inequalities
induction
vector
abstract algebra
invariant
algebra
polynomial

For each positive integer $ k$, find the smallest number $ n_{k}$ for which there exist real $ n_{k}\times n_{k}$ matrices $ A_{1}, A_{2}, \ldots, A_{k}$ such that all of the following conditions hold:

(1) $ A_{1}^{2}= A_{2}^{2}= \ldots = A_{k}^{2}= 0$,

(2) $ A_{i}A_{j}= A_{j}A_{i}$ for all $ 1 \le i, j \le k$, and

(3) $ A_{1}A_{2}\ldots A_{k}\ne 0$.

Looks like $ n_{k}=2^{k}$. First thing's first though :): I'm assuming we're working over some field $ L$.

$ n_{k}\ge 2^{k}\ (*)$

For $ k=1$ this is clear. Now we can prove the inequality by induction on $ k$. Suppose that $ k\ge 2$ and $ (*)$ is proven for all smaller values. Let $ V$ be an $ L$-vector space of dimension $ n_{k}$ on which the $ A_{i}$ act. The kernel of $ A_{1}$ has dimension $ \ge\frac{n_{k}}2$, since the sum of the dimensions of the range and the kernel of $ A_{1}$ is $ n_{k}$, and the range is contained in the kernel. Moreover, because all $ A_{i}$ commute, this kernel is invariant under all $ A_{i}$. Then $ A_{2},\ldots,A_{k}$ act on $ V'=V/\ker A_{1}$, an $ L$-linear space of dimension $ n_{k}-\dim\ker A_{1}\le\frac{n_{k}}2$. As endomorphisms of $ V',\ A_{j},\ j\ge 2$ clearly satisfy hypotheses (1) and (2) in Xevarion's message. They also satisfy condition (3): otherwise $ A_{2}\cdot\ldots\cdot A_{k}$ would map $ V$ into $ \ker A_{1}$, which means that $ \prod_{i\ge 1}A_{i}$ would map everything to zero. All of this means that $ \frac{n_{k}}2\ge n_{k-1}$, and the induction hypothesis finishes the proof of $ (*)$ above.

$ n_{k}\le 2^{k}\ (**)$

Assume we can find a unital $ L$-algebra $ \mathcal A$ of dimension $ 2^{k}$ with elements $ a_{i},\ i=\overline{1,k}$ satisfying (1), (2), and (3). The map sending each $ a\in\mathcal A$ onto the endomorphism $ x\mapsto ax$ of $ \mathcal A$ would then be an embedding of $ \mathcal A$ into $ \mathcal M_{2^{k}}(L)$, and $ (**)$ would be proven. Hence it suffices to find an algebra $ \mathcal A$ as above. The quotient of the polynomial ring $ L[x_{1},\ldots,x_{k}]$ by the ideal $ (x_{1}^{2},\ldots,x_{k}^{2})$ is such an algebra, with $ a_{i}$ being the image of $ x_{i}$ through the projection. A basis of size $ 2^{k}$ over $ L$ consists, for example, of all products of distinct $ a_{i}$ (including the empty product $ 1$, the multiplicative unit of $ L$).

An easier example for $ n=2^{k}$: Let $ v$ be the vector of all linear polynomials in $ k$ variables (there are clearly $ 2^{k}$ of these: each variable degree 0 or 1), and let $ A_{i}$ be the differential operator to the i-th variable. They clearly fulfill each of these properties. :)

Answer: $n_k = 2^k$.

Let $v$ be a nonzero $n_k$ length vector with $A_1 \cdots A_k v \neq 0$ and consider the $2^k$ vectors
\[A_1^{\varepsilon_1} \cdots A_k^{\varepsilon_k} v \qquad \text{for} \quad \varepsilon_1, \dots, \varepsilon_k \in \{0, 1\}.\]
We show that they are linearly independent. Suppose not, and let
\[\sum_{w = (\varepsilon_1, \dots, \varepsilon_k) \in \{0, 1\}^n} c_w A_1^{\varepsilon_1} \cdots A_k^{\varepsilon_k} v = 0\]
be a nontrivial relation among them. Let $w_0 = (t_1, \dots, t_k)$ be a binary vector with the least 1s with $c_{w_0} \neq 0$; then left-multiplying the above relation by $A_1^{1 - t_1} \cdots A_k^{1 - t_k}$,
\[c_w A_1 \cdots A_k v = 0,\]
a contradiction. Thus $n_k \ge 2^k$.

For $n_k = 2^k$, let $\mathbf{e}_0, \dots, \mathbf{e}_{2^k - 1}$ be a basis, and define $A_i$ by
\[A_i: \mathbf{e}_{t + 2^{i - 1}} \to \mathbf{e}_t \to 0\]
for all $t$ with $t \otimes 2^{i - 1} = 0$, where $\otimes$ denotes bitwise AND. It is easy to verify that all three conditions are satisfied.


I ask the following question , Lets see how you answer it :

[color=blue]Question[/color].Is it possible to take a Step and Travel in time ?

[hide="My answer"]I believe the answer is YES, and i can defend  it.[/hide]

PS.Please think before Answering :D, We all Know we.[i]cant[/i]  built time machine

If some time in the future the man can make such trips then [u]we should already know it[/u]

What I mean:
Suppose that in the year 10000 someone invents a way to make trips in the time. Then he will probably go to the year 2006 or earlier years.

Shouldn't we have a report from the past that some strange (future) man appeared?

Now, your question is a little different; you ask [u]if it is possible[/u], not if sometime will be. The difference is that I supposed that someone will go to the past

I have heard some other paradoxes.
The person x returns to the past and he kills his father when he was a child. Then the person x will never born, so he never killed his father, so he will born ...

[quote="pontios"]
I have heard some other paradoxes.
The person x returns to the past and he kills his father when he was a child. Then the person x will never born, so he never killed his father, so he will born ...[/quote]

I think that in that case, maybe a parallel universe would be created. In one universe, the father killer would be dead, and in one, he wouldn't. Or something like that. I don't think that it is possible for people to travel in time, simply because of this.

Some day, the world will end, due to natural disaster, or giant war. If people have the technology to travel in time at the end of the universe or the extinction of the human species, then EVERY SINGLE PERSON IN THE WORLD will want to travel back into time to save themselves. If that happened, I think that we would have noticed.

Well...technically, we're traveling through time right now. :lol:

Also, you could go on a really, really fast spaceship for a while, and you will "travel" into the future. :D

So is it a one way ticket then? You can't move back in time by going really really slow.......

With the Current technology We can travel at least one day in time :D

Consider the time in Vancouver at the Western Canada and the time in japan at the same moment .Japan Obviously Is ahead .
But if you keep going toward west from vancouver you will reach the japan .
Somewhere in the Pacific ocean , with is exactly Opposite Side of greenwich line , There is the international time line. the west side of the line has 24 hours differnce with the east side.

So stand on the east side of the line and take a step to west side. Tada, you are in future.

[quote="Ignite168"]Time moves slower for things going at extremely high speeds so the answer is yes.[/quote]
Actually , you argument in not enough.

Travelling in higher speeds slows down the time (if you manage to reach the speed of light it will [i]Stop[/i]) But
to go Back ward
you need to go faster than the speed of light Which is pretty [i]unlikely[/i]

[quote="now a ranger"]I have an idea. Read a book such as the L'Engle series with the Swiftly Tilting planet. or the Time Machine, which are easy books, and enjoy it. OR read some reallllllllllllly hard author's experiences back then or in the future[/quote]

I read both of those. They were really good, but they didn't exactly say anything legitamite about time travel. The time machine pretty much just said. "I did lots of research, and made a time machine"

Yes, I believe time travel is possible. My life's goal is to transcend space and time


Is there a way to do the problem without a calculator?


these are really sickening...

1\leq j\leq n gives: $1\leq j\leq n$
1 \leq j\leq n gives: $1 \leq j\leq n$
1 \leq j \leq n gives: $1 \leq j \leq n$
j gives: $j$
ij gives: $ij$
i j gives: $i j$
{ij} gives: ${ij}$
p_{ij} gives: $p_{ij}$
p _ { i j } gives: $ p _ { i j } $
p_1 gives: $p_1$
p_{1} gives: $p_{1}$
p_{} gives: $p_{}$
p _ 1 gives: $p _ 1$
p_ 1 gives: $p_ 1$
p _1 gives: $p _1$

is this just a problem of my account or do latex formulas generally don't work now?

  dg

no, you're right. I too have noticed that the number of unparsable messages has sensibly increased lately.
I don't know what is wrong, for the second thing it appears that freedom in spacing has gone :P
and for the first one, apparently the board wants to be safe that you can't latex  Dar[b]ij[/b] :D

Let's hope it's a temporary problem. I think that [b]every LaTeX formula somebody writes or edits today will be not displayed[/b].

  Darij

I was having a horrible time with it last night as well. I can't link to the post, though, because I never finished my solution and I closed the browser.

Actually, I think the LaTeX compiler doesn't work today. So my strategy is to post the messages with LaTeX as if nothing happened. What comes out are tons of "[unparseable and possibly dangerous latex formula]"s, of course, but I guess that after a while the problem will be fixed and all these formulas will pop up.

  Darij

no. you'll have to edit it afterwards.

From what I've seen, I think the machine works as follows:

there's a function $f(x)$ where $x$ is the text input between dollars and $f(x)$ is the outputted image. When pressing the "Sumbit" button, the $f(x)$ is generated for every $x$ in the message.

When displaying the topic, the board checks if there exists an image $f(x)$ for every $x$ in the message, but it doesn't generate it if that's not the case. If no $f(x)$ is found, the board assumes there was an error in making it and will display [unparseable or potentially dangerous latex formula].

This is only changed if a message with the missing $x$ in it gets submitted, so that the corresponding $f(x)$ gets generated.

Fixed now.

Not sure why it worked in the first place, but it works now.

Thanks!!!

Now you can delete the thread...

  darij

also: $great\;rrusczyk,\;thanks!!$

  dg

why do it simple if there's a hard way? :D


modular arithmetic
number theory proposed
number theory

Find all positive integers $ n$ such that $ n^2$ devides $ 3^n\plus{}1$.

obviously $ 2 \nmid n$.let $ p$ be the smallest prime dividing $ n$.Let $ ord_p(3) \equal{} d$ then we have $ 3^{2n} \equiv 1 \pmod {n^2} \implies 3^{2n} \equiv 1 \pmod {p},3^d \equiv 1 \pmod {p} \implies d|p \minus{} 1,d|2n \implies \mbox{if gcd(d,n) \equal{} k ,greater than 1, then n will have a divisor less than p which is a contradiction} \implies \gcd(d,n) \equal{} 1 \implies d|2 \implies d \equal{} 1,or,d \equal{} 2$
but in both cases u will get $ p \equal{} 2$ which means $ n$ is even.so the only solution is $ n \equal{} 1$.

$ p$ is minimum prime divisor of $ n$  $ \implies$
$ p|3^{p-1}-1,p|3^{2n}-1$  $ \implies$  $ p|3^{(2n,p-1)}-1$
$ (2,p)=1$ because if $ 2|n$ $ \implies$ $ 4|n^{2}$ but $ 4\nmid 3^{n}+1$ contradiction!
$ \implies$  $ (2n,p-1)=2$  $ \implies$  $ p|8$ contradiction!
$ \implies$  $ n=1$ is only solution $ \Box$


I was browsing through some math sites and found a few interesting problems (I altered the original problems that I found to the problems I have posted to make sure I understand why the steps were wrong :P). In these problems you need to look for what's wrong with the steps. I posted my solution to each of them. It would be nice if people can contribute different perspectives on explaining why a certain step in these problems is invalid.

[hide="Problem 1"]
\begin{align*} a & = b = 1 \\
a^2 - b^2 & = 0 \\
(a - b)(a + b) & = 0 \\
a + b & = 0 \\
a & = - b \\
1 & = - 1 \\
\end{align*}
(I think this one is the easiest out of all of them.)
[hide="My Solution"]
Simple, in the step $ (a - b)(a + b) = 0 \Longrightarrow a + b = 0$, there was a division by zero because $ a - b = 0$.
[/hide]
[/hide]
[hide="Problem 2"]
\begin{align*} \frac { - 1}{1} & = \frac {1}{ - 1} \\
\sqrt {\frac { - 1}{1}} & = \sqrt {\frac {1}{ - 1}} \\
\frac {i}{1} & = \frac {1}{i} \\
i^2 & = 1 \\
- 1 & = 1 \\
\end{align*}
This one is a bit more harder.
[hide="My Solution"]
$ \sqrt {\frac { - 1}{1}} \& = \sqrt {\frac {1}{ - 1}} \Longrightarrow \frac {i}{1} \& = \frac {1}{i}$
The step above is invalid. $ \sqrt {\frac {a}{b}} = \frac {\sqrt {a}}{\sqrt {b}}$ [u]only[/u] applies when $ \sqrt {a^2} = a$ and $ \sqrt {b^2} = b$. In the step above, $ \sqrt {( - 1)^2} \not = - 1$.
[/hide]
[/hide]
[hide="Problem 3"]
\begin{align*} (8 - 9)^2 & = (9 - 8)^2 \\
\sqrt {(8 - 9)^2} & = \sqrt {(9 - 8)^2} \\
8 - 9 & = 9 - 8 \\
- 1 & = 1 \\
\end{align*}

[hide="My Solution"]
The error was in the step $ \sqrt {(8 - 9)^2} = \sqrt {(9 - 8)^2}$. It is $ \pm \sqrt {(8 - 9)^2} = \pm \sqrt {(9 - 8)^2}$ because the square root of a number can be positive or negative (Ex: $ \sqrt {25} = \pm 5$). Now if we simplify from our fix:
\begin{align*} \pm \sqrt {(8 - 9)^2} & = \pm \sqrt {(9 - 8)^2} \\
\pm - 1 & = \pm 1 \\
\pm 1 & = \pm 1 \\
\end{align*}
[/hide]
[/hide]
[hide="Problem 4"]
\begin{align*} \left(\frac {3}{4}\right)^2 & > \left(\frac {3}{4}\right)^3 \\
\log{\left(\frac {3}{4}\right)^2} & > \log{\left(\frac {3}{4}\right)^3} \\
2 \log{\frac {3}{4}} & > 3 \log{\frac {3}{4}} \\
2 & > 3 \end{align*}
[hide="My Solution"]
In the step $ 2 \log{\frac {3}{4}} > 3 \log{\frac {3}{4}} \Longrightarrow 2 > 3$, $ \log{\frac {3}{4}}$ was divided, but the inequality sign wasn't changed. (If you have the log of a number such that $ 0 < n < 1$ to the base 10, the result is negative.)
[/hide]
[/hide]

Your answers to 1,3,4 are pretty much complete, but it turns out that the problem with #2 is the same as with #3. In particular, $ \minus{}1$ has two square roots, like any real number, namely $ i$ and $ \minus{}i$. Also, $ i\equal{}\frac{i}{1}$ (obviously) and $ \minus{}i\equal{}\frac{1}{i}$, so the third step there actually took the two different roots of $ \minus{}1$.

