State your thoughts.  You start the discusssion

Global politics so what do we believe? Well I can tell u I firmly believe in democracy and I see the why there is a monarchy but I dont understand why there is a ditartorship. I mean thats like liveing in a war based country that all u do in ur life is train to fight and for what. Its like they are asking to start or get involved in a war. I dont agree with it but that isnt going to help therfore I have to do is accept it. A monarchy I dont totally and completely agree with but its better than a dictatorship. I believe it should be the peoples vote that counts thats why I like a democracy. Now dont ask me to choose between being a democrat, republican, or whig because I refuse to. Im an optimist when it comes to national politics. I will support the president no matter what even though I may not like what hes doing or who he is but I will tell u I tend to favor the republican point of view. The fact is when it comes to politics Im not much of a talker because almost no one shares my point of view but like I say u cant judge a person when u dont know all the facts the same goes for any president or the person precideing over the country, state, ect. u arent in there shoes so u dont know everything they do. I also dont think u should gripe about the president or leader unless u voted then u have that right.

aalindsey -- what do you think the difference between a dictatorship and monarchy is?  The only thing I can think of is that the kings claim to be inherently higher in social status than the other members of society, and that monarchies tend to come along with established aristocracies.

Also, could you please try to work on your spelling and grammar?  I have difficulty reading what you write.

Have most of you adopted your own philosophies about politics and government, or have your parents corrupted you since Day 1? :)

It's interesting.. I find that most people living under democracy tend to get taught that monarchies, dictatorships, communism, etc. are cruel to the people somehow. Yet my parents (who lived in Communist China for 30+ years) worshipped Mao and were satisfied with what they had under Communist rule.

(Of course, once we immigrated here, their attitudes towards the Communist Party went from reverent to spiteful, but, eh, that's another story.)

I guess you could always say that the people living under "oppression" are unaware that they are indeed being oppressed.. but I hesitate to say that all non-democratic societies are necessarily unfair to their people. I personally am very grateful to be here instead of <I>there</i>, but if I had grown up in China maybe I wouldn't have minded so much.

wyra -- what was the social status of your parents in China?  That makes a very big difference in how they were treated and what their opinions would have been about the society.

aalindsey -- "I will support the president no matter what even though I may not like what hes doing ..."
So you think that it's better to step aside while your country does something that you find morally repugnant than it is to speak out about it, because it's your country?  No thanks -- I'll be at the peace rally.

wyra again -- "but I hesitate to say that all non-democratic societies are necessarily unfair to their people."
In addition, I would hesitate to say that democratic societies are ever fair to their people.  In order to get fairness, you have to trade away a certain amount of democracy, at least on larger scales.  (In small, homogenous situations this is not necessarily the case, but in anything resembling a modern nation I think this is fundamentally true.)

What was it Churchill said?  "Democracy is the worst form of government except for all those others that have been tried."

By the way, you might all want to try out http://www.nationstates.net -- it's a very interesting game that essentially lets you guide the policy for a nation.  It's a little simplistic sometimes, but it's fun.

What you said about social status.. that's very true.

My grandpa (mom's dad) actually invented some things for the government. Instead of paying him, of course, they gave him a tiny apartment to live in. My mom got off easy during the Cultural Revolution as well; she tended pigs for a few years. As a high school student (before the CR) she was an active young Communist.. though she shudders now to talk about it.

My grandpa (dad's dad) was a "capitalist," so once the Communists came to power my grandpa basically lived in fear of his life. My dad wasn't so much in support of the gov't as he was indifferent I guess. He walked across the eastern coast of China spreading the word of Mao during the CR, then went off to work on the mountainside (coal mines) for a good 6 years.

After that I guess you could say he was a little annoyed with the gov't.

I meant "satisfied" in the sense that they were happy with what they did have -- some food, some water, their lives. Well, alright, I am painting a very bleak picture after all.  :|

And yes, not all democracies are kind to their people.. or even have true democratic ideals. I am very tempted to talk about this country, for instance.. but I will control myself. Ahem.

By the way, Joel, this is Linda. (Hi!)

Hey!  Go figure.  Welcome to AOPS :)  Now you're following me around online, after showing up in all the classes I visited at Stuy ...

Okay, so humour has never been my strong suit.  What inspired you to join the forum?

Hahaha.

Well Long kept telling me about it in Sept., so I figured I should take a look. Yeah, it took me a few months to finally stop being lazy about it but I like the AoPS forum. A lot of nice problems and people with very nice solutions.

And now that I am dutifully slacking off in my classes, I have plenty of time to spare.  :D


If $ x < a < 0$ means that $ x$ and $ a$ are numbers such that $ x$ is less than $ a$ and $ a$ is less than zero, then:

$ \textbf{(A)}\ x^2 < ax < 0 \qquad\textbf{(B)}\ x^2 > ax > a^2 \qquad\textbf{(C)}\ x^2 < a^2 < 0$
$ \textbf{(D)}\ x^2 > ax\text{ but }ax < 0 \qquad\textbf{(E)}\ x^2 > a^2\text{ but }a^2 < 0$

clearly B.  Since $ a$ and $ x$ are both negative, $ ax$ is positive.  We also know $ a^2$ and $ x^2$ are positive, by trivial inequality, and $ x^2 > a^2$ since $ x < a < 0$ clearly gives $ x$ the larger absolute value.  Since these two squares are not equal and positive, we then know their geometric mean, $ ax$, is between them.  Thus $ B$ is the answer.


inequalities
geometry
incenter
trigonometry
geometry proposed

Given triangle $ABC$ with incenter $I,A'B'C'$ is cevian triangle of $I$ prove that
$IA+IB+IC\ge 2(IA'+IB'+IC')$

after a van aubel and some computations, if i am not mistaken, the inequality is equivalent with :

$\sum{\frac{p-a}{a}\cdot \sin{\frac{A}{2}}}\geq \sum{\sin{\frac{A}{2}}}$

Are you sure? can you show clearly ? and moreover how do you prove your posted inequality?

Can you show a solution pohoatza or everybody? actually I made a mistake when I creat this problem, I can prove the weaker inequality,only. Could anybody show a solution? :)

I took a look over it, and i realised I did a mistake, here is a complete proof:

From the bisector theorem in $\triangle{ABA'}$, we have that $\frac{AI}{IA'}=\frac{b+c}{a}$, so $IA'=\frac{a\cdot IA}{b+c}$.

But $AI = \frac{r}{\sin{\frac{A}{2}}}$, so we can rewrite the inequality as:

$\sum{\frac{1}{\sin{\frac{A}{2}}}}\geq  \sum{\frac{2a}{(b+c)\sin{\frac{A}{2}}}}$.

$\iff \sum{\frac{1}{\sin{\frac{A}{2}}}(1-\frac{2a}{b+c})}\geq  0$

$\iff \sum{\frac{1}{\sin{\frac{A}{2}}}(\frac{b+c-2a}{b+c})}\geq  0$

Thus it is enough to prove that $\sum{\frac{b+c-2a}{\sin{\frac{A}{2}}}}\geq  0$, which is [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=533[/url]  :)

[color=purple] Hi gemath ! This is a long time i don't meet you in Mathlinks and i miss you :P .How are you ? I hope you are sick :D .The above porblem is not hard but it's really a beautiful problem because the proff of mine using geometry and claasic inequality.
   Here is my proff :
 Letting $l_{a}, l_{b},l_{c}$ are three angle bisectors of triangle ABC. In triangle it's easy to check that $IA+ID=l_{a}$ thus your inequality is equipvalent to
    $IA+IB+IC\ge 2(l_{a}+l_{b}+l_{c}-IA-IB-IC) \longleftrightarrow IA+IB+IC\ge\frac{2}{3}(l_{a}+l_{b}+l_{c})(H)$
 It's not hard to show that $IA=\sqrt{\frac{bc(p-a)}{p}}and l_{a}=\frac{2\sqrt{bcp (p-a)}}{b+c}$ .Construct two similar relations you can see we need to prove
 $\frac{\sqrt{bc(p-a)}+\sqrt{ca(p-b)}+\sqrt{ab(p-c)}}{\sqrt{p}}\ge \frac{4\sqrt{p}}{3}[\frac{\sqrt{bc(p-a)}}{b+c}+\frac{\sqrt{ca(p-b)}}{c+a}+\frac{\sqrt{ba(p-a)}}{b+a}] \longleftrightarrow \sqrt{bc(p-a)}+\sqrt{ca(p-b)}+\sqrt{ab(p-c)}\ge\frac{4p}{3}(\frac{\sqrt{bc(p-a)}}{b+c}+\frac{\sqrt{ca(p-b)}}{c+a}+\frac{\sqrt{ba(p-a)}}{b+a})(2)$
Withuot loss of generality we assume that  $a\ge b\ge c$ .According to Steiner-Lehmus theorem we have $l_{a}\le l_{b}\le l_{c}$ thus $\frac{\sqrt{bc(p-a)}}{b+c}\le \frac{\sqrt{ca(p-b)}}{c+a}\le \frac{\sqrt{ba(p-a)}}{b+a}$ and $b+c \le c+a \le a+b$
 hence applying Trebyshev inequality we get
$(b+c)\frac{\sqrt{bc(p-a)}}{b+c}+(c+a)\frac{\sqrt{ca(p-b)}}{c+a}+(a+b)\frac{\sqrt{ba(p-a)}}{b+a}\ge \frac{1}{3}(a+b+b+c+c+a)(\frac{\sqrt{bc(p-a)}}{b+c}+\frac{\sqrt{ca(p-b)}}{c+a}+\frac{\sqrt{ba(p-a)}}{b+a})$ which is $(2)$. And the proff is finished.
  By similar method we can prove
  $\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le \frac{2}{3r}(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2})$
which is equipvalent to
$\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}}\le\frac{2}{3}(\frac{1}{IA}+\frac{1}{IB}+\frac{1}{IC}) (K)$
  Check $(H)and (K)$ please ! :D
 I hope you(gemath, pohoatza and other mathlinker..) can help me for this problem
$\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}\le \frac{2}{3r}(cosA+cosB+cosC)$
:D [/color]

I think there are some typos in your post, I think that the original inequality with $l_{a}$, $l_{b}$, $l_{c}$, is $\sum{\frac{1}{l_{a}}}\leq \frac{2}{3r \cdot \sum{\sin{\frac{A}{2}}}}$, posted by gemath here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=816658#816658[/url].

Now if in your proposed inequality with $\frac{1}{m_{a}}$, it is the same typo, and the $\sum{\cos{A}}$, is at the denominator, the same method works:

$\sum{\frac{1}{m_{a}}}\leq \sum{\frac{1}{h_{a}}}= \frac{1}{r}\leq \frac{2}{3r \cdot \sum{\cos{A}}}$.

If not, the inequality is really trivial, because
$\sum{\frac{1}{m_{a}}}\leq \sum{\frac{1}{h_{a}}}= \frac{1}{r}\leq \frac{2}{3r \cdot \sum{\cos{A}}}\leq \frac{2}{3r}\cdot (\sum{\cos{A}})$, the last one beeing equivalent with $(\sum{\cos{A}})^{2}\geq 1$, which is obvious, because $\sum{\cos{A}}= 1+\frac{r}{R}> 1$.

[color=purple] Sorry pohoatza the inequality you 've proved is fail.
  It must be $\sum\frac{1}{m_{a}}\le \frac{3}{2r}(\sum\cosh)$
  Sorry because i made you loss time. [/color]


symmetry
quadratics
LaTeX
algebra
algebra proposed

The problem is to find X,Y,Z where

X^2 = Y+a
Y^2 = Z+a
Z^2 = X+a

I found this to be an interesting problem.
It looks very easy but is a bit tricky.
What do you think?

Ray

What is the solution space $ S$ for $ x,y,z$ and what is the definition space $ D$ for $ a$ ? (is this the tricky part  :lol: ?)

E.g. for $ D\equal{}S\equal{}\mathbb{C}$ any solution of $ 0\equal{}((x^2\minus{}a)^2\minus{}a)^2\minus{}(x\plus{}a)$ and then $ y\equal{}x^2\minus{}a,z\equal{}y^2\minus{}a$ will do it.

If you meant for instance $ D\equal{}S\equal{}\mathbb{Z}$, then you could proceed as follows:

Case 1: Not all of the variables $ x,y,z$ are equal. Using symmetry wlog we assume $ y\not\equal{}x$.

Subtracting the original equations from each other we get $ x^2\minus{}y^2\equal{}y\minus{}z\,,\,y^2\minus{}z^2\equal{}z\minus{}x\,,\,x^2\minus{}z^2\equal{}y\minus{}x$. This yields after successive substitution:
$ y\minus{}x\equal{}(x\minus{}z)(x\plus{}z)\equal{}(x\plus{}z)(z^2\minus{}y^2)\equal{}\ldots \equal{}(x\plus{}z)(z\plus{}y)(y\plus{}x)(y\minus{}x)$ .i.e. $ 1\equal{}(x\plus{}z)(z\plus{}y)(y\plus{}x)\Longrightarrow x\plus{}z,z\plus{}y,y\plus{}x\in\{\pm 1\}$ and there can only be none or two expressions be equal to $ \minus{}1$, ie. we have at least 2 expressions of the same value. Subtraction  of two expressions of the same value yields then the identity of two elements and the remaining expression then that the sum of these 2 identical elements is $ \pm 1$, hence they are not integers, a contradiction.

