calculus
integration
limit
real analysis
real analysis unsolved

I was thinking about improper integrals, and the following question came up:

Let $ f: R\to R$ be a Riemann integrable function.
If  the $ \int_0^{\infty}f(x)dx$ integral is convergent then is it necessarily true that $ \lim_{\infty}f\equal{}0$?

An explicit counterexample: $ f(x) \equal{} \sin(x^2)$.

Absolute convergence is a bit trickier, but still possible- just make increasingly thin spikes.

[quote="jmerry"]Absolute convergence is a bit trickier, but still possible- just make increasingly thin spikes.[/quote]
Making that suggestion concrete:

Define $ \Lambda(x)\equal{}\max(1\minus{}|x|,0)\equal{}\begin{cases}1\plus{}x&\minus{}1\le x\le0\\1\minus{}x&0\le x\le 1\\0&\text{otherwise}\end{cases}$

Let $ f(x)\equal{}\sum_{k\equal{}2}^{\infty}k\Lambda\left(k^3(x\minus{}k)\right)$

Then $ \int_0^{\infty}f(x)\,dx\equal{}\sum_{k\equal{}2}^{\infty}\frac{k}{k^3}<\infty$

but $ \limsup_{x\to\infty}f(x)\equal{}\lim_{k\to\infty}f(k)\equal{}\lim_{k\to\infty}k\equal{}\infty.$


function
geometry
3D geometry
sphere
real analysis
real analysis unsolved

Let $f:\mathbb R^n\to\mathbb R^n$ be a continuous map with the property that $\langle f(x),x\rangle\ge 0,\ \forall x\in\mathbb R$ s.t. $\|x\|=1$. Show that there is some $x_0\in\mathbb R^n$ with $\|x_0\|\le 1$ s.t. $f(x_0)=0$.

Assume that this is not the case, so $f(x) \neq 0$ if $\|x\|\leq 1$. Then let $g(x) = f(x)/ \|f(x)\|$, $g: D^n \to S^{n-1}$. Moreover, denote by $h$ the restriction of $g$ to $\partial D^n$.
It exists by our assumption.
Now by $\langle f(x), x \rangle \geq 0$ we get that $h(x)$ and $x$ are in the same hyper-hemisphere and thus a unique shortest path exists between $x$ and $h(x)$.
So $h$ is homotopically equivalent to the identity and $deg(h)=1$. Now consider $S^{n-1}\to D^n \to S^{n-1}$, where the first arrow is the inclusion and the second arrow is $g$. The composition is $h$ and thus of degree 1, and yet $D^n$ is contractible so the composition is of degree 0. That's it.

[quote="jmerry"]Suppose $f(x)\neq0$ for all $x$ in the unit ball.
Let $g(x)=\frac{-f(x)}{\|f(x)\|}$. Then $g$ is a continuous function from the closed unit ball to itself, so it has a fixed point by the Brouwer theorem. Since $\|g(x)\|=1$ for all $x$, this fixed point must be on the boundary. Since $\langle g(x),x\rangle\le0$ for all $x$ on the unit sphere, this is impossible, and the contradiction proves that $f(x)=0$ for some $x$ in the closed ball.[/quote]

Yes, this is exactly what I did. Thank you both for replying :).


algebra
polynomial
arithmetic sequence
graph theory
combinatorics unsolved
combinatorics

We will prove the result in two steps:
Step 1: Suppose G contains no independent arithmetic progression of m terms.
Then $ E ( G_N ) \geq c N ^ 2 $ for some absolute positive constant c. Here
$ E ( G_N ) $  is the number of edges of our original graph G in its subgraph spanned by the first N positive integers.
Prove: We count the number of arithmetic progressions with m terms in the set
{ 1;2; ... N }. Each of those progressions is charactized by its first term a and the difference d, such that a+(m-1)d $ \leq $ N.
Thus the number of those progressions is M $ \geq  \sum N -( m-1 ) d $ , where d runs through all integers in [1; $\frac { N-1} { m-1 } $ ] (*). Thus clearly
M $ \geq  \frac {  N^2 } { 2m } $. Now for each those progressions there exist an edge of G between its terms, and by counting this way we have counted each edge of $ G_N $ at most $ m^2 $ times , and (*) holds, so step 1 follows. ( Each edge of G that joins i and j is counting at most $ m^2 $ times because their positions in a m-term arithmetic progressions will uniquely determine that progression. )
Step 2: If $ E ( G_N ) \geq c N ^ 2 $ then there is an sufficently large N for which  $ G_N $ contains a complete bipartie k * k subgraph.
Proof: Suppose the contratry, $ G_N $ does not contain a complete bipartie k * k subgraph for all N. (**)
Suppose $ d_1 \geq d_2 \geq . . . \geq d_N $ be the degrees of all the verices of $ G_N $. Suppose that $ d_J \geq k > d_{ J+1 } $.
Clearly $ 2 E ( G_N ) \leq  d_1 + . . . + d_J + N k $ , and
$ E ( G_N ) \geq c N ^ 2 $, so $ J D = d_1 + . . . + d_J \geq a N ^ 2 $ for some absolute positive constant a.
Now for each vertex of $ G_N $ with degree $ \geq k $ we consider all possible k- element subsets of the set of its neighbours. Since (**), each k- element sets in the list we just formed occurs at most k-1 times , thus :
$ { N \choose k }  \geq  \sum   { d_i \choose k }  \geq J  { D \choose k } $ (***) , which is false for sufficently large N because JD exceeds $ u N^2 $ and D exceeds
uN for some positive constant u and for large N ( because J does not exceeds N ).
Thus (***) can't happen for large enough N , because the left-most part is a polynomial of degree k of N, while the right-most part exceeds a polynomial of degree k+1 of N.
Thus we obtain a contrary and step 2 is proven.
From step 1 and step 2 we immediately obtain the desired result.

[quote="sam-n"]Let $ k$ denote an arbitrary positive integer. Consider a graph whose vertex set is the set of positive integers. Suppose that this graph does not contain any complete $ k\times k$ bipartite subgraph. Prove that there exist arbitrarily long arithmetic progressions of positive integers such that no two elements of the same progression are joined by an edge in the graph.

[i]Alternative formulation:[/i] Suppose that $ k$ is a natural number. Imagine a graph whose set of vertices is the set of all natural numbers. Suppose that this graph does not have any complete bipartite $ k \times k$ subgraph. Prove that for every positive integer $ n$, there exists an arithmetic progression of $ n$ natural numbers such that no two numbers in this progression are connected together (as vertices of the graph).[/quote]

can anyone explain what does "complete $ k \times k$ bipartite graph" means? i know what bipartite means, but i don't understand "complete" and $ k \times k$.

sorry if the question is dumb, but i'm very new to graph theory

it means that the graph has to partitions A and B such that |A|=|B|=k and all points in A are connected to all points in B or all points in B are connected to all points in A.
interesting problem though... i am trying it right now..

A 'complete $k \times k$ bipartite graph' (usually denoted by $K_{k,k}$) just means that $K_{k, k}$ can be split into two disjoint sets $A$ and $B$ each with $k$ vertices, such that [u]every[/u] vertex in $A$ is joined to [u]every[/u] vertex in $B$ by [u]exactly one[/u] edge. The word 'complete' just means that every vertex from $A$ and $B$ are connected, and that there are no loops (if we are talking about a conventional simple graph) so there will be a total of $k^2$ edges.

Oops! Sorry [i]manuel[/i]... Didn't see your post.


The odd positive integers $1,3,5,7,\cdots,$ are arranged into in five columns continuing with the pattern shown on the right.  Counting from the left, the column in which $ 1985$ appears in is the

[asy]
int i,j;
for(i=0; i<4; i=i+1) {
label(string(16*i+1), (2*1,-2*i));
label(string(16*i+3), (2*2,-2*i));
label(string(16*i+5), (2*3,-2*i));
label(string(16*i+7), (2*4,-2*i));
}
for(i=0; i<3; i=i+1) {
for(j=0; j<4; j=j+1) {
label(string(16*i+15-2*j), (2*j,-2*i-1));
}}
dot((0,-7)^^(0,-9)^^(2*4,-8)^^(2*4,-10));
for(i=-10; i<-6; i=i+1) {
for(j=1; j<4; j=j+1) {
dot((2*j,i));
}}
[/asy]

$ \textbf{(A)} \text{ first} \qquad \textbf{(B)} \text{ second} \qquad \textbf{(C)} \text{ third} \qquad \textbf{(D)} \text{ fourth} \qquad \textbf{(E)} \text{ fifth}$

$ 1985\equal{}1\plus{}16 \cdot 124 \Rightarrow 1985$ is in the sequence: $ 1,17,33,49,... \Rightarrow$ the answer is $ \boxed B$

[quote="shentang"]Notice how the numbers in the columns go up by 16, correspond to odd numbers mod 16? So 1985 is 1 mod 16. The column with all 1 mod 16 is in column 2.[/quote]

Sorry for reviving. But, this doesn't really make any sense to me. Only the fifth column goes up by 16.

In that case, look at every other number. If you look at every other row starting with the first, you notice that only the second column contains all $ n$ such that $ n\mod16\cong1$


Currently in a retarded state. Please help:

What is the probability, in selecting 2 cards from a 52-card deck, that you get a jack and a heart?

[hide="this is wrong"]
You either have the jack of hearts and another heart, or some other jack and an arbitrary heart.

There are 12 possibilities for the first case, and 39 for the second, for 51 possible combinations total. The answer is therefore

$ \displaystyle \frac {51}{\binom{52}{2}} \equal{} \frac {1}{26}$

I'm pretty sure that's right...
[/hide]
Edit: sorry I misunderstood the problem...

I don't think that accounts for all the cases:
Having the jack of hearts and ANY OTHER CARD simultaneously satisfies both conditions. (51 ways (all cards but the jack))
Having the jack of hearts and a different heart also satisfies both conditions. (12 ways)
Having any jack of (non-hearts) and any heart also does it. (36 ways)

I sum these to $ \frac{99}{\binom{52}{2}}$.

Having the JH and any other card certainly covers the case of having the JH and any other H.  Your first and third case (JH and any other card, J(not H) and (not J)H) are exhaustive.  (Myself missed the case that you have JH and (not J)(not H).)

Case 1: Having J and H in one card. Then the other can be arbitrary. There're $ 51$ ways.
Case 2: Not having J and H in one card. Then to satisfy the given, the two cards must be: J without H and H without J. There're $ (4 \minus{} 1)*(13 \minus{} 1) \equal{} 36$ ways.

Total: $ 51 \plus{} 36 \equal{} 87$ ways

The probability is $ \frac {87}{\binom{52}{2}}$


Find all priimes p such that both $ p\plus{}1$ and $ p^2\plus{}1$ is twice the square of an integer?

please include the complete proof, thank you

and also...try to give a good and easy explanation...so easy that a caveman can understand it.

i found that for all such p it holds that p(p-1)=4k+2 for some positive integer k.  Still thinking

i wonder if anyone can move this to preolympiad...people say its too hard for intermediate.

let p+1=2k^2

and p^2+1=2m^2.

and then i just used difference of squares

from this we can also see that p=3 mod 4

I found that (using a computer) the first such prime is 7 and next one, if there is any, must be greater than $ 10^7$ (!)

[quote="dingzhou"]uh...very helpful...but lemme make it easier for you...7 is the only answer.

but i just need a proof.[/quote]

why the hurry?

oh...i got this :

$ 2k^2\minus{}2k\plus{}1\equal{}b^2$

$ 2k\equal{}p\plus{}1$ btw

What the heck is it with homework incessantly being posted on this forum? I'm guessing the due date has gone by.

[hide="Solution"]Write $ p \plus{} 1 \equal{} 2m^2$, $ p^2 \plus{} 1 \equal{} 2n^2$. We have $ (p^2 \plus{} 1) \minus{} (p\plus{}1) \equal{} 2(n^2 \minus{} m^2)$, or $ (p^2 \minus{} p)/2 \equal{} (n\plus{}m)(n\minus{}m)$.

Therefore $ p \mid (n\plus{}m)(n\minus{}m)$, so either $ p \mid n\plus{}m$ or $ p \mid n\minus{}m$. If the latter, then $ n\minus{}m \geq p \Rightarrow 2n^2 \geq 2p^2 \plus{} 4pm \plus{} 2m^2 > p^2 \plus{} 1$, contradiction. So $ p \mid n\plus{}m$. If $ n \plus{} m > p$, then $ n \plus{} m \geq 2p$. Since $ n \geq m$, this implies $ n \geq p$, but then $ 2n^2 \geq 2p^2 \geq p^2 \plus{} 1$ with the last inequality being equal iff $ p \equal{} 1$, so that's not possible. This means that $ n \plus{} m \equal{} p$.

