algebra polynomial integration calculus derivative inequalities trigonometry

Suppose that $f$ is a complex poynomial of degree $n$ with $f(0)=1$ and the maximum value of $|f|$ in the (closed) unit disc is $M$. Then any root of $f$ in this disc has multiplicity at most $c\sqrt{n}$, where $c=c(M)$ is a constant depending on $M$ only.

I think I have a solution. It's rather ugly and long, maybe because it's based on the a bit weaker hypothesis that $\int_{D_{1}(0)}|f^{2}|\leq 2\pi M^{2}$.

Assume that a root $w$ has multiplicity $r$. By considering the polynomial $f_{1}=f(wx)$ which clearly satisfies the same condition we can suppose that $w=1$.

Let $f=1-a_{1}x-\ldots-a_{n}x^{n}$. As $1$ is a root of multiplicity $r$, by taking derivatives and after clear linear manipulations we deduce that

$a_{1}+a_{2}+\ldots+a_{n}=1$
$a_{1}+2a_{2}+\ldots+na_{n}=0$
$\ldots$
$a_{1}+2^{r-1}a_{2}+\ldots+n^{r-1}a_{n}=0$

or if we denote by $x_{k}=(1^{k},2^{k},\ldots,n^{k})$ and $v=(a_{1},a_{2},\ldots,a_{n})$

then $vx_{0}=1, vx_{1}=\ldots=vx_{r-1}=0$.

Let $X$ be the linear space spanned by $x_{1},x_{2},\ldots,x_{r-1}$.
We deduce that if $x\in X$ then $v(x_{0}-x)=1$. Therefore $\parallel v\parallel \parallel x_{0}-x\parallel \geq 1$. However the condition $\int_{D_{1}(0)}|f^{2}|\leq 2\pi M^{2}$ imposes $\parallel v\parallel \leq M-1$. The problem is solved if we show that for $r>c\sqrt n$ the distance from $x_{0}$ to $x$ is less than $\frac{1}{M}$.

Now let's rephrase this statement a bit, by switching from $r$ to $n$ to feel more confortable. So let $N<cn^{2}$ for suitably chosen $c$ and consider $x_{i}$ in $R^{N}$ as before, $X$ spanned by $x_{1},\ldots,x_{n}$. We show that

(1) there is $x\in X$ with $\parallel x_{0}-x\parallel <a$ where $a=\frac 1{2M}$.

to show (1) we use a more convenient Hilbert Space with similar properties, namely $\mathcal{L_{1}}[0;1]$. In this space, we define $f_{i}=x^{i}$, $(f,g)=\int_{0}^{1}f(x)g(x) dx$, $\parallel f\parallel =\sqrt{(f,f)}$. Clearly $(f_{i},f_{j})=\frac 1{i+j+1}$. Define $X$ the subspace spanned by $f_{1},f_{2},\ldots,f_{n}$ and we prove that

(2) $d(f_{0},X)=\frac 1{n+1}$.

Proof of (2):

Let $f\in X$, $f-f_{0}=d(f_{0},X)$. Then $f_{0}-f\bot X$. Set $f_{0}-f=1+a_{1}x+a_{2}x^{2}+\ldots+a_{n}x^{n}$.

The conditions $(f_{0}-f,f_{k})=0$ for $k=1,\ldots,n$ will imply that $g(x)=\frac 1{x+1}+\sum_{i=1}^{n}\frac{a_{i}}{x+i+1}$ satisfies $g(1)=g(2)=\ldots=g(n)=0$, therefore $g(x)=s\frac{(x-1)(x-2)\ldots(x-n)}{(x+1)(x+2)\ldots(x+n+1)}$. By multyplying this by $x+1$ and setting $x=-1$ we deduce $s=\pm \frac 1{n+1}$.

Therefore $g(x)=\pm \frac 1{n+1}\frac{(x-1)(x-2)\ldots(x-n)}{(x+1)(x+2)\ldots(x+n+1)}$. Then $(f_{0}-f,f_{0})=g(0)=\frac 1{(n+1)^{2}}$. Hence $\parallel f_{0}-f\parallel^{2}=(f_{0}-f, f_{0}-f)=(f_{0}-f,f_{0})-(f_{0}-f,f)=\frac 1{(n+1)^{2}}$ and (2) is proven.

Proof of (1):

Using (2) we deduce that there exist $a_{1},a_{2},\ldots,a_{n}$ s.t. $p(x)=1-a_{1}x-\ldots-a_{n}x$ satisfies $\int_{0}^{1}p^{2}(x) dx =\frac 1{(n+1)^{2}}$. But $\int_{0}^{1}p^{2}(x) dx=\int_{0}^{\frac 1{100n^{2}}}\sum_{i=0}^{100n^{2}-1}p^{2}(\frac i{100n^{2}}+x) dx$, and by Lagrange Theorem this equals $\frac 1{100n^{2}}\sum_{i=0}^{100n^{2}-1}p^{2}(\frac i{100n^{2}}+t)$ for $t \in [0;\frac 1{100n^{2}}]$.

Therefore for $q(x)=p(x+t)$ we have $\sum_{i=1}^{100n^{2}-1}q^{2}(\frac i{100n^{2}})=\frac{100n^{2}}{(n+1)^{2}}\in [99;100]$ for big $n$. Let's show that $q(0)>\frac{1}{2}$. Indeed, if $q(0)<\frac{1}{2}$ then $p(t)<\frac{1}{2}$ and as $p(0)=0$, by Lagrange we deduce $(p^{2})'(w)\geq \frac 1{2t}\geq 50n^{2}$. However if we denote by $F(x)=\int_{0}^{x}p^{2}(t) dt$ then by using the famous formula $max_{[0;1]}\parallel g'\parallel \leq deg^{2}(g) max_{[0;1]}\parallel g\parallel$ two times, we see that $max_{[0;1]}p^{2}(t)<4$ hence $max_{[0;1]}\parallel (p^{2})'(t)<16n^{2}$ contradiction. So $q(0)>\frac{1}{2}$. Therefore by denoting $h=\frac{q}{q(0)}$ we have found a polynomial $h=1+b_{1}x+\ldots+b_{n}x^{n}$ such that $h(0)=0$ and $\sum_{i=1}^{100n^{2}-1}h(\frac i{100n^{2}})\leq 200$.

If we put $c_{i}=\frac{b_{i}}{100n^{2i}}$ we see that we have actually obtained  $\parallel x_{0}-\sum_{i=1}^{n}c_{i}x_{i}\parallel^{2}\leq 200$.

Let $N=100n^{2}$. We say that $P$ is a one-polynomial that approximates zero up to $a$ on $R^{m}$ if $P(0)=1$ but $\sum_{i=1}^{m}P^{2}(i)<a$. Then (1) demands us to construct a one-polynomial of degree at most $n$ that approximates zero up to $\frac 1{2M}$ on $R^{cn^{2}}$. But above we constructed a one-polynomial of degree $n$ that approximates zero up to $200$ on $R^{100n^{2}}$. (1) then follows directly from the following lemma:

Lemma: If there is a one-polynomial $p$ of degree $n$ that approximates $0$ up to $200$ on $R^{N}$ then for some constants $c_{1},c_{2}$ depending just on $a>0$ there is a one-polynomial $q$ of degree at most $c_{1}n$ that approximates $0$ up to $a$ on $R^{Nc_{2}}$
Proof of lemma:
Fix $b=0.999$. Then for some $k$ we have $b^{2}+b^{4}+\ldots+b^{2k}>200$.

We claim there is a number $l$ such that $p(l)>b^{k}$ and $p(l+j)<bp(l)$ for any $j=1,2,\ldots,\frac{N}{k}$. Indeed, assume not. Then zero is not such a number so we have a $j_{1}\leq \frac{N}{k}$ such that $p(j_{1})>b$, then analogously there exists a $j_{2}\leq \frac{N}k$ such that $p(j_{1}+j_{2})>b^{2}$ and so on up to $j_{k}$ which contradicts the fact that $p$ approximates zero up to $200$.

So suppose we found such a number $l$. Consider $r(x)=\frac{p(l+x)}{p(l)}$. Then $r(0)=1$ but $r(i)<b$ for $i=1,\ldots,\frac{N}{k}$. We see that $\sum_{i=1}^{\frac{N}{k}}r^{2}(x)<\frac{100}{b^{2k}}$. Now choosing such a $c_{1}$ such that $b^{c_{1}}<\frac{b^{2k}}{100}a$ we see that $r^{c_{1}}(x)$ is the desired polynomial for $c_{2}=\frac{1}{k}$

QED

Yes, it's long, but it is extremely beautiful, not ugly! What a nice idea! I will try to post some other approach, much less natural, this week-end.

