Find all the monotone functions $f: N\Rightarrow{N}$ st:
$f(x^{2007}+y^{2007})=f^{2007}(x)+f^{2007}(y) \forall{x,y\in{R}}$
:)

Sorry ,$f: N\Rightarrow{N}$

[quote="Rust"][quote="Farenhajt"]And what if $f(0)=2^{-2006}$? :)[/quote]
Exactly $f(0)=2^{-1/2006}$?[/quote]
Sorry, I was wrong.
if $f(0)=2^{-1/2006}$,then $f^{2007}(x)+c=f(x^{2007})$
Let$x-\to x^\frac{1}{2007},y-\to y^\frac{1}{2007}$
then $f(x+y)=f(x)+f(y)-2c$
then $f(x+y)-c=(f(x)-c)+(f(y)-c)$
so $f(x)=x+c$
when$c\not= 0$ it doesn't satisfies the orginal condition.
Is it correct now?

[quote="Hawk Tiger"][quote="weiquan"][quote="Hawk Tiger"]Quite easy:$f(0)=0$
then $f(x^{2}007)=f^{2}007(x)$
Let$x-\to x^\frac{1}{2007},y-\to y^\frac{1}{2007}$
then $f(x+y)=f(x)+f(y)$
it is Chuachy function,the result is $f(x)=x$[/quote]

It is a Cauchy function, but the solution $f(x)=cx$ does not satisfy the original condition.

Actually the real function is harder, in the sense that there are a lot of little cases to consider (in the end all having no solution). If the function is over natural numbers, then does monotone mean strictly increasing/decreasing or non-increasing/decreasing? If it is strict then there's no solution because $f(1)=1$ and $f(2)=2$ thus $f(2^{2007}+1)=3$[/quote]
I think you get a mistake in calculating,weiquan.
I also wonder that if it is strictly monotone.
If it is strict,then $f(0)=0,f(1)=1,f(2)=2,f(2^{2}007+1)=2^{2}007+1$
then becuase $2=f(2)<f(3)<......<f(2^{2007}+1)=2^{2007}+1$
thus $f(i)=i$$1\leq i\leq 2^{2007}+1$
in this way we can prove by induciton $f(n)=n$
but if it is not strictly, I have no idea now.but I guess the result is the same.(I have no time to prove it now.)[/quote]

Hmm... is $f^{2007}(x)$ supposed to be $(f(x))^{2007}$? If so I have been doing it wrongly all along (I used the composite function $f^{2007}$.

Anyway I think there is still something wrong with the question. If it is over positive integers, $f(0)$ does not exist!

[quote="Hawk Tiger"]Oooops,you are right.
Does $N$ denote postive integer or non-negative integer?[/quote]
I'm really sorry :blush:   :oops:
$N$ denotes non-negative integer set.
Really,that sum is:
Find all function $f: Z\rightarrow{Z}$ st:
f: monotone and $f(x^{2007}+y^{2007})=(f(x))^{2007}+(f(y))^{2007}\forall{x,y\in{Z}}$ :)
I thinkl HaWk Tiger 's solution is correct :)

[quote="tmbtw"][quote="Hawk Tiger"]Oooops,you are right.
Does $N$ denote postive integer or non-negative integer?[/quote]
I'm really sorry :blush  :oops
$N$ denotes non-negative integer set.
Really,that sum is:
Find all function $f: Z\rightarrow{Z}$ st:
f: monotone and $f(x^{2007}+y^{2007})=(f(x))^{2007}+(f(y))^{2007}\forall{x,y\in{Z}}$ :)[/quote]

Just to clarify, does monotone functions include non-increasing and non-decreasing functions? If not then the proof by Hawk Tiger solves it.


How many positive integers have a value between $ \sqrt{8}$ and $ \sqrt{72}$?

$ \sqrt{8}\equal{}2\sqrt{2}<3$ and $ \sqrt{72}\equal{}3\sqrt{8}\equal{}6\sqrt{2}>8$.  So the integers between them are 3, 4, 5, 6, 7, and 8, so there are 6 numbers.


Let $ a,\ b,\ c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$.

Prove that $ \sqrt{a\plus{}bc}\plus{}\sqrt{b\plus{}ca}\plus{}\sqrt{c\plus{}ab}\leq 2$.

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$.

Prove that $ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ca} \plus{} \sqrt {c \plus{} ab}\leq 2$.[/quote]

$ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ca} \plus{} \sqrt {c \plus{} ab}$

$ \equal{}\sqrt {a(a\plus{}b\plus{}c) \plus{} bc} \plus{} \sqrt {b(a\plus{}b\plus{}c) \plus{} ca} \plus{} \sqrt {c(a\plus{}b\plus{}c) \plus{} ab}$

$ \equal{}\sqrt {(a\plus{}b)(a\plus{}c)} \plus{} \sqrt {(b\plus{}c)(b\plus{}a)} \plus{} \sqrt {(c\plus{}a)(c\plus{}b)}$

$ \le \frac{(a\plus{}b)\plus{}(a\plus{}c)}{2} \plus{} \frac {(b\plus{}c)\plus{}(b\plus{}a)}{2} \plus{} \frac {(c\plus{}a)\plus{}(c\plus{}b)}{2}$

$ \equal{} 2(a\plus{}b\plus{}c)$

$ \equal{}2$

Needless to say, that's correct.

I made the problem from $ (\sqrt{a\plus{}b}\minus{}\sqrt{b\plus{}c})^2\plus{}(\sqrt{b\plus{}c}\minus{}\sqrt{c\plus{}a})^2\plus{}(\sqrt{c\plus{}a}\minus{}\sqrt{a\plus{}b})^2\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$.


calculus
integration
logarithms
function
calculus computations

[quote="Moubinool"]Calculate with 5% relative error  $\int_{0}^{10}x^xdx$[/quote]

200

[quote="Myth"]I could repeat your phrase: are you mad? ;)
I hope, you understand me.[/quote]

yes i am resally mad:)))
i understand u u want to say that i must write how i did it
but i cant write in such lungagge lenux o something as all of u say
it vill x ustu x ln(x+1)
that mean i said 200 wrongit must be
100(ln10+1)

Whatever the answer is, it is at least $9^9$. Your answers are off by several orders of magnitude.

How do you think, jmerry, why did I say that:
[quote="Myth"]I could repeat your phrase: are you mad? ;)
I hope, you understand me.[/quote]

I understand what you said, but I'm not sure DON FAVOR did.

Some cheap estimates. This doesn't get us within 5% yet, but at least it nails down the order of magnitude.

$x^x$ lies above its tangent line, so on $[0,10),$ $\,x^x>10^{10}(1+(1+\ln10)(x- 10)).$

$\int_0^{10}x^x\,dx>\int_{10-\frac1{1+\ln10}}^{10}x^x\,dx >$

$>10^{10}\int_{10-\frac1{1+\ln10}}^{10}(1+(1+\ln10)(x- 10))\,dx= \frac{10^{10}}{2(1+\ln10)}\approx 1.51\times 10^9.$

On the other hand, $x^x<10^x$ on that interval, so

$\int_0^{10}x^x\,dx<\int_0^{10}10^x\,dx<\frac{10^{10}}{\ln10}\approx 4.34\times 10^9.$

Obviously, one could to better, but it's a start.

Instead of approximating $x^x$ itself by linear interpolation, better is to approximate its logarithm by linear interpolation.  Let $f(x) = x \ln x$.  Then f is convex, and so lies above its tangent line.  Thus
\[f(x) \ge f(10) + f'(10)(x - 10).\]
Exponentiating and then integrating shows that our integral is at least
\[\frac{10^{10} - e^{-10}}{1 + \ln 10} \, ,\]
which is bigger than 3,027,931,065.

For the other direction, let's draw a secant line from x = 9 to x = 10.  Then f will be below it for x between 9 and 10.  Thus
\[f(x) \le (10 - x) f(9) + (x - 9) f(10).\]
Again, exponentiating and then integrating from 9 to 10 shows that the portion of our integral from 9 to 10 is at most
\[\frac{10^{10} - 9^9}{10 \ln 10 - 9 \ln 9}.\]
For x between 0 and 9, we can use the simple bound $x^x \le 9^x$ to show that the portion of the integral from 0 to 9 is at most
\[\frac{9^9 - 1}{\ln 9}.\]
Adding up the two portions shows that our original integral is less than 3,133,284,515.

Using the average of our lower and upper bounds as our estimate, we get an error of less than 2%.

Excellent, Ravi B! It's what I should have been doing all along. Actually, I was trying something like that but had some algebraic mistakes that I hadn't gotten around to fixing.

There's a universal method dealing with such kind of integrals.You must have noticed that the function here has a sharp peak, so the main contribution comes from the region near that peak, we can use asymptotic expansion.
$\int_{0}^{10}x^x dx=\left(\int_{0}^{a}+\int_{a}^{10}\right)e^{x \ln x} dx$, where $a$ is some number to be determined s.t. $\int_{0}^{a}\ll \int_{a}^{10}$. $\int_{a}^{10}e^{x \ln x}dx=\int_{a}^{10}\frac{1}{1+\ln x}d e^{x \ln x}=\frac{e^{x \ln x}}{1+\ln x}\biggl\vert_{a}^{10}+\int_{a}^{10}\frac{e^{x \ln x}}{x\left(1+\ln x\right)^2} dx$
$=\left(\frac{e^{x \ln x}}{1+\ln x}+\frac{e^{x \ln x}}{x\left(1+\ln x\right)^3}\right)\biggl\vert_{a}^{10}+\int_{a}^{10}\frac{\left(4+\ln x\right)e^{x \ln x}}{x^{2}\left(1+\ln x\right)^4} dx=\cdots$
By throwing away all the annoying terms, we can easily make a pessimistic estimation that $\int_{0}^{10}x^x dx\approx \frac{10^{10}}{1+\ln 10}+\frac{10^{10}}{10\left(1+\ln 10\right)^3}\approx3.0557\times 10^9$ will give us the result within 5%. In fact, $ \frac{10^{10}}{1+\ln 10}\approx 3.028\times 10^9$ is good enough, the error is less than 1%. The first two terms give 0.06%.


@ carcul: Please post soluions to your earlier probelms if unanswered questions. Otherwise I am going on a no post strike.  :mad: :rotfl:  :D

Yes i too am of the same opinion  :D   :P

I don't see why Carcul [i]has[/i] to post the solutions to earlier posts. After all, what's the point? We really should be looking up resources (e.g. Wikipedia, Morrison and Boyd etc.) to find methods of solving these problems.

Needless to say, I totally agree with BanishedTraitor. The best I can do is to give some hints, but for that I have to see some work from you.

I agree with madness and chemrock............
PLEASE POST SOLUTIONS CARCUL
  :rotfl:  :rotfl:  :rotfl:

[quote="chemrock"]@ carcul: Please post soluions to your earlier probelms if unanswered questions.[/quote]

I understnd why you think this way (I want this as well) but isn't it ,much better for the sake of learning and sake of the art to explore synthesis and concepts by yourself?

When doing problems in a text book do you just go straight to the answer? No, you would generally try it yourself (or else look up on previous material). In real life, not everything has an answer. It is often about knowing the right things at the right time. :)

actually chemrock didn't know that carcul was a teacher...he thought he was a student like us :)

i too dont know. And it was me who typed from Chemrock's id u know it pardesi  :D  :D  :D

no i didn't :blush:
i thought chemrock was typing of his own from ur house :(

carcul, can you post more problems on predicting the products rather than synthesis. synthesis problems are interesting but are not that helpful for the exams.plz :)
[quote="pardesi"]
no i didn't
i thought chemrock was typing of his own from ur house
[/quote]
why would chemrock be typing from rohit's house

I have already posted some of that type of problems. Search the forum for them.

[quote="VARUNARASU"][quote="pardesi"]
no i didn't
i thought chemrock was typing of his own from ur house
[/quote]
why would chemrock be typing from rohit's house[/quote]

Why shouldnt he???  :D  Chemrock is my friend you know  :P


function
real analysis
topology
real analysis unsolved

are there examples of measurables subsets $S$ of $\mathbb{R}$ and continuous functions $f: \mathbb{R}\to \mathbb{R}$ such that $f(S)$ is non-measurable ?

I hope this is right:

Consider $F$ the standard Cantor-Lebesgue function on $[0, 1]$. Clearly the image of the Cantor set under $F$ is $[0, 1]$. Consider a nonmeasurable subset $N$ of $[0, 1]$ and look at the inverse image of it intersected with the Cantor set. That thing ($S$) is a subset of a measure 0 set so is measurable, but its image under $F$ is $[0, 1] \cap N = N$ which is not measurable.

I think it doesn't work if you only allow absolutely continuous functions.

Clearly, nothing is lost if we replace $f: \mathbb R\to\mathbb R$ with a homeomorphism between two compact intervals (with $S$ being a subset of one of the intervals).