Case 2: $ x\equal{}y\equal{}z$

Checking the resulting quadratic equation $ (2x\minus{}1)^2\equal{}4a\plus{}1$ this permits an integer solution, iff $ 4a\plus{}1$ is a perfect square

The tricky part comes after the 8th degree equation. This is a problem posed and solved by the late Ramanujan. He reduces the 8th degree equation to a 6th degree. After that (the tricky part) is to decompose the 6th degree equation into two equations of the third degree which can be solved by the usual methods. Perhaps this is a good high school algebra problem? Then based on the solution he comes up with various identities such as (sorry I didn't get a chance to get up to speed with LATEX yet): 1+4sin(10) = sqrt(11-2sqrt(11+2sqrt(11-2sqrt(11...etc) Definitely he was a genius with algebraic manipulation and coming up with all kinds of formulas and series. This particular problem comes from: "Collected Papers of Ramanujan" Chelsea Publishing Company, New York, reprinted 1962, Library of Congress Catalog Card No. 62-8326, page 327, Question 507 (v,240). These problems were published in the Journal of the Indian Mathematical Society. Perhaps this particular problem was one of the simpler ones but still is useful for a good high school level algebraic workout requiring a few insights here and there to produce the final solution. The mathematical package MAPLE only seems to produce a partial solution if one wants to use that for checking purposes.

The 8th degree equation
$ x^8 \minus{} 4x^6a \plus{} 2x^4(3a^2 \minus{} a) \minus{} 4x^2(a^3 \minus{} a^2) \minus{} x \plus{} a^4 \minus{} 2a^3 \plus{} a^2 \minus{} a \equal{} 0$

is divided by
$ x^2 \minus{} x \minus{} a$

to produce the 6th degree equation:
$ x^6 \plus{} x^5 \plus{} x^4(1 \minus{} 3a) \plus{} x^3(1 \minus{} 2a) \plus{} x^2(1 \minus{} 3a \plus{} 3a^2) \plus{} x(1 \minus{} 2a \plus{} a^2) \plus{} (1 \minus{} a \plus{} 2a^2 \minus{} a^3) \equal{} 0$

This is then reduced to two cubic equations and solved (I didn't find this part obvious).
Let $ \alpha, \beta, \delta, \alpha', \beta', \gamma'$ be roots of this equation. Then since $ y$ and $ z$ are also roots, we may suppose that

$ \alpha^2 \equal{} a\plus{}\beta, \alpha'^2 \equal{} a\plus{}\beta',$
$ \beta^2 \equal{} a\plus{}\gamma, \beta'^2 \equal{} a\plus{} \gamma',$
$ \gamma^2 \equal{} a\plus{}\alpha, \gamma'^2 \equal{} a\plus{}\alpha',$

Let $ \alpha\plus{}\beta\plus{}\gamma \equal{} u, \alpha'\plus{}\beta'\plus{}\gamma' \equal{}v;$
then we see that $ \alpha^2\plus{}\beta^2\plus{}\gamma^2 \equal{} 3a\plus{}u$
and
$ \alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha \equal{} \frac{1}{2}(u^2\minus{}u\minus{}3a)$

Then a few more steps and we get:

Then the sextic is identical with:
 $ (x^3\minus{}x^2u\plus{}\frac{1}{2}x(u^2\minus{}u\minus{}3a)\minus{}\frac{1}{6}(u^3\minus{}2u^2\minus{}7au\minus{}u\minus{}3a))(x^3\minus{}x^2v\plus{}\frac{1}{2}x(v^2\minus{}v\minus{}3a)\minus{}\frac{1}{6}(v^3\minus{}2v^2\minus{}7av\minus{}v\minus{}3a))$

Then eventually Ramanujan comes up with:
$ u\plus{}v\equal{}\minus{}1$ and $ uv \equal{} 2\minus{}a$

Therefore
u = $ \minus{}\frac{1}{2}(1\plus{}\sqrt(4a\minus{}7))$ , $ v \equal{} \minus{}\frac{1}{2}(1\minus{}\sqrt(4a\minus{}7))$

This is substituted into the equation above involving $ x,u,v$ and $ a$ to produce two cubic equations.

$ x^3\plus{}\frac{1}{2}x^2(1\plus{}\sqrt(4a\minus{}7))\minus{}\frac{1}{2}x(2a\plus{}1\plus{}\sqrt(4a\minus{}7))\plus{}\frac{1}{2}(a\minus{}2\plus{}a\sqrt(4a\minus{}7))\equal{}0$
$ x^3\plus{}\frac{1}{2}x^2(1\minus{}\sqrt(4a\minus{}7))\minus{}\frac{1}{2}x(2a\plus{}1\minus{}\sqrt(4a\minus{}7))\plus{}\frac{1}{2}(a\minus{}2\minus{}a\sqrt(4a\minus{}7))\equal{}0$

Ray

I think Ramanujan made an error in his solution.

Ray

Never mind he didn't make any error. Easy to make a mistake with a lot of tedious algebra. Definitely a good homework problem for highschool.

Ray


inequalities
function
inequalities theorems

I come the idea inspired by the method $siuhochung$ often used:
$theorem$,let, $f$ is a function increasing in $[0,+\infty)$
and $f(x) \geq 0$
let $a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n$
$g_1,g_2, \cdots,g_n$ are functions of $a_1 , a_2,a_3 , \cdots , a_n$
if we have $\sum_{i=1}^kg_i \geq 0$ and $g_1+g_2+\cdots+g_n \geq 0$
then we have:$f(a_1)g_1+f(a_2)g_2+\cdots+f(a_n)g_n \geq 0$
anybody who used this $theorem$ solved a inequality,please post here,
thank you :)

sorry,I find my theorem is easy and useless :blush:


linear algebra
matrix
function
vector
linear algebra unsolved

Let $A,B\in\mathbb{C}^{n\times n}$ with $\rho(AB - BA) = 1$. Show that $(AB - BA)^2 = 0$.

[quote="alekk"]we wil work with endomorphism.
Suppose that $Rank(uv-vu)=1$. One can find a basis where the matrix $M$ of $uv-vu$ is null
everywhere, except on the first line. (because $rank(M)=1$) But $Tr(M)=0$ so $M_{1,1}=0$.
Conclusion follows ...[/quote]

why trace $M$ equals $0$ ?

Or, in matrix language:

Since $\rho(AB-BA)=1$, we have $n-1$ eigenvalues equal to zero. Since $\Tr(AB)=\Tr(BA)$ we also have the sum of the eigenvalues zero, so all eigenvalues are zero.
Now consider the Jordan Form of $C=AB-BA$. This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\delta_{12}$, and thus is $C^2$ trivially the nullmatrix.

I'm sorry, but I don't understand which function is $\rho$...

Note that the result in this problem also hold for any field $ F $ , not only complex field.

Define $C = AB - BA $ .Since rank $C $ is 1 , then there exist a non zero column vectors $u $ and $v $ such that $uv^{t} $ = $C$

Since tr $ AB$ = tr$BA$ , we get that tr$C$ = $0$ , or we can say that $v^{t}u$ = $ 0$.

Hence we get that $C^2$ = $(uv^{t})^2$ = $uv^{t}uv{t}$ = $u0v^{t}$ = $0$

(assuming that $ \rho $ is a rank function :lol: )

[quote="alekk"]One can find a basis where the matrix $M$ of $uv-vu$ is null
everywhere, except on the first line. (because $rank(M)=1$)[/quote]  I am quite sure this is not true. In this particular problem, this is indeed the case but only due to the fact that all eigenvalues are zero (you can consider Jordan form as Peter has). However, you need to use that $tr M =0$. By the way, my solution is the same as removablesingularity's.

I understand upto the point where we know that the eigenvalue of $AB-BA$ are all 0. I didn't understand this.This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\delta_{12}$ (sorry I don't know this notation), and thus is $C^2$ trivially the nullmatrix.
also in solution, this is also written, In the Jordan canonical form there can only
be one 2 × 2 cage. I also didn't understand this. Someone please explain. Thank you.

[quote=Adamsawyer]I understand upto the point where we know that the eigenvalue of $AB-BA$ are all 0. I didn't understand this.This matrix has rank 1 with all diagonal elements zero, thus is of the form $C_{i,j}=\delta_{12}$ (sorry I don't know this notation), and thus is $C^2$ trivially the nullmatrix.[/quote]

I agree that the notation $C_{i,j}=\delta_{12}$ means nothing.
What the poster meant is that the Jordan form of a rank 1 nilpotent matrix must have only Jordan blocks (relative to the eigenvalue 0) of order 1 or 2 (in fact, only one Jordan block of order 2).

[url= https://www.imc-math.org.uk/imc2000/prob_sol1.pdf] problem 3 first day[/url]
https://www.imc-math.org.uk/imc2000/prob_sol1.pdf


For an unknown substance Cux(CO3)y, is it possible to determine the empirical formula if I only know the mass of CO2?

i'd say no b/c you'll need to convert them to mol and you can't do that with only mass of $CO_2$

Yeah, you gotta know the amount of the substance or the copper.


It's too interesting:
Find all triples $(f,g,h)$ satisfying
\[\frac{f(x)-g(y)}{x-y}=h\left(\frac{x+y}{2}\right)\] for all $x\neq y$.

[i]Edited by Myth[/i]

I mean find f,g,h.
I predict that f,g,h are poliminals; degf,degg<=2

Something strange is happening here....
If they would be continuous functions, making $x\rightarrow y$ we would get that $f(x)=g(x)$.
Also, it is very easy to get that $f(x)=f(-x)$ by putting $y=-x$.
A little work could be done to find the solutions :) :D
So I think $f(x)=kx^2$ would be a solution....
If the functions wouldn't be continuous, i don't know what to say, but it would be a very bad function.

Oh, the solution I find is more interesting: f(x)=g(x) for all x, and h=f'(x), (like Lagrange :) )
       So, we put x=-y, y=-x, x=0 ,y=0 we get f(x)-g(x)=const!
       Using this in the original equation (note that g(x)=f(x)-const), put x=-y, y=-x, we also get f(x)=g(x) (const=0).
       After that we can see the origin equation as the equation with h by getting
               f(x)-f(0)=xh(x/2) (put y=0),
       we have : (xh(x/2)-yh(y/2))/(x-y)=h((x+y)/2)
       Put H(x)=h(x)-h(0), we get
             (xH(x/2)-yH(y/2))/(x-y)=H((x+y)/2) and H(-x)=-H(x)
        Put x=-y, y=-x we get H((x+y)/2)=A(x+y)/2 or H(x)=Ax
        So h(x)= 2ax+b and f(x)=g(x)=ax^2 +bx+c


analytic geometry
graphing lines
slope
geometry
3D geometry
sphere
quadratics

Let there be the equation : $ x^{n}+y^{n}+z^{n}=2$ . Prove that for each $ \vee n \in \{1,2,3,4\}$ , these equation has an infinity of rational roots .
         I've posted this [url=http://www.mathlinks.ro/viewtopic.php?t=156176]here[/url]  before in romanian , but I've decided to translate it in English , too .

$ n = 1$ is trivial.

[hide="n=2"] First we must exhibit a rational solution.  Let $ z = 1$.  Then we are trying to solve $ x^{2}+y^{2}= 1$, which has its first nontrivial rational solution at $ x = \frac{3}{5}, y = \frac{4}{5}$.

The graph $ x^{2}+y^{2}+z^{2}= 2$ describes a sphere.

[b]Claim:[/b]  A line of rational slope passing through a rational point on the sphere intersects it at a second rational point.

Proof:  By "a line of rational slope" I mean a line that can be represented as the intersection of two planes

$ ax+by+cz = d$
$ ex+fy+gz = h$

With rational coefficients.  Substituting those equations into $ x^{2}+y^{2}+z^{2}= 2$ we can produce a quadratic in any of the variables (say $ x$) with rational coefficients.  Since it has one rational root (the original rational point) it must have another (by Vieta's, etc.).  For rational $ x$ we produce rational $ y, z$ from the first two equations.

So starting with our rational solution we can draw any line with rational slope to another rational solution.  QED. [/hide]

I'll appreciate any feedback as to how this solution would constitute as an acceptable proof:
[hide]
Claim: There is an infinite rational number between $ 0$ and $ 1$.
Proof: Let $ A_{n}= (\frac{1}{2})^{n}$. We know that $ \frac{1}{2}$ is a rational number, and the product of two rational number results in another rational number. Therefore the function $ A_{n}$ would always result in a rational number for $ n \in (0, infinity)$. For $ n \in (0, infinity)$, $ 0 \le A_{n}\le 1$ which concludes our claim.
------
Arbitrarily, let $ z=0$. Then the original equation becomes:
$ x^{n}+y^{n}= 1$
$ y^{n}= 1-x^{n}$
This relation shows that as x increases, y decreases. Similarly, as x decrease, y increase.

Regardless of any positive integer $ n$, two solutions to $ y^{n}= 1-x^{n}$ is $ (0,1) and (1,0)$. This shows that as the variable $ x$ goes from value $ 1$ to value $ 0$, the variable $ y$ goes from value $ 0$ to value $ 1$. What this means is that any number, ration or irrational, between $ 0$ and $ 1$ would be a solution for $ x$, and a corresponding number between $ 0$ and $ 1$ would likewise be a solution for $ y$.

