using complex analysis)
($ a\leq 1$)
is there an easy way?

$ I \equal{} \int_{\minus{} \pi}^{\pi} \frac{\cos x}{1 \minus{} a \cos x}\,dx \equal{} 2 \int_0^{\pi} \frac{\cos x}{1 \minus{} a \cos x}\,dx \equal{} \frac{2}{\minus{}a} \int_0^{\pi} \frac{1 \minus{} a \cos x \minus{} 1}{1 \minus{} a \cos x}\,dx$

$ \equal{} \frac{2}{\minus{}a} \left(\pi \minus{} \int_0^{\pi}\frac{1}{1 \minus{} a \cos x}\,dx\right)$.

But

$ \int_0^{\pi}\frac{1}{1 \minus{} a \cos x}\,dx \equal{} 2 \int_0^{\infty} \frac{1}{t^2 \plus{} 1 \minus{}a(1\minus{}t^2)}\,dt \equal{} 2 \int_0^{\infty} \frac{1}{(1\plus{}a)t^2 \plus{} 1\minus{}a}\,dt$

$ \equal{} \frac{2}{\sqrt{1\minus{}a^2}} \int_0^{\infty} \frac{\sqrt{\frac{1\plus{}a}{1\minus{}a}}}{1 \plus{} \left(t \sqrt{\frac{1\plus{}a}{1\minus{}a}}\right)^2}\,dt \equal{} \frac{\pi}{\sqrt{1\minus{}a^2}}$, for |a| < 1.

Therefore,

$ I \equal{} \frac{2}{\minus{}a} \left(\pi \minus{} \frac{\pi}{\sqrt{1\minus{}a^2}}\right) \equal{} \frac{2 \pi}{a} \left(\frac{1}{\sqrt{1\minus{}a^2}} \minus{}1\right)$.

If a = 1, its easy to see that the integral diverges.


For any $ a,b,c \ge 0$
\[ \sum \sqrt{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \ge \sqrt{\frac{2(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}}\plus{}\frac{1}{\sqrt{2}}
\]
:)

[quote="manlio"]A hint, please!!! :wink:[/quote]
Just squaring, my friend. :)

Another hint, please. After squaring what to do?

Thank you very much.

[quote="can_hang2007"]For any $ a,b,c \ge 0$
\[ \sum \sqrt {\frac {a^2 \plus{} bc}{b^2 \plus{} c^2}} \ge \sqrt {\frac {2(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}} \plus{} \frac {1}{\sqrt {2}}
\]
:)[/quote]
My solution. :)

Dear can_hang2007,

thank you very much for your solution. It is really far out !!!!

Congratulations  :)


Foreign nationals have always contributed significantly to the US economy; however, fewer foreign nationals are choosing to come.

Do you want a rating of immigrant friendly places for bright foreign nationals that intend to pursue a career in mathematics? I'll put the USA as #1.

Yea. USA #1.
France, Britain, or Germany might be #2.

But, definitely USA.

MIT is 35% Asian students, good for any school.

Any USA school with the word Technology in its full name and is in the Northeast or West Coast usually has good diversity.

I think there are more immigrants than ever coming to the United States of America, just not all of them coming through channels that involve issuing visas.

[quote="mdk"][quote="tongchen1226"]USA is definitely #1. I do not know of another country that comes anywhere close to the eclectic nature of American population. Some people in Europe are quite anti-immigrant, whereas most people in America have grown used to seeing Asians on the streets. My friend had lived in Belgium for 17 years when he was asked one day, "when are you going back to China?"  :huh:
Yeah, I agree, visa is a pain for most scholars who want to come. I've personally seen a line half a mile long in front of the American embassy in Beijing.[/quote]

those racist Europeans.[/quote]

Asia is smarter that the US

maybe so, or maybe not, but that was rather random.

Clearly this subject will soon lead to some flaming, therefore it's being closed.


[quote="solafidefarms"]Or at least a way to notify you of the new topics since your last visit by email/RSS.

You can do it for the blogs; why not for the forums?

Thank you
Billy[/quote]There is a watch forum and a watch topics option already ...

[quote="solafidefarms"]Or at least a way to notify you of the new topics since your last visit by email/RSS.

You can do it for the blogs; why not for the forums?

Thank you
Billy[/quote]

If you went beyond what Val suggested (using the "Watch topic" option), you would have to get an email for every new thread since the machines can't know the next time you [b]will[/b] log in (at least, not for a few decades they can't).

There are 40,000+ threads on AoPS.  That would be a lot of email.

[quote="MCrawford"][quote="solafidefarms"]There are 40,000+ threads on AoPS.  That would be a lot of email.[/quote][/quote]Times the potentially 10,000+ members that would use this, you get a very big number :)

An RSS feed for anouncements and happenings would be nice though...


I did some search but can't find it.  Where can I find the past years national competition problems for mathcounts?
Thanks in advanced!

For MathCounts Nationals tests and State/Chapters past a certain year, you have to  buy them. If you want to try and get some for free, either do a search on this forum or see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189]here.[/url]


geometry
ratio
calculus
integration
analytic geometry
geometry unsolved

Let $ a$ be the side of the square. Find the area of the geometric sector painted in gray and, eventually, show that it corresponds to aproximately $ \frac {17}{100}$ of the square's area.

[url=http://img22.imageshack.us/i/findthegrayarea.jpg/][img]http://img22.imageshack.us/img22/8640/findthegrayarea.jpg[/img][/url]

I get the ratio to be $ (1/4)\arccos(\sqrt{2}/4) \minus{} \arccos(5\sqrt{2}/8) \plus{} \sqrt{7}/8$, approximately $ 0.14638$, or $ \frac {15} {100}$.