On problem 3, the problem is that $ \sqrt{x^2}\equal{}|x|$
Therefore, $ \sqrt{(8\minus{}9)^2}\equal{}|8\minus{}9|\equal{}9\minus{}8$


The ordered pair of 4 digit numbers (2025,3136) has the property that each number in the pair is a perfect square and each of the second number is 1 more than the corresponding digit of the first number. Find, with proof, all ordered pairs of five-digit numbers and ordered pairs for six-digit numbers with eh same property: each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number.

Please tell me what to do?

Just ... solve the problem yourself, I guess?

Yes, you shouldn't really be posting problems which are so obviously from a competition which is currently running.

Nobody will answer questions about a problem from a contest while it is running. Ask for help after the round ends (Oct. 13).


Can anyone tell me what is a singular matrix?

We say that a matrix $A$ is singular if  $det(A)=0$   :P  ;)

Thanks for the answer, [b]cezar lupu[/b]. I didn't think it was so simple!

Do you know it's geometric significance???


Let a1,a2,..,an be real numbers so that

a1+..+na_n>=n(n+1)(2n+1)/6 for all n>=1,natural

Show that the equation x^2-12(a1^2+..+a_n^2)/((n(n+1)(2n+1))x+1=0 has real solutions.

unknown source

Trivial
 Apply Cauchy Schwartz to find that (1^2+...+n^2)(a_1^2+...+a_n^2)>=(1^2+...+n^2)^2, from where the conclusion follows.


Mainly, they have different problems.  They are roughly the same difficulty, and cover many of the same general subject areas, but with slightly different emphases depending on the problems that appear in each.  Basically, we had one AIME Problem Series, and then after running the Special AIME Problem Seminar a few times, we had enough material to build a second one.


function
limit
calculus
derivative
calculus computations

suppose we have a  real valued differentiable function f  such that both $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f'(x)$ exist and are finite. should it be true that the latter limit must be 0?

Yes. Try to assume the contrary and use integration.


Given an arbitrary prime P, show that we can find infinitely many positive integers n such that 2^n - n is a multiple of P.

2 can replaced by a, where a is one aleatory integer.

if p|a, let n = p.

now, suppose that p don't divide a. let $f(x) = a^{x}-x$.

see that $f(x+(p-1)) \equiv f(x)-1 (mod.p)$, by FLT.

so, $f(x+k(p-1)) \equiv f(x)-k (mod.p)$.

now, take k = f(x), for any integer x, and the results follow..

(POST 200  :D )

[hide="MySolution"]
$2^{p-1}\equiv 1 (mod p)$
$2^{(p-1)(mp-1)}\equiv 1 (mod p)$ where m is any integer
$2^{(p-1)(mp-1)}-(p-1)(mp-1) \equiv 0 (mod p)$
[/hide]


you know how there's like \documentclass{article}? I was just wondering what other types of \documentclass there are besides article.

If you look for class files in your LaTeX directory (they have extension .cls) you'll see how many there are on your computer. In fact, since any expert can write their own, there are potentially an infinite number. For example there's a modern CV class [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=moderncv]here[/url].

But the ones that you are most likely to see are:[list][*]Article: for, guess what, short articles
[*]Book: for books :)
[*]Report: for longer things than articles but shorter than books eg theses
[*]Letter: not to be confused with letter size paper, but for writing letters
[*]Beamer: for presentations (which can be stunning see [url=http://latex-beamer.sourceforge.net/beamerexample1.pdf]this example[/url]; when it opens in Adobe Reader press CTRL+L then go through it using -> key. Press Esc to finish)[/list]


Putnam integration calculus trigonometry logarithms derivative function

Here's a solution. It was about the fifth trick I tried, and it didn't work the first time I tried it:

Don't open this if you still want to do the problem on your own!
[hide="Solution"]

We have $\ln(x+1)=\int_0^x\frac1{t+1}\,dt$, so
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx= \int_0^1\int_0^x\frac1{t+1}\cdot\frac1{x^2+1}\,dt\,dx. \]
Triangles aren't very nice, so we should stretch to a square; substitute $t=xy$. This gives
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\int_0^1\int_0^1 \frac1{xy+1}\cdot\frac1{x^2+1}\,xdy\,dx. \]
Now reverse the order of integration:
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\int_0^1\int_0^1 \frac{x}{xy+1}\cdot\frac1{x^2+1}\,dx\,dy \]
We do the inner integral with partial fractions:
$\frac{x}{(xy+1)(x^2+1)}=\frac1{y^2+1}\cdot \left(\frac{-y}{xy+1}+\frac{y}{x^2+1}+\frac{x}{x^2+1}\right)$, so
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\int_0^1\frac1{y^2+1}\cdot\left[-\ln(xy+1) +y\tan^{-1}(x)+\frac12\ln(x^2+1)\right]_{x=0}^{x=1}\,dy. \]
Evaluating, we have
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\int_0^1\frac{-\ln(y+1)}{y^2+1}+ \frac{\pi}4\cdot\frac{y}{y^2+1}+\frac{\ln 2}2\cdot\frac1{y^2+1}\,dy. \]
The first term duplicates the integral we're trying to evaluate, so we pull it to the other side:
\[ 2\int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\left[\frac{\pi}8\ln(y^2+1)+\frac{\ln 2}2\tan^{-1}(y)\right]_{y=0}^{y=1}=2\frac{\pi\cdot\ln 2}{8} \]
and finally
\[ \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\frac{\pi\ln(2)}{8} \]
[/hide]

I got lucky, and the very first trick I tried worked.  I got the motivation for this from Larson's Problem Solving Through Problems.

Let $I(a) = \int^1_0 \frac{\log(ax+1)dx}{x^2+1}$

Differentiating under the integral sign, we get

\begin{eqnarray*} I'(a)&=&\int^1_0 \frac{xdx}{(ax+1)(x^2+1)} \\ &=&-\frac{\log(a+1)}{a^2+1}+\frac{\log 2}{2(a^2+1)}+\frac{\pi a}{4(a^2+1)}\end{eqnarray*}

By integrating over $[0,1]$ and using the fundamental theorem of calculus, we get

$I(1)-I(0)=-I(1)+\frac{\pi\log 2}{4}$

Obviously $I(0)=0$, so that

$I(1)=\frac{\pi}{8}\log 2$

Wow! This is the very the problem which I posted before.Here is my and [b]liyi'[/b]s solution.

Let $x=\frac{1-t}{1+t},$ we have $\frac{dx}{dt}=-\frac{2}{(1+t)^2}.$

Thus $\int_0^1 \frac{\ln (x+1)}{x^2+1}dx=\int_1^0 \frac{\ln \frac{2}{1+t}}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot \left\{-\frac{2}{(1+t)^2}\right\} dt$

$=2\int_0^1 \frac{\ln \frac{2}{1+t}}{(1+t)^2+(1-t)^2}\ dt$

$=\int_0^1 \frac{\ln 2-\ln (1+t)}{t^2+1}\ dt$

$=\int_0^1 \frac{\ln 2}{x^2+1}\ dx-\int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx\ (\because \ t \ is \ arbiterary \ real \ numbers)$

$\Longleftrightarrow 2\int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx=\ln 2\cdot \int_0^1 \frac{1}{x^2+1}\ dx=\frac{\pi}{4}\ln 2$

$\therefore \int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx=\frac{\pi}{8}\ln 2.$

P.S. I have found the post.Here is [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=149755#p149755[/url]

Somehow, I found this one easy possibly because I already knew the techniques to handle such problems!

First of all, it is easy to show that $\int_0^a\ f(x)dx = \int_0^a\ f(a-x)dx ... (1)$. Just use the substitution $y = a-x$ for the integral on the right hand side and simplify.

Now, let $I$ denote the given integral in the problem. Use the substitution $x = tan y$. Then $I$ = $\int_0^{\frac \pi 4}\ln(siny + cosy) dy - \int_0^{\frac \pi 4}\ln( cosy) dy$. Further, $I$ = $\int_0^{\frac \pi 4}\ln(\sqrt 2 (cos ({\frac \pi 4 - y})) dy - \int_0^{\frac \pi 4}\ln( cosy) dy$ = $\frac \pi 8 \ln 2 + \int_0^{\frac \pi 4}\ln(cos ({\frac \pi 4 - y}) dy - \int_0^{\frac \pi 4}\ln( cosy) dy$. Now, using the result $(1)$ obtained above, we note that the last two integrals cancel each other, so that $I$ = $\frac \pi 8 \ln 2$.

[quote="blahblahblah"]I got lucky, and the very first trick I tried worked.  I got the motivation for this from Larson's Problem Solving Through Problems.

Let $I(a) = \int^1_0 \frac{\log(ax+1)dx}{x^2+1}$

Differentiating under the integral sign, we get :?

\begin{eqnarray*} I'(a)&=&\int^1_0 \frac{xdx}{(ax+1)(x^2+1)} \\ &=&-\frac{\log(a+1)}{a^2+1}+\frac{\log 2}{2(a^2+1)}+\frac{\pi a}{4(a^2+1)}\end{eqnarray*}[/quote]

This isn't one of your 'Computation of integral 1xxx' threads - if you believe that there is a problem with my solution, then say what it is.  If you have a problem with me using $\log$ where the question uses $\ln$, it is something which I do consistently.

Please explain the calculation of $I'(a)$,blahblahblah. :)

kunny

Here's another question.  During the break, several of the people at my university were trying to calculate said integral using a contour in the complex plane.  My idea was to break it up into a rectangular contour, e.g, having vertices $(0,0)$, $(1,0)$, $(0,N)$, $(1,N)$, and letting $N\to\infty$.  Of course you have a pole at $i$, so you need to indent your contour.  We couldn't really make anything work (and I don't think we really cared that much) but does anyone have any ideas of how this might be done?

[quote="blahblahblah"]The derivative of the function

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

is precisely

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

since we are differentiating with respect to $a$, and hence treat $x$ as constant.  The technique of differentiating under the integral sign (and the associated conditions) gives us the equivalence

$I'(a)=\frac{d}{da}I(a)=\int^1_0 \frac{\partial}{\partial a} \left(\frac{\log(ax+1)}{x^2+1}\right)dx$[/quote]

Thank you! I can understand.

kunny

Just one coment. When I was a senior in high school (1989-1990) this problem was in the textbook as a homework problem.
Now I give this problem to my students and give them the entire semester to do it!

It was also used in Japan

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=389001#p389001

As already pointed out, let $I(a)=\int_{0}^{1}\frac{\ln(ax+1)}{x^{2}+1}$ and find $I(1)$.  I think letting $I(a)=\int_{0}^{a}\frac{\ln(ax+1)}{x^{2}+1}$ and finding $I(1)$ also works out nice.

I was skimming through the book Excursions in Calculus by R.M. Young and I found this problem in page 274 Prob 5 ...
Are the persons making the exam aware of some of these old sources for the problems, but think that a student will most probably not have seen them
or is it that the problems have to start circulating at some point :)

[quote="jmerry"][quote="blahblahblah"]
Let $I(a) = \int^1_0 \frac{\log(ax+1)dx}{x^2+1}$[/quote]
That's actually the first solution I saw- and when I saw it, I realized that my earlier idea would lead to the same computations. The only real difference is in motivating what to do.[/quote]

Would anyone mind tossing a link or explanation behind the motivation of differentiation under the integrand? I've never seen this before.

edit: I would say my first thought is that it somehow relates to the forms you should be looking at?

Very beautiful! My first Putnam Integral! :)

Using integration by parts, we may write $$ \int^1_0 \frac{\log (1+x)}{1+x^2}\, dx= \log(1+x)\cdot \tan^{-1}(x)-\int^1_0 \frac{\tan^{-1}(x)}{x+1}\, dx$$ since $$\int \frac{1}{1+x^2}\, dx=\tan^{-1}(x)$$

Thus, we only need to compute the definite integral $$I=\int^{1}_0 \frac{\tan^{-1}(x)}{x+1}\, dx$$ and the rest shall follow. Indeed, we do this by considering the following substitution: $x \longrightarrow \tan \theta$ where $0 \le \theta \le \frac{\pi}{4}$.

Since $$ dx=\frac{1}{\cos^2\theta}d\theta$$ we conclude that $$I=\int^{\frac{\pi}{4}}_0 \frac{\theta}{(1+\tan \theta)\cdot \cos^2\theta}\, d\theta$$ The trickiest step is this: substitute $\theta=\frac{\pi}{4}-\alpha$ where $\frac{\pi}{4} \ge \alpha \ge 0$. We get that \begin{align*} (1+\tan \alpha)\cdot \cos^2 \alpha=\left(1+\frac{1-\tan \theta}{1+\tan \theta}\right)\cdot \frac{1}{2}\cdot (\sin \theta+\cos \theta)^2=(1+\tan \theta)\cdot \cos^2 \theta \end{align*}

Therefore, as $$d\theta=-d\alpha$$ we conclude that $$I=-\int^{0}_{\frac{\pi}{4}}\frac{\frac{\pi}{4}-\alpha}{(1+\tan \alpha)\cdot \cos^2\alpha}\, d\alpha=\frac{\pi}{4}\cdot \int^{\frac{\pi}{4}}_0 \frac{1}{(1+\tan \alpha)\cdot \cos^2\alpha}\, d\alpha -I$$ and hence $$2I=\frac{\pi}{4}\cdot \int^{\frac{\pi}{4}}_0 \frac{1}{(1+\tan \alpha)\cdot \cos^2\alpha}=\frac{\pi}{4}\cdot \ln 2$$ where the last equality follows since re-substituting $t=\tan \alpha$ where $0 \le t \le 1$ we get $$dt=\frac{1}{\cos^2\alpha}d\alpha$$ and the antiderivative becomes $\ln (1+\tan\alpha)$ which integrating in the limits $0$ to $\frac{\pi}{4}$ gives $\ln 2$.

Hence, we conclude that the original definite integral is equal to $\frac{\pi}{8}\cdot \ln 2$.

