The sequence $ (a_n)_{n \geq 1},(b_n)_{n \geq 1},(c_n)_{n \geq 1}$ of positive numbers satisfy the conditions:

$ lim_{n \to \infty}a_1a_2...a_n\equal{}0$
$ lim_{n \to \infty}\frac{b_n}{1\minus{}a_n}\equal{}0$
$ x_{n\plus{}1} \leq a_nx_n\plus{}b_n \forall n \geq 1$

Prove that $ lim_{n \to \infty}x_n\equal{}0$

[quote="SUPERMAN2"]The sequence $ (a_n)_{n \geq 1},(b_n)_{n \geq 1},(c_n)_{n \geq 1}$ of positive numbers satisfy the conditions:

$ lim_{n \to \infty}a_1a_2...a_n \equal{} 0$
$ lim_{n \to \infty}\frac {b_n}{1 \minus{} a_n} \equal{} 0$
$ x_{n \plus{} 1} \leq a_nx_n \plus{} b_n \forall n \geq 1$

Prove that $ lim_{n \to \infty}x_n \equal{} 0$[/quote]

Could you check the statements

$ a(n)\equal{}b(n)\equal{}\frac{1}{n}$ and $ x(n)\equal{}\minus{}n$ verify the statements and $ x(n)$ tend to -oo

[quote="Moubinool"][quote="SUPERMAN2"]The sequence $ (a_n)_{n \geq 1},(b_n)_{n \geq 1},(c_n)_{n \geq 1}$ of positive numbers satisfy the conditions:

$ lim_{n \to \infty}a_1a_2...a_n \equal{} 0$
$ lim_{n \to \infty}\frac {b_n}{1 \minus{} a_n} \equal{} 0$
$ x_{n \plus{} 1} \leq a_nx_n \plus{} b_n \forall n \geq 1$

Prove that $ lim_{n \to \infty}x_n \equal{} 0$[/quote]

Could you check the statements

$ a(n) \equal{} b(n) \equal{} \frac {1}{n}$ and $ x(n) \equal{} \minus{} n$ verify the statements and $ x(n)$ tend to -oo[/quote]

Salut Moubinool!

I think $ x_n\equal{}c_n$ (typo error) and so must be positive.


How many grams of K2Cr2O7 in aqueous H2SO4 are required to oxidize 20 grams of cyclohexanol to cyclohexanone?

can anyone help me with this problem? I'm having trouble setting up the equation.

THanks

what I would do:
write a half-reaction for the reduction of Cr2O7 2- to Cr3+
write a half-reaction for the oxidation of cyclohexanol to cyclohexanone (pretty easy...just balance out the O's and H's)
combine the two
then its a stoichiometry problem.

equate the change in oxidation number in both cases by multiplying suitable coefficients.
1 mole of cr2o7 corr to change of 6 o.n and 1 mole of cyclohexanol corr to change of 2 in oxidation number.

there is a 6 electron transfer from Cr2O72- to Cr3+
and two electron transfer for Cyclohexanol to cyclohexanone write the half reactions and combine them!!!


for the first one see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49566]here[/url]

[quote="socrates"]for the first one see [hide="http://www.mathlinks.ro/Forum/viewtopic.php?t=49566"]here[/hide][/quote]
The code is :
[code][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49566]here[/url][/code]
Which will produce : [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49566]here[/url]. ;)

[quote="ali ergin"]Crux 3068  very difficult ????[/quote]
Yes.


Prove or disprove that $a_n = \frac{((\sqrt{3}-\sqrt{2})(\sqrt{24}+5)^n)^2 + ((\sqrt{3}+\sqrt{2})(\sqrt{24}-5)^n)^2 + 2}{12}$ is a perfect square for all positive integers n.

I don't have alot of time right now but maybe this can make the problem simpler:

If we multiply the whole thing by 3 we have a denominator of 36 which is square. So we need to find if the expression 3((:sqrt:3-:sqrt:2)( :sqrt:24+5)n)2 + 3((:sqrt:3+:sqrt:2)( :sqrt:24-5)n) 2 + 6 is square.

If thats another word for induction, then yes!

[hide]

Like I said we can simplify the fraction by multiplying both sides by 3.

$ a_n = 3((\sqrt{3} - \sqrt{2})(\sqrt{24}+5)^n)^2 + 3((\sqrt{3} + \sqrt{2})(\sqrt{24}-5)^n)^2 + 6 $

First we find out if a_1 is true.

$ a_1 = 3((\sqrt{3} - \sqrt{2})(\sqrt{24}+5))^2 + 3((\sqrt{3} + \sqrt{2})(\sqrt{24}-5))^2 + 6 $
$  3((\sqrt{3} - \sqrt{2})^2(\sqrt{24}+5)^2 + (\sqrt{3} + \sqrt{2})^2(\sqrt{24}-5)^2) + 6 $
$  3((5 - 2\sqrt{6})(49 + 20\sqrt{6}) + (5 + 2\sqrt{6})(49 - 20\sqrt{6})) + 6 $
$  3(5 + 2\sqrt{6} + 5 - 2\sqrt{6}) + 6 $
$  3(10) + 6 $
$  36 $ which is a perfect square.

Next we assume that any positive integer n will make the expression a perfect square. Then we need to find if n+1 works.

$ a_n = 3((\sqrt{3} - \sqrt{2})(\sqrt{24}+5)^n)^2 + 3((\sqrt{3} + \sqrt{2})(\sqrt{24}-5)^n)^2 + 6 $
$ a_{n+1} = 3((\sqrt{3} - \sqrt{2})(\sqrt{24}+5)^{n+1})^2 + 3((\sqrt{3} + \sqrt{2})(\sqrt{24}-5)^{n+1})^2 + 6 $

Ugh... anyone else want to take it from here, or post an easier way?

[/hide]

I highly suspect that the fact that $(\sqrt{3} - \sqrt{2})^2 = 5 - \sqrt{24}$ is helpful (as well as a couple analagous ones).  It would simplify your induction if nothing else.

[quote="seamusoboyle"]Actually, just looking at it, that looks like the general formula for a potentially easy recurrance equation.[/quote]
Can we transform it into a recurrence equation?

Sure! It turns out to be somethig like:

$6a_{n+2}=98a_{n+1}-a_n-16$

Unless im mistaken. Get the first two values $(a_0, a_1)$ off the origianal equation.

i think that this can be solved in another way as well,
if you multiply the whole thing by three (like before), then if you prove that the expression in the nominatore, has odd number of divisors then you can say that  the nominator is a perfect square.

(i'm working on it. :P )

$\frac1{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{3-2}= \sqrt{3}-\sqrt{2}$.
Therefore, the numerator $((\sqrt{3}-\sqrt{2})(\sqrt{24}+5)^n)^2 + ((\sqrt{3}+\sqrt{2})(5-\sqrt{24})^n)^2 + 2$ is $u^2+\frac1{u^2}+2$ where $u=(\sqrt{3}+\sqrt{2})^{2n-1}$.
The whole expression is then $\frac{\left(u+\frac1u\right)^2}{12}$.
The problem then reduces to showing that $u+\frac1u=2\sqrt{3}k_n$ for some integer $k_n$. Here we should rewrite $\frac1u$ as $(\sqrt{3}-\sqrt{2})^{2n-1}=(\sqrt{3}-\sqrt{2})(5-4\sqrt{6})^n$.
This can be done by expanding the product out and noting that every $\sqrt{2}$ term cancels while every $\sqrt{3}$ term doubles, or by finding the appropriate linear recursion ($k_{n+1}-10k_n+k_{n-1}=0$)


The outside faces of a large cube are painted. The large cube is then divided into unit cubes. There are exactly 64 unit cubes that have no paint. How many unit cubes were used to create the large cube?

Imagine that the painted part of the cube is sort of like a shell. When you remove that shell, you have 64 unit cubes. When we put the shell on, we find there are 125 cubes. But we're not done. We still have another layer, making 216 unit cubes.

Since you know the inner cube's side length is $4$, adding on another shell gives it an additional $2$ unit cubes to the lengths, width, and height. Hence $6$ is the side lengths of the large cube.

$6^{3}=\boxed{216}$


I feel that there is a lack of appreciation for contrapuntal music. True, it won't invoke profound emotions, but it still is very enjoyable to listen to the voices simultaneously going and yet still sounding good together. They're also fun to play... it always feels as though I'm going to mess up, but then I don't, and it's kind of a thrill - the same sort of feeling as when you're watching a Rube Goldberg machine in action.

What are your thoughts?

I think counterpoint music is pretty cool.  I've studied it a lot because I play a lot of Bach partitas and sonatas.  BUT IT'S SO HARD TO UNDERSTAND SOMETIMES IT KINDA MAKES ME MAD.   :mad:

yeah, counterpt is different, i've played some of bach's music, i haven't really decided if i like it or not, but it's a nice change from like a sonata

Omg piano sonatas are the BORINGEST type of piano music I have ever heard.  There're thousands of them and they all pretty much sound the same (well I'm a violinist, so...). Chopin's Fantasie Impromptu and Saint-Saens concertos are so much better. (even though the Saint Saens concerto is like 40 minutes long) :sleeping:

Bach is pretty boring too, even though the counterpoint sounds pretty cool at times.   :P

Excuse me?

I really like Beethoven's sonatas (Appasionata, Moonlight, Pathetique), and Chopin's. And FYI- I PLAY VIOLIN TOO!

Oh yeah pathetique is probably the only good sonata around. (my brother played that one)

COOL ARE YOU GOOOOOD AT THE VIOLIN???  DO YOU STUDY BACH SONATAS AND PARTITAS???  DO YOU KNOW WHAT COUNTERPOINT IS???  COUNTERPOINT IS COOL.   :rotfl:  :rotfl:  :rotfl:

[quote="dreambold22"]what's your point there?

what's counterpoint?[/quote]

Answer to question one:  I was arguing with qwerty the cucumber so...uh...yeah.

Answer to question two:  Counterpoint is basically when you're playing a double stop or triple stop or quadruple stop or quintuple stop (for you pianists) and all 2, 3, 4, or 5 notes each have their own melody.  (I told you, counterpoint is cool  :rotfl: )  It can also exist in singular notes but I'm too lazy to explain how that works.   :blush:  :lol:

Contrapuntal music never ceases to amaze me. Hell, [i]The Well-Tempered Clavier[/i] is even beautiful.
But, in the end, it boils down to "I wish I could write a fugue..."

"I wish I could write a fugue" indeed.

The thing with counterpoint and Bach these days is that audiences sadly don't want to listen to Bach anymore; it's simply not exciting enough. And look at the supposed new age of musicians (Lang Lang, Yundi Li, Maxim Vengerov, even Kissin); none of them play Bach because audiences don't want to hear it. All of them probably lack the capacity to play Bach well anyway (sorry to all you diehard Lang Lang, Yundi, Vengerov, or Kissin fans out there), I mean, concerts have supposedly gotten so boring these days that Andre Rieu has to resort to ridiculous cinematics in whatever he does - I don't like to call it music.

But yeah it is good stuff. I listened to all of Goldberg today.

It is quite a shame. Contrapuntal music is very difficult to write though...
At least in most competitions you're required to play some Bach.


hey did u looked at the ipho 2007 paper it had one full question on dimension and i solved the full question ..now that looks odd at one moment there are so tough questions and on the other hand there are easy questions on dimensions
eat an anomaly?

Which,Indian or international???? :maybe:

btw,I think,we are not going to do more dimensions(I don't think it'll reappear as many other topics will do)
moreover its an easy topic(I sholdn't say it, otherwise,I'll not be able to solve questions on it in my 1st term :oops: )
so,you were expected to answer it,especially sitting in home(I wouldn't be able to do so)

And if easy questions are there,they increase the cut-offs.

hey thats not the case
i just wanted to say that it was a topic which a eleventh grader having good knowledge on dimensions would be able to solve (just one part was tricky........)
well the olympiad is international physics olympiad and ofc cut off would be high. :)  :lol:

it was the First level problem of IPhO
they don't expect u to do it wrong..... :)

but still these questions were quite easy as compared to others!
hey about experimentation i wanted to ask to u all wen do students learn to do experiments. are they taught at the training camps?
or any other?

and I do think you are better stud aslo  :) ,so are expected to answer.

about experimentation,i only know one thing,I am not going there,so don't need to think  :rotfl:

for some of brilliant ones like you,I have some information(that may be outdated :oops: )
Divyanshu jha was a stud(brilliant one)from my city (AIR 10 in 2006,and IPhO bronze medallist)

i don't know him directly,but through some of my friends

After coming from IPhO,when my friend went to ask him about this,he said that he faced the equipment for the first time in his life
but,he said he was given instructions on how to handle it,still he performed badly in practical part.(yet he managed bronze,thats brilliance of him)
I think when he was not knowing the equipment,we(especially me)can't know it

but you needn't worry,you are good in experimentation(going for INTEL)I know :)
btw,they give training at hbcse that helps :)

thanx  :)
well hbcse is good (very good...)
even better than iit bombay's (lighter than air) lab
according to my experiences.............
 :blush:

have you gone here?????????? :)

Good
i feel so low :(

well
iit bombay doesn't has ONE lab :rotfl:
and it can't be compared to HBCSE (the field of view changes..........)

and well ,practicals r not too difficult--u r told abt almost everything (excpt what to take on x and y axis of graph :P )


Between what two consecutive integers on the number line is the graph of the sum $ \sqrt{30} \plus{} \sqrt{50}$ located?

sqrt{30} is a litte more than 5, and sqrt{50} is a little more than 7, so wouldn't the answer be 12 and 13

l don't know if "graph" is the right word here since we are talking about a point, but the sum should be located between $\boxed{\text{12 and 13}}.$

To check, you can just plug it into a calculator :)


1.prove that the product of 4 consecutive terms in arithmetic progression plus the fourth power of the common difference is always a square. verify this identity by incorporating symmetry into notation.
2.given positive real numbers $a$ and $b$ such that $a+b=1$,prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}}<ax+by$ $x,y>0$

[quote="srulikbd"]2.given positive real numbers $a$ and $b$ such that $a+b=1$,prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}}<ax+by$ $x,y>0$[/quote]
It's wrong: try $x=y=1$. I think it should be $\frac{1}{\frac{a}{x}+\frac{b}{y}}\leqslant ax+by$.
[hide="Proof for the corrected version"]It is equivalent to $(ax+by)\cdot\left(\frac{a}{x}+\frac{b}{y}\right)\geqslant 1$. After expanding the $\mathrm{LHS}$ we get $a^{2}+b^{2}+ab\cdot\frac{x}{y}+ab\cdot \frac{y}{x}= a^{2}+b^{2}+ab\cdot\left(\frac{x}{y}+\frac{y}{x}\right)\geqslant a^{2}+b^{2}+2ab= (a+b)^{2}=1$ which ends the proof.[/hide]


limit
logarithms
calculus
calculus computations

Are you sure?
$\ln{\sum (m^{2i})^{\frac{1}{2i}}}$???
trivially $...=\ln nm$ and problem is simplest

what a waste of space...