I will be glad if you show the method. Thank you in advance!

This is a simple one on integration ..

 take the x and y axis along the diagonals of the square

take the centre of the system as the centre of the square ..

now the coordinates of the centre of the orignial circle come out to be (-a/(2)^.5,0)

write down both equations for the two circles and find their point of intersection .. (comes out to be (5(2)^.5a/8,something not required)
now just find the area of the region which is made by the original circle and the x axis from x= 5(2)^.5a/8 to x=a-(a/(2)^.5)

further the area of region formed by the smaller circle with the x axis will be just the integration of ydx from x= 5(2)^.5a/8 to x=a/2 ..

subtract them .. that will be the answer..


Find all odd natural numbers like $\mathcal Y$ that can not be written as
$|2^{\mathcal J}-3^{\mathcal E}|$ where $\mathcal J,\mathcal E$ are natural numbers.

Well,I think this one is much more easier:
Find all natural numbers  like $\mathcal G,\mathcal T$ such that
$|3^{\mathcal G}-2^{\mathcal T}|=1$

I'm beginning to suspect that the answer to the first one is $\mathcal{Y}=6k+3$ for $k\in\mathbf{N}$.  I'm not at all sure how to give a complete proof though...

[quote="Diarmuid"]The second one is just another instance of Catalan's conjecture.[/quote]

Sure, but is far more easy than Catalan's conjecture, and already posted and solved on ML.

Pierre.


Triangle $\Delta GRB$  is dissected into $25$ small triangles as shown. All vertices of these triangles are painted in three colors so that the following conditions are satisfied: Vertex $G$ is painted in green, vertex $R$ in red, and $B$ in blue; Each vertex on side $GR$ is either green or red, each vertex on $RB$ is either red or blue, and each vertex on $GB$ is either green or blue. The vertices inside the big triangle are arbitrarily colored.

Prove that, regardless of the way of coloring, at least one of the $25$ small triangles has vertices of three different colors.

I edited  :oops: As you knowed, I am a joke man :P

one can prove that on each side of the big triangle,we have odd number of $2-colored$ segments...
now we will give the problem a graph as follow...
for each one of the $25$ triangles put a vertex in $G$ and also another vertex representing the outside of the big triangle name it $o$...
and also name the small triangles which adjacent to the outside $A$ and the rest of them $B$,now we will represent the edges of the graph:
for the $25$ small triangles,if the segment between two of them was $2-colored$ then put an edge between them otherwise no...
and for the triangles which are adjacent to the outside,if the segment which seperates it from the outside was $2-colored$ put an edge between it and the vertex from outside...
ok now we know that the number of edges between the vertex of $A$ and vertex $o$ is an odd number (its the sum of three odd numbers so its odd) and in the other hand we know that the sum of the degress of a graph is even,so the number of odd vertex must be even,so there exists an odd vertex in $B$,so the degree of this vertex must be $1$ or $3$,but it cant be $1$ (check it your self,its easy to see) so it must be $3$ so this triangle has $3$ $2-colored$ segments,so it has $3$ different colored vertices...

[quote="BaBaK Ghalebi"][quote="test20"]This problem is a very special case of the so-called Sperner lemma.
(Which is at the heart of modern proofs of Brouwer's fixed point theorem.)[/quote]
which you can find in page $148$ of the booke: [b]Martin Aigner-Gunter M.ziegler:Proofs from THE BOOK[/b] third edition [b]Springer[/b][/quote]
Yeah.
Or in Shashkin's book on fixed points.
Or in hundreds of other books that discuss Brouwer's fixed point theorem.
Or in West's book on graph theory.

[quote="BaBaK Ghalebi"]
 and in the other hand we know that the sum of the degress of a graph is even,so the number of odd vertex must be even,so there exists an odd vertex in $ B$[/quote]

Well, I think this is wrong. Even if there doesn't exist any odd vertex in $ B$, the sum of degrees is even anyway, just that there is an odd number of edges...  :maybe:

I agree, that the solution is not quite right

I think I have found a generalized solution. Someone please tell me how to attach an attachment. I clicked "Add an attachment" and put the msWord document. But when I try to see preview it says "The Extention docx is not allowed" :mad:

I downloaded the file, but my computer can't read it  :(

Can you post the solution in here?

Well, Akashnil, can you put the solution in here?
Because I cannot open the file.

Thank you

I am sorry for late reply. I overlooked your previous post. And I dont blame you... I cant open it now either :D.

[hide="Here it goes"]
Assume there is no small triangle of three colors.

Name each small triangle $ GR,RB,BG$ if it has only {Green and red},{Red and Blue},{Blue and Green} respectively.and name a all-green triangle $ GR$, a all-red triangle $ RB$ and a all-blue triangle $ BG$. Two adjacent triangles of different names must have their common side of their common color(It is understandable).

Now divide the big picture into some regions such that each region has triangles of same names. The boundaries of the regions must be monochromatic. And two boundaries can not intersect each other. Because three vertices of the graph have fixed color there is one boundary separating one fixed vertex(i.e. $ P,Z,C$ in the pic) from the others. It will intersect two of $ PC,CZ,ZP$. Say the boundary is red. Let $ P$ the Red Point. It will intersect $ PZ$ and $ PC$. Now cut away one side of the boundary where $ P$ is. Now Name Another point $ P$ on the boundary. It is a fixed Red point. We now have a similar picture where the same arguments can be applied. Again we reduce the picture and continue until we reach a small triangle whose vertices are of different colors.[/hide]

Thanks, Akashnil, but now I have a question  :D

Please clearly define your "boundary".

