calculus
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real analysis unsolved

Suppose we have a function $f: [a,b] \rightarrow \mathbb{R}$ which is integrable on its domain. Is the inequality $\int_{a}^{b}|f(x)|dx \geq \left| \int_{a}^{b}f(x)dx \right |$ necessarily true?

yes it's true when $f$ be continous on the domain.

Why wouldn't it be true if $f$ is not continuous?

$-|f| \le f \le |f|$

I think that seeing this inequality by summatories, turns it more clear

Here's a proof that assumes that the function is complex-valued and integrable on the set $E.$ (By whatever theory of integration you know; in my own mind, I'm assuming Lebesgue integrable.)

[b]Lemma 1:[/b] If $f(x)\le g(x)$ for all $x\in E,$ then

$\int_{E}f(x)\,dx\le \int_{E}g(x)\,dx.$

This particular lemma should be one of the first things you prove after you define the integral. And yes, in the Lebesgue theory, you could relax the "for all" in the hypothesis to "for almost all."

[b]Lemma 2:[/b] $\text{Re}\left(\int_{E}f(x)\,dx\right) =\int_{E}\text{Re}(f(x))\,dx$

This is also close the definition of, and the linearity of, the integral.

Now for the proof.

Let $w=\int_{E}f(x)\,dx.$ If $w=0$ the desired inequality is trivially true, so assume that $w\ne 0.$

Let $g(x)=\text{Re}\left(\frac{\overline{w}}{|w|}f(x)\right)$
Then $g(x)\le\left|\frac{\overline{w}}{|w|}f(x)\right|=|f(x)|.$

So $\int_{E}g(x)\,dx\le\int_{E}|f(x)|\,dx$ (by Lemma 1)

$\text{Re}\left(\frac{\overline{w}}{|w|}\int_{E}f(x)\,dx\right) \le \int_{E}|f(x)|\,dx$ (by Lemma 2)

$\text{Re}\left(\frac{\overline{w}w}{|w|}\right) =\text{Re}\left(\frac{|w|^{2}}{|w|}\right)=|w| \le\int_{E}|f(x)|\,dx$

q.e.d.


Mary is about to pay for five items at the grocery store.  The prices of the items are $ \$$7.99, $ \$$ 4.99, $ \$$2.99, $ \$$1.99, and $ \$$0.99.  Mary will pay with a twenty-dollar bill.  Which of the following is closest to the percentage of the $ \$$20.00 that she will receive in change?

$ \textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$

[hide="Answer"]Since one cent is 0.05% of $\$ 20$, five cents will not affect the approximate percentage. Therefore, we have $\frac{8+5+3+2+1}{20}=\frac{19}{20}=0.95$, so she receives 5% in change. $\boxed{A}$[/hide]


I already found that [hide]\[\sum_{i=1}^{\infty}\frac{i}{\sum_{j=1}^{i}\left(\frac{i(i-1)}{2}+j\right)^{2}}= \frac{36}{5}\left(\sum_{i=1}^{\infty}\frac{1}{3i^{2}+1}-\frac{1}{3i^{2}+6}\right)\][/hide]But I don't know how to evaluate...[hide]\[\sum_{i=1}^{\infty}\frac{1}{3i^{2}+1},\ \textrm{and}\ \sum_{i=1}^{\infty}\frac{1}{i^{2}+2}\][/hide] Someone? :dry:

well now that we know that, it is pretty clear that we need to do a taylor series or something


The number of four-letter words (not necessarily meaningful) with at least two A's that can be formed from the set $ \{A, B, C, D\}$ is

$ \textbf{(A)}\ 11 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 61 \qquad \textbf{(E)}\ 67$

Amount of four-letter words with at least two $ A$'s in them= Total amount of words that can be made minus amount of words that have at most one $ A$ in them.

Total amount of four-letter  words that can be made: $ 4^4$
Amount of four-letter  words with only one $ A$: $ 4\cdot (3\cdot 3 \cdot 3\cdot 1)$
Amount of four-letter  words with no $ A$'s in them: $ 3^4$.

So the amount of four-letter words with at least $ A$'s in them is $ 4^4\minus{}(4\cdot3^3 \plus{} 3^4)\equal{}\boxed{67}$

The answer is $ \mathbf{(E)}$


[quote="nebula42"]Has Art of Problem Solving/MathLinks accounted for Daylight Savings Time yet? The times I see for the posts still have the old times on them.[/quote]

just go to profile and you can change it yourself

The MathLinks time has been set (server wise) according to the Romanian DST settings. I dunno about the AoPS server time as I'm not sure about the US DST rules. Anyway that's only the time viewed while [b]not logged on[/b].

While logged on, you should set your timezone and DST preferences in the Profile >> Preferences >> Internationalization settings according to your own country DST / Time settings.


function
rotation
inequalities unsolved
inequalities

Anybody know how to solve the following ASU 1971?
(from http://www.kalva.demon.co.uk/soviet/sov71.html)

Let f(x, y) = x^2 + xy + y^2. Show that given any real x, y one can always find integers m, n such that f(x-m, y-n) <= 1/3. What is the corresponding result if f(x, y) = x^2 + axy + y^2 with 0 ≤ a ≤ 2?

(I thought of represent these function as regions in space, but what region in inside x^2+xy+y^2=1/3 for example)

It is a rotated ellipse. This problem is equivalent to saying that if we shift the region by one up or down, do we tile the whole plane. I think this is easy to show this by drawing and a little computation.


Find the sum of the absolute values of the roots of $ x^4\minus{}4x^3\minus{}4x^2\plus{}16x\minus{}8\equal{}0$.

A hint is preferred.
Thanks.

hello, one root is $ x\equal{}1\minus{}\sqrt{3}\minus{}\sqrt{2}$
Sonnhard.

Let $ y\equal{}x^2\minus{}2x$, then the equation becomes $ y^2\minus{}8y\minus{}8\equal{}0$
So $ y\equal{}4 \pm 2\sqrt{6}$
Sum of the two positive roots of $ x^2\minus{}2x\plus{}(2\sqrt{6}\minus{}4)$ is obviously $ 2$
After solving $ x^2\minus{}2x\minus{}(2\sqrt{6}\plus{}4)\equal{}0$, we get $ x\equal{}1 \pm \sqrt{2\sqrt{6}\plus{}5}\equal{}1 \pm (\sqrt{3}\plus{}\sqrt{2})$

So the answer $ \equal{}2\plus{}2(\sqrt{3}\plus{}\sqrt{2})$


Let  $a,b,k\in N$ . Then prove that    $a!b!|(a+b)!$

where  $|$  means  division

Basically the same question as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85865

we know a+b_C_a = frac(a+b)!/a!b!
and it is a integer


let be given an odd prime p and the polyminical Q(x) = $ {(p-1)x^p -x-1} $ .
prove that there exists an infinite number of positive integers a such that Q(x) is divisible by ${p^p}$

Let $a = -\frac 1 2 \mod [p]$ (note that $p$ is an odd prime, thus $2$ is invertible modulo $p$).
Using Fermat's little theorem, it is easy to verify that $Q(a) = 0 \mod [p].$
Moreover $Q'(a) = p(p-1)a^{p-1} - 1 = -1 \neq 0 \mod [p].$

Thus, from Hensel's lemma, we know that, for any $n$, there exists an integer $b$ such that $Q(b) = 0 \mod [p^n]$ (and $b = a \mod [p]$).
Just use this for $n=p$, and see that any $x$ which verify $x = b \mod [p^p]$ satisfies $Q(x) = 0 \mod [p^p]$.

Pierre.


I had problem with my computer on 20th so I wasn't able to send solution that day.  So I sent my solutions via computer on 21st and I haven't got reply whether you guys got my solution or not. On 20th, I wanted to send solutions and had them done couple days earlier but my computer just wouldn't work.. I don't know why but I solved 4 problems (all of them pretty sure to be 5/5).

So I wondered if I'm not getting reply because you guys are not counting it or..?

If that's the case, I'd be mad because it's not my fault that my computer was being unusual..  :maybe:

We have not yet logged all the solutions.

SilverFalcon

I'd appreciate it if you wouldn't use the word "gay" in that context.

JB

I echo the above.

And I just received my email.

I express my sincere apology on that one.. Sorry.. It's fixed now.

And thank you, I received mine too. Yay  :D


Is it true that if we have a number a multiple of m and a bijection g from {1,2,...} into the same set, then we can always find a function f from {1,2,...} to {1,2,..} such that $ f^m(x)=g(x)+a $ ? I don't know the answer. But I didn't think about it, I will let you try.

A nice question. I think we should level-up it :)


Millie drives from home to the Fixit Garage at a rate of 30 mph.
She leaves her car and rides a bus back home. The bus makes the
trip at a rate of 20 mph. If her total travel time for the round trip is
one hour, how many miles from the home is the Fixit Garage?

It's Countdown, so just try numbers that seem "right" (like 24, 12, 48, etc.).  We find that the answer is $ \boxed{12}$.

Although if we don't want to be that lazy...

Let $ d$ be the distance.  Then, \[ \frac{d}{30}\plus{}\frac{d}{20}\equal{}1\Rightarrow d\equal{}\boxed{12}\]


conics
ellipse
ratio
geometry proposed
geometry

Let F and F' be the focal pts of an ellipse. Let FM and F'N' be parallel s.t. M and N' are on the ellipse on the same side of FF'. FN' and F'M intersect at E. The parallel through E to FM (and F'N') intersects FF' at P. Prove that EP has a constant length.

sum up EP/MF=F'P/FF' and EP/NF=FP/FF', we get 1/EP=1/MF + 1/N'F.
we claim that 1/MF + 1/N'F is a constant.
Suppose F' is the right focus.
find a point N on the ellipse s.t. <NF'F=<MF'F and N and N' are on the same side of FF'. Suppose <MFx = <MF'x = \theta, then <NF' = \pi-\theta.
In polar system, we write the ellipse as ep/(1-cos\theta) and this makes F be the origin. thus,
N'F = ep/(1-cos \theta),
MF = F'N = ep/(1-cos(\pi-\theta)) = ep/(1+cos \theta).
hence,
1/MF+1/N'F = 2/ep which is a constant.

Cool! I never considered polar coordinates. I used that property which states that for a point X on the ellipse the ratio between d(X,dF) and d(X,F) is c/a=excentricity of the elipse, where dF is the directrice (is this the word?  :D ) corresponbing to F. It's just the main idea. No time for a complete soln right now  :(

you'll find polar system is sometimes very useful, especially when we are facing to the problem connected with 'focus chord'.. i don't know how to say it... 'focus chord', i mean, is a chord which passes through the chord.


I use MathLinks layout the entire time I have been in this forum but with MathLinks style, there is NO link at all for me to access AoPS Wiki. Mr.Vornicu, can you help me?

Doesn't possibly sound so, but that would be really helpful (I totally dislike the AOPS style).

There will be one soon. I am currently involved in other more pressing issues.

[quote="10000th User"]*Bumps* Just wanted to see if our beloved admin remembers about this :P[/quote]Still involved in pressing issues :P (somewhat related to the wiki stuff).


calculus
integration
function
real analysis
real analysis unsolved

Source:web

f a continuos function [0;1] --->R

n >0 fixed integer such that $ \int_{0}^{1}(1\minus{}x^n)f(x)dx\equal{}0$

Prove that $ (2n\plus{}2)(\int_{0}^{1}f)^2 \leq \int_{0}^{1}f^2$

By Cauchy-Schwartz,

$ \left( \int_0^1 f(x)dx\right)^2 \equal{} \left( \int_0^1 x^nf(x) dx\right)^2 \le \left( \int_0^1 x^{2n} dx\right) \left( \int_0^1 f(x)^2 dx\right) \equal{} \frac1{2n \plus{} 1}\left( \int_0^1 f(x)^2 dx\right).$

EDIT: This is not $ 2n\plus{}2$..

There are problems that haven't been solved for years, I guess a few days it's not too much. Anyway, if my computations are right, the problem is more or less trivial. Take $ u$ a real number and use Cauchy-Schwarz for $ f$ and $ u\plus{}x^n$, then choose $ u$ such that the thing given by Cauchy is minimal. I think $ u\equal{}\minus{}1/(2n\plus{}1)$ is the good choice.

Ah, harazi, you beat me to it! :) My solution is very similar to yours..

Consider $ \int_0^1 f(x)\frac{2n}{2n\plus{}1}dx$. By Cauchy-Schwartz,

 $ \left( \int_0^1 f(x)\frac{2n}{2n\plus{}1}dx\right)^2\equal{} \left( \int_0^1 (x^n\minus{}\frac1{2n\plus{}1})f(x) dx\right)^2\\ \le \left( \int_0^1 (x^n\minus{}\frac1{2n\plus{}1})^2 dx\right) \left( \int_0^1 f(x)^2 dx\right)\\ \equal{} \frac{2n^2}{(2n\plus{}1)^2(n\plus{}1)}\left( \int_0^1 f(x)^2 dx\right).$
Hence

$ \frac{4n^2}{(2n\plus{}1)^2} \frac{(2n\plus{}1)^2(n\plus{}1)}{2n^2} \left( \int_0^1 f(x)dx\right)^2 \\\equal{} (2n\plus{}2)\left( \int_0^1 f(x)dx\right)^2 \le \left( \int_0^1 f(x)^2 dx\right)$


calculus
integration
LaTeX
function
logarithms
calculus computations

Hello I am new here, can anyone solve this and show me how he did so?

integral of ln(root x) dx from 1 to 4

thank you :)

P.S.
I was woundering if anyone could explain to me where I can use the actual symbols to make my problem look moer readable?

On the sidebar, there is a link LaTeX with instructions on how to typeset. You can also click other people's formulas to see the code.