$ \frac{\left(\frac{m^{2}\left(m^{2 n}-1\right)}{m^{2}-1}\right)^{\frac{1}{n (n+1)}}}{\log (m n)}$

i'm not sure. but i thik it is equal to zero...

 a very well-know constant named in honor of Pascal, the pressure man how loves swimming


Let $ a,b \in (0,1]$. Prove that

$ \frac {(a\plus{}b)^{2a} \plus{}(a\plus{}b)^{2b}} { 2^{a\plus{}b\plus{}1} } \geq (ab)^{\frac{b}{a} \plus{}\frac{a}{b}}$.

I think it is easy to prove. :lol:

er.............. 4865550150  can u tell the question clearly...........i didnt understand what you meant by saying $ a,b \in (0,1]$..can u explain......pls  :huh:

$ a,b\in(0,1]$ means that $ a,b$ are real numbers, where $ 0<a,b\leq 1$.

[quote="Agr_94_Math"][hide="hint"]Very simple inequality.
I will just tell the hint:
Now write the LHS of the equation as $ \frac {(a^2 \plus{} b^2 \plus{} 2ab)^a \plus{} (a^2 \plus{} b^2 \plus{} 2ab)^b)}{2^{a \plus{} b \plus{} 1}}$. Now consider $ (a \plus{} b)^2 \equal{} t$.
We get that our inequality is equivalent to having the LHS as  that of  $ \frac {t^{a} \plus{} t^{b}}{2^{\sqrt {t} \plus{} 1}}$. We may also assume $ a\ge b$.
So all left is calculus. :wink:[/hide][/quote]

Would you offer the detailed proof?

[quote="4865550150"]Let $ a,b \in (0,1]$. Prove that

$ \frac {(a \plus{} b)^{2a} \plus{} (a \plus{} b)^{2b}} { 2^{a \plus{} b \plus{} 1} } \geq (ab)^{\frac {b}{a} \plus{} \frac {a}{b}}$.

I think it is easy to prove. :lol:[/quote]
Yes, it's easy to use AM-GM.
$ \frac {(a \plus{} b)^{2a} \plus{} (a \plus{} b)^{2b}} { 2^{a \plus{} b \plus{} 1} }\geq (ab)^{\frac {b}{a} \plus{} \frac {a}{b}}$
$ \Leftrightarrow \frac {(a \plus{} b)^{2a} \plus{} (a \plus{} b)^{2b}} { 2^{a \plus{} b } } \geq 2(ab)^{\frac {b}{a} \plus{} \frac {a}{b}}$
Applying AM-GM:
$ LHS \equal{} \frac {(a \plus{} b)^{2a} \plus{} (a \plus{} b)^{2b}} { 2^{a \plus{} b } } \ge \frac {4^a(ab)^{a} \plus{} 4^b(a \plus{} b)^{b}} { 2^{a \plus{} b } }$
$ \equal{} \frac {2^a}{2^b}.(ab)^a \plus{} \frac {2^b}{2^a}.(ab)^b \ge 2.\sqrt {\frac {2^a}{2^b}.(ab)^a.\frac {2^b}{2^a}.(ab)^b} \equal{} 2(ab)^{\frac {a \plus{} b}{2}}$
Since $ a,b \in (0;1] \Rightarrow (ab) \in (0;1]$     (1)
And we  have: $ \frac {a \plus{} b}{2} \le 1 < 2 \le \frac {a}{b} \plus{} \frac {b}{a}$         (2)
From (1) and (2) $ \Rightarrow (ab)^{\frac {a \plus{} b}{2}} \ge (ab)^{\frac {a}{b} \plus{} \frac {b}{a}} \Rightarrow 2(ab)^{\frac {a \plus{} b}{2}} \ge 2(ab)^{\frac {a}{b} \plus{} \frac {b}{a}} \Rightarrow LHS \ge RHS$
$ Q.E.D$
I think this inequality is very weak, equality holds when $ a\equal{}b\equal{}1$  :P


let $(m,n)=1$ prove that:
$x^m+y^m=z^n$
has infty answers in $Z$

Just take $x=y$ be powers of 2 and use Bezout theorem.


Omid, here is a solution to this nice but easy inequality

 Our inequality is equivalent to \[(a+b)^{2}(b+c)^{2}(c+a)^{2}\geq\frac{64}{27}\cdot\frac{(a+b+c)^{3}}{abc}.\] Using the following identity \[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\] it is not hard to see that \[(a+b)(b+c)(c+a)\geq\frac{8}{9}(a+b+c)(ab+bc+ca),\] so using this inequality and the obvious one \[ab+bc+ca\geq\sqrt{3abc(a+b+c)}\] the conclusion follows. :)

The problem is easy. This is my solution. Suppose $x=a+b,y=a+c,z=b+c$. Obviously $x,y,z$ are sides of triangle. So problem is equaivalent to $4p\leq3\sqrt[3]{\frac{x^{2}y^{2}z^{2}}{(p-x)(p-y)(p-z)}}\Longleftrightarrow 4p\leq\sqrt[3]{16R^{2}p}\Longleftrightarrow x+y+z\leq R\sqrt{27}$.
And the last inequality is famous.

Yes, it is famous and is due to Mitrinovic. :)
By the way, I have found two different solutions. I am a little bit lazy to post them now, but if there is anyone interrested I'll post them. :)


[b]Do anyone have selection exam (final)or preparatory questions of USA or CHINA or other countries?[/b]

i could not download files from the link you have given. but i found new link for how knows russian ,there is tasks for 9th and 10-11th grades:
http://74.125.45.101/translate_c?hl=en&sl=ru&u=http://bio.rusolymp.ru/default.asp%3FtrID%3D2662&prev=/search%3Fq%3D%2Bsite:bioolimp.narod.ru%2Bnarod.ru%2Bbioolimp%26hl%3Den&usg=ALkJrhgfoLG53IXqWfny44OVr5_0xTNz4g


I am as old as John will be when I am twice as old as John was when my age was half the sum of our present ages. John is as old as I was when John was half the age he will be $10$ years from now. How old am I?

Riddler: I am really not that old...

Okay, from what you've said, i've gotten that
If your current age is x years, and John's is y years, then
$x=4y-2x$ and $y=x+\frac{10-y}{2}$
That means
$x=\frac{4y}{3}$
And then
$y=\frac{5y+30}{6}$
$y=30$
which means that you're 40 years old...

Try this:

Tina and Louise both celebrate their birthdays today. In three years, Tina will be four times as old as Louise was when Tina was two years older than Louise is today. If Louise is a teenager, what is Tina's age?

Similiar problem
[hide="Answer"]
Let Tina's age=$t$
and Lousie's age=$l$
So the equation is
$t+3=8l-4t+8$
Which gives you,
$l=\frac{5(t-1)}{8}$
As $l$ is a teenager and $l$ relative to $t$ is integer so,
the $l$=15 and $t$=25
[/hide]
Sorry If this is wrong have to revise for French Sats.

You can easily check it by substituting back into the original problem. Assuming that these are the correct answers, when Tina was two years older than Louise is today, or 17 years old, then Louise was 7 years old. In three years, Tina will be 28 years old, which is exactly four times as old as Louise would have been at the given time. No contradictions, so your answer is correct. :)

Yep... by the way, does teenager mean 11-20 only? I thought ten would be allowable.

I would think that teenager is 13-19... i.e., all the ages ending in "teen."


Let $ a,b,c\geq0$.Prove that:
$ \sqrt{\dfrac{a}{4a\plus{}5b}}\plus{}\sqrt{\dfrac{b}{4b\plus{}5c}}\plus{}\sqrt{\dfrac{c}{4c\plus{}5a}}\leq1$

We have $ \displaystyle \left( \sqrt{\frac{a}{4a\plus{}5b}}\plus{}\sqrt{\frac{b}{4b\plus{}5c}}\plus{}\sqrt{\frac{c}{4c\plus{}5a}} \right)^2$
$ \equal{} \displaystyle \left( 1.\sqrt{\frac{a}{4a\plus{}5b}}\plus{}1.\sqrt{\frac{b}{4b\plus{}5c}}\plus{}1.\sqrt{\frac{c}{4c\plus{}5a}} \right)^2$
$ \le 3 \displaystyle  \left( \frac{a}{4a\plus{}5b}\plus{}\frac{b}{4b\plus{}5c}\plus{}\frac{c}{4c\plus{}5a}\right)$
$ \equal{} 3 \displaystyle \left( \frac{1}{4\plus{}\frac{5b}{a}}\plus{}\frac{1}{4\plus{}\frac{5c}{b}}\plus{}\frac{1}{4\plus{}\frac{5a}{c}}\right)$
Let $ x\equal{}\frac{b}{a}$, $ y\equal{}\frac{c}{b}$, $ z\equal{}\frac{a}{c}$ $ \Longrightarrow x, y, z > 0$ and $ xyz\equal{}1$.
So $ \frac{1}{4\plus{}\frac{5b}{a}}\plus{}\frac{1}{4\plus{}\frac{5c}{b}}\plus{}\frac{1}{4\plus{}\frac{5a}{c}}$ = $ \frac{1}{4\plus{}5x}\plus{}\frac{1}{4\plus{}5y}\plus{}\frac{1}{4\plus{}5z}$.
But if use $ x, y, z > 0$ and $ xyz\equal{}1$ easy we have $ \frac{1}{4\plus{}5x}\plus{}\frac{1}{4\plus{}5y}\plus{}\frac{1}{4\plus{}5z} \le \frac {1}{3}$.
Since $ \Longrightarrow$  $ \sqrt{\frac{a}{4a\plus{}5b}}\plus{}\sqrt{\frac{b}{4b\plus{}5c}}\plus{}\sqrt{\frac{c}{4c\plus{}5a}}\le1$.

[quote="thanhnam2902"]
But if use $ x, y, z > 0$ and $ xyz \equal{} 1$ easy we have $ \frac {1}{4 \plus{} 5x} \plus{} \frac {1}{4 \plus{} 5y} \plus{} \frac {1}{4 \plus{} 5z} \le \frac {1}{3}$.
[/quote]
Try $ x \equal{} y \equal{} \frac {3}{2}$ and $ z \equal{} \frac {4}{9}.$ :wink:
Inequalities Master, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150022

Thank. But why try $ x\equal{}y\equal{}\frac{3}{2}$ and $ z\equal{}\frac{4}{9}$, because $ xyz\equal{}1$.


Find all numbers which can simlify the fraction $ \frac{5k\plus{}6}{8k\plus{}7}$ with the known integer k.

um...your solution doesn't hold for n=0.
And does $ k\equal{}\minus{}1$ count as a solution?

[quote="Cronus"](3k+1,2k+5)=(2k+5,k-6)[/quote]

should be $ (3k\plus{}1,2k\plus{}5)\equal{}(2k\plus{}5,k\minus{}4)\equal{}(k\plus{}9,k\minus{}4)\equal{}(k\minus{}4,13)$.  Thus, the solution set is $ k\equal{}13n\plus{}4$ for $ n\in\mathbb{Z}$.

I have  a nice solution of this :

Supose that $ 5k\plus{}6$ and $ 8k\plus{}7$ are divisible by the number $ r$

1) Than $ (8k\plus{}7)\minus{}(5k\plus{}6)$ is  divisible by $ r$
2) Than $ (5k\plus{}6)\minus{}(3k\plus{}1)$ is divisible by $ r$
3) Than $ (3k\plus{}1)\minus{}(2k\plus{}5)$ is didisble by $ r$
4) Than $ (2k\plus{}5)\minus{}(k\minus{}4)$ is divisible by $ r$
5) Than $ (k\minus{}4)\minus{}(k\plus{}9)\equal{}\minus{}13$ is divisible by $ r$

So we have that -13 is divisible by $ r$, since 13 is prime  hence $ r\equal{}13$

The number wihch simplify the fraction is 13  .            :)

Oh, yeah.Sorry for that.The answer that you guys gave is right :blush: ....


Please help me with any/all of the problems by [i]explaining[/i] it specificially, I know the answers to each of the problems though. I just have trouble knowing how to do it. I really appreciate it! These are some similiar problems that will be on my test tomorrow, and I don't really understand how to do them. Oh and by the way, I've seen some people who posted in math format.. how do I do that? Thanks!!

1. How many positive integers less than 150 are there such that all the digits are odd?

2. Simplify: [img]http://i26.photobucket.com/albums/c135/starprincess1010/untitled.gif[/img]

3. [img]http://i26.photobucket.com/albums/c135/starprincess1010/untitled3.gif[/img]

Feel free to ask any more questions if you don't understand :) I personally think $\#1$ was a little too hard for just a school precalculus test though, it sounded more like a mathcounts question or something

[quote="Iversonfan2005"]1: We can do it by digits and cases. First case is that it's only one digit, so there are $5$ possible cases for that. Second case is that there are two digits, and so there are $5\cdot5$ for that and finally, in the third case, since it's less than $150$, there are $1 \cdot 2 \cdot 5$ for that, so we add everthing up and have $40$.