Going back to the equation $ (p^2 \minus{} p)/2 \equal{} (n\plus{}m)(n\minus{}m)$, we cancel $ p$ from both sides to get $ (p \minus{} 1)/2 \equal{} n\minus{}m$. Add the two equations for $ n\plus{}m$ and $ n\minus{}m$ to get $ p \plus{} (p\minus{}1)/2 \equal{} 2n$, or $ 3p \minus{} 1 \equal{} 4n$. Squaring both sides, $ 9p^2 \minus{} 6p \plus{} 1 \equal{} 16n^2$. Substituting $ 2n^2 \equal{} p^2 \plus{} 1$, $ 9p^2 \minus{} 6p \plus{} 1 \equal{} 8p^2 \plus{} 8$. This can be simplified to $ p^2 \minus{} 6p \minus{} 7 \equal{} (p\minus{}7)(p\plus{}1) \equal{} 0$, so it's clear that only $ 7$ will work.[/hide]


let me rewrite the inequality!
2($ \frac{a^2}{b}$+$ \frac{b^2}{c}$+$ \frac{c^2}{a}$)+3 ≥ 15(a²+b²+c²)

the stronger one holds:

find the greatest constant $ k$ such that the following inequality holds for all positive real numbers $ a,b,c$:

$ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\plus{}k(a\plus{}b\plus{}c)\geq 3(k\plus{}1)\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}$

this is from [b]Bach Ngoc Thanh Cong[/b] also in an article from [b]10maths[/b] he pointed that $ k_{max}\equal{}1, 5855400068...$

now let $ k\equal{}1.5$ and $ a\plus{}b\plus{}c\equal{}1$ we derive at your inequality. :wink:

[quote="NguyenDungTN"][quote="M.A"]and what is the solution? :help:[/quote]
See example 6.[/quote]

 Thanks a lot  :lol:

[quote="BaBaK Ghalebi"]also see example [b]1.35[/b] of the following document for another solution:[/quote]
 to BaBaK Ghalebi
 [hide]kheily mamnoon BaBaK [/hide]

[quote="NguyenDungTN"][quote="M.A"]and what is the solution? :help:[/quote]
See example 6.[/quote]
I don't think the solution of 10maths is correct.
See the following (I think we can't receive anything with it, and he just admitted the result)

[quote="can_hang2007"]

I don't think the solution of 10maths is correct.
See the following (I think we can't receive anything with it, and he just admitted the result)[/quote]

 I think your right can_hang2007
 thank you


MATHCOUNTS
ratio
geometric series
arithmetic series

this comes up in mathcounts and many other contests

the challenge is: what is the nth number in a geometric series

what is the nth number in an arithmetic series

what is the sum of an arithmetic series

what is the sum of a geometric series

what is the sum of an infinite geometric series

one important pont to make: know when your infinite geometric series will converge (i.e. when it does not keep on growing till it reaches infinity)
there, mr. whurm

[quote]one important pont to make: know when your infinite geometric series will converge [/quote]

When -1 < r < 1 or -1 < a < 1, where a is the first term and r is the multiplied difference, no?

[hide]

a + (n-1) d where a is the first number, and d is the common difference between two consecutive terms

a*r^(n-1) where a is the first term, and r is the common ratio between two consecutive terms

(2a+dn-d)n/2  where a is the first term, d is the common difference, and n is the nth term

a([r^[n+1]-1]/r-1)  where a is the first term, n is the nth term, and r is the common ratio between two consectuive terms

a(1/1-r) where a is the first term, and r is the common ratio

[/hide]

yeah, you were right the first time: when -1<r<1. The value of a makes no difference unless a=0, in which case it converges for any r. we'll see why after we get the formula for the sum of a geometric series.
BTW, if you answer this thread, please post a proof. It is more important for you to remember the method than the actual formula.

[quote="MysticTerminator"]The value of a makes no difference unless a=0, in which case it converges for any r. [/quote]

Yeah, my bad.


Well whenever you draw a graph, pick arbitrary values of $ x$, plug them into the function to get a $ y$-value, and plot a point at $ (x,y)$.  Do it multiple times until you can get a good picture.

This seriously looks like homework...

For #1 on scan 0001, graph both.


Hey,
Discussion of the test can begin tomorrow, right?

I think the tests had to be mailed in by Wednesday.

Ok I took this.  This is my first year of physics and I had no idea what was in store for me.  How did you do the bike problem?

I don't have the problems, but there was one about a bike skidding along the ground that I could not figure out.  You are given the distance between the center of the tires (w) and the height to the  center of gravity of the bike (h) and asked to find in terms of w and h the frictional force that will keep both tires on the ground.

(Something of this nature)

I think probably the bike was moving in a circular path of radius $r$ and the velocity of the bike $v$ was given.Then the problem is quite simple.

for no  vertical acceleration , we require -
${R_{1}}+{R_{2}}= Mg$     Where ${R_{1}}$ and ${R_{2}}$ are the normal reactions on the two tyres.

For centripetal acceleration
$F =$ $\frac{Mv^{2}}{r}$  where $F$ is the frictionsl force

For no overturning ,take torque about the centre of the mass
     $Fh+{R_{1}}w/2 ={R_{2}}w/2$

Solve the above three equations in ${R_{1}}$ , ${R_{2}}$ and $F$ to determine the frictional force in terms of the required parameters.

[quote="neelnal"]What do you guys think the cut-off will be for qualifying for the next round?

I think it will be around 60 points.[/quote]

prediction: the cutoff will be 65

could someone post the problems? Please :D

I think I got in the 90s this time, and I made the camp last year.
But then again, I was horribly sick that day.

Hmm, I predict the cutoff will be about 60-70

hey, i was on the us physics team last year and my coach is the academic director of the us physics team.

they're still going to take 200 people for the semifinal, but the way they determine who makes it will be a bit different this year.  they will compute a raw score for everyone based entirely on your multiple choice answers.  then, they select ~350-400 of the top papers to regrade.  based on the work you showed in the second part, your score will be adjusted up/down.  then, they will invite the top 200 scorers after that process to take the semifinals.

they changed the format of the exam in the hopes that it would produce a more spread out distribution (since the test was easier than normal), but from what i hear, the new test did not produce the long tail at the top end of the curve that they were looking for, and the mean score was still very low.   regrading will probably be quite important.   a score of ~60 post-regrading is probably going to be close to the cutoff.

i scored around an 80.  i was pretty stupid that day, but hopefully i will do better in the semifinals.  i'm better at semifinal/olympiad-style exams than i am at the rapid-fire, time-pressured exams anyway.

Hm... do you have any idea when  the 2nd round may be?
Oh, also, is the 2nd round test going to be easier as well, or will the difficulty be similar to that of past years' semi-final tests?

semifinal is probably going to be sometime during the second week of march.

the difficulty is expected to be about the same as usual.

Sorry to be persistent, but can someone explain how they score the exam?

Thanks

2 pts for a multiple choice correct, 0-5 [for the hardest, 7] points for the long response ones.

Okay, does anyone remember what they got for 32, 33, 34, and 37-38?

I don't even remember what the problems were since their were so many of them. Mixing "free response" and multiple choice didn't work out too well for me.

do they post the results/list of semifinalists on the 2007 AAPT Phyiscs Team website?

The bike problem, answering someone's question earlier, is simply done by knowing how a bike can flip.

The front tire can work as your hinge, and then it is really quite simple.

I think I like the new format for this test.

Out of the first non-showyourwork questions, I think I got a 22/25 cuz of two stupid mistakes and I completely forgot the values for various inertia for the third question.

Out of the last 13, I'm pretty sure I got 7 correct, with probably at least 3 point/problem.

Predicting my score: 65


a,b,c are lengths of a triangle
Prove: 59/40<a/(a+b)+b/(b+c)+c/(c+a)<61/40

I can bring this inequality to:

$-\frac{1}{20}<\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{20}$

which looks much better actually and I think it was posted somewhere on the forum before

For the other inequality we have

$\displaystyle2(x^3+y^3+z^3)+27(x^2y+y^2z+z^2x)+16xyz \ge13(xy^2+yz^2+zx^) $

I think also in the first we have $16xyz$ but it is ininfluent.


linear algebra
matrix
algebra
system of equations

The combined ages of Mary and Ann are 44 years, and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann. How old is Mary?

On time.com, I found the original problem is The combined ages of Mary and Ann are 44 years, and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann. How old is Ann?

Oops, I cut off a line when I was transcribing it... they all look the same!  :blush:

[hide="Here is my method"] Let m be mary's current age, a be ana's. We work backwards.
The last condition was when mary was 3 times as old as ann. Let that be n years ago. We have
m-n=3(a-n)
Now, it says when ann was three times as old as mary was at this time. We assign a new variable again. So we have
3(m-n)=a+p.
I will now just write out all the equations:
a+m=44
m-n=3(a-n)
3(m-n)=a+p
m-q=(a+p)/2
m=2(a-q).
We have five equations with 5 variables. Solving gives the answer (I haven't done it yet). [/hide]

Pwning your system of equations with a matrix, you get
[hide]a=16.5, m=27.5, n=11, p=33, q=2.75, which means that the answer is 27.5[/hide]

That is correct according to  [url]http://www.time.com/time/magazine/article/0,9171,857101-3,00.html[/url]  which has Sam Lloyd's solution, which is much easier, although I don't understand how he skipped to one part of it.


Let $ \alpha$ be the positive root of $ x^2\minus{}2008x\plus{}1$ and a sequence: $ x_1\equal{}2008,x_{n\plus{}1}\equal{}[\alpha x_n]$. Find the remainder of $ x_{2008}$ when divisibled by $ 2008$.

$ x_n$ satisfies the recursion

$ x_{n \plus{} 2} \equal{} 2008x_{n \plus{} 1} \minus{} x_n$

It remains to determine $ x_2$, which uniquely determines the rest of the sequence.  Note that $ \alpha$ is very close to (but slightly less than) $ 2008$.  In fact,

$ \alpha \equal{} \frac {2008 \plus{} \sqrt {2008^2 \minus{} 4}}{2} \equal{} 1004 \plus{} \sqrt {1004^2 \minus{} 1} \equal{} 1004 \plus{} 1004 \minus{} \frac {1}{2 \cdot 1004} \mp ...$

hence $ 2008\alpha \equal{} 2008^2 \minus{} 1 \mp ...$ and therefore $ x_2 \equal{} 2008^2 \minus{} 2$.  Now we have

$ x_1 \equiv 0 \bmod 2008$
$ x_2 \equiv \minus{} 2 \bmod 2008$
$ x_{n \plus{} 2} \equiv \minus{} x_n \bmod 2008$

hence $ x_{2008} \equiv x_4 \equiv \boxed{2} \bmod 2008$.

[quote="t0rajir0u"]$ x_n$ satisfies the recursion

$ x_{n \plus{} 2} \equal{} 2008x_{n \plus{} 1} \minus{} x_n$

  WHY? Can you show me! Please! Thank you very much

.[/quote]

A sequence of the form

$ s_{n \plus{} 2} \equal{} 2008s_{n \plus{} 1} \minus{} s_n$

can be given an explicit formula

$ s_n \equal{} a \alpha^n \plus{} b \beta^n$

where $ \alpha, \beta$ are the larger and smaller roots of $ x^2 \minus{} 2008x \plus{} 1$, respectively.  Now, $ \beta$ is on the order of $ \frac {1}{2008}$ and hence is quite small so we have $ s_n \approx a \alpha^n$ (it is either the integer above or below depending on the sign of $ b$).  It follows that if $ s_1 \equal{} x_1$ and $ s_2 \equal{} x_2$ then $ s_n \equal{} x_n$.  (Edit:  A minor calculation error.  When $ a \equal{} b \equal{} 1$ we have $ s_1 \equal{} 2008, s_2 \equal{} 2008^2 \minus{} 2$ as expected.)

Alternately:  $ x_n$ is the closest integer to $ \alpha^n$, and we can show easily that $ \alpha^n \plus{} \beta^n$ is the closest integer to $ \alpha^n$.


abstract algebra
superior algebra
superior algebra unsolved

Here's some nice, but not quite hard problem - I only post it since I love the proof with spectral sequences since the statement is completely trivial with them :) :

Suppose we have a collection of abelian groups $(L^{ij})$, $i,j\geq 0$ and homomorphisms $d_I^{ij}: L^{ij}\to L^{(i+1)j}$, $d_{II}^{ij}: L^{ij}\to L^{i(j+1)}$ such that $d_I^2=0$, $d_{II}^2=0$ and $d_Id_{II}=d_{II}d_I$ (fill in appropriate indices), so the whole diagram commutes and we can compute cohomology in rows and columns(set $L^{ij}=0$ for $i<0$ or $j<0$ with trivial maps in order to compute cohomology in the 0-th row or column). Denote by $H^p(L^{\cdot,q})$ resp. $H^q(L^{p,\cdot})$ the $p$-th coholomogy in the $q$-th row resp. the $q$-th cohomology in the $p$-th column. Assume that $H^p(L^{\cdot,q})=0$ for all $q>0$ and all $p$ as well as $H^q(L^{p,\cdot})=0$ for $p>0$ and all $q$; so all nontrivial cohomology is in the 0-th row or 0-th column. Prove that they are the same, that is $H^p(L^{\cdot,0})=H^p(L^{0,\cdot})$ for all $p$. (Please do not use spectral sequences :lol: ).
From here it follows immediately that $Tor_n(X,Y)=Tor_n(Y,X)$ for all $n$ - just take some free resolution of both and tensor these resolutions with each other to get groups $(L^{ij})$ which fulfil the above assumptions.

One normally considers double complexes which fulfil $d_Id_{II}+d_{II}d_I=0$ instead of $d_Id_{II}=d_{II}d_I$ - however, both things are the same by exchanging $d_{II}^{ij}$ with $(-1)^i d_{II}^{ij}$.

I hope someone here understands this :D

can we use $Tot_{\cdot}(L^{\cdot\cdot})$? i.e. the total complex associated to the bicomplex $L^{\cdot\cdot}$?
I don't know what spectral sequences are. I know that they exist and that I should learn them :)

hmm... i guess what you mean by using the total complex gets quite close to what i mean by using spectral sequences - the key ist that the cohomolgy of the total complex is the limit of both spectral sequences with terms $H^p(L^{q,\cdot})$ and $H^q(L^{p,\cdot})$ in the 2nd sheet. With our assumption, the 2nd sheet is already the limit, therefore the corresponding terms are the same.


calculus
integration
induction
LaTeX
combinatorics proposed
combinatorics

Write p =1/x-a      q=1/x-b

        u=a-b          v=b-a

Denote (C-r){ (vp)^n-r  + (uq)^n-r}  by A-r  where (C-r) is the coefficient of h^r in the expansion of 1/(1-h)^n. and n and r are positive integers.