Here is an alternative approach:as you observed, it works if the $L^{2}$ norm of $f$ is at most $M$. So, assume that the $L^{2}$ norm of $f$ is at most $M$ and suppose that $1$ has multiplicity $r$. Then for any polynomial $f$ of degree at most $r-1$ one has $g(0)=a_{1}g(1)+...+a_{n}g(n)$ (I keep your notations). Using Plancherel's inequality and the Cauchy-Schwarz one, we infer that $|g(0)|^{2}\leq M (|g(1)|^{2}+...+|g(n)|^{2})$. Now, I claim we can construct a polynomial $g$ of degree at most $C(M)\sqrt{n}$ such that the last relation fails, thus the conclusion wil follow. Consider $D_{p}(x)=\frac{\sin((p+1/2)x)}{\sin (x/2)}$ the Dirichlet kernel and observe that if you put $h(\cos(x))=1/2(1+D_{p}(x))$, this gives you a polynomial of degree at most $p$. However, it's clear that $|h(x)|^{2}\leq 2/(1-x)$ and so if we define $g(x)=h(1-2x/n)^{4}$, we get $|g(k)|^{2}\leq 16 n^{4}/k^{4}$ and thus $\sum_{k=1}^{n}{|g(k)|^{2}}\leq 20 n^{4}$. Also, $|g(0)|^{2}=|h(1)|^{8}=|D_{p}(0)|^{8}=(p+1)^{8}$. It's enough to ensure that $p^{8}>20 M n^{4}$, which is possible for some $p=O(C(M)\sqrt{n})$. [/code]

Thank you, it is shorter and simpler. Actually the proofs are the same up to the affirmation that there exists a polynomial $g$ of degree $c\sqrt N$ such that $|g(0)^{2}|\geq M(|g(1)^{2}+\ldots+g(n)^{2}|)$. I didn't see this elegant Dirichlet Kernel and involved in much more messy computations.


Does anyone have sat II math practice exams?

The College Board has SAT II practice exams (only one for math,  I think) in the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874475996/learninfreed]Real SAT II: Subject Tests[/url].

Most of the threads I have seen here and on College Confidential suggest that the Barrons cram book is useful for people who want even more SAT II math level 2 practice.

I have an old test which I'll post online when I get home.....

But don't worry.... I took the SAT II Math 2c without ever preparing for it (I meant to :blush: ) and it was EASY.

Barron's is pretty good for Sat Math 2.

But don't worry about SAT 2 Math since chances are if you're an AOPSer you shouldn't have any trouble with it :D

That's definitely not true. I know several pretty good students who found the Math SAT II to be pretty difficult.

I found it pretty easy myself, the Math IC and IIC (800 on both).

Math IIC apparently has a huge curve on it, people could miss 5 problems on that and still get an 800, whereas on the IC, one question wrong drops you instantly to a 790.

Both math iic and physics have amazingly huge curves on them(as opposed to many other courses)  personally, I think it's cuz of the "precal gap"- taking physics should presuppose a knowledge of functions and trig, and iic makes fair use of that.  Although you get some exposure to trig in Geometry, it's not that much.  However, the thing is, most people in the country take Precal or the equivalent of it much later(maybe 11th grade? after all, algebra 1 is on-level for freshmen), so they wouldn't do that well on iic.  At least, that's why I think there's such a big curve for these two tests, specifically

I did better on the Math IIC than the SAT Math.  (800 vs. 780)  Even though the math itself was harder, I thought the style of questions was actually easier.  They don't focus on tripping you up with units and such; they actually test your math ability (albeit on an easy level).  All of the questions I missed on the regular SAT Math, each time I took it, were due to some trick in the wording of the question.  (I'd solve for inches instead of feet, miss words like "not" or "except", etc.)  The SAT II math didn't have that "feel" to it.  They also didn't have those ridiculous questions like "If x=3+5, what is x+2?" :-P

The way I see it is, if you can do the AMC/AIME-style stuff, you should have absolutely no problem with the Math IIC.

Gauss202:  You have to know more than just the textbook basics.  It's about applying what you know, not memorizing a bunch of terms and conditions.  If all you do is memorize, you'll struggle.  If you attack the problems logically, the answer will show.  Most brilliant students that struggle in math do so because they just memorize their way through problems and don't truly understand the core concepts behind what they're doing.  If you do understand what's going on beyond the mundane memorization, you should not have a problem with this test at all.  (And when I say "you" I don't mean you, specifically... I mean anybody)

Oh, and... no, I didn't prepare for it either.  But from what I remember, it was mostly algebra; the stuff you'd learn in an 8th or 9th grade math course.

But kindly distinguish between algebra and pre-calculus... pre-cal has stuff like vectors. I left my vector question blank and still got 800. (Of course, now, in physics, I dare not avoid vectors lol)

Well, you know what I meant, right?  The level of knowledge is higher, which means a greater number of US students would be uncomfortable answering such questions


A pair of non-congruent triangles can have 5 congruent parts. For example, a triangle of sides 8, 12, 18 is not congruent to, but has 5 congruent parts with a triangle of sides 12, 18, 27 [2 sides and all 3 angles are congruent].

In a pair of non-congruent right triangles with 5 congruent parts, what is the ratio of the hypotenuse [of either triangle] to the shortest side [of that same triangle]?

This thread should be locked until today has ended in all parts of the US.


this might not be exactly helpful but here goes:
[url]http://imo.math.ntnu.edu.tw/Chinese/[/url]
its all in chinese.  I couldnt find a good english one.

I had a friend once who tried to convince me to learn how to read Chinese since here are apparently some great problem-solving tutorials or something like that on the web that are in Chinese. But I think it would a lot easier for me to learn how to do math than to learn Chinese and then learn math from the web on Chinese websites.

[quote="answerseeker95"]I can assure you that the website is about Taiwanese IMO in 1990s, not about China (Taiwan and China represent different teams).... I don't think China has an official website devoted directly to IMO (you know, a communist country like China usually keeps its national teams top secrets if possible)[/quote]

..............  plz dont bring politics into math discussions...

ok... I won't talk about politics then.......
Simon, where do you find those Chinese tutorial websites? I can read Chinese so hopefully I can learn something from those websites....

I don't know. Sorry. I didn't do any research since I was too lazy to learn how to read Chinese.

[quote="Fierytycoon"]Does China have a website for its olympiad and IMO team? If so, does anyone know what the url is?[/quote]

I'm not sure what the interest that prompts this question is, but if you would like to find Chinese-language books that help you learn a lot of math for doing better in math competitions, you could go to

九章出版社

http://ccmp.chiuchang.com.tw/

which is a bookstore and publishing company from which I got many of my best Chinese-language books about math. The publisher organizes competition teams in Taiwan, and he distributes EXCELLENT books in English and in Chinese about problem-solving competitions from other publishers.

Hope this helps! You can write to that Web site by email link in either English or in Chinese. Tell them the guy who used to take his kids to the store to play games and who moved to America in 2001 sent you.


If $ x$ and $ y$ are positive integers for which $ 2^x3^y \equal{} 1296$, what is the value of $ x\plus{}y$?

$ \textbf{(A)}\ 8\qquad
\textbf{(B)}\ 9\qquad
\textbf{(C)}\ 10\qquad
\textbf{(D)}\ 11\qquad
\textbf{(E)}\ 12$


1. For all real numbers x and y, f(xy) = f(x) + f(y)

Prove:
a. f(1) = ?*
b. When x is not equal to 0, f(1/x) = - f(x)
c. When y is not equal to 0, f(x/y) = f(x) - f(y)

*I know it is not a proof, but it was equal to something but I saw that it had the answer to Rep's Contest, so I made this the same problem as the one he has in the GS Contest. Sorry, Rep.

2. The domain of f(x) is (-  :inf: ,  :inf: )*.  If x_1 does not equal x_2, then
f(x_1) will not equal f(x_2), and for any x and y, f(x+y) = f(x)*f(y)

* (a, b) means a<x<b

Prove:
a. f(0) = 1
b. f(x) > 0 for all x

3. Functions f(x) and g(x) follow these rules:
(1) f(0) = 0
(2) For any x and y,
         g(x-y) = f(x)*f(y) + g(x)*g(y)

Prove: [f(x)]^1992 + [g(x)]^1992  :le:  1

Are you sure that this is Getting Started level? It looks accessible to most of the participants on Intermediate.

One thing you do by moving borderline problems to the higher level is to encourage people who consider themselves to be in the lower level not to look at the problems at all. At least if they remain in the lower level, someone will post a solution, and then more people who could learn from the solution will read the solution and hopefully become better problem solvers.

[quote="ComplexZeta"]One thing you do by moving borderline problems to the higher level is to encourage people who consider themselves to be in the lower level not to look at the problems at all.[/quote]

I'm not sure I agree. People move down by moving over to Fun and Games (or not staying regular lurkers on this site at all) if there isn't something approachable on what is, after all, the lowest-labeled level on the site. You've talked me into moving this thread, which isn't at the AMC 8/Mathcounts level that I take as a benchmark for the Getting Started Forum.

But there are plenty of problems in the Getting Started board that are approachable. Why shouldn't the Getting Started boards be like the other math boards, all of which have some problems that are solvable and some problems that are a bit (or in the case of Advanced Problems, much) too difficult for people at that level? That's the only way I know of to improve.

I thought the first one was GS level, and that doing a problem like the first one will help you figure out how to approach the second and the third... but ya, I wasn't sure whether I should've combined the three together.


How many ordered pairs of integers (a,b) satisfy all of the following inequalities?
a^2 + b^2 < 16
a^2 + b^2 < 8a
a^2 + b^2 < 8b

Please post your two problems in two separate topics.
I splitted your post.

Pierre.


[quote="white_horse_king88"]Latex.

$\sqrt{(n+\sqrt{(n+\sqrt{(n+ \ldots)})})}$[/quote]I've never seen it with the parentheses. \[\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}\] is a little neater.

As for the solution, [hide]Let $n=k(k-1)=k^2-k$ (by the hypothesis of the problem), where $k\in\mathbf{N}$ and $k>1$. Then $k=\sqrt{k^2-k+k}\Rightarrow{k}=\sqrt{k^2-k+\sqrt{k^2-k+k}}\Rightarrow{k}=\sqrt{k^2-k+\sqrt{k^2-k+\sqrt{k^2-k+k}}}\Rightarrow\ldots\Rightarrow{k}=\sqrt{k^2-k+\sqrt{k^2-k+\sqrt{k^2-k+\sqrt{k^2-k+\ldots}}}}\Rightarrow{k}=\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$.
Thus $\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=k$, so it is an integer.