Nice. :) By the substitution $x= \tan{\theta}$, we have $$\int_{0}^{1}{\frac{\ln(1+x)}{1+x^2} \ \mathrm{d}x} = \int_{0}^{\frac{\pi}{4}}{\ln{(1+\tan{\theta})}\ \mathrm{d}\theta}$$ We use the property $$\int_{a}^{b}{f(x) \ \mathrm{d}x} = \int_{a}^{b}{f(a+b-x) \ \mathrm{d}x}$$ So we have $$I = \int_{0}^{\frac{\pi}{4}}{\ln{(1+\tan{\theta})}\ \mathrm{d}\theta} = \int_{0}^{\frac{\pi}{4}}{\ln{(1+\tan{\left(\frac{\pi}{4} - \theta \right)})}\ \mathrm{d}\theta} = \int_{0}^{\frac{\pi}{4}}{\ln{\left(\frac{2}{1+\tan{\theta}}\right)}\ \mathrm{d}\theta}$$ Adding the two, $$2I = \int_{0}^{\frac{\pi}{4}}{\ln{2}\ \mathrm{d}\theta} \Rightarrow I = \boxed{\frac{\pi}{8} \ln{2}}$$

I feel like I saw this on the 2006 MIT integration bee

  Let
$$I=\int_{0}^{1}{\frac{\ln(1+x)}{1+x^2} \ \mathrm{d}x} $$
Subsituting $x=tan{\theta}$ and then using trignometric subsituting we get:
$$I=\int_{0}^{\frac{{\pi}}{4}}{\ln{(1+\tan{\theta})}\ \mathrm{d}\theta}$$
$$1=\int_{0}^{\frac{{\pi}}{4}}{\ln{(1+\tan(\frac{{\pi}}{4}-{\theta})}\ \mathrm{d}\theta}$$
Now using the fact that $tan(\frac{{\pi}}{4}-{\theta})=\frac{1-tan{\theta}}{1+tan{\theta}}$ we get:
$$ I=\int_{0}^{\frac{\pi}{4}}{\ln{\left(\frac{2}{1+\tan{\theta}}\right)}\ \mathrm{d}\theta}$$
$I=\int_{0}^{\frac{\pi}{4}}{\ln{2} \ \mathrm{d}\theta}-$ $ \int_{0}^{\frac{\pi}{4}}{ln{(1+tan{\theta})} \ \mathrm{d}\theta}$
But notice that the $2$nd integral is the same integral which we have found above. So,
$I=\frac{{\pi}}{4}{\ln{2}}-I$
$2I=\frac{{\pi}}{4}{\ln{2}}$
$$I=\frac{{\pi}}{8}{\ln{2}}$$

From what I can tell this solution is not present anywhere else in the thread. Let
\[I=\int_0^1 \frac{\ln(1+x)}{1+x^2}\,\mathrm dx.\]
Substitute $t=1+x$ to get
\[I=\int_1^2 \frac{\ln(t)}{t^2-2t+2}\,\mathrm dt.\]
Now substitute $u=\tfrac{2}{t}$ to get
\[I=\int_2^1 \frac{\ln(\frac{2}{u})}{\frac{4}{u^2}+\frac{4}{u}+2}\left(-\frac{2}{u^2}\right)\mathrm du=\int_1^2 \frac{\ln(2)-\ln(u)}{u^2-2u+2}\,\mathrm du=\int_1^2 \frac{\ln(2)}{u^2-2u+2}\,\mathrm du-I.\]
It's simple enough to evaluate
\[\int_1^2 \frac{\ln(2)}{u^2-2u+2}\,\mathrm du=[\ln(2)\arctan(u-1)]_1^2=\frac{\pi\ln(2)}{4},\]
so $I=\boxed{\frac{\pi\ln(2)}{8}}$.

Putnam 2005 A5 has been on my bucket list for a long time and I am so happy that I finally got it  :D

For this integral we are going to use a property which states that
$$\int_a^b f(x)\,dx=\int_a^bf(a+b-x)\,dx.$$
Substituting $x=\tan(\theta)$ with $dx=\sec^2(\theta)\,d\theta$ we get
\begin{align*}
\int_0^1\frac{\ln(x+1)}{x^2+1}\,dx&=\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(\theta)+1)}{\tan(\theta)^2+1}\left(\sec^2(\theta)\right)\, d\theta\\
&=\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(\theta)+1)}{\tan(\theta)^2+1}\left(\tan^2(\theta)+1\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(1+\frac{1-\tan(\theta)}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(\frac{1+\tan(\theta)+1-\tan(\theta)}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(\frac{2}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(2)-\ln(\tan(\theta)+1)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(2)\,d\theta-\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta.\\
\end{align*}
But we already had $I=\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta$ which means
\begin{align*}
I&=\int_0^{\frac{\pi}{4}}\ln(2)\,d\theta-I\\
2I&=\ln(2)\int_0^{\frac{\pi}{4}}\,d\theta\\
2I&=\ln(2)\left(\frac{\pi}{4}\right)\\
I&=\boxed{\frac{\pi\ln(2)}{8}}.
\end{align*}

So unoriginal by Putnam!!! This is proposed by [b]Serret[/b] in [url=https://mathworld.wolfram.com/SerretsIntegral.html]1844[/url]!

[hide=My work as others] Setting $x=(1-u)/(1+u)$ gives $$I=\int_0^1\frac{\ln(x+1)}{x^2+1}\, \text dx =\int_0^1 \dfrac{\ln 2-\ln (u+1)}{u^2+1} \text du =\ln 2\int_0^1 \dfrac 1{u^2+1} \text du-I =\frac 14\pi \ln 2-I$$ and so answer is $\tfrac 18 \pi \ln 2.$ [/hide]

[quote=Bibi]Here it is another nice solution.

Let $x=\tan t$. Then the integral becomes
\[ I=\int_0^{\pi/4} \ln (1+\tan t) dt. \]
Now make the change of variables $t=\frac{\pi}{4}-s$, and get
$I=\int_{\pi/4}^0 \ln \left( 1+\frac{1-\tan s}{1+\tan s}\right) (-ds)=\int_0^{\pi/4} \ln \left( \frac{2}{1+\tan s}\right) ds=\frac{\pi}{4}\ln 2 - I.$

Therefore $I=\frac{\pi}{8} \ln 2.$[/quote]

Wait, Newbie here! When we substitute $\tan t$ in, where did the denominator go?

[b]@above, this is how the denominator goes,[/b]

Setting $x=\tan \theta\implies dx=\sec^2\theta~ d\theta$
$$I=\int_0^1\frac{\ln(x+1)}{x^2+1}~ dx=\int_{0}^\frac{\pi}{4}\frac{\ln(\tan \theta+1)\sec^2\theta}{1+\tan^2\theta}~ d\theta=\int_{0}^\frac{\pi}{4}\frac{\ln(\tan \theta+1)\sec^2\theta}{\sec^2\theta }~ d\theta=\int_{0}^\frac{\pi}{4}\ln(\tan \theta+1)~ d\theta$$

Make the substitution $x=\tan\theta$. Then, the integral becomes
\begin{align*}
\int_0^1\frac{\log(x+1)}{x^2+1}dx &= \int_0^{\pi/4}\log\left[\frac{\cos\theta+\sin\theta}{\cos\theta}\right]d\theta \\
&= \int_0^{\pi/4}[\log[\sqrt{2}\sin(\theta+\pi/4)]-\log\cos\theta]d\theta \\
&= \frac{\pi}{8}\log 2 + \int_0^{\pi/4}\log\sin(\theta+\pi/4) - \int_0^{\pi/4}\log\cos\theta \\
&= \boxed{\frac{\pi}{8}\log 2}.
\end{align*}

1) set $I(a)=\int_{0}^{1} \frac{\ln(ax+1)}{x^2+1}dx$\\

differentiating under the integral sign we get:\\

$I'(a)=\int_{0}^{1} \frac{x}{(x^2+1)(ax+1)}dx$\\

we use partial fractions to solve this:\\

$\frac{A}{ax+1}+\frac{Bx+C}{x^2+1}=\frac{x}{(ax+1)(x^2+1)}$\\

which gives $A=\frac{-a}{1+a^2}, B=\frac{1}{1+a^2} , C=\frac{a}{1+a^2}$\\

now we have $I'(a)=\frac{-\ln(a+1)}{1+a^2}+\frac{\ln(2)}{2(1+a^2)}+\frac{\pi}{4}\cdot \frac{a}{1+a^2}$\\

gives $I'(a)=\frac{1}{1+a^2}\cdot \left(\frac{\ln(2)}{2}-\ln(a+1)+\frac{\pi a}{4} \right)$\\

now integrating from $0$ to $1$ we get:\\

$\int_{0}^{1} I'(a)da=\frac{\ln(2) \pi}{4}-I(1)$\\

from FTC we have:\\

$I(1)-I(0)=\frac{\ln(2) \pi}{4}-I(1)$\\

which gives $I(1)=\boxed{\frac{\ln(2) \pi}{8}}$\\

2) substitute $x=\tan(\theta)$ we get $dx=\sec^{2}(\theta) d\theta$\\

we have the integral as $I=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan(\theta)) d\theta$\\

now again $I=\int_{0}^{\frac{\pi}{4}} \ln\left(1+\tan\left(\frac{\pi}{4}-\theta \right) \right)d\theta$\\

$I=\int_{0}^{\frac{\pi}{4}} \ln(2)-I$ which gives $I=\boxed{\ln(2)\cdot \frac{\pi}{8}}$


articles
AoPSwiki
LaTeX
modular arithmetic
number theory
divisibility tests

On the AoPSwiki page on Fermat's Little Theorem, in the sample problem, could someone help me understand how the numbers in $ \mod{3}$ come about? I understand that $ 4 \equal{} n\mod{5}$, but not how $ 0 \equal{} n\mod{3}$

Thanks.

EDIT:[url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Link to Article[/url]

PS. How do you make the congruent symbol in Latex?

i don't know about the problem, but i do know about LaTeX

5\equiv0\pmod{5}$ \Longrightarrow5\equiv0\pmod{5}$

if u ever need help with LaTex this helped me a lot.

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php[/url]

the left hand side is divisible by 3, so n must also be a mutliple of 3.

I know what mod means...but I don't see how they found the LHS was divisible by 3, or how they knew to test divisibility by 3 in the first place...

[quote="Icy"]I don't see how they found the LHS was divisible by 3[/quote]
[b]Lemma:[/b]  $ a\equiv b\bmod m, c\equiv d\bmod m\implies ac\equiv bd\bmod m$

[b]Corollary:[/b]  $ a\equiv b\bmod m\implies a^{n}\equiv b^{n}\bmod m$

Hence because we have $ 110\equiv\minus{}1\bmod 3$ we have $ 110^{5}\equiv (\minus{}1)^{5}\equiv\minus{}1\bmod 3$ and so forth.

[quote="Icy"], or how they knew to test divisibility by 3 in the first place...[/quote]
It's not that they "knew."  It's simply that divisibility by $ 3$ is a convenient and easy test (see above), and that it narrows down the possibilities for $ n$ quickly to check divisibility by small primes.  All we are really doing is playing an educated guessing game where we try to eliminate as many possibilities as possible - ideally, we will only need to make two or three guesses before we find the right answer.


I can't seem to find a modular function on a TI-84+. I realize it isn't needed but during a program you can't implement another program you already made.
I can write 5-5*int(B/A) but it'll be much simpler if it was just A mod B (or mod(a,b)

[quote="moogra"]I can write 5-5*int(B/A) but it'll be much simpler if it was just A mod B (or mod(a,b)[/quote]
Alternatively, you can write it as B*fPart(A/B). (Not really necessary though.)
[quote="moogra"]I realize it isn't needed but during a program you can't implement another program you already made.[/quote]
Actually, you can.
[hide="hidden cause its really long"]For example, consider:
[code]:ClrHome
:Disp "OMG A MAGIC MOD"
:Disp "FINDER!"
:Disp "I FIND A MOD B."
:Prompt A,B
:B*fPart(A/B)
:Disp Ans[/code]
For something as simple (and pointless) as this, there isn't really a point of doing this, but for large programs, you can avoid typing out the formula by instead what's known as a subroutine. For example, make a prgmMOD with the code like:
[code]:B*fPart(A/B)
:Return[/code]
(The return command is found under CTL, "number" E.)
Then, for the original program, you can do something like:
[code]:ClrHome
:Disp "OMG A MAGIC MOD"
:Disp "FINDER!"
:Disp "I FIND A MOD B."
:Prompt A,B
:prgmMOD
:Disp Ans[/code]
You can either make prgm by going under CTL, D, then manually entering the name, or by selecting prgmMOD in the list under EXEC. Note that typing prgm in lowercase letters enabled my MirageOS will NOT work.
This command makes the calculator run the program (subroutine) specified. The return command in prgmMOD will make the calculator "return" to the original program, at the line it left off. (If it is used in a program that is not a subroutine, it is ignored... or it terminates the program... I forget.)
It may seem like a bunch of extra work just to do a simple function, but it can be really useful. For example, if you need to use a specific function multiple times in a program, you can use the program instead. Supposing you have few enough programs, the subroutine can easily be accessed through prgm-left arrow-number. (You can name your subroutines AAAAA or something to make sure of this.) Be sure to set the needed variables before you enter the subroutine. Also, you can use this technique to split a huge and potentially uneditable program into smaller, manageable parts.[/hide]
Hoped it helped... but I don't really have too much experience with any programming. I just figured out the stuff 'cause I was bored. :D

[quote="moogra"][quote="undefined117"]idk if the 84+ has it, but my 89, in addition to [b]fPart()[/b], has a [b]remain(a,b)[/b] function that returns the remainder when [b]a[/b] is divided by [b]b[/b].[/quote]

The 84+ Silver doesn't have that. I don't know if the 84+ has it but I doubt it.[/quote]

It does not.

Also, be careful when using the *fpart(a/b) code, because the floating point processors have a tendency to create rounding errors while dealing with such things (I have personally had this happen many times and it took me forever to figure out the reason; I thought it was bad code at first  :roll: )...

This is probably WAY more advanced than what you want, but if you learn z80 assembly, then you can create your own functions and integrate them into the OS  :D

erm buggy asm programs on my 89 results in me having to force-reset the RAM.. I've heard of cases where the OS got corrupted, but I've never experienced that one.

There are emulators available that lets you test your code on your pc before you get it on your calc. [url]http://www.ticalc.org[/url] has some.