What do you mean by "Because three vertices of the graph have fixed color there is one boundary separating one fixed vertex(i.e. $ P,Z,C$ in the pic) from the others."

[quote="Jumbler"]Please clearly define your "boundary".[/quote]
I hope  the picture is sufficient(If you can see it this time :D ). It suffices all conditions except That all the three sides of the big triangle have all the three colors. The segments joining two points of the boundary are given in bold. Note that all points on them are of the same color. And if you made a picture you couldn't avoid it because each small segment of the boundaries are owned by two small triangles of different names and as proved earlier their two common vertices would have same color, the triangles' common color.

If you still want a definition of a boundary:
Given two adjacent regions their boundary is the intersection of the points of the two regions.

[Definition of region: Collection of points such that you can travel between any two points of it through sides of small triangles of same names. And a region must be non-extendable.]
[quote="Jumbler"]What do you mean by "Because three vertices of the graph have fixed color there is one boundary separating one fixed vertex(i.e. $ P,Z,C$ in the pic) from the others."[/quote]
A single region can not contain all of $ P,Z,C$(a region contains points of only two colors). Thus there exists such a boundary

I think there is a little problem in the last part of the proof. We can not reduce the picture into a small triangle. But it can be like this-- We can not reduce the picture continuously. Thus our assumption that "there is no small triangle of three colors" was wrong.

P.S:
Please excuse me... I am not very good at defining and English isn't my mother tongue and i am only 14 years old.
Tell me if you can see the picture or not, understand the proof or not.

No worries, Akashnil. I really appreciate your effort. I'm trying to understand more now.

But, can you elaborate how you "reduce" the picture, because I don't understand this part. Maybe a picture will help  :)

[quote="massnet"]No worries, Akashnil. I really appreciate your effort. I'm trying to understand more now.[/quote]
Thanks.
[quote="massnet"]But, can you elaborate how you "reduce" the picture, because I don't understand this part. Maybe a picture will help  :)[/quote]
Now we have a boundary that separates one of $ P,Z,C$ from others. It will intersect two of $ PZ,ZC,CP$. Say the boundary is green. then it will intersect those two of $ PZ,ZC,CP$ where green is allowed. Say green is allowed in $ PZ,ZC$. Thus $ P$ is green and it is in one side of the boundary. Now cut away that portion of the picture(From Construction1-1 to Construction1-2). Then name a point on the boundary $ P$.

We will have another picture where same reasons hold good. But we cannot cut the picture infinitely many times.(Actually two boundaries will touch each at some point which is impossible).Thus we get contradiction. Actually I have assumed the picture quite large.

No one understood me?

PS. I am sorry to resume so old a topic. But I am unsure whether my solution is correct or not.


algebra
polynomial
superior algebra
superior algebra theorems

[b]Hello to all who can help me![/b]

We know that root of 2 in [size=150][b]R[/b][/size] [size=100]is algebraic of degree 2 over [/size][b]Q[/b] and of degree 1 over  [size=150][b]R[/b][/size].
Can vi tell that "3/2" is algebraic over [b][size=150]Z [/size][/b](integers)?
I know that (Z,+,*) is a ring, but not a field,...
Can someone explain this to me very-very simple? :blush:  :o

Thanks
Tanja

To your 2nd question: With the exception of 1 and -1, elements of Z do not have multiplicative inverses in Z.

I guess they just extrapolate the definition in the intuitive way.

Well, my guess would be that algebraic over $\mathbb{Z}$ means that there exists an $f(x)\in \mathbb{Z}[x]$ such that $f(3/2)=0$.

It is relatively easy to show something is algebraic since you just have to find the polynomial that it satisfies, transcendental (not algebraic) on the other hand is quite difficult since you have to show that no polynomial at all exists that your number in question is a zero of.

Of course all of this is in terms of fields, and since $\mathbb{Z}$ is only a ring, I don't think any of that applies (unless there is an equivalent definition of algebraic in rings).

Algebraic over a com. ring with unit often means that there is a monic polynomial having that element as a root.


AMC
USA(J)MO
USAMO
calculus
AwesomeMath
summer program

does anyone know about the book, [u]Problems from the Book[/u]?

i know it is very challenging but in terms of usefulness,material and difficulty, how does it fit in with other books?

also, what level is it compared to USAMO?

You can find every problem here:
http://www.mathlinks.ro/index.php?f=584

is that because the problems dont come with solutions?

i already have the book but i dont know how i want to use it just yet and i dont really know how it compares with other well known books  like PSS and such.

Engel is more of a collection of problems, while ACoPS is more instructional. I don't own PFtB so I can't say anything about it.

Hey, I remember both of us getting this book at AMP. :)

I've read through some of it, and quite a bit of it uses calculus and other higher-level topics, so if you haven't already taken a calculus course, it might not be very useful there. Also, calculus doesn't directly appear on most olympiads, but some of the techniques shown in the book could be helpful on some problems. To me, the first few chapters (and 2-3 of the later ones) are pretty good, use mostly elementary topics, and give lots of examples, so they might be comparable to engel.

ok, thank you

i dont have much experience in calc and i was thinking about waiting anyways till i get a little better at problem solving so i can benefit better from it.