Let $I=[0,1]$, and $A\subset I$ a nowhere dense closed subset of $I$ of positive measure. We can find a homeomorphism $h: I\to J=[0,m(A)]$ which sends $A$ onto a subset of $J$ of measure zero: simply define $h(x)=m((I\setminus A)\cap[0,x]),\ \forall x\in I$. Because $A$ has positive measure, it contains a non-measurable subset $T$. Now take $S=h(T)$ (it's measurable because it has measure zero, since it's a subset of $h(A)$), and $f=h^{-1}$. $f$ is a homeomorphism of $J$ onto $I$ which sends the measurable set $S$ onto the non-measurable set $T$.

Edit:

A bit too late I guess :).

Second edit:

Yes, it's not true if we consider absolutely continuous functions. A measurable set is the union between a $\sigma$-compact set and a set of measure zero, and absolutely continuous functions preserve both these properties of a set (i.e. the property of being $\sigma$-compact and that of having measure zero).


R=circumradius, ra=A-excircle radius.
the projection D of the vertex A on the opposite side BC. Then $ \boxed {R \equal{} r_a\Longleftrightarrow D\in OI}$ .

Construct three excenters $ I_a, I_b, I_c$ and the circumcenter $ O'$ of triangle $ I_aI_bI_c$.
We know that $ IOO'$ is [b]Euler's line[/b] of triangle $ I_aI_bI_c$.
Let $ DO'$ cut $ AI_a$ at $ I'$
We have $ O'I_a\perp BC$, it cuts $ BC$ at $ L$ and $ O'I_a\equal{}2R_{ABC}$
So $ R_{ABC}\equal{}R_a \Leftrightarrow LI_a\equal{}O'L (*)$
But $ AD//O'I_a$ then $ (*) \Leftrightarrow (AJI'I_a)\equal{}\minus{}1$
On the other side, $ (AJII_a)\equal{}\minus{}1$ thus $ (*)\Leftrightarrow I'\equiv I$
We complete the proof.


AMC
USA(J)MO
USAMO
geometry
3D geometry
tetrahedron

Posted this in the PreOlympiad Forum, but I'll post it here too.

2. A tetrahedron has opposite sides equal. Show that all faces are acute-angled.

I think i found a counterexample to this. 0.o
If you draw out the diagram and label congruent sides, you find that the tetrahedron's faces are just congruent triangles.

Well what if it was the congruent triangle with sides 1, 1, and the square root of 3? That's a 120-30-30 triangle, and it seems to contradict the theorem.

Did I mess up somewhere? What's wrong?

Have you ever tried to construct a tetrahedron out of 120-30-30 triangles?

[quote="Math Geek"]lol, Arvind, you just got pwnd.

HINT: Work on your 3D geometry visualization.

HINT2: Never doubt the USAMO writers from 36-37 years ago.[/quote]

False. Intel is inside. Not Linux.

Karthik, why quote my post if it had nothing to do with your "False" statement? :wink:

And anyway, my post was fact.  Linux is inside.  Not Intel. :lol:

oh...crap.

wow. >.<

someone shud put this on failblog, lol.


Solve $\cos 2\theta =\cos\theta+\sin\theta$ for $0\le\theta\le 2\pi$.

[hide=" my solution"]
Square both sides:
$\cos^{2}2\theta = \cos^{2}\theta+2\cos\theta\sin\theta+\sin^{2}\theta$
Pythagorean identity and double angle formulas:
$1-\sin^{2}2\theta = \sin 2\theta+1$
Solve the quadratic:
$\sin^{2}2\theta+\sin 2\theta = 0$
$\sin 2\theta = 0,-1$
$2\theta = 0, \pi , 2\pi , \frac{3}{2}\pi$
$\theta = 0, \frac{\pi}{2}, \pi, \frac{3}{4}\pi$
[/hide]

According to the solutions manual, I'm missing some solutions and $\frac{\pi}{2}$ isn't even a solution! What is wrong with me and this trig stuff that never seemed to give me problems before... Anyone?  :P

You can use $\cos 2\theta=\cos^{2}\theta-\sin^{2}\theta.$
Then factorize! Unit circle gives the solution simply.

Yes very true (didn't think of that one), but despite the fact that my solution may not be the best one, what is wrong with it?

EDIT: Basically, I understand why alternate solutions work, but I don't understand why my solutions don't work. I can't find the flaw in my logic. I know my solution is wrong because it produces answers that don't work, but I don't see why my solution is wrong.

I'm guessing I'm wrong (because I haven't learned trig in school yet, but then again... who learns math in school) but...

How do you go from $cos^{2}{2\theta}$ to $1-sin^{2}{2\theta}$?

One defining idea of trig is that $cos^{2}\theta+sin^{2}\theta = 1$.  This is just Pythagorean Theorem.

:oops_sign: Ha!!! Omg... I can't believe that I just made that mistake... hahaha... I knew that, truly, I did... Ooops... I guess my trig needs some more work (*takes AOPS vol I and II from shelf and starts re-reading*)  Ha... I knew that... Let's just pretend that I never asked that... now you know what happened at AIME...  :P

I can't really see what you did wrong... Hm...

Figured it out with a friend:

You can't just solve for $\sin 2\theta$ and divide the results by 2. The method of solution if I went through this route would have been much more involved than kunny's suggestion (which is the ideal solution). Nobody make the same mistake I did!

ok so in essence, here is what you did that makes your answer wrong

x=1

x^2=1

(x^2-1)=0

(x-1)(x+1)=0

x=-1,1

[quote="max_tm"]$\sin 2\theta = 0,-1$
$2\theta = 0, \pi , 2\pi , \frac{3}{2}\pi$[/quote]
Since you want $\theta$ to be between 0 and $2\pi$, $2\theta$ can be from 0 to $4\pi$.  So you should say $2\theta=0, \pi, \frac{3}{2}\pi, 2\pi, 3\pi, \frac{7}{2}\pi, 4\pi$, which gives $\theta=0, \frac{\pi}{2},\frac{3\pi}{4}, \pi, \frac{3}{2}\pi, \frac{7\pi}{2},2\pi$.

Then you have to check for extranneous roots, because of the squaring.

Altheman is right.
When you square an equation, you create another equation.
So when you get the answers from the squared equation, you have to plug them back to the original equation to check whether they are right.


modular arithmetic
Mandelbrot Competition

This year's Mandelbrot Competition has gotten under way. Good Luck everyone! Also remember not to discuss rounds until the testing window is over.  :)

mandelbrot.org/

can we now discuss the solutions or are people still taking the test.

Why don't you send an e-mail to Sam Vandervelde and ask for clarification?   :roll:

[quote="darkdieuguerre"][quote]"Students should refrain from discussing the contest problems until Monday following the end of the official testing week."[/quote]

Well, the testing week ended on Sunday. Therefore, we wait until the Monday after, which is next Monday, November 16th? It would be nice to have an explicit notification on when we can discuss problems, because there is ambiguity, since most of the testing week was last week.[/quote]

I'm pretty sure the "Monday" being referred to is November 9th. Also, results are posted on the website tomorrow.

Results are up now, you know...

I didn't take round 1, so this is the fourth year in a row I won't be taking all the Mandelbrot rounds.

uhh those definitely aren't all the results... lots of schools are missing. (No Enloe!)

Mandelbrot posts the results at the stated time (or earlier i guess?!). Those are the results. However, not all the schools have submitted their results yet, and the scoreboard will be later updated accordingly.

Go to mandelbrot.org/ and then look on the left side of the site. Under the heading of "Scorecenter", click "Results". There is then a link to national level results, and you can click on the map for regional results.

So how do you solve #7 in a reasonable timeframe?

Doesn't seem like there's a solution yet.

The largest discussion on AoPS so far:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311057

I came up with a casework method during the competition which seemed nice at first but unfortunately requires some stuff which makes it not appealing.

[hide="Solution"]Let each row have $ 1609$ squares and each column have $ 2009$ squares. Set up position $ (a,b)$ for column $ a$ and row $ b$, so they start at $ (1,1)$ and end at $ (1609,2009)$. Now consider the intersection points of red and green $ (a,b)$. The back and forth movement motivates casework on parity.

First we note the following (consider moves as including $ (1,1)$ and $ (a,b)$):
[list]
[*]if $ b$ is odd, then red moves $ (b \minus{} 1)1609 \plus{} a \equal{} 1609b \plus{} a \minus{} 1609$ squares
[*]if $ a$ is odd, then green moves $ (a \minus{} 1)2009 \plus{} b \equal{} 2009a \plus{} b \minus{} 2009$ squares
[*]if $ b$ is even, then red moves $ (b \minus{} 1)1609 \plus{} (1610 \minus{} a) \equal{} 1609b \minus{} a \plus{} 1$ squares
[*] if $ a$ is even, then green moves $ (a \minus{} 1)2009 \plus{} (2010 \minus{} b) \equal{} 2009a \minus{} b \plus{} 1$ squares.[/list]
Now we equate and do ugly stuff. :(

[i]Case 1: $ a$ odd, $ b$ odd.[/i]
\[ 2009a \plus{} b \minus{} 2009 \equal{} 1609b \plus{} a \minus{} 1609\implies2008a \equal{} 400 \plus{} 1608b \\
\implies251a \equal{} 201b \plus{} 50\implies50a\equiv50\pmod{201} \\
\implies a \equal{} 201n \plus{} 1.\]
Because $ a$ is odd, $ a \equal{} 402n \plus{} 1$, so $ b \equal{} 502n \plus{} 1$, for $ 5$ solutions when $ n \equal{} 0,1,2,3,4$ (remember $ 1\le a\le1609$ and $ 1\le b\le2009$).

[i]Case 2: $ a$ odd, $ b$ even.[/i]
\[ 2009a \plus{} b \minus{} 2009 \equal{} 1609b \minus{} a \plus{} 1\implies2010a \equal{} 1608b \plus{} 2010 \\
\implies5a \equal{} 4b \plus{} 5\implies a\equiv1\pmod4 \\
\implies a \equal{} 4n \plus{} 1.\]
Because $ a$ is odd, $ a \equal{} 8n \plus{} 1$ so $ b \equal{} 10n$, and there are $ 200$ solutions for when $ 1\le n\le200$.

[i]Case 3: $ a$ even, $ b$ odd.[/i]
\[ 2009a \minus{} b \plus{} 1 \equal{} 1609b \plus{} a \minus{} 1609\implies1610b \equal{} 2008a \plus{} 1610 \\
\implies805b \equal{} 1004a \plus{} 805\implies805b\equiv805\pmod{1004} \\
\implies b \equal{} 1004n \plus{} 1.\]
Because $ b$ is odd, $ b \equal{} 2008n \plus{} 1$ so $ a \equal{} 1610n$ and there are $ 0$ solutions for this case.

[i]Case 4: $ a$ even, $ b$ even.[/i]
\[ 2009a \minus{} b \plus{} 1 \equal{} 1609b \minus{} a \plus{} 1\implies2010a \equal{} 1610b \\
\implies201a \equal{} 161b\implies40a\equiv0\pmod{161} \\
\implies a \equal{} 161n.\]
Because $ a$ is even, $ a \equal{} 322n$ and $ b \equal{} 402n$ for $ 4$ solutions when $ n \equal{} 1,2,3,4$.

Thus we have $ 5 \plus{} 200 \plus{} 0 \plus{} 4 \equal{} \boxed{209}$ solutions in all.
[/hide]

So many places for screwing up...


linear algebra
matrix
linear algebra unsolved

Suppose we have a diagonal matrix $ A$ with distinct entries $ a_{1}> a_{2}>\dots > a_{N}$, and $ N\times N$ Hermitian matrix $ B \equal{} H*H^{\dagger}$ with N distinct eigenvalue $ b_{1}> b_{2}\dots > b_{N}$. Let $ C \equal{} A*B*A^{\dagger}$ with eigenvalue $ c_{1},c_{2},\dots,c_{N}$. The question is whether exists some relations between the eigenvalue $ a_{i}$, $ b_{i}$, and $ c_{i}$

The problem can actually be casted as the generalized eigenvalue problem, so the question becomes whether exists some identity between the generalized eigenvalue of matrix pensil$ (A^{\minus{}2},B)$and the eigenvalue of A and B


First, $2222^{5555}+5555^{2222}\equiv 3^{5555}+4^{2222}\pmod 7$.

The series of powers of a number modulo 7 is cyclic, with period 6.

For 3: 3, 2, 6, 4, 5, 1, 3, etc.

Since $5555\equiv 5\pmod 6$, $2222^{5555}\equiv 3^{5}\equiv 5 \pmod 7$.

Similarly, $5555^{2222}\equiv 2\pmod 7$.

Because $5+2$ is divisible by 7, the result follows.

This is a fun problem, where did you get it from?


The United Kingdom team at IMO 2005

UNK1 Saul Glasman
UNK2 Nathan Kettle
UNK3 Andre Kueh (IMO S 2004 [Singapore])
UNK4 Matthew Lee
UNK5 Martin Orr (IMO S 2003, B 2004 [UK]; IOI G 2003, S 2004 [both Ireland])
UNK6 Jack Shotton
The very best of luck to all participants.

Wow, Geoff joined the forums!  :D
A long time ago, at that.