Similarly $k^2-k=k(k-1)\Rightarrow{k-1}=\frac{k^2-k}{k}\Rightarrow{k}=1+\frac{k^2-k}{k}\Rightarrow{k}=1+\frac{k^2-k}{1+\frac{k^2-k}{k}}\Rightarrow{k}=1+\frac{k^2-k}{1+\frac{k^2-k}{1+\frac{k^2-k}{k}}}\Rightarrow\ldots\Rightarrow{k}=1+\frac{k^2-k}{1+\frac{k^2-k}{1+\frac{k^2-k}{1+\ldots}}}\Rightarrow{k}=1+\frac{n}{1+\frac{n}{1+\frac{n}{1+\ldots}}}$,
so the expression \[1+\frac{n}{1+\frac{n}{1+\frac{n}{1+\ldots}}}\] is also always an integer under the given conditions.

Edit: As you can see from santosh_blr's post below, mine is not a rigorous solution under any circumstances.
[/hide]

We shall apply the Bolzano-Weirstrass theorem to the problem.

The theorem, states that for any [b]monotonic[/b] (increasing or decreasing) [b]bounded [/b] sequence is [b]convergent[/b]

Examples consider the sequence x_n = 1/n

The sequence is decreasing and is bounded by the limits 0 and 1.

So by the above theorem ,it converges to some limit. By observation, the limit is nothing but Zero.

Now coming back to the original problem, we have

        x_m =  sqrt ( n+ sqrt(n+....)      [ m brackets]

       It follows that x_(m+1) =  sqrt ( n+ sqrt(n+....)   [ m + 1 brackets]

                                    x_(m+1)    =  sqrt ( n + x_m)  > x_m

        It clearly follows that the sequence is increasing.

    But the sequence is bounded below by sqrt(n) and above by some finite no. For the moment let's assume that x_m is bounded above too (This can be proved).

So by B-W theorem,

  there is some limit  L = lim x_m    as m -> infinity

   x_(m+1)    =  sqrt ( n + x_m)

Taking limits on both sides , and observing the fact that    lim x_(m+1) = lim x_m ( for any convergent sequence x_m),

 we have  L = sqrt (n + L)

    Squaring , we get

     L^2 = n + L ...(1)

   Let n= k(k+1)  where k is some integer.

   Then solving the quadratic in L (eqn. 1),
we get

    L = (1/2)*( 1 + or -(2k+1))

   rejecting the inadmissible negative root , we get L = k+1 , an INTEGER

[/b]

[quote="eryaman"]Prove that if  n is the product of two consecutive natural numbers, then :sqrt:(n+:sqrt:(n+:sqrt:(n+.... is an integer.

Can someone put this into latex?[/quote]

Call $n=a^2+a$ with $a$ a natural number.
Call also $b=\sqrt{n+\sqrt{n+\sqrt{\ldots}}}$

We can say:

$b = \sqrt{n + b}$
$b^2 = n + b$
$b^2 - b - n = 0$
$(b - \frac{1}{2})^2 = \frac{4n+1}{4}$
$b - \frac{1}{2} = \frac{\sqrt{4n+1}}{2}$
$b = \frac{1 - \sqrt{4n+1}}{2}$

Now to show that if $n = a^2+a$ then b is an integer:

$b = \frac{1 - \sqrt{4(a^2+a)+1}}{2}$
$b = \frac{1 - \sqrt{4a^2+4a+1}}{2}$
$b = \frac{1 - \sqrt{(2a+1)^2}}{2}$
$b = \frac{2a}{2}$
$b = a$

As we defined a as an integer then the expression is an integer.

Something tells me that was wrong.. any criticisms?

[quote]Prove that if $n$ is the product of two consecutive natural numbers,
then $\sqrt{n + \sqrt{n + \sqrt{n + \hdots}}}$ is an integer.[/quote][hide]
$\text{Let }x = \sqrt{n + \underbrace{\sqrt{n + \sqrt{n + \hdots}}}} $
$.\qquad\qquad\qquad\quad\text{all of this is }x$

$\text{So we have: }x = \sqrt{n + x} $

$\text{Square both sides: }x^2 = n + x\;\;\Rightarrow\;\;x^2 - x - n = 0$

$\text{Since }n\text{ is the product of two consecutive integers, say, }a(a+1),$

$.\qquad\quad\;\text{we have: }\;\;x^2 - x - a(a+1) = 0$

$.\quad\text{which factors: }(x + a)(x-[a+1]) = 0$

$\text{and has the positive root: }x = a+1$[/hide]


If a prime number greater than 340 and less than 380 is selected at random, what is the probability that the final two digits of the number also name a prime number? Express your answer as a common fraction.

The prime numbers are $ 347$, $ 349$, $ 353$, $ 359$, $ 367$, $ 373$, and $ 379$.  Of these $ 7$, $ 6$ of them have the property we want, so $ \boxed{\frac {6}{7}}$

Sorry for reviving, but is there any other way other than brute forcing? :)

Probably not, but its not very time-consuming to find 7 primes :)

I can't do that in 45 seconds though. :)

yeah, i dont know how this is possible in For The Win


inequalities
Cauchy Inequality
inequalities unsolved

Let $ a,b,c>0$ and $ \sqrt{ab} \plus{} \sqrt{bc}\plus{}\sqrt{ca}\equal{}1$.

Prove that $ \frac{a^3}{b(c\plus{}a)}\plus{}\frac{b^3}{c(a\plus{}b)}\plus{}\frac{c^3}{a(b\plus{}c)} \ge \frac{1}{2}$

My solution:

Aply Cauchy Inequality we get:
$ \frac {a^3}{b(c \plus{} a)} \plus{} \frac {b}{2} \plus{} \frac {c \plus{} a}{4} \ge \frac {3a}{2}$

$ \frac {b^3}{c(a \plus{} b)} \plus{} \frac {c}{2} \plus{} \frac {a \plus{} b}{4} \ge \frac {3b}{2}$

$ \frac {c^3}{a(b \plus{} c)} \plus{} \frac {a}{2} \plus{} \frac {b \plus{} c}{4} \ge \frac {3c}{2}$

$ \Longrightarrow \frac {a^3}{b(c \plus{} a)} \plus{} \frac {b^3}{c(a \plus{} b)} \plus{} \frac {c^3}{a(b \plus{} c)} \ge \frac {3a \plus{} 3b \plus{} 3c}{2}$ $ \minus{} \frac {a \plus{} b \plus{} c}{2} \minus{} \frac {2(a \plus{} b \plus{} c)}{4} \equal{} \frac {a \plus{} b \plus{} c}{2}$

But from $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$ we get $ 1^2 \equal{} (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})^2\le (a \plus{} b \plus{} c)(a \plus{} b \plus{} c) \Longrightarrow a \plus{} b \plus{} c \ge 1$

Hence $ \frac {a^3}{b(c \plus{} a)} \plus{} \frac {b^3}{c(a \plus{} b)} \plus{} \frac {c^3}{a(b \plus{} c)} \ge \frac {1}{2}$. Q.E.D.

Who can give me an other solution please !


There might be other solutions but I'll just post my half solution....
[hide="Half solution maybe all of it?"]Let $x=10^a$ and $y=10^b$. So then $10^a(10^b)^2=10^8 \Rightarrow a+2b=8$ and into the second, $\frac{10^{3a}}{10^b}=10^{10} \Rightarrow 3a-b=10$ solving the system gives $a=4, b=2$ so $x=10^4$ and $y=10^2$.[/hide]

[hide]
$y=\frac{x^3}{10^{10}}$
$\frac{x^7}{10^{20}}=10^8$
$x^7=10^{28}$
$x=10^4$
$\frac{10^{12}}{y}=10^{10}$
$y=10^2$
[/hide]

Are there more solutions?  $x$ and $y$ have to be positive, because if $x$ were negative,  $xy^2$ would have to be negative.  If $y$ were negative, $\frac{x^3}{y}$ would be negative.

[hide="Answer"]In the second equation, multiply both sides by $y$ and divide by $10^{10}$, so $y=\frac{x^3}{10^{10}}$. Substitute into the first equation to find $x^7=10^{28}\Rightarrow \boxed{x=10^4}$. Therefore, $\boxed{y=10^2}$.[/hide]

[quote="Silverfalcon"]Solve the following systems of equations:

\[ xy^2 = 10^8, \frac{x^3}{y} = 10^{10}  \][/quote]

I got X=10^4; Y=10^2.

Could you assume that both X and Y were powers of 10?

[hide]$x+2y=8$
$3x-y=10$

Solving gives $x=4, \ y=2$

So $X=10^4, \ Y=10^2$?[/hide]

[hide]Squaring both sides of the second equation gives $\frac{x^6}{y^2}=10^{20}$.

Multiplying that equation by $xy^2=10^8$ (the other given equation), we get $x^7=10^{28}\Longleftrightarrow x=10^4$

Then we get $y=10^2$, so $(x,y)=(10^4,10^2)$
[/hide]


inequalities
induction
inequalities proposed

Let $a,b,c\ge 0$ such $ab+bc+ca+6abc=9$. Prove that:$a+b+c+3abc\ge 6$
This problem come from "secret in inequalities" book (author: hungkhtn)
And it to fall within chapter: Mixing variable  :D But I can't solve it by mixing variable. Can you help me ?  :)
induction: In Viet Nam, people finded new method to talk about mixing variable.
It is GMV theorem  :D

Anynoe idea ? :(
Can you help me about determine condition mixing varible?  :D

It was solved in my book using MV too. I have to admit that it is not straightforward.