The equation $ y^{n}= 1-x^{n}$ shows that if any one of the variable is rational, then so is the other.

Because there is an infinite rational number between $ 0$ and $ 1$, there is an infinite rational solution for $ (x, y)$ in the equation $ y^{n}= 1-x^{n}$ , thus there is an infinte ration solution to $ x^{n}+y^{n}+z^{n}= 2$.
[/hide]

[quote="p4fn2w"]The equation $ y^{n}= 1-x^{n}$ shows that if any one of the variable is rational, then so is the other.[/quote]

Trivially false; take $ n = 3, x =-1$ (for example.)

[quote="p4n2w"]Because there is an infinite rational number between $ 0$ and $ 1$, there is an infinite rational solution for $ (x, y)$ in the equation $ y^{n}= 1-x^{n}$[/quote]

Er, no.  This would imply the falsehood of Fermat's Last Theorem.


[quote="mr. math"]How many times does the graph of the equation f(x) = x^5 - 3x^4 + 9x^3 -19x^2 +18x - 6 cross the x axis?

I solved the problem and got the correct answer by using descartes rule of signs and the rational root theorem. I was wondering if there is a faster way to do it...[/quote]
Yes, there's a very simple way. Find the degree of polynomial. In this case it is 5. So, it will have 5 roots. So, it will cross the x-axis at most 5 times when there are unique solutions. But if some of the roots happen to be equal then it will cross x-axis less than 5 times.

[quote="Lakshya"]Yes, there's a very simple way. Find the degree of polynomial. In this case it is 5. So, it will have 5 roots. So, it will cross the x-axis at most 5 times when there are unique solutions. But if some of the roots happen to be equal then it will cross x-axis less than 5 times.[/quote]
... or if there are imaginary roots.

[hide]Note that the coefficients add to $ 0$. That means $ x\equal{}1$ is a solution. Factor out $ (x\minus{}1)$:
$ (x\minus{}1)(x^4\minus{}2x^3\plus{}7x^2\minus{}12x\plus{}6)$

Hey, the coefficients of the quartic add to $ 0$. That means we have another factor of $ (x\minus{}1)$:
$ (x\minus{}1)^2 (x^3\minus{}x^2\plus{}6x\minus{}6)$

And again, the coefficients of the cubic add to $ 0$. So factor out another $ (x\minus{}1)$:
$ (x\minus{}1)^3 (x^2\plus{}6)$

Now we want the number of distinct real solutions to $ (x\minus{}1)^3 (x^2\plus{}6) \equal{} 0$. $ x^2\plus{}6 \geq 0$, so divide it out to get $ (x\minus{}1)^3 \equal{} 0$. The only solution is $ x\equal{}1$, thus the graph crosses the x-axis once.[/hide]


is it true (and if it is the prove it)  that for all positive $ a,b,c$
$ \frac {3}{a \plus{} b \plus{} c}\leq {\frac {1}{\sqrt {ab}}} \plus{} {\frac {1}{ \sqrt {bc}}} \plus{} {\frac {1}{\sqrt {ca}}}$ ?
Thanks

it is true...
$ (a+b+c)(\frac {1}{\sqrt {ab}} + {\frac {1}{\sqrt {bc}} + {\frac {1}{\sqrt {ca}})\geq3\sqrt[3]{abc}\times3\sqrt[3]{\frac {1}{\sqrt {ab}}\times\frac {1}{\sqrt {bc}}\times\frac {1}{\sqrt {ca}}}=9>3}}$

[quote="ali666"]it is true...
$ (a + b + c)(\frac {1}{\sqrt {ab}} + {\frac {1}{\sqrt {bc}} + {\frac {1}{\sqrt {ca}})\geq3\sqrt [3]{abc}\times3\sqrt [3]{\frac {1}{\sqrt {ab}}\times\frac {1}{\sqrt {bc}}\times\frac {1}{\sqrt {ca}}} = 9 > 3}}$[/quote]  IT IS TRY


Solve in $ \mathbb{ R}$:
$ \frac{x^2}{y}\minus{}\frac{4y^2}{x}\equal{}1$
$ 3x\minus{}10y\equal{}\frac{1}{2}$

After I equate 2 equations, the outcome is

$ x^3\minus{}6x^2y\plus{}20xy^2\minus{}4y^3\equal{}0$

One more:
Solve in $ \mathbb{R}:$
$ \frac {2x^3}{y} \plus{} \frac {3y^3}{x} \equal{} 1\minus{} 8xy$
$ 8x^2\plus{}9y^2\equal{} \sqrt2$

We change the  system of equations into a homogeneous form as follows

$ 2\sqrt{2}x^4\minus{}8x^3y\plus{}8\sqrt{2}(xy)^2\minus{}9xy^3\plus{}3\sqrt{2}y^4\equal{}0$

[quote="great math"]We change the  system of equations into a homogeneous form as follows

$ 2\sqrt {2}x^4 \minus{} 8x^3y \plus{} 8\sqrt {2}(xy)^2 \minus{} 9xy^3 \plus{} 3\sqrt {2}y^4 \equal{} 0$[/quote]
And how do you solve it?

[quote="JavaMan"]Once it's homogeneous let $ y \equal{} mx$ and solve for possible values of $ m$.  :)[/quote]
Yeah! Of couse it is!
But what I mean is to solve the equation of the fourth degree.

hello, to the first system, solving the second equation for $ y$ we get
$ y\equal{}\frac{3}{10}x\minus{}\frac{1}{20}$ inserting this in the first equation, simplifying and factoring we get
$ \frac{1}{100}\left(\frac{178x^3\minus{}492x^2\plus{}82x\plus{}1}{(6x\minus{}1)x}\right)\equal{}0$
so we have to solve
$ 178x^3\minus{}492x^2\plus{}82x\plus{}1\equal{}0$, with the following real solution
$ x\equal{}\minus{}{\frac {5}{66537402}}\,\sqrt [3]{3808521050288486841\plus{}
82272613989570141\,\sqrt {2409}}\plus{}91430\,{\frac {1}{\sqrt [3]{
3808521050288486841\plus{}82272613989570141\,\sqrt {2409}}}}\plus{}{\frac {41}{446
}}$
Sonnhard.

Hello, the second system has no real solutions.
Sonnhard.


Let $ a,b,c > 0$ such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$, and let
$ A \equal{} \displaystyle \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}, B \equal{} \displaystyle \frac {1}{1 \minus{} \left(\frac {a \plus{} b}{2}\right)^{2}} \plus{} \frac {1}{1 \minus{} \left(\frac {b \plus{} c}{2}\right)^{2}} \plus{} \frac {1}{1 \minus{} \left(\frac {c \plus{} a}{2}\right)^{2}},$ and
$ C \equal{} \displaystyle \frac {1}{1 \minus{} a^{2}} \plus{} \frac {1}{1 \minus{} b^{2}} \plus{} \frac {1}{1 \minus{} c^{2}}.$

Prove that (1) $ A \plus{} C\geq 9$, (2) $ 7B \plus{} 3C\geq 45$.

Yes, (2) has been unsolved for a long time. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81709]here[/url]. Have you found a solution for it ?  :roll:

(1) is easy (it can be proved by Cauchy-Schwarz), (2) is harder. My solution does not use any big calculations but it is rather long. See it here: http://can-hang2007.blogspot.com/2010/01/inequality-64-w-berndt.html

[quote="Shanku"]Will you please post the solution of (1), can_hang2007 ?[/quote]
Very please, of course. :-) See it here: http://can-hang2007.blogspot.com/2010/01/inequality-68-w-berndt.html


[quote="shyce"]What is a formula for the pattern shown below.

0,1,3,6,10,15[/quote]

[hide]
If you double each term in the sequence, you get 0, 2, 6, 12, 20, 30...

It's clear that each number in the sequence is $0\cdot 1, 1\cdot 2, 2\cdot 3, 3\cdot 4, ...$

So each term in the new sequence is $x(x-1)$

Divide by 2 to get $\frac{x(x-1)}{2}$
[/hide]

Basically, it's just

$\sum\limits_{i = 0}^n {i = \frac{{n(n - 1)}}{2}}$

EDIT:

BTW, it's the formula for the $n+1$th triangular number


[i]Let $ a_{1}$, $ a_{2}$ ,....,$ a_{n }$ , $ b_{1}$, $ b_{2}$,..., $ b_{n}$ be real numbers
   Prove that
      $ \sum_{i \equal{} 1}^n a_i.b_i \plus{} \sqrt { ( \sum_{i \equal{} 1}^n a_i ^2 )(\sum_{i \equal{} 1}^n b_i ^2) }$ $ \geq$ $ \frac {2}{n}.( \sum_{i \equal{} 1}^n a_i )(\sum_{i \equal{} 1}^n b_i)$[/i]
   Argady , Nayel , Jichen ... can you help me? :blush:

[i] [u]Problem 2[/u]
  Let a,b,c be nonnegative real numbers such that a+b+c =3[/i] Prove     that
  $ \frac{5 \minus{}3bc}{ 1\plus{}a} \plus{} \frac{5 \minus{} 3ca}{1\plus{}b} \plus{}\frac{ 5\minus{}3ab}{1\plus{}c} \geq ab \plus{}ba \plus{}ca$

[quote="nkht-tk14"]
[i] [u]Problem 2[/u]
  Let a,b,c be nonnegative real numbers such that a+b+c =3[/i] Prove     that
  $ \frac {5 \minus{} 3bc}{ 1 \plus{} a} \plus{} \frac {5 \minus{} 3ca}{1 \plus{} b} \plus{} \frac { 5 \minus{} 3ab}{1 \plus{} c} \geq ab \plus{} ba \plus{} ca$[/quote]
After expanding and homogenization we obtain:
$ \sum_{cyc}\left(\frac {5 \minus{} 3bc}{ 1 \plus{} a}\minus{}ab\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(5a^5\plus{}14a^4b\plus{}14a^4c\minus{}19a^3b^2\minus{}19a^3c^2\plus{}32a^3bc\minus{}27a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow5\sum_{cyc}(a^5\minus{}a^4b\minus{}a^4c\plus{}a^3bc)\plus{}19\sum_{cyc}(a^4b\plus{}a^4c\minus{}a^3b^2\minus{}a^3c^2)\plus{}$
$ \plus{}27\sum_{cyc}(a^2\minus{}ab)abc\geq0,$ which true by Schur and Muirhead.


geometry
3D geometry
tetrahedron
sphere
geometry theorems

hi ! i have question ,where is centre of ball  in tetrahedron?  :)

suppose the tetrahedron is $ABCD$ draw $6$ sphere with the diameters $AB,AD,BD ,BC,AC,CD$.

the intersection of these spheres will be a point which is the center of the tetrahedron...


MATHCOUNTS
number theory
prime factorization

[quote="frost13"]The prime factorization of a number X is $a^a$ $b^b$ $c^c$ where a, b, and c are unique prime numbers.  If X has 72 factors, what is the value of X?[/quote]

[hide="solution"]$(a+1)(b+1)(c+1)=72$

$72=2^33^2$

$a+1=6$
$b+1=4$
$c+1=3$

$a=5$
$b=3$
$c=2$

$x=2^23^35^5$[/hide]

So is there a formula, based on what you all wrote, for the number of factors a number has? Sorry I'm jw because I'm preparing for the States and want to know every formula that I can.

hockey,

there is a formula, and it can be manipulated for various problems.  I'm attaching a document of various formulas, shortcuts to common mathcounts questions for you.  Good Luck,

frost


USAMTS search LaTeX Asymptote geometry AMC AIME

Round 1 problems of Year 19 of the USA Mathematical Talent Search are now available at [url]http://www.usamts.org[/url].  The postmark deadline for solutions is October 9, 2007.  A copy of the rules is available with the problems; please read them carefully.

IMPORTANT REGISTRATION INFORMATION: If you have not already registered online for Year 19, please do so at http://www.usamts.org.  If you participated in Year 18, all you have to do to register for Year 19 is log in to [url]http://www.usamts.org[/url], then set your grade in the Profile section of the My USAMTS pages.

We hope you enjoy the problems, and we look forward to working with you on the USAMTS this year.

I noticed a few bugs in the LaTeX template for the solutions form.

1) The include file is labeled usamts, but the line in the solutions form reads "\include{usamts[b]_blank[/b]}".
2) Solution 1 has a "\ " on the line above the comment.

Are the templates the same as last year?

Great set of problems! I like having the sets during vacations because now I can write up solutions before school starts.

Um I faxed my entry form, under 13 permission form, and regular permission form to the designated location, but it still says that the people didn't receive it on my Profile page.  I faxed it like a week ago, so am I being hasty or is there some problem. My ID is 9524, if that helps.

P.S.   Looooooooove the questions. :lol:

[quote="LeFromage"]In the .tex file, it says to enter your AoPS username. Is it supposed to be your USAMTS username?[/quote]

i made them the same, just in case... but maybe its supposed to be your USAMTS username.

the problems are very nice. :)

Templates the same (but make sure you update the year).

That should be USAMTS username (I just fixed it on the site.)

We won't start filing the forms we receive until next month, so don't worry yet about it saying your forms are not received.

One more thing I noticed: on the fourth page of the USAMTS entry form (http://www.usamts.org/Forms/USAMTSEntryForm.pdf), it still says "Year 18" and "2006-7".