[quote="moogra"]Well does anyting bad happen if you screw up asm?[/quote]

It is VERY difficult to do anything worse than forcing yourself to clear your RAM (luckily, an application, Omnicalc allows you to recover your lost memory), but I did manage to mess up my first calculator somehow (I don't know why, but it wasn't anything I did, but I got a free replacement  :) )...

As long as you test everything on emulators, though, you will be fine.  And, TI has done a lot to make it difficult to write to the flash memory, which is the only thing that you can mess up, anyway...


calculus derivative integration logarithms

Base 10 or base e?

edit:  for any base b, the derivitive of log_b x is 1/(ln b *x), since
[tex]
\int x^{-1}\, dx = \ln x + C,
[/tex]

Letting x be x ln b, we get
[tex]
\displaystyle\int (x \ln b)^{-1} \, dx = \frac{1}{\ln b} \int \frac{1}{x}\, dx

\displaystyle= \frac{\ln x}{\ln b} = \log_b x.
[/tex]

Thus the answer is either [tex]\frac{1}{x}[/tex] if your logarithm is base e, or [tex]\frac{1}{x\ln 10}[/tex] if your log is base 10.


[hide]The slope-intercept form of the line is $y=2x+4.$ The perpendicular slope is $-\frac{1}{2}.$ The answer is $\text{B}.$[/hide]

[hide="Answer"]

Well to find the slope of a perpendicular line you find the opposite reciprocal of the original line.

so the slope of the original line was 2 so the slope of the perpendicular line is $-\frac{1}{2}$ or B.
[/hide]

Extension: What is the slope of the line parallel to 2y=4x+8?

To find the slope, simply refer to this point-slope formula:

y = mx + b (m is the slope, and b is the y-intercept)

You have this equation, 2y = 4x + 8, so if just rewrite it to the point-slope formula and find m, you'll get your slope.

Try it out! Answer is hidden if you need it.

[hide]

Isolate y

2y [÷2] = ( 4x + 8 ) ÷ [2]

y = 2x + 4

The value of m is 2, so the slope of this equation is 2. [/hide]

i would say that the slope should be -1/2 so B is the answer..

[quote="Kevin K."]What is the slope of a line perpendicular to the line 2y = 4x + 8?

(A) -1/4
(B) -1/2
(C) -1
(D) -2
(E) -4
[/quote]

[hide="answer"]
$2y = 4x+8$
$y = 2x+4$
m $= 2$
since the slope of a perpendicular is equal to the negative reciprocal of the original slope, the new slope has to be -1/2[/hide]


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Find the equation of the tangent line at the given point on the curve.
3x^2 + 4xy + y^2 + x - 2y = -7 ; ( -1 , 3)
Thanks

Step 1: Check to see whether that point is on the curve. It is.

Step 2: Use implicit differentiation to find an expression for $ \frac{dy}{dx}$ in terms of $ x$ and $ y.$

Step 3: Put $ x\equal{}\minus{}1$ and $ y\equal{}3$ in that formula to get the slope of the tangent line.

Step 4: You have the slope of a line, and a point it goes through. You know how to finish this, right?


For any three positive reals a, b, c, we have

$\frac{a^3+2abc}{b+c}+\frac{b^3+2abc}{c+a}+\frac{c^3+2abc}{a+b} \ge \frac{1}{2}(a+b+c)^2$

\[
\begin{gathered}
  \sum {\frac{{a^3  + 2abc}}
{{b + c}} - \frac{1}{2}a(a + b + c)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{2a^3  + 2abc - a^2 b - ab^2  - ac^2  - a^2 c}}
{{b + c}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{a(a - b)(a - c)}}
{{b + c}}}  + \sum {\frac{{a(a^2  + bc - b^2  - c^2 )}}
{{b + c}}}  \geqslant 0 \hfill \\
\end{gathered}
\]
First one immediately true by vornicu-schur.
For second one apply vornicu-schur again,
\[
\begin{gathered}
  \sum {\frac{{a(a^2  + bc - b^2  - c^2 )}}
{{b + c}}}  \hfill \\
   \geqslant \sum {\frac{{a(ab + ac - b^2  - c^2 )}}
{{b + c}}}  \hfill \\
   = \sum {\frac{{a(b(a - b) + c(a - c))}}
{{b + c}}}  \hfill \\
   = \sum {\frac{{ab(a - b)}}
{{b + c}} + \frac{{ac(a - c)}}
{{b + c}}}  \hfill \\
   = \sum {\frac{{ab(a - b)}}
{{b + c}} + \frac{{ba(b - a)}}
{{a + c}}}  \hfill \\
   = \sum {ab(a - b)\left( {\frac{1}
{{b + c}} - \frac{1}
{{a + c}}} \right)}  \hfill \\
   = \sum {\frac{{ab(a - b)^2 }}
{{(a + c)(b + c)}}}  \geqslant 0 \hfill \\
\end{gathered}
\]

[quote="xtar"]For any three positive reals $ a, b, c$, we have

$ \frac {a^3 \plus{} 2abc}{b \plus{} c} \plus{} \frac {b^3 \plus{} 2abc}{c \plus{} a} \plus{} \frac {c^3 \plus{} 2abc}{a \plus{} b} \ge \frac {(a \plus{} b \plus{} c)^2}{2}.$[/quote]$ \sum{\frac{a^3 \plus{} 2abc}{b \plus{} c}}\minus{}\frac {(a \plus{} b \plus{} c)^2}{2}$

$ \equal{}\frac{a\plus{}b\plus{}c}{2(b\plus{}c)(c\plus{}a)(a\plus{}b)}\sum{(b\minus{}c)^2\left[bc\plus{}(b\plus{}c\minus{}a)^2\right]}\geq0.$


limit
trigonometry
calculus
calculus computations

Why is the [b] limit [/b] of the following 1?
$\lim_{x \rightarrow \infty} x.\sin(\frac{1}{x})$

Why is not [b]0[/b] if :
$\lim_{x \rightarrow \infty} x.\sin(\frac{1}{x}) = \ \ \lim_{x \rightarrow \infty} (x) \ .\ \lim_{x \rightarrow \infty}\sin(\frac{1}{x}) \ = \lim_{x \rightarrow \infty} (x) . \ \ 0 = 0$


geometry
circumcircle
perimeter
IMO Shortlist
perpendicular

Let $O$ be an interior point of acute triangle $ABC$. Let $A_1$ lie on $BC$ with $OA_1$ perpendicular to $BC$. Define $B_1$ on $CA$ and $C_1$ on $AB$ similarly. Prove that $O$ is the circumcenter of $ABC$ if and only if the perimeter of $A_1B_1C_1$ is not less than any one of the perimeters of $AB_1C_1, BC_1A_1$, and $CA_1B_1$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let us denote by $p_{XYZ}$ the perimeter of a triangle $XYZ$.

"$\Rightarrow$" If $O$ is the circumcenter of $\triangle{ABC}$, then $A_{1},B_{1}, C_{1}$ are the midpoints of the sides of the triangle, therefore $p_{A_{1}B_{1}C_{1}}=p_{AB_{1}C_{1}}= p_{A_{1}BC_{1}}= p_{A_{1}B_{1}C}$.

"$\Leftarrow$"  Assume WLOG, that $p_{A_{1}B_{1}C_{1}}$ is greater than $p_{AB_{1}C_{1}}$,  $p_{A_{1}BC_{1}}$, $p_{A_{1}B_{1}C}$.

And let $\alpha_{1}$, $\alpha_{2}$, $\beta_{1}$, $\beta_{2}$, $\gamma_{1}$, $\gamma_{2}$ be $\angle{B_{1}A_{1}C}$, $\angle{C_{1}A_{1}B}$, $\angle{C_{1}B_{1}A}$, $\angle{A_{1}B_{1}C}$, $\angle{A_{1}C_{1}B}$, $\angle{B_{1}C_{1}A}$.

Now suppose $\gamma_{1}, \beta_{2}\geq  \alpha$. If $A_{2}$ is the fourth vertex of the parallelogram $B_{1}AC_{1}A_{2}$, then these conditions imply that $A_{1}$ is in the interior or on the sides of $\triangle{B_{1}C_{1}A_{2}}$, therefore $p_{A_{1}B_{1}C_{1}}\leq p_{A_{2}B_{1}C_{1}}\leq p_{AB_{1}C_{1}}$.

But, lets see that if one of the inequalitites $\gamma_{1}\geq \alpha$ and $\beta_{2}\geq \alpha$ is strict, then $p_{A_{1}B_{1}C_{1}}< p_{AB_{1}C_{1}}$, which contradicts our assumption.

Therefore $\beta_{2}\geq \alpha$ implies $\gamma_{1}\leq  \alpha$, and the analogues. $(\star)$

Now, because of the symmetry, we can suppose WLOG, that $\gamma_{1}\leq \alpha$, then since $\gamma_{1}+\alpha_{2}=180-\beta=\alpha+\gamma$, it follows that $\alpha_{2}\geq \gamma$.

From $(\star)$ we deduce that $\beta_{1}\leq \gamma$, thus $\gamma_{2}\geq \beta$, $\alpha_{1}\leq \beta$, and $\beta_{2}\geq \alpha$.

Now putting them all together, we obtain that:
$\gamma_{1}\leq \alpha\leq \beta_{2}$,  $\alpha_{1}\leq \beta\leq \gamma_{2}$,  $\beta_{1}\leq \gamma\leq \alpha_{2}$.

Now since $OA_{1}BC_{1}$ and $OB_{1}CA_{1}$ are cyclic quadrilaterals, we have that $\angle{A_{1}OB}=\gamma_{1}$ and $\angle{A_{1}OC}=\beta_{2}$.

Thus $\frac{BO}{CO}=\frac{\cos{\beta_{2}}}{\cos{\gamma_{1}}}$, so $BO\leq CO$. Now in a similar way prove that $CO \leq  AO$ and $AO \leq  BO$. Thus $OA=OB=OC$, therefore $O$ is the circumcenter.

Let $P$ be the circumcenter of $ABC$.  Let $D,E,F$ be the midpoints of $BC,AC,AB$, respectively and $G$ the foot of $A$ to $BC$.

Obviously if $O=P$, then the perimeters of all four triangles are equal.

Lemma 1: If $O\in AP, O\neq P$ then the perimeter $BA'C'$ is greater than the perimeter of $A'B'C'$.

Proof:  It is easily proven that $B'C'\|EF\|BC$.  Draw a line $l$ parallel to $A'B'$ such that $l$ passes through $C'$.  Let $Q=l\cap BC$.   We have that $Q$ is between $B$ and $A'$ because, \[A'B\ge BD>B'C'\] if $\angle C\ge \angle B$ and \[A'B=DB-DA'=DB-DG+DG(AO/AP)=DB-DG+DG(B'C'/FE)>DB(B'C'/FE)=B'C'\] if $\angle B\ge \angle C$.  By the triangle inequality $BC'+BQ>C'Q=A'B'$ so $BC'+BQ+B'C'=BC'+BA'>A'B'+B'C'$, as desired. $\boxed{}$.

Lemma 2: If $O\in APE$ with feet $A',B',C'$ and $O'\in AP$ with feet $A''B'C''$.  Then the $BA''+BC''-B'A''-B'C''> BA'+BC'-B'A'-B'C'$.

Proof: $BA''>BA'$ and $BC''>BC'$, obviously.  We also have that $-B'A''>-B'A'$ and $-B'C''>-B'C'$ because $90^{\circ}>\angle AC''B'>\angle AC'B'$ and$90^{\circ}>\angle CA''B'>\angle CA'B'$ .  Adding the four inequalities gives the desired. $\boxed{}$.

Thus, for any point $O$ in without loss of generality $APC$, we have that the perimeter of $A'B'C'$ is less than the perimeter of $BC'A'$, as desired.

Here's a much longer solution... I'm not certain that this is correct.
[hide]
Let $R$ be the circumradius of $\triangle ABC$.

[b]Lemma 1[/b]: $B_1C_1 = \frac{OA \cdot BC}{2R}$.
[i]Proof:[/i] $AC_1OB_1$ is cyclic, and has diameter $AO$, so by the law of sines $\frac{B_1C_1}{\sin A} = OA$, so $B_1C_1 = OA \sin A = \frac{OA \cdot BC}{2R}$.

[b]Lemma 2[/b]: $2R \geq \max(AB, BC, CA)$.
[i]Proof:[/i] This follows trivially from the fact that the length of a longest possible chord in a circle is the diameter.

[b]Lemma 3[/b]: Let $P$ be any point not in the exterior of $\triangle ABC$. Then $2R \geq PA + PB + PC - \max(PA, PB, PC)$.
[i]Proof:[/i] Let $f_1(P) = PB+PC$, $f_2(P) = PC+PA$, $f_3(P) = PA+PB$, and $f(P) = \min(f_1(P), f_2(P), f_3(P))$. We wish to show that $f(P) \leq 2R$.

We will first prove this when $P$ is on the perimeter of $\triangle ABC$. Without loss of generality, let it be on side $AB$. Then $f(P) = PA + PB + PC - \max(PA,PB,PC) \leq PA + PB = AB \leq 2R$

Suppose now that $O'$ is the circumcenter of $\triangle ABC$. If $P = O'$, then we are immediately done. Otherwise, without loss of generality suppose that $O'$ is the origin, let $u$ be the vector from $O'$ to $P$, and for any nonnegative real $t$ let $\alpha(t) = tu$. (The image of $\alpha$ would be the ray $\overrightarrow{O'P}$.) Let $g_i(t) = f_i(\alpha(t))$ for $i = 1,2,3$ and $g(t) = \min(g_1(t), g_2(t), g_3(t))$. We may suppose without loss of generality that $g_1(1) = g(1)$. We claim that $g_1(t) = g(t)$ for all nonnegative $t$. If $g_1(t) \neq g(t)$ for some $t$, then $g_j(t) < g_1(t)$ for some $j$, which we may suppose without loss of generality equals 2. Because $g_2(1) \geq g_1(1)$, the intermediate value theorem implies the existence of some $c \in [\min(t, 1), \max(t, 1)]$ such that $g_1(c) = g_2(c)$. Noting that $t \neq 0$ (since $g_1(0) = g_2(0)$ while $g_1(t) \neq g_2(t)$), we find that $c \geq \min(t, 1) > 0$.