[quote="Geoff Smith"]The United Kingdom team at IMO 2005
The very best of luck to all participants.[/quote]It seems Saul was faster at bragging :D http://www.mathlinks.ro/Forum/viewtopic.php?p=247159#p247159 :P
C'ya all in Mexico.

[quote="kueh"]You forgot UNK 0: Newt-on (the newt!)[/quote]Who's he_  :?  :)

Hi UNK 1:7 :)

I just saw the results from the IMO and was very impressed by your high scores. Well done to all of you!

Well, maybe Andre kicks himself because of the 4 missing points on the first problem (day 1) :D, which would have given him the gold and would have brought UK equaly positioned with Germany... but frankly speaking all of you boys have done a very good job and must be proud of yourselves!

Have a safe journey back home, guys.

All the best!


MATHCOUNTS
number theory
least common multiple

If a set of markers is placed in rows of 4 each, there are 2 markers left over; if in rows of 5 each,
there are 3 left over; and if in rows of 7 there are 5 left over.
What is the smallest number of markers that the set could contain?

( Source: Mathcounts '84 )

I'm sorry if this has been posted before..

[hide] Let be $k$ the number of markers. Then,
$k=4a+2=2*(2a+1); k=5b+3; k=7c+5.$

By the first equation we see that $k$ is even. Then, in the second equation, we see that $5b$ finishes with $5$ or $0$, and $k$ finishes with $8$ or $3$. As $k$ is even, $k$ finishes with $8$. Then, $b$ is odd. Looking at the third equation, we see that $7c+5$ finishes with $8$, and, as $c$ is an integrer, $c$ must finish with $9$. Going on, the least value of $c$ is $9$, but with $c=9$, $a=16,5$, but $a$ must be an integrer, so $c$ can't be $9$.So, let's put $c=19$. Then, $a=34$, $b=27$ and $k=138$, so the answer is $\boxed {138}.$ [/hide]

Both of you are correct. :)

[hide="Some input"]
This can easily be solved and visualised by using modular arithmetic.
[/hide]

Or, because of the specific numbers, it can be solved in one line.

[hide]If we had two markers more, then the number of markers would be divisible by $4,5,7$. Hence $N=\text{lcm}(4,5,7)-2=138$[/hide]


calculus
integration
calculus computations

In a [url=http://www.exampleproblems.com/wiki/index.php/IEVS3]page[/url] about integral equations, I've seen the following "fact":
$\int_{a}^{x}\int_{a}^{s}f(y) dy ds = \int_{a}^{x}f(y) (x-y) dy$

I didn't understand how this can be true  :huh: Anyone does?

It's just an application of Fubini's theorem. Switch the order; the inner integral is then trivial. The trickiest part is to figure out what the limits are when we interchange.

I still don't get it. I found the theorem you mentioned at wikipedia, but didn't help much. I tried a partial fraction sequence

$\int_{a}^{x}\left(\int_{a}^{s}f(y) dy\right) ds = \left(\int_{a}^{s}f(y) dy\right) s |_{a}^{x}-\int_{a}^{x}d\left(\int_{a}^{s}f(y) dy\right) s$
$= \left(\int_{a}^{x}f(y) dy\right) x-\int_{a}^{x}f(y) dy s$ (I'm sure sure about the 2nd term)
which would yield

$\left(\int_{a}^{x}f(y) dy\right) (x-s)$

OK, I give up. Can you work it out, apply Fubini's theorem?

[quote="rayne"]$\int_{a}^{x}\int_{a}^{s}f(y) dy ds = \int_{a}^{x}f(y) (x-y) dy$[/quote]

See [url=http://en.wikipedia.org/w/index.php?title=Fubini_theorem&redirect=yes]here[/url] for the statement of Fubini's Theorem.

Let $sOy$ be the plane. We are integrating over a triangle with the vertices $(a,a), \, (x,a), \, (x,x)$. Make the drawing. Now take $Oy$ as the "main axis". $y$ varies between $a$ and $x$, whilst $s$ varies between $y$ and $x$. Thus, \[\int_{a}^{x}\int_{a}^{s}f(y) \, dy \, ds = \int_{a}^{x}\int_{y}^{x}f(y) \, ds \, dy = \int_{a}^{x}(x-y) f(y) \, dy .\]


pigeonhole principle
greatest common divisor
number theory unsolved
number theory

This question has been bothering me:

#2 from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=739486#739486

Prove:
(1) If $2n-1$ is a prime number, then for any n distinct positive integers $a_{1},a_{2},...,a_{n}$, there exist $i,j\in\{1,2,...,n\}$ such that
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}\geq2n-1; \]
(2) If $2n-1$ is a composite number, then there exist n distinct positive integer $a_{1},a_{2},...,a_{n}$, such that for any $i,j\in\{1,2,...,n\}$,
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}<2n-1, \]
where $(x,y)$ denotes the greatest common divisor of positive integer $x,y$.

[color=red]Edit by Megus: added text of problem instead of just a link[/color]

(a) Let $p=2n-1$ and $A=\{a_{1},a_{2},...,a_{n}\}$. WLOG, $p \nmid z$ for some $z \in A$ (otherwise divide all elements of $A$ by a power of $p$). By pigeonhole principle, there are $x,y \in A$, $x < y$ such that $p|y-x$ or $p|x+y$. Suppose $p \nmid x$. Note that $p|y-x \Rightarrow y-x \geq p(x,y)$ and $p|x+y \Rightarrow x+y \geq p(x,y)$, so, in either case $x+y \geq p(x,y)$ and we are done. If $p|x$, $x+z \geq x \geq p(x,z)$.
(b) Let $p|2n-1$, $3 \leq p <2n-1$. The sequence $1,2,3,...,p,p+1,p+3,p+5,...,2n-1-p$ has the desired property.


I'm not so sure...I keep getting different answers for the cases $ \lim_{x\to0^\plus{}}$ and $ \lim_{x\to0^\minus{}}$...

oh... Sorry. Forgot to mention that. :oops:

Separate the cases and solve for both.

I get $ \minus{} 1$ and $ 1$. I'm not sure though.  :maybe:

[quote="math154"]I'm not so sure...I keep getting different answers for the cases $ \lim_{x\to0^ \plus{} }$ and $ \lim_{x\to0^ \minus{} }$...[/quote]
I think you're right! And I guess your approach likes the following:
We have: \limx\to{0}}{\sqrt{\frac{1}{x^2}-\frac{1}{x}}-\sqrt{\frac{1}{x^2}+\frac{1}{x}=lim sqrt(1-x)-sqrt(1+x)/(sqrtx^2)=lim -2x/sqrt(x^2).[sqrt(1-x)+sqrt(1+x)]
Now if x goes to 0 from RHS, then sqrt(x^2)=x.Therefore, limit of the original function in this case is: -2/2=-1
      if x goes to 0 from LHS, the sqrt(x^2)=-x.Therefore, limit of the oringinal function in this case is: 2/2=1
From these two limits, we can conclude that the limit of the original function when x goes to 0 is:[b] DNE[/b]
PS: Sorry if my post looks messy! Can anybody fix my LATEX for me? :oops:

I'll just tidy up your solution, but rephrase it.

$ \lim_{x\to0}{\sqrt {\frac1{x^2} - \frac1x} - \sqrt {\frac1{x^2} + \frac1x}}$

$ = \lim_{x\to0}{\frac {\sqrt {1 - x} - \sqrt {1 + x}}{\pm x}}$

$ = \lim_{x\to0}{\frac{-\frac1{2\sqrt {x + 1}} - \frac1{2\sqrt {1-x}}}{\pm1}}$

$ = \pm2$


linear algebra
matrix
combinatorics proposed
combinatorics

Quite interesting fact ,Which i dont know,who owns this :) :

Let $A_1,...,A_n$ be $n$  finitive and different sets such that :

$|A_i \cap A_j|=1,i \neq j$

Prove that $|\bigcup_{i=1}^{n} A_i|\ge n$

Answer to the next post :

Certainly ,|A| means the number of elements of The SET $A$

[quote="lomos_lupin"]Quite interesting fact ,Which i dont know,who owns this :) :

Let $A_1,...,A_n$ be $n$  finitive and different sets such that :

$|A_i \cap A_j|=1,i \neq j$

Prove that $|\bigcup_{i=1}^{n} A_i|\ge n$[/quote]
what is exactly $|\bigcup_{i=1}^{n} A_i|\ge n$ means ?
can you explain? :D

Here it is (not exactly the same problem; it's dual, i.e. we replace "sets" with "elements" and vice-versa):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Bourbaki&t=32408[/url]

A technical proof, consider the matrix $A$ of $n$ rows and $k$ columns ( $k=|\bigcup_{i=1}^{n}A_i|$), $A_{i,j}= 1$  iff element "$j$" is in  $A_i$ and $A_{i,j}= 0$ otherwise. Now $AA^t=B-I_n$,  where $B$ is the Matrix all filed with $1$. But $Rank (B-In)=n$, hence $rank A \geq n$, so $k \geq n$

Anyway, Grobber, ¿what Bourbaki have to do with that topic?

I am happy, its my first post with some "latex"...

Oops... this only work  when all sets have equal size.... sorry...


group theory
abstract algebra
Euler
number theory
superior algebra
superior algebra solved

please forgive me the utter stupidness of this question but... is this true: the order of an element in a group divides the order of the group?

Yeah, that's true :D Take some element $x$ in the group $G$, consider the subgroup generated by $x$ - it has size $n$, where $n$ is the order of $x$. But Lagrange's theorem tells us that the order of a subgroup divides the order of the whole group. :) So the order of $x$ divides the order of $G$.

an immediate consequence is that if G is a group of order n

a^n is trivial for every element in G

this gives a proof of fermat's little theorem and eulers generalisation


probability
expected value
probability and stats

Hi

I have been trying to understand poisson's distribution...but even after googling a lot and going through books, I have not been clear..

They only define the formula and say arrival of customers is a poisson process and such...

what really categorizes and differentiates poisson's process and what is the main typical speciality of it ??

The Poisson process and the Poisson distribution are two different things, although they are closely related.

Let's start with the Poisson distribution. It arises in the following situation:

1. The random variable is a count. It counts the number of times something happens. As such, it is a nonnegative integer.

2. There is no obvious upper bound, or if there is an upper bound, it is far above any reasonably likely values of the random variable.

3. We know one and only one piece of information, and that piece of information takes the form of a rate - an average rate. The shop sees an average of 6 customers per hour. The raindrops on the sidewalk are distributed as an average of 3.7 drops per square inch. The radiation counter is recording 33 clicks per second. The "hundred year flood" is an event that should happen on average once per century. Furthermore, we'd like for this single piece of numerical information to somehow characterize the entire distribution.

---

If we roll a six-sided die 12 times and count how many times the number 1 appears, then that is a binomial distribution. We have $ 12$ trials, and the probability of "success" is $ \frac16.$ This distribution is a count, and we know its expected value, but we really had two pieces of information: the number of trials, and the probability of success on each trial. There was an upper bound - the variable couldn't have been greater than 12.

For this binomial distribution, we have a mean of $ 2,$ a variance of about $ 1.667,$ and the probabilities of $ Y$ being $ 0,1,2,3$ or $ 4$ are approximately  $ 0.1122,0.2692,0.2961,0.1974,$ and $ 0.0888.$

Now choose $ 2000$ independent random integers between $ 1$ and $ 1000$, and count how many times the choice was the number $ 1.$ This is technically exactly the same as before - a binomial distribution. But the mean is $ 2$ and the upper bound is well out of practical reach. We have a mean of $ 2,$ a variance of $ 1.998,$ and the probabilities of $ Y$ being $ 0,1,2,3$ or $ 4$ are approximately $ 0.1352,0.2707,0.2708,0.1805,$ and $ 0.0902.$

Now, lets say that all we really know is that we did something, and on average, the occurrence we were looking should happen twice during the period we are observing. This could include our first two examples, but we have less information, and no sense that we are making any fixed number of "trials." We're just watching for something to happen, and it could happen at any time. So we model it with a Poisson distribution of mean $ 2.$ We have a mean of $ 2,$ a variance of $ 2,$ and the probabilities of $ Y$ being $ 0,1,2,3$ or $ 4$ are approximately  $ 0.1353,0.2707,0.2707,0.1804,$ and $ 0.0902.$ Notice how close these are to the previous example - that shows how an extremely skewed binomial distribution tends in the limit to a Poisson distribution.

I'll wait a day or so before saying anything about the Poisson process.


a new series of school textbooks,  say   ,  3 volumes.