Would it be possible to find the tex file of the problems, specifically the 18-gon diagram?

I think you should be able to build that diagram yourself in asymptote, with a little elbow grease.  Obviously we shouldn't post it on the forum, though.

By the way, do you prefer that we save asymptote images as separate files and send just their pdf, that we also send the .asy file, or that we keep the asymptote code within the .tex file?

The reason is that I don't want to do it in asymptote, because it takes too much time

[quote="roadnottaken"]I seriously doubt that graders will take off for a diagram being blurry, unless there is text that is so small as to be illegible.  But, if you are really worried about it, create your pictures in Asymptote.  That way you can just write the source code in your latex document, without any fuss.[/quote]
Dude, you're answering legitimate questions directed to the usamts people with things like "I think" or "I believe"

Let the question be answered, your reply didn't actually help anyone..

[quote="#H34N1"]Are the grader's going to print out the file and grade it? If I create a diagrma in geometry sketchpad, it becomes blurry in tex when on a 24" screen (don't know what screen resolution and all the grader's might use), so are they going to print it?[/quote]
Yes, they print it out.
I asked that once in a mathjam.

hey, just wondering. we're allowed to write everything by hand (in pen) and fax the answers to them right?

Just want to confirm.

thanks

[quote="Geeimatree"]hey, just wondering. we're allowed to write everything by hand (in pen) and fax the answers to them right?[/quote]Yes.  Alternatively, you can mail them.

I know for a fact that not all of their criteria are related to AoPS; the way I originally discovered this website was by being mailed a USAMTS form.

[quote="moogra"][quote="mustafa"][quote="moogra"]pretty interesting. I got one in the mail :)[/quote]

They send them out to people who took the AIME, or something like that. Or maybe people who bought some Aops books. Something like that...[/quote]

Well I also read on the USAMTS page that the don't send out the problems. I guess I was just lucky.[/quote]

Nah, I got the problems too

[quote="timwu"]I've just sent my solutions by fax; however, for some ridiculous reason, the secretary of my mother's office, where she faxed for me, did it thrice... And I'm also sorry for forgetting copying the USAMTS# on the coversheet (all other things are copied)... Do these things matter?

B.T.W.: I've registered on the USAMTS website and have never received its e-mail (I've verified it's correct) Does anybody know why?

Thanks.[/quote]

Don't worry about faxing too many times.  I'm assuming you have your USAMTS number on your solutions.  As for the emails, maybe your spam blocker nailed the.?

Thank you for reply;I did copy the number on every page, and I'm going to check if there's some spam blocker in my g-mail.
B.T.W.: The questions are very nice,  :)

Another question: when will the comfirmation check mark appear on my "My Scores" of the USAMTS website? Because it seems that my e-mail doesn't work (I've checked my e-mail, there's no spasm or trash e-mail at all.)

To 123456789: I mean the system e-mail from USAMTS website after a user registers.

We'll probably start filing the papers in a week or so.


1)Show that $a^{12}-b^{12}$ is divisible by 91 if a and b are both prime to 91.

2)Show that $n^{36}-1$ is divisible by 3374 if n is prime to 2,3,19 and 37.

3)If m is a prime number and a,b are less then m,prove that
$a^{m-2}+a^{m-3}b+a^{m-4}b^2+......b^{m-2}$
 is a multiple of m.

Just correcting a few typos :) :
[quote="shadysaysurspammed"]1)Show that $a^{12}-b^{12}$ is divisible by 91 if a and b are both prime to 91.

2)Show that $n^{36}-1$ is divisible by 3374 if n is prime to 2,3,19 and 37.
[/quote]


Something is wrong. No matter what I do my Karma stays at 1000!

Well it finally started working right. It really wouldnt change for a while, though.

You were probably in diagnostic mode; you hadn't done enough questions.  Are you a new user for Alcumus?  Look in the instructions.

Yeah I just started going on Alcumus. Its working now.

Please read the [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php]instructions[/url] as it contains information about such things.


Let $P_{n}$ be the number of ways to make $\$n$ by only using nickels, dimes, and quarters.  Compute the remainder when $P_{2007}$ is divided by $1000$.

Since I don't know all of the tricks with this method, i used an altered [hide] generating function argument. [hide="And the answer I got is:"] \[\boxed{115}\] [/hide] I'll write my solution up later, whether or not it's correct. I just don't have the time now. :) [/hide]


algebra
polynomial
logarithms
Functional Analysis
real analysis
real analysis unsolved

Here's a very beautiful problem: suppose than $ k_n$ is a sequence of positive integers with the following property: any series $ \sum_{n}{a_n e^{ixk_n}}$ that converges uniformly on $ \mathbb {R}$ converges actually normally, that is $ (a_n)_n$ is in $ l^1$. Then there are constants $ c>1$, $ d>0$ such that $ k_n\geq d c^n$ for all $ n$. Can you exhibit a series not having the above property if the sequence $ k_n$ is polynomial?

If I were to guess, I would rather attribute this result to Khintchin (though historically I may be completely wrong here, it is closer to his work in spirit). If you have a sum $ \sum_{j\equal{}1}^n\pm e^{ik_jx}$ with $ 0\le k_j\le N$, its $ L^\infty$-norm does not exceed $ C\sqrt {n\log N}$ with high probability. This (together with some standard tricks from functional analysis) allows to conclude that under our condition $ n\le C\log N$, which is equivalent to what harazi asked. To exhibit an explicit series for a given sequence $ k_j$ is a much harder problem.

[hide="2Fedor"]Я не понял связи :blush: Я так думаю, он бы не дал тут эту задачу, если бы у нее не было достаточно короткого решения, которое полностью можно тут изложить.[/hide]

Well, that book does not say who is the author of this theorem (which is a bit strange, because it seems to be rather accurate with all the other results presented...), but the solution it gives and which I have is exactly the one presented by Fedja. And I have to confess that I don't have explicit examples for practically no nontrivial situation.


Subiectele si solut de la etapa judeteana din 01 martie 2008 sunt postate pe site-ul http://www.mategl.com

Please ,
 can you send some other links, where I can find problems and solutions of Romanian competitions , like this nice Web site above ?
 Thanks - babis

Cum vi s-au parut subiectele la a unspea si a doispea? La a doispea mi s-au parut mai grele ca in anii trecuti.

Le puteti gasi si la http://mateforum.ro/viewtopic.php?t=1184


The related story can be found here: http://www.abc2news.com/news/local/story.aspx?content_id=caf271c6-f450-4499-84f8-0fa4fdd34d92&rss=702


$ S^1\equal{}\{(x,y)\in \mathbb{R}^2 | x^2\plus{}y^2\equal{}1\}$.
Let $ x\equal{}(x_1,x_2)\in S^1$,$ \minus{}x\equal{}(\minus{}x_1,\minus{}x_2)\in S^1$.
Define map $ r: S^1\to S^1$ as $ \forall \, x\in S^1, r(x)\equal{}\minus{}x$.
How to prove the continuity of $ r$?

thanks

Consider that map as a map between $ \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Clearly its a finite dimensional linear map. These maps are continuous. When restricted to $ S^1$ on the domain, then the map is also continuous. Then its trivial to check that the codomain is $ S^1$ because multiplication by $ \minus{}1$ doesn't change the absolute value.

More directly, $ \parallel{}f(x) \minus{} f(y)\parallel{} \equal{} \parallel{}\minus{}x \minus{} (\minus{}y)\parallel{} \equal{} \parallel{}x\minus{}y\parallel{}$. Any function for which there exists a positive constant $ \lambda$ (in our case $ \lambda \equal{} 1$) such that $ \parallel{}f(x) \minus{} f(y)\parallel{} \leq \lambda \parallel{}x\minus{}y\parallel{}$ can be easily checked by definition to be continuous (such functions are called lipschitzian; if $ \lambda < 1$ they are called contractions).

[quote="LkNsngth"]When restricted to $ S^1$ on the domain, then the map is also continuous.[/quote]This is not sufficient to be precise. Note that you have proven that the composition $ S^1\to S^1\to\mathbb R^2$ is continuous. But why is $ S^1\to S^1$ is continuous? There is one more argument needed.

@ water2011: Are all your postings homework? It smells like ...  :mad:  :wink:


geometry
3D geometry
sphere
calculus
calculus computations

This problem can be solved by using Calculus, but apparently there's an easy intuitive method as well.

If you cut a sphere through the centre with a cylindrical hole of diameter l, what is the remaining volume of the sphere (in terms of l)?

A hint is that the answer is independent of the radius of the sphere - in fact, the intuitive method uses this fact in its argument.

Oh, come on, isn't anybody going to help me???  :(

To get a volume independent of the radius of the sphere, its not the diameter of the hole that should be fixed, but rather the height of the cylinder. (Or if you think of the remaining object as something like a napkin ring, its standing height.)

So, if you ignore the hint and the volume is dependent on the radius of the sphere, what would the answer be???

Kent Merryfield is right. The volume of the remaining solid should be $\frac{4}{3}(R^2-I^2)^{\frac{3}{2}}\pi$, where $R$ is the radius of the sphere. I think the trick Melissa has in her mind is the use of the following theorem(by the way, this theorem was first discovered by Zu Heng, the son of the famous Chinese mathematician Zu Chongzi who obtained the numerical value of $\pi$ accurate to 7 decimal places):

Let $A$ and $B$ be two solids which stand on their bases on a plane $V$, with equal height. If the areas of the intersections of $A$, $B$ with any plane $V'$ parallel to $V$ are always equal, then $A$ and $B$ have the same volume.

In fact, the remaining solid in the problem has the same volume as a sphere of radius $\sqrt{R^2-I^2}$, by the above theorem.


probability
function
combinatorics unsolved
combinatorics

What is the probability that 10 coins selected randomly from 100 coins add up to $ 1?
Assume the 100 coins consist randomly of pennies, nickels, dimes, and quarters. Each coin has an equal probability of being selected.

I guess generating functions are required to find the number of ways.

By simple inspection we find the possible decomposition of 100 as the sum of 10 numbers out of 1,5,10,25: obviously, the task is impossible with at least 4 quarters, with 3 quarters you need less than 5 between nickels and dimes, so you need to have pennies, and you cannot have as many as 10, so you have exactly 5, leaving 20 cents in 2 coins, and the only possibility is $ 100 \equal{} 3\cdot25 \plus{} 2\cdot10 \plus{} 5\cdot1$; if you have exactly 2 quarters, and you have pennies, you need 45 cents in 3 coins which are nickels and dimes, absurd, so you have only nickels and dimes left such that $ 10d \plus{} 5n \equal{} 50$, $ d \plus{} n \equal{} 8$, and $ 100 \equal{} 2\cdot25 \plus{} 2\cdot10 \plus{} 6\cdot5$ is the only possibility; with exactly 1 quarter, you need at least 6 dimes or you don't reach 100, but you cannot have more than 7 or you have too much, so either 6 or 7 dimes, leaving 3 or 2 coins for 15 or 5 cents, the second case is impossible and the first case yields $ 100 \equal{} 1\cdot25 \plus{} 6\cdot10 \plus{} 3\cdot5$; with no quarters, only $ 100 \equal{} 10\cdot10$.

The total number of favorable cases is then
\[ \binom{25}{3}\binom{25}{2}\binom{25}{5} \plus{} \binom{25}{2}\binom{25}{2}\binom{25}{6} \plus{} \binom{25}{1}\binom{25}{6}\binom{25}{3} \plus{} \binom{25}{10},
\]
while the total number of possible cases is $ \binom{100}{10}$, since you need to have 25 coins of each one of the four types.  All cases are equally probable since all coins have equal probability of being selected.  Divide favorable into possible and you get the result...


A stock loses $ 10\%$ of its value on Monday.  On Tuesday it loses
$ 20\%$ of the value it had at the end of the day on Monday.  What is
the overall percent loss in value from the beginning of Monday to
the end of Tuesday?


number theory
least common multiple
superior algebra
superior algebra solved

Find the smallest n\in N* s.t. for any permutation x in S9 we have xn=e (the identity permutation). The same question for permutations in A9 (even permutations in S9).

If you need the lowest, I got the first part. It is known that the order of a permutation is the lcm of the lengths of its cycles. n is the lcm of all this orders. The number does not contain any prime greater than 9.

The power of 2 is at most 3 as we may have a cycle of length 8.
The power of 3 is at most 2 as we may have a cycle of length 9
The power of 5 is at most 1 as we may have a cycle of length 5
The power of 7 is at most 1 as we may have a cycle of length 7

So the number is 8*9*5*7 :)

Right! You need th smallest n, of course (my mistake). I edited the post. Sorry :).

For the case of An I will use the following result:

For any cycle we have (a1,...an)=(a1,an-1)(a1,an-2)...(a1,a2)

so one cycle is even if its length is odd and odd if its lengfth is even. A permutation is even only if it contains an even number of cycles of even length.

In the argument for Sn we should change only the power of 2 because it involves the even length cycles. The power of 2 is at most 2 when we have 2 cycles of length 4. So the number is 4*9*5*7 :)


geometry
circumcircle
congruent triangles
geometry proposed

Given a circle and its center $O$, a point $A$ inside the circle and a distance $h$, construct a triangle $BAC$ with $\angle BAC = 90^\circ$, $B$ and $C$ on the circle and the altitude from $A$ length $h$.

The circle is known, so let $R$ be its radius and $AD=h$ the altitude of $\triangle ABC$.