We sum the A-r for integral values from r=0 to r=n-1.

Prove that this sum is (-1)^n*(u)^2n*(pq)^n.

A proof by induction is very long. THere must be a shorter method.

Sorry Michael, but this is really  hard to read on the form you gave. Please try to use latex codes, or usual non latex code, as x_a instead of x-a for exemple (even me is able to do that, and I'm close to 0 about computers). So, I write it again (please check if it the right statement) :

Write $p = \frac 1 {x-a}$ and $ q= \frac 1 {x-b}$

$u= a-b$ and $ v=b-a$

Denote $A_r = C_r \cdot ( (vp)^{n-r} + (uq)^{n-r})$, where $C_r$ is the coefficient of $h^r$ in the expansion of $\frac1 {(1-h)^n}$, and $n,r$ are positive integers.

Prove that $\sum_{r=0}^{n-1} A_r =  (-1)^nu^{2n}(pq)^n. $

Pierre.

Pierre,

My apologies for confusing subtraction sign - and the subscript symbol_

You have written the problem correctly except for the definitions of p,q u andv which you have read as suffixed quantities. Infact they are diiferences  or quotients of differences.

By a-b I intended b subtracted from a.Similarly b-a means a subtracted from b

Again in the case of p and q,  p is 1 divided by x less a
that is 1/p= xminusa.A similar interpretation for q   : 1/q=x-b

Michael N.


linear algebra
matrix
group theory
abstract algebra
induction
linear algebra unsolved

Is the following version of a classical theorem true: any injective morphism from $ SL_n(Z)$ to $ SL_n(Z)$ is necessarily an isomorphism? The version with surjective was discussed some time ago and the answer turned out to be positive, although the result itself is far from being trivial. I suspect the answer to this question is positive too, but I think it's even more difficult. Maybe I'm wrong...

I think there is no nothing challenging
Answer is no,
 since I can construct  homomorphizm
\[ SL_2(Z)\to SL_2(Z/3Z)\to PSL_2(Z/3Z)\simeq A_4\to C_3\to SL_2(Z),
\]
where  generated $ C_{3}$ become in matrix :
 0  1
-1 -1

. but true is next:
Noncentral normal subgroup in $ SL_{n}Z)$ has a finite index,
such as homomorphizm which is not isomorphizm  go "after" finit group.

Sorry, I forgot to say that $ n>2$. Then try to see again if there's nothing challenging in it...

Especially it is not true for $ n > 2$. If there really is something challenging , than it was only for case $ n \equal{} 2$.
I don't think that you forgot to say $ n > 2$  :oops:

Well, think before you post, that would be so much better sometimes, Extremal. The answer is actually positive and the proof is far from being trivial.

:lol:  Indeed. Where  do you heard that  answer is positive?
Yes,  answer is positive:
Let's  prove that Homomorphizm is injective. from  $ SL_{n}(Z)$ to $ SL_{n}(Z)$

suppose $ H$ is kernel of this homomorphizm, . Then $ H$ is normal , and if he dont lie in center , then has finite index.
 and contains in $ SL_{n}(Z)$ for some prime $ p$.  (there is important that $ n > 2$ (for $ n \equal{} 2$ it don't works.)), hence
Factor group by $ H$ contain elemets , more or equal then  $ SL_n(Z,Z/pZ)$. and last group has order :
\[ \frac {(p^n \minus{} 1)(p^n \minus{} p)\ldots (p^n \minus{} p^{n \minus{} 1})}{p \minus{} 1}
\]
on the other hand , it's wellkown that finet subgroup in $ SL_n(Z)$has order less or equal $ (2n)!$
it's simple to prove by induction that for big $ n$ or more then $ p$ , that expression is more then $ (2n)!$
. we need to consider simle cases:
$ p \equal{} 2$ and $ n \equal{} 3$ or $ n \equal{} 4$
however finet subgroups in $ SL_{3}(Z)$ and $ SL_{4}(Z)$ is wellknown. and they have small order (less then $ PSL_2(3)$ and $ PSL_4(3))$.)

Cases when kernel is central , but is nontrivial.
it's possible when $ n$ is like $ 2k$, and in this case kernel is same as centre. (=).(consist scalar matrix 1 and -1)

We need to prove that $ PSL_n(Z)$ can be lie in $ SL_{n}(Z)$.
then will be proved that it's injective.
Now I'm too lazy to explain also why it will be surjective. :sleeping:

[quote="harazi"]Is the following version of a classical theorem true: any injective morphism from $ SL_n(Z)$ to $ SL_n(Z)$ is necessarily an isomorphism? [/quote]

Yes, for $ n>2$, it is called a co-hopfian group.
I posted the question a couple of years ago here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106156 but no success.
The usual proof is pretty evolved (bounded co-homology).
Using super rigidity (Margulis), you can find a proof. It is useful to read http://arxiv.org/abs/math/0612736v2 from Pansu.
I asked several experts in this field for a "simple proof" (for instance the fact that SL_n(Z) is hopfian has more than one "simple proof"), and no one found any, as far as I know.
If you find one, please share it, I've tried for a long long time :(


Is there some quick way to see that if $ f(x)={\rm log}(e^{x}-x)$ and $ f^{n}= f \circ f \circ ... \circ f$ with $ f^{0}$ denoting the identity...then \[ \sum_{n=0}^{\infty}{f^{n}({\rm log}2)}= 1?\]

Fix $ 0 < x < 1$ and let $ \sum_{i \equal{} 0}^n f^i(x) \equal{} s_n(x)$.  Note that $ e^{f^n(x)} \equal{} e^{f^{n \minus{} 1}(x)} \minus{} f^{n \minus{} 1}(x)$, so

$ f^n(x) \equal{} \log(e^{f^{n \minus{} 1}(x)} \minus{} f^{n \minus{} 1}(x)) \equal{} \log(e^{f^{n \minus{} 2}(x)} \minus{} f^{n \minus{} 2}(x) \minus{} f^{n \minus{} 1}(x)) \equal{} \ldots \equal{} \log(e^x \minus{} s_{n \minus{} 1}(x)).$

It's clear that if $ s_n(x)$ converges we must have $ f^n(x) \to 0$, but this implies $ \log(e^x \minus{} s_{n \minus{} 1}(x)) \to 0$ and so $ s_{n \minus{} 1}(x) \to e^x \minus{} 1$.  So if our series/sequence converges, this is the only value it can converge to and in particular if we prove convergence of $ s_n$ we get the requested result.  Suppose $ f^n(x) \equal{} r$ with $ 0 < r < 1$.  Then we have $ e^r \leq 1 \plus{} r \plus{} r^2$ and so $ f^{n \plus{} 1}(x) \equal{} \log(e^{f^n(x)} \minus{} f^n(x)) \leq \log(1 \plus{} r \plus{} r^2 \minus{} r) < r^2$, so $ s_n(x)$ converges by comparison to $ \sum_{n} r^{2^n}$.  (Actually, I imagine someone else could prove convergence for $ x \in \bf R$.)


When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole number quotient with a remainder of 1. How many distinct values are possible for this three-digit dividend?

[hide]
Basically this is saying that the number below the one with units digit 2 has a factor of 2,3,5,7. So it has a units digit of 1.

It cant have 2 (odd)
It cant have 5 either
For 3, the lowest is clearly 111.
Then, every time we add 30, we get another.
Count them out to get 30.

For 7, the lowest is 151.
Again, add 70 to get another.
Count them out to get 12.

Add 30+12=42.

But we have to subtract multiples of 21
They are 231, 441, 651, 861.

So the total is 42-4=38[/hide]


This question may look hard but if you think really good, you'll get it.

The least prime factor of $a$ is 3; th least prime factor of $b$ is 7. Find the least prime factor of $a+b$.

[hide="Hint"]Try to find out what are the parities of $a$ and $b$.
[/hide]
[hide="Answer"] We know that odd + odd = even
Therefore $a+b$ is an even number, so the least one is 2.[/hide]

Your first answer worked for this example but it has no logical sense. Why? Because $a,b$ can be integers other than $a=3$ and $b=7$. I only find the second answer convincing.

Hint: Your hint is more confusing than the actual answer.

Silverfalcon I've edited your post to make it more easily readable, and more logical. Everyone else, just don't mind this post and/or 3xlich's post, and try to solve the problem.

this is easy

[hide]since 3 is the smallest factor, a must be odd because 2 does not go into it.  C is also odd since 2 does not go into it.  Therefore, odd + odd = even so a+b has a smallest factor of 2
[/hide]

demji -- you showed that must be a factor, but you never actually showed it's the smallest one -- how do you know that?

[hide="err"]
Unless I misunderstood you, 3 would have to be the smallest factor if 2 isn't a factor...
[/hide]


You have 45 identical-looking boxes, one of which is slightly
heavier than the rest.  With one balance scale, what is the fewest weighings needed to identify the heavier box?

[hide="This is the easy version"] Since we know the different box is heavier, at any stage we can divide the boxes we have doubts about into three groups and weigh two of them.  If one is heavier, we know that one contains the heavy box, and if they weigh the same we know the third group contains the heavy box.

In other words, before every weighing we know that the heavy box must be one of $n$ boxes.  After every weighing we know that the heavy box must be one of $\lceil \frac{n}{3}\rceil$ boxes.

$\frac{45}{3}= 15$
$\frac{15}{3}= 5$
$\frac{5}{3}< 2$
$\frac{2}{3}< 1$

Hence we require $\boxed{4}$ weighings. [/hide]
[b]Harder version:[/b]  One of the boxes is different, but you don't know whether it's heavier or lighter.  What is the fewest number of weighings needed to identify this box now?


This is a current problem (September 2009):
[url]http://www.fen.bilkent.edu.tr/~cvmath/Problem/problem.htm[/url]


linear algebra
matrix
linear algebra unsolved

Let $ \alpha$ be nonzero real number and suppose that $ F$ and $ G$ are linear maps (operators) from $ \mathbb{R}^n$ into $ \mathbb{R}^n$ satisfying $ F \cdot G \minus{} G \cdot F \equal{} \alpha F$.

Show that there exists $ k \geq 1$ such that $ F^k \equal{} 0$.

It must be on this forum already, but here is a hint: use the trace.

We can prove that for any $ A,B \in M_{n}(\mathbb{R})$ such that $ AB\minus{}BA$ commutes with $ A$ then $ AB\minus{}BA$ is nilpotent
(i.e $ (AB\minus{}BA)^k\equal{}0$ for some $ k$ )

first prove that if $ C \in M_{n}(\mathbb{R})$ such that $ \text{Tr}(C^i)\equal{}0$ for $ i\geq 1$ then $ C^k\equal{}0$ for some $ k$.
Suppose that $ \lambda_1,\lambda_2,\cdots,\lambda_m$ are distinct nonzero eigenvalues of $ C$ (possiblely complex) with algebraic multipicity $ n_1,n_2,\cdots,n_m$ respectively, then we have the system $ V(\lambda_1,\cdots,\lambda_m) x^T\equal{}0$ where $ x\equal{}(n_1\lambda_1,\cdots,n_m\lambda_m)$ and $ V(\lambda_1,\cdots,\lambda_m)$ is the Vandermonde matrix. Since all $ \lambda_i$ are all distinct and nonzero then $ V(\lambda_1,\cdots,\lambda_m)$ is nonsigular and then $ (n_1\lambda_1,\cdots,n_m\lambda_m)\equal{}0$, contradiction. Hence all eigenvalues of $ C$ is zero, that is $ C$ is nilpotent.

For any $ m >1$, we have $ (AB\minus{}BA)^m\equal{}AS\minus{}SA$ where $ S\equal{}(AB\minus{}BA)^{m\minus{}1}B$ thus $ \text{Tr}(AB\minus{}BA)^m\equal{}0$ for $ m \geq 1$, hence $ AB\minus{}BA$ is nilpotent.

the argument also works for matrix $ \sum_{i\equal{}1}^m A_iB_i \minus{} B_iA_i$.


Well then you should show your method instead of simply disagreeing.

Um, how did you get $ \frac {1} {1\plus{}1\plus{}2}$?  I think that Kouichi Nakagawa (I think that's how you spell it) is right.  His/her method was how I did it.

[quote="asterday1"]1 divided by 1/2 equals 2
1 divided by 1 is 1
so, 1 divided by 1+1/2 is 1+2
didnt feel like using LaTeX[/quote]
$ \frac{1}{1\plus{}\frac{1}{2}}\ne\frac{1}{1}\plus{}\frac{1}{\frac{1}{2}}$


integration
trigonometry
real analysis
real analysis unsolved

$ \int_0^{2a} \frac{x}{\sqrt{2ax\minus{}x^2}}\ dx\equal{}\int_{\minus{}a}^a \frac{x\plus{}a}{\sqrt{a^2\minus{}x^2}}\ dx$

$ \equal{}2a\int_0^a \frac{1}{\sqrt{a^2\minus{}x^2}}\ dx$

$ \equal{}2a\left[\sin ^{\minus{}1} \frac{x}{a}\right]_0^a\equal{}\pi a$.


Prove that $\frac{a-b}{a-c}-\frac{b}{c},\frac{b-c}{b-a}-\frac{c}{a},\frac{c-a}{c-b}-\frac{a}{b}$ cannot be all integers if $a,b,c\in Z^$

You can check that:

$\frac{a-b}{a-c}-\frac{b}{c}=\frac{a(c-b)}{c(a-c)}=x_1$
$\frac{b-c}{b-a}-\frac{c}{a}=\frac{b(a-c)}{a(b-a)}=x_2$
$\frac{c-a}{c-b}-\frac{a}{b}=\frac{c(b-a)}{b(c-b)}=x_3$

By contrary suppose that $x_1,x_2,x_3 \in \mathbb{Z}$

but $x_1x_2x_3=1$ hence $x_1=x_2=x_3=1$ or two of $x_i=-1$ and one equals 1. This leads to direct checking when you get that some of $\{a,b,c\}$ must be equal which is a contradiction.