You want to integrate the function $f(x) = \ln(\sqrt{x})$? All you have to do is recall that $\sqrt{x} = x^{1/2}$ and $\ln(x^a) = a\ln(x)$. Then it is a simpler function, for which I suggest integration by parts.


determine all  odd integers  n  such  that  if  a and  b  are  distinct prime divisors  of n  then  a+b-1 also  divides  n

There are infinetely many solutions.
Let $n=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}, \ 2<p_{1}<...<p_{m}$ is solution. Then \[n_{1}=\prod_{i=1}^{j}p_{i}^{r_{i}}, \ j\le m, \ r_{i}\ge k_{i}\] is solution.
Therefore we can find properties for new prime $p_{m+1}$, when $np_{m+1}$ is solution too. It give \[\forall i\le m \ \ p_{m}+p_{i}-1=\prod_{j=1}^{m}p_{j}^{r_{j}}, \ r_{j}\le k_{j}.\] In this way we can find all solutions.
We have
1) $n=1$ - solution,
2) $n=p^{k}$ - solution for any odd prime.
3 $n=p_{1}^{k_{1}}p_{2}, \ p_{1}<p_{2}$ is solution if and only if $p_{2}=p_{1}^{r}-p_{1}+1, \ 1<r\le k_{1}$, et.c.
For example $7=3^{2}-3+1$, $79=3^{4}-3+1$ give solutions $3^{k}7^{l}, \ k\ge 2$, $3^{k}79^{l}, \ k\ge 4$.

yes  but can  you  find all  solutions??
  also   i  think  russia  2001 says that  a  and b   are  prime relative  not  necesarily  primes  no?

I don't undestand, why $n=63$ is not solution?
1. n is odd
2. n had only 2 distinct prime divisors a=3,b=7 and a+b-1=9, and 9|n.

As I wrote in that other thread, the problem 9.8 from Russia, 2001 asked for all odd $n$ with the property that whenever $a,b$ divide $n$ and are coprime, $a+b-1$ also divides $n$. Problem 10.8 asked for all such $n$ (including even ones). Even though Marko wrote that $a,b$ are prime divisors of $n$, I think he actually meant to say that they are coprime divisors (his second message suggests this). $63$ is not a solution for this modified version: $9|63, 7|63$, but $15=9+7-1$ does not divide $63$.

[quote="grobber"] I think he actually meant to say that they are coprime divisors (his second message suggests this). $63$ is not a solution for this modified version: $9|63, 7|63$, but $15=9+7-1$ does not divide $63$.[/quote]
He asked, when a,b distinct prime divisors, $9$ is not prime divisor.

yes.  my  bad. i should check  my  writing better.  i  meant to say  prime relative divisors.  :oops:


[quote="spanferkel"]$a,b,c\ge0$. For $n\in\mathbb{N}$ and $0\le k \le 2n-1$, prove
$\sum_{cyc}a^k(b-c)^{2n}(a-b)(a-c)\ge 0$.[/quote]

Wlog, assume that $a \geq b \geq c$. As ${f(x)= {\frac{(x+t)^k}{x^{2n}}} = {\frac{{{n}\choose{0}}}{x^{2n-k}}} + \cdots + [\frac{{{n}\choose{n}} t^k}{x^0}}$ is decreasing when t is positive, $f(a+c) \leq f(b+c)$. This means that $a^k(b-c)^{2n} \geq b^k(c-a)^{2n}$, and this completes the proof by Vornicu-Schur ineq.


a,b,c>0
a+b+c=3
prove that:
a)$ \frac{1}{ab}\plus{}\frac{1}{ac}\plus{}\frac{1}{cb}\geq a^2\plus{}b^2\plus{}c^2$

[quote="tuandokim"]a,b,c>0
a+b+c=3
prove that:
a)$ \frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{cb}\geq a^2 \plus{} b^2 \plus{} c^2$[/quote]
It 's  Romania TST 2006 :wink::
$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2} \ge a^2\plus{}b^2\plus{}c^2$

[quote="phymax"]$ \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \geq\ (\frac {a \plus{} b \plus{} c}{3})^2$

$ a^2 \plus{} b^2 \plus{} c^2 \geq\ 3$

but ur ineq. is
$ \frac {a \plus{} b \plus{} c}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

$ \frac {3}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

the max. of $ abc \equal{} 1$

so,
$ 3 \geq\ a^2 \plus{} b^2 \plus{} c^2$

a contradiction
so its wrong.. :wink:  :ninja:

if the other way, plz show mistake in my proof   :huh: :?:[/quote]

I fully agree with you. The inequality is only right when $ a\equal{}b\equal{}c\equal{}1$.

[quote="phymax"]$ \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \geq\ (\frac {a \plus{} b \plus{} c}{3})^2$

$ a^2 \plus{} b^2 \plus{} c^2 \geq\ 3$

but ur ineq. is
$ \frac {a \plus{} b \plus{} c}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

$ \frac {3}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

the max. of $ abc \equal{} 1$

so,
$ 3 \geq\ a^2 \plus{} b^2 \plus{} c^2$

a contradiction
so its wrong dude.. :wink:  :ninja:

if the other way, plz show mistake in my proof   :huh: :?:[/quote]
are you sure ?
because $ \frac{3}{abc}\geq 3$
now try it

[quote="tuandokim"][quote="phymax"]$ \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \geq\ (\frac {a \plus{} b \plus{} c}{3})^2$

$ a^2 \plus{} b^2 \plus{} c^2 \geq\ 3$

but ur ineq. is
$ \frac {a \plus{} b \plus{} c}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

$ \frac {3}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

the max. of $ abc \equal{} 1$

so,
$ 3 \geq\ a^2 \plus{} b^2 \plus{} c^2$

a contradiction

ooops...Sorry I made a dumb mistake. :blush:
so its wrong dude.. :wink:  :ninja:

if the other way, plz show mistake in my proof   :huh: :?:[/quote]
are you sure ?
because $ \frac {3}{abc}\geq 3$
now try it[/quote]

[quote="phymax"]thats what Stephen said

see that $ a^2 \plus{} b^2 \plus{} c^2 \geq\ 3$

and
$ 3\geq\ a^2 \plus{} b^2 \plus{} c^2$ cannot simultaneously hold true until and unless both are equal at the most

see carefully[/quote]
I don't understand what do you mean.tell me more .
but I'm sure that my ineq is right and I can prove it.

ok
this is my solution :
we need to prove that:
$ 2\sum ab\plus{}\sum \frac{1}{ab}\geq 9$
which is equivalent:
$ 2(\sum ab)abc\plus{}3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$
but this is obvious true
Q>E>D
p/s : ok?

[quote="tuandokim"]ok
this is my solution :
we need to prove that:
$ 2\sum ab \plus{} \sum \frac {1}{ab}\geq 9$
which is equivalent:
$ 2(\sum ab)abc \plus{} 3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$
but this is obvious true
Q>E>D
p/s : ok?[/quote]
It 's false  :wink: ,because by your way,we must prove that:
$ ab\plus{}bc\plus{}ca \ge 3$,which 's obvious false :wink:

[quote="quykhtn-qa1"][quote="tuandokim"]ok
this is my solution :
we need to prove that:
$ 2\sum ab \plus{} \sum \frac {1}{ab}\geq 9$
which is equivalent:
$ 2(\sum ab)abc \plus{} 3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$
but this is obvious true
Q>E>D
p/s : ok?[/quote]
It 's false  :wink: ,because by your way,we must prove that:
$ ab \plus{} bc \plus{} ca \ge 3$,which 's obvious false :wink:[/quote]

$ (ab\plus{}bc\plus{}ca)\equal{}\frac{1}{3}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\ge 3abc$

[quote="Duy_VN"][quote="quykhtn-qa1"][quote="tuandokim"]ok
this is my solution :
we need to prove that:
$ 2\sum ab \plus{} \sum \frac {1}{ab}\geq 9$
which is equivalent:
$ 2(\sum ab)abc \plus{} 3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$
but this is obvious true
Q>E>D
p/s : ok?[/quote]
It 's false  :wink: ,because by your way,we must prove that:
$ ab \plus{} bc \plus{} ca \ge 3$,which 's obvious false :wink:[/quote]

$ (ab \plus{} bc \plus{} ca) \equal{} \frac {1}{3}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)\ge 3abc$[/quote]
I want to speak,If you use this inequality,the inequality 's false :wink:

[quote="tuandokim"]ok
this is my solution :
we need to prove that:
$ 2\sum ab \plus{} \sum \frac {1}{ab}\geq 9$
which is equivalent:
$ 2(\sum ab)abc \plus{} 3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$
but this is obvious true
Q>E>D
p/s : ok?[/quote]

Yeah, now I see what you mean

[quote]
$ 2(\sum ab)abc \plus{} 3\geq 9abc$
but $ 3\geq (\sum ab)abc$
so we will prove that :$ \sum ab \geq abc$[/quote]

This problem is difficult a little bit for it looks.

[quote="phymax"]$ \frac {a^2 \plus{} b^2 \plus{} c^2}{3} \geq\ (\frac {a \plus{} b \plus{} c}{3})^2$

$ a^2 \plus{} b^2 \plus{} c^2 \geq\ 3$

but ur ineq. is
$ \frac {a \plus{} b \plus{} c}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

$ \frac {3}{abc} \geq\ a^2 \plus{} b^2 \plus{} c^2$

the max. of $ abc \equal{} 1$

so,
$ 3 \geq\ a^2 \plus{} b^2 \plus{} c^2$

a contradiction
so its wrong dude.. :wink:  :ninja:

if the other way, plz show mistake in my proof   :huh: :?:[/quote]
We have $ \frac{3}{abc} \ge 3$. But that does not mean that $ 3 \ge a^2\plus{}b^2\plus{}c^2$...

Essentialy what you are saying is: $ x \ge y$ can not be true because $ x \ge z$ and $ z \le y$.... ;)

[quote="phymax"][quote="Duy_VN"][quote="tuandokim"]a,b,c>0
a+b+c=3
prove that:
a)$ \frac {1}{ab} + \frac {1}{ac} + \frac {1}{cb}\geq a^2 + b^2 + c^2$[/quote]
[i]
Solution:[/i]

Easily seen : $ abc \le 1$

$ Inq < = > \frac {a + b + c}{abc} \ge (a + b + c)^2 - 2(ab + bc + ca)$

$ < = > f(a,b,c) = 2(ab + bc + ca) + \frac {3}{abc} \ge 9$

Suppose that : $ c = max\{a,b,c\}$ , then we have : $ (a + b)\le 2$

${ f(a,b,c) - f(\frac {a + b}{2},\frac {a + b}{2},c) = (a - b)^2.(\frac {3}{abc(a + b)^2} - \frac {1}{2}}) \ge 0$

Set : $ \frac {a + b}{2} = t$ , then : $ c = 3 - 2t$ and $ t\le 1$

$ f(t,t,3 - 2t) - 9 = - 6(t - 1)^2 + \frac {3(t - 1)(2t^2 - t + 1)}{t^2(3 - 2t)} \le 0$

We've done.[/quote]

i dont understand why $ a + b \leq\ 2$?
explain clearly[/quote]

Assume that $ \max\{a,b,c\}=c\ ,\ a+b\ge 2$ . Since $ a+b+c=3\Rightarrow c\le 1$ .

Since $ a+b\ge 2$ , we can't have $ a\le 1,b\le1 \Rightarrow$ $ a+b\le 2$ , false . This means at least one of $ a,b\ge 1\Rightarrow c$ is not the maximum false $ \Rightarrow a+b\le 2$

[quote="alex2008"][/quote][quote="phymax"][quote="Duy_VN"][quote="tuandokim"]a,b,c>0
a+b+c=3
prove that:
a)$ \frac {1}{ab} + \frac {1}{ac} + \frac {1}{cb}\geq a^2 + b^2 + c^2$[/quote]
[i]
Solution:[/i]

Easily seen : $ abc \le 1$

$ Inq < = > \frac {a + b + c}{abc} \ge (a + b + c)^2 - 2(ab + bc + ca)$

$ < = > f(a,b,c) = 2(ab + bc + ca) + \frac {3}{abc} \ge 9$

Suppose that : $ c = max\{a,b,c\}$ , then we have : $ (a + b)\le 2$

${ f(a,b,c) - f(\frac {a + b}{2},\frac {a + b}{2},c) = (a - b)^2.(\frac {3}{abc(a + b)^2} - \frac {1}{2}}) \ge 0$

Set : $ \frac {a + b}{2} = t$ , then : $ c = 3 - 2t$ and $ t\le 1$

$ f(t,t,3 - 2t) - 9 = - 6(t - 1)^2 + \frac {3(t - 1)(2t^2 - t + 1)}{t^2(3 - 2t)} \le 0$

We've done.[/quote]

i dont understand why $ a + b \leq\ 2$?
explain clearly[/quote][quote="alex2008"]

Assume that $ \max\{a,b,c\} = c\ ,\ a + b\ge 2$ . Since $ a + b + c = 3\Rightarrow c\le 1$ .

Since $ a + b\ge 2$ , we can't have $ a\le 1,b\le1 \Rightarrow$ $ a + b\le 2$ , false . This means at least one of $ a,b\ge 1\Rightarrow c$ is not the maximum false $ \Rightarrow a + b\le 2$[/quote]
Or just: $ a + b = 3 - c \le 3 - \frac {a + b + c}{3} = 2$...

sorry for my stupidity :| (although i was wrong but my ineq is still true.Lucky:D)
and thank you Duy_Vn for a nice solution.

[quote="tuandokim"]a,b,c>0
a+b+c=3
prove that:
a)$ \frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{cb}\geq a^2 \plus{} b^2 \plus{} c^2$[/quote]

The Inequality equivalent to
\[ 3\ge abc(a^2 \plus{} b^2 \plus{} c^2)
\]
By AM-GM we have
\[ 3^4abc(a\plus{}b\plus{}c)(a^2 \plus{} b^2 \plus{} c^2)\le3^3(ab\plus{}bc\plus{}ca)^2(a^2 \plus{} b^2 \plus{} c^2)\le\left((a\plus{}b\plus{}c)^2\right)^3\equal{}3^6
\]
done!