The triangle $ABC$ is right so we have $AD^2 = BD \cdot DC$

But $-BD \cdot DC$ is also the power of the point $D$ with respect to the circle $(O,R)$

$-AD^2 = OD^2-R^2 \Rightarrow OD^2 = R^2-h^2$

$\hline$

So, we construct the circles $w_1 = (O, \sqrt{R^2-h^2})$ and $w_2 = (A,h)$.

If the two circles have common point(s), then take $D\in w_1 \cap w_2$.

The perpendicular line to $AD$ intersects the circle $(O,R)$ at the points $BC$.

$\hline$

Now we should check if the constructed figure has the desired properties

From it's construction, $AD$ is the altitude of the triangle $ABC$

$OD=\sqrt{R^2-AD^2} \Rightarrow OD^2 = R^2-AD^2 \Rightarrow -AD^2 = OD^2-R^2 \Rightarrow$
the power of $D$ w.r.t. $(O,R)$ is $-AD^2,$ but it is also $-BD \cdot DC \Rightarrow$

$AD^2=BD \cdot DC \Rightarrow \frac{BD}{AD} = \frac{AD}{DC} \Rightarrow$

The right triangles $BDA,ADC$ are similar $\Rightarrow \angle DBA = \angle DAC$

$\angle BAC = \angle BAD + \angle DAC \Rightarrow$
$\angle BAC = \angle BAD + \angle DBA \Rightarrow$
$\angle BAC = 90^o$

$\hline$

At first we should notice that $h$ must be less than (or equal to) $R$ because $OD^2 = R^2-h^2$
(if $h=R$ then $BC$ is a diameter)

Also the circles $w_1,w_2$ must have common point(s).

So we must have $|OA-h|\leq OD\leq OA+h \Leftrightarrow$

$|OA-h|\leq \sqrt{R^2-h^2}\leq OA+h$

If one of the equalities holds, then the circles $w_1,w_2$ are touching, so we have one solution for $D$.

If $|OA-h|< \sqrt{R^2-h^2} < OA+h$, then the circles have two common points $D,D'$ and we get two congruent triangles $ABC$ and $AB'C'$

Draw the circle with center $A$ and radii $h$. We denote it by $(A)$.

Let $X$ and $Y$ be the intersections of $(A)$ and $(O)$. Let $D \equiv AO \cap XY$. Clearly $AO \perp XY$.

Now construct a circle $T$ with center $A$ and radii $\frac{AD\cdot AO}{h}$. I think we can do this, otherwise, we can just construct a segment with this length and proceed.

Notice that the radii of the circle $T$ is greater than $AD$ since we obviously have $AO \ge h$.

Hence circle $T$ has at least one intersection with $XY$, let it be $E$. Let $E'$ be the intersecion of $AE$ with the circumcircle of   $EDO$. We have, $\angle AE'O = \angle EDO = 90$. Also, $AE \cdot AE'  = AD \cdot AO \implies AE' = \frac{AD \cdot AO}{AE} = h$. Let $B$ and $C$ be the intersections of $E'O$ with $(O)$. Triangle $ABC$ satisfies the conditions.

If we had $AO = h$ then just draw the perpendicular to $AO$ at $O$.

[quote="sunken rock"]Construct the circle $(A,h)$, draw the tangents from $O$ to it, they are the two diameters $BC$.

Best regards,
sunken rock[/quote]
But we can't draw tangents when $O$ is in the circle $(A,h)$.


If $ a,b,c \in \mathbb{Z}$ so that $ 8a\minus{}10b\plus{}27c\equal{}0$ then prove that:

$ \frac{(2a\minus{}b)(2b\minus{}3c)(3c\plus{}2a)}{60} \in \mathbb{Z}$.

[hide="Solution"]
From the given equation we get that:

$ 2a-b=3(3b-9c-2a)$

$ 2b-3c=8(a-b+3c)$

$ 3c+2a=10(3c+a-b)$

Multiplying these three together we get and dividing by $ 60$:

$ \dfrac{(2a-b)(2b-3c)(3c+2a)}{60}=4(3b-9c-2a)(a-b+3c)(3c+a-b)$

And since $ a,b,c\in \mathbb{Z}$ then so is $ 4(3b-9c-2a)(a-b+3c)(3c+a-b)$.
[/hide]


limit
calculus
integration
function
calculus computations

These were from a math test at school, and I'm not completely sure I got the right answers:

Evaluate the following limits:

$ \lim_{n\to{\infty}} (\minus{}1)^n * cos((\frac{2n\plus{}1}{2})\pi)$

$ \lim_{n\to{\infty}} (\minus{}1)^n * sin((\frac{2n\plus{}1}{2})\pi)$

I guess that depends where are you from... in Croatia we learn it in high school :)

[quote="t0rajir0u"]Both sequences are periodic, so just evaluate the first few terms.

Limits do not belong in the High School forums.[/quote]
I'm sorry, we learn limits at our high school so I assumed everyone did =/.

I'm still a bit confused as to how evaluating the first few terms gets the limits for those, though.

You'll understand once you evaluate the first few terms.

Alright, I see how for all positive integer values of n, the first sequence always evaluates as 0, and the second sequence always evaluates as 1. But in between integer values of n, the evaluation jumps around a lot, so doesn't that mean that there's no limit?

Hmm.  On second thought, the problem is somewhat ambiguous.  Usually, though, when we use the letter $ n$ we are talking about sequences.  $ (\minus{}1)^n$, in particular, is very poorly defined at non-integers.

Forgive me; I am but an 8th grader, but I know some stuff about limits.
But perhaps my knowledge is inadequate.

I thought that ${ \lim_{n\to\{\infty}} ( - 1)^n$ does not exist because any integral value of n will result in -1 or 1, so there is no definite limit.  How can you even evaluate the problem if the problem involves a limit that does not exist?

Unless you mean:
$ \lim_{n\to{\infty}} [( - 1)^n * cos((\frac {2n + 1}{2})\pi)]$
and
$ \lim_{n\to{\infty}} [( - 1)^n * sin((\frac {2n + 1}{2})\pi)]$

[quote="MathAndKnowledge"]
Unless you mean:
$ \lim_{n\to{\infty}} [( \minus{} 1)^n * cos((\frac {2n \plus{} 1}{2})\pi)]$
and
$ \lim_{n\to{\infty}} [( \minus{} 1)^n * sin((\frac {2n \plus{} 1}{2})\pi)]$[/quote]

I never thought it could be something else.. It's quite unusual for a part of an expression to be evaluated using the $ \lim$ operator, if actually not incorrect to assume it does. It's like setting a variable of a function to a value but only for certain places in the same expression. If $ n\rightarrow \infty$ then it's commonly (and I repeat, if not actually definitely) for all $ n$ in the expression to be evaluated in such manor.

[quote="Flame"]
I guess that depends where are you from... in Croatia we learn it in high school Smile
[/quote]

Bome da..

[quote="Lulze"]These were from a math test at school, and I'm not completely sure I got the right answers:

Evaluate the following limits:

$ \lim_{n\to{\infty}} ( \minus{} 1)^n * cos((\frac {2n \plus{} 1}{2})\pi)$

$ \lim_{n\to{\infty}} ( \minus{} 1)^n * sin((\frac {2n \plus{} 1}{2})\pi)$[/quote]

$ \lim_{n\to{\infty}} ( \minus{} 1)^n * cos((\frac {2n \plus{} 1}{2})\pi) \equal{} \lim_{n\to{\infty}} ( \minus{} 1)^n.cos( \frac {\pi}{2} \plus{} n \pi) \equal{} 0$

$ \lim_{n\to{\infty}} ( \minus{} 1)^n * sin((\frac {2n \plus{} 1}{2})\pi) \equal{} \lim_{n\to{\infty}} ( \minus{} 1)^n. sin( \frac {\pi}{2} \plus{} n \pi) \equal{} 1$


geometry
3D geometry
sphere
Support
MIT
college
Harvard

Goin by the amt of money on stake for research ppl, looks like ill make more money marketing for chinna vayasu sema talent than doing research in mathematics...

@chappli ...
people don't do research for money ofc whatever comes in the way won't be or shudn't be denied :D
they do that for fun or atleast that's what i think it's os good to be pais for what u like doing the most  :D

[quote="pardesi"]@chappli ...
they do that for fun or atleast that's what i think it's os good to be pais for what u like doing the most  :D[/quote]

Perfect point! Sadly researchers dont get paid too much today. Neither do teachers.... Thats why we are going to the dogs and progressing narrowly. :(

[quote="pardesi"]but things have certainly improved what they were 2-3 years ago just look at the number of peopl applied for NEST :D ...
yes money is imp.....but u shud do something for ur nation(ofc - paying taxes :P )[/quote]

No, If our teachers are paid more? Then surely they will be encouraged to give their heart and soul to take class. I personally know that in sum schools middle school teachers are being paid paltry amounts and in some even high school teachers.

Money is imp. Only if you have some stability by the side, you will do research happily giving your heart and soul knowing that you are earning bread for your family...And majority go for job after 4 year courses except the focussed few.

Does India not have the quality to produce scientists with superior technology? Does India not have the tech skills to produce a desi version of Google, which can go one up? Or can we not produce people who can solve the Yang Mills Mass-Gap problem or Birch & ... Dyer??? Surely we can.

Pardesi, start a new thread, this will be an interesting topic.

Guys cool down.. Plz understand wat I meant.. I jus meant researchers ought to be paid more than wat they are given.. Its such a sad sight in India.. As a matter of fact, I will only do research in my career as I personally feel job satisfaction is much more important for me as a human being..I want to work for myself, and not for another guy and his country..
And I think shre very well knows what my plans in life are, so I thot u wud atleast understand what I meant..Im disappointed by ur interpretations!!

I think this problem would change as more and more people are getting to understand the valid point which Chappli put forward......
I think money for resarch would not be a problem from now on, keeping in mind so many are ready to sponsor researchers

yes money is not a problem..problem is mindset...even a millionare can be 'poor' :D

I dont think its a question of getting paid at all.

But as others have pointed out, the research potential at India is very very minimal, and to top that, the recognition and FUNDING for research is barely sufficient, so it productively prevents more people from going into research. If you work for a top notch robotics or bio tech corp, YOU get paid well enough, its that your project is put on hold until and unless sponsors deem it fit to go into production

On the other hand, abroad, wider ranging research opportunities are infinitely more available, and in all spheres, but then again, that could be because they are more developed and the govt can afford to fund such research.

Our institutions are doing pretty well at the outset, but then if only they were to accept colloborations with the institutes of the likes of those abroad, and by that I dont only mean US, even places like Singapore, where resources are technically not abounding, we could do much better

IIT is definitely on par with the top technical universities in the world, just that, the exposure to students is very minimal, and placements are becoming an issue day by day, with more people preferring to do an MBA to settle in the corporate grind rather than go into research.....

And as someone who lectured in our school mentioned, iit has lesser of practical training than many other foreign univs which are @ lower standards than iit  :(

i knwo the research oppurtunities outside are far far far better than in india...but i wud prefer anyday doing it here because doing research outside is as good as selling pancakes(Corporate) in India :)
it's a duty we must do something for our nation....
if it so happens that say 'x' loves do research on 'y' then he/she shud do it here otherwise that find will go 'waste'..
our society has given us many things and it deserves much more respect....
one  C.V.Raman in a century is better than 20 Haragobind Khoronas  :rotfl:

@Rohit: yes, that is one of the predominant cause of concern here

@pardesi: You completely misinterpreted what I said... I never talked about opportunities at all, nor do I undermime the intellect of all those who want to pursue research. But fact is, funding, very clearly monetary support for taking up research isnt available. Ultimately you may argue that we dont need support if we've got the will, but I say that complete rubbish.... You need equipment, you need samplings, and you need testing outlets(for all branches of science perhaps, maybe not math) and for that you need funding, and for that your govt should support research or you should have ties with other sponsors who are willing to fund such research.

I said India lacks INITIATIVE... Like what baby said, IIT's are far superior to colleges like say, Michigan School of Engineering, Georgia Tech or Viriginia Tech, but they give hands on exp.. No guesses as to which one potential research scientists would prefer dont you think?

And I completely resent that statement about Raman and Khurana. No scientist is less than the other, quite literally. It doesnt do good to downsize one.

As a biologist, I think that Khurana's contribution is equally important, for without that, stem cell research of today would be damn near impossible. So never undermine anyone..