2: Knowing that logarithms are exponents, we know that $\log_{a}{b}$ is $a^x=b$, so we have $3+3(1)-0+(-1)=5$

3. Are we evaluating $g(f(x))$? because I can't understand what you wrote above[/quote]

For #3, yes that I what I mean.

[b]EDIT[/b] Other people beat me to it.
The math format thing is called LaTeX and can be found [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php]here[/url].
For number $2$, you have to use properties of logarithms. To simplify $\log_28+3\log_33-\log_21+log_4\frac 14$ use the properties $\log_aa=1, \log_aa^b=b$. Think in terms of exponents. For $\log_28$ use $2^x=8$ and solve for $x$, so you simplify it to become $\log_22^x$ which is just $x$.
For number three, you have to know that composite form, $(f \circ g)(x) = f(g(x))$ and plugging in $3$ for $x$ and then performing the equations on it in order will give you your answer. For example, $Let f(x)=2x, g(x)=x+2 \Rightarrow (f \circ g)(x) = f(g(x)) = f(x+2) = 2(x+2) = 2x+4$.
Oh, and $log_a1 = 0$.
Good luck on your test.

Thank You Iversonfan2005, frt, and Hokkage!! All of you are so awfully kind, and bright! I wish I can just go to each and every one of you and give you a big hug.  :lol:

Put $\$$ (dollar sign) at the beginning and end of the stuff that you want to be LaTeX'd. \circ is for the little circle. You can hover over any of the things we've typed to see what the code is. \frac{a}{b} gives $\frac{a}{b}$

[quote="Hokkage"][quote="Altheman"]number 1 is not that hard...
odd numbers are in the form 2n-1 for n=1,2,3...
so 149=2n-1
n=75, so there are 75 odd numbers[/quote]
Yes, but the question was all the numbers where all the [b]digits[/b] are odd. So 147, for example, would not count.[/quote]

Yeah, so you have to solve it the way I did

woops...i did not really read the question  :blush:


algebra
binomial theorem
Divisibility
number theory
IMO Shortlist

By the binomial theorem is easy to prove that if $ x \equiv 1 (mod m^k)$ then $ x^{m^n} \equiv 1 (mod  m^{k\plus{}n})$ for every integers positive $ x, m$, and if $ m$ is odd and $ x$ satisfying $ x \equiv \minus{}1 (mod m^k)$ then $ x^{m^n} \equiv \minus{}1 (mod  m^{k\plus{}n})$.
Thus $ 1992^{1991^{1990}} \equiv 1 (mod 1991^{1991})$ and $ 1990^{1991^{1992}} \equiv \minus{}1 (mod 1991^{1993})$.
Hence, the highest degree is $ 1991$.

[quote="shinichiman"]More simple!
We let $1991=a$ then $a$ is a [u]prime.[/u] [/quote]
Umm nope since $1991=11 \times 181$.

I did this by letting $1990 = 1991 -1 $ and $1992 = 1991 +1 $ and then expanding using the binomial theorem. The "ones" cancel out and we are left with giant powers of $1991$. Observing the second to last term tells us that the minimal degree of a $1991$ is indeed $1991$.

[quote=shinichiman]More simple!
We let $1991=a$ then $a$ is a prime. By Lifting The Exponent lemma \[\begin{aligned} v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) & = v_a \left( (a-1)^{a^2})^{a^{a-1}}+ (a+1)^{a^{a-1}} \right) \\ & = v_a \left( (a-1)^{a^2}+a+1 \right)+ v_a(a^{a-1}) \\ & = a-1+v_a \left( (a-1)^{a^2}+a+1 \right)  \end{aligned} \]
We have $v_a \left( (a-1)^{a^2}+1 \right)= v_a(a)+v_a(a^2)=3$ so $v_a \left( (a-1)^{a^2}+a+1 \right) =1$.
Thus, $v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) = a$. We obtain $\max k =a= \boxed{1991}$.[/quote]

1991 isn't a prime.

[b]LEMMA[/b]:- For every odd number $a\ge 3$ and integer $n\ge 0,$ the following holds.
$$a^{n+1}||(a+1)^{a^n} - 1, a^{n+1}|| (a-1)^{a^n} + 1$$.

These lemma can be proved by induction on $n$.
The problem is trivial from here.

The answer is $\boxed{1991}$. We can rewrite the expression as $$\nu_{1991}(1990^{1991^{1992}} + 1992^{1991^{1990}}) = \nu_{1991}([1990^{1991^2}]^{1991^{1990}} + 1992^{1991^{1990}}$$$$= \nu_{1991}([1990^{1991^2}]^{1991^{1990}} - (-1992^{1991^{1990}})).$$ By the LTE lemma, this is equal to $$1990 + \nu_{1991}(1990^{1991^2} + 1992).$$ The finishing step is to realize that $$\nu_{1991}(1990^{1991^2} + 1) = 1 + 2 = 3,$$ which implies that the second $\nu_{1991}$ must be equal to exactly 1, hence the result.

[b][color=purple]Remark.[/b][/color] Note that 1991 is not prime, so some of our early manipulations might be handwavy directly. However, we can simply replace 1991 by the product of its two distinct prime factors then run the algorithm similarly.

Cool, but I just wanted to mention out that after getting $\nu_{11}(A)=1991$, we don't actually need to check it for $\nu_{181}(A)$ since it will also be $\ge 1991$ as $181 \mid 1990^{1991^2}+1992$. So we automatically get $k=1991$. Many of the above posts actually calculated it.

[b]Two main realizations here:
- We should make the power the same to use LTE
- LTE can be used even in addition by negating the second term[/b]

With this in mind,
$\nu_{11}(A^{1991^{1990}} + B^{1991^{1990}}) = \nu_{11}(A + B) + \nu_{11}(1991^{1990}) = 1 + 1990 = 1991$ ($A = 1990^{1991^2}, B=1992$), $\nu_{11}(A + B) = 1$ by binomial expansion.

Now, we don't even have to compute $\nu_{181}(A^{1991^{1990}} + B^{1991^{1990}})$, as we know it will be $\ge 1991$.

Regardless, our final answer is $\boxed{1991}$.

[quote=S.Das93]Just write $1991=11 \times 181$ and use LTE on both of those numbers.[/quote]

yea i know I just needed to go right then


Find all the sets $A$ and $B$ stisfying :
1 - $A\cup B=Z$   ($Z$ is the set of integers)
2 - if $x\in A$ then $\left( x-1\right) \in B$
3 - if $x,y\in B$ then $\left( x+y\right) \in A$

a)$0 \in B$
then $A=B=Z$
b)$0 \in A$
so $-1 \in B$
so if $x\in A$ then $x-2 \in A$
so if $x \in B$ then $x-2 \in B $
so every minus even $x$ is in $A$ and every minus odd $x$ is in $B$

ba)$3 \in A$ so $1 \in A$ so $0 \in B$ look at a)
bb)$3 \in B$ so every even $x$ is in $A$ and every odd $x$ is in $B$
let assume that exist even $x$ in $B$ then we have $B=A=Z$
let assume that exixt odd $x$ in $A$ then we have $B=A=Z$

answer: $A=B=Z$, $A-even$ $B-odd$

This problem is used on the french correspondance program and has to be solved by 29th March.

Please don't discuss this problem before....

Can you delete your solutions please ?

Isn't this a problem from the Romanian TST by Laurentiu Panaitopol? It seems to me it is... Anyway, I do not see why we should delete the solutions. A responsible guy will not copy the solution from mathlinks.  ;)

Hum...
Then why aren't discussed the problems of the actual Tournament Of Towns ?

Simply because they are quite original problems, not taken from past olympiads.

Lol but a problem from France can be original too  ;) and therefore mustn't be posted.

Dude, do not consider a romanian problem already discussed on the forum and given in a TST as french trademark!  :D . Probably there are other original problems in the set, but not this one.

Yes, but to know if a problem if original or not, you have to post it.
It can be original, so it's better not to post it  :lol:

I think it is better to stop this meaningless discussion :P :)  :lol:

Me too. But do not delete posted solution. :P  :P  :P

Just a thing :

I use some problems appearing on ML for our correspondance program. Thus, clearly, solutions can be found here.
I think the main question is that anyone should take care about the source (the one the proposer has, not necessarly the original one) of the problems he is willing to post : Tournament of Town or anything, if it is a current 'competition' it would be nice to avoid posting them here, even if they already appeared.

Pierre.


In the following problems, consider each op amp as ideal.

1. In each of the following circuits find the current intensity across resistor $ R_{L}$:

2. Compute $ V_{out}$ in the two following circuits:

3. Consider the following op amp circuit.

(a) Find the voltage at the output of the op amp.

(b) What is the current read on A?

(c) Find the power dissipated at resistor $ R_{4}$.

4. Consider the following op amp circuit.

(a) In what type of configuration is this op amp? Why?

(b) Compute the current intensities that pass through $ R_{4}$ and $ R_{L}$.

5. Consider the following circuit, composed by an ideal op amp, two dc sources, five resistors, and one capacitor.

(a) What is the current intensity through $ R_{5}$ if the capacitor is [i]discharged[/i]?

(b) What will be the current intensity through $ R_{5}$ if the capacitor is [i]charged[/i]? Compare with your answer in (a) and comment.

6. In order to measure the membrane voltage of a squid's axon it was employed the op amp circuit shown bellow. The electrode, with resistance $ R_{elect}\equal{} 80\Omega$, was introduced in the axon, and the value obtained for its potential (relative to ground) was $ V_{memb}\equal{}\minus{}60\, mV$.

(a) Find the output voltage of the op amp.

(b) What is the current intensity through $ R_{L}$?


and ARML's official website only show last year's question.
That doesn't help with people who want to train for the contest  :(

But for practicing under "live" conditions during practice meetings, we coaches do like having actual ARML material that our students [i]haven't[/i] seen.

http://www.arml.com/2006contest.htm

http://www.arml.com/2005contest.htm

right 1989 -2003 are in 2 books. and idk about 2004. but that is a good idea i think some mod should look into that.

Check out the ARML Index. It has a list of topics if you want practice for ARML.

It's the first sticky  in the ARML forum :D

Iirc, ARML doesn't let us post their problems in the resources section unlike other contests.  So that's the reason that AoPS doesn't them.  The reason that ARML doesn't want us posting them...?  Perhaps what Dr. Merryfield said or perhaps they want to make their books the only place to get them.  If you don't have the book, chances are that your ARML coach or fellow teammate has at least one of them.  So this should encourage to you practice together.  And that's what ARML's really about...getting the best math students from each region to come together and interact for a weekend.

We've tried emailing them and they won't let us post the problems.

See my post at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1130542#1130542[/url].  Everything from 2004, 2005, 2006, and 2007 can be found on the ARML website.


function
integration
algebra
polynomial
trigonometry
complex analysis

Determine the Fourier Transform of the function $ 3te^{\minus{}9t^{3}}$.

My first impulse was simply to plug into the Fourier transform formula to obtain $ \hat{f}(w) \equal{} \int_{\minus{}\infty}^{\infty} 3te^{\minus{}9t^{3}}e^{\minus{}iwt} dt$.

However, now I feel stuck since I 'm not given any interval for $ f$. I have read some things on the "window function" but I 'm not terribly sure on how to use it.

Any help? Thanks.

The function $ f(t) \equal{} 3te^{ \minus{} 9t^3}$ on the whole real line is not integrable on $ \mathbb{R},$ so we can't defined the Fourier transform directly by using integral. But worse than that: since the function grows (as $ t\to \minus{} \infty$) faster than any polynomial, it cannot in any useful way be considered a tempered distribution.

I'm afraid I see no good way to define its Fourier transform.

On the other hand, consider the function $ g(t) \equal{} \begin{cases}3te^{ \minus{} 9t^3}, & t \ge 0 \\
0, & t < 0\end{cases}.$

This is an integrable function - in fact, any polynomial times this function is still integrable. And it is continuous, and has an integrable (but not continuous) first derivative. Hence, $ \hat{g}$ can be defined directly by the integral. Furthermore, $ \hat{g}$ will be $ C^{\infty},$ and $ \hat{g}(\omega) \equal{} O(\omega^{ \minus{} 2})$ as $ |\omega|\to\infty.$

Thanks Kent, what you said really helps to clarify a few things for me. A few questions though:

for $ g(t)$ did you mean $ g(t) \equal{}0$ when $ t\le 0$ and for $ g(t) \equal{} 3te^{\minus{}9t^{3}}$ when $ t > 0$?

Also, what does $ C^{\infty}$ and $ O(w^{\minus{}2})$ mean?

Yes, that's what I meant. I've edited.

$ C^{\infty}$ means infinitely differentiable, and $ O(\omega^{ \minus{} 2})$ means that it decays at infinity at least as fast as a constant times $ \omega^{\minus{} 2}.$

For instance, $ \frac {1 \minus{} \cos\omega}{\omega^2}$ (extended at zero in the obvious way) is would qualify as $ C^{\infty}$ and $ O(\omega^{ \minus{} 2})$ as $ |\omega|\to\infty.$


$\sum |b_n | $is converges.
and $\frac {a_n}{a_{n+1}}=1+\frac{1}{n}+O(b_n)$
prove $\sum a_n$ is disverges.
thanks.this problem ends
and if $\frac {a_n}{a_{n+1}}=1+\frac{u}{n}+O(b_n),u > 1$
can we prove $\sum a_n$ is converges?

For the first one note that $\frac{a_{n+1}}{a_n}\geqslant \frac n{n+1}e^{- C |b_n|}$ for all $n$. It follows that $a_n\geq e^{-CB}\frac {a_1}n$ where $B=\sum_j|b_j|$, so the series $\sum_n a_n$ dominates the harmonic series, which diverges.