Factor completely. If the polynomial is prime, say so: 3z^2 + 4z + 1

A) (3z + 1)(z – 1)   B) (3z + 1)(z + 1)   C) (3z – 1)(z + 1)

D) (3z – 1)(z – 1) E) none of these

This method is called Civil War (or at least that's what my teacher calls it).  In a polynomial $ ax^2 \plus{} bx \plus{} c$.  You have to find factors of $ ac$ that differ by $ b$, split the two, and then factor by grouping.


Hello,
 Hope someone can help with this. I have a large martix made up using \begin{array}.
When the document is built you cant see the right hand side of the matrix as its been built off page.
Thers gotta be a basic fix, i just cant find it.
 Cheers 4 any help,
Sam

sorry, minimal example is:

{\tiny\begin{equation}{cccccccccccc}
\begin{array}
%array elements 12x12
\end{array}
\end{equation}

have tried using \begin{sideways} etc but it wont fit any better landscape.

It's impossible to tell since you haven't provided a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url]. Is it really a matrix, in which case why aren't you using a matrix environment like pmatrix? Is it a large table where the entries need to be on more than one line using p{[i]width[/i]} instead of l, r or c in the environment's parameters. Maybe you need to use a smaller font or change the margins or use \smallmatrix? Who knows without a [b]complete compilable[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url]?


Our company's contests contain a number of problems that may be of interest to this community.  One of our contests is a fifteen-problem-in-one-week all-school contest, which allows us to incorporate some very challenging problems, such as

"Two people play a game in which they alternate turns and the probability of the player whose turn it is winning on turn n is n/(n+2).  What is the probability that the second player to take a turn ends up winning the game?  Note that the first player's first turn is turn 1, the second person's first turn is turn 2, etc."

While at first glance this problem is similar to many you've seen on other tests, the solution to this one was quite unexpected to us; hopefully you'll find it interesting.

We offer contests in a number of formats, the easiest of which serves to expose less-experienced students to a broad variety of topics, and is somewhat like piano scales for the more advanced student: a collection of problems that should be second-nature to you if your goal is mathematical excellence.  An example of one of its harder problems is

"Express the sum of 467 (base 8) and 152 (base 8) in base 8."

Because of the short time limit on this test, it is important that competitors be able to do the operation quickly in base 8, rather than converting to base ten and back.

     Our next contest is sort of a combination of these two formats, with one hundred problems to be solved in 30 minutes by your entire team.  There are some old standards represented, as well as a number of interesting new problems we've never seen before.

Max -- Because most of our competitions are new formats, we do not have sample tests for all of them.  However, we do have a sample of my favorite, the Collaborative Problem-Solving Contest, which is fifteen problems for your entire school to collaborate on for one week.  We also have an excerpt of last year's test for the Fall Startup Event, which is the other end of our testing spectrum... it is 100 problems in 30 minutes, basically testing whether you've mastered all of the fundamentals to the point that they are second nature.  I'll e-mail you the files, which are due to go on our web site later this month, I think.


geometry
incenter
circumcircle
trigonometry
angle bisector
similar triangles
geometry proposed

In a non equilateral triangle $ABC$ the sides $a,b,c$ form an arithmetic progression. Let $I$ be the incentre and $O$ the circumcentre of the triangle $ABC$. Prove that

(1) $IO$ is perpendicular to $BI$;

(2) If $BI$ meets $AC$ in $K$, and $D$, $E$ are the midpoints of $BC$, $BA$ respectively then $I$ is the circumcentre of triangle $DKE$.

Let $R$ be the circumradius, and $\alpha, \beta, \theta$ angles $A,B,C$ respectively. Recall Briggs' formulas:
1) $\sin \frac{\alpha}{2}=\sqrt{\frac{s(s-a)}{bc}}$
2) $\cos \frac{\theta}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$, where $s=\frac{a+b+c}{2}$
and Euler's theorem
3) $IO^2=R^2-2Rr$
Then, we want to show that
$BI^2+IO^2=BO^2 \Leftrightarrow BI^2=2Rr$

But from the sine law on $\triangle AIB$ we find $BI=\frac{AB\sin\frac{\alpha}{2}}{\cos \frac{\theta}{2}}=\sqrt{\frac{ac(s-b)}{s}}=\sqrt{\frac{ac}{3}}$.
And from $sr=\frac{abc}{4R}$ we obtain $2Rr=\frac{abc}{a+b+c}=\frac{ac}{3}=BI^2$, hence $(1)$ is proven.

Now, recall $AK=\frac{bc}{a+c}$, then $AK=\frac{c}{2}=AE$, so by $l-a-l$ we have $\triangle EAI\cong \triangle KAI \Rightarrow EI=KI$. In the same way we show that $KI=DI \Rightarrow EI=KI=DI$, so we conclude $I$ must be the circumcenter of $\triangle DKE$.

True, here trigno gives an efficent and direct solution!!! Geometry takes a while to figure out!!

1)

Let D be the intersection of BI with c(ABC). From Ptolomeus we have that  AB*DC + BC*AD = AC*BD. But we have that 2AC = AB+BC and, as one can prove, AD=DC=DI. Then I is the midpoint of BD, i.e. OI is orthogonal to BD.

[quote="Vihang"]I used pure Geometry (no Trignometry)
Try it using Pure Geometry.(I liked your method though.It's Efficient)[/quote]
Just Curious, can you post your pure Geometry solution?
BTW - when I looked at the problem, what just 'jumped out' was, $IB = \frac r {\sin (B/2)} = \frac{(s-b)}{\cos (B/2)} = \frac b {2 \cos(B/2)}$
so $IB^2 = \frac{br}{2 \sin (B/2) \cos(B/2)} = r \frac b {\sin B} = 2rR$  and of course $(R^2-2rR)+2rR = R^2$

2)
From similitude IE =ID, as PA=PC.  And from AK/CK = AB/CB = AE/CD it follows that AE = AK (and DC=KC) then by simmetry EI=KI.

[hide="Hint for a pure geometry solution"]
Well use property of an angle bisector to get AK=1/2*AB
and CK=1/2*BC[/hide]

[hide="The solution"]
Then we can get triangles AEK and CDK to be isosceles
But AI and CI are angle bisectors of \[ \angle EAK  \]
and\[ \angle DCK  \]
thus giving triangles IEK and IDK also to be isosceles
From which we get IE=ID=IK proving I as circumcentre of triangle EDK

Also using a bit of angle chasing we can show \[ \angle DIE  \]to be supplementary to \[ \angle DBE  \]

Thus BDIE is cyclic
But BDOE is cyclic because O is the circumcentre.
Thus BIOD is cyclic.
Thus\[ \angle BIO = \angle BDO = 90^\circ  \]
[/hide]

I used pure geometry in the exam. :D
Wonder how I could think of such a solution. :?

My solution for this problem is very simple, although intuitive, very poor. I was wondering to show that

$OI^2+IB^2=OB^2$

Call $H_1,H_2,H_3$ the foot of the perpendiculars to the sides $a,b,c$, and $I_1,I_2,I_3$ the points where the internal bissectors cut the oposite sides. It's easy to compute (although it's a boring work), the segment $H_2I_2$, using the bissector theorem and simply pithagoras' theorem. Easy to calculate $H_2B$, and by pithagoras, we determine all the sides of $\Delta BH_2I_2$. By similar triangles $\Delta AH_2O$ and $\Delta OBH_1$ we can determine the side $OH_2$ and so $OB$. By using the bissector theorem on triangle $ABI_2$ we can calculate $II_2$ and so $BI$, now using the cosine law, we can find $OI$, and now it's just check the pithagoras' theorema for triangle $OBI$. It has some nice points about this solution, although it has many computations, if everything is right in this problem, it has to be ok! Now looking here, I'll try to find one solution using quadrilaterals, since $OIH_1B$ has to be cyclic if $\angle OIB =\frac{\pi}{2}$. I'll put the computing when I finish the work.

I'll be glad if you correct me if I commited any mistake
thanks

I think that all proofs above are for $O$- the circum center of the triangle. But in the problem it is given that $O$ is the ortocenter of the triangle. Who is wrong?

For part 1, use the formula:
$OI^2$ = $R^2$ - $2Rr$.
Then with a bit of algebra solving, we get
$OB^2$ = $OI^2$ + $BI^2$.
This implies the result.

Here's a solution that uses pure geometry for the first part, and almost pure geometry for the second part. It is really astonishing how everything neatly cancels out! It might seem like heavy computation, but bear with me, it is really easy!
We have the condition: $a+c=2b$. We first compute the area($\triangle$), the inradius($r$), and the circumradius($R$). Firstly note that $s=\frac{a+b+c}{2}=\frac{3b}{2}$.
Using the Heron's formula,
$$\triangle^{2}=\left( \frac{3b}{2} \right)\left( \frac{3b}{2}-a \right)\left( \frac{3b}{2}-b \right)\left( \frac{3b}{2}-c \right) \Rightarrow \triangle=\frac{\sqrt{3}bk}{4}$$
Where $k=\sqrt{(3b-2a)(3b-2c)}$.
Hence,
$$r=\frac{\triangle}{s}=\left( \frac{\sqrt{3}bk}{4} \right)\left( \frac{2}{3b} \right)=\frac{k}{2\sqrt{3}}$$
$$R=\frac{abc}{4\triangle}=(abc) \left(\frac{1}{4} \right) \left( \frac{4}{\sqrt{3}bk} \right)=\frac{ac}{\sqrt{3}k}$$
Also, $$BK^{2}=ca \left(1- \left(\frac{b}{c+a}\right)^{2} \right)=\frac{3}{4}ca$$
Now, using the angle bisector theorem,
$$\frac{BI}{IK}=\frac{AB}{AK}=\frac{BC}{CK}=\frac{AB+BC}{AK+CK}=\frac{a+c}{b}=2 \Rightarrow BI=\frac{2}{3}BK$$
And thus $$BI^{2}=\frac{4}{9}BK^{2}=\frac{4}{9}.\frac{3}{4}ca=\frac{1}{3}ca$$
The last two elements that we evaluate are
$$BO^{2}=R^{2}=\frac{a^{2}c^{2}}{3k^{2}}$$
$$OI^{2}=R(R-2r)=\frac{ac}{\sqrt{3}k}\left( \frac{ac}{\sqrt{3}k}-\frac{k}{\sqrt{3}} \right)=\frac{ac}{3k^{2}}(ac-k^{2})$$
And now, the first result that we desire is trivial! Just use pythagoras theorem:
$$IO \perp BI \Leftrightarrow BO^{2}=BI^{2}+IO^{2} \Leftrightarrow \frac{a^{2}c^{2}}{3k^{2}}=\frac{ac}{3k^{2}}(ac-k^{2})+\frac{1}{3}ca$$
$$\Leftrightarrow ac=k^{2}+(ac-k^{2})$$
which is clearly true.

For the second part, since $IK=BK-BI=\frac{1}{3}BK$, hence
 $$IK^{2}=\frac{1}{9}BK^{2}=\frac{1}{9}.\frac{3}{4}ca=\frac{ca}{12}$$
To evaluate $IE$, we use the cosine rule in $\triangle BEI$. Note that
$$cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}}=\sqrt{\frac{\left( \frac{3b}{2} \right) \left( \frac{b}{2} \right)}{ca} }=\frac{\sqrt{3}b}{2\sqrt{ca}}$$
Thus,
$$IE^{2}=BI^{2}+BE^{2}-2BE.BI.cos\frac{B}{2}$$
$$\Leftrightarrow IE^{2}=\frac{ca}{3}+\frac{c^{2}}{4}-2 \left( \frac{c}{2} \right) \left( \sqrt{\frac{ca}{3}} \right) \left( \frac{\sqrt{3}b}{2} \right) \left( \frac{1}{\sqrt{ca}} \right)$$
$$=\frac{ca}{3}+\frac{c^{2}}{4}-\frac{bc}{2}$$
$$=\frac{ca}{3}+\frac{c}{4} (c-2b)$$
$$=\frac{ca}{3}-\frac{ca}{4}=\frac{ca}{12}$$
Similarly $ID^{2}=\frac{ca}{12}$.
Hence $IK=ID=IE$ and hence $I$ is the circumcenter of $\triangle KDE$ and we are done!

[hide=A very bashy solution]
We will use barycentric coordinates with the reference triangle $ABC$. Hence set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Also wlog let $a = BC = b - d, b = AC , c =AB = b+d$.