I think for certain people, Parental Pressure would be a  problem, They might just want them to be like the rest

[quote="anitsahu"]in india in every decade research hits new lows :|
not a single indian research paper published in "nature" or "science" since 2 or 3 decades
indias best journal has an impact factor of 0.6 against the world average of 8.45
in india one research paper of citation above 1000 is something like 1 in 100000 :roll:
for the matter of fact not a single indian research paper has had citation above 500 in this decade :([/quote]

Yes that's what govi would say, that any indian scientist would rather put his work in a foreign journal rather than put his work in an indian journal and not get recognition  :(

EDIT: For once anit we agree on a topic  :P   :rotfl:

hi i am rt.
wanna study nuclear physics or molecular exobiology from MIT or Harvard or Princeton. :rotfl:  :rotfl:

btw shruti is at Penn State and Ashwath at Princeton......keep it up people....u all r geniuses........

dont worry,u'll be among them by next year :)

And my views abt Hitlerism (:P),really it can do good
i endorse it fully,but dont u think some of us may land up in concentration camps then :rotfl:

Its time for India to attack US:those bloody hells attract Indians like *******:

no one discusses abt lord hitler  :omighty:  without me pokin my nose

we wudn't land up in conc camps but
[hide]
lok sabha, vidhan sabha etc wud have to be disinfected using zyklon b  :rotfl:
[/hide]

attacking america is not the solution, its suicide

i think dis discussion shud be moved to metathread

[quote="rt_08"]Ashwath at Princeton[/quote]

Not yet, sadly. I have four more months of waiting to survive before I leave for Princeton.

what rank is the cutoff for the Integrated MSc in chemistry ????? can I get it in IIT-B ????? heard it's got best faculty for chem.

some advice please

my marks are[hide]220 around that :blush:  :blush:  [/hide]

[quote="ashwath.rabindranath"]Not yet, sadly. I have four more months of waiting to survive before I leave for Princeton.[/quote]
ofc u'll survive:rofl:
best of luck for that!!!

and plz plz do come back to INdia :o

A tradition doesn't bind you with a compulsion that you can't be different from others.
When I say I WILL come back, I mean it. Ofc, the time period may vary in some highly improbable scenario like me getting the opportunity to do some pathbraking research that benefits ALL mankind; Not a particular country.

Ashwanath? Can you post your full profile? Besides INMO TC.

[quote="rt_08"]yes ashwath do come back to india.......v need u all........[b]JAI HO[/b][/quote]

What for?

And for the record, my name is Ashwath. Not Asswat. Not Asvanath. Not Arshavd (yes, people have been very cruel to my name before). Not Aswath. ASHWATH.

Let's talk about the future later. Right now, I don't see myself returning to India for the next 10-15 years once I leave, except for holidays and stuff.

[quote="rituraj007"]there is no use coming back after say 50yrs :rotfl:  coz ppl will then call u NRA (Non reliable americans)[/quote]
I'll be glad if I even live that long :P
So the stipulated time period is much lesser than that  :P
And don't worry, I'm a poetic idiot! Never manged to take life;'s practicality as it should be.  :blush:
Reminds me of a long discussion with rt once. Some interview ques he got. He'll probably elaborate with more clarity :)

well interview s***** big time and i am in no mood of elaborating that acriminous  session....... :mad:

GRRRRRRRRRR!!!!!!!!!!!
Now I've been rejected from both Summer Science Program and International Summer School for Young Physicists. Just got my rejection letter for the latter. Some South Indian Champu got in instead. http://www.perimeterinstitute.ca/Outreach/Students/ISSYP_2009_Participant_List/ (Note that 50% of the international participants have surnames starting with k/l/m) From what google shows, he doesn't have any big physics olympiad/kvpy/ntse etc. I thought my application had enough passion in it to get through :(
If I can't even get into a good summer program, what hope do I have left for top US Univs.....  :(

Dont worry yaar,US admission procedures are far more different and in depth in nature.Rejection from a Summer Program nvr means that ur chances are low at an ivy or a good US university  :)


Find all the continuous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x\plus{}y\minus{}xy)\equal{}f(x\plus{}y)\minus{}f(x)f(y)$, for all $ x,y \in \mathbb{R}$

$ P(x,0): f(x)f(0)\equal{}0$. $ f(x)\equal{}0$ is one solution.
Now assume $ f(0)\equal{}0$ and let $ a\equal{}f(1)$.
$ P(x,1): f(x\plus{}1)\equal{}a(f(x)\plus{}1)$ (*).
$ P(1,\minus{}1): a(1\plus{}f(\minus{}1))\equal{}0$
If $ a\equal{}0$, (*) implies $ f(x)\equal{}0$. So $ f(\minus{}1)\equal{}\minus{}1$.
$ P(x,\minus{}1): f(2x\minus{}1)\equal{}f(x\minus{}1)\plus{}f(x)$ (**)
So $ f(3)\equal{}f(2)\plus{}a$,
(*) gives $ f(2)\equal{}a^2\plus{}a$ and $ f(3)\equal{}a^3\plus{}a^2\plus{}a$
Hence, $ a^3\plus{}a^2\plus{}a\equal{}a^2\plus{}2a\Rightarrow a\equal{}0,\pm1$. We already exclude $ a\equal{}0$.
If $ a\equal{}\minus{}1$, (*) gives $ f(x\plus{}1)\plus{}f(x)\equal{}\minus{}1$ and then (**) implies $ f(2x\minus{}1)\equal{}\minus{}1$, contradiction with $ f(0)\equal{}0$.

Hence, $ a\equal{}1$ and $ f(x\plus{}1)\equal{}f(x)\plus{}1$.
We have $ f(n)\equal{}n,\forall n\in\mathbb Z$ and $ f(x\plus{}n)\equal{}f(x)\plus{}n$.
$ P(x,\minus{}n\plus{}1): f(nx)\equal{}nf(x)$. So $ f(r)\equal{}r,r\in\mathbb Q$ and $ f(rx)\equal{}rf(x)$
$ f(x\plus{}r)\equal{}rf(\frac xr\plus{}1)\equal{}rf(\frac xr)\plus{}r\equal{}f(x)\plus{}r$ (F1)
$ P(x,\minus{}x): f(x^2)\equal{}\minus{}f(x)f(\minus{}x)\equal{}f(x)^2$ (F2)
So $ f(x)\ge0,\forall x\ge0$.
For any $ x$, we can find rational sequences $ r_n\uparrow x,R_n\downarrow x$
$ f(x)\equal{}f(r_n)\plus{}f(x\minus{}r_n)\ge f(r_n)\equal{}r_n$ and similarly $ f(x)\le f(R_n)\equal{}R_n$.
Hence, $ f(x)\equal{}x$.

Either $ f(x)\equal{}0$ or $ f(x)\equal{}x$. We do not need continuity.


prove that for all positive integers $n,m$ so that $m$ is odd we have
       $ 3^{m} n \mid\sum_{k=0}^{m}\binom {3m}{3k} \ (3n-1)^k. $

it's Romanian ROM TST 2004: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=3%5Em%2An&t=5487


limit
trigonometry
function
calculus
calculus computations

There is a classic example to show that if the limit produced by application of L'Hopital does not exist, then the original limit may or may not exist: $ \lim_{x\to\infty} \frac {x \plus{} \sin x}{x}.$

I wish to disprove: Suppose $ f$ and $ g$ are differentiable functions on an open interval containing $ x \equal{} a$, except possibly at $ x \equal{} a$, and that $ \lim_{x\to a} f(x) \equal{} \lim_{x\to a} g(x) \equal{} 0$. If $ \lim_{x\to a} \frac {f'(x)}{g'(x)}$ does not exist, then $ \lim_{x\to a} \frac {f(x)}{g(x)}$ does not exist.

Try $ \frac{x^2 \sin (1/x)}{x}$ as $ x \rightarrow 0$.

Thanks. Should have thought of that, using $ \minus{}1 \leq \sin(1/x) \leq 1.$

Or just turn $ x$ upside-down in the first example and use ${ \frac{\frac{1}{x} + \sin \frac{1}{x}}{\frac{1}{x}}}$.  (Of course, these all work because $ \sin x$ is like $ 1$ but squiggly, while $ \frac{d}{dx} \sin x$ is still like $ 1$ but squiggly instead of being like $ 0$.)

(The terminology "like 1 but squiggly" is a fabulous way of thinking about things that I'm either paraphrasing or directly quoting from Kent Merryfield.)

[quote="JBL"](The terminology "like 1 but squiggly" is a fabulous way of thinking about things that I'm either paraphrasing or directly quoting from Kent Merryfield.)[/quote]
I think I usually say "wiggly" or "wigglier" rather than than "squiggly" - but it works with either word.


I love doing math but I have to take AP classes to get into a good college. But I thought maybe colleges would be impressed by having a high score on the AMC/AIME/USAMO so I could possibly drop my AP and honors classes and have more time to train in mathematics for these competitions.  I've heard that it is more rare to make a perfect score on the AMC12 and only 10,000 people are invited to the AIME. I'm sure the people who qualify for the USAMO or IMO must sacrifice their grades in order to be that good. Would it be wise to concentrate on something I love and will major in college or take honors and AP classes and have less time to do math.

[quote="jobro_gold"]I'm sure the people who qualify for the USAMO or IMO must sacrifice their grades in order to be that good.[/quote]

That's the part I am not at all sure about. I rather suspect that there is a sufficient number of USAMO qualifiers who are really good on AP tests (including AP tests in non-math subjects) that it would be a dangerous strategy to concentrate only on USAMO to the exclusion of taking a challenging general curriculum in high school.

In general, I advise parents and young people not to view college admission requirements as a reason NOT to do something, but as a reason to DO something. So the thinking I would apply to the issue you are concerned about is, "Hey, lots of kids in my high school are taking AP classes, and I can keep up with them, but I'm also going to prepare for the AMC series of contests and try my best to make MOP." Note that it's at least theoretically possible to STILL get dinged in college admission to your favorite school if you are an AP scholar with distinction and an IMO team member and someone who scored a pair of 800s on the SAT I (although that wouldn't be likely  :D  ). Colleges aren't required to admit anybody--not in the United States they aren't, if they are in the top tier of privately operated colleges.

Especially because many of the mistakes I read about here on AoPS that cause people to crash on the AIME are essentially reading mistakes, I think you are best advised doing whatever you can, throughout your high school years, to build up your reading ability. One can never read too well to be good college material. If an AP English course helps you read better than you have ever read before, that would be a good reason to take the course, whether or not you think you need the score on the AP exam.

Pick the path that fits you, then work hard to move forward along it.

Good luck!

I honestly think that no English class is going to help you read the questions on the AIME more accurately. The only thing that can do that is not being in such a hurry. You get twelve minutes per question if you want to get a perfect score, so if you're trying to solve each problem in two minutes, that's not so good. (Wow, I really should listen to my own advice.)

[quote="jobro_gold"]I have to take AP classes to get into a good college.[/quote]

I think the concept, which I have complained repeatedly about before, that anyone "has" to do anything to get into a "good" college is just wrong.  There are a wide variety of things which will help a student get into selective universities, and many different students have many different combinations of these studies, attributes, activities, and whatnot.  Perhaps more annoying, there are many different students, who had very similar combinations of activities, studies, attributes, and whatnot, who did [i]not[/i] get into the same selective universities.  It is highly unlikely that by picking between various activities, you will increase your chances of getting into college; more annoyingly, there is no one, anywhere, who would be able to tell you definatively which combinations would be best.  This, I guess, is just a corollary to tokenadult's mantra that "college admission requirements should not be a reason NOT to do something, but should be a reason to DO something."  Frankly, you're probably best off doing whatever you prefer to do, because doing what you like reverberates in other areas: it makes for something to talk about at interviews, it makes for something to write about in essays, and it makes for something that you are more likely to do in your free time and thus more likely to do well in.

As for my personal interaction with the things, I took 5 AP classes/tests, counting 2 half-year classes and both Phsyics C exams, during high school.

What JBL said is exactly what I was thinking. Not everyone has to take the same course of action (CoA :D ) to be accepted into a decent college. Anyway, I enjoy doing math and I'm extremely stressed out by the AP class I have now, AP World History. So I say, the only way I'm going do work ( I'm a moderate slacker and procrastinator) is if I'm having fun. Then, it doesn't seem like work anymore.  But most people don't even understand why I do math for fun so how can they understand my ideology. But, if I could get some opinions from college staff members - I know their must be some who go to AoPS - that would be EXTREMELY :shock: helpful.


It is impossible to find two positive integers such that the square of one is exactly twice the square of the other. However, there are integers
such that the square of one is one less than twice the square of the other.
Find the product of the smallest two positive integers where this is true.

Testing leads to (5,7), so the product is 35.

quick question here
why can't it be (1,1)?
the question doesn't state that they have to be different


hey i saw this qsn in a test i wrote. i had no idea how to do it. :(

If 2 moles of A and 3 moles of B are mixed to form an ideal soln, vap pressure of A and B are 120 and 180 mm of Hg respectively.

1. What are the compositions of A and B in the vapour phase when the first traces of vapours are formed??

2. At what pressure will the last traces of liquid disappear?

Any ideas on how to proceed?

Its very easy
[hide="idea"]
1.separately find xa and xb and use raoult's law to find Pressure in vapour phase
2.Then use Dalton's law of partial pressures then that's all
[/hide]

[quote="valeriummaximum"]have u given the vapour presuure of the pure component or the pressures at equilibrium[/quote]

pure component.

chemrock-- you will only get the conc of the two components at equilibrium i.e the equlibrium vapour pressure

[quote="chemrock"]
2.Then use Dalton's law of partial pressures then that's all
[/hide][/quote]

can u explain a bit more? how do i connect that to the composition when the first traces of vapours are formed??

It's just a twist that's all:
Let Pa0 & Pb0 be the pressures of pure solvent.
P=Pa0.xa + Pb0.xb
P.x'a=Pa0.xa
and then x'a is the compostion in vapour phase of A.

Vp of the pure component is given
Also you are not given the Vp at equilibrium..
So,I think some data is missing???

@chemrock: your formulae are right no doubt. :) But then how do the answers to the 2 qsns i asked differ??