Find all natural $ n$ such that $ \frac {2^n \plus{} 1}{n^2}$ is integer.


AMC
AIME
AMC 10
AMC 12
MATHCOUNTS
probability
conics

I'm very inexperienced in AMC10 and AIME
I've only taken these tests in 7th Grade (last year)
I scored a 120.5 on the AMC10
and
I scored a 2 on the AIME (haha i know i suck)
AMC12 Preparation is not my main focus
so if you guys feel the need to post different preparation guides/tips for AMC12
Please do so

So due to my inexperience
This guide may not be the best
but I'm trying
I would like any of you guys to also post your own preparation guide/tips in this topic too

Preparation Guide:

April:
Finish Doing the Art of Problem Solving Book Volume 2
Do zetamac.com/arithmetic/# everyday in order to keep your mind sharp

May:
Start Working on past AMC10's
Make sure you do all the AMC 10's till the point that if you redo it again and again, you will score a perfect
Basically. Don't move on to the next year until you have completely mastered the previous years problems

June:
Continue working on past AMC10's
Start doing zetamac.com again in order to keep sharp

July:
You should finish doing all past AMC10's this month

August:
Start working on past AIME's
Remember to master each year before you move on to the next year
Don't let YOUR GRADES SLIP in high school because of this preparation
Start doing zetamac.com again

September:
Continue working on past AIME's
Don't forget to prepare for other tests
Such as the SAT (10th to 12th graders), ACT (10th to 12th graders, PSAT (9th to 10th graders) and other standarized tests
I recommend saab.org for practice on the vocab part of the SAT

October:
Continue working on AIME's
Start doing zetamac.com again

November:
Finish working on AIME's
Finish Doing the First Steps for Math Olympians

December:
You can start doing USAMO's if you want
Start doing zetamac.com again
Review problems you found difficult on the past AMC10's and AIME (problems you missed the first time)
Don't forget to prepare for other contests

January:
Review Art of Problem Solving Volume 2
You may supplement your knowledge by ordering Intermediate Art of Problem Solving Problems once they came out

February:
Continue reviewing
Start doing zetamac.com again

I hope this preparation guide helped you
Please post your own preparation guide/tips

[quote]Um so like what's the point of this? You're basically telling us exactly how to prepare for everything?[/quote]

Well... people post topics asking for something like this all the time. While I don't agree with everything here, this is a lot more comprehensive than the replies in those topics.

[quote="Aznwithabrain"]
Remember to master each year before you move on to the next year[/quote]

Sometimes you might come across something you aren't ready for and have to step aside and actually learn a concept eg, when I first saw $e^{\pi \theta}$ on an AIME. Since you advocate finishing AoPS Vol 2 in the first month, one would have all the necessary curriculum. However, I think going through AoPS and actually doing all the work takes longer unless you're spending a huge amount of time on it.

I did alright with this strat:
- Relax.
- Occasionally do problems in class when it's boring.
- Spam AoPS to build confidence.

Long term plans are the easiest to make and the hardest to carry out.

You have made some very ambitious aspirations, and unless you are currently functioning close to that level, I have reason to expect that you will not work to that level.

For example, you plan to finish all of the previous AIME contests in the span of 3 months.  There are currently 33 official contests available on AoPS.  11 contests per month is a huge jump from doing 0 a month.

I'm not saying what you planned is impossible; just set realistic goals and gradually work up to where you want to be.

Some of you are more talented in math than others (like me)

I prepared alot for MATHCOUNTS
and I still didn't do better than
my teammate who really didn't prepare as much as me

Even though everyone in my school agrees that i'm better at math than that certain teammate,
that certain teammate has always had a natural ability for math, but he doesn't prepared it enough
I've beaten and destroyed him in almost every single math competition except yesterday's State competition

I personally need to prepare year round for AMC's in order to do good

Explanation for fourierseries:
I actually do prepare at this level
I did 20 national mathcounts packets in a span of two weeks
even though i didn't do so good at state
That's why I still think if I had begun my preparation earlier
I would've done better

Edit:
Actually its 4 months to do all past AIME's
I forgot to put a space between october and november
sorry for the confusion

[quote="rnwang2"][quote="Ignite168"]Who does math year round? I mean seriously, a little cramming in January-May should be fine.[/quote]
Umm...I thought most people prepare for AMC/AIME/USAMO the entire year.  :|[/quote]

Why you alwyas reading books? I always see you are reading some.. what.. Zietz, or.. what.. Apostol. You are doing too much reading, not enough thinking! You thinking, "Oh, I read Zeitz, now I can get better at AIME" when you not even have fundamentals yet. You need to prcatice AT LEAST AIME level before exploring those more advanced topics. Why you are always give up on porblem after 2 minutes? It isnot like SAT reading comprahension, it is not enough to know what those words mean... you only reading words, not understanding mathematcal concepts!

?
wtf probability 1.01

you could make your posts more understandable

[quote="probability1.01"]Why you alwyas reading books? I always see you are reading some.. what.. Zietz, or.. what.. Apostol. You are doing too much reading, not enough thinking! You thinking, "Oh, I read Zeitz, now I can get better at AIME" when you not even have fundamentals yet. You need to prcatice AT LEAST AIME level before exploring those more advanced topics. Why you are always give up on porblem after 2 minutes? It isnot like SAT reading comprahension, it is not enough to know what those words mean... you only reading words, not understanding mathematcal concepts![/quote]
I'm sorry. Teach me how to be wise like you.  :(

And what is up with the grammatical structure of your post??

What Zhai is trying to say is that if you're reading AoPS Volume 2 and you haven't worked up to a level where you can confidently attempt AMC/AIME problems, then there's a problem.

The essence of problem solving is THINKING.  Reading books can only take you so far. You must be willing to spend hours laboring over a problem.

oh i see
well thats why i have all those tests you have to do in the preparation guide :lol:

Maybe probability was trying to mimic zuming? That sounds exactly like something he might say...

By the way, you probably shouldn't spend too much time on the arithmetic game or whatever it's called.

oh
alright

i just do it during the school day
when i'm bored or during my free period
cause
it really does keep my mind sharp and ready

your goals are too strict...just do stuff...always challenge yourself...and that arithmetic game is nice and all, but you are wasting your time, if you can get above say...25 correct (which is what i got my second time), then you are fine...

amc s are not like mathcounts...you don't need to do lots and lots of computation

Can anyone give me their opinion on the this issue:

I just finished reading Chapter 1 of AOPS Volume Two
I finished working on the Chapter Problems
I only got 3/13 totally correct on the first try
I got 7/13 after spending 40 more minutes (my mom checked them so I couldn't see the solutions)
I finally understood 8/13 of them after looking at the solutions
I can't understand 5/13 of them yet even with the solutions :ninja:

Is this good enough for me to move on to the next chapter?

I'll comeback to those five problems after finishing the whole book

Dude.. don't worry about being so systematic? Just make a note of problems you don't know how to do and review them later when you get the chance. Although you should also try to get used to working on problems for longer stretches of time... like maybe 30 minutes a problem for AIME-level...

The problems were mostly from AHSME and Mu Alpa Theta...

We know. We have the books.
Well, I don't think chapter 2 is related to chapter 1 at all, so go for it, I guess, and come back to chapter 1.

Skimming is good. Studying is bad. It's more important to develop problem solving skills and a general better understanding of math, and that is accomplished better by skimming a bunch of different things than studying something to death. Of course, you should come back later and tie up the loose ends. Okay, maybe I should clarify. You can skim the material, as long as you not only know it, but you understand it. But DON'T SKIM THE PROBLEMS! And you don't need to memorize everything at first. Later on, as you work problems that involve those things, come back and you will learn them more easily.

[quote="junggi"][quote="Aznwithabrain"]alright

skimming ch.2
all this stuff is covered in my pre-cal book and trig book from last year
so it should be a breeze :lol:[/quote]
errrr no
Never skim ANY PART of the book...
Unless your pre-cal book actually proves all the theorems, facts, etc, which I really doubt.[/quote]

Mine does.

But I second you on the not skimming.

About the number of problems you get right: It's a good sign when you're not getting all of the problems, perhaps even with the solutions. This kind of comes from the idea that the best kid in a class has the least to learn. Here's another suggestion for you: Don't make an hourly schedule and do your math studying that way. It might work for science or history, but with math, I think the best way to study is to pick it up once in a while and then spend as much time as you want to on it. Some days, I don't feel like doing anything but math, and on others, rote memorization sounds a lot more rewarding. Speaking of which, don't study math on its own. Read other things too, especially science textbooks, or anything that gets you thinking. A lot of times, with AIME and USAMO, it isn't a matter of applying the standard techniques; you have to approach problems from a totally different angle. To continue this train of partially connected thoughts, I honestly would study AoPS chapters backwards: Read through the problems, work on them, and if you get stuck on any, read the relevant sections of the chapter. That way, you don't get bored reading and you also become interested in the material as explaining a problem's solution to you. Plus, you skip the stuff you already fully comprehend.

Anyway, the biggest message is that studying should be fun. You should never prepare for a contest or competition because you think you "have to" or "won't make USAMO if you don't." If math isn't fun right now, study science. If that doesn't interest you either, study history. If you're into music, play some music. You can always go to Wikipedia and read featured articles. No matter what [i]engage your mind[/i]. For the weak-minded, that can mean Sudoku or a novel. For those of us who see Sudoku as a matter of applying standard algorithmic techniques, it usually means textbooks, so make sure to check out a backpack-full or two of interesting textbooks (just don't trust the titles of any books called "Modern Physics" written before World War Two).

Hopefully you can pull something out of that.

I've learned to enjoy working math problems especially if they're new and interesting and they stand as a challenge

Maybe it's because I'm very competitive (even though I don't get very depressed when I don't win)
Maybe it's because I did MATHCOUNTS in middle school
Maybe I just like challenges

but
I think your advice is very good
I'll try my best to follow that

<bump>

since all the drama of writing the USAMO has probably cooled down by now (the list of qualifiers is up) :D

Anyone have any preparation guidelines for next year

um way to triple post?

Do what you think is best, if you can't get any problems clearly you don't understand the material that well or are just not thinking enough, because the first few are usually trivial if you actively read through the material.

What's with this preparation guidelines thing? It's pretty ridiculous, in my opinion, you'll end up spending more time drafting and re-drafting your "preparation guidelines," talking about/planning what you will do, instead of doing it.

Do what you want to do, what you're having fun doing.

[quote="Aznwithabrain"]I just finished Chapter Two
I got very little of the problems
should I move on?[/quote]

Huh, I should think the answer is obvious. NO.
If you read through the chapter and still can't get almost all the problems, then there's a problem.
Not to mention that ch. 2 is one of the more important ones.

btw, @ your sig, you can't make USAMO with a 5 AIME.

eh true

I just forgot all the trig I did over the summer

 :rotfl:


The graphs of the equations
$y=k$, $y=\sqrt{3}x+2k$, $y=-\sqrt{3}x+2k$
are drawn in the coordinate plane for $k=-10,-9,-8,...,9,10$. These $63$ lines cut part of the plane into equilateral triangles of side $2/\sqrt{3}$. How many such triangles are formed?

hmm, is the answer:
[hide]
$682$
[/hide]
???

Kalva's answer is [hide]660, that's what I got too [/hide]

Do you have a solution...because I need one  :| :dry:


function
inequalities unsolved
inequalities

[quote="chien than"]Prove that:$(a+b-c-1)(b+c-a-1)(c+a-b-1) \leq 8$
where $a;b;c$ are positive real numbers satisfying $ab+bc+ca=abc$[/quote]
Yes, it's true! I found ugly proof.

I found a solution too :) , and I hope it is not so ugly :blush: .

[color=blue]Solution:[/color]
Setting $2x=b+c-a-1,2y=a+c-b-1,2z=a+b-c-1$ and considering $x,y$ and $z$ to be possitive(the case with negative ones is easy) we get to the following equivalent problem

[color=blue]Prove that if $(x+y)(x+z)(y+z)=2(x+y+z+1)$ and $x,y,z\geq 0$ then
\[xyz\leq 1\][/color]

Let's assume that $xyz>1$
then $xy>1$ and $z>\frac{1}{xy}$ ($xy>1$ condition should be true for at least two of these numbers and we suppose that it is true for $x$ and $y$).
Now, we have that
\[z^{2}(x+y)+z((x+y)^{2}-2)+((x+y)(xy-2)-2)=0\]
So,concidering that $(x+y)^{2}\geq 4xy>4$ =>
\[\frac{x+y}{x^{2}y^{2}}+\frac{(x+y)^{2}-2}{xy}+((x+y)(xy-2)-2)<0\]
<=>
\[x+y+xy((x+y)^{2}-2)+x^{2}y^{2}((x+y)(xy-2)-2)<0\]
<=>
\[(x+y)(x^{3}y^{3}-2x^{2}y^{2}+1)+xy(x+y)^{2}-2xy(xy+1)<0\]
Setting $x+y=m$ and $xy=n$ we get that

\[m^{2}n+m(n^{3}-2n^{2}+1)-2n(n+1)<0\]

Let's look at the function $f(x)=x^{2}n+x(n^{3}-2n^{2}+1)-2n(n+1)$.
We have that $f'(x)=2xn+n^{3}-2n^{2}+1$=> $f$ is increasing in $[2;\infty)$
We also have that $m=x+y\geq 2\sqrt{xy}>2$=>$f(m)>f(2\sqrt{n})$
=>
\[f(2\sqrt{n})=(\sqrt{n}-1)(n^{2}\sqrt{n}+3n^{2}+n\sqrt{n}-1)<f(m)<0\]

But the laste is an obvious contradiction because we assumed that $n=xy>1$.