Letting $X = \alpha(c)$, we find that $BX + CX = f_1(X) = f_2(X) = AX + CX$, so $AX = BX$. It follows that $X$ must be equidistant from $A$ and $B$, so $X$ must lie on the perpendicular bisector of $AB$. Because $X$ also lies on the line through $\alpha(t)$ and $O$, we find that either $X = O$ or $\alpha(t)$ lies on the perpendicular bisector of $AB$. The former case is impossible since $c \neq 0$. In the latter case, we find that every point on the image of $\alpha$ is equidistant from $A$ and $B$, so $g_1(t) = g_2(t) < g_1(t)$, which is impossible. Hence, it is safe for us to assume that $g(t) = g_1(t)$ for all nonnegative $t$.

We now observe that $g = g_1$ is convex for all nonnegative $t$; this can easily be verified from the fact that as $P$ varies along $\overrightarrow{O'P}$, the distances $PB$ and $PC$ are convex in the $x$-coordinate of $P$ (which can easily be verified with coordinates), and the fact that the sum of convex functions is convex.

If $O'$ lies inside or on the perimeter of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect a side of the triangle at $M$, and let $P = \alpha(m)$. Because $g = g_1$ is convex over $[0,m]$, the maximum value of $g$ is attained at either $0$ or $m$. Since $P$ is between $O'$ and $M$, $0 \leq 1 \leq m$, so $f(P) = g(1) \leq \max(g(0), g(m)) = \max(f(O), f(P)) \leq 2R$.

If $O'$ lies in the exterior of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect two sides of the triangle at $Q$ and $R$. Let $Q = \alpha(q)$, $R = \alpha(r)$, and suppose without loss of generality that $q < r$. Because $g = g_1$ is a convex function over $[q,r]$, the maximum value of $g$ is attained at either $q$ or $r$. Since $P$ is between $O'$ and $R$, $q \leq 1 \leq r$, so $f(P) = g(1) \leq \max(g(0), g(r)) = \max(f(O'), f(R)) \leq 2R$, which completes the proof of this lemma.

[b]Lemma 4[/b]: For any point $P$ in the interior of $\triangle ABC$, $\min(AP, BP, CP) \leq R$.
[i]Proof:[/i] Let $O'$ be the circumcenter of $\triangle ABC$, and let $O'_A$, $O'_B$, and $O'_C$ be the projections of $O'$ onto $BC$, $CA$, and $AB$, respectively. Without loss of generality, let $P$  lie in the interior of or on the perimeter of $AO'_B O' O'_C$. Then $AP$ must be less than or equal to the length of the diameter of the circumcircle of $AO_B' O' O_C'$.

--------------------------------------------------------------------------------------------------------------------

If $O$ is the circumcenter of $\triangle ABC$, then the only if direction of this problem follows immediately from lemma 2.

Without loss of generality, let $OC = \max(OA, OB, OC)$. We will show that if $O$ is not the circumcenter, then the perimeter of $\triangle A_1 B_1 C_1$ is less than the perimeter of $\triangle A1B1C$.

By lemma 2, this statement is equivalent to $\frac{AO \cdot BC}{2R} + \frac{BO \cdot AC}{2R} < CB_1 + CA_1 = OC(\cos B_1 CO + \cos OCA_1)$. By the law of cosines, $\cos B_1 CO = \cos ACO = \frac{OC^2 + AC^2 - AO^2}{2OC \cdot AC}$ an $\cos A_1 CO = \cos BCO = \frac{OC^2 + BC^2 - OB^2}{2OC \cdot BC}$, so the desired inequality can be rewritten as
\[ BC \cdot \frac{AO}{R} + AC \cdot \frac{BO}{R} < \frac{OC^2 - OB^2}{BC} + \frac{OC^2 - OA^2}{AC} + BC + AC.\]

By lemma 4, we can suppose without loss of generality that $OA < R$. Then $BC \cdot \frac{AO}{R} < BC$. Furthermore, by the triangle inequality $OC + OB > BC$ and $OC+OA > AC$, so it is enough to show that
\[ AC\left(\frac{BO}{R} - 1\right) \leq 2OC - OB - OA. \]
If $\frac{BO}{R} - 1 \leq 0$, then we are done since the right-hand side is nonnegative. If $\frac{BO}{R} - 1> 0$, then by lemma 2, $AC\left(\frac{BO}{R} - 1\right) \leq 2BO - 2R$, so it is enough to show
\[ 2OC + 2R \geq 3BO + OA. \]
But $2OC \geq 2OB$, and by lemma 3, $2R \geq OB + OA$; adding these inequalities yields the desired inequality.
[/hide]

One direction is obvious, so now assume that all three perimeter inequalities hold. Let $X$ be the orthocenter of $A_1B_1C_1$; we will show $O=X$. If this is not the case, $O$ must without loss of generality lie within triangle $B_1OC_1$ (where $O$ can lie on the sides but can't be $X$). This is true since it's easy to show  $O$ lies within triangle $A_1B_1C_1$ (if $A_1B_1C_1$ is obtuse, we choose $B_1OC_1$ with $\angle A_1$ obtuse). Now let $x=\angle C_1B_1O$ and let $y=\angle B_1C_1O$. We know $x\le \angle C_1B_1X=90-\angle C_1$ and $y \le \angle B_1C_1X=90-\angle B_1$. This also implies that if $z=180-x-y$, then $z\ge 180-\angle A_1$ (and since $O$ lies in $A_1B_1C_1$, $z\ge \angle A_1$. Now we note the following:
\[
\frac{AB_1}{B_1C_1}+\frac{AC_1}{B_1C_1}=\frac{\sin( \angle AC_1B_1 )+\sin(\angle AB_1C_1)}{\sin \angle A} =\frac{\cos(x)+\cos(y)}{\sin z}
\]
Meanwhile.
\[
\frac{A_1B_1}{B_1C_1}+\frac{A_1C_1}{A_1B_1}=\frac{\cos(90-\angle B_1)+\cos(90-\angle C_1)}{\sin(\angle A_1)}
\]
Now $0<x\le 90-\angle C_1<180$ so $\cos(x)\ge \cos(90-\angle C_1)$. Similarly, $\cos(y)\ge \cos(90-\angle B)$. Meanwhile, $z\ge \text{max}(180-\angle A_1,\angle A_1)$, so $\sin(z)\le \sin(A_1)$ in all cases. Putting these together yields
\[
AB_1+AC_1\ge A_1B_1+A_1C_1
\]
with equality when $x=90-\angle C_1$, $y=90-\angle B_1$, or equivalently when $O=X$.


Let x,y,z be positive real numbers. Prove that:
x/(x+ \sqrt (x+y)(x+z)) + y/(y+ \sqrt(y+z)(y+x)) + z/(z+ \sqrt(z+x)(z+y))  \leq 1

I think this was posted before.

Main idea is to prove that

x/(x + \sqrt(x + y)(x + z)) \leq \sqrtx/(\sqrtx + \sqrty + \sqrtz)

for all x, y, z. Then you just have to add these inequalities.

Another approach: sqrt((x+y)(x+z))>=x+sqrt(yz). The inequality reduces to proving that for all x,y,z>0 with xyz=1 we have sum 1/(x+2)<=1, which is simple.


trigonometry
inequalities proposed
inequalities

Let $ x,y,z \in \mathbb{R}^\plus{}$ and satisfy the condition $ x\plus{}y\plus{}z\equal{}xyz$ prove that \[ xy\plus{}yz\plus{}zx \geq 3\plus{}\sqrt{x^2\plus{}1}\plus{}\sqrt{y^2\plus{}1}\plus{}\sqrt{z^2\plus{}1}\]

We have $ x\equal{}\tan A,y\equal{}\tan B,z\equal{}\tan C$ with $ A\plus{}B\plus{}C\equal{}180^0$ and $ 0<A,B,C<90^0$ then we must to prove:
$ \sum\tan A\tan B\geq 3\plus{}\sum\frac{1}{\cos A}$
$ \Leftrightarrow \tan A\tan B\tan C\sum\frac{1}{\tan A}\geq 3\plus{}\sum\frac{1}{\cos A}$
$ \Leftrightarrow \sum\tan A\sum\frac{1}{\tan A}\geq 3\plus{}\sum\frac{1}{\cos A}$
$ \Leftrightarrow 3\plus{}\sum\frac{\tan B\plus{}\tan C}{\tan A}\geq 3\plus{}\sum\frac{1}{\cos A}$
$ \Leftrightarrow \sum\frac{\tan B\plus{}\tan C}{\tan A}\geq \sum\frac{1}{\cos A}$
$ \Leftrightarrow \sum\frac{\sin (B\plus{}C)}{\cos B\cos C\tan A}\geq \sum\frac{1}{\cos A}$
$ \Leftrightarrow \sum\frac{\cos ^2A}{\cos A\cos B\cos C}\geq \sum\frac{\cos A\cos B}{\cos A\cos B\cos C}$
$ \Leftrightarrow \sum \cos^2A\geq\sum\cos A\cos B$. OK!

[quote="tdl"]We have $ x \equal{} \tan A,y \equal{} \tan B,z \equal{} \tan C$ with $ A \plus{} B \plus{} C \equal{} 180^0$ and $ 0 < A,B,C < 90^0$ then we must to prove:
$ \sum\tan A\tan B\geq 3 \plus{} \sum\frac {1}{\cos A}$
$ \Leftrightarrow \tan A\tan B\tan C\sum\frac {1}{\tan A}\geq 3 \plus{} \sum\frac {1}{\cos A}$
$ \Leftrightarrow \sum\tan A\sum\frac {1}{\tan A}\geq 3 \plus{} \sum\frac {1}{\cos A}$
$ \Leftrightarrow 3 \plus{} \sum\frac {\tan B \plus{} \tan C}{\tan A}\geq 3 \plus{} \sum\frac {1}{\cos A}$
$ \Leftrightarrow \sum\frac {\tan B \plus{} \tan C}{\tan A}\geq \sum\frac {1}{\cos A}$
$ \Leftrightarrow \sum\frac {\sin (B \plus{} C)}{\cos B\cos C\tan A}\geq \sum\frac {1}{\cos A}$
$ \Leftrightarrow \sum\frac {\cos ^2A}{\cos A\cos B\cos C}\geq \sum\frac {\cos A\cos B}{\cos A\cos B\cos C}$
$ \Leftrightarrow \sum \cos^2A\geq\sum\cos A\cos B$. OK![/quote]
really nice  :clap2:
This is My solution :
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2} \geq \frac{1}{xy}\plus{}\frac{1}{yz}\plus{}\frac{1}{zx} \equal{} 1$

$ \rightarrow  x^2y^2\plus{}y^2z^2\plus{}z^2x^2 \geq x^2y^2z^2$

$ \rightarrow  (xy\plus{}yz\plus{}zx)^2 \geq 2xyz(x\plus{}y\plus{}z)\plus{}x^2y^2z^2 \equal{} 3(x\plus{}y\plus{}z)^2$

$ \because  (xy\plus{}yz\plus{}zx\minus{}3)^2 \equal{} (xy\plus{}yz\plus{}zx)^2\minus{}6(xy\plus{}yz\plus{}zx)\plus{}9 \geq 3(x^2\plus{}y^2\plus{}z^2)\plus{}9$

$ \therefore xy\plus{}yz\plus{}zx \geq 3\plus{}\sqrt{3(x^2\plus{}y^2\plus{}z^2)\plus{}9} \geq 3\plus{}\sqrt{x^2\plus{}1}\plus{}\sqrt{y^2\plus{}1}\plus{}\sqrt{z^2\plus{}1}$


Let $a, b, c$ three positive real numbers such that $abc=1$. Show that \[a^{2}(b+c+1)+b^{2}(c+a+1)+c^{2}(a+b+1)\geq 4\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right).\]

Is this the right form?

We have that

$LHS = \sum \left( \frac{a}{c}+\frac{a}{b}\right)+\sum a^{2}$
$\geq \sum \left( \frac{a}{c}+\frac{a}{b}\right)+3$
$\geq 2 \sum \sqrt{\frac{a}{b}+\frac{a}{c}}$
$\geq 4 \sum \sqrt \frac{a}{b+c}$.

The first are AM-GM's and the last one is AM-HM (or Cauchy).

But there's no equality case... Maybe when $a \to 0$? I'm not sure.

PS: I was trying to solve it, and I reached the (well-known) inequality I posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=628497#628497]here[/url].  Jan's idea helped me solve this one. We can also use the inequality posted there, but in this way it became more obvious to replace $\sum a^{2}$ by $3$.


Harvard
college
calculus
vector
Princeton
Yale
number theory

I was wondering whether the math department (specifically pure math) of University of Chicago is comparable to that of Priceton (first choice right now), Harvard, or Yale.  I'm pretty sure I'll get into the U of C, and I was wondering whether it would be beneficial to attend for college.

Also, how does a great variety of math classes affect college admissions for someone planning on being a math major?  I go to ASMSA, a math and science school in Arkansas, which has allowed me to take the following classes:

10th:
AP BC Calculus -- B's both semesters (horrible teacher =/ )
Number Theory -- A for semester (one semester class)
Topics in Pure Mathematics -- A's in both semesters
Linear Algebra -- A for semester (one semester class)

I am taking in the fall the following classes:

Vector Calculus
AP Statistics
Math Modeling (one semester)
Differential Equations
and possibly a self-study of Group Theory

Other than BC Cal, I made two B's in AP Physics C, but those have been the only B's in my academic career, and I think I'll make A's in all other classes in the future.  Is there a change of me getting into Princeton or Harvard with this sort of stuff?

Your chance for getting into the very most selective schools may depend on how many of your own high school classmates outrank you, but what is "outranking" depends very much on what extracurricular accomplishments each of you have besides your course grades. Based on what you have written, it would be worthwhile to apply to all the schools you are interested in, but I don't predict people's chances, because I have found that I am usually wrong in such predictions.  :blush:

The University of Chicago definitely has a very good mathematics department. It's top of the line for pure mathematics, although the other schools you mention are good too. (Yale is mentioned less often as a strong undergraduate mathematics program than the other schools, on this forum.)

An issue you might want to consider is how to learn some math outside your school classes, through self-study, summer programs, or whatever else is available to you in your community.

Good luck in your applications when the time comes.

You should get involved in some extracurricular math programs because it seems you enjoy it and are rather good at it.  Perhaps try the AMC Competitions.

By the way, my class is not ranked, but I think I'm number two anyway.