I think the second is related to boiling point that is vapour pressure =atmospheric pressure :maybe:  :maybe:  :maybe:


limit
real analysis
real analysis unsolved

Assume that the series $ \displaystyle\sum_{n\equal{}1}^{\infty}a_n$ is convergent and $ a_n>0$. Show that there exist a sequence $ b_1\leq b_2\leq\ldots\leq b_n\leq\ldots$ such that $ \displaystyle\lim_{n\to\infty}b_n\equal{}\infty$ and the series $ \displaystyle\sum_{n\equal{}1}^{\infty}a_nb_n$ is convergent.


let $ x,y\in\mathbb{R}_\plus{}$ such that $ x^n\plus{}y^n\equal{}1(n>1,n\in\mathbb{N}_\plus{})$,

Then find the maximum of $ x\plus{}4y$

write the n-th power of x+4y on the left side, and the things only left is to fill in the blank

What is the generalized form of Cauchy-Schwarz inequation?

Well, not every inequality member know it. I'll show it here.

For positive integers m,n and positive reals from x11 to xmn(Failed to use subscripts here, you know what I say, don't you?)
(x11^m+x12^m+……+x1n^m)(x21^m+x22^m+……+x2n^m)……(xm1^m+xm2^m+……xmn^m)
is not less than
(x11x21……xm1+x12x22……xm2+……+x1nx2n……xmn)^m

let $ a\equal{}2^{\frac {2n}{n\minus{}1}}$.
with AM-GM:
$ \frac {(n\minus{}1)a}{a\plus{}1} \plus{} y^n \geq 4ny.(a\plus{}1)^{\frac {\minus{}n\plus{}1}{n}}$
and
$ \frac {n\minus{}1}{a\plus{}1} \plus{} x^n \geq nx.(a\plus{}1)^{\frac {\minus{}n\plus{}1}{n}}$
so
$ (n\minus{}1)\plus{}x^n\plus{}y^n \geq n.(x\plus{}4y).(a\plus{}1)^{\frac {\minus{}n\plus{}1}{n}}$
so
$ x\plus{}4y \leq (a\plus{}1)^{\frac {n\minus{}1}{n}}$.


Triangle ABC is acute-angled

Prove that SinA +SinB+SinC > CosA+ CosB +CosC

There are many possible proofs of this result

Here is one of the proofs.
Triangle $ABC$ is acute-angled so the both sides of the given inequality are positive.Using this we can square this inequality.So we have to prove that
$\sum_{cyclic}{(\cos^{2}{A}-\sin^{2}{A})}$$+\sum_{cyclic}(\cos{A}\cdot\cos{B}-\sin{A}\cdot\sin{B})<0$
Using that $\sin^{2}{A}+\cos^{2}{A}=1$ we get
$\sum_{cyclic}{(2\cdot\cos^{2}{A}-1)}+2\cdot\sum_{cyclic}{\cos(A+B)} <0$ or
 $t=2\cdot\sum_{cyclic}{\cos{A}\cdot(\cos{A}-1)} <3$.
But $0<\cos{x}<1$ there $0<x<\frac{\pi}{2}$ so $t<0$.

With some rewriting and use of identities, we get that the inequality is equivalent to proving:

$\cos(a+\pi/4)+\cos(b+\pi/4)+\cos(c+\pi/4) < 0$

because of the identity $\sin a - \cos a = -\sqrt{2}\cos(a- \pi/4)$

with $a+b+c = \pi$

Now at least two of $a,b,c$ are greater than $\pi/4$

Write $a = \pi/4 + A, b = \pi/4 + B, c = \pi/2 - A - B$

with $A,B < \pi/4$

We now have to prove

$\sin A + \sin B > \cos(3\pi/4 - A - B)$

The RHS is $\cos(3\pi/4)\cos(A+B) + \sin(3\pi/4)\sin(A+B)$

The first term is clearly negative, so we need to prove that

$\sin A + \sin B > \sin(3\pi/4)\sin(A+B) $

But

$\sin(3\pi/4)\sin(A+B) = \sqrt{2}/2 \sin A \cos A + \sqrt{2}/2 \sin B \cos B$

Since $\cos A, \cos B \leq 1$, we are done.

I really wanted a proof by convexity/concavity, though  :(

One more note.  The LHS is > 2, the RHS is < 2.  Anyone have a simple way of proving this, which trivializes the result?

Jensen's was (of course) the first thing I tried.

$f(x) = \sin x - \cos x$ is not convex over $(0,\pi]$

However, there's a cool trick used when finding the minimum of $\cos x +\cos y + \cos z$ where $x,y,z$ are angles of a triangle, I wonder if it will work here.

Blahblahblah, You wanted to apply convexity/concavity.In this problem you can only apply Jensens inequality to suitable bounding functions as follows:

SinA =2SinA/2.CosA/2 >2(SinA/2)^2 , as angles are acute.

Therefore SinA +SinB +SinC>2[ (SinA/2)^2 +(SinB/2)^2 +(SinC/2)^2 ]

The function (SinX/2)^2 is convex on the interval 0<X<PI/2

It follows that the expression in  [     ] above  is greater than or equal to

3 {SIn(A+B+C/6)}^2 =  3/4. by Jensen's Inequality.So finally we get that SinA+SinB +SinC>3/2

Now we find an upper bound for the sum of the cosines

CosA +Cos B =2cosA+B/2.CosA-B/2
                    <=2 sinC/2

THerefore  CosA +COsB +CosC <=SInA/2 +SinB/2 +SinC/2

The function SinX/2 is concave on the intereval 0<X<PI/2

It follows that the sum of the cosines is <=3SIn(A+B+C/6)= 3/2.Using  Jensen again.

Actually there is a very simple proof that SinA +Sin B +SinC >2 in an acute angled triangle
SinA +SinB+SinC >SinA^2 +SinB^2 +SinC^2 = 2 +2CosACosBCosC >2 since the cosines of acute angles are positive

At first, I just wanted to note that in Armo's nice solution

[quote="Armo"]Here is one of the proofs.
Triangle $ABC$ is acute-angled so the both sides of the given inequality are positive.Using this we can square this inequality.So we have to prove that
$\sum_{cyclic}{(\cos^{2}{A}-\sin^{2}{A})}+\sum_{cyclic}(\cos{A}\cdot\cos{B}-\sin{A}\cdot\sin{B})<0$
Using that $\sin^{2}{A}+\cos^{2}{A}=1$ we get
$\sum_{cyclic}{(2\cdot\cos^{2}{A}-1)}+2\cdot\sum_{cyclic}{\cos(A+B)} <0$ or
 $t=2\cdot\sum_{cyclic}{\cos{A}\cdot(\cos{A}-1)} <3$.
But $0<\cos{x}<1$ there $0<x<\frac{\pi}{2}$ so $t<0$.[/quote]

the formula

$\sum_{cyclic}{(\cos^{2}{A}-\sin^{2}{A})}+\sum_{cyclic}(\cos{A}\cdot\cos{B}-\sin{A}\cdot\sin{B})<0$

should actually be

$\sum_{cyclic}{(\cos^{2}{A}-\sin^{2}{A})}+2\sum_{cyclic}(\cos{A}\cdot\cos{B}-\sin{A}\cdot\sin{B})<0$.

Then, indeed, as Michael Niland showed, $\sin A + \sin B + \sin C > 2$ and $\cos A + \cos B + \cos C \leq \frac32$. The first of these two inequalities was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=5379 ; the second one is well-known and holds not only for acute triangles.

  Darij

This is easy! :D
$\cos A+ \cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}=2\sin \frac{C}{2}\cos \frac{A-B}{2}\leqq2\sin \frac{C}{2}$, similarly $\cos B+\cos C\leqq2\sin \frac{A}{2}$, $\cos C+\cos A\leqq2\sin \frac{B}{2}$,
adding up these inequality and dividing by 2 both of sides,we have $\cos A+\cos B+\cos C\leqq\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}<\sin A+\sin B+\sin C$,because when $0<\frac{A}{2}<\frac{\pi}{4}$, we get $\frac{1}{\sqrt{2}}<\cos \frac{A}{2}<1$.Thus we obtain $\sin\frac{A}{2}<\sin A$ etc.

kunny


inequalities
number theory proposed
number theory

[quote="stvs_f"]prove that :$ [1,2,3,...,n] \geq 2^{n \minus{} 1}$[/quote]

Hemmm, what's the meaning of this inequality ? interval greater than integer ?  :huh:

[quote="stvs_f"]($ n \in N$)what is the problem?[/quote]

I'm sorry, but I dont understand the notation $ [1,2,...,n]$. What is the meaning of this expression ?

then prove that lcm(1,...,n)<4^(n-1) and deduce that 1/(a) + 1/(a+b) + ... + 1/(a+nb) isn't an integer(a and b are positive coprime numbers)

[quote="math10"][quote="stvs_f"]prove that :$ [1,2,3,...,n] \geq 2^{n \minus{} 1}$[/quote]
that mean: prove that: $ lcm(1,2,...,n)\geq 2^n$[/quote]
 $ lcm(1,2,...,n)\geq 2^{n\minus{}1}$


integration
logarithms
MIT
college
real analysis
real analysis unsolved

Here are three series including binom.
(1)$ \sum_{n\equal{}1}^{\infty }\frac{(\minus{}1)^{n}}{n\binom{2n}{n}}$
(2)$ \sum_{n\equal{}1}^{\infty }\frac{(\minus{}1)^{n}}{n^{2}\binom{2n}{n}}$
(3)$ \sum_{n\equal{}1}^{\infty }\frac{(\minus{}1)^{n}}{n^{3}\binom{2n}{n}}$

hello, do you want to know the values of your sums?
Sonnhard.

Use the identity $ \int_{0}^{1} t^n (1 \minus{} t)^{n\minus{}1} \, dt \equal{} \frac{n! (n\minus{}1)!}{(2n)!} \equal{} \frac{1}{n {2n \choose n}}$.  This takes care of the first two sums but the third I'm not completely sure about.

hello, for the first sum i have got
$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n}{n\binom{2n}{n}} \equal{} \minus{} \frac {2}{5}\sqrt{5}arsinh(\frac {1}{2})$.
Sonnhard.

Thanks for your replies.I solved (2),and it is $ \minus{}\ln^{2} \frac{\sqrt{2}\minus{}1}{2}$.
But (3) is still difficult.

hello, for 2) i have got
$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n^2\binom{2n}{n}}\equal{}\minus{}2arsinh(\frac{1}{2})$
and for 3)
$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n^3\binom{2n}{n}}\equal{}\minus{}\frac{2}{5}\zeta(3)$.
Sonnhard.

Allow me to change the sign operator, after all it makes no difference.

I can only confirm Sonnhard's statement, that the first series $ \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^{n \minus{} 1}}{n\, {2n\choose n}}$

converges to $ \frac25 \, \text{arsinh}\left(\frac12\right)\cdot \sqrt {5}$, or equivalently to $ \frac2{\sqrt {5}} \, \text{arsinh}\left(\frac12\right)$ .

You also get this answer from Maple and Mathematica.

The second series $ \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^{n \minus{} 1}}{n^2\, {2n\choose n}}$  is $ 2\, \text{arsinh}^2\left(\frac12\right) \equal{} 2\log^2\phi$ ,

where $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}$ stands for the golden ratio.

This value is also stored in both computer algebra systems.

The third series $ \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^{n \minus{} 1}}{n^3\, {2n\choose n}}$ seems to be the most difficult of them.

Here the value $ \frac25 \zeta(3)$ is only available in Mathematica.

If you want to know, how to prove, that these series match with their corresponding values,

you can try to read this [url=http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Unendliche_Reihen#Reihen_mit_zentrierten_Binomialkoeffizienten]part[/url] of the German wikibooks formula collection.

For one method of proving (3), check equations 36, 37 of this
http://mathworld.wolfram.com/AperysConstant.html


function
integration
calculus
real analysis
complex analysis
complex analysis unsolved

Fix $a > 0$, $p > 1$, and consider any function $f$ s.t. the Fourier inversion formula holds. Also define the Fourier transform $\hat{f}: t \mapsto \int_{-\infty}^{\infty}f(x) e^{-2\pi i t x}\, dx$, and assume $\hat{f}(t) = O\left(e^{-\frac{2\pi a^{p}}{p}|t|^{p}}\right)$ as $|t| \to \infty$. Now let $q = \frac{p}{p-1}$ so that $\frac{1}{p}+\frac{1}{q}= 1$. Prove that $f$ can be extended to a entire fct. (easy) and for $b > \frac{2\pi}{qa^{q}}$, there exists $A_{b}> 0$ s.t. $|f(z)| \le A_{b}e^{b|z|^{q}}$ for all $z \in \mathbb{C}$ (hard).

Hint: $|ty| \le |t|^{p}\frac{a^{p}}{p}+|y|^{q}\frac{1}{qa^{q}}$ by AM-GM.

Aww... no takers? :) I guess this forum doesn't like Fourier analysis very much.

[quote="fairyofwind"]

Hint: $|ty| \le |t|^{p}\frac{a^{p}}{p}+|y|^{q}\frac{1}{qa^{q}}$ by AM-GM.[/quote]

Dont you mean "Holder inequality".
Ha ha!

Ignore sylow_theory, he's just trolling.  You'll get more responses on Fourier analysis problems in the Calculus - Real Analysis section though; it gets a lot more traffic than does this forum.