It can be easily shown that $I = (a:b:c)$
Hence $(a+b+c) \overrightarrow{IO} = a \overrightarrow{AO} + b \overrightarrow{BO} + c\overrightarrow{CO}$
Hence shifting the origin to $O$, we get $\overrightarrow{IO}=(b-d:b:b+d)$
Also $\overrightarrow{IB} = \overrightarrow{I}-\overrightarrow{B} = (b-d:-2b:b+d)$

As $\overrightarrow{IB}$ is a displacement vector, we can use [tip=Strong EFFT][img]https://media.discordapp.net/attachments/805662581330804736/811855316161265684/unknown.png[/img][/tip] on

\begin{align*}\overrightarrow{IB} &= (b-d:-2b:b+d)  \text{ and }\\ \overrightarrow{IO} &= (b-d:\;\; \;\; b:b+d)\end{align*}

Hence we need to show that
$(b-d)^2[(-2b)(b+d)+b(b+d)] + b^2 [(b-d)(b+d)+(b-d)(b+d)]+(b+d)^2[b(b-d)+(-2b)(b-d)]$ is identically equal to zero

that is , we need to show that $(b^2-d^2)(d-b)+2(b^3-bd^2)+(b+d)(d^2-b^2)=0$ which is obviously true.
Hence $IO \perp BI$

Now if $K = (\alpha,\beta,\gamma)$, then as $K \in AC, \beta = 0$ and as $K \in AI, \frac{\alpha}{\gamma} = \frac{a}{c} = \frac{b-d}{b+d}$

hence $K = (b-d:0:b+d)$
Also $D=(0:1:1),E=(1:1:0)$

We need to show that $DI^2=EI^2=KI^2$

So we consider the displacement vectors \begin{align*}
\overrightarrow{ID} &= \left(\frac{b-d}{3b},\;\; \;\;\;\frac{-1}{6},\frac{2d-b}{6b}\right)\\
\overrightarrow{IE} &= \left(\frac{-b-2d}{6b},\;\frac{-1}{6},\frac{b-d}{3b}\right) \\
\overrightarrow{IK} &= \left(\frac{d-b}{6b},\;\; \;\;\;\;\;\frac{1}{3},\frac{-b-d}{6b}\right) \\
\end{align*}

Hence after some simplification we have $|\overrightarrow{ID}|^2 = (b-d)^2\frac{(b-2d)}{36b}+b^2\frac{(3bd-b^2-2d^2)}{18b^2}+(b+d)^2\frac{d-b}{18b} = \frac{b^2-d^2}{12} = \frac{ac}{12}$
By symmetry $|\overrightarrow{ID}|^2 = \frac{ca}{12} = \frac{b^2-d^2}{12}$

Now $|IK|^2 = -\left[\frac{-(b-d)^2(b+d)}{18b}+\frac{b^2(b+d)(b-d)}{36b^2}+\frac{(b+d)^2(b-d)}{18b} \right]= \frac{(b^2-d^2)(-2d+2b-b+2b+2d)}{36b} = \frac{b^2-d^2}{12}$

Hence $I$ is the circumcenter of $KED$
[/hide]


this one looks quite easy

\[ b^2 + c^2 \geq 2bc \]

so lhs \[ \geq 2 \sqrt{a^2 + b^2 + c^2} \geq 2\sqrt3 \]
by \[ am \geq  gm \] :lol:

oh , I didnt see that it is that easy .Haha. :P
Anyway can we find the best constant such that

$\frac{b^2+c^2}{\sqrt{a^2+kbc}}+\frac{a^2+c^2}{\sqrt{b^2+kac}}+\frac{a^2+b^2}{\sqrt{c^2+kab}}\geq \frac{6}{\sqrt{1+k}}$

where $abc=1$ , $a,b,c>0$

I think this inequality holds for any positive $k$. ;)

[quote="arpit goel"]well for k>=2 i do hv a proof but nothin for k<2
so lhs \[ \geq \sum {\frac{b^2 + c^2}{ \sqrt{p + (k-2)bc}} \geq \frac{2}{3} \sum {a^2} \sum { \frac{1}{ \sqrt{p + (k-2)bc)}}  }}\]
by chebbychev[/quote]
Chebyshev works for $k \leq 2$.


inequalities
function
integration
calculus
calculus computations

if f: [1,+oo)-->R is differentiable function with f(1)=0 and also satisfies $ f'\left( x \right) \equal{} \frac{1}{{{x^2} \plus{} 3{f^2}\left( x \right)}},\forall x \in \left[ {1, \plus{} \infty } \right)$
then proove that $ 0 \le f\left( x \right) \le \sqrt {\frac{{x \minus{} 1}}{2}} ,\forall x \in \left[ {1, \plus{} \infty } \right)$

[quote="mathxl"]

if f: [1,+oo)-->R is differentiable function with f(1)=0 and also satisfies $ f'\left( x \right) \equal{} \frac {1}{{{x^2} \plus{} 3{f^2}\left( x \right)}},\forall x \in \left[ {1, \plus{} \infty } \right)$
then proove that $ 0 \le f\left( x \right) \le \sqrt {\frac {{x \minus{} 1}}{2}} ,\forall x \in \left[ {1, \plus{} \infty } \right)$[/quote]

$ f$ is increasing because $ f' > 0.$ So $ f(x) \geq 0, \ \forall x \geq 1,$ since $ f(1)\equal{}0.$ We also have $ (1\plus{}3f^2(t))f'(t) \leq (t^2 \plus{} 3f^2(t))f'(t)\equal{}1, \ \forall t \geq 1,$ which after integrating

in the interval $ [1,x]$ gives us: $ f(x) \plus{} f^3(x) \leq x \minus{} 1.$ But $ f(x) \plus{} f^3(x)\equal{}f(x)(1 \plus{} f^2(x)) \geq 2f^2(x),$ and we're done.

More general : $ f'(x) \leq \frac 1{x^2} \Longrightarrow\int_1^xf'(x)dx \leq \int_1^x\frac {dx}{x^2}\Longrightarrow f(x) \leq 1 \minus{} \frac 1x \leq \sqrt \frac {x \minus{} 1}2$
:cool:

Thanx for your replies
My solution is the one of Διογένης
What can we say about the type of f? Can we find it?

So...errr..anyone ? Can we find type of f?


analytic geometry
graphing lines
slope
calculus
calculus computations

Let f(x) = x^2 - 3x - 4.

Find the slope, m(h), of the secant line between the points A(1, f(1)) and B(1 + h, f(1 + h)).

This is what I've done so far:

f(1) = 1^2 - 3(1) - 4
f(1) = -6
So, A is the point (1, -6)

f(1 + h) = (1 + h)^2 - 3(1 + h) - 4
f(1 + h) = 1 + 2h + h^2 - 3 - 3h - 4
f(1 + h) = h^2 - h - 6
So, B is the point (1 + h, h^2 - h - 6)

What is the slope??


[quote="SplashD"][quote="b-flat"]All shirts sell at the same price, and all caps sell at a different price than the shirts (but the same as other caps).  Bob pays $\$104$ for 6 shirts and 11 caps.  Joe pays an integral amount of dollars for 1 shirt and 2 caps.  Janie buys 1 shirt and 1 cap.

A) How much does Janie pay?

B)  Do you even need to know anything about Joe?[/quote]

[hide="A"]$10+4=14$ dollars[/hide]
[hide="B"]
N0[/hide][/quote]

Can you please explain how you got your answers, and why they are correct (and the only solution for part A)?

Are you sure we don't need to know anything about Joe? It is not stated that the prices of 1 cap or 1 shirt are integers.  :maybe:

I did not say that SplashD's answers were correct; I asked him to explain why he thought that his answers were correct.

Since no one has posted both a response and a solution in a week, I will post a hint.

[hide="hint"]Find the range of how much Joe could pay.[/hide]

6 shirts and 11 caps cost $104$ dollars, so 1 shirt and $\frac{11}{6}$ caps cost $17 \frac{1}{3}$ dollars, by proportion. Therefore Joe paid $\geq 18$ dollars for his purchase of 1 shirt and 2 caps.

If Joe paid $\geq 19$ dollars for 1 shirt and 2 caps, then 6 shirts and 12 caps would cost $\geq 114$ dollars and 1 cap would cost $\geq 10$ dollars (through comparison with Bob's purchase). But this is impossible as 1 cap would cost $10$ dollars and the shirt would have a negative price! So Joe must have paid $18$ dollars. So 6 shirts and 12 caps then cost $108$ dollars and 1 cap costs $4$ dollars. 11 caps cost $44$ dollars and 6 shirts then cost $(104-44) = 60$ dollars, whence 1 shirt costs $10$ dollars.

So Janie pays $(10+4) = 14$ dollars. But we need to know something about Joe, i.e. that he paid an integral amount for the 1 shirt and 2 caps (this is crucial!). And the hint was crucial too, b-flat. Not a bad problem. Should have seen it from the start (*slap myself on the face*).

Yay, you got it.  Nice work  :)

My dad and I were talking about extending this MOEMs problem for the math group at my school, and I did some stuff with limits and we got around to this problem.

About Joe, lets say he bought $x$ caps and $y$ shirts.  We must know he payed an integral amount for them, but is there anything we must know about $x$ and $y$?

Also, here is the way I did it:
[hide]We know that the cost of each shirt/cap is greater than $\$0$.  To find a range for how much Joe payed, we need to find the greatest amount he could have spent, and the least (looking at the maximum price of each item).  A shirt is $>\$0$, so the maximum price of a cap is $\frac{104}{11}$.  Joe bought 2 caps, so one of the end points of the amount he payed is $\frac{208}{11}=18\frac{10}{11}$.  A cap is also $>\$0$, so caps cost at maximum $\frac{104}{6}=\frac{52}{3}=17\frac13$, so that is the other end of his price range.  The only integer in between is 18, so he payed $\$18$.  The rest follows.   :D  :) [/hide]

Seems you have a nicer solution. :)  And I guess you have to know how many caps and shirts Joe bought? If not we cannot set up the limits or actually deduce how much 1 cap costs. Not absolutely sure though.  :blush:

Well, you don't want both of them to be even, because then you could have amounts ending in $\$x.50$, which could change the problem.

Also, the method of solving the problem could change considerably based on the information about Joe.


Prove that $\frac{a}{b+c-a} + \frac{b}{a+c-b} + \frac{c}{a+b-c} \geq 3$

where a, b, c are sides of a triangle.

hint
[hide] let $b+c-a=x$ and $a+c-b=y$ and $a+b-c=z$ where x, y, z are positive real numbers as a+b>c, b+c>a, a+c>b in any nondegenerate triangle

[/hide]

solution
[hide]

let $b+c-a=x$ and $a+c-b=y$ and $a+b-c=z$ where x, y, z are positive real numbers as $a+b>c$, $b+c>a$, $a+c>b$ in any nondegenerate triangle

$\therefore$ we obtain, $\frac{y+z}{2x}+\frac{x+z}{2y} +\frac{x+y}{2z}$

$=\frac{y}{2x}+\frac{z}{2x}+\frac{x}{2y}+\frac{z}{2y}+\frac{x}{2z}+\frac{y}{2z}$

$=( \frac{x}{2y}+\frac{y}{2z}+\frac{z}{2x}) + (\frac{x}{2z}+\frac{y}{2x}+\frac{z}{2y})$

The AM-GM inequality gives:

$\frac{\frac{x}{2y}+\frac{y}{2z}+\frac{z}{2x}}{3}\geq{\sqrt[3]{\frac{x}{2y}\cdot \frac{y}{2z}\cdot \frac{z}{2x}}}$

$\implies \frac{x}{2y}+\frac{y}{2z}+\frac{z}{2x} \geq 3\sqrt[3]{\frac{1}{8}} = \frac{3}{2}$

Similarly $\frac{x}{2z}+\frac{y}{2x}+\frac{z}{2y} \geq \frac{3}{2}$

adding gives the desired result.

$\frac{y+z}{2x}+\frac{x+z}{2y} +\frac{x+y}{2z} \geq 3$

[/hide]

Just AM-HM then you get

LHS $\geq a/b + a/c + b/a + b/c + c/a + c/b - 3$

Then AM-GM $a/b + a/c + b/a + b/c + c/a + c/b - 3\geq 6-3=3$


AMC
AIME
geometry
3D geometry
prism
Euler
trigonometry

Ok, here's the deal: I haven't had time to prepare a nice solutions packet, so take this contest in[b] 3 hours [/b]and PM me your answers by 11:59PM (Eastern U.S. Time) on Sunday.  We will have a lively solutions/results/discussion thread on Monday night.

All answers are integers between 000 and 999 inclusive.  If you find an error in the answer formatting, just post your concern here and PM the precise answer (in simplest form) in case I don't get back to you quickly.  Good luck!

[hide]
1. The length of a diagonal connecting opposite vertices of a rectangular prism is 47. Determine its volume, given that one of its dimensions is 2 and that the other two dimensions differ by $\sqrt{2005}$.

2. Compute the sum of the prime divisors of $1^2 + 2^2 + 3^2 + \cdots + 2005^2$.

3. $x$, $y$, and $z$ are positive integers. Let $N$ denote the number of solutions of $2x + y + z = 2004$. Determine the remainder obtained when $N$ is divided by 1000.

4. The walkway to the front of Euler Elementary School is a $6 \times 15$ grid. ACME Sidewalk company has been chartered to tile it with 15 non-overlapping $2 \times 3$ concrete slabs. Compute the number of tilings ACME can choose from for this task.

5. If $\frac{1}{0!10!} + \frac{1}{1!9!} + \frac{1}{2!8!} + \frac{1}{3!7!} + \frac{1}{4!6!} + \frac{1}{5!5!}$ is written as a common fraction reduced to lowest terms, the result is $\frac{m}{n}$. Compute the sum of the prime divisors of $m$ plus the sum of the prime divisors of $n$.

6. Let $F$ be a sequence of positive integers defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for integers $n \ge 1$. Determine the greatest common factor of $F_{1495}$ and $F_{1989}$.

7. Line segments $\overline{AB}$ and $\overline{CD}$ intersect at $P$ such $AP = 8, BP = 24, CP = 11$, and $DP = 13$. Line segments $\overline{DA}$ and $\overline{BC}$ are extended past $A$ and $C$ respectively until they intersect at $Q$. If $\overline{PQ}$ bisects $\angle{BQD}$, then $\frac{AD}{BC}$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m+n$.

8. Determine the smallest positive integer $n$ for which
\[
\frac{2\sin\left(16^{\circ}\right)\cos\left(32^{\circ}\right)}{\sin\left(29^{\circ}\right)} = \sqrt{2} - 2\sin\left(n^{\circ}\right) \]
9. Three spheres $S_1$, $S_2$, and $S_3$ are mutually externally tangent and have radii 2004, 3507, and 4676 respectively. Plane $P$ is tangent $S_1$, $S_2$, and $S_3$ at $A$, $B$, and $C$ respectively. The area of triangle $ABC$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Determine the remainder obtained when $m+n$ is divided by 1000.