When does equally hold, pvthuan? Can you post your solution in here  :) Thank you very much  :wink:
My opinion:
$f(a,b,c) \geq f(a+c+6ac+\frac{ac}{b},b,0)$
It's mean equality hold when one variable $c=0$
case generalization of above problem:
Let $a,b,c$ are positive real number such that  $ab+bc+ca+6abc=9$. Find the best $k$ constant to follow inequality is true:
$a+b+c+kabc \geq k+3$
For above problem, solution only use Schur inequality.  :)

[quote="zaizai-hoang"]
My opinion:
$f(a,b,c) \geq f(a+c+6ac+\frac{ac}{b},b,0)$
It's mean equality hold when one variable $c=0$
[/quote]
Let $f(a,b,c)=a+b+c+3abc-6.$ Then
$f(a,b,c)-f(a+c+6ac+\frac{ac}{b},b,0)=$
$=a+b+c+3abc-a-c-6ac-\frac{ac}{b}-b=\frac{ac(3b^{2}-6b-1)}{b}.$
Why $3b^{2}-6b-1\geq0$ $?$

Your solution is very nice Arqady .But this is not in the earth  :o
[quote="arqady"] Hence, $G(a,b,c)=G\left(a+c+6ac+\frac{ac}{b},b,0\right).$
[/quote]

Could this be proved by Langrange Multipliers ?

I don't think so  :D  Maybe argady mistaken in $\ge$ and $=$.
And it is condition mixing variable  :P

Let $a=b=3,c=0$, we have $k \leq 3$.
$a+b+c+3abc \geq 6$
Let $p=a+b+c,q=ab+bc+ca,r=abc$ then  $q+6r=9 \Rightarrow p,q \geq 3$ above inequality equivalent
$p+3r \geq 6 \\ \Leftrightarrow 2p-q \geq 3$
If$p \geq 6$ above ineq is true.
If $3 \leq p \leq 6$ then we have 2 cases happen:
+, $p^{2}\geq 4q$. In here, we have
$2p-q \geq 2p-\frac{p^{2}}{4}$
But$2p-\frac{p^{2}}{4}\geq 3 (*)$. It true because:
$(*) \Leftrightarrow p^{2}-8p+12 \leq 0 \\ \Leftrightarrow (p-2)(p-6) \ leq 0$
+, $4q \geq p^{2}$. Therefore, by Schur inequality, we have $r \geq \frac{4pq-p^{3}}{9}\geq 0$. Then,  $q+\frac{2(4pq-p^{3})}{3}\leq 9 \Leftrightarrow q \leq \frac{2p^{3}+27}{8p+3}$. So, to obtain
$2p-q \geq 2p-\frac{2p^{3}+27}{8p+3}$.But, $2p-\frac{2p^{3}+27}{8p+3}\geq 3 (**)$. Then,
$(**) \leftrightarrow 2p(8p+3)-2p^{3}-27 \geq 3(8p+3) \\ \leftrightarrow p^{3}-8p^{2}+9p+18 \leq 0 \\ \leftrightarrow (p+1)(p-3)(p-6) \leq 0$
It is true.
Complete :) Solution of toanhocmuonmau (in diendantoanhoc.net forum)

Have fun!!!!!!!!!!!!!!!!!!!!! :)

Complete  Solution of toanhocmuonmau (in diendantoanhoc.net/forum) or canhang_2007 (in mathlinks.ro)


function
real analysis
geometry
trapezoid
real analysis unsolved

Let $ f(t)$ be a continuous function on the interval $ 0 \leq t \leq 1$, and define the two sets of points \[ A_t\equal{}\{(t,0): t\in[0,1]\} , B_t\equal{}\{(f(t),1): t\in [0,1]\}.\] Show that the union of all segments $ \overline{A_tB_t}$ is Lebesgue-measurable, and find the minimum of its measure with respect to all functions $ f$. [A. Csaszar]

We first show that the set $ A \equal{} \bigcup_{0 \le t \le 1} \overline{A_tB_t}$ is closed, hence Lebesgue-measurable. Let $ P_n$ be a convergent sequence of points in $ A$, $ P_n \to P_0$. By the definition of $ A$, to every $ n$ there is a $ t_n$ such that $ P_n \in \overline{A_{t_n}B_{t_n}}$. According to the Bolzano-Weierstrass theorem, the sequence $ t_n$ contains a convergent subsequence: $ t_{n_k} \to t_0$. The distance from the point $ P_{n_k}$ to the segment $ \overline{A_{t_0}B_{t_0}}$ is not greater than $ \max \{|f(t_{n_k} \minus{} f(t_0))|, |t_{n_k} \minus{} t_0|\}$, which tends to $ 0$ by the contains the limit point of any convergent sequence of its points, $ A$ is closed.

A simple calculation shows that the point of the segment $ \overline{A_tB_t}$ that lies on the straight line $ y\equal{}c$ has abscissa $ (1\minus{}c)t \plus{} cf(t)$, which, by the continuity of $ f(t)$, is a continuous function of $ t$. Consequently, if two points of the set $ A$ lie on the line $ y\equal{}c$, then $ A$ contains the segment that joins these points.

Now we can determine the minimum of the measure of $ A$. If $ f(t)$ is constant, then $ A$ is a triangle of unit base and unit altitude, so it has measure $ 1/2$. If the segments $ \overline{A_0B_0}$ and $ \overline{A_1B_1}$ do not intersect, then the trapezoid with vertics $ A_0$, $ B_0$, $ A_1$, and $ B_1$ is a subset of $ A$, so the measure of $ A$ is not less than $ 1/2$. If the segments $ \overline{A_0B_0}$ and $ \overline{A_1B_1}$ intersect at some point $ C$, then the triangles $ A_0CA_1$ amd $ B_0CB_1$ are subsets of $ A$, so the measure of $ A$ is not smaller than
\[ t(d) \equal{} \frac{1}{2}\left( \frac{1}{1\plus{}d} \plus{} \frac{d}{1\plus{}d}d \right) \equal{} \frac{1}{2}\frac{1\plus{}d^2}{1\plus{}d},\]
where $ d$ denotes the distance from $ B_0$ to $ B_1$. By a simple calculation, we obtain that the minimum of $ t(d)$ on the positive half-axis is $ \sqrt{2} \minus{} 1$. Thus, the measure of $ A$ cannot be less than $ \sqrt{2} \minus{} 1$. On the other hand, for $ f(t) \equal{} (\sqrt{2} \minus{} 1)(1 \minus{} t)$, the measure of $ A$ is exactly $ \sqrt{2} \minus{} 1$.


geometry
geometric transformation
vector
analytic geometry
combinatorics unsolved
combinatorics

A polygon $\prod$ is given in the $OXY$ plane and its area exceeds $n.$ Prove that there exist $n+1$ points $P_{1}(x_1, y_1), P_{2}(x_2, y_2), \ldots, P_{n+1}(x_{n+1}, y_{n+1})$ in $\prod$ such that $\forall i,j \in \{1, 2, \ldots, n+1\}$, $x_j - x_i$ and $y_j - y_i$ are all integers.

Divide the plane into unit squares. Let $O$ be the origin.
Let $U$ be one of the unit squares, with vertices $XYZT$ where $X$ is the lower-left one. Then consider the translation $t_U$ with vector $XO$. It maps $U$ onto the unit square $S$ with lower-left vertex $O$. Note that for each point $M$ in $U$, the vector $Mt(M)$ has integer coordinates (since it is $XO$).
The common part of the given polygon and $U$, say $P_U$, which is a polygon, is map to a congruent polygon $t(P_U)$, so that it has the same area.
Now, except for a part with area $0$ (the intersection of the given polygon with the lines with equation $x=k$ of $y=k$, where $k$ is an integer), the area of the given polygon is the sum of the areas of the polygons $P_U$, where the sum is over all the unit squares $U$. But all the $t(P_U)$ are in $S$, which has area $1$. Since the sum of the areas is greater than $n$, the pigeon-hole principle ensures that a point, say $M$, of $S$, is covered at least $n+1$ times by the $t(P_U)$. It means that there exists $M_1, \cdots, M_{n+1}$ pairwise distinct points in the given polygon such that each of them has image $M$ by the appropriate translation. Since all the translations have vectors with integer coordinates and that a composition of translation is a translation obtained by summing the vectors, it follows that the vector $M_iM_j$ has integer coordinates for all $i<j$.

Pierre.

Translate all unit squares of the integer lattice in the plane onto one particular one, With overlaps or not, the portions of the interior of the polygon will cover (at least) one point (at least) $ n\plus{}1$ times.
The back-translates of that point in the interior of the polygon will yield the required points.

Is it possible to state solution like this? Divide the plane into unit squares.Take some random unit sqare.Then color that polygon with red.Now start packing
squares with color on top of chosen square (translate it on top).When we are finished volume of red color is bigger than n(cm).Take random point in chosen  square and if there are more than n points of red color that are above (and collinear with it ) then we are done.Otherwise  there are less than n points above every point so volume is less that or equal to n contradiction.

Nice problem!
Solved with [b]Dimath27[/b]  :-D .