The   author  is   planning   to   write  the  contents  of  all    the   three  volumes  into   the  contents  of  the   first   volume,   so   the  contents  of  the   first   volume   will  be   like  this:

                                  contents

                        [color=green]volume I[/color]

chapter  1 ............................. [color=red]I [/color] 1
chapter  2 ............................. [color=red]I [/color] 100
chapter  3 ............................. [color=red]I [/color] 200

                           [color=green]volume II[/color]

chapter  1 ............................. [color=red]II [/color] 1
chapter  2 .............................[color=red] II[/color]  100
chapter  3 ............................. [color=red]II [/color] 200

                             [color=green]volume III[/color]

chapter  1 ............................. [color=red]III [/color] 1
chapter  2 ............................. [color=red]III[/color]  100
chapter  3 ............................. [color=red]III [/color] 200

How to  get  this  output?   Packages   titlesec   and   titletoc   will   be   of   any   use?   How  to   insert  the  word "[color=green]volume I,
volume II, volume II [/color]"  into  the  middle  of  the   contents?
Thanks!

For all of your Volumes, use the code:
[code]\section{Volume <insert number here>}[/code]
and for all your chapters, use:
[code]\subsection{Chapter <insert number here>}[/code]
Then at the beginning of your document (or whatever page the contents will be on), type:
[code]\tableofcontents[/code]
You have to compile your document twice before you can see the results.

Thank you very much.
I don't understand you method!

3  volumes  are  3  different   books.

The differences    between  the   first  book  ,   that is    volume    I,    and     a  ordinary book   is  the  contents.

The  first  book's  contents contains    contents  of  3 different books,  volume I, II, III,         but   the content  of  the   volume II,III  is  not   in  the   first book.


Ok, this idea will be in effect now.

[b]Pool A[/b]
ernie

[b]Pool B[/b]
isabella2296

Please sign up here.

i just wanted to note that in the board member discussion thread, you said that the people eliminated from the tournaments will be placed into the pools
so why are we doing this now?

I changed my mind.  I presume you are in a pool.

[b]Pool A[/b]
ernie
vallon22

[b]Pool B[/b]
isabella2296
billybobjoe95

I'll PM you when it starts, k?

[quote="ernie"]Ok.  Since the CTC board members don't care about my idea (and mewto :P), I am putting my idea now.

This is how my idea works.

The first person who signs up (obviously me) will be placed in Pool A.  The second person in Pool B.  The third person in Pool A, etc.

You must play three games with someone in your pool.  After those three games, you may play one "interpool" game, where you can select a player from the other pool to play.  If that selected player hasn't played their three pool games yet, that interpool game will count for one of their three games.  Questions?  Comments?  PM me.

You do not have to be a CTC member to join.[/quote]

I am confused by this. At first you say the intention is to change CTC ratings. Then you open it to non-CTC members. And I do care, I was deliberating and trying to understand what you were asking with your idea. I always thought the pool idea was too confusing, and instead people should be allowed to randomly challenge other CTC members to a rated best 3 out of 5 cds and post it on a specially designed thread (after first tournament is almost over).

you know, i think mewto's idea might be a little better
with his way, you don't have to remember how many games a person played and see if he/she is eligible to play someone out of his/her pool

just a question
for mewto's idea, do both people have to agree to the cd?

I would say both have to agree (unless it is a tourney game).

[quote="ernie"]And I do not see any problem with non-CTC members joining.  That's called [b]exclusion[/b].  My first-grade teacher told me not to do that. :P[/quote]
True, but then people can PM mewto and say they want to be a part of CTC too

Some people want tourneys but not CTC. :P

why? there's nothing wrong with joining CTC. Anyway, by letting non-CTC members in, you are breaking the rules of CTC.

Erm...trying not to spam, but splatyango, don't revive month old threads :D


I put this topic in the NJ forum a couple weeks ago, but apparently noone who does mathcounts in New Jersey visits that forum. So, just as a last resort, does anyone know the average score needed to get to nationals from New Jersey? My scores range from 37 on a bad day to average 39 and 41 on a lucky day.

I average about 39 a round.  Once I got a 42, but I get a 38-39 almost every time.  (Note: I'm in New Jersey, so I'll see you in States (hopefully)
=) )

Well on easy tests, like the school or chapter tests, I'm basically in the 36+ range.
Harder tests...
 :wink:

I only got 41 on a really easy year. I usually get 38 and 39. Does anyone know if NJ is using countdown to decide rankings this year?

[quote="ZzZzZzZzZzZz"]NJ is an above average state, so a 41 should be okay to get you in. However, in my state of Missouri, I got first last year with a 36, but got a 37 at nats.[/quote]

that's a great score. who are you btw?

[url]http://www.mathcounts.org/webarticles/articlefiles/815-Winners_List_2007.pdf[/url]
Look for his state.

In 6th grade, I got something in the lower 30s. But when I'm practicing now I seem to get perfects in chapter. Hopefully I should improve this year with my new AoPS books.

Darkmatter, and anyone else on here from NJ, hi, I'm from NJ too, 7th grader - how 'bout you? What grade(s) are you in?  did you make states or nat last year? most of the people on last yr's team were 8th graders...

I'm only in 6th grade, but I seem to have an idea that since my state (TX) includes First Colony, you need something like a 42 to get to Nats because First Colony has lots of 42+ scorers. I've only been in 2 competitions (and both were minicomps), but I have 37 in one, 31 in the other. In the minicomps, First Colony had 3 44's and a 46!

P.S. My school, TH Rogers, has made it to the state almost every year, but is that just because we were lucky and had a really good person, or that we are a good school?


Counting is as Easy as 24, 28, 32

We count the number of numbers in three different lists in which the numbers change by a number besides 1 each time.

i love these vids  :lol: they teach a lot. for new people.

[quote="nsun48"]For example:

24, 28, 32, 36... 40:

easier way: $ \dfrac{40 \minus{} 24}{4}$ $ \plus{} 1 \equal{} 5$[/quote]

It isn't necessarily easier because it is virtually the same thing, just with the $ 4$ to be divided non-distributed. Dojo's method is probably more intuitive.

Although i was confused at first this was a better video (in my opinion) then the first as it went deeper into the subject :starwars:  :yoda:

I love all these videos.
Not only are they entertaining to watch, they motivate people to learn math.
Especially in this safe, clean, AoPS environment.


search
advanced fields
advanced fields unsolved

Yesterday, I have found on the Internet a strange article about Riemannn Hypothesis on ArXiv. This guy claims that he has proved R-H by quantum knots theory. Here is the artcle:

 http://arxiv.org/PS_cache/math/pdf/0603/0603275.pdf

 I'm waiting for your opinions and comments. :)

Looks like just one more crazy topologist... :roll: If you go to arXiv and do an author search by his name, you'll discover that he also claims having proved the Twin Prime conjecture, the Goldbach conjecture, the Poincare conjecture and having disproved the geometrization conjecture (all by pretty much the same method). My education in the field is not good enough to understand what exactly he talks about in his papers and pinpoint the errors but I just don't believe in such miracles. I'll be most happy to be proved wrong, of course :).


Prove that ${\frac{a^3+abc}{b+c}+\frac{b^3+abc}{a+c}}+\frac{c^3+abc}{a+b}\geq ab+ac+bc$

Forgive me if I make an obvious mistake

$\sum \frac {a^3+abc}{b+c}\geq \frac {1}{3}\left(\sum \frac {1}{b+c}\right)\left(\sum a^3+abc\right)\geq \frac {1}{3}\left(\sum \frac {1}{b+c}\right)\left( \sum a^2(b+c)\right)\geq \frac {\left(\sum a\right)^2} {3}\geq \sum ab$

In order, by Chebyshev + Schur + Cauchy + AM-GM

Wow!!! Nice solutions  :)
$\frac{a^3+abc}{\sqrt{b^2+c^2}}+\frac{b^3+abc}{\sqrt{a^2+c^2}}+\frac{c^3+abc}{\sqrt{a^2+b^2}}\geq \sqrt{2}(ab+ac+bc)$ Nice inequality  :)

aren't they being posted by xtar before?

Slightly shorter solution:

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

by Chebychev, Nesbitt, and Rearrangement/Trivial.

I think the following nice and stronger inequality is valid:

${\frac{a^3+3abc}{b+c}+\frac{b^3+3abc}{a+c}}+\frac{c^3+3abc}{a+b}\geq 2(ab+ac+bc)$.

You are right, xstar.
In that topic, I solved it myself. :P


ratio
number theory proposed
number theory

Let $ f_n\equal{}[2^n\sqrt{2009}]\plus{}[2^n\sqrt{2010}]$
Prove that
1. $ \exists$ infinite $ n$ such that $ f_n$ is even

2. $ \exists$ infinite $ n$ such that $ f_n$ is odd

[quote="Algadin"]Let $ f_n \equal{} [2^n\sqrt {2009}] \plus{} [2^n\sqrt {2010}]$
Prove that
1. $ \exists$ infinite $ n$ such that $ f_n$ is even

2. $ \exists$ infinite $ n$ such that $ f_n$ is odd[/quote]

$ \sqrt{2009\cdot 2010}\notin\mathbb Q$ $ \implies$ $ \sqrt{2010}\minus{}\sqrt{2009}\notin \mathbb Q$ and $ \sqrt{2010}\plus{}\sqrt{2009}\notin \mathbb Q$ $ \implies$ the binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ have infinitely many common digits (in same positions) and infinitely many different digits (in same position).

Hence the result.

[quote="Algadin"]I don't understand  :([/quote]
It's quite useless to say "I dont understand :(". A good reaction would be to say where is the first part you dont understand in order to help poster to explain.

1) $ \mathbb Q$ is the set of all rational numbers. If you dont understand this, just accept : it's the current definition of $ \mathbb Q$
===============================

2) $ \sqrt{2009\cdot 2010}\notin \mathbb Q$
============================
 This means that $ \sqrt{2009\cdot 2010}$ is not a rational number, which means that this number can not be written as a ratio of two integers :
$ 2009\cdot 2010\equal{}2\cdot 3\cdot 5\cdot 7^2\cdot 41\cdot 67$ and it's supposed to be a well known result that a natural number is a perfect square if and only if all its prime divisors has an even power in its prime factor representation.
Q.E.D.

3) $ \sqrt{2010}\minus{}\sqrt{2009}\notin \mathbb Q$
=============================
This means that $ \sqrt{2010}\minus{}\sqrt{2009}$ is not a rational number, which means that this number can not be written as a ratio of two integers.
Suppose that $ a\equal{}\sqrt{2010}\minus{}\sqrt{2009}$ is a rational number.

Then $ a^2\equal{}4019\minus{}2\sqrt{2009\cdot 2010}$ is a rational too (consider $ a\equal{}\frac pq$, and then $ a^2\equal{}\frac {p^2}{q^2}$ and, since the square of an integer is again an integer, you get the result).

Then $ \frac{4019\minus{}a^2}2$ is a rational too (addition, subtraction and division of rationals are rationals).

But $ \frac{4019\minus{}a^2}2\equal{}\sqrt{2009\cdot 2010}$ and $ \sqrt{2009\cdot 2010}$ is not a rational (previous point).

So $ a$ cant be a rational number.
Q.E.D.

4) $ \sqrt{2010}\plus{}\sqrt{2009}\notin \mathbb Q$
==============================
This  means that $ \sqrt{2010}\plus{}\sqrt{2009}$ is not a rational number, which means that this number can not be written as a ratio of two integers.

Suppose that $ c\equal{}\sqrt{2010}\plus{}\sqrt{2009}$ is a rational number.

Then $ c^2\equal{}4019\plus{}2\sqrt{2009\cdot 2010}$ is a rational too (consider $ c\equal{}\frac pq$, and then $ c^2\equal{}\frac {p^2}{q^2}$ and, since the square of an integer is again an integer, you get the result).

Then $ \frac{c^2\minus{}4019}2$ is a rational too (addition, subtraction and division of rationals are rationals).

But $ \frac{c^2\minus{}4019}2\equal{}\sqrt{2009\cdot 2010}$ and $ \sqrt{2009\cdot 2010}$ is not a rational (see point 2 above).

So $ c$ cant be a rational number.
Q.E.D.

5) Binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ have infinitely many different digits in same positions
============================================================================
Suppose the contrary :  binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ have a finite number of positions in their binary representation with different digits .

This would mean that all their digits are the same beyond a given position.

Hence binary representation of $ \sqrt{2010}\minus{}\sqrt{2009}$ is made of zeroes beyond a given position.

So $ 2^k(\sqrt{2010}\minus{}\sqrt{2009})$ is an integer for some $ k$

So $ \sqrt{2010}\minus{}\sqrt{2009}$ is a rational number, which is wrong, according to point 4 above.
Q.E.D.

6) Binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ have infinitely many common digits in same positions
===========================================================================
Suppose the contrary :  binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ have a finite number of common digits in same positions.

This would mean that all their digits are different beyond a given position.

Hence binary representation of $ \sqrt{2010}\plus{}\sqrt{2009}$ is made of ones beyond a given position.

So $ 2^k(\sqrt{2010}\plus{}\sqrt{2009})$ is an integer for some $ k$

So $ \sqrt{2010}\plus{}\sqrt{2009}$ is a rational number, which is wrong, according to point 4 above.
Q.E.D.

7) $ [2^n\sqrt{2009}]\plus{}[2^n\sqrt{2010}]$ is even for infinitely many $ n$
===============================================
Consider a position $ k$ in binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ such that the digits in these two representations are the same.