10. $p$, $q$, and $r$ are positive real numbers such that

$p^2 + pq + q^2 = 211$
$q^2 + qr + r^2 = 259$
$r^2 + rp + p^2 = 307$

Compute the value of $pq + qr + rp$.

11. $x_1, x_2$, and $x_3$ are complex numbers such that

$x_1 + x_2 + x_3 = 0$
$x_1^{\phantom{0}2} + x_2^{\phantom{0}2} + x_3^{\phantom{0}2} = 16$
$x_1^{\phantom{0}3} + x_3^{\phantom{0}3} + x_3^{\phantom{0}3} = -24$

Let $\gamma = \mbox{min}\left(\left|x_1\right|,\left|x_2\right|,\left|x_3\right|\right)$, where $|a + bi| = \sqrt{a^2+b^2}$ and $i = \sqrt{-1}$. Determine the value of $\gamma^6 - 15\gamma^4+\gamma^3+56\gamma^2$.

12. $ABC$ is a scalene triangle. The circle with diameter $\overline{AB}$ intersects $\overline{BC}$ at $D$, and $E$ is the foot of the altitude from $C$. $P$ is the intersection of $\overline{AD}$ and $\overline{CE}$. Given that $AP = 136$, $BP = 80$, and $CP = 26$, determine the circumradius of $ABC$.

13. Let $T_n$ be defined by
\[
T_n := \sum_{i=1}^{n-1} \left\lfloor \frac{i}{n-i} \right\rfloor
\]
where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Compute $T_{2009} - T_{2005}$.

14. In acute triangle $ABC$, $BC = 10$, $CA = 12$, and $AB = 14$. $\omega_1, \omega_2$, and $\omega_3$ are circles with diameters $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively. Let $B$ denote the boundary of the region interior to the three $\omega_i$. $\Omega$ is the circle internally tangent to the three arcs of $B$. The radius of $\Omega$ can be expressed as $\frac{m-p\sqrt{q}}{n}$, where $m, p$, and $n$ are positive integers with no common prime divisor and $q$ is a positive integer not divisible by the square of any prime. Compute $m+n+p+q$.

15. Let $O = (0,0)$ and $A = (14,0)$ denote the origin and a point on the positive $x$-axis respectively. $B = (x,y)$ is a point not on the line $y = 0$. These three points determine lines $l_1, l_2$, and $l_3$. Let $P_1, \dots, P_n$ denote all of the points that are equidistant from $l_i$ for $i = 1, 2, 3$. Let $Q_j$ denote the distance from $P_j$ to the $l_i$ for $j = 1, \dots, n$. If

$Q_1 + \cdots + Q_n = \frac{103}{3}$
$\frac{1}{Q_1} + \cdots + \frac{1}{Q_n} = \frac{2}{3}$

then the maximum possible value of $x+y$ can be expressed as $\frac{u}{v}$, where $u$ and $v$ are relatively prime positive integers. Determine $u + v$.[/hide]

Oooo, thanks. I would do it now, but I'm supposed to be doing chinese hw...

Four-and-a-half hours?  Am I missing something?  I thought these fifteen questions were supposed to take us just three hours--for that matter, this sounds like the SAT all over again...

But I was planning on taking this on the road (cause I will be for three  hours), and it will seriously crimp my game plan to not have sufficient time without tacking an additional couple--at 5:00 in the morning??  Doubtless, I won't hear anything before I leave at your 7:30.  Still, the clarification would be nice.  (And thanks for the AIME.)

Am I being extremely dumb, or does #1 really have an answer greater than 1000?  :?

Thank you very much for the AIME.

mock 7 already? what did i miss in just  one day?

this is not due until sunday right?

i have music testing tomorrow ^^

I would just take it in 3 hrs.  It seems like Mildorf has been practicing so much for USAMO that he got the time limits mixed up...just as long as he didn't get the difficulties mixed up also  :D :P  :)

Ah No! I got a 7. 3 careless mistakes :( , 2 of the them were because of laziness to check:( .

4 1/2 hours = Too much USAMO practice. I know the feeling.

Just as a point: I don't think anyone who gets to 11-15 on these mock AIMEs can reasonably say they aren't "difficult" and just "normal".

[quote="filletwho"]What if the exact same problems show up on the real AIME?[/quote]

Then you should consider yourself lucky. I know I would! :lol:

[quote] What if the exact same problems show up on the real AIME? [/quote]

Then we should consider Mildorf to be both psychic and extremely generous.

[quote="filletwho"]solutions...neone? I want to know how to do the fibonacci one.[/quote]

If n|m, the F_n|F_m, not sure how to prove, but I got it by doing patterns.

#7. Consider triangle PAD, PBC. Drop altitude from P to AD, BC, since those altitudes are equal because P lies on the angle bisector, [PBC]/[PAD] = BC/AD.

[PBC]= 11*24sinCPB /2,   [PAD] = 8*13 sinAPD /2, so

[PBC]/[PAD] = 11*24/(8*13) = 33/13=BC/AD. m+n=46

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3742684_2005_mildorf_mock_aime_55]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Just a friendly reminder: [b]please, write in English unless you are in a national subforum or sending a PM[/b].:)


modular arithmetic
algorithm
number theory
greatest common divisor
pigeonhole principle
AMC
USA(J)MO

Is given the set of natural numbers $ 1,2,3......................2n$ prove that if we choose $ n\plus{}1$ of this numbers . There will be at least  2 relative primes in our numbers.

[hide="probably wrong and pretty informal"]
One requirement for a list of natural numbers to have a common factor is that they all must be more than one apart (ie, any two numbers can't be represented in the form $ a$ and $ a \plus{} 1$, where a is a natural number, because $ \gcd(a,a\plus{}1) \equal{} 1$). If we pick $ n$ numbers from the set of integers from 1 to $ 2n$, the only way to make sure that none of the numbers are one apart would be to make them all 2 apart (all odd or all even). Therefore, picking one more number would have to be of the opposite parity and make at least two numbers one apart, which means that they'll be relatively prime.

That probably didn't make any sense...
[/hide]

First we have this "lemma":

For any positive integer $ m$, $ m$ and $ m \plus{} 1$ are relatively prime.

Proof: For every proper factor $ p$ of $ m$, $ m \equiv 0 \pmod p$. However, $ m \plus{} 1 \equal{} 1 \pmod p$, which implies that all proper factors of $ m$ will NOT be proper factors of $ m \plus{} 1$, and thus they are relatively prime.

(Alternatively, we could use the Euclidean Algorithm to determine that $ \text{gcd}(m, m \plus{} 1) \equal{} \text{gcd}(m, 1) \equal{} 1$.)

Now that we have that lemma solved, look at the original problem.

Pair the terms $ (1, 2), (3, 4), (5, 6), \dots, (2n \minus{} 1, 2n)$. There are $ n$ pairs. By the Pigeonhole Principle, choosing $ n \plus{} 1$ natural numbers implies that there must be at least one pair in which both elements of that pair are selected. Since they are 1 apart, they are relatively prime by the above lemma.

[hide="One Sentence"]
There are $ n$ evens and $ n$ odds in that set so, even if all the evens are chosen, one more member of the set needs to be chosen and thus making a pair of relatively prime numbers.
[/hide]

Hope that is sufficient.  If it is not, say so.  :)

I think you mean "even if all the evens are chosen"... since 3 and 7 are already relatively prime, for example...

Edit: darn, you edited it already!

This is what he said before:

[quote="AstroPhys"]There are  evens and  odds in that set so, even if all the odds are chosen, one more member of the set needs to be chosen and thus making a pair of relatively prime numbers.[/quote]

I just changed it before you posted.  :)

And just another thing, your proof falls at this point.

"... even if all the evens were chosen..."

The phrase "even if" shows that you haven't proved all cases... even though it's hopelessly trivial, USAMO graders wouldn't give you a full score... not that this would appear on a USAMO... lolz :D

Yeah, I thought it was missing a few cases. However, I considered it more of picking "optimal" numbers from the set, numbers that are obviously not relatively prime. :lol:  Thanks.


For every nonnegative integer $ n$ and every real number $ x$ prove the inequality:

$ |\cos x|\plus{}|\cos 2x|\plus{}\ldots\plus{}|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$

[url=http://www.artofproblemsolving.com/community/c6h1357679p7435504]Proof of RagvaloD:[/url]
$|\cos x|+|\cos 2x| \geq \frac{\sqrt{2}}{2}$
If $n=2k-1$ then $|\cos x|+...+|\cos 2^n x|\geq k\frac{\sqrt 2}{2}=\frac{n+1}{2\sqrt{2}}>\frac{n}{2\sqrt{2}}$
If $n=2k$ then $ |\cos x|+...+|\cos 2^n x|\geq k\frac{\sqrt 2}{2}+|\cos x| \geq \frac{n}{2\sqrt{2}}$

For every real number $ x$ prove the inequality:
$$|(\sin x-1)(\cos x+1)-\sin^2x|\leq \frac{9+3\sqrt 3}{4}$$


How do different people actually study maths?

Do you guys take notes or read carefully and try to keep in memory etc?

If you have an excellent memory,of course you can study maths with only reading notes.
For example,i can't study maths without writing,if i read a solution for some problem,i'll forget solution after some time,that's why i make an abstract.Before reading any solution you must at first try to solve it yourself,and only then (of course,if you can't solve it), read a solution.

The Way I Learn:

1. Read/be lectured on topics and see some example problems.
2. Try problems
3. Get stuck on all or most of them after working on them for hours
4. Gather questions
5. Ask someone smart about it
6. Ask questions until you understand how to the problem OR if they just give you advice, try to solve problem again
9. (repeat until understand the problems)
10. Cover up the solutions
11. repeat step 2

Learning is difficult. Trial and error accompanied with intuition, analysis, motivation, and logic force me to learn.

You can't truly know something unless you understand it. It's best to understand a topic and apply that to a general problem.

What I find helpful for me is to familiarize myself with what I intend to learn. This can mean simply flipping through a book about it, looking at things, trying to pick up some general things. A few weeks later, I'll find that coming back to the material, I can learn a lot easier. It's like my mind has processed the material over time, then I can learn the details. :D


Princeton
college
Harvard
Yale
Stanford
University of Chicago
geometry

I was wondering weather anybody here is interested in economics? Maybe studying it in college?

yes, i am. i'm planning to take a course in economic science at a local university this summer.. unless i get accepted to TASS (a humanities seminar at ann arbor).. which is highly unlikely, since the seminar only accepts 18 sophomores. it's a little early to make a sure decision, but economics is an appealing major.. to me and about 4853475 of my peers. it's an insanely competitive field, i've heard. i'm considering schools such as princeton, harvard, yale, and georgetown for undergrad, and stanford and UChicago for grad. and yourself?

Oh. I am taking an Economics course in college, Microeconomics. I am a business major I am planning to concentrate in Accounting and Economics. I think its an interesting subject. I am in Drexel University in Philadelphia. It's not one of the Ivy league schools but its one of the better schools in the area (located right beside UPENN) and has a CO-OP program (which is when you work for several terms). Why are you saying Economics is a competitive field? Do you think it requires a lot of Math?

There are many brilliant and talented people in each and every field. Choosing whether or not to pursue a certain major based on "other people" in the industry would eliminate just about every good choice out there. Economics is indeed quite popular these days, but in my opinion the more the merrier. I'm also looking at something along the lines of business management and financial services.

In any sort of business, it's not just about what you know. I'm sure mathematical problem solving will help, if in no other way than the fact that it's trained you to think about a problem from different angles and look for unique solutions. But more importantly, consider WHO you know. The right contacts, the people skills, the connections...that's what should get you miles ahead of everyone else - not the MBA or Ph.D. from Princeton. Besides, I'd like to have fun in college. High school is too stressful, working won't be much better...so what's left of life to enjoy?

This is somewhat off topic, but I saw Joseph Stiglitz in my university sveral months ago. He was talking about how disastrous was Bushenomics and how Kerry could do better. (It was before the election, so he was trying to get kerry elected). He was a nobel prize winner in econ and a head econ advisor to Clinton for some years.


[quote="Number1"]Construc a triangle $ ABC$ given two foots $ A'$ and $ B'$ of altitudes
from vertex $ A$ and $ B$ and midpoint $ M$ of side $ AB$.[/quote]

The triangle is not uniquely defined.

[geogebra]c5b36dc2acc25cbae0202d1040a9af5a300e6fd0[/geogebra]

Use $ OB'$ as radius and $ O\equal{}M$ as center to construct the circle. Then $ A,B$ can be any points as long as $ AB$ is a diameter of the circle and $ \triangle ABC$ is a triangle.


Poles A,B, and $P_{1}, P_{2}, P_{3},. . .$ are vertical line segments with bases on the x-axis. The tops of poles A and B are (0,1) and (200,5), respectively. A string S connects (0,1) and (200,0) and intersects another string connecting (0,0) and (200,5) at point T. Pole P1 is constructed with T as its top point. For each integer i > 1, pole $P_{i}$ is constructed so that its top point is the intersection of S and the line segment connecting the base of $P_{i-1}$ (on the x-axis) and the top of pole B. Find the height of pole $P_{100}$.