I got that book for free from Awesome Math. It's a bit challenging, and you need to know random stuff  :)

Hey, I got this at Awesomemath too. I personally like it more than Zeitz and Engel, but that may be because I had class with the author for a few weeks :) The practice problems are somewhat high level though, so I can't really do many of them. A lot of them are taken from various olympiads though, so it shouldn't be too hard to find solutions, though you should finish the problem before you look at the solution. It is a fine instructional text by any means though.


A 3-brick is a $ 1 \times 1 \times 3$ block of 3 unit cubes in a row. In a $ 7 \times 7 \times 7$ box $ B$ there is room for $ 7^3 \equal{} 343$ unit cubes. Since $ 343$ is not divisible by $ 3$, $ B$ cannot be packed exactly with just 3-bricks. However, 114 3-bricks can be packed into $ B$ so as to occupy all the unit cells except one; that is, there will be a single unit-cell hole $ H$ someplace. Prove that if $ H$ is not on the outside surface, but buried somewhere in the $ 5 \times 5 \times 5$ inner core, it must be the center cell itself.

[hide="Solution"]Coordinatize each cell as $ (x,y,z)$ for $ 0 \leq x,y,z \leq 6$. Give $ (x,y,z)$ a number $ c$, where $ c \in \{0,1,2\}$ and $ x\plus{}y\plus{}z \equiv c \pmod{3}$. Notice that every 3-brick covers a $ 0$, $ 1$, $ 2$, and that there is one more $ 0$. So the remaining cell must be a $ 0$. Now notice this argument works even if the cube is flipped or rotated. When enough symmetries are considered, the only possibilites will be points with all components divisible by $ 3$. The $ 5 \times 5 \times 5$ interior is all points without a $ 0$ or $ 6$ component. It follows that $ (3,3,3)$, the center, is the only remaining point.[/hide]


limit
trigonometry
function
calculus
calculus computations

$ \lim_{x\to 0} \frac {e^{x^{2}} \minus{} \cos x}{x^{2}}$
[hide="my approach"]
$ e^{x^{2}} \equal{} 1 \plus{} x^{2} \plus{} \frac {x^{4}}{2!} \plus{} f(x^{6},x^{8}.....)$
$ \cos x \equal{} 1 \minus{} \frac {x^{2}}{2!} \plus{} g(x^{4},x^{6}....)$
where $ f$ and $ g$ are the functions of powers higher than $ x^{4}$ and the powers are integral...
so the limit should be $ \frac {3}{2}$...
but my book says the limit does not exist :o [/hide]

sorry i saw the solutions in the wrong page  :rotfl:
mod please delete this :(

[quote="pardesi"]$ e^{x^{2}} \equal{} 1 \plus{} x^{2} \plus{} \frac {x^{4}}{2!} \plus{} f(x^{6},x^{8}.....)$
$ \cos x \equal{} 1 \minus{} \frac {x^{2}}{2!} \plus{} g(x^{4},x^{6}....)$
where $ f$ and $ g$ are the functions of powers higher than $ x^{4}$ and the powers are integral...[/quote]
This is a place where [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31517]big-O notation[/url] is very useful in terms of elegance of presentation.

$ \lim_{x \to 0} \frac{e^{x^2} \minus{} \cos x}{x^2} \equal{} \lim_{x \to 0} \left(\frac{e^{x^2} \minus{} 1}{x^2} \plus{} \frac{1 \minus{} \cos x}{x^2}\right) \equal{} 1 \plus{} \frac{1}{2}$.


Let $a,b,c$ be nonnegative real numbers. Prove that
\[\underset{cyc}{\sum}\sqrt{\frac{a^{2}+bc}{b^{2}+c^{2}}}\ge 2+\frac{1}{\sqrt{2}}\]
:)

this is a very nice problem, so i hope it will be interesting for you, and someone will solve it. (before i post my solution, i have a very short solution for it) :)

[color=purple].I've post this problem for \sqrtk yesterday in topic easy.But it couldn't read Latex.My solution of mine is very easy and nice as you said.I used Mixing Variables.How about you Can?
  Anh Quy' cung~ noi' la` 3 dong` thoi anh Can? a[/color]

how about this one?
\[\sum\sqrt[5]{\frac{a^{2}+bc}{b^{2}+c^{2}}}\ge 2+\frac{1}{\sqrt[5]{2}}\]
@evarist: i don't know what is the method which i used in my proof but i used it to solve the generation of this. :) And i think "vô chiêu thắng hữu chiêu" :D

[color=purple]Can you tell me your general? I've said my solution can use for k_root k is a interger.Thus k=5 the proof complete.
   Do you want i post my solution Can?[/color]

I have solved it already, so you don't need to send your solution to me. :)

[quote="evarist"][color=purple] Oh The following problem is interresting isn't it?[/color]
  $\sum\sqrt [k]{\frac{a^{2}+bc}{b^{2}+c^{2}}}\geq 2+\sqrt [k]{\frac{1}{2}}$ ($k\in N$ k>0)[/quote]
hm, $k$ don't need to be an integer, we only need $k>0$. :)

Anh Can? a em ko biet' nhan' tin tren Mathlink anh cho em nick YIM em co' viec nho`.Khas' wan trong.Lien wan den' danh du.Thanks


A dealer sold 100 books for $100, selling some for $2, some for $1 and the rest for 50 cents apiece. Compute the greatest number of two-dollar books he could have sold...