We are done :wink:  :lol:.

[quote="chien than"]Who have other solution?[/quote]
You don't like the one above? :P


MATHCOUNTS calculus geometry Harvard college USAMTS AMC

[i]I'm not sure if I should post this, but I feel like it's unfair to ask people about their achievements without context.  It is a bit like taking the SAT.  And I spent a long time writting this post.  I'd hate to delete it.[/i]

[quote]I thought it would be cool for all of U.S. students to give their math profile (good for future reference). Here is the format, (and my profile) and please, don't me modest with your achievements; be very specific as well.[/quote]

Here is my response to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17518[/url]:

I think it would be interesting to see how oportunities at schools around the country (and world) vary.  No format is required.
--
In my (public) school system, the standard procedure is to apply the Iowa Algebra Aptitude Test (no, I don't live in Iowa) to all students at the end of the seventh grade.   Those students who do "well enough (33% or so, two sections)" move on to Algebra in the eighth.  I got around this by specially requesting to take the IAAT a year early.  Some students are kept out of algebra until the second semester of their freshman year in highschool.  All the students who take algebra in the eighth grade pass.  They get twice the time to learn as a typical high school student.  Maybe the minimum test score is a little too selective?  None of the seventh graders know this.  I only know from experiance.  The middle school has a mathcounts team.  The high school first offered the AMC last spring.  The AMC B was only offered because my parents paid for it.  High School mathematics extends to Calculus BC.  Most of this year's Calculus AB students are dropping BC.  We recently added Discrete Mathematics and AP Statistics to the ciriculum because the school system was able to hire a highly gifted new teacher.  I'm a sophmore and president of the math club (No one else would do it.  Litterally.)  I am the only person I know of ever to take calculus in their sophmore year at my school.  By May, I will have taken every math class my school offers.  Yes, that is only possible due to block scheduling.
--
I mention my own achievements not to show that I have unusual talent (around here, I don't) but to show how much the "average" student is left out.  Everything I have done should be required.

Does anyone here go to a school that is too small/backward to have Calculus?

Well, I am in Round Rock ISD in the Austin area, and basically we have a TAG program (talented and gifted) starting from elementary school in English and Math.  6th grade TAG students take a test determining whether they take algebra I in 7th or 8th grade.  There is another program called Excel.  I'm not exactly sure how they spell it, but its similar to TAG except for being easier and 1 year slower.  High school wise, the courses are basically just algebra I/II, gemetry, pre-cal, calc AB/BC, stat, and some independent study junk that no one except for the not so smart people take.  The math program at our school is pretty weak, we don't have very good teachers or very good math students.  For math standardized testing, we just take TAKS, as required to graduate in Texas.  Our school does not offer any math competitions except U.I.L. academics, in which UT basically "governs" all school competitions, be it band, athletics, math, writing, etc.  The UIL math consists of three different events you can participate in: Number Sense (Mental Math), Calculator Applications (Problems with a calculator) and Mathematics (which is similar to a general math subject test).  Our school, or as far as I know, our district does not compete in any other competitions such as AMC, though my middle school and one or 2 others in our district do compete in mathcounts.  Our school does have a Mu Alpha Theta chapter.

Well I go to school in Ontario, Canada in an affluent suburb of Toronto, but the school is Public-Roman Catholic. Our school is nothing special although we do have a good rep in our district. Because of the standard curriculum implemented in our school, we only have one academic math class in grades 9 to 11. Finally in grade 12 we have 3, Discrete, Introductory, Calculus and Data Management. The DM class is nothing special so few 'higher' level math people take it. We are only offered U of Waterloo math contests (Cayley, Fermat, Pascal and Euclid) although I was able to convince our head of mathematics to offer the COMC this year just for me:). Otherwise, our school really does pretty bad relative to what everyone on this forum does, as we get one or two peopl with certificates on the math contests and that with minimal scores that barely qualify. I actually got scores that completely shocked everyone when I got 136 and 138 in my first two contests. Back to the curriculum, I'm one of only two people who have gone a year ahead as I took gr 10 and 11 ias a sophmore and am now able to do calculus as a junior, although my schedule is too full to do the discrete as well this year.
Otherwise we have really bad placement, and got really lucky this year getting two kids into Harvard, both as recruited athletes (one was a star football player, while the other is a legacy who does crew). We have no math club or anything along those lines as people here are, well, allergic to math and our only academic clubs are a book club and school reach (which I'm on and we are an awesome team who won first place in regionals in intermediate where regionals are the highest level and we went undefeated 19-0)

The high school that I will attend next year only goes to Calculus AB, and has no clue what mathcounts the AMC's or the USAMTS is.  The only math team that they have is UIL and a thingy called tmsca which is rediculously easy.  So I'd say high schools are pretty strange as far as what some do and what some don't.

The high school I go to next year goes all the way up to Calculus.  The highest math a person can take in 9th grade is ALgebra II.  It used to be Pre-Calulus, but they had to change stuff becuase of the Ohio Graduation Test.  I know one kid who had Algebr 1 in 4th grade, but his mom was on the school board.

We don't really have any math contest at the high school though.  At our school, we were going to have a MATHCOUNTS team, but that didn'y work out, but we get to take the AMC8 instead.

My school goes up to Multivariable Calculus. We take AMC10 and 12 A, AIME1, math team that does GNHML (Greater New Haven Math League), CSAML, NEML, and NEAML. Guess what school I'm in. JSRosen3 and yasha can't answer.

My middle school had mathcounts, AMC 8 and AMC 10B for the all the kids who got over 20 on the AMC8. They only went up to Algebra I, but 8th graders who were ahead could go up to Geometry to take it at the high school. If you could do well enough on the AMC10B, then you could take the AIME with the high schoolers.

I go to a public school in upstate NY.  My school normally goes up to calc BC, but they didn't have it last year because they supposedly don't do classes with less than 15 people.  In my opinion, this is a horrible excuse because my Latin class last year only had 12 people.  I think it's because my school hates math.  Occasionally, people take calc in 11th grade.  I probably should have taken it in 8th, but my school blocked me from doing math in 6th grade, for some dumb reason.  My school doesn't want to pay for online classes or anything, so I get to teach myself multivariable calc, etc.  My school does AMC 10/12 A and AIME.  It also does Monroe County Math League, a local competition that sends people to NYSML.

Well, pretty affluent suburban math program with the IB program. Sadly, my school uses Core-Plus (I had to endure 2 years of it). Highest level math course is HL Math but blech; I'll just take a Stanford EPGY course my senior year; self-education is an important skill to have.

I started out with the first year high school course (blech, Integrated I) in 8th grade. There were some people taking it in 7th grade.  Then  I decided that I must get this Integrated program over with ASAP so I took the SAT and took Integrated II over the summer. Then I had to endure Integrated III in 9th. THen I took Pre-Calculus over the summer so I'm now in the equivalent of Calculus AB in 10th. Yeah, IB Math Methods II is not BC Calc. But blech, Calc in 10th I guess is good enough. I need some time to self-study some APs anyways (like Physics C).

Of course, public school + International Baccalaureate = almost no flexibility. I actually like IB but don't like its inflexibility, especially when dealing with profoundly motivated unidirectioanl students (I don't use gifted for a reaon; I believe that the word gifted is discriminatory; we judge too much on test scores and there are those with a strong subconscious who will never have their abilities picked out in timed tests).

Problems come in the junior high, which goes to 9th grade. Thus, I'm denied opportunity. *sigh*

Math Team isn't all that great (and people there mostly want to have fun) although this year we had a genius student who got in USAMO in 9th. Sadly, I started so late that I'll probably never catch up to him (And my ADD symptoms certainly don't help). School sends around 5-6 every year to AIME, which isn't all that exceptional. I'm determined to get AIME this year. But I'll seriously need to brush up my work ethic.

Furthermore, I was defeated for the vice-president election. *sighs* I made the longest speech while the other contenders made 10 second speeches, I was the only person who seemed to really WANT to be vice-president, I had the passion, I had the knowledge of various math competitions. You don't have to be a genius in math to be vice-president. But since I'm not very likeable and probably due to my Asperger's, I was defeated. Gosh our world is a popularity contest. Democracy is evil, especially in small groups.

Off with the rant. Anyways... Suffice to say that it's really individual motivation that matters. Not all Nobel-prize winning physicsits go to Thomas Jefferson High School for Science and Technology (yeah, these kids are lucky enough to have parents that are willing to move to that place just for them. It's uncommon to have so many naturally highly motivated students to just happen to be in one place).

I go to a public school in Chicago suburbs. We do a series of books called Connected Math in 5th through 7th grades. They teach you math through examples and emphasize real world connections to math. I think the example part is bad because it doesn't tell you a general way to do things.
In 8th we do Algebra 1. Then as a freshman it's Algebra 2, then geometry, pre calc, and finally as a senior we do actual calculus. Since a couple kids and I are year ahead of the program, we do stuff a year ahead. So next year we'll be in a sophomore math class! :D

our school only goes up to calculus and it is only for seniors.  I didnt take the class junior year, but passed the test with a five which i self studied, which was impressive.  we do not offer amc's, or aime's although i wish we did (i'll take it somewhere else though).  geometry is taken as a 9th grader to a very small pool of people.  and then 10th grade is algebra two, where some people start to realize that they suck in math, and then comes pre-calc, where only the smart ones make it.  those who do good in this go on to calculus as a senior.

Eversince the summer between eighth grade and ninth grade, i've been really interested in mathematics, especially geometry problems.  It all happened when i was proposed a problem that summer, in eighth grade, by my dad.  He said, "if i cut the top off of a sphere (with some radius) with some height, what will be the volume of the top.  You may laugh, but without calculus, i found that i was working hours on the problem, even to the point of inscribing polyhedra inside the sphere.  it was ridiculous.  Other brain twisting problems were finding the roots of a cubic functions.  you guys probably think these are easy, but i sure didnt.


probability
algorithm
linear algebra
matrix
analytic geometry
vector
modular arithmetic

If I start out from one vertex of a regular $n$-gon, and in each move I move from one vertex to an adjacent vertex with equal probability, create a recursion that gives the probability that after $k$ moves, I will be back to my starting vertex.

I'm not sure if a recursion really works for the general case.

However, I can outline an algorithm:

[hide]
The problem is the same as finding how many ways you can add a total of $k$ 1's and -1's such that the end result is $0 \mod n$.  So let there be $x$ 1's meaning there are $k-x$ -1's.  So we have $x+(k-x)(-1) \equiv 0 \mod n$ or $2x\equiv k\mod n$.

I don't really feel like going through with the whole generalization here so I'll just show the process for $n=7$:

So we want $2x\equiv k\mod 7$.

For $k\equiv 0\mod 7$, we must have $x\equiv 0\mod 7$.  So $x$ can be $0,7,14,\ldots$.  However, the 1's and -1's can be added in multiple ways.  For $x=0$, there $\binom{k}{0}$ ways to add them. For $x=7$, there are $\binom{k}{7}$ ways to add them.  And so on.

So we have \[P_{k\equiv 0}= \frac 1{2^{k}}(\binom{k}{0}+\binom{k}{7}+\binom{k}{14}+\cdots.\] For $k\equiv 1$, we need $x\equiv 4\mod 7$.

For $k\equiv 2$, we need $x\equiv 1\mod 7$.

For $k\equiv 3$, we need $x\equiv 5\mod 7$.

For $k\equiv 4$, we need $x\equiv 2\mod 7$.

For $k\equiv 5$, we need $x\equiv 6 \mod 7$.

For $k\equiv 6$, we need $x\equiv 3\mod 7$.

Thus, we have: \[P=\{\begin{array}{cc}\frac 1{2^{k}}[\binom{k}{0}+\binom{k}{7}+\binom{k}{14}+\cdots] & k\equiv 0 \mod 7\\ \frac 1{2^{k}}[\binom{k}{4}+\binom{k}{11}+\binom{k}{18}+\cdots] & k\equiv 1 \mod 7\\ \frac 1{2^{k}}[\binom{k}{1}+\binom{k}{8}+\binom{k}{15}+\cdots] & k\equiv 2 \mod 7\\ \frac 1{2^{k}}[\binom{k}{5}+\binom{k}{12}+\binom{k}{19}+\cdots] & k\equiv 3 \mod 7\\ \frac 1{2^{k}}[\binom{k}{2}+\binom{k}{9}+\binom{k}{16}+\cdots] & k\equiv 4 \mod 7\\ \frac 1{2^{k}}[\binom{k}{6}+\binom{k}{13}+\binom{k}{20}+\cdots] & k\equiv 5 \mod 7\\ \frac 1{2^{k}}[\binom{k}{3}+\binom{k}{10}+\binom{k}{17}+\cdots] & k\equiv 6 \mod 7 \end{array}\] Someone please tell me if there's a flaw in my reasoning...[/hide]

[quote="joml88"]Someone please tell me if there's a flaw in my reasoning...[/quote]
There's no flaw in your reasoning, it's a very good work.
I wrote a huge (60 pages) paper about algebraical examining the behavior of such random processes while $k\to\infty$. While solving this problem and many of its generalisations I created some methods for describing such situations. I share some of my ideas which can solve this problem and many of its other versions.