The second one is similar, though the technical details are slightly more complicated. One can show that $a_n$ are dominated by $n^{-u}$, so the series converges.


Prove that if isoceles trapeziod ABCD ( AB parallel to CD) is orthodiagonal, then its altitude is equal to (AB + CD)/2

Middle schoolers do [b]NOT[/b] know what orthodiagonal means probably. ;)

this should be in getting started in the high school forum

Orthodiagonal - the diagonals form right angles.

No offense, but I find it easy after finding out what orthodiagonal means.

[hide="The Proof"]Note that in the isosceles trapezoid, the angle where a diagonal intersects a base is congruent to the angle formed by where the other diagonal and the same base intersect (that's both acute angles). i.e., <BAC is congruent to <ABD, and <BDC is congruent to <ACD.  Lets call $O$ the intersection of lines $AC$ and $DB$.  Since <AOB and <DOC are right angles and triangle DOC and triangle AOB are isosceles, then the triangles are both 45-45-90 triangles.  Note that a median and an altitude of an unequal side of an isosceles triangle are the same, and median to the hypotenuse of a right triangle is $\frac {1}{2}$ the hypotenuse.  The medians (altitudes) are length $\frac {AB}{2}$ and $\frac {CD}{2}$, and the altitudes are the same line.  Thus, the height is $\frac {AB+CD}{2}$[/hide]

Yeah...you can think of it as two perpendicular lines, the points are drawn such that they form an isosceles trapezoid.  Then just draw the two bases rather than the whole trapezoid.  The altitude clearly goes through the intersection of the two lines, and you clearly have two 45-45-90 triangles.

orthogonal only means, prependecular...
that's all  ;)


trigonometry
quadratics
geometry
angle bisector

Prove that $\sin{18^o}=\frac{\sqrt{5}-1}{4}$ without using a calculator!

Since no one seems to want to answer; a clue:
For shame if you look at this witout trying it first:

[hide]$ \cos{x}=\sin{(90-x)}$ therefore $\cos(2.18)=\sin(3.18)$,[/hide]

How would you do that??? If you split my one up and write it in terms of $ \sin{x}$ you can get it as a root...

kunny is right, you can find it using some similarities in regular pentagon.

Lets see a solution! Could be interesting! :)

I have to imagine that there's a solution involving complex numbers.  There always is with trig problems.

I'm thinking of solving $x^5-1=0$ which gives us the fifth roots of one, which in the complex plane form a regular pentagon. And the angle of a regular pentagon happens to be $108^\circ$, which gives us the answer. Now I just need to work out the details.

Well, since you asked for it, we may as well find $\cos 72^{\circ}$.  Now, $e^{2\pi i/5}$ is clearly a root of the equation $z^4+z^3+z^2+z+1=0$.  So solve this equation (via an appropriate substitution) and equate the resultant roots with the roots in trigonometric form.

That's right...that equation can be solved by dividing by z^2, and then using the substitution u=z+1/z.

Notice that 5.18 =90 and write X=18 degrees then
 Cos5X =16(CosX)^5 -20(CosX)^3 +5CosX =0

 Solving This equation  reduces to solving a quadratic in (CosX)^2,from which you can easily obtain sinX.

[hide="hint"]
Construct a triangle which safisfies $\angle{A}=36$ and $\angle{B}=\angle{C}=72$.
Then draw the angle bisector of $\angle{B}$ or $\angle{C}$ and notice the $\frac21\angle{B}$ is still 36...[/hide]

Prove that:
$\sin 18^0  = \frac{{\sqrt 5  - 1}}{4}$

[u]Solution:[/u]
We have:
$\sin \frac{{2\pi }}{{10}} = \cos \left( {\frac{\pi }{2} - \frac{{2\pi }}{{10}}} \right) \Leftrightarrow \sin \frac{{2\pi }}{{10}} = \cos \frac{{3\pi }}{{10}} \Leftrightarrow $
$ \Leftrightarrow 2 \cdot \sin \frac{\pi }{{10}} \cdot \cos \frac{\pi }{{10}} = 4 \cdot \cos ^3 \frac{\pi }{{10}} - 3\cos \frac{\pi }{{10}}\mathop  \Leftrightarrow \limits_{\mathop {\cos \frac{\pi }{{10}} \ne 0}\limits^ \uparrow  } $
$ \Leftrightarrow 2 \cdot \sin \frac{\pi }{{10}} = 4\cos ^2 \frac{\pi }{{10}} - 3 \Leftrightarrow $
$ \Leftrightarrow 4 \cdot \sin ^2 \frac{\pi }{{10}} + 2 \cdot \sin \frac{\pi }{{10}} - 1 = 0 \Leftrightarrow $
$ \Leftrightarrow \sin \frac{\pi }{{10}} \in \left\{ {\frac{{ - 1 - \sqrt 5 }}{4},\frac{{ - 1 + \sqrt 5 }}{4}} \right\} \cap \left( {0, + \infty } \right) \Leftrightarrow $
$ \Leftrightarrow \sin \frac{\pi }{{10}} = \frac{{\sqrt 5  - 1}}{4}$


Prove that for each positive integer $n$, there exist $n$ consecutive positive integers none of which is an integral power of a prime number.

Choose $k \equiv -i \mod p_i q_i$ for some distinct primes $p_i, q_i$ by the chinese remainder theorem, $i=1,...,n$.
Then $k+i$ is divisible by $p_i q_i$, thus surely not power of a prime.

A bit stronger: we also don't allow the numbers to be a nontrivial power of any integer. Then the proof can be adapted to:

Choose $k \equiv p_i-i \mod p_i^2$ for some distinct primes $p_i$ by the chinese remainder theorem, $i=1,...,n$. We can assume $k>p_i$ for all such $i$, too.
Then $k+i$ is divisible by $p_i$, but not by $p_i^2$. Since also $k+i \neq p_i$, $k+i$ is neither nontrivial power of any integer, nor prime itself.

another proof: these numbers satisfies the condition:$ (2(n \plus{} 1))! \plus{} 2,(2(n \plus{} 1))! \plus{} 3,...,(2(n \plus{} 1))! \plus{} (n \plus{} 1)$


calculus
integration
trigonometry
quadratics
inequalities
calculus computations

Let $a,\ b$ are real numbers such that $a+b=1$.

Find the minimum value of the following integral.
\[\int_{0}^{\pi}(a\sin x+b\sin 2x)^{2}\ dx \]

[hide="Answer"]$I = \int^{\pi}_{0}(a\sin x+b\sin 2x)^{2}\,dx = \int^{\pi}_{0}a^{2}\sin^{2}x\,dx+\int^{\pi}_{0}2ab\sin x \sin 2x\,dx+$
$+\int^{\pi}_{0}b^{2}\sin^{2}2x\,dx$.

Evaluating each integral we get

$I = \frac{\pi(a^{2}+b^{2})}{2}= \frac{\pi}{2}[a^{2}+(1-a)^{2}] = \pi(a^{2}-a+1/2) = \pi \cdot f(a)$

Now, $f'(a) = 2a-1$ and $f'(a) = 0 \Rightarrow  2a-1 = 0 \Rightarrow  a = 1/2$. This value of $a$ is a minimum, because $f'(a) < 0$ for $a < 1/2$ and $f'(a) > 0$ for $a > 1/2$. So the minimum value of the integral is for $a = \frac{1}{2}$, for which

$I = \frac{\pi}{4}$
[/hide]

That's right, but you don't need to differentiate $f(a)$, because $f(a)$ is simply a quadratic function.

Another solution:

You can use Jensen's inequality.

How can we use Jensen's inequality to solve this problem?

I know this is old, but I am bored. Sorry if this is too old.

$ a+b=1$

$ \Rightarrow{a=1-b}$

$ \Rightarrow\int_0^{\pi}\left(a\sin(x)+b\sin(2x)\right)^2\,dx$

$ =\int_0^{\pi}\left((1-b)\sin(x)+b\sin(2x)\right)^2\,dx$

$ =I(b)$

$ I'(b)=\int_0^{\pi}\frac{\partial}{\partial{b}}\left(\left((1-b)\sin(x)+b\sin(2x)\right)^2\right)\,dx$

$ =\int_0^{\pi}\bigg[2b\sin^2(2x)+(2-4b)\sin(x)\sin(2x)+(2b-2)\sin^2(x)\bigg]\.dx$

$ =(2b-1)\pi$

$ \therefore\quad{I}(b)=0\,\text{when}\,{b=\frac{1}{2}}\Rightarrow{a=\frac{1}{2}}$

Therefore

$ max\left\{\int_0^{\pi}\left(a\sin(x)+b\sin(2x)\right)\,dx\right\}$

$ =\int_0^{\pi}\frac{1}{4}\left(\sin(x)+\sin(2x)\right)^2\,dx$

$ =\frac{\pi}{4}$

Here is my solution.

$ \int_0^{\pi} (a\sin x + b\sin 2x)^2\ dx$

$ = a^2\int_0^{\pi} \sin ^ 2 x\ dx + 2ab\int_0^{\pi} \sin x\sin 2x\ dx + b^2\int_0^{\pi} \sin ^ 2 x\ dx$.

For a integer $ m\neq 0$, $ \int_0^{\pi} \sin^ 2 mx\ dx = \int_0^{\pi} \frac {1 - \cos 2mx}{2}\ dx = \frac {1}{2}\left[ x - \frac {1}{2m}\sin mx\right]_0^{\pi} = \frac {\pi}{2}$ and ${ \int_0^{\pi} \sin x\sin 2x\ dx = 2\int_0^ \pi} \sin ^ 2 x\cos x\ dx = \frac {2}{3}\left[\sin ^ 3 x\right] _0^{\pi} = 0$, so

$ \therefore \int_0^{\pi} (a\sin x + b\sin 2x)^2\ dx$

$ = a^2\int_0^{\pi} \sin ^ 2 x\ dx + 2ab\int_0^{\pi} \sin x\sin 2x\ dx + b^2\int_0^{\pi} \sin ^ 2 x\ dx$

$ = \frac {\pi}{2}(a^2 + b^2)\geq \ \pi \left(\frac {a + b}{2}\right)^2 = \boxed{\frac {\pi}{4}}$, the equality holds when $ a = b$ and $ a + b = 1\Longleftrightarrow a = b = \frac {1}{2}$.

[color=darkblue][b]I"ll present to my dear friend Kunny an easy extension of his nice exercise.[/b][/color]

[quote="Virgil Nicula"] [color=darkred]Given are $ \{m,n\}\subset\mathbb Z$ so that $ m \plus{} n$ is an odd integer. Find $ \min_{\begin{array}{c} \{a,b\}\subset\mathbb R \\
\ a \plus{} b \equal{} 1\end{array}}\int_{0}^{\pi}(a\sin mx \plus{} b\sin nx)^{2}\ \mathrm {dx}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] $ I(a,b)\equiv\int_{0}^{\pi}(a\sin mx \plus{} b\sin nx)^{2}\ \mathrm {dx} \equal{}$ $ \int_{0}^{\pi}\left[a\sin m(\pi \minus{} x) \plus{} b\sin n(\pi \minus{} x)\right]^{2}\ \mathrm {dx} \equal{}$ $ \int_{0}^{\pi}\left[( \minus{} 1)^{m \plus{} 1}a\sin mx \plus{} ( \minus{} 1)^{n \plus{} 1}b\sin nx\right]^{2}\ \mathrm {dx}$ . Therefore, $ 2\cdot I(a,b) \equal{} \int_0^{\pi}\left\{(a\sin mx \plus{} b\sin nx)^{2} \plus{} \left[( \minus{} 1)^{m \plus{} 1}a\sin mx \plus{} ( \minus{} 1)^{n \plus{} 1}b\sin nx\right]^{2}\right\}\ \mathrm {dx}$ $ \implies$

$ I(a,b) \equal{} \int_0^{\pi}\left(a^2\sin^2mx \plus{} b^2\sin^2nx\right)\ \mathrm {dx} \equal{}$ $ \frac 12\cdot\int_0^{\pi}\left[\left(a^2 \plus{} b^2\right) \minus{} \left(a^2\cos 2mx \plus{} b^2\cos 2nx\right)\right]\ \mathrm {dx}$ . Thus,

$ I(a,b) \equal{} \frac {\pi}{2}\cdot \left(a^2 \plus{} b^2\right)\ge\frac {\pi}{4}$ because $ a^2 \plus{} b^2\ge \frac 12\cdot (a \plus{} b)^2 \equal{} \frac 12$ . We have equality iff $ a \equal{} b \equal{} \frac 12$ .[/color]


For the function f, f'(x) = 2x +1 and f(1) =4, what's the approximation for f(1.2) found ny using the line tangent to the graph of f at x=1?

We use a first-order approximation: $ f(x)\approx f(a)\plus{}f'(a)(x\minus{}a)$.  $ x\equal{}1.2$ and $ a\equal{}1$.


vector
analytic geometry
pigeonhole principle
combinatorics unsolved
combinatorics

Prove that every set of 2^m +1  integer lattice points in R^m contains a pair of
points whose centroid (mean vector) is also an integer lattice point.

Reduce the coordinates modulo 2, and use the pigeonhole principle.

Each point is of the form $ (a_1,a_2,..., a_m)$, where each $ a_i$ is an integer. There are $ 2^m$ different ways to assign a parity to each component. Since there are $ 2^m\plus{}1$ points, there must be two in which each component has the same parity, by the pigeonhole principle. The sum of those two vectors then has all components even, so the midpoint has all integer components.


modular arithmetic
number theory unsolved
number theory

Given a positive integer $ n$, consider a subset $ A$ of $ \{1,2,...,n\}$ in which the congruence $ x \plus{} y \equal{} u \plus{} v \pmod n$ has no other solutions but the trivial $ x \equal{} u, y \equal{} v$, and $ x \equal{} v, y \equal{} u$.