[quote="Mario_KaRt"]A dealer sold 100 books for $\$100, selling some for $\$2, some for $\$1 and the rest for 50 cents apiece. Compute the greatest number of two-dollar books he could have sold...[/quote]
For sure I am half slept. I am editing an stupid statment I did.
[hide="Solution"]
As it says that he sold books for $\$1, then, he should sold at least one. Then you have to solve the equation system $2x+\frac12y=99$ and $x+y=99$, that has solution for $x=33.$ therefore he sold 33 books of $\$2.
[/hide]
[hide="Stupidicity before"]
I am sure you are missing some hypothesis, otherwise the problem is too easy. As it says that he did sold books for $\$1 and for 50 cents, then, at least $\$1.50 were sold in non $\$2 books, but, logically, at least $\$2 though had to be solved in other books, then, at most 49 $\$2 books were sold.
[/hide]

Best,

I think the total number of books has to equal 100.
[hide]
Therefore x+y+z=100
2x+y+.5z=100
maximize for x
I think the answer is 33, but I'm not sure how to prove it (might be wrong).
[/hide]

[hide]Well, notice that it is possible for all 100 books to be $\$1. If we increase the price of one book to $\$2, that means we need to balance it out by replacing $\$1 books with two $\$.50 books. In this way, one $\$2 book and two $\$.50 books balance out three $\$1 books.
This means that there can be at most $33$ $\$2 books, $33\cdot2=66$ $\$.50 books and of course, one $\$1 book left over, so your answer was correct.

[b]djimenez[/b] way is probably more mathematically correct. They basically both translate to the same thing.[/hide]

Well, we have x+y < or = 100, and 2x+.5 y=100. the max x is [hide]33[/hide][/u]

[hide="Solution"]I started with two basic formulas \[x+y+z=100  ,  2x+y+.5z=100\]If you solve for $z$ in the first solution you get $z = 100-x-y$ , if you plug this formula into the second formula for $z$ you get (after simplifying) \[3x+y = 100\]  Maximizing $x$ in this case leads to an obvious answer of $33$.  I'm not sure if I did this correctly or if I got the answer by chance?[/hide]


[quote="rofler"]Recent COMC question 8, contest too recent.[/quote]

Have you some thread to solve?

But, I have got some threads to do.

[quote="vinskman"]求參數r的所有整數和
若x^3-rx+r+11=0
x,至少有一正整數解

[color=blue]Translate:[/color]
Find the sum of all possible integer values of $ r$ such that
Equation: $ x^3 \minus{} rx \plus{} r \plus{} 11 \equal{} 0$ ,for each $ r$
obtains at least one[color=red] positive [/color]   integer solution for $ x$ .

[color=darkblue]Sorry, it revised[/color][/quote]

[hide="Hints"]
$ x^3 \minus{} rx \plus{} r \plus{} 11 \equal{} 0$
$ \implies x^3  \plus{} 11 \equal{} r(x\minus{}1)$
$ \implies \frac{x^3\minus{}1}{x\minus{}1}\plus{}\frac{12}{x\minus{}1}\equal{}r$
[/hide]

Oh really? There's no requirement that $ \frac{12}{x\minus{}1}$ be an integer. It's seems possible to get rational values for both terms on the left that add to an integer.

But $ \frac{x^3\minus{}1}{x\minus{}1}\equal{}x^2\plus{}x\plus{}1$, which is an integer if $ x$ is, which completes the proof.

[quote="ManuelS"]using vinskman's last hint,
we notice that we need to have "x-1" equal to a integer divisor of 12 and we get 5 possible x's
{2,3,5,7,13}
we evaluate the values for "r" and we get
{19,19,59,34,184} (if we substitute for x with the values above)
so
the sum of all possible integer r's is 296[/quote]

[color=blue]You missed the case for x=4.[/color]

[quote="athunder"]I got r is 324 you might have missed a case or i did it wrong[/quote]

[hide="ans."]
$ \sum {r}\equal{}321$
[/hide]


quadratics
search
function
algebra
polynomial
Rational Root Theorem

[hide="1"]Rational root theorem and guessing and checking tells us that 3 and -2 are solutions. Dividing by $(x-3)(x+2)$ gives us the quadratic equation $x^2+x+10$. The solutions of this are $\frac{-1\pm\sqrt{39}i}{2}$. [/hide]

[quote="Kurt Gödel"]Finding the square roots = solving, right?  :lol:

[hide="Solution of 2"]
We have to find $x,y,z$ such that \[ \begin{aligned} x+y+z+2\sqrt{xz+yz} & = 0 \\ x+y+z & = -2\sqrt{xz+yz}\end{aligned}  \]
We have to claim that $xz+yz = (x+y)z \geq 0$ for the equation to exist and $x+y+z \leq 0$ for squaring, which yields \[ \begin{aligned} (x+y+z)^2 = x^2+y^2+z^2+2xy+2xz+2yz & = 4xz+4yz \\ x^2+y^2+z^2+2xy-2xz-2yz & = 0 \\ (x+y-z)^2 & = 0\end{aligned}  \]

So, $x+y=z$. But we had two claims: $(x+y)z = z^2 \geq 0$, so that's ok, and $x+y+z = 2z \leq 0$, which results in $z \leq 0$.
So the solution is $\{(x,y,z)\ |\ z \leq 0 \vee x+y=z\}$.[/hide][/quote]
I think he meant "Simplify $\sqrt{x+y+z+2\sqrt{xz+yz}}$".