We will solve the general case (arbitrary $n$) using the matrix representation of the random events you mentioned (it's the Pre-Olympiad forum, so if you don't understand the matrix notation or some steps of my reasoning, just PM me and I will explain it). Let's number the vertices with numbers modulo $n$, i.e. $0,\ldots,n-1$ and we identify the $i$-th and the $i+n$-th vertices. We will use this numeration also for elements of matrices and vectors. Symbol $[v]_{i}$ denotes the $i$-th coordinate of vector $v$, $[M]_{i,j}$ denotes the $i,j$-th element of matrix $M$ and $I$ denotes the identity matrix.
Let the starting vertex have number $0$ and let's consider a matrix $M$ such that the element $[M]_{i,j}$ is the probability of going from the $i$-th vertex to the $j$-th vertex. Then we have \[M=\left[\begin{array}{cccccc}{0 }&{\frac{1}{2}}& 0 & \cdots & 0 &{{\frac{1}{2}}}\\{{\frac{1}{2}}}& 0 &{\frac{1}{2}}& \cdots & 0 &{ 0}\\{0 }&{\frac{1}{2}}& 0 & \cdots & 0 &{ 0}\\{\vdots }& \vdots & \vdots & \ddots & \vdots &{ \vdots}\\{0 }& 0 & 0 & \cdots & 0 &{{\frac{1}{2}}}\\{{\frac{1}{2}}}& 0 & 0 & \cdots &{\frac{1}{2}}&{ 0}\\ \end{array}\right].\] Now we will consider a vector whose $i$-th coordinate is the probability of being in the $i$-th vertex in the current while. In the moment corresponding with $k=0$, i.e. when we didn't do any move, this vector is \[x^{(0)}=\left[\begin{array}{c}{1}\\{0}\\{\vdots}\\{0}\end{array}\right]^{T}.\] Now let the vector of such probabilities after the $k$-th move be \[x^{(k)}= \left[\begin{array}{c}{x_{0}^{(k)}}\\{x_{1}^{(k)}}\\{\vdots}\\{x_{n-1}^{(k)}}\end{array}\right]^{T}.\] We will create a recurrence expressing $x^{(k+1)}$ by $x^{(k)}$ and $M$.
Let's notice that \[x_{i}^{(k+1)}= P(\text{we are in the }i\text{-th vertex after }k+1\text{ moves})= \sum_{0\leqslant j\leqslant n-1}P(\text{we are in the }j\text{-th vertex after }k\text{ moves})\cdot\]  \[P(\text{we move from the }j\text{-th vertex to the }i\text{-th vertex})= \sum_{0\leqslant j\leqslant n-1}x_{j}^{(k)}\cdot [M]_{j,i}= \left[x^{(k)}\cdot M\right]_{i}.\] This gives $\left[x^{(k+1)}\right]_{i}=\left[x^{(k)}\cdot M\right]_{i}$ for any $i$, so $x^{(k+1)}= x^{(k)}\cdot M$. This recurrence gives an immediate consequence: $x^{(k)}= x^{(0)}\cdot M^{k}$.
We want to express the $k$-th power of $M$ in an explicit form. Let's notice a very important property of this matrix: $[M]_{i,j}= [M]_{i-j\bmod{n},0}$. We will name such matrices [i]cyclic matrices[/i]. Of course $I$ also is cyclic. For any cyclic matrix $A$ and $\alpha,\beta\in\mathbb{Z}$ let $c_{\alpha,\beta}(A)$ be a matrix defined as follows: $\left[c_{\alpha,\beta}(A)\right]_{i,j}= [A]_{i+\alpha,j+\beta}$. It's easy to see that $[c_{0,z}(I)]_{i,j}= \begin{cases}{1\text{ for }i-j\equiv z\pmod{n}}\\{0\text{ for }i-j\not\equiv z\pmod{n}}\end{cases}$.
The functions $c_{\alpha,\beta}$ are important for two reasons. The first is $M= c_{1,0}\left(\frac12\cdot(I+c_{0,2}(I))\right)$. The second is very nice  and easy to check formula $c_{i,j}(A)\cdot c_{i',j'}(B)= c_{i+i',j+j'}(AB)$ which works for any cyclic matrices $A$, $B$. Now we can easily write $M^{k}$ in an explicit form: \[M^{k}= c_{1,0}\left(\frac12\cdot(I+c_{0,2}(I))\right)^{k}= \frac{1}{2^{k}}\cdot c_{k,0}\left(I+c_{0,2}(I)\right)^{k}= \frac{1}{2^{k}}\cdot c_{k,0}\left(\sum_{a_{1},\ldots,a_{k}\in \{0,1\}}c_{0,2a_{1}}(I)\cdot\ldots\cdot c_{0,2a_{k}}(I)\right)= \frac{1}{2^{k}}\cdot c_{k,0}\left(\sum_{a_{1},\ldots,a_{k}\in \{0,1\}}c_{0,2(a_{1}+\ldots+a_{k})}(I)\right)= \frac{1}{2^{k}}\cdot c_{k,0}\left(\sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{0,2a}(I)\right)= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{k,2a}(I).\] Now we want to find an explicit form of $[M^{k}]_{i,j}$: \[[M^{k}]_{i,j}= \left[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{k,2a}(I)\right]_{i,j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}\left[c_{k,2a}(I)\right]_{i,j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}[I]_{k+i,2a+j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop i-j\equiv 2a-k\pmod{n}}\binom{k}{a}[I]_{k+i,2a+j}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop 2a\equiv i-j+k\pmod{n}}\binom{k}{a}.\] And now the final step. The probability of the event that we will be in the $j$-th vertex after $k$ moves equals $\left[x^{(k)}\right]_{j}= \left[x^{(0)}\cdot M^{k}\right]_{j}$. The vector $x^{(0)}\cdot M^{k}$ is just the $0$-th row of the matrix $M^{k}$, so $\left[x^{(0)}\cdot M^{k}\right]_{j}= \left[M^{k}\right]_{0,j}$. Applying the formula for $[M^{k}]_{i,j}$ we get that the probability of event that we will be in the $j$-th vertex is just \[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop 2a\equiv k-j\pmod{n}}\binom{k}{a}.\] In particular, the probability that we will return to the starting vertex equals this expression for $j=0$, i.e. \[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop 2a\equiv k\pmod{n}}\binom{k}{a}.\] It proves that $\mathbf{joml88}$ was right - we can just substitute $n=7$ and see that $2a\equiv k\pmod{7}$ can be replaced by $a\equiv k'\pmod{7}$ where $k'=\frac{k}{2}$ in the field $\mathbb{Z}_{7}$.

In someone is intersted in theorems about limits of such probabilities while $k\to\infty$ and their proofs without the ergodic theorem, we can discuss it. I proved them in much more general cases than the case we considered. Here is an example:
Let $n\in\mathbb{N}^{+}$ and $m\in\mathbb{Z}$ be two constant numbers. Then in Bernoulli decomposition with $k$ samples the probability of event that the number of sucesses is congruent to $m$ modulo $n$ is running to $\frac1n$ while $k\to\infty$.

[quote="mna851"]Wow...haha how about 3-D ?[/quote]
I also considered polidimensional verions of this problem.
The problem we are considering is moving through elements of $\mathbb{Z}_{n}$. It can be generalised to moving through elements of $\mathbb{Z}_{n_{1}}\times\ldots\mathbb{Z}_{n_{m}}$ for some $n_{1},\ldots,n_{m}\in\mathbb{N}^{+}$. So before considering 3-D you mentioned, i.e. $\mathbb{Z}_{n_{1}}\times \mathbb{Z}_{n_{2}}\times \mathbb{Z}_{n_{3}}$ we should consider 2-D, i.e. $\mathbb{Z}_{n_{1}}\times \mathbb{Z}_{n_{2}}$. To explain it we need the definition of vectors and matrices in other dimensions. A plain vector can be called one-dimensional, because we numerate its elements using the elements of one -dimensional $\mathbb{Z}_{n}$. We will define $m$-vector as a mapping $v: \mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}\rightarrow \mathbb{R}$ and $m$-matrix as a mapping $M: \left(\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}\right)^{2}\rightarrow \mathbb{R}$. It's easy to see that a $2$-vector is a non-necessairly square matrix.
The two-dimensional case can be written as following: we start form the $(0,0)$-th element of a $2$-vector. When we are in the $(i,j)$-th element, we can go to the $(i+1,j)$-th and $(i+1,j+1)$-th with probabilities equal $\frac12$. But in this case the matrix corresponing to the matrix $M$ from my previous post will be four-dimensional, because it has to give a real value for any $i,j\in\mathbb{Z}_{n_{1}}\times \mathbb{Z}_{n_{2}}$.
In all these operations the matrix of probabilities of going from state $i$ to state $j$ is cyclic, i.e. its $i,j$-th element depends only on $i-j$. It can be easily explained by fact that while defining the probabilities we choose the current state (vertex of a polygon or an element of a matrix - generally an element of $\mathbb{Z}_{n_{1}}\times\ldots\mathbb{Z}_{n_{m}}$) as the starting state and the situation is the same as at the beginning. In other words, when we are in $j$, the probability of being in $i$ is described only by the position of $i$ when $j$ is choosen as $0$, i.e. by $i-j$. For example the zeroth column of matrix describing the sample $2$-dimensional process (i.e. the vector whose $i$-th coordinate is the $i,0$-th elemnt of the matrix; $0$ means here the zero element of group $\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}$) is a $2$-vector whose $(1,0)$-th and $(1,1)$-th coordinates are equal $\frac12$ and the rest equals $0$. Let's notice that while defining the cyclic matrix we defined only its zeroth column. We could do this because if we know $[M]_{i,0}$ for any $i$, we know $[M]_{i,j}$ for any $i,j$: just $[M]_{i,j}= [M]_{i-j,0}$ and we already know it.
I hope that the idea of polidimensional generalisation of our problem is now clear.

Thank you for these kind opinions. I'll demonstrate the power of my method in $2$ dimensions with following example:

[i]Given $n_{1},n_{2}\in\mathbb{N}^{+}$. We start from the $(0,0)$-th element of a $2$-vector of dimension $(n_{1},n_{2})$ (i.e. a matrix $n_{1}\times n_{2}$). When we are in the $(i,j)$-th element, we can move to the $(i+1,j)$-th and the $(i,j+1)$-th with probabilities equal $\frac12$. Find the probability of being in the $(i,j)$-th element after $k$ moves.[/i]