Let $ f(n)$ denote the maximum number of elements in such sets.
a) Prove that $ f(n) < \sqrt {n} \plus{} 1$
b) Find infinitely many $ n$ such that $ f(n) > \sqrt {n} \minus{} 1$


Find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all $x,y\in \mathbb R$
\[f(f(x+y))=f(x+y)+f(x)f(y)-xy.\]
I have the solution but it's long. I want the another solution :D  :D  :D  :D

[i]Edited by Myth[/i]

Pfff...... :(  :(  :(
Please Nhat, be careful while posting in the appropriate section...
This one must be in the proposed and own forum....
Pff..let's move it...

Pierre.

I think, long and boring solution is a usual situation for such kind of problems...

Pierre! If you moved it to an appropriate section, why you didn't edit it to make the problem LateX readable?  ;)  :)

:blush:  I didn't think about it...

Pierre.

Nice problem. :)
Put $x=y=0$, so $f(f(0))=f(0)+f^{2}(0)$.Put $x=f(0), y=-f(0)\Rightarrow{f(0)+f^{2}(0)=f(0)+f(f(0))\cdot{f(-f(0))}+f^{2}(0)\Rightarrow f(f(0))=0}$ or ${f(-f(0))=0 }$.If $f(f(0))=0$, then put $x=0, y=f(0)\Rightarrow f(0)=0$.If $f(-f(0))=0$, do with the same way.We got that $f(0)=0$.Put $y=0$ to obtain $f(f(x))=f(x)$ for all $x$.Using this we can write $f(f(x+y))=f(x+y)$ in original equality, then we get $f(x)f(y)=xy$ (*) for all $x,y$.Put $y=f(x) ,x\ne{0}$ in (*) (for each $x$ we can denote $f(x)$ by $y$ )$\Rightarrow f(f(x))=x$, but $f(f(x))=f(x)$, so $f(x)=x$ for all $y$.
I hope my solution is enough short for you [b]nhat [/b] .   :)

oh it's very short and nice. your solution is well than my solution.here are some same function.i think  they aren't difficult:
1)find all functional f(x) such that f:\mathbbR\to\mathbbR and for all x,y in real  we have
                  f(x^5-y^5)=x^3*f(x^2)-y^3*f(y^2)
2) find all functional f(x) such that f:\mathbbR\to\mathbbR and for all x,y in real we have
                  f(x-f(y))=f(x+y^n)+f(f(y)+y^n)+2004    (n is natural  >= 1)
two problems are quite easy . :D  :D  :D  :D

[quote="nhat"]oh it's very short and nice. your solution is well than my solution.here are some same function.i think  they aren't difficult:
1)find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all x,y \in \mathbb$  we have
                  f(x^5-y^5)=x^3*f(x^2)-y^3*f(y^2)
2) find all functional $f(x)$ such that $f:\mathbbR\to\mathbbR$ and for all $x,y \in \mathbbR$ we have
                  f(x-f(y))=f(x+y^n)+f(f(y)+y^n)+2004    (n is natural  >= 1)
two problems are quite easy . :D  :D  :D  :D[/quote]

[quote="nhat"][quote="nhat"]oh it's very short and nice. your solution is well than my solution.here are some same function.i think  they aren't difficult:
1)find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all $x,y \in \mathbb R$
                 \ [f(x^5-y^5)=x^3*f(x^2)-y^3*f(y^2)\]
2) find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all $x,y \in \mathbb R$
                 \ [f(x-f(y))=f(x+y^n)+f(f(y)+y^n)+2004\]    (n is natural  >= 1)
two problems are quite easy . :D  :D  :D  :D[/quote][/quote]

[quote="nhat"][quote="nhat"][quote="nhat"]oh it's very short and nice. your solution is well than my solution.here are some same function.i think  they aren't difficult:
1)find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all $x,y \in \mathbb R$
                   \[f(x^5-y^5)=x^3*f(x^2)-y^3*f(y^2).\]
2) find all functional $f(x)$ such that $f:\mathbb R\to\mathbb R$ and for all $x,y \in \mathbb R$
                  \[f(x-f(y))=f(x+y^n)+f(f(y)+y^n)+2004.\]    (n is natural  >= 1)
two problems are quite easy . :D  :D  :D  :D[/quote][/quote][/quote]

sorry because i have some problem with the latex


geometry
geometric transformation
rotation

Five distinct boys and five distinct girls are seated around a table. In how many ways can this be done such that every person is sitting between two members of the opposite sex (alternating boy -girl - boy - girl etc.) and no arrangement is a rotation of another arrangement?

Im not sure but heres a guess. [hide="Guess"] This is really just a guess but isn't it something like how many ways can you rearrange the letters in the word [i]parallel[/i]? Like there are 8 letters in the word but 3 l's and 2a's so it would be $ \frac{8!}{2!3!}$. Not really a permutation.  But in this problem, it would be $ \frac{20!}{10!10!}$ since the boys and girls repeat ten times each.  So then it becomes $ \frac{(20)(19)(18)(17)(16)(15)(14)(13)(12)(11)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}$.  Then that would simplify to be $ \boxed {184756}$ This is just a guess so Im not sure if its correct. [/hide]

OH, I'm sorry! I made a typo! It was 5 each. I'll go edit it. :blush:

[quote="azn6021023"]Ketvoman, your method is used to eliminate repetitions of the same elements.  The problem states that the boys and girls are distinct.  Also, boys and girls have to sit next to each other; there is exactly one way to do this if the boys and girls are indistinguishable.

[hide]Let's imagine that the girls are all in a circle.  There are $ 9!$ ways to organize them.  There are now ten gaps to choose for the boys.  There are $ 10!$ ways to organize them.  Therefore, the solution is $ 1316818944000$, which is really humongous...[/hide][/quote]

Okay whoops I messed up too with that 5 each thing and I thought it was 10 each.  But I realize my mistake.  But why would it be $ 4!$ spots for girls and then $ 5!$ spots for boys? Why isn't it $ 5!$ each?

When doing circular permutations, there are actually $ (n\minus{}1)!$ permutations possible, where $ n$ is the number of elements arranged.  This is because rotations are not allowed.

When dealing with circles with distinct objects (or people), you can "lock" the position of one person to account for the rotation. In this case, we can position a girl in chair 1, then permute the other $ 4$ in $ 4!$ ways. The boys can then be permuted in $ 5!$ ways, for a total of $ 4!\cdot5!\equal{}\boxed{2880}$.

this is how i like to think of it:
So let there be 10 chairs to seat 10 kids. Let a boy be $ A$ and a girl $ B$. So what if we fuse a boy and a girl together and two chairs together, we have 5 chairs and 5 "fuse-kids". Let $ AB$ represent a fuse-kid then we can place $ AB$ in 5 chairs with $ 5!$, so we have $ 720$ (i don't know how to do the box thing :oops: ) why am i getting different answer?  :?:

[quote="Poincare"](i don't know how to do the box thing :oops: ) [/quote]

To put answers in boxes, use this:
[code]$\boxed{1729}$[/code]
The output is:
$ \boxed{1729}$

[quote="Poincare"]this is how i like to think of it:
So let there be 10 chairs to seat 10 kids. Let a boy be $ A$ and a girl $ B$. So what if we fuse a boy and a girl together and two chairs together, we have 5 chairs and 5 "fuse-kids". Let $ AB$ represent a fuse-kid then we can place $ AB$ in 5 chairs with $ 5!$, so we have $ 720$ [...] why am i getting different answer?  :?:[/quote]

There are two things wrong with this:

(1) First of all, there are only $ 4!\equal{}24$ ways to circularly permute the fuse-kids, since they are arranged in a circle, since each arrangement is overcounted by a factor of $ 5$ due to rotation.

(2) Second, there is more than one way to form the fuse-kids! Specifically, I can permute/pair the boys with the girls in $ 5!\equal{}120$ ways.

Putting this together, the total number of permutations is $ 4!\cdot5!\equal{}24\cdot120\equal{}\boxed{2880}$.


calculus integration LaTeX quadratics Vieta algebra polynomial

Find the integral values of a for which the equation,
(a+2)x^2 + 2(a+1)x + a=0 will have both integral roots.??? :wink:

$ \text{\LaTeX}$-ing it makes it \[ (a+2)x^2+(2a+2)x+a=0\]. Using the quadratic formula, the roots are \[ \frac{-(2a+2)-\sqrt{(2a+2)^2-4a(a+2)}}{2(a+2)}\]
\[ \frac{-(2a+2)+\sqrt{(2a+2)^2-4a(2a+2)}}{2(a+2)}\]

For both of these to be integers, we notice \[ \sqrt{(2a+2)^2-4a(a+2)}\] must be an integer. Then we'll worry about the denominator later.
Expanding and simplifying gives \[ \sqrt{-4a+4}=2\sqrt{1-a}.\] Obviously, $ a\le1$. Now rewriting the solutions gives \[ \frac{-a-2+\sqrt{1-a}}{a+2}\]\[ \frac{-a-2-\sqrt{1-a}}{a+2}\].

There are two ways for $ \sqrt{1-a}$ to be an integer. They are $ a=0, 1$ etc.

Case 1: $ a=1$
The solution is \[ \frac{-1-2\pm0}{1+2}=\frac{-3}{3}=-1\]

Case 2: $ a=0$
The solutions are \[ \frac{-2+1}{0+2}=\frac{-1}{2}\] \[ \frac{-2-1}{0+2}=\frac{-3}{2}\]

Therefore, the only possible value of $ a$ is $ \boxed{1}$

root(1-a) can be integer for infinite values of a condition is that a<1.. but how to prove that threre exists no integral a other than 1 which can give integral values of x ..??

u cant prove it by just substituting only two values of a , you have to proove this fact that there exist no other a except 1 which can give integral roots....

oh wow fail... didn't read the "a can be any integer"  :oops:

ok then.
umm...

I found $ a\equal{}\minus{}3$ works, but I don't know how to prove it for the other solutions    sorry

Let $ x$ and $ y$ be the integer solutions of the quadratic equation. Then $ xy$ also is integer and using Vieta's formula we obtain $ \frac {a}{a \plus{} 2}$ is an integer. Obviously a=0 is a solution.Now, if a is nonzero, we have $ gcd(a,a \plus{} 2)|2$.If $ gcd(a,a \plus{} 2) \equal{} 1$, we need $ a \plus{} 2 \equal{} 1\Rightarrow a \equal{} \minus{} 1$. If $ gcd(a,a \plus{} 2) \equal{} 2$, let $ a \equal{} 2s,a \equal{} 2s \plus{} 2$, then $ \frac {a}{a \plus{} 2} \equal{} \frac {s}{s \plus{} 1}$, which doesn't have nonzero solution.Now, checking for $ a \equal{} \minus{} 1,0$ we conclude that both are solutions to the problem. Q.E.D

$ a\equal{}\minus{}4$ is also a solution.

By Vieta, $ x_1\plus{}x_2\equal{}\minus{}{2a\plus{}2\over a\plus{}2}\equal{}\minus{}2\plus{}{2\over a\plus{}2}$ and $ x_1x_2\equal{}{a\over a\plus{}2}\equal{}1\minus{}{2\over a\plus{}2}$

Hence $ a\plus{}2\equal{}\pm 1,\pm 2$, which yields $ a\in\{\minus{}4,\minus{}3,\minus{}1,0\}$. Checking shows that all the values work.

$ (a\plus{}2)x^2\plus{}(2a\plus{}2)x\plus{}a\equal{}0$

Rewrite this as $ ax^2\plus{}2ax\plus{}a\equal{}\minus{}2x^2\minus{}2x$ or $ a(x\plus{}1)^2\equal{}\minus{}2x(x\plus{}1).$
Obviously $ x\equal{}\minus{}1$ is a solution for all a.

Assuming $ x\ne\minus{}1 \Rightarrow a\equal{}\minus{}\frac{2x}{x\plus{}1}\equal{}\minus{}2\plus{}\frac{2}{x\plus{}1}$

As $ a$ is integer $ \frac{2}{x\plus{}1}$ must be integer and as we a looking for integer values of x this is true when
$ x\equal{}\minus{}3;\minus{}2;0;1$.

So $ a\equal{}\minus{}4;\minus{}3;0;\minus{}1$


Let $ n$ be an integer greater than $3$. Prove that all roots of the polynomial :
\[P(x)\equal{}x^n\minus{}5x^{n\minus{}1}\plus{}12x^{n\minus{}2}\minus{}15x^{n\minus{}3}\plus{}a_{n\minus{}4}x^{n\minus{}4}\plus{}\cdots\plus{}a_0\]
can't be both real and positive.

[quote="math10"]Let $ n$ be an integer greater than 3.Prove that all roots of the polynomial :
$ P(x) \equal{} x^n \minus{} 5x^{n \minus{} 1} \plus{} 12x^{n \minus{} 2} \minus{} 15x^{n \minus{} 3} \plus{} a_{n \minus{} 4}x^{n \minus{} 4} \plus{} ... \plus{} a_0$ can't be both real and positive.[/quote]

I surely misunderstand something.
Are you asking to show that $ P(x)$ has no real positive roots ?

If so, it's obviously wrong : Choose any $ a_1,a_2, ....a_{n\minus{}4}$ and $ a_0<0$ and you obtain at least one positive real root.

Could you be more precise ?

[quote="pco"][quote="math10"]Let $ n$ be an integer greater than 3.Prove that all roots of the polynomial :
$ P(x) \equal{} x^n \minus{} 5x^{n \minus{} 1} \plus{} 12x^{n \minus{} 2} \minus{} 15x^{n \minus{} 3} \plus{} a_{n \minus{} 4}x^{n \minus{} 4} \plus{} ... \plus{} a_0$ can't be both real and positive.[/quote]

I surely misunderstand something.
Are you asking to show that $ P(x)$ has no real positive roots ?

If so, it's obviously wrong : Choose any $ a_1,a_2, ....a_{n \minus{} 4}$ and $ a_0 < 0$ and you obtain at least one positive real root.