Fie $x,y,z \geq  0$ cu $xyz=x+y+z+2$. Sa se demonstreze ca:

$\sum{xy}\geq 2\cdot\sum{x}$

PS: m-ar interesa o rezolvare nefolosind substitutiile $x=\frac{a+b}{c}$, $y=\frac{b+c}{a}$ si $z=\frac{c+a}{b}$

Pai ideea asta cu substitutiile este esentiala. Dupa ce faci aceste substitutii, care verifica ipoteza( asta era si ideea), treci calare pe inegalitatea aia. :) Pai uite la ce se reduce(Slava Domnului ca nu e Muerhead :P):

 \[a(a+b)(a+c)+b(a+b)(b+c)+c(c+a)(b+c)\geq 2\left(ab(a+b)+bc(b+c)+ca(c+a)\right)\Leftrightarrow\]
 \[a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a),\]
 asta dupa mici calcule. :wink:

O aplicatie frumoasa este urmatoarea:
Cu aceeasi conditie,$xyz=x+y+z+2$,sa se demonstreze inegalitatea:\[ (x-2)^{2}+(y-2)^{2}+(z-2)^{2}+\frac{18}{x+y+z}-3\geq0\]
Ce parerea aveti?

Pai asta a ta este o aplicatie a iengalitatii postate de cosmin plus un Schur.
Pai, hai sa vedem cum vin calculele:

Ce-ai scris tu este echivalent cu

        \[\sum x^{2}+12-4\sum x+\frac{18}{\sum x}\geq 3\Leftrightarrow\]
       \[\sum x^{2}+9+\frac{18}{\sum x}\geq 4\sum x,\]

 dar cum din ipoteza avem $18=9xyz-9\sum x$, rezulta ca

               \[\sum x^{2}+9+\frac{9xyz}{\sum x}-9\geq 4\sum x.\]

  Acum folosind inegalitatea postata de cosmin si Schur sub forma asta

             \[2\sum xy-\sum x^{2}\leq\frac{9xyz}{\sum x},\]

   totul este clar. :)

O observatie:Inegalitatea $\sum xy\geq 2\cdot\sum x$ este echivalenta cu \[\boxed{(x-1)(y-1)(z-1)\leq 1}\]
Si asta se restrange,folosind substitutia,in $ \prod(b+c-a)\leq abc$


Find all pairs of integers $ (x, y)$ satisfying

$ 1\plus{}x^{2}y \equal{} x^{2} \plus{}2xy \plus{}2x\plus{}y$.

[hide]Writing this as $ y\equal{}\minus{}{x^2\plus{}2x\minus{}1\over x^2\minus{}2x\minus{}1}\equal{}\minus{}1\minus{}{4x\over x^2\minus{}2x\minus{}1}$, we must find all integer $ x$ such as the last fraction is integer.

First let's check the domain where the denominator is negative. Since the roots of $ x^2\minus{}2x\minus{}1$ are $ 1\pm\sqrt{2}$, those values of $ x$ are $ x\in\{0,1,2\}$. For all of them we get an integer $ y$, so the first three pairs are $ (0,1),(1,\minus{}1),(2,\minus{}7)$

Now we check $ x\geqslant 3\lor x\leqslant \minus{}1$. We have two inequalities:

1. $ {4x\over x^2\minus{}2x\minus{}1}\geqslant 1\iff 4x\geqslant x^2\minus{}2x\minus{}1$ (note: the denominator is positive in the examined domain), and that yields $ x^2\minus{}6x\minus{}1\leqslant 0$. The corresponding values for $ x$ are $ x\in\{3,4,5,6\}$. Checking shows that only $ x\equal{}3$ yields an integer $ y$, so another solution is $ (3,7)$

2. $ {4x\over x^2\minus{}2x\minus{}1}\leqslant \minus{}1\iff 4x\leqslant \minus{}x^2\plus{}2x\plus{}1\iff x^2\plus{}2x\minus{}1\leqslant 0$. The corresponding values for $ x$ are $ x\in\{\minus{}2,\minus{}1\}$, but only $ x\equal{}\minus{}1$ yields an integer $ y$, hence another solution is $ (\minus{}1,\minus{}1)$

Therefore the complete set of solutions is $ \{(\minus{}1,\minus{}1),(0,1),(1,\minus{}1),(2,\minus{}7),(3,7)\}$[/hide]


Find all the groups of 2005 positives integers, satisfying the next condition:
    abcd/a^4+b^4+c^4+d^4    for anyone four elements a, b, c , d in the group.

[quote=conejita]If my proof is correct, there are two solutions: $a_1 = a_2 = \dots = a_{2005} = k$ and $a_1 = a_2  = \dots = a_{2004} = k$, $a_{2005} = 3k$, where $k$ is any positive integer.[/quote]
how do you get this?

My proof is a bit long, but let me share a sketch of it:

Assume WLOG that $a_1 \leq a_2 \leq \dots \leq a_{2005}$. Also, assume that $gcd(a_1, a_2, \dots, a_{2005}) = 1$ (if not equal to 1, then divide all those numbers by its gcd). First, show that $a_{2005} | a_i^4 - a_1^4 $ for $i = 2, \dots, 2004$ (you can do this by selecting appropriate indexes and subtracting some of them). Now, let $gcd(a_1, a_{2005}) = k$. If $k >1$, then let $p$ any prime that divides $k$. Therefore, $p$ divides all terms $a_i$, which contradicts the assumption that $gcd(a_1, a_2, \dots, a_{2005}) = 1$. Hence, $k = 1$.

Now, show that $a_{2005} | 3a_1^4$. This implies that $a_{2005} | 3$. If $a_{2005} = 1$, then all values are equal to 1, leading to the first solution. If $a_{2005} = 3$, then all terms in the sequence belong to the set $\{1,2,3\}$. To find the second solution, use mod3 to conclude that you can only have one term different than 1, which is $a_{2005} = 3$.

Let me know what you think!


calculus
integration
logarithms
calculus computations

Hi,
This integral below may be simple to many here, im just having trouble with it.
If you can briefly tell me the steps to solve it - id greatly appreciate your help.

[img]http://www.egesi.com/question.gif[/img]

[hide="answer"]$\int{\frac{1}{x+x^{\frac{1}{3}}}dx= \frac{3\ln({\sqrt[3]{x^2}+1)}}{2}}$ now just plug in values for the defenite integral[/hide]

kunny is right, of course, but I think there's some pedagogical value to doing this in a less polished fashion.

The key is the substitution $u=x^{1/3}$. This isn't a "natural" subsitution so it will have to be "forced." The way to force a substituion is to write it backwards:

$x=u^3$ so $dx = 3u^2du.$

I like to go ahead and change the limits of integration. As $x$ ranges from 1 to 8, $u$ ranges from 1 to 2. Our integral becomes

$\int_1^2\frac{3u^2du}{u^3+u}=\int_1^2\frac{3u\,du}{u^2+1}$

Now we have a "natural" substitution: let $v=u^2+1,\,dv=2u\,du.$ As $u$ ranges from 1 to 2, $v$ ranges from 2 to 5. The integral becomes

$\frac32\int_2^5\frac{dv}v =\frac32\ln v|_2^5=\frac32\ln(\frac52).$

I suppose this following way isnt bad also, it's along the lines of kunny's idea:

$\int_{1}^{8} \frac{1}{x + x^{1/3}} = \int_{1}^{8} \frac{x^{-1/3}}{x^{2/3} + 1} \,dx$

let $u = x^{2/3} \rightarrow du = \frac{2}{3} x^{-1/3} \,dx \rightarrow \frac{3}{2} \frac{\,du}{x^{-1/3}} = \,dx$

so we get $\frac{3}{2} \int_{1}^{4} \frac{1}{u+1} \,du$

and so on, it comes out pretty easily


pigeonhole principle
number theory
relatively prime

Show that every prime other than 2 and 5 divides infinitely many of the integers 1, 11, 111, 1111,...

Oh this one is easy.

Just use BISSP...

BISSP = Because I Said So Postulate

[hide]We only need to prove that it divides some number in the form 1111...111, because if we double that string, that number is divisible by the prime as well.

We must then show that there exists some k such that 9pk = 10^n - 1, or 10^n :equiv: 1 mod 9p. Obvious here, right?[/hide][/hide]

Why obvious?  Doesn't look obvious to me.  I've not seen this problem done that way, that I can remember -- doesn't mean it isn't doable, though.

10^n :equiv: 1 mod 9p <-- since 10^n :equiv: 1 mod 9, it suffices to show that there exists n such that 10^n :equiv: 1 mod p. n = p-1 works.

Oh, okay.  Except that you have to make a special allowance for p = 3, but that's pretty trivial.  The way I've seen this done is to use [hide]the pigeonhole principle.[/hide]

Hm I was thinking about some very similar things this morning and considered this problem and solved it the same way as masamune2387 did. Great minds think alike?

Yes, I've seen the pigeonhole principle too for problems like these involving exponents and remainders and such.

Like this one!

"Prove that there exists n such that 3^n ends in 001."

[edit]-Lol, yeah, gmta indeed.

Yeah pigeonhole principle makes it nice and simple.  But masamune's idea was pretty cool.

Pigeonhole Solution:

let a_k = the repunit [tex]\underbrace{111...111}_k[/tex]

Consider prime p not equal to 2 or 5.

Then there are only p residues mod p, so there are two repunits a_i and a_j with same residue mod p.

WLOG, j > i. Then a_j - a_i == 0 (mod p). But a_j - a_i = 10^i * a_(j-i). Since p is relatively prime to 10, p | a_(j-i) and we're done.

"Prove that there exists n such that 3^n ends in 001."

I am assuming that you are not allowing the trivial solution n=0.

Consider the 1001 numbers 3^0, 3^1, ..... , 3^1000
There are only 1000 residue classes mod 1000, so we pick the two numbers m  :equiv: n mod 1000, where m > n.

Question: How do we prove from here that 3^(m-n)  :equiv: 1 mod 1000?


algebra
polynomial
superior algebra
superior algebra unsolved

is there a known formula for the number of irreducible polynomials of a given degree with coefficients in {0,1,2,3,4,5,6,7,8,9} ?

Not that I know of, but you can certainly place various upper bounds.  For example, if the $ a_{0}$ coefficient is 0, you know it's not irreducible (hence lowering your upper bound).


Let the following operations are the caried out:
  (a,b,c,d)->(|a-b|,|b-c|,|c-d|,|d-a|)
i)Prove that if a,b,c,d in Q then after a finite number of above operations we can obtain (0,0,0,0)
ii) What do we obtain if a,b,c,d in R? :P

I remember orl posting a problem which generalizes (a): for which $n$ do all the numbers eventually become $0$? I think I solved that by answering $n=2^k$. The problem is around here somewhere.. :)

[Edit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1073&highlight=]Here[/url] it is.]

Well, this is a really really well-known problem, even if the case of real numbers is less classical.

M.Lotan has killed it in the AMM (1949, p.535-541) by proving that :

Let $a,b,c,d$ be four real numbers. After a finite number of steps, we will always obtain the configuration $0,0,0,0$ except in the case which the first step has led to the configuration $1,q,q^2,q^3$, where $q$ is the positive solution of th equation $q^3-q^2-q-1=0.$

Pierre.


$ 7.$ Twenty five of King Arthur's knights are seated at their customary round tables.  Three of them are chosen--all choices of three being equally likely--and are sent off to slay a troublesome dragon.  Let $ P$ be the probability that at least two of the three had been sitting next to each other.  If $ P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

$ 1983$ AIME

To make this problem more interesting, I'm looking for $ 3$ separate solutions:
$ (1)$ Where you assume the position of $ 1$ person on the table and do the problem accordingly.
$ (2)$ Where you DON'T assume the position of $ 1$ person on the table.
$ (3)$ Using complementary counting.

I was able to do this problem using $ (1)$ and $ (2)$, but I had trouble with $ (3)$.

Continue on the thread that is designated for AMC/AIME problems so others can look at it if they have same question in the future.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=73151#73151

Locked.


induction
number theory unsolved
number theory

EDIT: Oh...just realized $p$ isn't necessarily prime.  That makes it harder.  I'll get back to you with a complete solution...

[quote="cuenca"]$K81o7$, you are wrong, p is not a prime number, you cant say that $5^{k_{1}}-3^{n_{1}}= 1$, this is for example a solution: $(k,n,p)=(2,2,4)$[/quote]

Yeah, that's the trivial solution i came up with after realizing $p$ is not prime...and I believe for any other existing solution, $n \equiv 10\mod{20}$...

If $k$ is odd: $p^{2}=5^{k}-3^{n}\cong-1mod(3)$ and that's absurd.
Then we have $k=2m$ for some positive integer $m$.
In particular, $(5^{m}+p)(5^{m}-p)=3^{n}$, and then, there exist nonnegative integers $x>y$ such that $5^{m}+p=3^{x}$; $5^{m}-p=3^{y}$ and $x+y=n$.
Now we have $2.5^{m}=3^{y}(3^{x-y}+1)$, and then $y= 0$, $x=n$, and $2.5^{m}=3^{n}+1$.
$3^{n}=2.5^{m}-1\cong-1mod(5)$ this implies that $n=2s$ for some odd positive integer $s$.
Let $s=t.5^{r}$ for nonnegative integers $t$, $r$ such that $(t,5)=1$.
Since $t$ is odd, we can prove by induction that $2.5^{m}=3^{n}+1=9^{t.5^{r}}+1\cong 2t.5^{r+1}mod(5^{r+2})$ wich implies $m\leq r+1$.
If $r\geq 1$: $2.5^{r+1}\geq 2.5^{m}=9^{t.5^{r}}+1>9^{r+2}>2.5^{r+1}$ wich is absurd.
So $r=0$ and $m\leq 1$.
If $m=0$ we obtain $p=0$  nonpositive.
If $m=1$ we obtain $(k,n,p)=(2,2,4)$

[quote=matheus]Find all numbers $(k,n,p)$ in the positive integers such that
$5^{k}-3^{n}=p^{2}$[/quote]

is it solvable using gausian prime ?


geometry
trigonometry
Asymptote
trig identities
Law of Sines

can u make it a bit clear sine of what? :?:

OH, I GOT IT :lol:
BEC= 2BAC
BED+DEC=2BAC
2BED=2BAC
BED=BAC
So, BD=RsinA
BC=2RsinA SIMILARLY ALL OTHER SIDES CEN BE PROVED

[quote="sharkman"]Triangle ABC is inscribed in a circle with centre E and radius
R.

a. Prove that the length of any side of Triangle ABC is equal to the product of
the diameter and the sine. Where is the angle subtended by the side
selected?

b. Prove the area of Triangle ABC is equal to 2R^2(sinA * sinB *sinC).[/quote]

First one's just the extended law of sines...
[hide="Second one"]From extended law of sines,

$a=2R \sin A$
$b=2R \sin B$

$[ABC] = \frac{ab\sin C}{2}= \frac{4r^{2}\sinh \cdot \sinh \cdot \sin C}{2}= 2r^{2}\cdot \sinh \cdot \sinh \cdot \sin C$[/hide]

[quote="sharkman"]parijat,

How did you make that lovely geometric picture?[/quote]

Probably paint.