Before solving this problem we will define $m$-dimensional analogons of the $1$-dimensional objects I used in my first post. We just defined $m$-matrices and $m$-vectors. To clarify the situation, we will name a $m$-matrix $M$ [i]cyclic [/i]iff for any $i,j\in\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}$ we have $[M]_{i,j}= [M]_{i-j,0}$, where the subtraction of indices is of course subtraction in the group $\mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}$ (notice that if we consider an element of a $m$-matrix, we write the two indices from $\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}$ without parenthesis, but if we consider an element of $\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}$ which is a set of $m$ integers, we write it in parenthesis). If a $m$-vector $v$ is the zeroth column of a $m$-matrix $M$ (i.e. $[v]_{i}= [M]_{i,(0,\ldots,0)}$), we mark $M=\mathbf{M}(v)$. The identity matrix is defined as in the one-dimensional case: $[{I}]_{i,j}=\begin{cases}{1\text{ for }i=j}\\{0\text{ for }i\neq j}\end{cases}$ and satisfies $IA= AI= A$ for any matrix $A$.
We should define linear operations on $m$-vectors and $m$-matrices, i.e. adding and multiplying by a real number. It isn't difficult, we just add corresponding elements or multiply all elements by a specified real number. Multiplying two $m$-matrices or a $m$-matrix and a $m$-vector is defined in a following way: $\mathop\forall_{i,j\in\mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}}[AB]_{i,j}= \sum_{s\in\mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}}[A]_{i,s}[B]_{s,j}$, $[Av]_{i}= \sum_{s\in\mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}}[A]_{i,s}[v]_{s}$, $[vA]_{i}= \sum_{s\in\mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}}[v]_{s}[A]_{s,i}$. The basic laws of matrix operations are still true: $(A+B)C=AC+BC$, $A(B+C)= AB+AC$, $(AB)C= A(BC)$, $(\lambda A)B= A(\lambda B)= \lambda (AB)$.
For $m$-vectors and $m$-matrices we also define functions $c_{\alpha}$ and $c_{\alpha,\beta}$ respectively (they can be called [i]cyclic translations[/i]) analogous to cyclic translations of $1$-matrices: for any $m$-vector $v$, $m$-matrix $M$ and $\alpha,\beta\in \mathbb{Z}_{n_{1}}\times\ldots \times\mathbb{Z}_{n_{m}}$ we have $[c_{\alpha}(v)]_{i}= [v]_{i+\alpha}$, $[c_{\alpha,\beta}(A)]_{i,j}= [A]_{i+\alpha,j+\beta}$. There is a very nice formula for any cyclic $A,B$: $c_{i,j}(A)\cdot c_{i',j'}(B)= c_{i+i',j+j'}(AB)$. It gives in particular $\left(c_{i,j}(A)\right)^{k}= c_{ki,kj}(A^{k})$. It's also easy to prove that $c_{0,i}= c_{-i,0}$ and $\mathbf{M}(c_{i}(v))= c_{i,0}(\mathbf{M}(v))$.
Now we can solve the problem. Let $x^{(k)}$ be the  $2$-vector whose $(i,j)$-th element is the probability of being in the $(i,j)$-th element of the $2$-vector we move through after $k$ moves. It's easy to see that $\left[x^{(0)}\right]_{(i,j)}= \begin{cases}{1\text{ for }i=j=0}\\{0\text{ otherwise}}\end{cases}$. Now we will find matrix $M$ such that $[M]_{i,j}$ is the probability of moving from the $i$-th element of a $2$-vector to its $j$-th element. The zeroth column of this matrix is a  $2$-vector whose only nonzero elements are the $(0,-1)$-th and $(-1,0)$-th and both are equal $\frac12$, because we can move to $(0,0)$ only from $(-1,0)$ and $(0,-1)$. Thus \[M= \frac12\cdot \mathbf{M}\left(c_{(-1,0)}(v_{I})+c_{(0,-1)}(v_{I})\right)= \frac12\cdot \left( c_{(-1,0),(0,0)}(I)+c_{(0,-1),(0,0)}(I)\right)=\]  \[\frac12\cdot c_{(-1,0),(0,0)}\left(I+c_{(1,-1),(0,0)}(I)\right),\] where $\mathbf{M}(v_{I})= I$. The formula $x^{(k+1)}= x^{(k)}\cdot M$ giving $x^{(k)}= x^{(0)}\cdot M^{k}$ has as easy proof as for $m=1$ - only the limits of summation change (of course here we have $i\in\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$): \[x_{i}^{(k+1)}= P(\text{we are in the }i\text{-th element after }k+1\text{ moves})=\]  \[\sum_{j\in\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}}P(\text{we are in the }j\text{-th element after }k\text{ moves})\cdot\]  \[P(\text{we move from the }j\text{-th element to the }i\text{-th element})=\]  \[\sum_{j\in\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}}x_{j}^{(k)}\cdot [M]_{j,i}= \left[x^{(k)}\cdot M\right]_{i}\] and form $\left[x^{(k+1)}\right]_{i}=\left[x^{(k)}\cdot M\right]_{i}$ for any $i$ we get that $x^{(k+1)}= x^{(k)}\cdot M$ and $x^{(k)}= x^{(0)}\cdot M^{k}$ as claimed. Let's notice that $\left[x^{(k)}\right]_{i}= \left[x^{(0)}\cdot M^{k}\right]_{i}= \sum_{s\in\mathbb{Z}_{n_{1}}\times\ldots\times\mathbb{Z}_{n_{m}}}\left[x^{(0)}\right]_{s}\left[M^{k}\right]_{s,i}$ and the only nonzero constituent in this sum corresponds with $s=(0,0)$, so the sum equals $\left[x^{(0)}\right]_{0}\left[M^{k}\right]_{0,i}=\left[M^{k}\right]_{0,i}$. Now we can raise $M$ to power $k$: \[M^{k}= \left(\frac12\cdot c_{(0,0),(1,0)}\left(I+c_{(1,-1),(0,0)}(I)\right)\right)^{k}= \frac{1}{2^{k}}\cdot c_{(0,0),(k,0)}\left(\left(I+c_{(1,-1),(0,0)}(I)\right)^{k}\right)=\]  \[\frac{1}{2^{k}}\cdot c_{(0,0),(k,0)}\left(\sum_{a_{1},\ldots,a_{k}\in \{0,1\}}c_{(a_{1}+\ldots+a_{k},-(a_{1}+\ldots+a_{k})),(0,0)}(I)\right)=\]  \[\frac{1}{2^{k}}\cdot c_{(0,0),(k,0)}\left(\sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{(a,-a),(0,0)}(I)\right)= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{(a,-a),(k,0)}(I)\] and get an explicit formula for $\left[x^{(k)}\right]_{i}= \left[M^{k}\right]_{0,i}$ (we set $i=(i_{1},i_{2})$, where $i_{1}\in\mathbb{Z}_{n_{1}}$ and $i_{2}\in\mathbb{Z}_{n_{2}}$): \[\left[M^{k}\right]_{0,i}= \left[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}c_{(a,-a),(k,0)}(I)\right]_{0,i}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}\left[c_{(a,-a),(k,0)}(I)\right]_{0,i}=\]  \[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k}\binom{k}{a}[I]_{(a,-a),(k,0)+i}= \frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop (a,-a)= (k,0)+i}\binom{k}{a}[I]_{(a,-a),(k,0)+i}= \frac{1}{2^{k}}\cdot \sum_{\tiny{\begin{array}{c}{0\leqslant a\leqslant k}\\{a\equiv k+i_{1}\pmod{n_{1}}}\\{a\equiv-i_{2}\pmod{n_{2}}}\end{array}}}\binom{k}{a}.\] So finally the probability of being in the $(i_{1},i_{2})$-th element of the vector we move through after $k$ moves is \[\frac{1}{2^{k}}\cdot \sum_{\tiny{\begin{array}{c}{0\leqslant a\leqslant k}\\{a\equiv k+i_{1}\pmod{n_{1}}}\\{a\equiv-i_{2}\pmod{n_{2}}}\end{array}}}\binom{k}{a}.\] In particular, the probability that we return to the the $(0,0)$-th element is \[\frac{1}{2^{k}}\cdot \sum_{0\leqslant a\leqslant k\atop a\equiv k\pmod{n_{1}},\ n_{2}\mid a}\binom{k}{a}.\]

dude ur scarin me   :o
that is waaaaay to complex for my teeny little head  :lol:


function
integration
inequalities
probability and stats

If $ X$ and $ Y$ are independent continuous random variables with the hazard rate functions $ \lambda_{X}(t)$ and $ \lambda_{Y}(t)$, and $ W \equal{} \min(X,Y)$, what is the distribution function of $ W$?

I noted that the distribution of W: $ F(w) \equal{} P(W \le w) \equal{} P(\min(X,Y), \le w) \equal{} ? \equal{} \sum p(w)$ ?

At this point, I couldn't figure out how the independent distribution functions of $ X$, ($ F_{X}(t) \equal{} 1 \minus{} e^{\lambda_{X}(t)}$ ) and similarly for $ Y$, were going to help ..

It's ugly if you base it on densities, and nice if you work with cumulative distribution functions. $ P(W\le t)\equal{}P(X\le t)\cdot P(Y\le t)$.

How would that factor in the definition of $ W$?

[b]jmerry[/b] has made a mistake.

$ F_W(w)\equal{}\mathbb{P}[min(X,Y)<w]\equal{}1\minus{}\mathbb{P}[min(X,Y)\geq w]\equal{}1\minus{}\mathbb{P}[X \geq w, Y \geq w]\equal{}1\minus{}\mathbb{P}[X \geq w]\mathbb{P}[Y\geq w]\equal{}1\minus{}(1\minus{}F_X(w))(1\minus{}F_Y(w)).$
Then you can easily rewrite the latter in terms $ \lambda _X, \lambda _Y$

And then you should have:

$ 1\minus{} (e^{\int \lambda_{X}(w) dw})(e^{\int \lambda_{Y}(w)dw})$

But how did you know that it was $ P(W < w)$ and not $ P(W \le w)$?

The mistake was that I gave you the cumulative distribution function for the maximum. Reversing all of the inequalities (to $ P(W\ge t)\equal{} P(X\ge t)\cdot P(Y\ge t)$) is the simplest fix.

If your distributions are "continuous" and have densities, the difference between $ <$ and $ \le$ is completely irrelevant, because "=" has zero probability.  Otherwise... it still works either way, as long as you use the same type of inequalities everywhere. $ P(W>t)\equal{}P(X>t)\cdot P(Y>t)$ is also true.

Both versions come from the simple logic of what a minimum is. The minimum is bigger than $ t$ if and only if each of the terms that goes into it is bigger than $ t$. With independence, "and" is multiplication of probabilities.

[quote]But how did you know that it was P(W < w) and not P(W \le w)?[/quote]
I was taught this definition of cumulative d.f.. $ F_X(x)\equal{}\mathbb{P}[X<x]$.
In different books you can find different definitons. If you've understood the main idea, you can rewrite my post .


The value of $ x$ at the intersection of $ y\equal{}\frac{8}{x^2\plus{}4}$ and $ x\plus{}y\equal{}2$ is:
$ \textbf{(A)}\ \minus{}2\plus{}\sqrt{5} \qquad\textbf{(B)}\ \minus{}2\minus{}\sqrt{5} \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \text{none of these}$

First, isolate $ y$ in the second equation:

$ x\plus{}y\equal{}2$
$ y\equal{}2\minus{}x$

Now that $ y$ is isolated in both equations, we could set the two quantities given to $ y$ equal to each other and then solve.

$ \frac{8}{x^2\plus{}4}\equal{}2\minus{}x$
$ (2\minus{}x)(x^2\plus{}4)\equal{}8$
$ 2x^2\plus{}8\minus{}x^3\minus{}4x\equal{}8$ (FOIL)
$ \minus{}x^3\plus{}2x^2\minus{}4x\plus{}8\equal{}8$ (associative property of addition)
$ \minus{}x^3\plus{}2x^2\minus{}4x\plus{}8\minus{}8\equal{}8\minus{}8$
$ \minus{}x^3\plus{}2x^2\minus{}4x\equal{}0$

$ 0$ is clearly an answer to the last form of the equation reached, so $ \fbox{C}$ is the answer.


geometry rectangle function calculus derivative quadratics perimeter

A 20-meter string is cut into 2 pieces. The first piece is used to form a rectangle while the other piece is used to form a triangle. What should be the length allotted to form the triangle in order to maximize the total area of the two geometric figures?

Oh, well, I had a similar problem, but with a circle and a square...
Anyway, it's the same. I only have some doubts about that rectangle: even if you don't know the properties of the triangle, there exists Heron's formula. with which you can calculate the triangle's area; but with the rectangle... I suppose you need some more informations about its sides.

Once you'll have this extra-information, call $x$ the length of the piece of string used for the triangle, consequently $20-x$ will be the other piece. Of course $0<x<20$...
With this notation, you can express the sum of the areas of the two figures as a function of $x$, let's say $S(x)$. That's the expression you have to maximize.
Take the derivative of $S(x)$ and calculate the intervals where it's positive (not strictly).
I think you'll obtain something like $0<x\leq a$ positive, $a<x<20$ negative.
That means $S(x)$ it's increasing in $(0,a]$, decreasing in the other interval. Thus the searched maximum will be $a$.

Hope I've been clear...  :)

[quote="vishalarul"]Let's try it out!
Click the smiley to reveal hidden content.
[hide=" :lol: "]$\textrm{We have to maximize}$ $\frac{x^{2}\sqrt{3}}{4}+(20-x)^{2}$ $\textrm{.}$

$\textrm{Simplifying,}$ ${(1+\frac{\sqrt{3}}{4})(x^{2}})-40x+400$ $\textrm{.}$  $\textrm{We have to find the x that maximizes this expression. This is just a normal quadratic, though. So we can use the formula}$ $-\frac{b}{2a}$:

$\frac{40}{2(1+\frac{\sqrt{3}}{4})}$

$\textrm{Simplifying this ugly expression,}$

$=\frac{80}{4+\sqrt{3}}$

$=\frac{80(4-\sqrt{3})}{16-9}$

$=\frac{320-80\sqrt{3}}{7}$

$\textrm{Is this answer right (}$$\frac{320-80\sqrt{3}}{7}$ $\textrm{)?}$[/hide][/quote]

[hide="Almost..."]
If the string is cut at length $x$, then the side of the triangle [i]won't[/i] just be $x$. It will be $\frac{x}{3}$. Similarly, the side of the square is $\frac{20-x}{4}$.

And also, it's trivial to prove but i still think you need to show that of all rectangles, a square has maximized area for a set perimeter. Just for completeness. [/hide]

I'll do the rectangle to a square one:
Perimeter of a rectangle is fixed, call it G.
Call each side x and y respectively.

By AM-GM
$\frac{x+y}{2}\ge \sqrt{xy}$
$\frac{x^{2}+2xy+y^{2}}{4}\ge xy$
$x^{2}+y^{2}\ge 2xy$
$\frac{x^{2}+y^{2}}{2}\ge xy$
$(y-x)^{2}\ge 0$
$(y-x) \ge 0$
$y \ge x$

Where equality holds iff x = y, i.e. the rectangle is a square.

[quote="Carcul"][quote="Mathgeniuse^ln(x)"]As well as an equilateral triangle, but it is harder to prove for an equilateral triangle than a square.[/quote]

So we have to prove that, for a given perimeter, a triangle will have maximum area if their sides are all equal, i.e., if the triangle is equilateral. Let me have a try.

For a triangle with sides [i]x[/i], [i]y[/i], and [i]z[/i], its area [i]A[/i] is given by Heron's formula,
\[A = \sqrt{p(p-x)(p-y)(p-z)}\]
where [i]p[/i] is equal to half the perimeter. So, for a given [i]p[/i], the area will be maximum if the product (p-x)(p-y)(p-z) is also a maximum. But this can happen only if x = y = z, because
\[\sqrt[3]{(p-x)(p-y)(p-z)}\le \frac{p}{3}\Rightarrow (p-x)(p-y)(p-z) \le \frac{p^{3}}{27}\]
and the equality holds only when x = y = z. So, for a given perimeter, in order for a triangle to have maximum area it must be equilateral.

Carcul[/quote]

Oh, wow, I didn't realize that.  Thanks So Much.

I also realized that that is true by AM-GM.


what is the side length, in inches, of the largest square that can be inscribed in a circle of radius 8 inches? express your answer in simplest radical form.

i think a 45-45-90 triangle property can be used here. can someone explain that property?

[hide]The center of the square will be the center of the circle. So the diagonal of the square is the diameter of the circle. Now let the side of the square be $a$. By pythagoras,
$a^{2}+a^{2}=2a^{2}=16^{2}=256$
$a^{2}=128$ and
$a=8\sqrt{2}$[/hide]

is it just me or can you only inscribe [b]only one square[/b] in every circle

so why would they ask for the largest square when there's only one square?


function
limit
number theory proposed
number theory

Let $ m$ and $ n$ denote integers greater that $ 2$. Prove that if the equation $ \frac{n}{\pi(n)}\equal{}m$ has a solution then so does the equation $ \frac{n}{\pi(n)}\equal{}m\minus{}1$.