Then $ [2^k\sqrt{2009}]$ and $ [2^k\sqrt{2010}]$ are both even (if the common digit is $ 0$) or both odd (if the common digit is $ 1$)

In both cases $ [2^k\sqrt{2009}]\plus{}[2^k\sqrt{2010}]$ is even.

And since, according to point 6 above, infinitely many such $ k$ exist, we got the required result.

8) $ [2^n\sqrt{2009}]\plus{}[2^n\sqrt{2010}]$ is odd for infinitely many $ n$
===============================================
Consider a position $ k$ in binary representations of $ \sqrt{2009}$ and $ \sqrt{2010}$ such that the digits in these two representations are different.

Then :
either $ [2^k\sqrt{2009}]$ is even and $ [2^k\sqrt{2010}]$ is odd

either $ [2^k\sqrt{2009}]$ is odd and $ [2^k\sqrt{2010}]$ is even

In both cases $ [2^k\sqrt{2009}]\plus{}[2^k\sqrt{2010}]$ is odd.

And since, according to point 5 above, infinitely many such $ k$ exist, we got the required result.

I'm sorry if I missed some explanation but your "I dont understand :(" is not quite clear for me and i tried to explain each step. Dont hesitate to point to the precise step you still dont understand.

You're welcome. Glad to have helped you.


The real numbers $a_1,a_2,\ldots,a_{100}$ satisfy the relationship
\[ a_1^2+ a_2^2 + \cdots +a_{100}^2 + ( a_1+a_2 + \cdots + a_{100})^2 = 101. \]
Prove that $|a_k|\leq 10$, for all $k=1,2,\ldots,100$.

Hmm.. vinoth didn't you post the solution 11 or 12 times already?
what happened with this site?:):):)

   Palosh alias Downito

_______________________________
"Our days are never coming back... " (Highway Song -> System of a down)

I see this thread has 11 more posts than it should ;)

crap im very sorry
im such an idiot  :blush:  :blush:
my internet connection was very slow and the page didn't load when i pressed submit ... so i think i must have pressed it again not knowing it would actually submit twice , etc.

the server was down for a couple of hours this morning, that is why the site was going slow.

Here is a nice solution for this nice problem.

 Let us assume by contradiction that $|a_{k}| >10,$ for some $k$. WLOG, let $k=1.$ So $a_{1}^{2} >100.$
 and $a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2} <1,$ where $s=a_{1}+a_{2}+...+a_{n}.$
 But, from Cauchy-Schwarz inequality we have:

                                        \[ a_{1}^{2}\leq 100(a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2}) <100. \]
 We get a contradiction. :)

[quote=Cezar Lupu]
 But, from Cauchy-Schwarz inequality we have:

                                        \[ a_{1}^{2}\leq 100(a_{2}^{2}+a_{3}^{2}+\ldots+a_{100}^{2}+s^{2}) <100. \]
 We get a contradiction. :)[/quote]

I think your reasoning was
(1+1+…+1)(a_2^2+a_3^2+…+a_100^2+s^2) >= (a_2+a_3+…+a_100+s)^2 >= (a_1)^2, where the last step follows from |2a_2+2a_3+…+2a_100+a_1| >= |a_1|, but this could not be true if a_2+a_3+…+a_100 has opposite sign of a_1, as it would make the first expression closer to 0 than a_1 is. @above nice solution! A stronger version means (a_100)^2 <= 101/2, or a_100 <= +- sqrt(101/2).


trigonometry
geometry
calculus
calculus computations

1-Lawn Sprinkler : A lawn sprinkler is constructed in such a way that dz/dt (z is the angle of the sprinkler and the ground) constant, where z ranges between 45 degree and 135 degree. The distance the water travels horizontally is
\[ x \equal{} \frac{{v^2 \sin (2z)}}{{32}}
\]

Where v is the speed of the water. Find dx/dt and explain why this model of lawn sprinkler does not water evenly. What part of ther lawn receives the most water?

2-Honeycomb : The surface area of a cell in a honeycomb is
\[ S \equal{} 6hs \plus{} \frac{{3s^2 }}{2}(\frac{{\sqrt 3  \plus{} \cos z}}{{\sin z}})
\]
where h and s are positive constants and z is the angle at which the upper faces meet the altitude of the cell. Find the angle z(30 degree<=z<=90 degree) that minimizes the surface area S

$ \frac{dx}{dt}\equal{}\frac{v^{2}}{32}\left(2\frac{dz}{dt}\cos(2z)\right)\equal{}\frac{v^{2}C}{32}\cos(2z)$

$ \frac{dS}{dt}\equal{}\frac{3s^{2}}{h}\left(\minus{}\sqrt{3}\csc{z}\cot{z}\minus{}\csc^{2}{z}\right)$
Setting to 0,
$ \minus{}\sqrt{3}\csc{z}\cot{z}\equal{}\csc^{2}{z}\Rightarrow\minus{}\sqrt{3}\cot{z}\equal{}\frac{\sqrt{3}\cos{z}}{\sin{z}}\equal{}\csc{z}\equal{}\frac{1}{\sin{z}}\Rightarrow\cos{z}\equal{}\minus{}\frac{\sqrt{3}}{3}\Rightarrow z\equal{}\cos^{\minus{}1}\left(\minus{}\frac{\sqrt{3}}{3}\right)$

Check my math.


What is your favorite subject? Mine is math ;)

gee, i wonder how many people will say math.................. ;)

my favorite is math, of course. second favorite is history or latin

I ofcourse like math, but I like like basically all of my subjects. I don't alsways like english. Science is fun.

I voted on how our classes at school go.  I like math, but in school, I just play with my graphing calculator the whole time.  Or something else.  Anyways, my favorite "subject" at school is recess/lunch.  To bad I can't have it this year   :mad:

... You forget to include one of the most important subjects taught at school.  Physical education.  And yes, thats my favourite subject, considering math class is nothing more than a tragedy of elementary topics taught my teachers who couldnt solve a mathcounts problem if their life depended on it.

this is aops isnt it then whats the point without math

i see that math is prevailing :D lets see how much longer the other subjects can stand it ;)

lol you get the feeling that 14 and 1 aren't the same...

Math has 14 votes.  Vocab has 1.  14 doesn't equal 1. Or is my math wrong :huh: ?

wow..you people are slow

he obviously meant that his favorite subject is a tie between math and vocab

Wow, Math is crushing all its opponents :)

well, no wonder. you wouldn't be on AOPS if you didn't like math. ;)

[quote]you are a genius![/quote]

a genius :rotfl: how am i a genius?

h78reg, this is why you are a geniuos, click to reveal why [hide][hide][hide]i have more hides than andiruh! :)i admire you, you are a professional clicker! [/hide][/hide][/hide]

you do know that you can find the message at the end of a lot of hidden contents just by quoting it in the post reply section? therefore, i'm not a professional clicker. :P

i love science  :D
no more recess... :mad:

[quote="Spencer"]NO DUH many people would like math.
i like it too.[/quote]

Well, not the majority of the world like math.. (not even like 10% lol)

but the majority of AoPS likes math... :D

i like a lot of subjects. math (of course), biology, chemistry, and history are my favorites.

and wow at the life skills option on the poll.


A question asked me to draw a phasor diagram and show that it was in resonance and then asked:
If we increase the capacitance, will the phase angle increase, decrease, or stay the same?

Assume the answer is increase and it is, how would i explain that? What is the explanation? wh would it increase and under what conditions would it do so?

It depends of which phasor are represented; if it's the impedance or the voltage, its OK cause the capacitive reactance goes with the inverse of the capacitance an it's contribution to the phase angle is negative.
Editing: That was for a series circuit , but the final conclusion holds for a parallel as well; capacitive susceptance increases with  increasing capacitance and it's contribution is positive to the phase angle.


Here are the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293724]rules[/url]

We start at:

[b]15[/b]

Have fun!

16 no double post ('cause that can end a game)

16. If there are 2 guys/girls on at the same time, won't that make this game trivial?

15. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1591379#1591379

Please read.

[b]48[/b]

Girls are going down, boys are going up.

49

Dude i did not do that my friend prob.

0 and girls win because my vote is worth 51!!! :D

This thread is over a month old.
Please do not post here.
Thanks.


inequalities
quadratics
inequalities unsolved

I believe it refers to Inequality section, not Algebra. Though...

I don't understand it completely. I have just expanded it in MathCad to get quadratic equation. It cannot be factorized (contains many many terms), so answer is something monstrous.

I think we need a section for BAD problems :)


limit
function
calculus
derivative
real analysis
real analysis unsolved

Let $ L$ be linear isomorphism of space $ R^{k}$ into $ R^{k}$. Prove that $ \lim_{\mathbf{x}\rightarrow \infty}\parallel{}L\mathbf{x}\parallel{}\equal{}\infty$. We define $ \phi(\mathbf{x})\equal{}\frac{\mathbf{x}}{\parallel{}\mathbf{x}\parallel{}^{2}}$ for $ \mathbf{x}\neq 0$ We define also function f as $ f(\mathbf{x})\equal{}\phi(L\phi(\mathbf{x}))$ and $ f(\mathbf{0})\equal{}\mathbf{0}$. For which isomorphisms $ L$ function $ f$ has derivative in $ \mathbf{0}$


Aqui discutiremos los resultados una vez vallan saliendo, ya los equipos estan listos, solo falta ver que tal lo hacen!

Hola Pascual, los resultados de Vzla son

VEN 1 Sofia Taylor        7 1 0 2 0 0
VEN 2 Rafael Guedez    7 0 0 2 0 0
VEN 3 Andres Guzman  7 1 0 3 0 0
VEN 4 Victor Villamizar  2 0 0 2 0 0

Como veras nos regresamos a Vzla con 3 menciones honorificas, gracias por todo y por el entrenamiento, nos vemos!

Los felicito de verdad! estuvieron muy parejos con nuestro equipo: como veran:

COL1 7 1 0 3 1 0 12
COL2 7 1 0 7 0 0 15
COL3 7 1 0 6 0 0 14
COL4 5 1 0 2 0 0 8
COL5 7 1 0 3 1 0 12
COL6 7 1 0 7 0 0 15

Felicitaciones a todos dado que era su primera IMO!!!

En realidad estamos muy contentos con los resultados!

Quién es el que sacó Oro para México???

Resultados argentinos:

ARG1 22 PLATA
ARG2 22 PLATA
ARG3 15 BRONCE
ARG4 7
ARG5 14 MENCIÓN
ARG6 15 BRONCE

Felicitaciones!!!! Un gran resultado, sobre todo si cosideramos que había 5 debutantes.

Bueno a colombia no le fue tan bien... solo dos bronces.... esperamos oros el otro ano... saludos a todos los de la comunidad latina desde slovenia....

Pablo Soberón. Plata en OIM 2005, Bronce en IMO 2005 Oro en OMCC 2004

Gracias.

Un dato curioso: Miren la cantidad de participantes que sacaron oro con 28, pero de la forma 7-7-0-7-7-0 (yo encontré bastantes)

Peru
PER1  7  0	0 7 5  0	:19
PER2  7  1	0 7 0  0	:15
PER3  7  0	0 7 0  0	:14
PER4  7  0	0 7 0  0	:14
PER5  7  0	0 7 0  0	:14
PER6  7  0	0 2 0  0	:9

Me parece que se obtuvo la segunda plata  :wink:

Américo

Buenos dias a todos ..
En primer lugar .. felicitar a cada uno de los chicos q han tomado parte de la IMO este año en Slovenia...

Como tambien a los entrenadores y lideres de cada delegacion que han estado en la IMO 2006 ... y claro tambien a todas las personas que han apoyado a los chicos de manera directa o indirecta para que cada seleccion tome parte de este gran evento ...

Gracias a cada persona que ha colaborado con la comunidad en Español (y portugues, claro esta :) ), con la publicacion de la informacion y los resultados de este evento.

Y claro esta, enviarles un especial saludo a los chicos que han obtenido un desempeño destacado en la IMO. Esperemos que cada vez sean más chicos latinoamericanos que se ubiquen entre los primeros .. , hoy solo fueron unos pocos los que lograron obtener una medalla de oro o de plata o bronce, pero estoy convencido que el nivel latinoamericano puede mejorar si colaboramos y empujamos juntos en una misma direccion q es el de mejorar el nivel de las selecciones de esta region, apoyando las iniciativas de entrenamiento, competiciones locales, nacionales, regionales, publicacion de materiales, difusion de informacion de eventos de otras regiones del mundo, financiamiento de las selecciones, etc.

Saludos,
Carlos Bravo ;)
Lima -PERU

[quote="Dave Vather"]México

MEX1	7-7-0-7-7-0-28 Primer oro Mexicano
MEX2	7-1-0-6-0-0-14
MEX3	2-0-0-6-0-0-8
MEX4	7-7-0-6-4-0-24 4a plata
MEX5	7-0-0-6-7-0-20 5a plata
MEX6	7-1-1-6-1-0-16 29o bronce
	Total	110
Jejeje, les fue genial. El mejor equipo latino si no me equivoco  :D  :D  :D  ¡¡¡Excelente equipo!!![/quote]

Felicitaciones MEXICO, .........
nos han dado una gran leccion de esfuerzo en esta edicion :)

Carlos Bravo :)
Lima -PERU

Hola a todos!