I prefer Geometer's Sketchpad.

YES I used paint. Actually i dont have geometer's sketchpad. :blush:

just wondering - me@home - what state is this?

[quote="me@home"]Geometers Sketchpad is awesome!! Assuming that I have no skill, it single handedly won first in the state math category of ISEF! Seriously, my board was basicaly one huge sketchpad file printed on a large format printer at kinko's...

---

sorry for that outburst.
but you should check out my gallery of images on this site, it is almost entirely sketchpad (all 50 images)[/quote]
which program is the best for drawing geometry pictures?
Sketchpad or asymtote?
The main problem to draw with paint is that the pictures r not sharp if made by paint.
And is sketchpad a freeware like asymptote?
If then how to get it?

well , I think GSP is much better .
but it seems that we should pay money to buy it .
all the GSP software on the net can only be used for a few days .


Well, usually, the moderators would create a thread that contains all the problems by week.

What do you mean by "copy and paste"? If you don't want to use the masterlists, just go to each problem separately and solve them...

[quote="swimfreak07059"]I know there are masterlists of problems... but in some forums...
they disappear
does this mean i have to go to each problem separately and copy and paste?

PLZ PLZ say no... :huh:[/quote]
Basically it's the mod's job.
But you don't need to copy&paste anything


1. Let $ABC$ be a triangle and $M$ be a point on $AB$. Suppose $r_1$, $r_2$ and $r$ are respectively the inradii of triangles $AMC$, $BMC$, $ABC$. Call $p_1$, $p_2$, $p$ respectively the C-exradii of triangles $AMC$, $BMC$, $ABC$. Prove that $\frac{r_1}{p_1} . \frac{r_2}{p_2} = \frac{r}{p}$

2. (Gergonne's theorem) Let $P$ be a point inside triangle $ABC$ and $A', B', C'$ be the intersection of $AP$ and $BC$, $BP$, and $CA$, $CP$ and $AB$ respectively. Then show that:

$\frac{AB}{AA'} + \frac{BP}{BB'} + \frac{CP}{CC'} = 2$

3. All angles of triangle $ABC$ are acute. Let triangle $ABC'$ be isosceles, right-angled at $C'$, with $C'$ and $C$ on the same side of line $AB$. Define triangle $BCA'$ and triangle $CAB'$ analogously. Show that together triangles $ABC'$, $BCA'$ and $CAB'$ cover all of triangle $ABC$.

4. Suppose $ABCD$ is a convex quadrilateral. Let $O$ be the intersection of $AC$ and $BD$. Find the area of $AOB$ if we know $S_{BOC} = 15$, $S_{COD}=9$, $S_{DOA}=3$

[quote="vincent_five"]for question #4  answer is 5 ?[/quote]

Please show your work  :)

Problem 2: $\frac{AP}{AA'}=\frac{S_{BACP}}{\triangle{ABC}}$, sum up and done.

For question #4, triangle AOD and DOC have the same height. The ratio of their area is 1:3 (3:9). By using the formula A=1/2 a*h, we can get AO:OC=1:3. Similarly, the ratio of AOB and BOC is 1:3. Area of BOC is 15, so area of AOB is 5

Hi guys;

Thanks for the solutions for questrions 2 and 4. Anyone has any ideas how to do  1 and 3?


Interested students should note that another MATHCOUNTS Math Jam will be held tomorrow night.  For information, visit out Math Jams page.

[quote="crakow"]Will the transcript be available for this jam?[/quote]

Yes.  Due to the fact that we tested out the new classroom software last night we are playing around with the transcripting process, but it should be up soon.

[quote="jobro_gold"]Hey! I won't be able to attend the next math jam ( I was really wanting to check out that USAMO techniques jam.) so how do I get a transcript of what happened?[/quote]

We post transcripts of all the Math Jams in the Math Jams area.  Just click the transcript button and you can see the whole history of them.


Let a,b,c be positive real of numbers such that abc<=8/27.PROVE THAT:
                 1/1+a  +  1/1+b  +   1/1+c   >=9/5

Can you post the detail solution(ANH GIAI CU THE DI,DUNG NOI CHUNG CHUNG THE)

let$a=b$,$c=\frac{8} {27ab}$
and $a \to +\infty$
$\sum \frac 1{1+a} \to 1$
then it is false....


Well, I personally don't have a book, but I have seen several posts exposing typos within the texts.... so maybe it's a few clarifications?

Yeah...usually when that book is updated, just a few mistakes or typos are fixed, not really any of the content or the problems...The look may be different, idk...

[quote="benzi455"]I borrowed my friend's AoPS Volume 2, 6th ed. book, and now I want my own copy.  It would be most convenient to just give him money to order a new book.  What would I be missing?  What's new to the 7th edition of AoPS Volume 2?  New examples, or what?[/quote]

Just typo fixes.  You won't be missing notable content changes.

There actually is a typo in my 7th edition. On the To Students page (page -11), the seventh paragraph just starts with "to great lengths to compile the end-of-chapter problems". I assuming they meant to start it with We went to great lengths to compile the end-of-chapter problems. Did anyone else notice this?

More important than grammatical issues, there might have been a few mathematical conceptual typos as well? See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207380]here[/url].

Obviously you can't fix [i]all[/i] the typos in one edition but I believe they made some small fixes.

There are still several typos in the 7th edition, but almost none that will make you miss the gist of the text (one notable exception is the typo in the main problem of section 7.3.4; it should be $ f \left( \frac{1\plus{}a}{2} \right)$ and not $ f \left( 1 \plus{} \frac{a}{2} \right)$, I think.)

[quote="archimedes1"]There are still several typos in the 7th edition, but almost none that will make you miss the gist of the text (one notable exception is the typo in the main problem of section 7.3.4; it should be $ f \left( \frac {1 \plus{} a}{2} \right)$ and not $ f \left( 1 \plus{} \frac {a}{2} \right)$, I think.)[/quote]

Funny, that typo is actually not present in the sixth edition.


Prove that if x,y,z>0 then:

[tex] \sum \sqrt[3] {(\frac{x}{y+z})^{2}} \geq \sqrt[3]{\frac{27}{4}}[/tex]

Are you sure?  Consider x=y=z=1

[tex]
\displaystyle \sum_{i=1}^1 \sqrt[3]{\left(\frac{1}{2}\right)^2}
[/tex]

This is not greater than (27/4)^(1/3)

nr1337, I think the sum on the LHS that Vialli wrote is a symmetric sum, so the LHS = RHS when x=y=z=1.

the sum is cyclic, but your computaion is wrong..

I thought [b]Mildorf[/b] was going to answer your question. I guess he is too busy at MOP to check into AoPS or he never look at posts in the [b]Intermediate Topics[/b] :-). He proved a more general case in the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12763&highlight=]UNSOLVED - ineq[/url]. He also mentioned that the case  [tex]k=\frac{2}{3}[/tex] (which is exactly this problem) was a problem at MOSP 2002. Here is a recap of his proof:

By AM-GM, we have
[tex]\displaystyle x+y+z=x+\frac{y+z}{2}+\frac{y+z}{2}\ge 3 \cdot \sqrt[3]{ x \cdot \left(\frac{y+z}{2} \right)^2 }[/tex] [tex]\displaystyle = 3x^{\frac{1}{3}}\left(\frac{y+z}{2} \right)^\frac{2}{3} } =3x\left(\frac{y+z}{2x} \right)^\frac{2}{3} } [/tex].

So [tex]\displaystyle \sum_{cyc} \left(\frac{2x}{y+z}\right)^{\frac{2}{3}} \ge \sum_{cyc}\frac{3x}{x+y+z} = 3[/tex] which is equivalent to the desired result.


calculus
integration
trigonometry
calculus computations

Substitute $ u \equal{} x^2$ to get the integral $ \frac {1}{2} \int_0^1 \! \frac {1}{u^2 \plus{} 1} \ du$. Then we evaluate $ \frac {\tan^{ \minus{} 1}1 \minus{} \tan^{ \minus{} 1}0}{2}$ to get $ \frac {\pi}{8}$.


Stanford college email AMC USA(J)MO USAMO geometry

Anyone going????? Which tests are you taking?????

( I know this was started in another forum but many people didn't reply, I just thought more people might reply here)

I'm going, for Gunn. I'll be taking Adv. topics and Geo. I think.

i got an email from Ted Alper about it but I don't think i'm going

I'm probably going with Paly, taking advanced topics and geometry. We still need another team member, which is really annoying.

heh, so got a team for paly, bookaholic?
ha now that's some good motivation for gunn
i'll bet you 2 dollars that at least one gunn team beats the paly team

What the...how many teams do you have, and why is Gunn so much more motivated than we are about EVERYTHING?!
(Yeah, we have 7 people willing to go right now. Just need one more. You guys ought to take pity and donate some Gunn person to us.)

rofl we'll probably end up having one full team and a couple people left over if we're lucky....there were a lot of interested people while back, but I somehow think half of them changed their minds...

HAH! We found our eighth team member AND two individuals! So THERE!

oh poo
gunn is gonna get pwned by paly
especially with adrian on your team...ugh
I don't think we'll have that many people come from gunn....about 8-10 people probably

Adrian's not going :( Actually, none of the people who made USAMO last year are going. So our team isn't very strong at all.

holy.....
ahahaha finally gunn MIGHT be able to pwn paly then
naw let's not get to hopeful

I was just wondering (since this is my first time going and I am an 8th grader) what kinds of things are asked on the power tests?

I think it's the first year they're doing a power round, so no one knows....

Oh yeah, I'm taking General Topics.

Beat that.

(More proof that I suck at math).

Hahaha and Paly beats Gunn into the ground at the Field Day! No thanks to me, of course. (Paly: 2nd place. Gunn: 5th place. 1st place: Lynbrook *big surprise*.)

Watch me get a 0 combined on the SMT individual tests.

[quote="bookaholic"]Hahaha and Paly beats Gunn into the ground at the Field Day! No thanks to me, of course. (Paly: 2nd place. Gunn: 5th place. 1st place: Lynbrook *big surprise*.)[/quote]
oh dude i totally forgot to ask steven about how it went...
Hmm how did gunn do so bad....

Yeah dude how did Gunn get beat by some random crappy middle school with random crappy people on it.

Wait, which subjects did you take?

Yeah right Lynnelle, a 0 combined :roll: that's how you scored three points higher than everyone else on Geometry.

Wait I heard that there was only one middle school team and that was Redwood.........
Aw man I want to know what i got on the tests although I bet I failed Geometry and failed Algebra even more.

WOO GO GUNN WE GOT 1ST what do you mean we got beat by seom random crappy middle school? well we did get beaten on the team round...but that's because steven made a stupid mistake, and this guy made a stupid mistake on 13. AND, redwood isn't a random crappy middle school and doesn't have random crappy people. dude you got 1st in general. Congrats!(just for the record, none of the gunn people took general lol)

Congrats to Lynnelle too, for getting 1st! ccy too, he got 2nd in Geo(although he wouldn't have gotten 2nd, since 4 other people were tied for it. I'm sure if they were there at the awards, ccy would have been pwned....lol jk).
And woo for me, I got 6th in adv.(well that's actually a boo, but still...)

My school didn't even have 8 people show up!
And I thought Redwood was a good school?

Just curious, for the people who stayed for awards, do you guys remember who won for Algebra, Advanced Topics, & Calculus?

Steven Ge(gunn) won Algebra, Dan Li(gunn) won calculus. Forgot about Adv, I think it was some Lynbrook guy

Yeah, Paly was the one that got beaten by Redwood :blush: Therefore demonstrating that Redwood's team *is* as strong as ever despite what Ubemaya keeps trying to make everyone believe.

Congrats Ubemaya and ccy!

6th place isn't bad. What place would a 4 have been? That's apparently what I got, but I didn't see the sheet, someone told me. 4 is such an unhappy score. I should have done algebra.

[quote="junggi"]Steven Ge(gunn) won Algebra[/quote]
nevermind.
I think I heard wrong lol

Who exactly is ccy?

Despite the fact that I won first in General Topics, my score was horrible even considering the fact that I'm POSITIVE that I deserved a lower score. There were 2 problems where the wording/meaning of the problems was unclear; on one of them almost everyone understand the correct meaning except me, and on another of them I think that the test writers didn't see an obvious loophole that I missed and everyone else got, so I got the point and everyone else didn't when it should be the other way around. And I finished the whole test except one problem in 50 minutes, had an hour left, and decided to spend that entire hour working on ONE gay problem that made NO sense that NO ONE understood instead of checking the rest of the test, and I made tons of careless mistakes. So I failed the test. And there's no way I'm ever doing General again.

Redwood cheated because everyone took General, which was much easier... so we received undeserved points. So in short, we failed. Although the team round was kind of surprising... how did we manage to beat everyone with a 11? (It was all thanks to Quatto that we didn't fail the team and power rounds).

I suck at math :( .

[quote="Ubemaya"]Who exactly is ccy?
[/quote]
The most hyper person at the awards ceremony.
[quote]gibberish.[/quote]
ew emo
Dude, you did awesome, don't be emo about first place.  :|

i failed smt.
our team totally bombed...............


logarithms
irrational number
algebra unsolved
algebra

Prove there arte irrational numbers a and b such hat a to the power of b is rational

There are many such examples. A nice one is the following: take $a=b=\sqrt{2}$. If $a^b$ is rational, we are done. If not, take $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$. Then $a^b=\left( \sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=\sqrt{2}^2=2.$
(Actually, it can be shown (not elementary) that $\sqrt{2}^{\sqrt{2}}$ is an irrational number.)

Let $x$ be an irrational number and $y$ a rational number not being a power of $x$. Then $\log_x y$ is irrational.
Consider $x^{\log_x y}=y$ is rational.

[quote="leepakhin"]and y a rational number not being a power of x.[/quote]

Not clear enough. Actually, every positive y is a power of a given positive x $(x\neq 1)$, namely $y=x^{\log_{x}y}$.

I think you assumed from the beginning that $\log_{x}y$ is irrational.