[quote="Max D.R."]Let $ m$ and $ n$ denote integers greater that $ 2$. Prove that if the equation $ \frac {n}{\pi(n)} \equal{} m$ has a solution then so does the equation $ \frac {n}{\pi(n)} \equal{} m \minus{} 1$.[/quote]
The following problem is also true.
Let $ f(n)$ is a non decrease function, $ f(n) > 0$ and $ \lim_{n\to \inf} \frac {f(n)}{n} \equal{} 0$. In addition, there is an positive integer $ n$ such that $ n/f(n) \equal{} m$ (m>2) then there also exists $ k$ such that $ k/f(k) \equal{} m \minus{} 1$.
Proof : Consider function $ g(t) \equal{} t \minus{} (m \minus{} 1)f(t)$ where $ t\in N$. We have $ g(t \plus{} 1) \minus{} g(t) \equal{} 1 \minus{} (m \minus{} 1)(f(t \plus{} 1) \minus{} f(t))$. Because f is non decrease so $ f(t \plus{} 1)\geq f(t)$. It means that if $ g(t \plus{} 1) > g(t)$ then $ g(t \plus{} 1) \minus{} g(t) \equal{} 1$.
Besides, we also have $ g(n) \equal{} n \minus{} (m \minus{} 1)f(n) \equal{} f(n) > 0$ (because $ n \equal{} mf(n)$ and $ g(2) \equal{} 2 \minus{} mf(2)\leq 0$
By these facts, the range of g(t) has negative elements, positive elements and if at every increase step ($ g(t \plus{} 1) > g(t)$) we always have $ g(t \plus{} 1) \equal{} g(t) \plus{} 1$. Therefore it must have 0 as an element.
Proof is completed.
Note: We also see that $ g(t)$ tends to infinite when n tends to infinite. Therefore, it also exists $ h$ such that $ n/f(h) \equal{} m \plus{} 1$. And of course, the function $ n/f(n)$ covers the set $ [m, \inf)$


algebra
polynomial
Vieta
calculus
derivative
greatest common divisor

hello.

Let $ P(x)\equal{}x^4\plus{}kx^3\plus{}10x\plus{}7$.   Find all possible values of $ k\in\mathbb{R}$ so that $ P(x)$ will have a root (real or complex) of multiplicity $ 2$.  Show a complete proof that those are the only possible values for $ k$.

 :lol:

[hide="Partial Solution"]Let $ a$ be the root of the polynomial with multiplicity 2, and let $ b, c$ be other roots.

Then, we have
$ x^4+kx^3+10x+7$
$ =x^4-(2a+b+c)x^3+(a^2+2ab+bc+2ca)x^2-(2abc+a^2b+ca^2)x+a^2bc$ using Vieta's relations.

Comparing coefficients, we have \begin{align}1 &=1 \\k &=-(2a+b+c)\\ 0 &=a^2+2ab+bc+2ca\\ 10 &=-(2abc+a^2b+ca^2)\\ 7 &=a^2bc \end{align}
$ a\cdot(3)+2\cdot(4)=a^3+2a^2b+abc+2ca^2-(2a^2b+4abc+2ca^2)=20$$ \implies a^3-3abc=20$

Noting that $ a^2bc=7\implies abc=\frac{7}{a}$, we have $ a^3-3abc=a^3-\frac{21}{a}=20\implies a^4-20a-21=0$.

Hence, $ a=-1, 3, -1-i\sqrt{6}, -1+i\sqrt{6}$.[/hide]

Ugh, we have to check all 4 values of $ a$ which will be quite tiring...

hello fantasy lover.  the real roots that you obtained are correct, i haven't checked out the complex roots really.
but remember that since $ k\in\mathbb{R}$, i think that it is not possible for $ P(x)$ to have complex roots of multiplicity $ 2$...since if it were the case, then the conjugate of this complex root must also be a root of $ P(x)$, and also necessarily of multiplicity $ 2$ also.  That would imply that $ P(x)\equal{}(ax^2\plus{}bx\plus{}c)^2$, for some real numbers $ a,b,c$...proceeding further i will leave it to you to find a contradiction.    :)

I think that from here, $ k$ can easily  be found using substitution or maybe long division??  but i haven't tried long division..  :)

can you now complete the solution?

hello.

can someons finish the problem and find all possible values for $ k$ ?

thanks.   :)

[hide="My approach"]We have that $ r$ is a double root of $ P$ if and only if $ P(r) \equal{} P'(r) \equal{} 0$, where $ P'(x)$ is the derivative of $ P$ at $ x$. Thus, we only need to find all $ k$ such that the greatest common factor of $ P$ and $ P'$ is nonconstant. This is easy with the euclidean algorithm. [/hide]

hello.

can anyone give me an answer for all the possible values of k ?    I came up with only 2.
One for each real number $ a$ described by FantasyLover below.   :)


Vanessa will graph the lines $ y \equal{} \frac{1}{3}x \plus{} 7$ and $ y \equal{} 2x\plus{}5$. What is the sum of the $ x$-coordinates of the two lines' $ x$-intercepts? Express your answer as a decimal to the nearest tenth.

x intercept -> y=0

$ \frac{1}{3}x\plus{}7\equal{}0, x\equal{}\minus{}21.$

$ 2x\plus{}5\equal{}0, x\equal{}\minus{}2.5$

answer : -23.5


Let n be a natural number such that $ n\equal{} 2^{2007}.k\plus{}1$ with k is odd. Prove that
$ n$doesn't divide$ 2^{n\minus{}1}\plus{}1$.

Let $ p|2^{n\minus{}1}\plus{}1\equal{}2^{2^{2007}*k}\plus{}1$ and $ d$ is minimal natural, suth that $ p|2^d\minus{}1$, then $ 2^{2008}|d|p\minus{}1$.
Because from $ n|2^{n\minus{}1}\plus{}1$ we get for any prime divisor p of n $ 2^{2008}|p\minus{}1$. It give $ 2^{2008}|n\minus{}1$ - contradition.
Posted before, that $ n|2^{n\minus{}1}\plus{}1$ had unique solution $ n\equal{}1$.

[quote="Rust"]Let $ p|2^{n \minus{} 1} \plus{} 1 \equal{} 2^{2^{2007}*k} \plus{} 1$ and $ d$ is minimal natural, suth that $ p|2^d \minus{} 1$, then $ 2^{2008}|d|p \minus{} 1$.
Because from $ n|2^{n \minus{} 1} \plus{} 1$ we get for any prime divisor p of n $ 2^{2008}|p \minus{} 1$. It give $ 2^{2008}|n \minus{} 1$ - contradition.
Posted before, that $ n|2^{n \minus{} 1} \plus{} 1$ had unique solution $ n \equal{} 1$.[/quote]
why $ 2^{2008}|d$


Let be $ a,b,c\in \mathbb{R}$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} a^2 \plus{} b^2 \plus{} c^2$ . Prove that $ a,b,c$ can't all be in $ \mathbb{Z}_\minus{}^*$ .

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} a^2 \plus{} b^2 \plus{} c^2$ . Prove that $ a,b,c$ can't all be nonpositives .[/quote]

How about $ a\equal{}b\equal{}c\equal{}0$, all of them are non-positive.

[quote="stephencheng"][quote="alex2008"]Let be $ a,b,c\in \mathbb{R}$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} a^2 \plus{} b^2 \plus{} c^2$ . Prove that $ a,b,c$ can't all be nonpositives .[/quote]

How about $ a \equal{} b \equal{} c \equal{} 0$, all of them are non-positive.[/quote]

Yes , but 0 is a positive number . I reffered to the strictly non-positives numbers .

Mods , this is to easy for Pre-Olympiad , please move it back . Thanks .
[color=green]We'll be the judges of that, if you don't mind.[/color]

[quote="stephencheng"][quote="alex2008"]Let be $ a,b,c\in \mathbb{R}$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} a^2 \plus{} b^2 \plus{} c^2$ . Prove that $ a,b,c$ can't all be nonpositives .[/quote]

How about $ a \equal{} b \equal{} c \equal{} 0$, all of them are non-positive.[/quote]
hello stephencheng! with all respect but you must reread the problem,
for alex! your problem is nice
[hide="click here to see my solution"] suppose that $ a;b;c$ are all strictly non-positive then $ abc<0$  (*)
we have, $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}a^2\plus{}b^2\plus{}c^2$
$ \Leftrightarrow 2(a^2\plus{}b^2\plus{}c^2)\minus{}2(ab\plus{}bc\plus{}ca)\equal{}2abc$
$ \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\equal{}2abc$
but from (*) we have $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\geq 0 >2abc$ wich is a contradiction where we deduct that $ a,b,c$ can not be all strictly non-positive. [/hide]

Nice solution , peine !

Here is my solution , even it is similar with your solution  :)

Suppose that $ a,b,c$ are all strictly non-positives numbers $ \Rightarrow abc<0$

We have that $ a^2\plus{}b^2\plus{}c^2\equal{}ab\plus{}bc\plus{}ca\plus{}abc\le a^2\plus{}b^2\plus{}c^2\plus{}abc \Rightarrow abc\ge 0$ , which is a contradiction since $ abc<0$ .

[quote="Farenhajt"][quote="alex2008"]Yes , but 0 is a positive number . I reffered to the strictly non-positives numbers .[/quote]

Zero is neither positive nor negative, and there's no such thing as "STRICTLY non-positive" - either strictly positive or strictly negative. Thus stephencheng has a point.[/quote]

Ok , ... i changed . I think everybody must be content now .


$PA, PB, PC$ are three rays in space. Show that there is just one pair of points $B', C$' with $B'$ on the ray $PB$ and $C'$ on the ray $PC$ such that $PC' + B'C' = PA + AB'$ and $PB' + B'C' = PA + AC'$.

Just a very brief sketch:

Try to prove that the condition implies the existence of a ball tangent to $AB',B'C',C'A$ inside these segments, and to $(PA,(PB',(PC'$ in points which are separated from $P$  by $A,B',C'$ respectiely, and draw the conclusion fom here.


Hi everybody.
Im new here and im very bad in the math subject.
I wish to get better at the subject by learning from everybody here. :)
Recently my teacher gave me a algebra problem i cant seem to solve.
I have to hand it in at the end of the week and my deadline is up!
First I dont expect people to give me the answers but i would like some tips and methods to solving my problems.
Thank you.
Heres the math problem.

I have 5 bags of rice. The bags of rice are being weighed 2 at a time.
The outcome of each pair is 80,82,82,84,85,86,87,88,90, and 91.
Find the individual weight of each bag of rice.

I understood the problem.
I first listed the way each bag of rice is put.
If i give each bag of rice a name than it would be easier to understand.
Bag A,B,C,D, and E.
The combinations goes something like this
A+B, A+C, A+D, A+E
B+C, B+D, B+E
C+D, C+E
D+E.
After that i figured that there would that each bag would have been used before.
If I average the combinations of 80,82,83,84,85,86,87,88,90, and 91 i would get 214.
So I understood that A+B+C+D+E=214.
I got stuck up to here.
Can anyone here be kind enough to help me continue?
Some tips or some pointers?
Thank for looking.

hint or tip
[hide]you have got the weights and the corresponding combinations. so, u can form equations. now solve them by choosing right pairs.[/hide]

Welcome to AoPS!
This would go into the High-school Basics forum, but I will still give you some pointers.
[hide="lame solution"]Reduce-row Echelon form[/hide]
[hide="algebraic pointer"]You can add A+B, A+C, A+D, and A+E to get A+A+A+A+B+C+D+E. How can you use this if you know the sum?[/hide]
Also, when you don't know how to do a problem, most experienced problem solvers(a.k.a. people who can do problems that less than 1000 people in the world can do, and less than 500 in the time limit that they use) play with the problem, to try to get somewhere. For instance, here you would try to add stuff to find things to cancel out. It may take a while, but it should lead you to the solution.

Food for thought: Now how about if you were given that A+B=80, B+C=82, C+D=82, D+E=84, and E+A=85?

[hide="Hint"]You can assume without loss of generality that $ A\leq B\leq C\leq D\leq E$.[/hide]

Label the bags $ A \le B\le C\le D\le E$

$ A\plus{}B\equal{}80$

$ D\plus{}E\equal{}91$

And you know $ A\plus{}B\plus{}C\plus{}D\plus{}E$ By summing up all the bags and dividing through. So you know C.

But you also know that $ A\plus{}C$ is going to be the next smallest value, and $ C\plus{}E$ the next largest.


Let be $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$ . Show that :

$ \sqrt{x\plus{}yz}\plus{}\sqrt{y\plus{}zx}\plus{}\sqrt{z\plus{}xy}\le 2$

By Cauchy, $ LHS^2 \leq (\sum(x\plus{}yz))(\sum 1)\equal{}3(x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx)$. If we can prove that $ 3(x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx)\leq 4$, then we are done. Putting in $ x\plus{}y\plus{}z\equal{}1$, we see this is equivalent to $ 3(xy\plus{}yz\plus{}zx)\leq 1$. Now we substitute $ 1$ for $ (x\plus{}y\plus{}z)^2$, getting $ 3(xy\plus{}yz\plus{}zx)\leq (x\plus{}y\plus{}z)^2$, which, after expanding, is equivalent to $ xy\plus{}yz\plus{}zx\leq x^2\plus{}y^2\plus{}z^2 \iff 0\leq \frac{1}{2}((x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2)$, which is true.
Equality occurs when $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$.

[hide="Even easier"]$ \text{LHS}\le \sqrt {3 \plus{} 3(xy \plus{} yz \plus{} xz)}\le \sqrt {3 \plus{} 3(1/3)} \equal{} 2$, the first inequality by RMS-AM, the second by AM-GM. [/hide]

Or :

$ LHS\equal{}\sum_{cyc}\sqrt{x(x\plus{}y\plus{}z)\plus{}yz}\equal{}\sum_{cyc}\sqrt{(x\plus{}y)(y\plus{}z)}\le \sum_{cyc}\frac{x\plus{}y\plus{}x\plus{}z}{2}\equal{}2$

nice,
by Cauchy or Jensen,
$ S\equal{}\sqrt{x\plus{}yz}\plus{}\sqrt{y\plus{}zx}\plus{}\sqrt{z\plus{}xy}\le \sqrt{3(x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx)}\equal{}\sqrt{3\plus{}3(xy\plus{}yz\plus{}zx)}$
and we have, $ 3(xy\plus{}yz\plus{}zx)\leq (x\plus{}y\plus{}z)^2\equal{}1$ then , $ S\leq\sqrt{4}\equal{}2$


modular arithmetic
number theory proposed
number theory

Show that for any $n$ we can find a set $X$ of $n$ distinct integers greater than 1, such that the average of the elements of any subset of $X$ is a square, cube or higher power.