Could you be more precise ?[/quote]
No,polynomial root can't both $ REAL$ and $ POSITIVE$(If all roots is real then can't all positive)

[quote="pco"]
I surely misunderstand something.
Are you asking to show that $ P(x)$ has no real positive roots ?[/quote]

I think that he just got his quantifiers wrong.
He stated that ................ $ \forall$ roots $ r$: Not($ r$ real and $ r$ positive)
But he meant that ......... Not$ \big($ $ \forall$ roots $ r$: ($ r$ real and $ r$ positive) $ \big)$

@test20 : you're right, thanks

@math10 :
We have :
$ S_1 \equal{} \sum_ir_i \equal{} 5$
$ S_2 \equal{} \sum_{i\neq j}r_ir_j \equal{} 12$
$ S_3 \equal{} \sum_{i,j,k\text{ pairwise different }}r_ir_jr_k \equal{} 15$

So $ \sum_ir_i^3 \equal{} S_1^3 \minus{} 3S_1S_2 \plus{} 3S_3 \equal{} \minus{} 10 < 0$ and so $ P(x)$ cant have $ n$ real positive roots.


trigonometry
geometry
ARML
trig identities
Law of Cosines

[quote="PenguinIntegral"]1. The measure of the vertex angle of isosceles triangle ABC is $\theta$. The sides of the triangle have length $sin \theta$, $\sqrt{sin\theta}$ and $\sqrt{sin\theta}$. Compute the area of triangle ABC.[/quote]

[hide="Im not very sure"]If you tilt the triangle sideways, you get that $\sqrt{\sin\theta}$ is the base.

$\sin\theta=\frac{x}{\sqrt{\sin\theta}}$ where $x$ is the new height.

$x=\sin\theta\sqrt{\sin\theta}$.

Now multiplying that by the base $\sqrt{\sin\theta}$ and dividing by $2$ gives $\frac{\sin^2\theta}{2}$.[/hide]

EDIT: fixed hide tags

[quote="blackwellbc"]using the formula A= 1/2*a*b*sinc this is simple
A=1/2*$\sqrt{sin\theta}$*$\sqrt{sin\theta}$*$sin\theta$
A=$(sin\theta)^2$/2[/quote]

But, then you have to find $sin\theta$, which involves a little more work.

[quote="mathgeniuse^ln(x)"][quote="PenguinIntegral"]1. The measure of the vertex angle of isosceles triangle ABC is $\theta$. The sides of the triangle have length $sin \theta$, $\sqrt{sin\theta}$ and $\sqrt{sin\theta}$. Compute the area of triangle ABC.[/quote]

This was an ARML problem, solution from my coach.

[hide]"By the Law of Cosines, $sin^2\theta=sin\theta+sin\theta-2sin\theta cos\theta$.  Since $sin\theta \neq 0$, divide by $sin\theta$ to obtain $sin\theta=2-2cos\theta$, which implies that $1-cos^2\theta=4cos^2\theta-8cos\theta+4$, which implies that $5cos^2\theta-8cos\theta+3=0$.  Then $cos\theta=1$ or $cos\theta=\frac{3}{5}$, by factoring that last equation.  We disregard 1 since it makes $\theta=0$.  Thus, $sin\theta=\frac{4}{5}$ and the area of triangle ABC is equal to $\frac{1}{2} \sqrt{sin\theta}*\sqrt{sin\theta}*sin\theta=\frac{1}{2}*\frac{4}{5}*\frac{4}{5}=\frac{8}{25}$."

$\frac{8}{25}$ [/hide][/quote]
What number was it on the ARML test?

This was number 7 on the 2000 ARML.  Was my answer correct, even though this is my coach's answer.


calculus
derivative
algebra
polynomial
real analysis
function
vector

Recently, I was talking to John Conway* about Fourier transforms, and he told me that he was interested in analysis for a time, but he stopped working in that field because he felt that he didn't understand the theorems. Not that he couldn't see what was going on, that he couldn't see why the equations worked out and why the error estimations all went to 0, but that he didn't really see why certain fundamental properties were true. In particular, he noted that he still doesn't have any idea why the inverse Fourier transform should be so similar to the original transform.

This got me thinking. What is the Fourier transform, anyway? I can look at the expression, I can follow the proofs of convergence and the relationships between derivative and multiplication by polynomial, the relationships between convolution and product. But why?? I don't understand any more than Conway!

I realized that you can think of the function $k(x) = e^{ix\theta}$ as a corkscrew spiral in three dimensions. The function is always distance 1 from the x-axis, spiraling around, and the "tightness" of the "threads" of the "screw" is determined by $\theta$.

This led me to think of the Fourier transform as if you take a graph of your function $f(x)$ in the xy plane and then revolve the function around the x-axis. Then try to wrap a string around the function, spiraling around the weird tubular thing along the whole real line. $f$ has to drop off at infinity, so you can approximate by finite intervals. Then you can divide your finite interval into a large number of subintervals, and add up the complex (or vector, if you like) distances from the x-axis to your string. Then take the limit as the number of points you use gets large.

So there: a fairly decent visual explanation of the Fourier transform. But here's why I'm still asking about this: I don't see why the imaginary (or horizontal, if you prefer) parts of the vectors will always cancel out when your function $f$ is real-valued. And this makes the fact that the inverse transform is the exact same thing but with the chirality of the spiral reversed even more perplexing.

Does anyone have any insights about how the theory of Fourier transforms works? Why these things [i]should[/i] be true, as opposed to [i]proof[/i] that they're true?

* (Yes, [i]the[/i] John Conway)

There are other John Conway's in math, you know, e.g. the one who wrote the Springer-Verlag GTMs on Complex Analysis and Functional Analysis.

I concede that he's not nearly as famous as 'the' John (H) Conway though.

A couple of ideas:

First idea: The simple harmonic motions are an interesting class of functions. The set of all simple harmonic motions of frequency $s$ is the set of all (bounded) solutions to the differential equation $y''+4\pi^2s^2y=0.$

$s=0$ is an exception, so assume $s\ne0.$ The set of all simple harmonic motions of  frequency $s$ is a two dimensional linear space. We need two free parameters in which to express this, but there are several good ways to do this:

In cosine-sine (real) form: $y=a\cos(2\pi sx)+b\sin(2\pi sx).$

In amplitude-phase form: $y=A\cos(2\pi sx-\varphi).$

In complex exponential form: $y=ce^{2\pi isx}+c^*e^{-2\pi isx}.$

In cas (Hartley) form: define $\text{cas}(\theta)=\cos\theta+\sin\theta.$

Write $y=d\text{cas}(2\pi sx)+d'\text{cas}(-2\pi sx).$

You should be able to find the algebraic expressions for any of $(a,b),\,(A,\varphi),\,(c,c^*),$ or $(d,d')$ given any other form.

The stuff about spiralling this way or that way - I see that as just algebra. The complex exponential form is the nicest form for the Fourier transform for some uses, but you could write the whole theory in any of these other forms.

You can view the Fourier transform as an effort to write an arbitrary function as a linear combination of these simple harmonic motions.

The second idea: take the hard-core analysis out of it while you look for understanding. Let $N\in\mathbb{N}$ and use the symbol $\mathbb{P}_N$ for the quotient group $\mathbb{Z}/N\mathbb{Z}.$

Let $f: \mathbb{P}_N\mapsto\mathbb{C}.$ The set of all such functions is an $N$-dimensional complex vector space. There's a natural basis - the functions $e_j$ where $e_j(n)=1$ if $n=j$ and zero otherwise.

Now consider the simple harmonic motions (as above). Define $h_k(n)=e^{2\pi ikn/N}$ for $k\in\mathbb{P}_N.$ These functions form an ortho-almost-normal basis of the $N$-dimensional vector space.

Can we express our original function in terms of this basis? Of course we can; that's a change of basis in a vector space.

$\widehat{f}(k)=\frac1N\sum_{n=0}^{N-1}f(n)e^{-2\pi ikn/N}= \frac1N\sum_{n=0}^{N-1}f(n)\omega^{-kn}$ where $\omega=e^{2\pi i/N}.$

We can recover the original function by

$f(n)=\sum_{k=0}^{N-1}\widehat{f}(k)e^{2\pi ikn/N}= \sum_{k=0}^{N-1}\widehat{f}(k)\omega^{kn}.$

Look at it this way: it's a change of basis. Now since the basis (what I've called "simple harmonic motions") are actually characters of the group that is the domain, this change of basis relates very nicely to such group ideas as translation (and convolution).

The Fourier transform on $\mathbb{R}$ can be approached, at least formally, as the result of a couple of limiting processes starting with this finite case.

If you want to understand formal aspects, like the similarity of the formula for the Fourier transform and the inverse Fourier transform, try looking here, at the finite case - nearly everything you're interested in is on display, and the demands of the analysis don't get in the way.

Thanks, Dr. Merryfield, this is definitely an interesting perspective that I haven't seen before. That does make the reason for the inverse transform looking like the original transform more apparent.

Perhaps I need to understand algebra a little better to see what you meant by this. (I haven't had any class except linear algebra, so
[quote] Now since the basis (what I've called "simple harmonic motions") are actually characters of the group that is the domain, this change of basis relates very nicely to such group ideas as translation (and convolution). [/quote]
doesn't really make sense to me. But I'm learning algebra next semester (from a great professor too, so I've heard), and I can be patient. I think it is beautiful how the diverse branches of math are all interconnected by this sort of relationship.

Actually, this might give some insight into the derivative $\rightarrow$ polynomial thing, too. But I'm going to have to put a lot more thought into this before I really feel like I have a feeling for what's going on.

Again, thanks very much, Dr. Merryfield.

By the way, blahblahblah, I've heard of that John Conway too. One of my classmates said he felt sorry for him, because whenever people say "John Conway" they think of John Horton Conway, because he overshadows the other Conway so much.

In addition to what Dr. Merryfield said I believe there are several very deep reasons for why Fourier Series and Integrals have such "nice" symmetries. Its great that you're asking such questions and looking for connections! That's a very good sign as a mathematician.

This question can probably be answered on many different levels depending on how deep of an answer you're prepared to handle. You should ask your question to either Stein, Fefferman, or Daubechies if you can catch them. They are all experts in this area. You may even get different perspectives from each of them.

Thanks! Professor Stein is off campus for this and next semester, and Professors Daubechies and Fefferman don't come to the afternoon tea very often, but I'll see if I can catch one of them some time. Professor Sinai is teaching my class; I suppose I could ask him as well.

I was thinking about my "visual explanation" and its benefits and disadvantages. I realized that it does give a bit of intuitive feeling (though not a real visual proof!) of why a function being smoother means the transform decays faster. But I don't see any way to explain why the inverse transform should be the same sort of operation, at least using only this "wrapping a string around a solid of revolution" visualization.

To be honest I don't really understand your string analogy. A picture might help me to visualize it a little better.

The best way for me, personally, to think of the Continuous Fourier Transformation is to think of it as a way of expressing the "frequency content" of a function.

So if you think of it this way, a smooth function in the time domain will have very little "high frequency" components in the frequency domain. I've never been completely satisfied with this interpretation myself, but its worked for me (most of the time).

Right, that's pretty much the same idea. I'll try to make a picture, but it's a 3D idea, and I'm not so good at art... We'll see....

Obviously, I understand this far less than Conway, but...

Intuitively, I like to think of the exponentials as sort of like an uncountable "basis" for some big space of functions, and the Fourier transform is just a formula for the coefficients.  The standard "basis" is made up of delta functions, and therefore the formula for the Fourier transform, as well the Fourier inversion formula, follows from the idea that the delta function can be written as an integral of exponentials.  Once these ideas are in place, one can turn it into the usual rigorous theory that we know and love.  This way of looking at things (for me) was motivated by physics.

Actually, this is the standard physicist explanation of Fourier transform and Fourier inversion, which is intuitively nice:  First, look at Fourier series, where the exponentials really do form a complete orthogonal set.  Then take the limit as the length of your interval goes to infinity.  This is the non-rigorous part, or at least I've never seen a rigorous treatment of Fourier transform using this idea.


probability
function
probability and stats

Question I'm working on:
A system has two components: a main component and a back-up. The back-up starts working only after the main component fails. Let X be the lifetime in years of the first component and let Y be the the lifetime in years of the second component.  X and Y have joint probability density function f(x,y)=2(2-x)/8 for 0<x<y<2, and 0 elsewhere. The system fails when the back-up component fails.

Find the probability that the system is still working 18 months from now.

What I've done:

Found the Probablity that X+Y>1.5
The solution finds the probability that Y>1.5, but I don't know if that's right so that's why I'm posting to ask if someone can either confirm or explain to me why I'm wrong. Thanks!

This is, as you said, the question of interpretation. I would go your way assuming that the backup unit starts to deteriorate only when it is put in service, not when it is just staying in reserve. But the alternative interpretation also has right to exist.


It  are known for natural some  n  $ 5^n\plus{}3^n\plus{}1$ he  is  prime  number.Show that 12 is divides by n.  :oops:

Do you mean 12 divides $ n$ or $ n$ divides 12 ?

A.K.A.
$ 12|n$ or $ n|12$ ?