It suffices to consider a concrete example. Take $a=\sqrt{10}$ and $b=\log{9}$(decimal logarithm).
Then $a^b=3$. It is easy to see that $\log{9}$ is irrational, because if $\log{9}=\frac{m}{n}$ for some positive integers $m,n$, then $9^n=10^m$, a contradiction.


Show that each point of the Cantor set is a cluster point of itself.

the Cantor set is the set of all numbers in $ [0,1]$ whose base 3 representations don't require the digit 1. to see that every such number is a limit of a sequence of infinitely many such numbers: (1) if the number has infinitely many non-zero digits in its representation, take the sequence of $ n$-digit approximations, (2) if the number has finitely many non-zero digits, form the sequence of numbers obtained by replacing one of the trailing 0's, starting with the first, with a 2. e.g., for the number 0.2, take the sequence 0.22, 0.202, 0.2002, 0.20002, ...


Given $ a, b, c > 0$. Prove that:
[color=orange]$ \frac {a^7}{bc} \plus{} \frac {b^7}{ca} \plus{} \frac {c^7}{ab} \plus{} \frac {2}{abc} \geq\ \frac {3(a^4 \plus{} b^4 \plus{} c^4) \plus{} 11}{4}$[/color]
PS: My solution is only AM-GM  :)

i  also  use Cauchy:
Let: $ X \equal{} \frac {a^7}{bc} \plus{} \frac {b^7}{ca} \plus{} \frac {c^7}{ab}$,$ Y \equal{} \frac {1}{xyz}$ and$ Z \equal{} x^4 \plus{} y^4 \plus{} z^4$
By Cauchy's ineq  we have:
 $ X \plus{} 3xyz \geq 2Z$  and $ xyz \plus{} Y \geq 2$
 Hence, $ X \plus{} 2Y \geq 2Z \plus{} 4 \minus{} 5xyz \geq \frac {3Z \plus{} 11}{4}$
or $ Z \plus{} 1 \geq 4xyz$( obvious by Cauchy's ineq)


MATHCOUNTS
probability
algebra
polynomial
geometry
AMC
AMC 8

This is a progressive difficulty test I made up by myself.  I know the answers, so lets see if you can come up with them!  If you can get all ten right, you are definitely prepared for the Mathcounts chapter and even Mathcounts state competition.
1) What is the value of $ 3^5 \minus{} 3^x$ if x equals 3?
2) If 2x+5=29, what is the value of $ x(x \minus{} 1)$?
3) If 3/5 of the students at a certain middle school are taking gym class,  75% are taking Pre-Algebra, and 10% are taking both.  If there are 1,000 students in the middle school, how many are taking neither?
4) In an arithmetic sequence, the first term is 3 and the eighth term is 45, and the ninth and tenth terms are defined as a and b, what is the value of $ b! \minus{} a!$?
5) Sherman is taking a multiple choice test.  There are 4 choices for each problem, and there are 20 problems.  If he randomly guesses on each problem, what is the probability that he will get at least a 50%?  Express your answer as a common fraction.
6) Expand the polynomial $ x(2x^2 \plus{} 3x^5 \minus{} 2x)$
7) What is the least common multiple of 5!, 4!, 3!, and 2!?
8) What is the units digit of 3 to the ninety ninth power?
9) A man has a certain age.  If you reverse the two digits of his age, multiply the result by two, and subtract 4, the result is his age.  What is the man's age?
10) What is the area of a figure enclosed by the equations y=x+1, y=5x+3, and y=x-2?
Good luck. :lol:

[quote="cgyao15"]1) if u know the answers, then why post them here? cuz u should be asking about probs u don't know.[/quote]
The problem set was merely for fun and a simulation for MathCounts.  Anyways, the problems I made up was partially to see if other people can get the same solutions I did, and to make sure I'm not wrong........which I may be

[quote="cgyao15"]If this is a simulation for mathcounts, it shoudn't be here ne way.
It should be in the MathCounts, MOEMS, and AMC8 forum.[/quote]
OK, it's moving.

Ummm.... I got in trouble for doing something similar... Mods and admins don't really like it when you make marathons in sections such as middle school and high school math... There are already lots of marathons in mathcounts and MOEMS so check those out!

PLS DON'T SPAM. (i am sorry for responeding)
it is a bad thing to do. i think the fun factory would be best for this.

1)  $ 3^5\minus{}3^3\equal{}243\minus{}27\equal{}216$
2)  $ 2x\plus{}5\equal{}29$
     $ 2x\equal{}24$
     $ x\equal{}12$
     $ x(x\minus{}1)\equal{}12(12\minus{}1)\equal{}12\times11\equal{}132$

That"s all my answer thanks :lol:

6) Expand $ x(2x^{2}\plus{}3x^{5}\minus{}2x)\equal{}3x^6\plus{}2x^3\minus{}2x^2$

Does number 10 "10) What is the area of a figure enclosed by the equations y=x+1, y=5x+3, and y=x-2?" has a solution? Because $ y \equal{} x \plus{} 1$ and $ y \equal{} x \minus{} 2$ are parallel.

7)...um... obviously 120 because 5! contains the previous factorials...

Yeah, number 10 has no solution, since nothing is "enclosed".

I disagree with the spamming criticism. This thread is just fine, in my opinion. Anyway, to make this not spam, I'll answer a prob-

#8 What is the units digit of $ 3^{99}$ ?

Find a pattern in the units digit-
$ 3^0 \equal{} 1$
$ 3^1 \equal{} 3$
$ 3^2 \equal{} 9$
$ 3^3 \equal{} 7$
$ 3^4 \equal{} 1$
$ 3^5 \equal{} 3$
$ 3^6 \equal{} 9$
$ 3^7 \equal{} 7$
We have a four-block pattern. 99 is 3 mod 4. All powers of three 3 mod 4 have units digit ending in $ \boxed{7}$.

[quote="hasan4444"]1) What is the value of $ 3^{5} \minus{} 3^{x}$  if x equals 3?
$ 216$

2) If 2x+5=29, what is the value of  $ x(x \minus{} 1)$ ?
$ 12(11) \equal{} 132$

3) Are you sure about it?

4) Is it $ 4.0526919503326 \cdot 10^{76}$?[/quote]

Number 3 was a typo:  Here's what I meant to say:  If 25% of the kids at a school are taking Pre-Algebra, 35% of the kids are taking gym, and 10% are taking both, and there are 600 students at the school, how many students are taking neither?

4)  Yes, but lets just say the problem asked for b-a rather than b!-a!(I can no longer edit this post).

[quote]If 25% of the kids at a school are taking Pre-Algebra, 35% of the kids are taking gym, and 10% are taking both, and there are 600 students at the school, how many students are taking neither? [/quote]
Taking both=60
Taking Pre-Algebra only=90
Taking gym only=150
Neither=$ 600\minus{}(90\plus{}60\plus{}150)\equal{}300$ student

[quote="jin thynj"][quote="maybach"]10) What is the area of a figure enclosed by the equations y=x+1, y=5x+3, and y=x-2?
[/quote]

make integrals..
or find intersection point and fin d area by determinants
or just a simple way to the area of such triangle[/quote]
Can you do it please because I think it can't be  :wink:

What of this 3 methods would you like to see?

that's not possible. y=x+2 and y=x-1 don't intersect, so the figure will look like two parallel lines being crossed by a third line. you need a fourth line of reference.

I have answered too quickly..i'm very sorry.

Take $ (a_1,a_2,a_3,b_1,b_2,b_3) \in \mathbb{R}^6$ and consider the system $ y\equal{}a_ix\plus{}b_i$. If $ \prod_{cyc}{(a_1\minus{}a_2)^2}>0$ then a bounded triangle exists and my previous method works..otherwise there'll be 4 unbounded figures in the plane

Sorry again for the mistake  :oops:


hello can you prove this identity without uses of induction ?

\[ 1^3\plus{}2^3\plus{}\cdots\plus{}n^3\equal{}\frac{n^2(n\plus{}1)^2}{4}\]


can someone show me how

|r+ce^(i*beta)|
---------------- = 1   ???
|c+re^(i*beta)|

where 0<r<c  belongs to complex number, and  0<=beta<=2pi....???
Thanks a lot

Try examining it geometrically.

The statement "0 < r < c belongs to complex numbers" doesn't make sense.  If you have the order relation, then must be real (although the statement is true for if they are complex numbers not both 0, as well).

If you want an algebraic solution, try to figure out how to apply the fact that the product of a complex number and its conjugate is equal to its norm, squared.


prove that :

$ \sin(x)\equal{}\frac{\sin(a)\plus{}\sin(b)}{1\plus{}\sin(a)\sin(b)}    \Leftrightarrow \tan^2(\frac{\pi}{4}  \plus{}\frac{x}{2}  )\equal{}\tan^2(\frac{\pi}{4}  \plus{}\frac{a}{2})\tan^2(\frac{\pi}{4}  \plus{}\frac{b}{2})$

[quote="greatestmaths"]prove that :

$ \sin(x) \equal{} \frac {\sin(a) \plus{} \sin(b)}{1 \plus{} \sin(a)\sin(b)} \Leftrightarrow \tan^2(\frac {\pi}{4} \plus{} \frac {x}{2} ) \equal{} \tan^2(\frac {\pi}{4} \plus{} \frac {a}{2})\tan^2(\frac {\pi}{4} \plus{} \frac {b}{2})$[/quote]

$ tan(\frac{\pi}{4}\plus{}\frac{y}{2})\equal{}\frac{1}{cos^{2}(\frac{\pi}{4}\plus{}\frac{y}{2})}\minus{}1\equal{}\frac{1}{(1\plus{}cos2(\frac{\pi}{4}\plus{}\frac{y}{2}))/2}\minus{}1\equal{}\frac{2}{1\plus{}cos(\frac{\pi}{2}\plus{}y)}\minus{}1\equal{}\frac{2}{1\minus{}siny}\minus{}1\equal{}\frac{1\plus{}siny}{1\minus{}siny}$

$ \tan^2(\frac {\pi}{4} \plus{} \frac {x}{2} ) \equal{} \tan^2(\frac {\pi}{4} \plus{} \frac {a}{2})\tan^2(\frac {\pi}{4} \plus{} \frac {b}{2})$
$ \Leftrightarrow$  $ \frac{1\plus{}sinx}{1\minus{}sinx}\equal{}\frac{1\plus{}sina}{1\minus{}sina}\cdot\frac{1\plus{}sinb}{1\minus{}sinb}$

$ (1\plus{}sinx)(1\minus{}sina)(1\minus{}sinb)\equal{}(1\minus{}sinx)(1\plus{}sina)(1\plus{}sinb)$

$ (1\minus{}sina)(1\minus{}sinb)\plus{}(1\minus{}sina)(1\minus{}sinb)sinx\equal{}(1\plus{}sina)(1\plus{}sinb)\minus{}(1\plus{}sina)(1\plus{}sinb)sinx$

$ ((1\minus{}sina)(1\minus{}sinb)\plus{}(1\plus{}sina)(1\plus{}sinb))sinx\equal{}(1\plus{}sina)(1\plus{}sinb)\minus{}(1\minus{}sina)(1\minus{}sinb)$

$ (2\plus{}1sinasinb)sinx\equal{}2sina\plus{}2sinb$

$ sinx\equal{}\frac{sina\plus{}sinb}{1\plus{}sinasinb}$


Any general method we can apply for these kind of questions:-

Find the largest 3 digit factor of $\binom{3000}{1000}$.

I'm not sure exactly, but...

First you do $\binom{3000}{1000}=\frac{3000!}{2000!1000!}=\frac{3000\cdot 2999\cdot 2998\cdot ... \cdot 2001}{1000!}$

Then you cancel stuff and get $\frac{3\cdot 2999\cdot 2998\cdot 3\cdot...\cdot 2002\cdot 3}{666!}$

Ignore the 3's and since all the 3's have been taken out, try dividing the largest term (divisible by 4) by 4.  That happens to be 2996, so the answer is $\frac{2996}{4}=749$.

Again, I'm not sure exactly, but I think you get the idea.


LaTeX
abstract algebra
geometry
group theory
number theory unsolved
number theory

Prove that the equation $x^3 + y^3 = 7z^3$ has infinitely many solutions in
integers $x, y, z$ with $gcd(x, y, z) = 1$.

This is equvalent
rang elliptic curve x^3+y^3=1 is not equal 0.

Let x/z=, y/z=v. u and v rational and belongs elliptic curve
u^3+v^3=7. If infinelely solution, then rang>0. You need find one solution t=(u,v) with not finite order, then t,t+t,t+t+t,... give infinetely solution, were t1+t2 is summ in elliptic curve.

We have trivial solution t1:x=2,y=-1,z=1. By formulas to summ can be find 2t1,3t1,4t1,...

Just for note:

It's from KOREAN MATHEMATICAL OLYMPIAD

[quote="Rust"]If infinelely solution, then rang>0. You need find one solution t=(u,v) with not finite order, then t,t+t,t+t+t,... give infinetely solution, were t1+t2 is summ in elliptic curve.[/quote]

Can you explain better using Tex? Thanks

This is one of the rare Rust-posts I don't think $\LaTeX$ is really necessary...

He does the following (a well known procedure indeed), the addition on elliptic curves:
If we find infinetely many rational solutions to $x^3+y^3=7$, we are done.
Now consider the curve given by this in the plane.
Now any line through two (by the note following not necessary distinct) points $A,B$ on this curve intersects (again see the following note) it again in another point now called $(A \oplus B)$. Also when two of these points are rational, then the third is so, too.
Note: by seeing it from an limital point we can have that some of these points coincide, we then have tangents, which will not matter, just do as in the other cases.

Now by choosing some fixed rational (can also be infinity which can also occure but is easy to avoid for our purposes) point $O$, we can define $(A+B) : =( (A \oplus B) \oplus O)$ and we see that by this we get (most times) an abelian group (neutral element will be $O$) of (rational) points on the curve, thus we can now 'add' points and also use commutativity.
Now when choosing some 'good' rational point $A$ (just guess some or use that from Rust), all the points $A , A+A , A+A+A , ...$ will be different giving us infinetely many solutions.

How can we use elliptic curves here? I think elliptic curves are of the form $y^2=x^3+ax+b$, not $y^3=(-x)^3-7z^3$. I don't see how we can reduce $y^3$ to $y^2$.

No they aren't necessary of that form, but it's possible to write them this way (see http://mathworld.wolfram.com/EllipticCurve.html ).
And even iff it wouldn't be so: what is the difference¿

I see. I just heard from one of my friend once that elliptic curves are curves of the form I gave. I think I need to find some good book to study more about algebraic geometry and elliptic curves.