The key idea is divided the plane into unit squares and after this, select the squares that cut the polygon, turn them into paper and put one on top of the other so we are left with an unit square divided in finite areas.
If exists an area that was in at least $n$ squares we can go through this area with a pin and we get at least $n$ point that satisfy the problem (because when we put the unit squares back in their places, the point distances are compositions of vertical or horizontal translations of size $1$).
But this area always exists since otherwise each area is repeated at most $n-1$ times and adding the areas of the unit squares we obtain that the area of the polygon is at most $n-1$. Contradiction! $\blacksquare$

[hide=solution]
Translate the lattice unit squares all onto one square then the squares with parts of the interior of the polygon will hit one point at least $n+1$ times.

Have the lattice lines seperate the polygon into different sections and merge them all into a $1\times 1$ cell. By Pigeonhole, there exists a point $P$ that is overlapped by $n+1$ different regions. After translating the regions back, the $n+1$ points created by this operation would satisfy the given conditions. $\square$

For a point $P_0=(x_0,y_0)$ in the coordinate plane, define $\bf{F} (P_0)$ to be $(\{x_0\},\{y_0\})$, where $\{r\}$ of a real number $r$ denotes the fractional part of $r$. The problem is asking us to prove that given a polygon $\Pi$ with area greater than $n$, there is a point $P_0\in [0,1]\times[0,1]$ such that are $n+1$ points $P_1$, $P_2$, $\dots$, $P_{n+1}$ in $\Pi$ such that $P_0=\bf{F} (P_1)=\bf{F} (P_2)=\dots=\bf{F} (P_{n+1})$. \\\\
Assume for the sake of contradiction that there is no such point $P_0$. This means that over all points $Q$ in $\Pi$, the values of $\bf{F}(Q)$ "cover" each point in the unit square $[0,1]\times[0,1]$ at most $n$ times each, meaning that the area of $\Pi$ is at most $1*1*1=n$, a contradiction by problem conditions. Therefore, there must exist at least $n+1$ such points as described in the problem, and we are done.

Consider the mapping of polygon $\Pi$ under $(x,y) \mapsto (\lfloor x \rfloor, \lfloor y \rfloor)$, where we also take multiplicities into account. By Pigeonhole, there exists at least one point covered at least $n+1$ times, which we set as our origin. $\blacksquare$


[quote="Treething"]Simply, I'm tryin' to create a legal version of Kazaa through the members of the AoPS. This gets you free music, similarly to how Kazaa did it, and it's perfectly legal.

The main reason Kazaa was outlawed was because it was a [b]clientside server[/b]. That got me thinking - what about a private network of people?

If you upload music on a private server, sorta like mass email to all of the members, then this is by definition not clientside, which makes this legal.

If this is illegal, somebody stop me, as I'm willing to do it even if it's immoral, as long as it's legal.[/quote]

Hmm... This seems like a great idea! I'm not too sure if its illegal or legal though. I would use it all the time if it was legal.

[quote="djimenez"][quote="GoBraves"]Kaaza was cool.......you could even get movies on it.[/quote]

Yes, you can, but if the RIAA trace you, you will have a strong fine, at least $2K is what I have heard!! Remember that in many of the States here in the US, to download any copyrighted material that was not uploaded by the copyright owner himself is ilegal, even when it is on a P2P!

I believe that the Canadian law is better. It is illegal to upload that information (even to P2P) but it is not illeagal to download it  :lol:

Best,[/quote]

Lol...I still have like 6 movies and a bunch of music from kaaza.

[quote="cardwacko9999"]why would RIAA go after a random user? they would probably go after users that download or upload huge amounts of music, movies, etc.[/quote]

OK, define huge amounts. I have read cases of people that downloaded not that much, but they got cought, Of course, that is, around 500 files in a month or less period, but I really think that it is not that much. How much do you download.

Also, I have heard that some of those organism put some "trash files" (a file that contains probably nothing more than a house film, and says "MATRIX REVOLUTIONS" or something like that) and trace who download that file, and that is exactly how they find their "victims". Of course, much of it may be just paranoia distributed around the web, but I know that at least eight or nine students have been fined by RIAA here at Tech in the last six months  :? , how much had they download, I don't know, but it is better try to be careful.

Best,

Please do not start start a filesharing system on AoPS. After all, we don't want lawsuits being filed on the company, even if it is on the borderline of legality.

This discussion is now suspended. You may discuss filesharing on a separate thread if you wish.


Is this equation valid? If it exists, does it have a specific name?

x^2k =  :Sigma:{m= 0 to k}       (k!/(m!k-m!)) (x :^2: -1)^m

I think it's true.

Just expand (1 + (x^2-1))k with the Binomial Theorem(see http://mathworld.wolfram.com/BinomialTheorem.html for instance)


inequalities
quadratics
inequalities proposed

Take a sequence of numbers where $a_j \geq 0$ and $\sum a_{j}^2 < \infty$.
Construct another sequence $b_j=\frac{1}{2} \left(a_j+a_{j+1}\right)-\frac{1}{8} \left(y_{j+1}-2y_j+y_{j-1}\right)$ where $y_j=min(a_j,a_{j+1})$.

Prove that $\sum b_{j}^2 \leq \sum a_{j}^2$

Still no attempts :blush: I will admit, this is a very tough inequality, not even suitable for IMO or Putnam.

My dad posed this problem to his grad students.

I have just proved it for the very special case $a_1\geq a_2\geq a_3\geq ...$. It is not that hard, but anyway...

That proof, or the proof of the problem in general? Btw, would you mind posting your solution to the decreasing sequence?

I used well known trick to minimize $S=\sum a_i^2-\sum b_i^2$. We see that it is quadratic trinomial wrt $a_1$, where $a_\in[a_2,+\infty)$. We can show that it takes minimum for $a_1=a_2$. Then we will minimize $S$ varying $a_1=a_2\in[a_3,+\infty)$. We show that minimum reaches for $a_1=a_2=a_3$. It means that we can throw out $a_1=b_1$. And so on. So acting this way we decrease $\sum a_i^2$ and $\sum b_i^2$ to 0, so $S$ decreases to 0. It means that $S\geq 0$ initially.

The proof of this case is trivial...the rest is very difficult.


Ring Theory
group theory
abstract algebra
Units
linear algebra
invariant

It must be simple to find but I didn't find the answer yet.
Find a finite ring $ R$, such that its unit group $ U(R)$ contains exactly 26 elements.

Can $ R$ be a field? There is definitely a field with $ 3^3$ elements. :wink:

You can take R as the direct product of K and Z/2Z,where K is the(up to isomorphism) field with 27 elements,as well.
Then U(R) is the direct product of U(K) and U(Z/2Z) and hence isomorphic to U(K) which has 26 elements.
Clearly R is not a field,because it has proper zero divisors.

Patrick

Thanks, I was looking for a ring that is not a field, and it was not that difficut :blush:
What if a replaced $ 26$ by $ p\plus{}1$ where $ p$ is prime?

If $ p \equal{} 2$, then the field of order 4 works.  If $ p$ is odd, then
$ \mathbb{Z}\times\mathbb{Z}\times$...
 $ \frac {(p \plus{} 1)}{2}$ times works.

Steve

[quote="SD5225272"]... If $ p$ is odd, then
$ \mathbb{Z}\times\mathbb{Z}\times$...
 $ \frac {(p \plus{} 1)}{2}$ times works.
Steve[/quote]
But that's not a finite ring :wink:

I don't understand your question. One can prove that up to isomorphism there is only one field of a given prim power degree p^n. If you use the existence part of the above theorem you can prove: Let p be a prime and n a natural number,then there is a Ring R which is not a field so that card(U(R))=p^n -1. Take R as the product of a field K with
p^n elements with Z/2Z for example. You ask if for a given prime p one can construct a Ring R which is not a field so that card(U(R))=p+1?

patrick

[quote="Patrix"]You ask if for a given prime p one can construct a Ring R which is not a field so that card(U(R))=p+1?

patrick[/quote]
Yes patrick.

And if you try to look for counter example, don't bother for $ p$ smaller than $ 37$, because it works.

Sorry.  Take $ \mathbb{Z}/3\mathbb{Z}$
$ \frac{(p\plus{}1)}{2}$ times then.

Steve

sorry,but U((Z/3Z)^m)=U(Z/3Z)^m so card(U(R)) would be 2^(p+1)/2

Patrick

Nobody for a finite ring such its unit group has $ p\plus{}1$ elements?

$ U\!\left(\mathbb Z/8\mathbb Z\right)$ has $ 3 \plus{} 1$ elements, $ U\!\left(\mathbb Z/9\mathbb Z\right)$ has $ 5 \plus{} 1$ elements, etc. They’re not that hard to find. ;)

Finding a few examples and proving a general result for all odd primes $ p$ (or even better, all odd numbers $ p$), are two very different things :wink:

[quote="ValentineA"]You should’ve said you wanted to prove the result for all odd primes. I thought your question was: show that it is possible for finite rings such that $ \left|U(R)\right|$ is of the form $ p \plus{} 1$ to exist.[/quote]
When there is some ambiguity in question phrasing, a common custom is to pick the quantifiers that make it as difficult as possible.  This is done for the purpose of excitement.

I'll take a look at this $ |U(R)| \equal{} 38$ problem...

[quote="JC_math"][quote="LydianRain"]Made some progress.  If $ R$ has exactly 38 units, then $ \rm{char}\ R \equal{} 4$.[/quote]

I made some (little) progress too, $ R$ needs to have a least one non trivial nilpotent element.[/quote]
That follows from $ \rm{char}\ R \equal{} 4$ :)

Below is an outline of a proof for my claim, by the way.  Notice that it breaks down somewhat when the characteristic is 4, because the ideal structure of $ \mathbb{Z}/4[x]$ is not quite as simple as we might like.  It may be possible to extend my method to this case with more careful analysis, but it's also possible that it's a dead end.