$ 12|n$; for example, try $ n\equal{}12,36,48,72 \ldots$ [hide="hint"]test modulo small(est) primes[/hide]

if $ n$ is odd then
$ 5^{2k \plus{} 1} \plus{} 3^{2k \plus{} 1} \plus{} 1 \equiv ( \minus{} 1) \plus{} (0) \plus{} 1 \equal{} 0$ mod 3,
but $ 5^n \plus{} 3^n \plus{} 1 > 3$.

if $ n \equal{} 4k \plus{} 2$ then
$ 5^{4k \plus{} 2} \plus{} 3^{4k \plus{} 2} \plus{} 1 \equal{} (0) \plus{} 9(1) \plus{} 1 \equal{} 10 \equal{} 0$ mod 5,
 but $ 5^n \plus{} 3^n \plus{} 1 > 5$.

if $ n \equal{} 3k \plus{} 1$, then
$ 5^{3k \plus{} 1} \plus{} 3^{3k \plus{} 1} \plus{} 1 \equal{} 5(125)^k \plus{} 3(27)^k \plus{} 1 \equal{} 5( \minus{} 1)^k \plus{} 3( \minus{} 1)^k \plus{} 1 \equal{} ( \minus{} 1)^k(8) \plus{} 1 \equal{} ( \minus{} 1)^k \plus{} 1$ mod 7
but if $ k$ is even $ n$ is odd as in previous case, so $ n$ is odd, and $ 5^n \plus{} 3^n \plus{} 1$ is divisible by $ 7$.

if $ n \equal{} 3k \plus{} 2$ then
$ 5^{3k \plus{} 2} \plus{} 3^{3k \plus{} 2} \plus{} 1 \equal{} 25(125)^k \plus{} 9(27)^k \plus{} 1 \equal{} 25( \minus{} 1)^k \plus{} 9( \minus{} 1)^k \plus{} 1 \equal{} ( \minus{} 1)^k(25 \plus{} 9) \plus{} 1 \equal{} 34( \minus{} 1)^k \plus{} 1 \equal{} ( \minus{} 1)^{k \plus{} 1} \plus{} 1$ mod 7.
 if $ k$ is odd then $ n$ is odd as in previous case. then $ k$ is even, so $ ( \minus{} 1)^{k \plus{} 1} \plus{} 1 \equal{} \minus{} 1 \plus{} 1 \equal{} 0$ mod 7.

so $ n$ must be divisible by $ 12$.


The sequence $ a_1$, $ a_2$, $ a_3$, $ \dots$ satisfies $ a_1 \equal{} 19$, $ a_9 \equal{} 99$, and, for all $ n \ge 3$, $ a_n$ is the arithmetic mean of the first $ n \minus{} 1$ terms. Find $ a_2$.

$ \textbf{(A)}\ 29\qquad
\textbf{(B)}\ 59\qquad
\textbf{(C)}\ 79\qquad
\textbf{(D)}\ 99\qquad
\textbf{(E)}\ 179$

[hide]
Adding the arithmetic mean to a sequence of numbers doesn't change that mean.

[b]Proof:[/b] For a sequence $ a_1,a_2...a_n$ the arithmetic mean is $ \frac {a_1 \plus{} a_2 \plus{} a_3... \plus{} a_n}{n}$

If we add the arithmetic mean to the origional sequence and take the mean again then we get;

$ \dfrac{(a_1 \plus{} a_2 \plus{} a_3... \plus{} a_n) \plus{} \frac {a_1 \plus{} a_2 \plus{} a_3... \plus{} a_n}{n}}{n \plus{} 1}$

$ \Rightarrow\dfrac{\frac{(n\plus{}1)(a_1\plus{}a_2\plus{}a_3...\plus{}a_n)}{n}}{n\plus{}1} \Leftrightarrow \frac{a_1\plus{}a_2\plus{}a_3...\plus{}a_n}{n}$

So using this lemma the question becomes $ \frac{19\plus{}a_2}{2} \equal{} 99 \Rightarrow a_2 \equal{} 179 \longrightarrow \boxed{D}$ [/hide]

[hide="Actually..."]
For $ n \ge 3$, $ a_n\equal{}\frac{a_1\plus{}a_2\plus{}\dots \plus{} a_{n\minus{}1}}{n\minus{}1}$.

Thus $ (n\minus{}1)a_n\equal{}a_1\plus{}a_2\plus{}\dots\plus{}a_{n\minus{}1}$.  It follows that

$ a_{n\plus{}1}\equal{}\frac{a_1\plus{}a_2\plus{}\dots\plus{}a_{n\minus{}1}\plus{}a_n}{n}\equal{}\frac{(n\minus{}1) \cdot a_n\plus{}a_n}{n}\equal{}a_n$,

for $ n \ge 3$.  Since $ a_9\equal{}99$ and $ a_1\equal{}19$, it follows that $ 99\equal{}a_3\equal{}\frac{19\plus{}a_2}{2}$,

so $ a_2\equal{}179$, or $ \boxed{\textbf{(E)}}$.
[/hide]

[hide="My solution"]
Because we have,
$$a_{n}=\frac{a_1+a_2+\cdots+a_{n-1}}{n-1}$$
for every positive integer $n \geq 3.$ we deduce that,
$$a_{n+1}=\frac{(a_1+a_2+\cdots+a_{n-1})+a_n}{n}=\frac{(n-1)a_n+a_n}{n}=a_n.$$
for every positive integer $n \geq 3.$ Because, $a_3=\frac{a_1+a_2}{2}$ we have, $a_2=2a_3-a_1=2 \times 99 -19=179.$
which corresponds to answer choice $ \textbf{(E)}$.
[/hide]


I have just found it :"If exist a  function f:N* ->{1,2,3,..,9} such that f(f(a)*f(b))=f(a*b)
 I found f defined as:  with x=a1a2..an (base10) then define y=a1+a2+..+an=b1b2..bm (base10).we continued until y<10.then f(x)=y(y<10).
  I don't know  why  :?: but it's true!

sorry for me ! It's ...stupid question :blush: !  f(x)=r, with r=x(mod 9),r<10!
  then it's obvious.


geometry
3D geometry
sphere
search
geometry unsolved

It is easy to someone to tell me, what is the volume of the Euclidean Sphere $ S^{n\minus{}1}$, and a formula, which connects, the Euclidean sphere with the euclidean ball?

Easy search on Google yields [url]http://en.wikipedia.org/wiki/N-sphere[/url]

Thanks a lot listening REMBETIKA... :lol:


Mechanics of Materials is a basic subject commonly given at the third semester of several engineering courses - Civil Engineering, Mechanical Engineering, and Aeronautic Engineering, among others. For those interested, you can find on this post a selection of some exercises and problems.

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

[b]Problem 1.[/b] A solid body has the following state of stress, defined relative to a certain Oxyz frame. Compute the forces applied along the body.

$ \sigma \equal{}
\begin{pmatrix}3xy & 5y^2 & 0 \\5y^2 & 0 & 2z \\0 & 2z & 0 \end{pmatrix}$

[b]Problem 2.[/b] Consider a point in the interior of a solid body whose state of stress is described by the following stress tensor (relative to a Oxyz frame):

$ \sigma \equal{}
\begin{pmatrix}100 & \minus{}200 & \minus{}300 \\\minus{}200 & 200 & 100 \\\minus{}300 & 100 & \minus{}100 \end{pmatrix}\,\,(MPa)$

(a) Compute the shear and normal components of the stress acting on a plane equally inclined relative to the xyz axes. Determine also the direction of the shear component of the stress.

(b) Find the principal stresses, and the corresponding directions, for this stress tensor.

(c) Compute the maximum shear stress and the orientation of the planes where it occurs.

[b]Problem 3.[/b] A point in the interior of a steel body is subjected to a state of stress defined by the following values: $ \sigma_x \equal{} 50\,MPa$, $ \sigma_y \equal{} \minus{}12\,MPa$, $ \sigma_z \equal{} 40\,MPa$, $ \tau_{xy} \equal{} 35\,MPa$, $ \tau_{xz} \equal{} \tau_{yz} \equal{} 0$.

(a) Find the principal stresses and the corresponding directions.

(b) Compute the maximum shear stress.

(c) Express the stress tensor as the sum of an isotropic stress tensor with a pure shear stress tensor.

[b]Problem 4.[/b] A certain state of stress is given by the following tensor (relative to Oxyz):

$ \begin{pmatrix} \sigma & a \cdot \sigma & b \cdot \sigma \\ a \cdot \sigma & \sigma & c \cdot \sigma \\ b \cdot \sigma & c \cdot \sigma & \sigma \end{pmatrix}\,\,(MPa)$

where a, b, c are constants and $ \sigma$ is a stress. Find the values of a, b, and c such that the stress in a plane equally inclined relative to xyz is zero.

[b]Problem 5.[/b] A certain state of stress is given by the following tensor (relative to a Oxyz frame):

$ \sigma \equal{}
\begin{pmatrix}3 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{pmatrix}\,\,(MPa)$

(a) Compute the principal stresses and corresponding directions.

(b) What is the stress tensor relative to the principal axes frame?

[b]Problem 6.[/b] In a certain point of a material the plane state of stress is defined by the following stress tensor (relative to Oxy):

$ \sigma \equal{}
\begin{pmatrix}80 & 20\\ 20 & \minus{}40\end{pmatrix}\,\,(MPa)$

(a) Find the normal and shear components of the stress acting on a plane whose normal makes an angle of 36.87º (in the direct sense) with the x direction.

(b) Compute the principal stresses and directions. What is the value of the maximum shear stress?

(c) Find the stress (normal and shear components) on a plane whose normal makes a (clockwise) angle of 60º with the maximum principal direction.

(d) Solve the previous questions by using Mohr's Circle.

[b]Problem 7.[/b] Consider the plane state of stress represented in picture 1 bellow. Find the normal and shear components of the stress acting on a plane equally inclined relative to xy; compute also the principal stresses and directions.

[b]Problem 8.[/b] Picture 2 bellow represents two planes A and B defined in the neighbourhood of a point, the shear and normal stresses on plane A, and the normal stress on plane B.

(a) Find the shear stress on plane B.

(b) Compute the principal stresses and directions.

[b]Problem 9.[/b] In a point subjected to a plane state of stress, the normal stresses on two perpendicular planes and one of the principal directions are known, as picture 3 bellow shows. Find the principal stresses.

[b]Problem 10.[/b] The principal stresses on a given plane state of stress have the values $ \sigma_1 \equal{} 500\,MPa$ and $ \sigma_2 \equal{} \minus{}100\,MPa$, occurring the first one on a plane that makes 30º with the x direction. Determine:

(a) The normal and shear components of stress that act on a plane that makes 90º with the x direction.

(b) The orientation of the planes where the stress is purely shear.

(c) The orientation of the planes for which $ \tau \equal{} 200\,MPa$, and the normal stresses on those planes.

[b]Problem 11.[/b] In the neighbourhood of a point belonging to a simply connected solid body, the displacement field is given, on a Oxyz frame, by

$ u \equal{} (3x^4 \plus{} 2x^2y^2 \plus{} x \plus{} y \plus{} z^3 \plus{} 3) \times 10^{\minus{}4}$
$ v \equal{} (3xy \plus{} y^3 \plus{} y^2z \plus{} z^2 \plus{} 1) \times 10^{\minus{}4}$
$ w \equal{} (x^2 \plus{} xy \plus{} yz \plus{} zx \plus{} y^2 \plus{} z^2 \plus{} 2) \times 10^{\minus{}4}$

(a) Find the deformation state in a point P with coordinates (1, 3/2, 2).

(b) Verify the conditions of compatibility of the deformations in a general point (x,y,z).


A positive number $ x$ has the property that $ x\%$ of $ x$ is $ 4$. What is $ x$?

$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 10 \qquad
\textbf{(D)}\ 20 \qquad
\textbf{(E)}\ 40$

[hide="Click for solution"]
We know that $ \frac{x}{100} \times x\equal{}4 \implies x^2\equal{}400$.  Because $ x>0$, $ x\equal{}20$, or $ \boxed{\textbf{(D)}}$.
[/hide]

x % of x = 4
So, x/100 * x =4
So, x^2/100=4
So, x^2 = 400
So, x =√400
So, x= 20

So, correct option is option(D) i.e 20
Hope you like, understand and appreciate this solution.

[quote=michaelchang1][hide=Solution]

We set up an equation:

$$\frac{x}{100}x=4\implies \frac{x^2}{100}=4\implies x^2=400$$
$$\implies x=\pm 20$$
Therefore, $\boxed{\textbf{(D)}}$ is our answer.[/quote]

Why did you make a 12 year bump....?
The solution is the same as the second post.

lol everyone's just bumping super old threads

[hide=Solution to 2005 AMC10B #2]Using the given information, we can make the equation;

\begin{align*}
x\% \cdot x = 4 &\implies \\
\frac{x}{100} \cdot x = 4 &\implies \\
\frac{x^2}{100} = 4 &\implies \\
x^2 = 400 &\implies \\
x = 20. \\
\end{align*}

So the value of $x$ is $20$ or option D.

[b][color=#f00]Remarks:[/color][/b] At the $x^2=400$ part of our simplification, we could take the square root of both sides because in the information given, it says $x$ is positive.

[/hide]


Let $a_1\geq a_2\geq ...\geq a_n>0$ and $M_n=\min_{k=1,n}{\frac{b_1+b_2+...+b_k}{k}}$. Prove that \[\sum_{k=1,n}{a_kb_k}\geq{M_n}{\sum_{k=1,n}a_k}.\]

P.S. It is stated in http://www.mathlinks.ro/Forum/viewtopic.php?p=183536#p183536
I think it is a good idea to make separate topic for this useful inequality.
It is highly possible it appeared on the forum before, but I don't know how to find it.
Let's prove it again :)

This, if the partial sums $b_1 + \cdots + b_k$ are all non-negative. Must they be?

we can increase all b_k by the same positive value.


If I'm thinking about mailing my solutions, does the mail have to arrive there by the final submitting date or do I have to send the mail by that date?

Thanks.

[quote="zanttrang"]The date given is the postmark date - so as long as your mail is initially processed by that date, you're fine.[/quote]

This is correct.  Thanks!


A bag contains 20 black balls and 30 white balls. Four balls are randomly drawn without replacement. What is the probability of selecting a white ball followed by a black ball, and then selecting a black ball followed by a white ball? Express your answer as a common fraction.