I want write a theorem with this i discovered in the high school when i was 16 years old
\[ \begin{array}{l} {\rm x}^{\rm 2} \plus{} (y \minus{} 1)^2 \equal{} 4{\rm } \\
{\rm (0}{\rm ,1)} \\
\left\{ {_{\frac {{\partial (y \minus{} 1)^2 }}{{\partial y}} \equal{} \frac {{\partial (y^2 \minus{} 2y \plus{} 1)}}{{\partial y}} \equal{} 2y \minus{} 2 \equal{} 0 \Leftrightarrow y \equal{} 1}^{\frac {{\partial x^2 }}{{\partial x}} \equal{} 2x \equal{} 0 \Leftrightarrow x \equal{} 0} } \right. \\
\end{array}
\]
:(

[b]Claim:[/b]  Let $ (a, b)$ be a point on a circle $ (x \minus{} h)^2 \plus{} (y \minus{} k)^2 \equal{} r^2$.  The equation of the tangent line at $ (a, b)$ is $ (a \minus{} h)(x \minus{} h) \plus{} (b \minus{} k)(x \minus{} k) \equal{} r^2$.

[b]Generalization:[/b]  Let $ \mathbf{P}$ be a point in the zero set of a  [url=http://en.wikipedia.org/wiki/Quadratic_form]quadratic form[/url] $ \mathbf{XAX}^T \equal{} 0$.  The equation of the tangent space at $ \mathbf{P}$ is $ \mathbf{PAX}^T \equal{} 0$.

Edit:  I seem to have misread your post.  I'm not sure what you're trying to say.  It's true that $ \frac {d(x \minus{} h)^2}{dx} \equal{} 0 \Leftrightarrow h \equal{} 0$ because the rate of change of a quadratic is zero only at its vertex.  There's nothing else going on here.


AMC
AMC 8
AMC 10
AIME
MATHCOUNTS
AMC 10 A

how come you guys never talk about AMC8 in this forum ????

and never make mock AMC8????

and another problem....

my brother got the first place in school last year in AMC8 but never got invitation for AMC10....

[quote="shinwoo"]
and another problem....

my brother got the first place in school last year in AMC8 but never got invitation for AMC10....[/quote]

First place in school or a perfect score - you don't need an invitation to do the AMC10.  Someone in the school may suggest/recommend.

In our school, we recommend AMC10 to students who we think have some ability to do it - especially if they are in middle school.  Based on the response we inform the high school about the number of test takers so they get included in the count for ordering tests.   We do not want students to feel overwhelmed or have a bad experience.

In high school they sign up if interested.

Naga

[quote="shinwoo"]how come you guys never talk about AMC8 in this forum ????

and never make mock AMC8????

[/quote]

Here's a link to a mock AMC8.  Those discussions were hot before November.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14666

Now you will hear more about AIME and then USAMO.  Simply because the AIME is coming up and that will be followed by the USAMO.

Naga

huh??

they are AMCB

not AMC 8

B  :!=:  8

The mock AMC tests here have letters, to indicate that they are mock tests.

The interest in each AMC test level is seasonal. Now we have just been in the AMC 10/12 season, so everyone was talking about those tests, and people are beginning now to talk about the AIME, which comes next. Stay here for a year and you will see each test mentioned in turn.

but

last year, we never talked about AMC8 and I screwed up the test pretty good... :(

Try [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878[/url] for the Mock AMC 8.  And I think part of the reason that people aren't talking about it is because not that many people are trying to do well on it.  If you want practice, just keep doing the problems on the website, and look at some of the MathCounts resources too.

AMC8 is not big at all. The people who score high on the AMC8 are usually more focused on the AMC10. That's why it isn't very popular.

[quote]The people who score high on the AMC8 are usually more focused on the AMC10.[/quote]

Not only focused on AMC10.  The top MATHCOUNTS students can usually score well in AMC10.  I am sure that you will see  a few perfect scorers from middle school in this year's AMC10A because it was too easy.

IMO, if you participate MATHCOUNTS in middle school, AMC8 and AMC10 will come along like a piece of cake...

-My 22


Let $a_1=\sqrt2$. For $n\ge2$, define $a_n=\sqrt{2+a_{n-1}}$. Then $a_4=2\cos\theta$ for which of the following values of $\theta$?

(a)$\frac{\pi}{12}$

(b)$\frac{\pi}{16}$

(c)$\frac{\pi}{18}$

(d)$\frac{\pi}{24}$

(e)$\frac{\pi}{32}$

[hide]$a_n=2\cos \frac{\pi}{2^{n+1}}$

Base case: $a_1=\sqrt{2}=2*\frac{\sqrt{2}}{2}=2\cos \frac{\pi}{4}$.

Inductive step: Let $\theta=\frac{\pi}{2^{n+1}}$. Then $a_{n+1}=\sqrt{2+2\cos \theta}=2\sqrt{\frac{1+\cos \theta}{2}}=2\sqrt{\cos^2 \frac{\theta}{2}}=2\cos \frac{\theta}{2}$.

Setting $n=4$, $a_4=2\cos \frac{\pi}{32}\Rightarrow \boxed{E}$.[/hide]

[hide="Answer"]Well, I have it memorized that $2\cos \frac{\pi}{2^n}$ is the sum of $n-1$ $\sqrt{2}$'s successively inside the previous one. So, $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}=2\cos \frac{\pi}{32}\Rightarrow \boxed{E}$.

For a more in-depth solution, read scorpius'. He probably explained it better than I can. :P[/hide]


inequalities
AMC
USA(J)MO
USAMO
algebra proposed
algebra

strange problem ;)  well,what does[color=red] "adapted" by harazi [/color]mean?

I know it is strange and especially I think difficult. "Adapted" is a word I like to put in front of those "creations" of mine which are easily derived from some famous problems, but which, strangely, cannot be seen so easy even by experienced people. This one is an example.

Too hard (at least for me!)
Dear harazi,could you give me some hints?

I believe you made a mistake. Check the case where $a_1=1$ and $a_2=0.5$.

I didn't notice that $i,j$ need not be distinct; sorry.

Harazi, did you adapt the result from the inequality in USAMO 2000 Problem 6?

Of course!  ;)    And you know, I really hoped someone who does not see that will come with an easy solution. But I guess this is impossible.  :D

Your clever plan may have worked.  Your question got me thinking about the USAMO 2000 inequality.  I just posted a proposed simple solution to that inequality in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15292]The hardest USAMO inequality problem[/url].

My plan always works but we must have some patience and the results quickly follow. I said it there and I say it here: this is the most beautiful solution to an inequality that I have ever seen.

Coming from you, a master of inequalities, that is high praise.  Thanks!

I haven't finished my solution yet but here is my idea :
 The hypothesis is equivalent to :
 $ \sum_{1\ leq i \leq n}|1-x_i^2| = 2.(\sum_{1 \leq i < j \leq n}|x_i-x_j| - \sum_{1 \leq i < j \leq n}|1-x_ix_j|) $
[b] Case 1 : [/b]  $ x_i \geq x_j \geq 1 $
 [b] Case 2:[/b]  $ 1 \geq x_i \geq x_j $
 In 2 cases , we will prove :
    $ |1-x_i^2| + |1-x_j^2| \geq 4.(|x_i-x_j| - |1-x_ix_j|) $
   It's easy  ;)
 [b] Case 3 :[/b] $ x_i \geq 4 , x_j \leq 1 $ , we can easily show : $ x_i^2-x_j^2 \geq 4(x_i-x_j) $
  [b]Case 4 :[/b] $ 4 \geq x_i \geq 1 \geq x_j >0 $ : not yet  :)
[/b]

in fact ,I still do not know how to solve it,help!

[hide="Hint"]Use the identity
\[
  \left| p - q \right| = p + q - 2 \min(p, q)
\]
to get rid of all the absolute values.  You can then use the USAMO 2000 inequality.
[/hide]

Moubi asked me to write a more complete solution, so here goes.  By replacing the absolute values with minimums as my hint recommended, we get the equation
\[
  \sum_{i,j} [x_i + x_j - 2 \min(x_i, x_j)]
     = \sum_{i, j} [1 + x_i x_j - 2 \min(x_ix_j, 1)].
\]
By grouping the terms not involving minimum together, we get the equation
\[
  -2 \sum_{i, j} \min(x_i, x_j)
    = \sum_{i,j} (x_i-1)(x_j-1) - 2 \sum_{i, j} \min(x_i x_j, 1).
\]
But the first sum on the right-hand side is just the square of $\sum_i (x_i - 1)$.  Thus we get
\[
  -2 \sum_{i,j} \min(x_i, x_j)
    = \left(\sum_i x_i - n\right)^2 - 2 \sum_{i, j} \min(x_i x_j, 1).
\]
Now let's apply the USAMO 2000 inequality with $a_i = x_i$ and $b_i = 1$.  We get the inequality
\[
   \sum_{i, j} \min(x_i, x_j) \ge \sum_{i, j} \min(x_i x_j, 1).
\]
Doubling the last equation and adding it to the second-to-last equation, all the minimum terms cancel out, and we are left with the inequality
\[
  0 \ge \left(\sum_i x_i - n \right)^2.
\]
The only nonpositive square that I know of is zero.  Thus we get
\[
  \sum_i x_i = n.
\]
That's what we wanted.

Harazi this is the greatest problem I have ever seen :lol: But the solution of Ravi B is unpredictable. It is all the money.WOW :first:

Thanks.

Yes, harazi did come up with a cool problem.  Without the hint that his inequality was derived from another famous problem, it would have been a lot harder.


source: Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu

Prove that if $a_{1}<a_{2}< \ldots <a_{n}$ and $1 \leq k < n$ $(k,n \in N)$ then: $\displaystyle \frac{a_{1}+a_{2}+ \ldots +a_{k}}{k} < \frac{a_{1}+a_{2}+ \ldots +a_{n}}{n} < \frac{a_{k+1}+a_{k+2}+ \ldots +a_{n}}{n-k}$.

Let me prove first the left inequality. Be back soon with the right :).

Lemma:

[tex]\frac{a_1+...+a_k}{k} < \frac{a_1+...+a_{k+1}}{k+1}[/tex]. This reduces to [tex](k+1)(a_1+...+a_k) < k(a_1+...+a_{k+1})[/tex] or [tex]a_1 + ... + a_k < ka_{k+1}[/tex], which is obviously true.

By applying the lemma repeatedly, we get the left hand side.

For the right hand side, we could apply same idea, but we can look at it this way. If we multiply by [tex]n \times (n-k)[/tex], we get  [tex]n \times (n-k)[/tex] summands on both sides. It is obvious that after eliminating some equal terms on both sides, we get left with a sum of lower terms on the left than on the right.


geometry
trigonometry
AMC
AIME
area of a triangle
Heron\u0027s formula
perpendicular bisector

Right triangle ABC has AC=7, BC=24, and right angle at C. M is the midpoint of AB, and D is on the same side of AB as point C such that AD = BD = 15. What is the area of triangle CDM?

Well, after applications of the Law of cosine, Stewart's Theorem, etc..., I managed to find the sides of triangle CDM:

DM = 5/2 * Sqrt(11)
CM = 25/2
CD = Sqrt(225 - 168/5 * Sqrt(11))

One could apply Heron's Formula now, but with the nasty numbers and no calculator, that could be very tough. Does anyone know of a way around this (or even an easier way to solve the problem)?

[quote]I don't really understand how AD=BD and for that side to be 30 (15+15) if the other two sides of the right triangle are 24 and 7 shouldn't the third side be 25 and also isn't point C the same as point M?[/quote]

No, you see, it is given in the problem that AD = BD--I didn't derive it. D is just some other point off somewhere that satisfies that equality. Yes, the third side is 25, but no one said it wasn't.  8-) And how is C the same as M? M is the midpoint of AB, and C is the vertex opposite that hypotenuse AB, so unless triangle ABC is degenerate, which it isn't, then C cannot be M.

Wow, I guess Fierytycoon likes the 2003 AIME B. Anyway, I distinctly remember this to be a hard problem, becasue I couldn't figure out on which side of CM point D was. I tried the side of B, but i guess that didnt' work out. Anyone help me out here?

By application of mind power, it is possible to determine that DM = 5sqrt(11)/2, CM = 25/2, angle CMD = pi/2 - 2arcsin(7/25), hence sin CMD = cos(2arcsin(7/25)) = 1 - 2*(7/25)^2 = 527/625. By a forgotten formula of the ancients, one determines that the area of triangle CMD is equal to 1/2 * 5sqrt(11)/2 * 25/2 * 527/625 = 527sqrt(11)/40.

Obviously, D lies on the perpendicular bisector of AB and being as AC is shorter than BC, this is the side of CM that B is on.

You have allowed the people to make the golden calf and for this, you will die before you see the promised land.

Hmm, teratomato, so how did you obtain angle CMD?

[quote]Wow, I guess Fierytycoon likes the 2003 AIME B[/quote]

Yeah, I didn't get quite a few on there...  :-x

[quote="teratomato"]

Obviously, D lies on the perpendicular bisector of AB and being as AC is shorter than BC, this is the side of CM that B is on.

[/quote]

Omg, i can't believe i missed that. Shoot, that would have made the problem a lot easier.....

[quote]Hmm, teratomato, so how did you obtain angle CMD?
[/quote]

nm, I managed to derive the sine of the angle in a different way. Thanks to all who helped.