[hide="Big Hint"]
$\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}$
(To prove it, simply square both sides)[/hide]

[quote="shadysaysurspammed"]1)Find x:-
$x^4+3x^2=16x+60$

2)Find sqroot of:-

$x+y+z+2\sqrt{xz+yz}$[/quote]

[hide]$\sqrt{x+y+z+2\sqrt{z(x+y)}}$

Let $a=x+y$

$a+z+2\sqrt{z\cdot a}=\sqrt{a}+\sqrt{z}=\boxed{\sqrt{x+y}+\sqrt{z}}$[/hide]

[quote="frt"][quote="ch1n353ch3s54a1l"][quote="shadysaysurspammed"]A proper method for the 1st one has still not been given.All of you are right for the sqroot one,except Kurt[/quote]

[hide]i got 3 and -2.

so first i found (x-3) as a factor by guess and check then i used synthetic division

3|  1  0     3  -16    -60

_____3_   9_36__60_____
      1  3   12   20    0

so then it's (x-3)($x^3 + 3x^2 +12x+20$)

by guess and check for the second factor i got (x+2) as a factor, then i used synthetic division again

-2|   1  3  12  20

           -2   -2  -20
_____________
       1    1   10  0

so now it is (x-3)(x+2)($x^2+x+10$)
finding the solutions for $x^2+x+10$ we get irrational solutions, which i didnt count in the solution set..
so the solution is
[size=200]x=3, x=-2[/size][/hide][/quote]
That's pretty much what 236factorial did ;)

shadysaysurspammed, what do you mean by "a proper method"?
I believe the method that 236factorial and ch1n353ch3s54a1l used  is probably the easiest approach...[/quote]

oh. hehe ..didnt see that post  :)

is "rational root theorem" just finding factors of the constant that make the equation equal zero?

[quote="ch1n353ch3s54a1l"]oh. hehe ..didnt see that post  :)

is "rational root theorem" just finding factors of the constant that make the equation equal zero?[/quote]
You could have searched for the answer to your question if you used the SEARCH in this forum. In case you do not have time to do that, here is a link:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=122430#p122430]http://www.mathlinks.ro/Forum/viewtopic.php?p=122430#p122430[/url]

This should clarify you when & how to use rational root theorem

[quote="10000th User"]You could have searched for the answer to your question if you used the SEARCH in this forum. In case you do not have time to do that, here is a link:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=122430#p122430]http://www.mathlinks.ro/Forum/viewtopic.php?p=122430#p122430[/url]

This should clarify you when & how to use rational root theorem[/quote]As a moderator, I realize the importance of the search function and the importance of spreading awareness of the search function.

However, I do not see any need to be so harsh. ;)

[quote="shadysaysurspammed"][hide="give up?"][b]ADDING[/b] $x^2+4$ to both sides ;) [/hide][/quote]
I see, that's a good approach, too. I didn't think of that.

[quote="shadysaysurspammed"]Yes the rational root theorem does work but i obviously meant something other than that method though it is the easiest but it does not work in all the questions.Anyone else wirth another method?

Here's my m,ethod if you don't get any other...awesome and obvious method
....

[hide="give up?"][b]ADDING[/b] $x^2+4$ to both sides ;) [/hide][/quote]

WOW. that's a great move. mad props dude.


geometry
circumcircle
trigonometry
similar triangles
geometry unsolved

Let triangle $ABC$ is acute. Let $BB'$,$CC'$ is perpendicular to $AC$,$AB$ such that $B',C'$ in segment $AC,AB$ ,respectively.
Let $O,O'$ is the circumcenter of $(ABC),(AB'C')$.Let a line through $A$ and meet $BC,B'C'$ at $K,K'$ respectively . Let $L$ in segment $OO'$ such that $LAK$ is a right angle. Prove that:
      $\frac{LO}{LO'}$ = $\frac{KO}{KO'}$

The point K' does not appear to be used and without it, the claim is absurd. (If K $\longrightarrow \infty,$ $\frac{KO}{KO'}\longrightarrow 1,$ AL is approaching the A-altitude which does not cut OO' anywhere close to its midpoint.) Assuming that the claim is $\frac{LO}{LO'}= \frac{KO}{K'O'},$ this is true only if AK is the external and AL the internal bisector of $\angle A.$ BB', CC' are the B- and C-altitudes, $AB' = AB \cos A,\ AC' = AC \cos A,$ triangles $\triangle AB'C' \sim \triangle ABC$ are oppositely similar by SAS with similarity coefficient $k = \cos A.$ O is on the A-altitude of $\triangle ABC$, because AO, AO' are isogonals with respect to $\angle A.$ K', K are the feet of the common external bisector of $\angle A$ of the oppositely similar triangles $\triangle AB'C' \sim \triangle ABC,$ $\frac{KO}{K'O'}= \frac{AO}{AO'}.$ Since AL bisects the angle $\angle OAO'$ of $\triangle OAO',$ $\frac{LO}{LO'}= \frac{AO}{AO'}= \frac{KO}{K'O'}.$


abstract algebra
calculus
integration
Ring Theory
algebra
function
domain

Let $ R$ be a normal, Noetherian integral domain. Fix a nonzero element $ f\in K(R)$, the field of fractions of $ R$, and define $ Z\equal{}\{P\in SpecR|f \notin R_P\}$ be the closed set where $ f$ is not a regular function. Show that for all irreducible components of $ Z$, the corresponding prime ideal $ P$ of $ R$ has codimension = 1.

I have no idea where to start on this so any input would be very helpful. Thanks in advance!

Ideal quotient: For a subring $ R \subset R' \subset K(R)$, define $ (R': f) \equal{} \{ r \in R| rf \in R' \}$ which is an ideal of $ R$.

Lemma: $ \displaystyle R \equal{} \cap_{\mathrm{ht p} \equal{} 1} R_p$ (because $ R$ is an integrally closed noetherian domain)

Claim 1: $ Z \equal{} V((R: f))$
Proof: $ f \notin R_p$ iff $ \forall a \notin p, af \notin R$ iff $ af \in R \Rightarrow a \in p$ iff $ (R: f) \subset p$.