First, in the case that the group of units of a ring is cyclic and generated by $ r$, we can consider the subring generated by $ 1$ and $ r$.  In this way, we can assume WLOG that $ R$ is a quotient of $ \mathbb{Z}/k [x]$, where $ k$ is the characteristic of $ R$.

Also, since $ U(\mathbb{Z}/k)$ is a subgroup of $ U(R)$, we must have $ \phi(k) \big| |U(R)|$.  This alone restricts $ k$ to $ 2$, $ 3$, or $ 4$.

If $ k\neq 4$, $ R \equal{} K[x]/I$ for $ K$ a finite field and $ I \equal{} (f)$ an ideal.  Furthermore, since $ r^{38} \equal{} 1$, $ f$ is a factor of $ x^{38} \minus{} 1$.  When $ k \equal{} 3$, this has no repeated roots, so by the Chinese Remainder Theorem $ R$ is a product of fields, so $ |U(R)|$ is a product of integers of the form $ 3^m \minus{} 1$, which doesn't work.

When $ k \equal{} 2$, $ x^{38} \minus{} 1 \equal{} (x^{19} \minus{} 1)^2$ has no triple roots, so $ R$ is a product of fields and rings of the form $ \mathbb{Z}/2[x] / (g^2)$, where $ g$ is irreducible.  We can show that the number of units in such a ring is of the form $ 2^m (2^m \minus{} 1)$.  We cannot write $ 38$ as a product of numbers of this form and numbers of the form $ 2^m \minus{} 1$, so this doesn't work either.

Now, it's possible that $ U(R)$ is not cylic.  If this is the case, it still has an element $ r$ of order $ 19$.  If the ring generated by $ 1$ and $ r$ contains all of $ U(R)$, we can proceed as above.  Otherwise, we have constructed a ring with exactly $ 19$ units, which we can show is impossible using methods directly analogous to the above.

[quote="MeKnowsNothing"]For 38 I think the solution might be on the paper http://schmidt.nuigalway.ie/~goetz/pub/unitburn.pdf . I just skimmed the paper but it looks like a suitable ring is $ J_2$ in GAP notation. Unfortunately I don't know enough ring theory to say if this is what was asked.[/quote]

I believe they are saying the Burnside ring of the sporadic simple group J2 has a unit group with exactly 2^38 elements (rather than 38).  Since the unit group is a subgroup of an elementary abelian group (the unit group of the "ghost ring"), all that is needed to describe it up to isomorphism is the rank as a vector space.  That table gives the ranks, rather than the actual sizes.

　　　some results: if R is comutative, and contains only 38 units than, $ R\equal{}mB_2\oplus B_4$ as abel group, $ \forall v\in U(R), 2(v\minus{}1)\equal{}0$

Denote $ G_{38}$ the multiply group formed by all units. $ G_38$ has only one sylow subgroup of order 19, which is an abel group. For the sake of simplicity, assume $ G_{38}\equal{}G_2*G_{19}$, which is cyclic. Denote g the generater, we have $ g^{38}\minus{}1\equal{}0$ . set r=char(R). since (m,r)=1 implies m being unit. So $ \phi(r)\equal{}2$. If r=3, it will be a field of much more than 38 elements. So r=4 as LydianRain said.

$ R\equal{}Z_4[u]/(u^{19}\minus{}3)$.  $ u^i$ are supposed to be all unit. set $ R\equal{}mB_2\plus{}nB_4$ as abel group,
$ I_2$={x|2x=0},     $ 2R$={$ y| \exists x\in R, y\equal{}2x$}.       $ |I_2|\equal{}2^{m\plus{}n}$,
since $ I_2$ is an ideal, so $ |2R|\equal{}\frac{|R|}{|I_2|}\equal{}2^n$.
Consider group (with multiplication in R) $ A_2$={$ 1\plus{}2x| x\in R$}, all $ y\in A_2$ are reversible, which means $ |A_2| | |U(R)|\equal{}38$,
however $ |A_2|\equal{}|2R|\equal{}2^n$, so n=1. (if n=0 R became a field)

set $ u^k\equal{}u\plus{}2$,   $ u^{2k}\equal{}(u\plus{}2)^2\equal{}u^2$,   so $ 38|2(k\minus{}1), k\equal{}20, 1$, the latter implies $ u\equal{}u\plus{}2$ which makes R a field.
since $ u^{19}\equal{}3$,   $ u^{20}\equal{}3u\equal{}u\plus{}2$,  2(u-1)=0.  similarily $ 2(u^i\minus{}1)\equal{}0$.

Could you explain in more detail?  What are $ B_2$ and $ B_4$?  And why do you declare $ R \equal{} Z_4[u]/(u^{19} \minus{} 3)$--isn't it possible that $ R$ is a quotient of this ring?

EDIT: On further consideration, I think that even concluding $ u^{19} \equal{} \minus{} 1$ is suspect.

EDIT 2: Actually, now I agree that $ u^{19} \equal{} \minus{} 1$ when $ U(R)$ is cyclic. (the element of order 2 is unique)  Furthermore, this observation allow us to eliminate the dihedral case: we can let $ u$ and $ \minus{}1$ generate $ U(R)$, but $ u$ and $ \minus{}1$ commute, contradiction.

OK, I still have no idea what $ B_2$ and $ B_4$ are, but I've managed to extract the interesting fact that $ \forall x\in R$, $ 2x\in \{0,2\}$.  In particular, $ 2x\equal{}2$ if $ x\in U(R)$.

The observation that $ (2) \equal{} \{0,2\}$ seems to solve the problem:

Look at the quotient $ S \equal{} R/(2)$.  We claim that $ \overline{u}$ has mutiplicative order $ 19$.  For certainly $ \overline{u}^{19} \equal{} \overline{1}$, but if $ \overline{u} \equal{} \overline{1}$, then $ u \minus{} 1\in\{0,2\}$, so $ u^2 \equal{} 1$, contradiction.

Furthermore, the induced map $ U(R)\to U(S)$ is surjective, because if $ \overline{x}\overline{y} \equal{} \overline{1}$, then either $ xy \equal{} 1$ or $ xy \equal{} \minus{} 1$, and in either case $ x$ is a unit.

Hence we have constructed a ring of characteristic 2 with 19 units, which is impossible by the previous discussion.

We conclude (I think) that there are no finite unital rings with exactly 38 units.  The same technique should work for $ |U(R)| \equal{} 2p$, under the condition that $ p \plus{} 1$ is not a power of 2 or 3, and $ 2p \plus{} 1$ is not prime.

Whoops, both myself and mathiq seem to have overlooked the case $ \rm{char}\ R \equal{} 6$.  The problem's still open...

$ B_i$ denote the cyclic group of order i.

OK, I didn't know this fact before, but it seems (feel free to correct me!) that any ring $ R$ with characteristic $ n \equal{} \prod_i p_i^{k_i}$ can be written as a direct sum $ \oplus J_i$, where $ J_i \equal{} \{p_i\ ^{k_i}\cdot x \equal{} 0\}$.  This direct sum also seems to respect unity.

It follows that we have a ring of characteristic 2 or 3 with either 19 or 38 units, which we've already shown to be impossible.  This should resolve the problem.

If anyone expresses interest, I'd be happy to compile the scattered statements here into a full proof.  There is no ring with exactly 34, 38, or 94 units!

[quote="JC_math"][quote="LydianRain"]There is no ring with exactly 34, 38, or 94 units![/quote]
You meant no [b]finite[/b] ring, right?[/quote]
I don't think I used the assumption that $ R$ was finite.

[quote="LydianRain"]
I don't think I used the assumption that $ R$ was finite.[/quote]
Then I have a doubt. I think we can prove easily that for each even integer $ n$ there exists a ring $ R$ (infinite) such that $ |U(R)|\equal{}n$.

[quote="JC_math"][quote="LydianRain"]
I don't think I used the assumption that $ R$ was finite.[/quote]
Then I have a doubt. I think we can prove easily that for each even integer $ n$ there exists a ring $ R$ (infinite) such that $ |U(R)| \equal{} n$.[/quote]
Do tell.

Oh I'm sorry!  Yes, I did in fact assume that $ R$ has finite characteristic.  Still curious about your example though.

Let $ 2n$ be an even integer and $ (G,\plus{})$ an abelian group of order $ n$.
Define $ A\equal{}\mathbb{Z}\oplus G$, with "usual + operation" and $ \otimes$ defined by $ (p\plus{}x)\otimes (q\plus{}y) \equal{} pq \plus{} (py \plus{}qx)$. $ A$ is a ring and $ |U(A)|\equal{}2n$.

[quote="JC_math"]Let $ 2n$ be an even integer and $ (G, \plus{} )$ an abelian group of order $ n$.
Define $ A \equal{} \mathbb{Z}\oplus G$, with "usual + operation" and $ \otimes$ defined by $ (p \plus{} x)\otimes (q \plus{} y) \equal{} pq \plus{} (py \plus{} qx)$. $ A$ is a ring and $ |U(A)| \equal{} 2n$.[/quote]

If it helps others, this is also known as a "trivial extension" of the ring Z by its module G.  To verify multiplication behaves nicely, it can help to view "p+x" as the 2×2 matrix [p,x;0,p].  The associativity, and formula for the inverse follows easily.