[quote="sonny"][quote="veryfriendly"][hide]I got $\frac{1}{21}$ as my answer is that correct?[/hide][/quote]

Yes your answer is correct however can you please provide a solution?[/quote]

My way was a mixture of brute force and guessing so it isn't the best way. I just found the equations of the lines and and their intersections. I found a pattern of the heights which went like 5/6, 5/7, 5/8.........So my way is long and extremely time consuming.....sorry I couldn't find another way...


inequalities
calculus
integration
function
trigonometry
real analysis
real analysis unsolved

Find the best constants $a,b$ such that for any continuously differentiale function $f$ with real values we have
   $\int_{0}^{1}{f^2(x)dx}\leq a\cdot (\int_{0}^{1} f(x)dx)^2+b\cdot\int_{0}^{1}{ (f'(x))^2dx}$

Let
\[k=\int_0^1 f(x)dx.\]
Then $f(x)=k+g(x)$, where the average of $g$ in $[0,1]$ is 0. Now
\[\int_0^1 f(x)dx=\int_0^1 (k^2+2kg(x)+g^2(x))dx=k^2+\int_0^1 g^2(x)dx.\]

By a well known inequality, the integral of $g^2$ is less or equal than $\pi^{-2}$ times the integral of $(g')^2$ (I'm not so sure of the constant, but it should be realized by $g(x)=\cos(\pi x)$).

This gives the desired inequality with $a=1$ and $b=\pi^{-2}$.

On the other hand, $a=1$ cannot be improved (use $f=$ constant), and also $b$ cannot be better that in the known inequality (use $f$ with 0 average).


analytic geometry
inequalities
induction
rearrangement inequality
inequalities proposed

Let $ i,\ j$ be integers such that $ 1\leq i\leq n,\ 1\leq j\leq n$. A number $ ij\ (\text{the product of i and j})$ is recored in the coordinate of $ (i,\ j)$.
Denote $ S_1$ the sum of the numbers recorded in the line segment passing through two points $ (1,\ 1)$ and $ (n,\ n)$, and $ S_2$ the sum of the numbers recorded in the line segment passing through two points $ (1,\ n)$ and $ (n,\ 1)$. Let $ p_1,\ p_2,\ \cdots ,\ p_n$ be any permutations of $ 1,\ 2,\ \cdots n$. If $ S$ is the sum of the numbers recorded in $ n$ points $ (1,\ p_1),\ (2,\ p_2),\ \cdots (n,\ p_n)$, then prove that $ S_2\leq S\leq S_1$.

Is it of [i]that[/i] much use to post a particular case of the rearrangement inequality and to bump it two times?

  darij

$ S_1 \equal{} \sum_{k \equal{} 1}^n k\cdot k \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6},\ S_2 \equal{} \sum_{k \equal{} 1}^n k(n \plus{} 1 \minus{} k) \equal{} \frac {n(n \plus{} 1)(n \plus{} 2)}{6}$

Since $ \{1,\ 2,\ \cdots ,\ n\} \equal{} \{p_1,\ p_2,\ \cdots ,\ p_n\}$, we have $ S \equal{} \sum_{k \equal{} 1}^n kp_k\leq \sum_{k \equal{} 1}^n \frac {k^2 \plus{} p_k^2}{2}\leq \frac 12 (S_1 \plus{} S_1) \equal{} S_1$

I'm stuck here.

Let's better prove the rearrangement inequality in its generality. (The proof below is longer than the standard one, but it avoids trouble with equal variables.)

[color=blue][b]Definition.[/b] A sequence $ \left(a_1,a_2,...,a_n\right)$ of reals is said to be [i]nondecreasing[/i] if and only if $ a_1\leq a_2\leq ...\leq a_n$. A sequence $ \left(a_1,a_2,...,a_n\right)$ of reals is said to be [i]nonincreasing[/i] if and only if $ a_1\geq a_2\geq ...\geq a_n$.[/color]

[color=blue][b]Theorem 1, the rearrangement inequality.[/b] Let $ \left(a_1,a_2,...,a_n\right)$ and $ \left(b_1,b_2,...,b_n\right)$ be two nonincreasing sequences of reals. Let $ \pi\in S_n$ be a permutation of the set $ \left\{1,2,...,n\right\}$. Then, $ \sum_{i = 1}^{n}a_ib_{n + 1 - i}\leq\sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n}a_ib_i$.[/color]

[i]Proof of Theorem 1.[/i] We will first prove that $ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n}a_ib_i$ by induction over $ n$. The induction base, $ n = 1$, is obvious.

Induction step: Fix some natural $ n$. We want to prove that

[color=green][b](1)[/b][/color] $ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n}a_ib_i$ for any two nonincreasing sequences $ \left(a_1,a_2,...,a_n\right)$ and $ \left(b_1,b_2,...,b_n\right)$ of reals and any permutation $ \pi\in S_n$.

We assume (as the induction assumption) that a similar assertion holds for $ n - 1$ instead of $ n$, i. e. that

[color=green][b](2)[/b][/color] $ \sum_{i = 1}^{n - 1}A_iB_{\psi\left(i\right)}\leq\sum_{i = 1}^{n - 1}A_iB_i$ for any two nonincreasing sequences $ \left(A_1,A_2,...,A_{n - 1}\right)$ and $ \left(B_1,B_2,...,B_{n - 1}\right)$ of reals and any permutation $ \psi\in S_{n - 1}$.

So let's prove [color=green][b](1)[/b][/color]. Two cases are possible:

[i]Case 1: The permutation $ \pi$ leaves the element $ n$ fixed.[/i] In this case, $\pi\left(n\right) = n$. Thus, $ \pi$ can be seen as a permutation of the set $ \left\{1,2,...,n - 1\right\}$, i. e., as an element of $ S_{n - 1}$. Thus, [color=green][b](2)[/b][/color] (applied to $ \psi = \pi$, $ A_i = a_i$ and $ B_i = b_i$ for every $ i\in\left\{1,2,...,n - 1\right\}$) yields $ \sum_{i = 1}^{n - 1}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n - 1}a_ib_i$. Hence,

$ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)} = \sum_{i = 1}^{n - 1}a_ib_{\pi\left(i\right)} + a_nb_{\pi\left(n\right)}\leq\sum_{i = 1}^{n - 1}a_ib_i + a_nb_n$     (since $ \sum_{i = 1}^{n - 1}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n - 1}a_ib_i$ and $ \pi\left(n\right) = n$)
$ = \sum_{i = 1}^{n}a_ib_i$,

and thus [color=green][b](1)[/b][/color] is proven, i. e., the induction step is complete in Case 1.

[i]Case 2: The permutation $ \pi$ doesn't leave the element $ n$ fixed.[/i] In this case, $ \pi\left(n\right)\neq n$, and thus $ \pi\left(n\right) < n$. Let $ \psi = \left(n,\pi\left(n\right)\right)\circ\pi$ (where, $\left(n,\pi\left(n\right)\right)$ denotes the transposition in $S_n$ which switches $\pi\left(n\right)$ with $n$). Then, the permutation $ \psi$ leaves the element $ n$ fixed, so it can be seen as a permutation of the set $ \left\{1,2,...,n - 1\right\}$, i. e., as an element of $ S_{n - 1}$. Thus, [color=green][b](2)[/b][/color] (applied to $ A_i = a_i$ and $ B_i = b_i$ for every $ i\in\left\{1,2,...,n - 1\right\}$) yields $ \sum_{i = 1}^{n - 1}a_ib_{\psi\left(i\right)}\leq\sum_{i = 1}^{n - 1}a_ib_i$.

Note that $ \psi\left(i\right) = \pi\left(i\right)$ for every $ i\in\left\{1,2,...,n\right\}$ satisfying $ i\neq n$ and $ i\neq\pi^{ - 1}\left(n\right)$ (because of $ \psi = \left(n,\pi\left(n\right)\right)\circ\pi$). Also, $\pi^{-1}\left(n\right) \neq n$ (since $ \pi\left(n\right)\neq n$).

Also note that $ a_{\pi^{ - 1}\left(n\right)}\geq a_n$ (since $ \pi^{ - 1}\left(n\right)\leq n$, and since the sequence $ \left(a_1,a_2,...,a_n\right)$ is nonincreasing) and $ b_n\leq b_{\pi\left(n\right)}$ (since $ n\geq \pi\left(n\right)$, and since the sequence $ \left(b_1,b_2,...,b_n\right)$ is nonincreasing). Thus,

$ \left( a_{\pi^{ - 1}\left( n\right) }b_{n} + a_{n}b_{\pi\left( n\right) }\right) - \left( a_{n}b_{n} + a_{\pi^{ - 1}\left( n\right) }b_{\pi\left(n\right) }\right)$
$ = \underbrace{\left( a_{\pi^{ - 1}\left( n\right) } - a_{n}\right) }_{\geq0\text{ (since }a_{\pi^{ - 1}\left( n\right) }\geq a_{n}\text{)} }\underbrace{\left( b_{n} - b_{\pi\left( n\right) }\right) }_{\leq0\text{ (since }b_{n}\leq b_{\pi\left( n\right) }\text{)}}\leq0$,

so that

[color=green][b](3)[/b][/color] $ a_{\pi^{ - 1}\left( n\right) }b_{n} + a_{n}b_{\pi\left( n\right) }\leq a_{n}b_{n} + a_{\pi^{ - 1}\left( n\right) }b_{\pi\left(n\right) }$.

Now,

$ \sum_{i = 1}^{n}a_{i}b_{\pi\left( i\right) } = \sum_{i\in\left\{ 1,2,...,n\right\} }a_{i}b_{\pi\left( i\right) }$
$ = \sum_{\substack{i\in\left\{ 1,2,...,n\right\} ; \\
i\neq n;\ i\neq\pi^{ - 1}\left( n\right) }}a_{i}\underbrace{b_{\pi\left( i\right) } }_{\substack{ = b_{\psi\left( i\right) }\text{ (since} \\
\psi\left( i\right) = \pi\left( i\right) \text{)}}} + a_{\pi^{ - 1}\left( n\right) }\underbrace {b_{\pi\left( \pi^{ - 1}\left( n\right) \right) }}_{ = b_{n}} + a_{n} b_{\pi\left( n\right) }$
(since $n$ and $\pi^{-1}\left(n\right)$ are two distinct elements of $\left\{1,2,...,n\right\}$)
$ = \sum_{\substack{i\in\left\{ 1,2,...,n\right\} ; \\
i\neq n;\ i\neq\pi ^{ - 1}\left( n\right) }}a_{i}b_{\psi\left( i\right) } + a_{\pi^{ - 1}\left( n\right) }b_{n} + a_{n}b_{\pi\left( n\right) }$
$ \leq \sum_{\substack{i\in\left\{ 1,2,...,n\right\} ; \\
i\neq n;\ i\neq\pi ^{ - 1}\left( n\right) }}a_{i}b_{\psi\left( i\right) } + a_{n}b_{n} + a_{\pi^{ - 1}\left( n\right) }b_{\pi\left(n\right) }$

(by [color=green][b](3)[/b][/color]). Since

$ \sum_{i = 1}^{n}a_{i}b_{\psi\left( i\right) } = \sum_{i\in\left\{1,2,...,n\right\} }a_{i}b_{\psi\left( i\right) }$
$ = \sum_{\substack{i\in\left\{ 1,2,...,n\right\} ; \\
i\neq n;\ i\neq\pi^{ - 1}\left( n\right) }}a_{i}b_{\psi\left( i\right) } + a_{n}\underbrace{b_{\psi\left( n\right) }}_{\substack{ = b_{n}\text{ (since} \\
\psi\left(n\right) = n\text{)}}} + a_{\pi^{ - 1}\left( n\right) }\underbrace{b_{\psi\left( \pi^{ - 1}\left( n\right) \right) }}_{\substack{ = b_{\pi\left( n\right) }\text{ (since} \\
\psi\left( \pi^{ - 1}\left( n\right)\right) = \pi\left( n\right) \text{)}}}$
(since $n$ and $\pi^{-1}\left(n\right)$ are two distinct elements of $\left\{1,2,...,n\right\}$)
$ = \sum_{\substack{i\in\left\{ 1,2,...,n\right\} ; \\
i\neq n;\ i\neq\pi^{ - 1}\left( n\right) }}a_{i}b_{\psi\left( i\right) } + a_{n}b_{n} + a_{\pi^{ - 1}\left( n\right) }b_{\pi\left( n\right) }$,

this becomes

$ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq \sum_{i = 1}^{n}a_ib_{\psi\left(i\right)} = \sum_{i = 1}^{n - 1}a_ib_{\psi\left(i\right)} + a_nb_{\psi\left(n\right)}$
$ \leq\sum_{i = 1}^{n - 1}a_ib_i + a_nb_n$     (since $ \sum_{i = 1}^{n - 1}a_ib_{\psi\left(i\right)}\leq\sum_{i = 1}^{n - 1}a_ib_i$ and $ \psi\left(n\right) = n$)
$ = \sum_{i = 1}^na_nb_n$.

Therefore, [color=green][b](1)[/b][/color] is proven, i. e., the induction step is complete in Case 2.

Now, the induction is complete in both cases, and the inequality

[color=green][b](4)[/b][/color] $ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n}a_ib_i$

is proven.

Let us now prove the inequality $ \sum_{i = 1}^{n}a_ib_{n + 1 - i}\leq\sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}$.

Define a permutation $\rho$ of $\left\{1, 2, ..., n\right\}$ by

$\rho\left(i\right) = n+1-\pi\left(i\right)$ for all $i \in \left\{1, 2, ..., n\right\}$.

Thus,

$n+1-\underbrace{\rho\left(i\right)}_{=n+1-\pi\left(i\right)} = n+1 - \left(n+1-\pi\left(i\right)\right) = \pi\left(i\right)$

for every $i \in \left\{1, 2, ..., n\right\}$.