Therefore, $ \displaystyle V\left((R: f)\right) \equal{} V\left(\cap_{\mathrm{ht p} \equal{} 1} (R_p: f)\right) \equal{} \cup_{\mathrm{ht p} \equal{} 1} V(R_p: f)$

Claim 2: For height 1 prime ideal $ p$ of $ R$, $ \sqrt{(R_p: f)}$ is prime.
Proof: $ R_p$ is a DVR (because it is a 1-dimensional integrally closed noetherian domain). Let $ v: K(R)^* \rightarrow \mathbb{Z}$ be the corresponding valuation.
Suppose that $ ab \in \sqrt{(R_p: f)}$, i.e. there exists $ n$, such that $ a^nb^nf \in R_p$ implying that $ a^{2n}b^{2n}f^2 \in R_p$. This implies that $ v(a^{2n}f) \plus{} v(b^{2n}f) \equal{} v(a^{2n}b^{2n}f^2) \ge 0$, so at least one of $ a^{2n}f$ or $ b^{2n}f$ belongs to $ R_p$. This shows that at least one of $ a \in \sqrt{(R_p: f)}$ or $ b \in \sqrt{(R_p: f)}$.

Claim 3: $ (R_p: f) \subset p$ if $ (R_p: f) \neq R$.
Proof: If $ (R_p: f) \neq R$, then $ f \notin R_p$, i.e. for all $ s \notin p$, $ sf \notin R$.
Suppose that there exists $ s \notin p$, but $ s \in (R_p: f)$. This means that $ sf \in R_p$, i.e. there exists $ t \notin p$, $ tsf \in R$. But $ ts \in R\minus{}p$, contradiction.

Claim 4: $ V(R_p: f) \equal{} V(\sqrt{(R_p: f)})$ is of codimensional 1 if $ p$ is of height 1.
Proof: It suffices to show that $ \sqrt{(R_p: f)} \subset p$ is a 1-dimensional prime ideal. But $ p$ is of height 1, so $ \sqrt{(R_p: f)} \equal{} p$ or $ 0$, and we only need to reject the second case. This is apparent, for there must exists some nonzero element $ s \in R$ such that $ sf \in R$. (because $ f \in K(R)$)

Therefore by ignoring those $ p$ such that $ (R_p: f) \equal{} R$, we obtain a union $ V\left((R: f)\right) \equal{} \cup V(R_p: f)$, with each $ V(R_p: f) \equal{} V(\sqrt{(R_p: f)})$ being codimensional 1. This completes the proof.


Putnam
number theory
relatively prime
number theory unsolved

For a given positive integer m, find all triples (n,x,y) of positive integers with m and n relatively prime which satisfy the relation $ (x^2\plus{}y^2)^m\equal{}(xy)^n$

[quote="mkn19"]For a given positive integer m, find all triples (n,x,y) of positive integers with m and n relatively prime which satisfy the relation $ (x^2 \plus{} y^2)^m \equal{} (xy)^n$[/quote]

Let $ x\equal{}au$ and $ y\equal{}av$ with $ a\equal{}\gcd(x,y)$ and so $ \gcd(u,v)\equal{}1$

The equation becomes $ a^{2m}(u^2\plus{}v^2)^m\equal{}a^{2n}(uv)^n$.

Since $ x^2\plus{}y^2>xy$, we get $ m<n$ and the equation becomes $ (u^2\plus{}v^2)^m\equal{}a^{2n\minus{}2m}(uv)^n$ and since $ \gcd(uv,u^2\plus{}v^2)\equal{}1$, we get $ u\equal{}v\equal{}1$ and $ a\equal{}2^{\frac m{2n\minus{}2m}}$

And so $ x\equal{}y\equal{}2^{\frac m{2n\minus{}2m}}$ and we need $ 2n\minus{}2m|m$ and so $ n\equal{}\frac{(2k\plus{}1)m}{2k}$ where $ 2k$ is any even divisor of $ m$ but, since $ \gcd(m,n)\equal{}1$, we get $ m\equal{}2k$ and $ n\equal{}2k\plus{}1$

Hence the answer :
If $ m$ is odd, no solution.
If $ m$ is even, a unique solution $ (x,y,n)\equal{}(2^{\frac m2},2^{\frac m2},m\plus{}1)$


function
search
algebra
polynomial
real analysis
real analysis unsolved

Hi,

here

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1085635078&t=116654

Kent Merryfield states that

4. If $ f\in L^{p},\ 1\le p < \infty,$ then $ \sigma_{N}(f)\to f$ in the $ L^{p}$ sense. (This bootstraps on Fejer's work.)

I am searching for quite a long time for some $ f\in L^{\infty}(\mathbb{T})$ for which
the Fejér sums $ \sigma_{N}(f)\to f$ do not converge to $ f$ in the $ L^{\infty}(\mathbb{T})$
sense, but cannot find one.

What do you think the most instructive example for this?

Note that the Cesaro means of the Fourier series are trigonometric polynomials, and hence continuous. Convergence in $ L^{\infty}$ is (essentially) uniform convergence. The uniform limit of a sequence of continuous functions must be continuous.