You could do this with any ring.  Z is inconvenient because it has two units, hence the even order of the unit group.  A ring like Z/2Z has only one unit, but now the choice of G is restricted to be Z/2Z module, so the unit group has order 1*2^n, still even.  Indeed, any ring of finite characteristic runs into this terrible restriction on G, and any ring of infinite characteristic contains -1 ≠ 1, and so has an even number (or infinitely many) units.

[quote="LydianRain"]OK, I didn't know this fact before, but it seems (feel free to correct me!) that any ring $ R$ with characteristic $ n \equal{} \prod_i p_i^{k_i}$ can be written as a direct sum $ \oplus J_i$, where $ J_i \equal{} \{p_i\ ^{k_i}\cdot x \equal{} 0\}$.  This direct sum also seems to respect unity.[/quote]

I believe this is called the primary decomposition.  The Sylow subgroups of an abelian group are extremely invariant under morphisms of various sorts, so it should have lots of good properties like you mentioned.  You can generalize this from rings viewed as abelian groups (so Z-algebras), to R-algebras over nice enough rings R.  PIDs like k[x] work well, and there the primary decomposition is looking at minimal polynomials of elements acting on the underlying (hopefully behaving as if it were finite dimensional) vector space.

In characteristic 0, you only get a nice decomposition of the torsion subgroup.  Even when the additive group is finitely generated, there is no nicely characteristic "torsion-free" part, so that the decomposition there may behave very badly in the part that contains the infinite cyclic subgroup generated by 1.

Do you think you could write up some of the results you guys have worked out about sizes of unit groups?  It seems like there has been enough worked out here to write a nice enough essay about it.

[quote="JC_math"][quote="LydianRain"]
I don't think I used the assumption that $ R$ was finite.[/quote]
Then I have a doubt. I think we can prove easily that for each even integer $ n$ there exists a ring $ R$ (infinite) such that $ |U(R)| \equal{} n$.[/quote]
Do you mean $ |U(R)| \equal{} 2n\,?$ :maybe:

[quote="mathiq"]R's comutativity is assumed. :)[/quote]
I don't think it is.  The important properties are derived by looking at the subring generated by $ 1$ and $ r$, where $ r$ is a suitable unit, and this ring is always commutative.

[quote="ValentineA"][quote="JC_math"]
Then I have a doubt. I think we can prove easily that for each even integer $ n$ there exists a ring $ R$ (infinite) such that $ |U(R)| \equal{} n$.[/quote]
Do you mean $ |U(R)| \equal{} 2n\,?$ :maybe:[/quote]
To show that statement $ X(n)$ holds for all positive even integers is the same as showing that statement $ X(2n)$ holds for all positive integers.  In an ideal world, we'd avoid changing the meaning of variables between posts, but there are only so many letters!

Sorry, my mistake. I misread [b]JC_math[/b]’s post. :oops:

[quote="mathiq"][quote="LydianRain"][quote="mathiq"]R's comutativity is assumed. :)[/quote]
I don't think it is.  The important properties are derived by looking at the subring generated by $ 1$ and $ r$, where $ r$ is a suitable unit, and this ring is always commutative.
[/quote]

but U(R) in general might not be cyclic.[/quote]
I covered this case.  $ U(R)$ always contains an element $ r$ of order 19.  The subring generated by $ 1$ and $ r$ either has $ 19$ units, which we show is impossible, or $ 38$ units, in which case $ U(R)$ must have been commutative anyway.

By the way, I'm not ignoring the request to write this up, but a lot of things are in flux for me at the moment.  Hopefully I'll have a nice summary in the coming week.


Given that A1B1C1D1 is a square with A1B1=14.  A2B2C2D2 is formed by connecting the midpoints of A1B1C1D1.  This pattern is continued to form an infinite sequence of squares.  Find the sum of the areas of all the squares if this process is continued indefinitely.

[quote="Temperal"][hide]It can be shown that each square is $ \frac {1}{2}$ the area of each preceding square (if I'm required to show this, just say so), so the sum of the area is
\[ \sum^{\infty}_{i \equal{} 1}14^2\cdot \left(\frac {1}{2}\right)^{i \minus{} 1}
\]
which converges to 392.[/hide][/quote]
so would the answer be the equation or 392?

From the way the question is typed it seems to want it as $ 392$.

Okay, I'm bored, so in case you need to show that each square is half the area of the last one, here you go:

Let ABCD an arbitrary square, and EFGH be the square formed by connecting the midpoints of AB, BC, CD, and AD.

It looks sort of like this, except for the fact that I'm horrible at asymptote and can't make a neat drawing for my life:

[asy]size(100); draw((0,0)--(100,0)--(100,100)--(0,100)--(0,0)); draw((50,0)--(100,50)--(50,100)--(0,50)--(50,0));[/asy]

Anyways...

divide ABCD into four congruent squares by connecting opposite midpoints, so that the sides of EFGH are the diagonals of the squares.

Now, since the two triangles in each square share a side, and their adjacent angles are congruent (because the sides of EFGH are the diagonals of the squares), each triangle is congruent to its pair, so all the triangles are congruent. ABCD is formed by 8 congruent triangles, and EFGH by 4, so ABCD is twice the area EFGH. (I skipped some steps here, ones I thought were obvious, like basic postulates and all that)

You can also you Pythagoras to solve this, which is actually easier, but that's no fun.


vector
category theory
linear algebra
linear algebra unsolved

With V and W F-vector spaces, and f:V->W a linear map.
The dual linear map is defined as f*:W*->V*.

How to show that the following holds for f1:V1->V2, f2:V2->V3:
(f2○f1)*=f1*○ f2*

thnx

I would apply the definition of the dual map.

It's a canonical arrow-chasing sort of thing; $ f^*(\phi)\equal{}\phi\circ f$.

The proof of the result is similar abstract nonsense*:
$ \left(f_2\circ f_1\right)^*(\phi)\equal{}\phi\circ(f_2\circ f_1)\equal{}(\phi\circ f_2)\circ f_1$
$ \equal{}f_1^*\circ(\phi\circ f_2)\equal{}f_1^*\circ(f_2^*\circ\phi)\equal{}(f_1^*\circ f_2^*)(\phi)$

*No disrespect intended. Abstract nonsense is very useful, and it's a standard name. The thing is, this isn't really linear algebra; the exact same proof applies to dual structures in many contexts. In that field's language, what does this say? The dual map is a contravariant functor.


calculus
integration
real analysis
real analysis unsolved

who can help me to draw the diagram here?

[img]http://farm4.static.flickr.com/3314/3269084114_3b1980d8f0_o.jpg[/img]

The second is a classic problem. I first saw it in Art and Craft of Problem Solving, and it has appeared in at least [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=226729]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=156070]three[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=34446]threads[/url]. There are many different proofs in those threads, but I like the calculus proof best- although I would rather use complex algebra and integrate $ e^{2\pi i(x\plus{}y)}$. With that version, we don't have to worry about where the corner is.

[quote="jmerry"]The second is a classic problem. I first saw it in Art and Craft of Problem Solving, and it has appeared in at least [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=226729]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=156070]three[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=34446]threads[/url]. There are many different proofs in those threads, but I like the calculus proof best- although I would rather use complex algebra and integrate $ e^{2\pi i(x \plus{} y)}$. With that version, we don't have to worry about where the corner is.[/quote]

thanks for the suggestion. I have worked it out.

Please don't remove content that has been responded to. It makes threads incomprehensible.


Let $ 1 < a_1 < a_2 < .......$ be a sequence of positive integers.Show that
$ \frac{2^{a_1}}{{a_1}!} \plus{}\frac{ 2^{a_2}}{{a_2}!} \plus{} .......$
is irrational
Help me

Locked: http://www.mathlinks.ro/viewtopic.php?t=135914
And see PEN.


let
P(x,y)=x^2(x^2-y^2)(x^2-1)+y^2(y^2-x^2)(y^2-1)+(1-x^2)(1-y^2)

How can one prove that P(x,y) cannot be written as
tha sum of squares of any number of polynomials ???

the solution for this one can be found here:
http://www.mathlinks.ro/phpBB/viewtopic.php?t=190


The Calculation of simple yield:

The Return*
(Winnings)
__________ X 100= ROI (%)

Investment*
(Stakes)

so for example: I have 1000 bets with odds 2,00
500 of them are winning and 500 losing. So i am on 0 and my ROI is zero

Would like to get right odds to have 15% yield,
How to calculate on what odds should i be to have ROI 15% ?
Because odds are not always 2,00 - they are different but i gave 2,00 as example...

So, i know what are odds to have 0% yield, but would like to get with some formula odds to have 15% yield (ROI) according to that 0% ROI odds!?

Thanks very much for help!
Ivica

i recieved one answer from some other forum, but don't understand it all really
here it is:

Let p = 1/odds be the probability of success, and R = 2 be the amount you win on a success (as a multiple of the amount you put down). Suppose you play n games, where n is large (in your case n = 1000, but it doesn't really matter).

Expected winnings are npR, and expected cost is n. ROI is (npR - n)/n = pR - 1.

If you want ROI to be 0.15, just solve pR - 1 = 0.15. If R is fixed at 2.00, solve 2p - 1 = 0.15.