Si, verdad, estoy muy contento con el resultado de los países latinos.

Brasil podría ir un poco mejor (6 bronces es muy bueno y me deja contento, pero era posible ganar plata tambien!), pero tengo mucho gusto en saber que Mexico ganó una medalla de oro y como otros países como Argentina, Colombia (considerando que son estudiantes jovenes!) y Perú fueran bien.

Y Paraguay ganó su primera medalla en la IMO, y una de plata! Portugal ganó algunas medallas también!

Felicitaciones a todos y que sigamos trabajando para el progreso de los países latinos!

PD: perdón por mi español que no hablo tan bien...

El Salvador

SLV1  7 1 0 2 0 0   10
SLV1  7 1 0 2 0 0   10
SLV1  5 1 0 1 0 0   7

Es nuestra segunda participación en IMO ya tenemos 4 menciones,  :P

FELICITACIONES A TODOS LOS EQUIPOS IBEROAMERICANOS
Gran papel el realizado.

En especial para MEXICO ganaron su primer ORO :winner_first:
PARAGUAY su primer medalla una PLATA :winner_first:
y a COSTA RICA su segunda IMO este año su primera medalla un BRONCE :winner_first:


Two ships move with constant speeds and directions. At noon, 2pm, and 3pm the distance between them was respectively 5,7, and 2 miles. What was the distance between them at 1pm?

A)6
B)10
C)$\sqrt{50}$
D) $\sqrt{56}$
E) None of the above

[hide]I made a parabola with those values and guessed; there has to be a better way![/hide]

This method is rigorous, but not necessarily better.
Out at sea, we can assume the water level is constant, so we have a plane the two boats travel on.
As the boats are not accelerating, a generic boat will have an x-distance (from some origin) to be $x = x_0 + V_xt$
Likewise the y-distance will be $y=y_0 + V_yt$
The distance between the two boats along the x-axis will then be written as $\Delta x = \Delta x_0+\Delta V_xt$
Likewise the y separation is $\Delta y = \Delta y_0+\Delta V_yt$
The distance between the boats for a given time, where noon represents $t = 0$
$D(t) = \sqrt{\Delta x_0^2+\Delta y_0^2+t^2(\Delta V_x^2+\Delta V_y^2)+2t(Deltax_0\Delta V_x+Deltay_0\Delta V_y)}$
This notation will be rewritten (in order) as:
$D(t) = \sqrt{K+t^2(V)+tC}$
From the problem:
$D(0) = 5$
$D(2) = 7$
$D(3) = 2$
From this we know:
$K = 25$
$K+4V+2C = 49$
$K+9V+3C = 4$
By black magic, $V = -19$ and $C = 50$
$D(1) = \sqrt{K+V+C} = \sqrt{56}$
Answer choice D.


I haven't checked the solutions, but the official answers (barring appeals) are:

1. B
2. B
3. D
4. B
5. D
6. A
7. A
8. B
9. C
10. C
11. E
12. E
13. A
14. D
15. D
16. D
17. E
18. A
19. E
20. E
21. C
22. D
23. A
24. D
25. A

I'll upload the solutions later today or tomorrow.

Rick Yanco
Worcester Academy

Spent too much time fishing for typos and computing annoying stuff (notably on #12, but for some reason the first page took me like 10 minutes, but also #22, which was just annoying) and got two wrong. And #25 is almost definitely 211, which isn't a choice. But t3h tru3 h4x0rz guessed A anyway.

[quote="CatalystOfNostalgia"]And #25 is almost definitely 211[/quote]

Confirmed.  I'll pass it along.

Yeah, I had trouble with 25's answer choices as well.  I spent about half an hour trying to get one of the answer choices with no luck.  Anyways, 144 for me.  I felt that this MAML was more computational than other ones, with not particularly nice geometry problems.  How'd you guys do?

ah blast, only 100. Did get 25 wrong :( Think I may get to level 2? or not so likely?

argh, 132 (not counting #25) this year for me.

I got 144.
(forgot to change to base 9 on #22)

Wait... what were the cutoffs last year?

i got 122 with 18 right, spent too much time on 22 and should have looked at 24 which i thought was easy

i thought geometry questions 20 and 24 were much easier than 17 although i didn't do any of these three during the test

can someone post a solution for 25 please?

Sry for asking the same thing again but, does anyone know if the choice A for 25 was 231? Cuz I remember i picked 231 but I dont remember the choice letter. Thanks

Yes, Answer choice A is 231.

Last year's cutoff was 105 i believe.

[quote="perfect628"]Darn, I missed 22 evidently (that was just mean), and I misread 24 on the test thinking it was much more difficult than it was, and proceeded to skip it. 140. I might have caught these things had I not spent so long on 25, convinced my answer of 211 was wrong.[/quote]

wait, so does anyone know if they're gonna throw #25 out or add 6 to everyone's score?? A lot of people i talked to are also convinced 211 was the right answer.

I feel like if they threw out question 25, it would be very unfair for people who took the time to solve it and picked A because it was the closest answer to 211 (it seems like most people who got 211 picked A).  I mean, if the question were to be thrown out, that means the people who didn't spend time on it at all will receive the points just like the people who did waste precious time on it...

[quote="baozhale"]how was the cutoff 105? I'm pretty sure that wasn't a possible score last year.[/quote]

The cutoff was 106. just off by 1 point

Can someone PLEASE post a solution to 25 now?

The correct answer to #25 is indeed 211.  The original solution had a typo in it-- :blush: --that was not caught; that typo in turn caused 231 to be a choice and 211 to not be a choice.  Our bad.

It is unclear how the board will handle the scoring; there are good reasons for each of several approaches.

Anyway, here is the solution; the full set will be posted by Mr. Yanco at matharchive.

MM

You people are so lucky! If I asked my teacher to order the MAML, I'd be the only one to take it... :|

FYI, the meeting to set the finalist and semifinalist cutoffs is tomorrow, 12/4.  I'll put the results online tomorrow night after the meeting (once I know the cutoffs).

Rick Yanco

Results are online:  http://www.wocomal.org/statistics/olympiad/200910/200910o_level1.html
(Students from WOCOMAL schools are highlighted, since it's the WOCOMAL website.)

All students were given credit for question 25, for which the answer was not an option.  Two students had perfect papers.  The finalist cutoff was 112 points (115 students).  The semifinalist cutoff was 100 points (an additional 126 students).

As with previous years, some of the names seem wrong to me.  If you spot any errors (and know how to fix them), let me know.

Thanks.

From Roxbury Latin: Edwin Wing should be Edwin Ewing, Christoph Balthazard should be Christopher Balthazard, and Franklin Li is in grade 9.

For Newton South, Kim Jong should be Kim Jongwook

From Boston Latin: Middleton, Christophd should be Middleton, Christopher


I was wondering if anyone had this year's AMC 10 B.  Or, could someone show me where to look for them, since my school doesn't offer the B?

Thanks.


function
trigonometry
real analysis
real analysis solved

let f :[0,1] -->IR a continuos fonction , is there an interval (not impty) so that f is monoton over  it ?

Original questions should not be asked in "Solved Problems". This is where threads get moved, not where they start.

A classic example of a continuous function which is not monotone on any interval: the Weierstrass nowhere differentiable function.

[quote]... the Weierstrass nowhere differentiable function.[/quote]
To clarify the argument: a function which is monotone on an interval must be differentiable "almost everywhere" on that interval. Hence, a function which is nowhere differentiable must have no intervals on which it is monotone.

There are several variations on the exact form of the Weierstrass function. I usually use this:

\[f(x)=\sum_{n=1}^{\infty}(\frac34)^n\cos(4^nx)\]

thank you Jemery and Kent, i know this fonction witch is no where differentiable (but i didn't know that's for Weirstrass ), i know also that a monotone fonction over on interval is differentiable at last in a point on this interval ,but how we can prove that result ?  :oops:

[quote="yassinus"]i know also that a monotone fonction over on interval is differentiable at last in a point on this interval ,but how we can prove that result ?  :oops:[/quote]

Yes. Furthermore, a monotone function is differentiable almost everywhere. (Lebesgue's Theorem) The proof is fairly long and uses the Vitali's Covering Theorem.

can any one told me how to prove that the Weierstrass fonction is not differentiable anywhere , i know that there is a classic proof , but i want special one's :D


inequalities
probability
number theory unsolved
number theory

The next problem, is medium theory of numbers, and medium inequality!!

Find all the pairs (a,b) of positive integers, if
   a^2- 4b and b^2-4a are both perfect squares!!

Is a pretty problem!!

Well we can safely pretend that a < b (if a = b, then it's easy to check). Then we only have a few possibilities:

$b^{2}-4a = b^{2}-4b+4$
or
$b^{2}-4a = b^{2}-2b+1$

Clearly the second one fails by parity, so I guess it has to be the first. Then $a = b-1$, so the second thingy is

$a^{2}-4b = b^{2}-2b+1-4b = b^{2}-6b+1 = (b-3)^{2}-8$

Ehhh also pretty easy to see $b = 6, a = 5$... is this from APMO?

Thanks for your solution probability!
I dont know, maybe is an APMO... but is not difficult....

 :D


vector
combinatorics proposed
combinatorics

Does there exist a convex polyhedron with the property that every plane which does not pass through any of its points but cuts it does so along a polygon with

(a) an even number of sides?

(b) an odd number of sides?

Call $Q$ is the polyhedron, $D$ is the set insists of $Q$'s peaks.
Because of finiteness of lines joining two arbitrary points of the polyheron, there is an unit vector $e$ that never perpendicular to any those lines.
Let the plane $P$ with normal vector is parallels to $e$. Let the plane move in the space such that never change the normal vector. Then at any instant time t, the plane $P$ cut $f(t)$ sides of the polyhedron. Since $P$ moving continuously, and $P$ is contains at most one pont of $D$, then the change of $f(t)$ is only at time it throws one point. The changing is always even. Easily deduce that at every peak of $Q$, there are even side of $Q$.
Easily to show that a, can and b, cant happen


Two hundred fifty eighth-graders took a test. Eighty percent of them passed. One-fifth of those who did not pass received a score below 60. One-half of those below 60 scored below 50. How many
students scored below 50?

To be more clear:

$ \frac{1}{5} \times 250\equal{}50$ did not pass.
$ \frac{1}{5} \times 50\equal{}10$ got less than $ 60$.
$ \frac{1}{2} \times 10\equal{}\boxed{5}$ got less than $ 50$.


When I type
[code][asy]dot((0.5,$\frac{\sqrt{3}}{2}$));[/asy][/code]
it comes out as
[asy]dot((0.5,$\frac{\sqrt{3}}{2}$));[/asy]

What's the problem?

try:

[code][asy]dot((0.5,(sqrt(3))/2));[/asy][/code]

[asy]dot((0.5,(sqrt(3))/2));[/asy]

I think that LaTeX is only used for displaying, so it is only used in labeling points. For drawing points, you just use (2)/(3) for fractions and sqrt(2) for square roots.

Does anyone know how to label angles, like a small little arc to show that two angles are congruent?  Also, how do I label angle measures without making the font really tiny, because numbers don't really wanna fit into such a small place..

For the angle measure thing, I just use arcs. There might be some command that allows you to do that. However, it would not be any easier than making the arcs themselves, for it will require as many arguments as the Arc() command uses.

As for the labeling angle thing, use arrows.

Ok, then how do you make arcs?  And arrows?

Could you label angles BAE and CDE congruent, and mark angle AED 60 degrees?  Also, mark AB, BC, CD, and AD congruent.  Just so I can see how it's all done..

Arcs:

[asy]draw(Arc(origin, 5, 120, 180), Arrows);[/asy]

[code]draw(Arc(origin, 5, 120, 180), Arrows);[/code]

Arrows obviously makes 2 arrows and both ends. You make want to see the asy guide which explains arrows much better than I do.

In the arcs command, the first argument decides the center of the "circle," and the second is the radius (just like [b]Circle()[/b] command). The third and fourth decide the endpoints in terms of degrees. If you want to make an anlge under 180 from 350 to 10 degrees, be sure you use -10 degrees instead of 10 degrees. Did I say that very clearly?

For congruent angles, you can just use two arcs for now.

I nevver bothered with congruent sides, so you might want to ask Ph Ivaldi or fedja about that.

I also know there is a command [b]rightanglemark()[/b] that obviously makes right angles, but I never figured out how they work. Hope that helped!