Hence we can say that $ \sigma_nf\to f$ uniformly if and only if $ f$ is continuous on $ \mathbb{T}.$


Let be $ a,b \in \mathbb{N^*}$. Prove that:

$ a)$ If $ a^2\plus{}4b$ and $ b^2\minus{}4a$ are  simultan squares, then $ b\equal{}a\plus{}1$
$ b)$ Thare are no $ a,b$ so that $ a^2\plus{}2b$ and $ b^2\minus{}3a$ are simultan squares.

"simultan" = "simultaneously."  So, the question is asking about when those are both perfect squares.


function
quadratics
algebra
polynomial
conics
parabola
analytic geometry

Define p(x) = ax2 + bx + c. If p(x) = x has no real roots, prove that p( p(x) ) = 0 has no real roots.

This is false. For example take p(x)=x^2+4x+3. Then there is no real root of the equation p(x)=x, but p(p(-2))=p(-1)=0.

Perhaps you want to say that the equation p(p(x))=x has no solution either.

I think so, because I tried, but can't solv

Well this true for any continuous function and not just a quadratic polynomial, by the Intermediate Value Theorem. However, let's give an elementary proof for this case where p(x)=ax^2+bx+c.

We want to prove that either p(x)>x for ALL x or p(x)<x for all x.
Indeed, since the equation p(x)=x has no real solution we have

(b-1)^2-4ac<0.

This is equivalent to

p( (-b+1)/2a)> (-b+1)/2a when a>0

and to

p( (-b+1)/2a)< (-b+1)/2a when a<0.

Let us examine only the case where a>0 (similarly the other case). Then
from our work above, the point P of the parabola with x-coordinate
(-b+1)/2a  lies above the line y=x.
Moreover, note that this is the point on the parabola of minimum distance from the line y=x and so every other point of the parabola lies above the line y=x. (The tangent line at the parabola at P is parallel to the line y=x).
Hence p(x)>x for all x, and so p(p(x))>p(x)>x for all x. Similarly, the other case implies that when a<0, then p(p(x))<x for all x.
The proof is complete.


[quote="Mircea Lascu"]Happy New Year !
Mircea Lascu[/quote]
And to you as well. :)
I still have to wait till 12:00 tonight. :D

Happy New Year to everyone

ya i also have to wait until 12:00 tonight..

Happy new year on behalf of the greek community

$2006+1$ wishes for a math-challenging year.

Don't forget to prime factor the year!
$2007=3^{2}*223$
Should produce some fun contest questions...good luck guys!

What are everyone's plans for 2007? Any Biggies?

I have nothing much, except that I am starting my freshman year in August, and lots of math junk , etc. and a summer trip to my native land. and a few camps

Happy 2007

My plans are to make the 2007 Canadian IMO team.

I didn't make it to Winter Camp.  So not the best start.

eeey Happy New Year!!

My goals are:
-To own the Euclid to the best of my ability.
-Get into Waterloo with a president's scholarship/ whatever other scholarships and bursaries I can get my hands on.

Also, another un-math related goal (if we're expected to post them?) is to get my CASI level 2.

....oh, and finish highschool  :P

Happy new Year, My goal is to Get into MathPath :), get a score above 11 on the AMC 10 and and to finish the AoPS volume 1 book, other goals, to get better at Klingon, French, Russian, C++, Basic, and make .5 percent each week including commissions :)

My plan is to type up all the problems that I brought from home. I got 8kgs of problems + some booklets (which is pretty heavy too). Should take a while.... :stretcher:

My plan is to get into UW with a little bit of scholarship money..also to do well on various contests (math, science, computing..)


LaTeX
trigonometry
complex numbers
AMC 10
AMC

Would anyone be interested in taking a mock AMC test, written by me? I haven't written any problems yet, but I have a bunch of ideas. I might have to handwrite the test, and then use a scanner, because my Latex is pretty much non-existent.

I'm interested in a mock AMC test.
Will you give us a couple of days to work on it (like a competition) or just an hour. A long period is better since more people might be able to submit it.

I'm thinking of giving people a week to do it, and having them time themselves.

I'd be interested, but someone should move this to the AMC forum.

I am interested beyond belief. Also, as b-flat said, is it going to be a 10 or 12?

[quote="b-flat"]Yes, to both the contest idea & leetspeak.

What would it be, AMC 10 or 12?[/quote]

I'm not sure, I haven't written the test yet, and haven't taken many practice AMC's. It would probably be somewhere in between. You can ask junggi after he proofreads it :rotfl:

I wrote about 22 problems last night, so the test will probably be pretty soon.

[quote="rnwang2"][/quote][quote="chess64"][quote="lyra"][quote="chess64"]I am not interested.[/quote]

Then why did you post, to be rude?[/quote]

oui, bien sur[/quote][quote="rnwang2"]

sois gentil.
sinon, je dois te manger  :icecream:[/quote]

C'ecoute amusant.  Je veux le regarder...

Can someone say what difficulty it is compared to the 12 and 10?

I think it's more like the AMC 10 but has some material similar to the AMC 12. Maybe the log question should be taken out? I don't know, I don't even know what it is.

If you are interested, just take it and send me the answers

umm i must really be losing my eyesight but where is it?

It's here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122126

[quote="mustafa"]Due to lack of answers received, I'll probably extend the deadline, unless I get a ton of answers tomorrow.[/quote]

I'm going to do it late tonight/early tomorrow morning...

Ok, I am officially moving the deadline to midnight on Monday night. (Honestly, if you sent it in before I woke up on Tuesday, I won't really care). Could a mod please make this obvious?

Where is the AMC forum? "Someone should move this to the AMC forum".

This is the AMC forum :wink:   AMC = American Mathematics Competitions.  Evidently, the topic was originated in some other forum, but was moved here.