Now, $b_1 \geq b_2 \geq ... \geq b_n$ (since the sequence $ \left(b_1,b_2,...,b_n\right)$ is nonincreasing). Thus, $-b_n \geq -b_{n-1} \geq ... \geq -b_1$. In other words, the sequence $ \left( - b_n, - b_{n - 1},..., - b_1\right)$ is nonincreasing. Hence, we can apply the inequality [color=green][b](4)[/b][/color] to the sequence $ \left( - b_n, - b_{n - 1},..., - b_1\right)$ and the permutation $\rho$ instead of the sequence $ \left(b_1,b_2,...,b_n\right)$ and the permutation $\pi$. Thus we obtain

$ \sum_{i = 1}^{n}a_i\left(-b_{n+1-\rho\left(i\right)}\right)\leq\sum_{i = 1}^{n}a_i\left(-b_{n+1-i}\right)$
$= - \sum_{i = 1}^{n}a_i b_{n+1-i}$.

Hence,

$- \sum_{i = 1}^{n}a_i b_{n+1-i} \geq \sum_{i = 1}^{n}a_i\left(-b_{n+1-\rho\left(i\right)}\right)$
$= - \sum_{i = 1}^{n}a_i \underbrace{b_{n+1-\rho\left(i\right)}}_{\substack{= b_{\pi\left(i\right)} \\ \text{(since } n+1-\rho\left(i\right)=\pi\left(i\right) \text{)}}} = - \sum_{i = 1}^{n}a_i b_{\pi\left(i\right)}$.

Hence, $ \sum_{i = 1}^{n}a_ib_{n + 1 - i}\leq\sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}$.

Thus, both inequalities $ \sum_{i = 1}^{n}a_ib_{n + 1 - i}\leq\sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}$ and $ \sum_{i = 1}^{n}a_ib_{\pi\left(i\right)}\leq\sum_{i = 1}^{n}a_ib_i$ are proven, and Theorem 1 ensues.

  darij


Prove that if $n$ is an integer greater than $1$, then
\[ \frac{4^n}{2\sqrt n}<C_{2n}^n<4^n \]

right side:

for the initial cases, is easy to see that the inequality holds.

suppose that for n<=k the inequality holds.
we have (2k)! < (k!)² . 4^n.
is easy to see that (2k+1).(2k+2) < 4(k+1)². multiply the two inequalities, and the result for n = k+1 is valid.

left side:

for the initial cases, is easy to see that the inequality holds.

suppose that for n<=k the inequality holds.
we have (2k)! . 2 . sqrt(k) > (k!)² . 4^k
is easy to see that (2k+1).(2k+2). (sqrt(k+1)/sqrt(k)) > (k+1)² . 4. multiply the two inequalities and the result for n = k+1 is valid.  :)


I invite everybody, who is interested in the modern neo-surrealism art movement and surreal fantasy artists, to take a look at http://fantasyartdesign.com/3d-fine-art-surrealists/modern-surrealism01.htm  The site also has the best 3d artist gallery, free 3d art fantasy wallpapers and extremely cool, surreal  pop-art optical illusions exhibition. In addition, there are absolutely amazing flash animated desktops and free digital 3d art software.
[img]http://www.fantasyartdesign.com/3d-art-surrealism/02artfantasy3d/artworks/3d-fantasy-art/183d-the-three-graces-s.jpg[/img] [url]http://fantasyartdesign.com[/url]

it seems that the entire site was divided to several parts. Modern surrealism gallery has moved to http://www.neosurrealismart.com/

The section Flash animated desktops has been moved to http://www.artdigitaldesign.com  The site still has  interactive animated desktop wallpapers, and desktop mind and puzzle games.


Two different natural numbers are selected from the set {1, 2, 3,…. 6}. What is the probability that the
greatest common factor of these two numbers is one? Express your answer as a common fraction

Pairs possible:

1 and 2
1 and 3
1 and 4
1 and 5
1 and 6
2 and 3
2 and 5
3 and 4
3 and 5
4 and 5
5 and 6

Probability is $ \frac{11}{C(6,2)}\equal{}\boxed{\frac{11}{15}}$


geometry
circumcircle
incenter
geometry proposed

Given is a non-isosceles triangle $ \triangle ABC$. Let $ M$ be a point in its interior. Denote by $ A'$, $ B'$ and $ C'$ projections of point $ M$ onto lines $ BC$, $ CA$ and $ AB$ and define $ S(M)\equal{}AC'\plus{}BA'\plus{}CB'$. Prove that for two interior points $ M$ and $ N$ of triangle $ \triangle ABC$, $ S(M)\equal{}S(N)$ if and only if $ MN$ is perpendicular to $ OI$, where $ O$ is circumcenter and $ I$ incenter of triangle $ \triangle ABC$.

I think is $ MN \parallel OI$.

the attach was my italian solution.


The championship from the 4th COP is:

(drum rolllllllllllllllllllll :D )

The Original Pi with 10 points!

Congratulations!

The runner-up was BornForMath with 4 points.

Good job, Original Pi Guy!

Are you going to post solutions, Silverfalcon?

I'm working on it but now, I'm trying to write a book from 1st COP to 5th COP (5th COP problems will be added after the contest). The book I'm writing will contain all problems from 1st COP to 5th COP and not only that, it will contain COPSD problems.

I haven't really decided what to do with it when I'm done though. Sell it for cheap price or give to people by free?  :P

[quote="Silverfalcon"].
I haven't really decided what to do with it when I'm done though. Sell it for cheap price or give to people by free?  :P[/quote]

Sell it on E-bay  :rotfl:
It can be a pretty good book though. How many problems?

Well... There are 15 problems in COP 1st (It was different format) and from COP 2nd, we have 29 problems (2 are proof). So, 15+29(4) = 131. But in that book, there will be 3 COPSD so 131+30 = 161 problems.

I will probably add one or two Olympiad problems as "Solvable problem with simple solution." That will be like "Big Picture" in AoPS book.

If many people think it's good idea to get this book (I haven't decided how much the price will be), I might add Mock AMC-12 and/or Mock AIME (if the buyer wants Mock Mathcounts or Mock AMC-10, it can be changed to it as well) and raise the price a bit (not sure how much) :D .

Again, I'm not trying to build fortune on this thing anyway. I like writing problems and having my own contest. The book will contain full solutions on each problem as well. It won't be "Answers ONLY" but "Answers and Solutions."

You guys can talk about this issue more.. Maybe discussion on this matter can help determine what the price will be.

One thing for sure: It will be cheaper than AoPS book and eh.. around 10 dollars maybe?

[quote="PenguinIntegral"]Distrubute it free on AoPS, on the condition they must worship you eteranlly.[/quote]

I don't think that will work  :roll:  but nice idea  :D

On average, a problem book is no more than 15 dollars (paperback), so 10 dollars is a good price to start with. After all, you're not trying to use this as a main source of income, right?

I think many people will appreciate your book, given not only it has real competition probelms, but also full solutions  :)

I think 10 dollars seem pretty good deal. I'm pretty sure that I'll add Mock AMC/MC/AIME because that will attract more people and luckily, my lateness Mock AMC-12 was closest to the level of the real one  :lol: (which was what I planned. It was little bit easier in some way though).

What I might do is first, finish working and second, give the free copy to who want them (maybe about 2 or 3 people). And using the opinions I get, I'll edit and finally post the final copy of it.

If you have any more ideas or suggestions, post here. This seem to be a good place to discuss it.

[quote="Silverfalcon"]What I might do is first, finish working and second, give the free copy to who want them (maybe about 2 or 3 people)[/quote]

I want it. :lol:

I'm thinking of giving it to 4 people.

I guess two people are already talking about it since 236factorial and chess64 are interested. I would be happy to have at least one high school wanting it and especially one of the people who are active on the forum. But it doesn't matter much.  :D

Anyone want it?

(236factorial and chess64 are not the ones who will get them YET. I might just randomly pick four people)

I would like to have one, but if someone is desperate for the book, I'll let them have it instead.

[quote="236factorial"]....if someone is desperate for the book...[/quote]

Oooh!! Me! Me! :D :D By "active on the forum", do you mean: "runs competitions, is a mod, has 1654 posts, and 9.24 posts per day"?? Because I fit that description exactly. :)

whoa, I just realized i had over 15.9 posts a day  :o

I think 9 posts a day is pretty active  :D

Yeah, that's pretty active. Wow.. I'm glad to see many people are interested.

[quote="shobber"]I am also interested in it.  :)[/quote]
I give up. There are more people who need it more than I do.

I'd like to have it, and I'm in highschool, but I dunno if I'm considered active, since I only have 2.6 posts a day.

I don't quite need it (unless for future uses), because i'm still in middle school, so the COP's won't help a lot.

I'm still interested in having it :)
Although I have only 4.70 posts per day..  :roll:


[hide]Err... if you use...umm....very complicated mathematics (namely guess and check) you get that x=10!  :)
[/hide]

Pay no attention to that... i made a mistake...  :blush: sorry.

[quote="Drunken_Math"][hide]Err... if you use...umm....very complicated mathematics (namely guess and check) you get that x=10!  :) [/hide][/quote]

Umm... somehow I don't think $0 = -90$ works quite right... ;)

well the equation results into $x^2+x-10=0$ and $x^2-x-9=0$. this took me a while to factor $x^4-20x^2+x+90$. i find MysticTerminator's method interesting :)

Thats how i do it. I thought it was really cool, so i posted. Solve for 10. Wild! We were given all these equations with loads of variables, and we had to find the easiest one to work it. This was the best one though.


number theory
modular arithmetic
\/closed

I don't know modular arithmetic, however I do have AoPS Volume 1 which teaches modular arithmetic. However, I know pretty much the rest of the material. Should I take the Intro. to Number Theory Class?

-rts2007

Introduction to Number Theory is pretty basic, and won't go into much more mods then you would get form Volume 1 (at least it didn't when I took the class).

Intro to Number Theory goes as far as to teach solving systems of congruences and the Chinese Remainder Theorem.  Additionally, there's lots of challenging problems in the problem sets and weekly practice problems.  The feedback you receive while taking a class is really helpful, and might make taking the class worth it for you (yes, I did take it).

If you have a pretty good handle on mods, you should be OK moving on to the intermediate level of number theory.  Check out the diagnostic test -- if you have trouble with just a couple problems, you can probably get by just working through the challenge problems in the text.


Let $x$ be a postive integer. Let $a$ and $b$ be positive integers such that $a\ge b$. Prove that for some $(a,b)$, $ax+b$ is prime.

isn't this just a weaker case of dirichlet's theorem?

[quote]Take the product of every prime number that is not a factor of x, up to the largest prime factor of x. Then let b = 1. ax + b is then prime[/quote]

This is incorrect. The product of all primes up to x plus one is not necessarily prime. For instance:

$2*3*5*7*11*13+1=30031=59*509$

[hide="Er..."] Actually, this result is much weaker than Dirichlet's Theorem.  A construction:

Take the first prime number greater than $x^{2}$.  The quotient upon division by $x$ must be at least $x$, whereas the remainder must be less than $x$.  Let the quotient be $a$ and the remainder $b$.  Then $a \ge x > b \implies a > b$ and $ax+b$ is prime. [/hide]


function
combinatorics unsolved
combinatorics

Two positive integers are written on the blackboard. Initially, one of them is $2000$ and the other is smaller than $2000$. If the arithmetic mean $ m$ of the two numbers on the blackboard is an integer, the following operation is allowed: one of the two numbers is erased and replaced by $ m$. Prove that this operation cannot be performed more than ten times. Give an example where the operation is performed ten times.

Define the function $ f$ of the two numbers written on the board as: $ f(a,b)\equal{}a\minus{}b$

$ (a,b) \mapsto (a,\frac{a\plus{}b}{2})$ with $ f(a,\frac{a\plus{}b}{2})\equal{}a\minus{} \frac{a\plus{}b}{2}\equal{}\frac{a\minus{}b}{2}$

$ (a,b) \mapsto (\frac{a\plus{}b}{2},b)$ with $ f(\frac{a\plus{}b}{2},b)\equal{}\frac{a\plus{}b}{2}\minus{}b\equal{}\frac{a\minus{}b}{2}$

Then $ 2^{10}\equal{}1024 < 2000$ and $ 2^{11} \ge 2000$ then max amount of operations are $ 10$ with only example for $ m\equal{}2000\minus{}1024\equal{}976$

Moreover it is not difficult to generalize to any $ n\neq 2000$ or to give the set of numbers for having the possibility of performing $ k$ operations with $ k<10$.


probability
real analysis
probability and stats

Hi, I have a question here
http://tinyurl.com/6p2fut (Include both question and solution)

For (vii)
May I know why the solution claim that (X-Y, Y) is bivariate normal?

Yes,,, here is a result.
Suppose that $ (X_i: i\in I)$ are jintly Gaussian, and suppose $ (Y_i: i\in J)$ are random variables indexed by some set $ J$ such that each $ Y_i$ is finite linear combination of $ (X_i: i\in I)$. Then the random variables $ (Y_i: i\in J)$ are jointly Gaussian too.
The proof is very trivial. If you want, I can give you.

Dear [b] jack1234 [/b],

What [b] limsup [/b] means is just (finite) linear combinations of multi-variate Normal Random Variables is also multi-variate Normal.

But I think that he missed pointing out that the transformation must be reversible. In such case, the new multi-variate normal variables are degenerate. Take this for example: $ X_1, X_2$ bivariate, $ Y_1 \equal{} Y_2 \equal{} X_1 \plus{} X_2$.

P/S: You have the solution. Why not ask the one who gave it to you for clarification?

Thanks for your help. I have one more question:
For (vi),
I try to connect the standard form of bivariate formula and the formula we get, but failed to do so
this is my derivation:
http://tinyurl.com/6aknd3

May I know where is my mistake?

>>You have the solution. Why not ask the one who gave it to you for clarification?
Hi, the solution I am getting from my friend, not the author.