[asy]import markers;
import geometry;

size(8cm);
draw(unitsquare, dot);
pair Ep=(0.5,(sqrt(3))/2);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0);
dot(Ep);
draw(A--Ep--D^^Ep--B^^Ep--C);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", Ep, N);

pen RedPen=0.8*red;
marker cong=marker(markinterval(1,stickframe(2,4mm,RedPen),true));

real radius=15mm;
markangle(radius=radius,A,Ep,D,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,D,A,Ep,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,Ep,D,A,ArcArrow(2mm), RedPen, cong);

markangle("${"+(string)(degrees(B)-(360-degrees(Ep-B)))+"}^\circ$",
          radius=radius,A,B,Ep,ArcArrow(2mm), green);

perpendicular(A,NE,D,size=radius/4, 1mm+blue);[/asy]

[code][asy]
import markers;
import geometry;

size(8cm);
draw(unitsquare, dot);
pair Ep=(0.5,(sqrt(3))/2);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0);
dot(Ep);
draw(A--Ep--D^^Ep--B^^Ep--C);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", Ep, N);

pen RedPen=0.8*red;
marker cong=marker(markinterval(1,stickframe(2,4mm,RedPen),true));

real radius=10mm;
markangle(radius=radius,A,Ep,D,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,D,A,Ep,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,Ep,D,A,ArcArrow(2mm), RedPen, cong);

markangle(radius=radius,B,A,Ep,ArcArrow(2mm), BluePen, cong);
markangle(radius=radius,Ep,D,C,ArcArrow(2mm), BluePen, cong);

markangle("${"+(string)(degrees(B)-(360-degrees(Ep-B)))+"}^\circ$",
radius=radius,A,B,Ep,ArcArrow(2mm), green);

perpendicular(A,NE,D,size=radius/4, 1mm+blue);
[/asy][/code]

When I do that, it comes like this:

[asy]import markers;
import geometry;

size(8cm);
draw(unitsquare, dot);
pair Ep=(0.5,(sqrt(3))/2);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0);
dot(Ep);
draw(A--Ep--D^^Ep--B^^Ep--C);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", Ep, N);

pen RedPen=0.8*red;
marker cong=marker(markinterval(1,stickframe(2,4mm,RedPen),true));

real radius=10mm;
markangle(radius=radius,A,Ep,D,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,D,A,Ep,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,Ep,D,A,ArcArrow(2mm), RedPen, cong);

markangle(radius=7,B,A,Ep,ArcArrow(2mm), BluePen, cong);
markangle(radius=7,Ep,D,C,ArcArrow(2mm), BluePen, cong);

markangle("${"+(string)(degrees(B)-(360-degrees(Ep-B)))+"}^\circ$",
radius=radius,A,B,Ep,ArcArrow(2mm), green);

perpendicular(A,NE,D,size=radius/4, 1mm+blue);[/asy]

All the code under the 3 congruent angle markers doesn't work...

Also, when I tried to add another perpendicular, it came out as unparseable code...

Help!

If you install Asymptote on your hard disk and compile yourself the code, you obtain this message:
[quote]file.asy: 438.47: no matching variable 'BluePen'[/quote]
In my code I have defined the pen RedPen like this:
[code]pen RedPen=0.8*red;[/code]
I let you imagine how define BluePen

[asy]import markers;
import geometry;

size(8cm);
draw(unitsquare, dot);
pair Ep=(0.5,(sqrt(3))/2);
pair A=(0,0), B=(0,1), C=(1,1), D=(1,0);
dot(Ep);
draw(A--Ep--D^^Ep--B^^Ep--C);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", Ep, N);

pen RedPen=0.8*red;
marker cong=marker(markinterval(1,stickframe(2,4mm,RedPen),true));

pen BluePen=0.8*blue;
marker acong=marker(markinterval(1,stickframe(1,4mm,BluePen),true));

real radius=10mm;
markangle(radius=radius,A,Ep,D,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,D,A,Ep,ArcArrow(2mm), RedPen, cong);
markangle(radius=radius,Ep,D,A,ArcArrow(2mm), RedPen, cong);

markangle(radius=12mm,B,A,Ep,ArcArrow(2mm), BluePen, acong);
markangle(radius=12mm,Ep,D,C,ArcArrow(2mm), BluePen, acong);

markangle("${"+(string)(degrees(B)-(360-degrees(Ep-B)))+"}^\circ$",
radius=radius,A,B,Ep,ArcArrow(2mm), green);

perpendicular(A,NE,D,size=radius/4, 1mm+blue);[/asy]

I feel so smart :D

How did you get angle ABE to say $ 75^\circ$?


function
inequalities proposed
inequalities

X,Y,Z aare positive real numbers and XYZ=1

Find the extreme values of

(X+1)/(XY+X+1) +(Y+1)/(YZ+Z+1) +(Z+1)/(ZX+Z+1)

it seems there is no extremum...
but the infimum is 1.

It's interesting because I found that this expression always equals . Put $ x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c} $. Probably I'm wrong.

ah..there must be some serious mistakes in my calculations...  :(

Another interesting problem of finding the extreme value of a [b]constant[/b] expression is found here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11924


analytic geometry
calculus
derivative
search
calculus computations

A particle moving along a curve in the xy-plane has position (x(t), y(t)) at time t>0 where

dx/dt = ((6/t)-3)^(1/3) and dy/dt = te^-t

At time t = 3, the particle is at the point (5,4)

Is there a time t at which the particle is farthest to the right? If yes, explain why and give the value of t and the x-coordinate of the position of the particle at that time. If no, explain why not.

Describe the behavior of the path of the particle as t increases without bound.

Farthest to the right means that x-coordinate is at a maximum.  In other words, set the derivative of x with respect to t to zero and search for a critical time as you would normally.

yea i did that but how can you tell if its farthest left or right?
would  you have to do the intervals and see if its positive/negative  or negative/positive?


logarithms
number theory open
number theory

I know there are estimates of number theoretic sums such as $ \sum_{p \le x} \log p$ and the likes. I tried to estimate $ \sum_{p \le x} \frac {1}{\log p}$ and it got nasty. Are there any known bounds for it? (hopefully using something simple, like considering a binomial coefficient or double counting...)

It seems a very interesting question. I guess a nice formula can be obtained by Selberg's sieve method, but I didn't manage to find anything. I would be happy if you write down your 'nasty' conclusions.

I find following formula by using Partiol summation and The prime number theorem: $ \sum_{p\le x}\frac{1}{\log p} \equal{} x/(log x)^{2} \plus{}O(x/(log x)^{3})$


linear algebra
matrix
algebra unsolved
algebra

we have a matrice with n rows and m columns such that each entry is a nonnegative real and there is a positive number in each row and each column.if the intersection of a row and a column is positive,then the elements in the row and the column  has equal sum.prove that m=n.

Partition the positive elements of the matrix into disjoint subsets using the relation 'is in the same row or column as.'  For a subset S define a path through S as a sequence of moves between positive elements in the same row or column.  If a path exists between two elements, the totals for each of the rows and columns included in the path must be equal.  Call this total T.  Define A = number of distinct rows that some element of S is present in and B = number of distinct columns.   Then for every subset S, we require for the sum of all the elements, AT = BT, so that A = B.  Since the sets S constitute a partitioning, and each set contributes the same number of rows and columns to the matrix as a whole, it follows that the union of all the S is itself square.


invariant
topology
advanced fields
advanced fields unsolved

From the article [url=http://terrytao.wordpress.com/2008/01/10/254a-lecture-2-three-categories-of-dynamical-systems/]Three categories of dynamical systems[/url].

[b]Lemma 1.[/b] Every topological dynamical system $ (X,\mathcal{F},T)$ contains a minimal dynamical system (i.e. minimal nonempty closed invariant subset) (assume $ (X,\mathcal{F})$ is compact metric space and $ T: X\to X$ is a homeomorphism.). It is easily proved with Zorn's lemma.

[b]Exercise 2.[/b] Every compact metric space is second countable and thus has a countable base. Suppose we are given explicit enumeration $ V_1, V_2, \ldots$ of such a base. Then find a proof of Lemma 1 which avoids the axiom of choice.

Let $ W_i \equal{} \cup_{k\equal{}\minus{}\infty}^{\infty} T^k V_i$ for each $ i$=1,2,3,...

If $ A$ is a non-minimal invariant closed subset of $ X$, there is some $ V_i$ that can be used to carve out $ A$ to give a smaller non-empty closed invariant subset, i.e. there is $ i$ such that $ A \setminus W_i$ is nonempty (and thus $ A \setminus W_i$ is a smaller nonempty closed invariant subset.

A minimal one would be somehow a fixed point of this carving out operations, or the result of iterations of these operations. The construction of a minimal system should involve carving out with $ V_i$ for all $ i$.

Define $ X_0 \equal{} X$ and
$ X_i \equal{}
\begin{cases}
  X_{i\minus{}1} & \text{ if \(X_{i\minus{}1} \subseteq W_i\)} \\
  X_{i\minus{}1}\setminus W_i & \text{ otherwise}
\end{cases}$

$ X_i$ are nonempty closed invariant subsets of $ X$. $ X_\infty \equal{} \cap_{i\equal{}1}^{\infty} X_i$ is nonempty because $ X$ is compact. For all $ i$ we have either $ X_\infty \subseteq W_i$ or $ X_\infty \cap W_i \equal{} \emptyset$. Therefore $ X_\infty$ is minimal.


Suppose $  K \in {\mathbb N}\;  $  , $ K= \overline{ a_0a_1a_2\ldots a_n} \; $  be written in base $ 10 . $ Find the best constant  $ \displaystyle {\mathbf c}_{\mathbf n} \; $ in inequality
\[ \ds  K \le\displaystyle  {\mathbf c}_{\mathbf n}\cdot \ds
\sum\limits_{k=0}^n a_k \; \; .\]

I really think that there is a type in the statement, because in this form the problem is trivial.

i think there is some mistake there in the $C_n$
bye

I try to change the original text of proposed question. Thanks, Alex


[quote="scpajmb"]
\[ \begin{array}{l} {\rm{Let }}x{\rm{,y,z,}}\lambda {\rm{,}}\mu {\rm{ > 0, if}}\frac {\lambda }{\mu } \in (\frac {1}{8},8),{\rm{then}} \\
 \\
\sqrt {\frac {x}{{\lambda y \plus{} \mu z}}} \plus{} \sqrt {\frac {y}{{\lambda z \plus{} \mu x}}} \plus{} \sqrt {\frac {z}{{\lambda x \plus{} \mu y}}} > 2\sqrt [4]{{\frac {1}{{\lambda \mu }}}}{\rm{ }}.....(1); \\
 \\
{\rm{if}}\frac {\lambda }{\mu } \in (0,\frac {1}{8}] \cup [8, \plus{} \infty ),{\rm{then}} \\
 \\
\sqrt {\frac {x}{{\lambda y \plus{} \mu z}}} \plus{} \sqrt {\frac {y}{{\lambda z \plus{} \mu x}}} \plus{} \sqrt {\frac {z}{{\lambda x \plus{} \mu y}}} \ge \frac {3}{{\sqrt {\lambda \plus{} \mu } }}.....(2) \\
\end{array}
\]
[/quote]
Try $ \lambda\equal{}1,$ $ \mu\equal{}3$ and $ x\equal{}y\equal{}z\equal{}1.$ :wink:


Let $ a, b$ and $ c$ be positive real numbers. Prove that: $ (\frac{a}{a\plus{}2b})^2\plus{}(\frac{b}{b\plus{}2c})^2\plus{}(\frac{c}{c\plus{}2a})^2\geq \frac13$

we have $ 3\sum_{cyc}\left(\frac{a}{a\plus{}2b}\right)^2\ge\left(\sum_{cyc}\frac{a} {a\plus{}2b}\right)^2$.

Hence it remains to show that $ \sum_{cyc}\frac{a}{a\plus{}2b}\ge 1$

By Cauchy-Schwarz inequality $ \sum_{cyc}\frac{a^2}{a^2\plus{}2ab}\ge\frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca} \equal{}1$

This was posted a number of times before I'm sure.


Let circle O1 and circle O2 be tangent internally to circle O at A.B, respectively. Line g1 and line g2 are internal tangent of circle O1 and O2. Line g3 is external tangent of circle O1 and O2. The circle O1,O2 touches g3 at D,E, respectively. AD intersects g2 at P, BE intersects g1 at Q. Prove that PQ is parallel to g3.

Sorry if I am mistaken, but it seems by the drawing, that the problem is not true.

Kostas Vittas.

[quote="vittasko"], but it seems by the drawing, that the problem is not true.

Kostas Vittas.[/quote]
I don't think so
why you say the problem not true??? :roll:  :roll:

I'm pretty sure it's incorrect (see image, please click twice)
[img]6906[/img]


find, with proof, the number of positive integers \leq{2006} that can be written as the sum of three distinct squares.

Will it be too difficult?
I think we have first to get the number of solutions to the general case $x^2+y^2+z^2=k$, where x,y,z and k are positive integers.
But this question is already not easy  :!:
Maybe someone knows other advanced technique in Number Theory to handle this problem?

Well, the discussion of this problem is already on the Intermediate forum...

For me I just brute forced it..

Do you mean using computer programming?
I just got the result of 8008 by mathematica. Is it the correct answer?