We multiply the probabilities together. We also have to take into consideration the fact that we aren't replacing the balls (we aren't putting them back after we choose them). $ P(\text{white ball 1st})\times P(\text{black ball 2nd})\times P(\text{white ball 3rd})\times P(\text{black ball 4th}) \equal{}  \frac{30}{50}\times\frac{20}{49}\times\frac{29}{48}\times\frac{19}{47}$.
Fortunately, some numbers factor out and we are left with $ \frac{19\times29}{4\times47\times49} \equal{} \boxed{\frac{551}{9212}}$.

It is actually another black ball 3rd, but the results you get would be the same.


algorithm
number theory
greatest common divisor
relatively prime
Euclidean algorithm

Find n belong to N satisfying $\frac{n-37}{n+43}$ is a squared of a rational number. :lol:

[hide]
Subtracting $43$ from the perfect squares starting from $49$, I get $38$ that works: $\frac{38-37}{38+43} = \frac{1}{81} = \displaystyle \left(\frac{1}{9}\right)^2$.

I don't know how I would go about finding the solution if the numbers were changed to any other arbitrary integers.
[/hide]

I think $n=101, n=398, n=38$ are all possible sol...

hey can u explain the euclidean algorithm to me? the mathworld thing is pretty involved

euclidean algorithm uses the fact that the gcd of two numbers is unchanged if you add multiples of  one to another, and the gcd is the minimum number resulting from linear combination of the two, therefore, If we keep on subtracting multiples of one number from the other, then eventually we can find the gcd.


AMC
USA(J)MO
USAMO
induction
modular arithmetic

Find all integer solutions to $ a^2\plus{}b^2\plus{}c^2\equal{}a^2b^2$.

We first find the obvious solution, (0,0,0).

If the right side was odd, then both a and b would be odd. Therefore, in order to make the left side odd, c would also have to be odd. However, we know that all odd perfect squares are congruent to 1 (mod 4). The left side, a sum of three odd perfect squares, would be 3 (mod 4), contrasting to the right side. The right side, therefore, cannot be odd.

If the right side was even, then either one of a or b would be even, or they would both be even. Without loss of generality, let a be even and b odd (to test our first case). Then, in order to make the left side even, c would have to be odd. Taking both sides modulo four as before, we find the right side is 0, and the left side is 2 (there are two odd perfect squares, each 1 modulo 4). Therefore, both a and b must be even, implying c is even.

The only possible way that the equation could be satisfied is if all three variables are even.

Let a=2x, b=2y, and c=2z for some x, y, and z. Then, substituting and simplifying, we have $ x^2\plus{}y^2\plus{}z^2\equal{}4x^2y^2$.

Using the same arguments as above, we find that x,y, and z also must be even. We can substitute another set of variables  like we did above, but we find that a, b, and c must be divisible by an infinitely high power of two. The only solution to fit this claim is (0,0,0).

QED

Please feel free to pwn my proof. :D

Your argument seems nice until the [quote]We can substitute another set of variables like we did above[/quote].  You could probably clarify the point (perhaps through induction?).

Then again, I've never made USAMO, so I wouldn't know.

hmmm...ok :)

I found out that this method was called infinite descent (right? xD). So would it help if I said that in there? Maybe if I said "we proceed by infinite descent" (or something like that, could someone provide better wording) there I could avoid the whole induction thing altogether? Or do the induction thing anyways, but still mention the infinite descent?

Are there any other problems you could see?

Also, if someone could critique my proof not just based on accuracy of the argument itself, but style writing, that would be helpful.

If I want you guys to edit more of my proofs, should I keep posting here, or make new topics?

You did not actually show that there is infinite descent! However, this is easy with induction, so let's construct an induction argument systematically.

An inductive proof is actually an infinite number of proofs that follow the same format, which are executed in order, much like a loop. You must ensure that:
1. The preconditions of the base case are met [i]before[/i] any of the inductive argument has been executed
2. The preconditions of each subsequent step are met by either the postconditions of previous steps, or material outside of/before the inductive argument, or some combination of these.

[b]Step 1:[/b] Determine the preconditions and the post-conditions for each step:

[hide="Preconditions and postconditions"][b]I. BASE CASE:[/b]

[b]Precondition:[/b] We are given that $ a^2 + b^2 + c^2 = a^2 b^2$ for an ordered triple $ (a,b,c) \in \mathbb{Z}^3$

[b]Postcondition:[/b] There exists $ (a_1, b_1, c_1) \in \mathbb{Z}^3$ such that:
$ a_1 ^2 + b_1 ^2 + c_1 ^2 = 4 a_1 ^2 b_1 ^2$
$ (a,b,c) = 2 (a_1, b_1, c_1)$

[b]II. INDUCTIVE STEP:[/b] For each $ n \in \mathbb{N}$, execute the following argument:

[b]Precondition:[/b]
There exists $ (a_n, b_n, c_n) \in \mathbb{Z}^3$ such that:
$ \begin{array}{c} a_n ^2 + b_n ^2 + c_n ^2 = 4^n a_n ^2 b_n ^2 & (a,b,c) = 2^n (a_n, b_n, c_n) \end{array}$

[b]Postcondition:[/b]
There exists $ (a_{n + 1}, b_{n + 1}, c_{n + 1})$ such that:
$ a_{n + 1} ^2 + b_{n + 1} ^2 + c_{n + 1} ^2 = 4^{n + 1} a_{n + 1} ^2 b_{n + 1} ^2$
$ (a,b,c) = 2^{n + 1} (a_{n + 1}, b_{n + 1}, c_{n + 1})$[/hide]

[b]Step 2:[/b] Using the roadmap that we've set up, we now determine how to work from the precondition of each step to the postcondition. This may be different for the base case and for the inductive step, though the two procedures are usually similar.

[hide="Base case"]Note that for any integer $ n$, $ n^2 \equiv 0 \pmod{4}$ iff. $ n$ is even and $ n^2 \equiv 1 \pmod{4}$ iff. $ n$ is odd. Consider two cases:

[u]Case 1:[/u] [i]a and b are both odd[/i]
$ a^2 \equiv b^2 \equiv 1 \pmod{4}$, so taking the original equation modulo 4, we get:
$ 2 + c^2 \equiv 1 \pmod{4}$
However, since $ 2 + c^2$ is congruent to either 2 or 3 modulo 4, the above cannot be met. So, a and b cannot both be odd.

[u]Case 2:[/u] [i]At least one of a and b is even[/i]
$ a^2 b^2 \equiv 0 \pmod{4}$, so taking the original equation modulo 4, we get:
$ a^2 + b^2 + c^2 \equiv 0 \pmod{4}$
This can only happen if $ a^2 \equiv b^2 \equiv c^2 \equiv 0 \pmod{4}$, which means $ a,b,c$ are all even.
Thus, there exists $ (a_1, b_1, c_1) \in \mathbb{Z}^3$ such that $ (a,b,c) = 2 (a_1, b_1, c_1)$[/hide]

[hide="Inductive step"]Taking the equation $ a_n ^2 + b_n ^2 + c_n ^2 = 4^n a_n ^2 b_n ^2$ modulo 4, we get:
$ a_n ^2 + b_n ^2 + c_n ^2 \equiv 0 \pmod{4}$

This only holds if $ a_n ^2 \equiv b_n ^2 \equiv c_n ^2 \equiv 0 \pmod{4}$, or when $ a_n$, $ b_n$, and $ c_n$ are all even.

Thus, there exists $ (a_{n + 1}, b_{n + 1}, c_{n + 1}) \in \mathbb{Z}^3$ such that:
$ (a_n, b_n, c_n) = 2 (a_{n + 1}, b_{n + 1}, c_{n + 1}) \implies (a,b,c) = 2^{n + 1} (a_{n + 1}, b_{n + 1}, c_{n + 1})$

Substituting in $ 2a_{n + 1}, \; 2 b_{n + 1}, \; 2 c_{n + 1}$ for $ a_n, b_n, c_n$, respectively yields:
$ 4a_{n + 1} ^2 + 4b_{n + 1} ^2 + 4c_{n + 1} ^2 = 4^{n} \cdot 4a_{n + 1} ^2 \cdot 4 b_{n + 1} ^2$
$ \iff a_{n + 1} ^2 + b_{n + 1} ^2 + c_{n + 1} ^2 = 4^{n + 1} a_{n + 1} ^2 b_{n + 1} ^2$[/hide]

[b]Step 3:[/b] Put the parts together to get a complete proof. Add an introduction and a conclusion so that the reader knows where you are going with your proof right from the start, and knows what you have proven when you are done.

------------------------------------
[hide="SOLUTION"]By inspection, $ (0,0,0)$ is a solution to the equation. The following infinite descent argument shows that there are no other solutions.

[b]0. INTRODUCTION:[/b]
The inductive argument below will prove that, for each $ n \in \mathbb{N}$, each of $ a,b,c$ must be divisible by $ 2^n$. This will result in an instance of infinite descent.

[b]I. BASE CASE:[/b]

[b]Precondition:[/b] We are given that $ a^2 + b^2 + c^2 = a^2 b^2$ for an ordered triple $ (a,b,c) \in \mathbb{Z}^3$

[b]Proposed Postcondition:[/b] There exists $ (a_1, b_1, c_1) \in \mathbb{Z}^3$ such that:
$ a_1 ^2 + b_1 ^2 + c_1 ^2 = 4 a_1 ^2 b_1 ^2$
$ (a,b,c) = 2 (a_1, b_1, c_1)$

[b]Proof:[/b] Note that for any integer $ n$, $ n^2 \equiv 0 \pmod{4}$ iff. $ n$ is even and $ n^2 \equiv 1 \pmod{4}$ iff. $ n$ is odd. Consider two cases:

[u]Case 1:[/u] [i]a and b are both odd[/i]
$ a^2 \equiv b^2 \equiv 1 \pmod{4}$, so taking the original equation modulo 4, we get:
$ 2 + c^2 \equiv 1 \pmod{4}$
However, since $ 2 + c^2$ is congruent to either 2 or 3 modulo 4, the above cannot be met. So, a and b cannot both be odd.

[u]Case 2:[/u] [i]At least one of a and b is even[/i]
$ a^2 b^2 \equiv 0 \pmod{4}$, so taking the original equation modulo 4, we get:
$ a^2 + b^2 + c^2 \equiv 0 \pmod{4}$
This can only happen if $ a^2 \equiv b^2 \equiv c^2 \equiv 0 \pmod{4}$, which means $ a,b,c$ are all even.
Thus, there exists $ (a_1, b_1, c_1) \in \mathbb{Z}^3$ such that $ (a,b,c) = 2 (a_1, b_1, c_1)$

[b]II. INDUCTIVE STEP:[/b] For each $ n \in \mathbb{N}$, execute the following argument:

[b]Precondition:[/b]
There exists $ (a_n, b_n, c_n) \in \mathbb{Z}^3$ such that:
$ \begin{array}{c} a_n ^2 + b_n ^2 + c_n ^2 = 4^n a_n ^2 b_n ^2 & (a,b,c) = 2^n (a_n, b_n, c_n) \end{array}$

[b]Proposed Postcondition:[/b]
There exists $ (a_{n + 1}, b_{n + 1}, c_{n + 1})$ such that:
$ a_{n + 1} ^2 + b_{n + 1} ^2 + c_{n + 1} ^2 = 4^{n + 1} a_{n + 1} ^2 b_{n + 1} ^2$
$ (a,b,c) = 2^{n + 1} (a_{n + 1}, b_{n + 1}, c_{n + 1})$

[b]Proof:[/b] Taking the equation $ a_n ^2 + b_n ^2 + c_n ^2 = 4^n a_n ^2 b_n ^2$ modulo 4, we get:
$ a_n ^2 + b_n ^2 + c_n ^2 \equiv 0 \pmod{4}$

This only holds if $ a_n ^2 \equiv b_n ^2 \equiv c_n ^2 \equiv 0 \pmod{4}$, or when $ a_n$, $ b_n$, and $ c_n$ are all even.

Thus, there exists $ (a_{n + 1}, b_{n + 1}, c_{n + 1}) \in \mathbb{Z}^3$ such that:
$ (a_n, b_n, c_n) = 2 (a_{n + 1}, b_{n + 1}, c_{n + 1}) \implies (a,b,c) = 2^{n + 1} (a_{n + 1}, b_{n + 1}, c_{n + 1})$

Substituting in $ 2a_{n + 1}, \; 2 b_{n + 1}, \; 2 c_{n + 1}$ for $ a_n, b_n, c_n$, respectively yields:
$ 4a_{n + 1} ^2 + 4b_{n + 1} ^2 + 4c_{n + 1} ^2 = 4^{n} \cdot 4a_{n + 1} ^2 \cdot 4 b_{n + 1} ^2$
$ \iff a_{n + 1} ^2 + b_{n + 1} ^2 + c_{n + 1} ^2 = 4^{n + 1} a_{n + 1} ^2 b_{n + 1} ^2$

[b]III. CONCLUSION:[/b]
Therefore, for each $ n \in \mathbb{N}$, there exists integers $ a_n, b_n, c_n$ such that:
$ a = 2^n a_n \implies 2^n | a$
$ b = 2^n b_n \implies 2^n | b$
$ c = 2^n c_n \implies 2^n | b$
This establishes infinite descent, as desired. The only $ (a,b,c) \in \mathbb{Z}^3$ that satisfies the above for all $ n$ is $ (a,b,c) = (0,0,0)$. We have already noted that this solves the original equation, so it must be the unique solution. $ \square$
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Note that the explicit preconditions, postconditions, introduction, and conclusion in the above took up more than half of the space.

In practice, you don't have to designate the preconditions and postconditions of the base-case and inductive step so explicitly in your final solution. In theory, it is not a problem if you do, but it might take up time if you are a slow writer.

Do not sacrifice clarity to save a few minutes -- remember, on the USAMO, getting the right idea earns you at most 2 out of 7 points! It is more optimal to finish a problem that is almost done than to start a new problem.

In this case, because there were two preconditions and two corresponding postconditions in the inductive step, they served as a nice summary of the assumptions and results of each step.