For every subset $S$ of $[n]=\{1,\ldots,n\}$, let $m_S$ be a positive integer such that the $m_S$ are pairwise coprime. We'll find a set of distinct positive integers $x_i,\ i\in[n]$ such that the average $A_S$ of the numbers $x_j,\ j\in S$ is an $m_S$ power.

First let $x_i$ be distinct powers ($x_i$ is an $m_i$ power for each $i$). Now we order the subsets $S\subset[n]$ with more than one element: $S_1,\ldots,S_{2^n-1-n}$. At step $1\le t\le 2^n-n-1$, let $A_{S_t}$ be equal to $p_1^{a_1}\cdot\ldots\cdot p_k^{a_k}$. We multiply all current $x_i$ by a positive integer of the form $p_1^{mm_1\ldots m_n\prod_{u<t}m_{S_u}}\ (*)$, with $m$ chosen such that the exponent in $(*)$ be congruent to $-a_1\pmod m_{S_t}$. This is possible because of the fact that all $m_S$ are pairwise coprime. Then we do something similar for $p_2$, and so on up to $p_k$. At the end of the $t$'th step, all averages $A_{S_u},\ u\le t$ will be $m_{S_u}$ powers respectively. We replace the old $x_i$'s with these new ones, and go to step $t+1$.

I claim that for every positive integer $\ell$, there exists a positive integer $d$, such that $\{d,2d,\dots,\ell d\}$ are all perfect powers. Let us first prove this lemma. Let $d=2^{\alpha_2}3^{\alpha_3}\cdots \ell^{\alpha_\ell}$, for $\alpha_2,\dots,\alpha_\ell\geqslant 1$. The condition holds, provided we show the following:
(a) There is a positive integer $p\geqslant 2$, such that $p\mid {\rm gcd}(\alpha_2,\dots,\alpha_\ell)$.
(b) For every $j\in\{2,\dots,\ell\}$ there is a $p_j\geqslant 2$, such that   $p_j\mid {\rm gcd}(\alpha_2,\dots,\alpha_{j-1},\alpha_j+1,\alpha_{j+1},\dots,\alpha_\ell)$.

Now, the idea is to fix primes $p>p_1>\cdots >p_\ell$, and design $\alpha_j$ such that, $\alpha_j\equiv 0\pmod{p}$, $\alpha_j\equiv 0\pmod{p_i}$ for every $i \neq j$; and $\alpha_j\equiv -1\pmod{p_j}$. The existence of such $\alpha_2,\dots,\alpha_\ell$ follow immediately by the Chinese remainder theorem.

Having proven this, now return to the problem. We construct $A$ of form $A=\{d\cdot n!,2d\cdot n!,\cdots, nd\cdot n!\}$. Observe that, for any subset $S\subseteq A$, $\Sigma_S\triangleq \sum_{x\in S} x$ is of form, $\frac{1}{|I|}(\sum_{k\in I}k)n!d$, where $I\subset \{1,2,\dots,n\}$. Applying the previous result by taking $\ell$ large enough (e.g., $\ell>n!(1+2+\cdots+n)$ works), we finish the problem.

why there is a geometry, 3D geometry tags on this problem ?


I used the distributive property.
$ a\plus{}a\equal{}a(1\plus{}1)\equal{}a(2)$
$ 2^3\plus{}2^3\plus{}2^3\equal{}2^3(1\plus{}1\plus{}1)\equal{}2^3(3^1)\equal{}3^12^3$

[quote="Fraenkel"]I used the distributive property.
$ a \plus{} a \equal{} a(1 \plus{} 1) \equal{} a(2)$
$ 2^3 \plus{} 2^3 \plus{} 2^3 \equal{} 2^3(1 \plus{} 1 \plus{} 1) \equal{} 2^3(3^1) \equal{} 3^12^3$[/quote]
Or $ \ 3(2^3)\equal{}3^1\cdot 2^3$

I get it.. That's one smart way of looking at it.
Nice. :lol:


number theory
least common multiple
prime factorization

What is the least natural number that will have a remainder of 3 when divided by any of the numbers 4, 5, 6, 8 or 10?

Since the least common multiple of 4, 5, 6, 8, and 10 is 120, you just add 3 to get an answer of 123.

the numbers are 4,5,6,8, and 10.
first of all, we do not have to worry about the 4 and the 5, because these two numbers will be covered by the 8 and 10, respectively.

we need to find the lcm of 6,8,10
prime factorizations:
6: 3*2
8:2*2*2
10:5*2

the numbers prime factorization MUST be 2*2*2*3*5 to cover all three of these factorizations.
therefore, it is 15*8, or 120. but, we must add three so that the remainder, when dividing by these numbers, is 3.
ANSWER:123

EDIT:beaten to it...=[
my answer is more detailed though....

123 is way too high!

If you divide $ 3$ itself by any of those, you also get remainder 3 :)

:o That's true.  The answer in the game is actually 123, but... obviously it's wrong.

I sure hope they're not using this problem as a template.

Whenever they ask "leaves a remainder of k for a,b,c,..." they'll have to choose a number less than all of a,b,c,.... and then immediatley the answer will be "k".

This looks like a dumbening down of a Chinese Remainder Theorem problem.  An example of a better problem would be: "What is the first positive natural number to give a remainder of 2 when divided by 3, remainder of 1 divided by 4 and a remainder of 3 divided by 5?"


Find the positive integers $ a_{1},a_{2},...,a_{n}$, such that $ \sqrt[k] {a_{1}^{k} \plus{} a_{2}^{k} \plus{} ... \plus{} a_{n}^{k}}$ is rational for every $ k > 2$.

[quote="Cristi Sav"]Find the positive integers $ a_{1},a_{2},...,a_{n}$, such that $ \sqrt [k] {a_{1}^{k} \plus{} a_{2}^{k} \plus{} ... \plus{} a_{n}^{k}}$ is rational for every $ k > 2$.[/quote]

Clearly, $ \sqrt [k] {a_{1}^{k} \plus{} a_{2}^{k} \plus{} ... \plus{} a_{n}^{k}} \in \mathbb{Q}$
$ \Leftrightarrow \sqrt [k] {a_{1}^{k} \plus{} a_{2}^{k} \plus{} ... \plus{} a_{n}^{k}} \in \mathbb{Z}$

WLOG,  $ a_1 \le a_2 \le ... \le a_n$.

$ a_n \equal{} \sqrt [k]{a_n^k} \le \sqrt [k] {a_{1}^{k} \plus{} a_{2}^{k} \plus{} ... \plus{} a_{n}^{k}} \le \sqrt [k]{na_n^k} < a_n \plus{} 1$ for any large enough $ k$.
Equality holds for the left inequality $ \Rightarrow n \equal{} 1$,

So $ n \equal{} 1, a_1 \in \mathbb{N}$, is the only solution after checking.

P.S. Mistake corrected.

[quote="stephencheng"] $ \Rightarrow n \equal{} 1$

$ \sqrt [k]{a_1} \in \mathbb{Z}$ for all $ k > 2$ $ \Leftrightarrow a_1 \equal{} 1$

So $ n \equal{} 1,a_1 \equal{} 1$ is the only solution.[/quote]

I do agree with $ n\equal{}1$ but then we get $ \sqrt [k]{a_1^k} \in \mathbb{Z}$ for all $ k > 2$ and not $ \sqrt [k]{a_1} \in \mathbb{Z}$ for all $ k > 2$ and so any $ a_1$ matches the equation.


linear algebra
matrix
search
LaTeX
IMO Shortlist
algebra unsolved
algebra

Consider all $n\times n$ real matrices $(a_{jk})_{ 1 \leq j \leq n, 1 \leq k \leq n }$ , where $n$ is even , $|a_{jk}| \leq 1$ and $\sum_{j=1}^{n} \sum_{k=1}^{n} a_{jk}$ = 0. Determine the least positive number $C$ such that every such matrix has a roe or a column with the sum of its entries not exceeding $C$ in its absolute value.

Thanks a lot for posting these problems, but I think you could have first checked which of them are from the IMO Shortlist 2004 (in fact, there is quite a number of such problems, ca. 10 I believe). This one, for instance, is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41119]IMO Shortlist 2004 combinatorics problem 4[/url] (see proposed solutions at http://www.mathlinks.ro/Forum/viewtopic.php?t=15622 ).

  darij

Sorry!!! I didn't realise that. Just aa question: How do you search for the problem without knowing where its from? Does the search engine supprt searching for entire problem statements or $ \text{\LaTeX} $keywords


P(3,-5) is the midpoint of a chord of the circle C: x2 + y2 = 203. Find the length of the chord.

This is an interesting problem.

I think this is the solution:

[hide]You have (3, -5) being the midpoint of the chord, so from the origin to the midpoint, it is $\sqrt{34}$.

If a chord perpendicularly intersects a chord, $ab = cd$, where $a$ and $b$ are segments of one chord and $c$ and $d$ are segments of the other chord.

So, $x^2 = (\sqrt{203} + \sqrt{34})(\sqrt{203} - \sqrt{34})$.

This is $x^2 = 169 \\ \Rightarrow x = 13$.

Finally, a simple solution. My old one just looked incorrect...

[/hide]

white_horse_king88, you used a radius instead of a chord

my method
[hide]if (3, -5) is the midpoint, then the segment from the origin (the center of the circle) to (3, -5) will be perpendicular to it.
this forms a right triangle with hypotenuse equal to the radius, and one leg equal to the distance between the origin and (3, -5), which is :sqrt: 34 by the distance formula

the radius of the circle is :sqrt: 203 because
x 2 + y 2 = :sqrt: 203

the last leg squared = 203 - 34 = 169
the last leg is 13, which is 1/2 of the chord
therefore length of chord = 26[/hide]


This is here because it is easy.

Prove that in a 2x2 matrix, in order to find the inverse of $\text{A}$, $\text{A}^{-1}$, you interchange $a_{11}$ with $a_{22}$, and multiply $a_{12}$ and $a_{21}$ by $-1$.

Masoud Zargar

oops...I forgot to mention that.  Yes, you're right.

jsut use the basic defenitions.
very easy. ;)

yup.  Since the LaTeX will require lots of codes, I'll do this later...I'm a little lazy.  :lol:

Well, a very simple method will be to find the inverse of $\left(\begin{array}{ccrr}a_1&b_1\\a_2&b_2\end{array}\right)$, which can be figured out in an a lot easier method.  However, nice job. :)

Masoud Zargar

Actually it helps a lot to know what matrix inversion really means, and under what conditions a linear transformation can have an inverse.  This gives a lot of insight on why methods like this work out.

[i]Proof[/i].

If $\text{A}=(\begin{array}{ccrr}a_1&b_1\\a_2&b_2\end{array})$
$\text{A}^{-1}\implies(\begin{array}{ccrr}a_1&b_1\\a_2&b_2\end{array}|\begin{array}{ccrr}1&0\\0&1\end{array})$
We multiply the first row by $b_2$ and teh second by $b_1$, and subtract the second from the first.
$(\begin{array}{ccrr}a_1b_2-a_2b_1&0\\a_2b_1&b_1b_2\end{array}|\begin{array}{ccrr}b_2&-b_1\\0&b_1\end{array})$
We now divide the first row by $a_1b_2-a_2b_1$.
$(\begin{array}{ccrr}1&0\\a_2b_1&b_1b_2\end{array}|\begin{array}{ccrr}\frac{b_2}{a_1b_2-a_2b_1}&\frac{-b_1}{a_1b_2-a_2b_1}\\0&b_1\end{array})$
Then, we multiply teh first row by $a_2b_1$ and subreact it from the second row.
$(\begin{array}{ccrr}1&0\\0&b_1b_2\end{array}|\begin{array}{ccrr}\frac{b_2}{a_1b_2-a_2b_1}&\frac{-b_1}{a_1b_2-a_2b_1}\\\frac{-a_2b_1b_2}{a_1b_2-a_2b_1}&\frac{a_1b_1b_2}{a_1b_2-a_2b_1}\end{array})$
We then divide the second row by $b_1b_2$.
$(\begin{array}{ccrr}1&0\\0&1\end{array}|\begin{array}{ccrr}\frac{b_2}{a_1b_2-a_2b_1}&\frac{-b_1}{a_1b_2-a_2b_1}\\\frac{-a_2}{a_1b_2-a_2b_1}&\frac{a_1}{a_1b_2-a_2b_1}\end{array})$
We now notice that $\text{A}^{-1}=\frac{(\begin{array}{ccrr}b_2&-b_1\\-a_2&a_1\end{array})}{|\begin{array}{ccrr}a_1&b_1\\a_2&b_2\end{array}|}$
The pattern is now obvious.
$\mathbb{QED}$

Masoud Zargar


Diophantine equation
number theory unsolved
number theory

Prove that there exist infinitely many integers$ n$ such
that $ n, n \plus{} 1, n \plus{} 2$ are each the sum of the squares of
two integers. [Example:$ 0 \equal{} 0^2 \plus{} 0^2, 1 \equal{} 0^2 \plus{} 1^2,
2 \equal{} 1^2 \plus{} 1^2.$]

[hide]Take $ x$ as a solution to the Pell equation:

$ x^2 \minus{} 2a^2 \equal{} 1$ (for which there exist infinitely many values of $ x$)

Then $ n \equal{} x^2 \Rightarrow n\minus{}1 \equal{} x^2\minus{}1 \equal{} a^2\plus{}a^2, n \equal{} x^2\plus{}0^2, n\plus{}1 \equal{} x^2\plus{}1^2$[/hide]

Also, see [url=http://www.mathlinks.ro/viewtopic.php?&t=116256]here[/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=254064&start=20]here[/url].

Best of luck.