There aren't 8008 integers less than or equal to 2006 though...
Maybe if you try mod 4, it would only work for numbers congruent to 0 or 3 (mod 4) where the numbers on the left are all congruent to 0 or 1 respectively :idea:  :huh:

Sorry, the number of solutions for distinct values of x, y and z should be 41598 :!:
I forgot to set the constraint of "distinct" and forgot to permute the solutions.  :blush:

[quote="t0rajir0u"][hide="Hint"]
Try $\bmod 8$.
[/hide]

[hide="Answer without proof"]
Precisely the numbers of the form $4^k (8n + 7)$ [i]cannot[/i] be written as the sum of three squares.  (This is given that $0$ is a square.)
[/hide][/quote]

the problem talks about distinct squares, i thought of that question too!, (i posted that question a while ago, although it has been posted many times)

that reasoning seems like it would still work though

[quote="t0rajir0u"]
Precisely the numbers of the form $4^k (8n + 7)$ [i]cannot[/i] be written as the sum of three squares.  (This is given that $0$ is a square.)
[/quote]

Is there any link to the proof of this fact?

Any proof of this fact (the Three-Squares-Theorem) requires harder theorems like Dirichlet's theorem on primes in arithmethic progression.
One attack to this kind of problems would be to find the number of representations as sum of three squares (can be e.g. be computed by class numbers) and the representations as $x^2+2y^2$ (can be computed by some sum over the divisors/prime factors of the number), because the second one counts the number of representations where not all numbers are different. Indeed both is not trivial to calculate.

The easiest way seems to be to calculate all such sums by computers (should need as good as no time)...


Let $ S\equal{}\{1,2,3,\ldots,11\}$. How many subsets of $ S$ have two elements that are relatively prime?

We count the number of sets with no pair of elements relatively prime.

First, there is the set $ \{1\}$, and $ \varnothing$.
Next, we count all subsets of the set $ \{2,4,6,8,10\}$, of which there are $ 2^5$ subsets. However, we have counted the empty subset, so there are in fact $ 2^5 \minus{} 1 \equal{} 31$ for this case.
Then, we count the subsets of $ \{3,6,9\}$--$ 2^3 \minus{} 1 \equal{} 7$. We have already counted $ \{6\}$ as well, so it is $ 6$.
$ \{5,10\}$--$ 2^2 \minus{} 1 \equal{} 3$. We have already counted $ \{10\}$, so it is $ 2$.
$ \{7\}$--$ 1$.
$ \{11\}$--$ 1$.

There are $ 2 \plus{} 31 \plus{} 6 \plus{} 2 \plus{} 1 \plus{} 1 \equal{} 43$ bad sets, for $ 2^{11} \minus{} 43 \equal{} \boxed{2005}$ good sets.


inequalities
quadratics
function
limit
triangle inequality

If $ x,y,z>0$ and $ x\plus{}y\plus{}z\equal{}4$, what is the maximum value of $ xy\plus{}yz\plus{}xz$?

You can use the inequality $ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx$.

[hide]Well, we can see that:

$ (x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} xz \plus{} yz)$

$ xy \plus{} xz \plus{} yz \equal{} \frac {16 \minus{} (x^2 \plus{} y^2 \plus{} z^2)}{2}$

So all we need to do is minimize $ x^2 \plus{} y^2 \plus{} z^2$

We see that $ x^2 \plus{} y^2 \plus{} z^2\ge3\sqrt [3]{x^2y^2z^2}$ by AM-GM.

So the minimum for $ x^2 \plus{} y^2 \plus{} z^2$ is $ 3\left(\frac {4}{3}\right)^2 \equal{} \frac {16}{3}$.

Since the maximum of $ xy \plus{} yz \plus{} xz$ is equal to the minimum of $ x^2 \plus{} y^2 \plus{} z^2$, the answer is $ \frac {16}{3}$[/hide]

I don't quite understand what you're doing when you minimize $ x^2 \plus{} y^2 \plus{} z^2$ by AM-GM -- to me, it would seem more natural if you had used RMS-AM. In particular, where did you get information about the value of $ xyz$?

In any case, after your first line, you can say:

$ xy \plus{} xy \plus{} yz \leq x^2 \plus{} y^2 \plus{} z^2$

$ \therefore 3(xy \plus{} xy \plus{} yz) \leq 2(xy \plus{} xz \plus{} yz) \plus{} (x^2 \plus{} y^2 \plus{} z^2) \equal{} 16 \implies \boxed{xy \plus{} xy \plus{} yz \leq \frac {16}{3}}$

You get the information about xyz because AM-GM has equality when x=y=z.

I understand -- it just seems a bit slimy to me because on one hand you're maximizing $ xyz$ based on your knowledge of $ x \plus{} y \plus{} z$, and on the other hand, minimizing $ x^2 \plus{} y^2 \plus{} z^2$ with given $ xyz$, which you're in turn maximizing given $ x \plus{} y \plus{} z$ via another AM-GM.

Maybe I'm just being slow -- ignore me if I am. [i](but make sure I am before doing so)[/i]

[quote="Carbon57"]Well, we can see that:

$ (x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} xz \plus{} yz)$

$ xy \plus{} xz \plus{} yz \equal{} \frac {16 \minus{} (x^2 \plus{} y^2 \plus{} z^2)}{2}$

So all we need to do is minimize $ x^2 \plus{} y^2 \plus{} z^2$ [/quote]

which follows by Cauchy-Schwarz, $ 3(x^2 \plus{} y^2 \plus{} z^2)\geq (x \plus{} y \plus{} z)^2$. :lol:

Remark that $ \boxed{3(x^2 \plus{} y^2 \plus{} z^2)\geq (x \plus{} y \plus{} z)^2\geq 3(xy \plus{} yz \plus{} zx)}$, which is very useful.

Which is identical to $ 9[\text{RMS}]^2 \geq 9 [\text{AM}]^2 \iff \text{RMS} \geq \text{AM}$  :lol:

I personally wrote this problem with RMS-AM in mind  :lol: .

RMS, or [b]root-mean-square[/b], denotes the quadratic mean $ M_2(x_1, x_2, ... x_n) \equal{} \sqrt{ \frac{x_1^2 \plus{} ... \plus{} x_n^2}{n} }$.

RMS denotes the root of the mean of the squares. The RMS-AM inequality states that for positive reals $ x_1,x_2\text{...}x_n$

$ \sqrt {\frac {x_1^2 \plus{} x_2^2 \plus{} \text{...}x_n^2}{n}}\ge\frac {x_1 \plus{} x_2 \plus{} \text{...}x_n}{n}$

Thank you!  :) Now I have 6 tools of solving Ineqaulity, AM-GM-HM, RMS, Cauchy-Schwarz, Jensen, Triangle Inequality, Tchevichev.

Do I need to acquire Schur, Muirehead?

How do you pronounce them?

I am no good at inequalities, but I know that instead of Muirhead, you can use AM-GM many times.

RMS-AM-GM-HM are special cases of the Power Mean Inequality, which states that

$ (\frac{1}{n}\sum_{i\equal{}1}^{n} k_{i}^{r})^{\frac{1}{r}}$ is a decreasing function of $ r$.

Of those four, the GM may seem a bit out of place. An interesting thing is to try to show that:

$ \lim_{r \to 0} \sqrt [r]{\frac {1}{n} \left( \sum_{i \equal{} 1}^n k_i ^r \right) } \equal{} \sqrt [n]{\prod_{i \equal{} 1}^n k_i } \equal{} \text{GM}[k_i]$

That is, for the purposes of the Power Mean Inequality, the geometric mean takes the role of the power mean with exponent 0. This is sometimes incorporated into the definition of the inequality.

See my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117509&search_id=1260662465]here[/url].


geometry
3D geometry
pyramid
Pascal\u0027s Triangle

How is it that the n-th term of the m-th row of Pascal's triangle comes out to be $ \binom{m}{n}$? I know how Pascal's triangle may be used in this way, but I still don't see how a number pattern made with unrelated methods (adding the two above to get the one below) yields the same pattern as the pattern of number of ways of choosing a certain number of things from a group.

Can someone explain the connection?

The pascal's triangle is defined so that each number is the sum of the two above it.  Then we need to show that
\[ {n\plus{}1\choose k}\equal{}{n\choose k}\plus{}{n\choose k\minus{}1}
\]
which can be shown by brute force.

[quote="jli"]The pascal's triangle is defined so that each number is the sum of the two above it.  Then we need to show that
\[ {n \plus{} 1\choose k} \equal{} {n\choose k} \plus{} {n\choose k \minus{} 1}
\]
which can be shown by brute force.[/quote]
What? Have I been called here? :P
[hide="Brute Force"]$ RHS\equal{}\frac{n!}{k!(n\minus{}k)!} \plus{} \frac{n!}{(k\minus{}1)!(n\minus{}k\plus{}1)!}\equal{}\frac{n!(n\minus{}k\plus{}1)}{k!(n\minus{}k\plus{}1)!}\plus{}\frac{n!k}{k!(n\minus{}k\plus{}1)!}\equal{}\frac{n!(n\minus{}k\plus{}1\plus{}k)}{k!(n\minus{}k\plus{}1)!}\equal{}\frac{(n\plus{}1)!}{k!((n\plus{}1)\minus{}k)!}\equal{}\binom{n\plus{}1}{k}\equal{}LHS$[/hide]
[hide="Fine, prettier method"]Consider $ n\plus{}1$ marbles, from which we choose $ k$ of them. Let $ n$ of the marbles be boring dull gray marbles, and let $ 1$ marble be a cool shiny one. Now we can either choose only from the dull marbles, in which case there are $ \binom{n}{k}$ ways to choose it. Or we can choose the shiny marble and then $ k\minus{}1$ marbles, in which case there are $ \binom{n}{k\minus{}1}\cdot\binom{1}{1}\equal{}\binom{n}{k\minus{}1}$ ways to choose. Adding these up gives the total amount of combinations possible, which is $ \binom{n\plus{}1}{k}$.[/hide]

Ok, this is pretty late a reply, but I feel that no one really explained the nice, intuitive, way. (Everyone just proved Pascal's identity, but here is how I view this connnection):

Each number in Pascal's triangle represents the number of ways to "travel" from the top of the pyramid to the position of that number if you only go down-left and down-right. For example, to get the the number 10 (the left one, so $ n\equal{}2$ since the left most term is $ n\equal{}0$, and not $ n\equal{}1$) on the row where the second entry is 5 (the 5th row, so $ m\equal{}5$, one possible way to reach the 10 is left, left, left, right, right (I did not write out the down's for brevity)( this should go from the top 1 to 1,1,1,4,10, and as another example, left,right,left,right,left should go from 1 to 1,2,3,6,10).
However, note that we can model this as LLLRR. Thus the problem becomes the number of ways to arrange the letters of LLLRR. You can think of this as the number of ways to CHOOSE 2 positions out of 5 positions (_ _ _ _ _) to places the the 2 L's. Thus, the number of ways to get to the left 10 on the 5th row is $ \dbinom 52$  since we are choosing 2 positions out of 5.

Now, of course, this is quite easy to extend to $ \dbinom mn$.

Also, this is a great explananation of why $ \dbinom mn \equal{} \dbinom {m}{m\minus{}n}$ since, for example, in the previous example, there are clearly an equivalent number of ways to 2 positions to place the R's as there are to choose 3 positions to place the L's, so $ \dbinom 53 \equal{} \dbinom 52$.

Thinking about the number of paths it takes to reach a certain number on the pascal triangle is another way to prove Pascal's identity (which is the identity Brute Force proved above, by brute force). Let's look at a specific case first again. Pascal's Identity let's us say that $ \dbinom 41 \plus{} \dbinom 42 \equal{}\dbinom 52$. This is clear, since all 10 paths to the $ \dbinom 52\equal{}10$ must pass through the two entries above it, which are $ \dbinom 42$ and $ \dbinom 41$.

We can generalize quite easily again.

I guess Pascal's Identity can be used as an explanation to Text's problem, if you notice that the first two rows are $ \dbinom 00$ and $ \dbinom 10, \dbinom 11$, respectively, and you can build the triangle off of this row one (the one with two entries, not row 0), if you pretend that to the outside of the triangle (beside the outer 1's) are $ \dbinom m0$ and $ \dbinom mm$ for the mth row.


Let $ f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ satisfying that:
1)$ f(x_1)>f(x_2)$ when $ x_1>x_2$
2)$ f\Big(f(n)\Big)\equal{}kn$
prove that:
$ f(n)\geq\frac{2kn}{k\plus{}1}$

from $ 2)$ $ kf(n)\equal{}fofof(n)\equal{}f(kn)$ and $ f(f(n))\equal{}kn$
from $ 1)$ we see that if $ x>y$ then $ f(x)\minus{}f(y)\ge x\minus{}y$
we take $ x\equal{}kn$ and $ y\equal{}f(n)$
$ f(kn)\minus{}f(f(n))\ge kn\minus{}f(n)\iff kf(n)\minus{}kn\ge kn\minus{}f(n)\iff f(n)\ge \frac{2kn}{k\plus{}1}$