Is this right? i don't understand how to get this npR...  :huh:


[quote="moogra"]In how many ways can six dollars (U. S. currency) be changed into dimes and quarters if you must have at least one dime and one quarter in each grouping?[/quote]
[hide="solution"]
this is not the fastest way, but itll work
since every 2 quarters are multiples of 10 cents, each way has to be a multiple of 2 quarters.
since every 5 dimes are multiples of 25 cents, each way has to be a multiple of 5 dimes.
starting with 2 quarters, goin up: (q=quarters, d=dimes)
$2q+55d$
$4q+50d$
$6q+45d$
$8q+40d$
$10q+35d$
$12q+30d$
$14q+25d$
$16q+20d$
$18q+15d$
$20q+10d$
$22q+5d$

in other words, find the amt of multiples of 2 between 0 and 24, not including 0 or 24.

Or, find the amt of multiples of 5 between 0 and 60, not including 0 or 60.

There are $11$ ways
[/hide]

[hide]Well if we factor out the dime and quarters than we need to make 565.
$10x+25y=565$
$2x+5y=113$
For every 2q there are 5 d.
The maximum number of quarters is 21. The minimum is 1. You can only have odd numbers of quarters so there are 11 EDIT:ways.[/hide]

I think you meant 11 ways, not 11 quarters.  Just saying.

[quote="moogra"]In how many ways can six dollars (U. S. currency) be changed into dimes and quarters if you must have at least one dime and one quarter in each grouping?[/quote]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=642437#642437[/url]

[hide]since there needs to be 1 quarter and 1 dime in each group, it is the same as figuring how many ways(with no constraints)that quarters and dimes add to $5.65$. With the max. # of quarters here it would appear that there would be 21 quarters and 4 dimes. Keep substituting 5 dimes for 2 quarters until, well, you can't anymore. These other manipulations are
21q, 4d
19q, 9d
17q, 14d
15q, 19d
13q, 24d
11q, 29d
9q, 34d
7q, 39d
5q, 44d
3q, 49d
and
q, 54d

In all, there are 11 arrangements possible.[/hide]


When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$ \textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$

[quote="jli"][quote="mysmartmouth"]When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$\text{(A) }0.0002\qquad\text{(B) }0.002\qquad\text{(C) }0.02\qquad\text{(D) }0.2\qquad\text{(E) }2$[/quote]

[hide]$\frac{x}{10000}=\frac{4}{x}$

$x=\frac{2}{100}=.02$[/hide][/quote]

I think you mean this equation instead:

$10000x=\frac{4}{x}$
$10000x^{2}=4$
$x^{2}= \frac{4}{10000}$
$x=\frac{2}{100}$
$\boxed{C}$

Can anyone give the answer for this question

 Which of the following is greater than its reciprocal?

$ \textbf{(A) }3\qquad\textbf{(B) }-3\qquad\textbf{(C) }0\qquad\textbf{(D) }-1\qquad\textbf{(E) }1$

[hide="Solution to @above"]We eliminate C,B,D and E because $0$ has no reciprocal. $1$'s reciprocal is itself, $-3$'s reciprocal is $\frac{-1}{3}$ which is less than it, and $-1$'s reciprocal is also itself.

So the answer is A)
[/hide]

[hide="Generalized solution to second problem"]If a number is greater than its reciprocal, then we have $x > \frac{1}{x}$, or, moving everything to the LHS, getting a common denominator, and factoring, $\frac{(x+1)(x-1)}{x} > 0$.

We have $3$ terms on the LHS that are being multiplied:  $x+1$, $x-1$, and $\frac{1}{x}$.  Note that $\frac{1}{x}$ will always have the same sign as $x$, except when $x=0$ - and that's undefined.

For the expression to be positive, we must have either:

A)  All three terms are positive.  Since $x -1 < x < x+1$, having $0 < x-1$ is sufficient (and necessary), so we have $(1, \infty)$ as part of the solution set.

B)  Exactly one term is positive.  This means that $x-1 < x <0< x+1$, which has a solution set of $(-1, 0)$.

So, our solution set is $(-1, 0) \cup (1, \infty)$, and the only answer in that range is $\boxed{3}$.

Note that because $x-1, x, x+1$ are strictly ordered, I can take a shortcut and not bother with sign charts.  If I were trying to solve something like $(x - 2)(2x + 3)(5-x) > 0$, I'd probably want to make a full blown sign chart.[/hide]


conics
ellipse
inequalities
inequalities proposed

For real numbers $ x,\ y$ satisfying $ \frac{x^2}{a^2} \plus{}\frac{y^2}{b^2}>1$,
Prove the following inequality.

\[ \frac{x^4}{a^2}\plus{}\frac{y^4}{b^2}>\frac{a^2b^2}{a^2\plus{}b^2}\]

[quote="kunny"]For real numbers $ x,\ y$ satisfying $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} > 1$,
Prove the following inequality.
\[ \frac {x^4}{a^2} \plus{} \frac {y^4}{b^2} > \frac {a^2b^2}{a^2 \plus{} b^2}\]
[/quote]
It is easy by Cauchy-Schwartz! The inequality is eqiavalent to
\[ \left(\frac{x^4}{a^2}\plus{}\frac{y^4}{b^2}\right)\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\right)>1\mbox{,}\]
which is obvious by Cauchy and the condition.  :)

Hmmm...what do you mean, pablo221092?

[quote="kunny"]Thank you, but could you show how you used Cauchy-Schwarz in detail?[/quote]
By Cauchy-Schwartz, we get
\[ \left(\frac {x^{4}}{a^{2}} \plus{} \frac {y^{4}}{b^{2}}\right)\left(\frac {1}{a^{2}} \plus{} \frac {1}{b^{2}}\right)\geqslant\left(\frac {x^2}{y^2} \plus{} \frac {y^2}{b^2}\right)^2\mbox{,}\]
and then it is clear.  :)


AMC
USA(J)MO
USAMO
AMC 10
AIME
AMC 12
AIME I

Vote for the likely floor score! Also, do you guys think index will be required for <=10th graders this year.

It all depends on the floor. If anyone gets 150/6 (or maybe 145.5/6 or 144/6), the floor will be a 6 and index will come into play. I'm not sure it will if the floor is a 7.

[quote="1=2"]Somebody got a 144/6.[/quote]

on 10 or 12? i know this is not valid but i do know someone who got a 150/6 with 10.

I would bet it is 7.

I heard someone got 144/7 in AMC12/AIME. If we have someone else getting 150/6 in AMC12/AIME, that would make the USAMO index (with AMC10 score) a lot higher than 202.5 this year as there will definitely a lot of students taking AMC10 meeting the AIME floor score.

Yeah, so I got owned...
150 on AMC 10 A
Excess celebration led to no study --> 6 on AIME I (7 last year)
I'm really, REALLY hoping for that 144/6 or 150/6 on the 12, but it's still kinda shameful to depend on others. >_>

Dang it all.

*edit*
Wait, 1=2, is that confirmed, or just a myth?? Because that would be really nice if it was true.

[quote="1=2"]Somebody got a 144/6.[/quote]

NO!!!!!!!!!!! Is the 144 on the AMC12 or the AMC10?

[quote="simo14"]please be on the amc 10!! i got an 8 but a 204.5 amc10 index (don't ask, slept rlly late) and am hoping the floor is 7 or 8 so there won't be an index tiebreaker (or a low one, if any).[/quote]

Ditto(sorta):

I got a 7 on the AIME, but i phailed the AMC10(129 :o ), so my index is only 199. I'm hoping the floor is exactly 7, so that (hopefully) the index does not matter.

If the floor was a 6 but the 10 tiebreaker Index was actually a... 220, say, then would they just move the floor to 7 and recompute a different 10 Index?

Well, if there was a 150/6, then the floor would probably be 6, but on a different note, if someone got a 150/5, would there be any chance of a floor of 5?

There's probably no chance of a 150/5.

First of all, someone scoring 150 on the 12 would have to have a really really bad day to get a 5 on AIME.  Second, a 200 index with the AMC 12 might not even be enough to make USAMO.

Oh, i just realized...

The AMC 12 cutoff index will most likely be higher than last year(204). Thus, I doubt that the 144/6 person will qualify this year. I suppose that the only chance for the index to be below 7 is for someone to get 150/6( and NOT 144/6). Otherwise, the floor will be 7 (as 8 is very unlikely, also someone got 144/7), which probably (at least i hope!) means that the index won't be used.

[quote="mcsquared"]Argh. I'm not gonna make USAMO this year. I'm regressing![/quote]

What was your index? are you <=10th grade?

According to Kent Merryfield, there was a 150/6 in California. So the floor will probably be 6.

Edit: Ignore this. I'm not paying sufficient attention.

worthawholebean, you misread my post. That was 2008 data, not 2009.

You made me wake up my mom because I shouted "YES!!!!!!" :mad:

Haha, AIME15 you should change ur USAMO goal of Didn'ty Qualify/Qualify to 50%/Qualify

Will AIME II scores affect this speculation at all?  Do they have the same USAMO index or do they recalculate another one for that?

I have a question...

What is the importance of the floor score besides being used when the index gets nullified?

That is, why are you guys so concerned about the floor score?

ok. 500 people qualify for usamo.
1st cut: top 330 indices of all amc 12 takers
of the 330, the lowest aime score is the floor score
2nd cut: all <=10th graders who have >=floor score
if the number of people meeting this is >160, then indices are used to weed it down to 160.

Oh... Now I understand...

I should be concerned about the floor score too from now on  :P

Thanks, simo14 :)

i got a 141 on the AMC 10A which was a total suprise
but i thought i might have had a decent chance of making usamo

BUT.............
on the aime i screwed up ------ i solved 8 problems (from 1-9, except for some reason i didnt know how to do #4)
but missed the first one (because 964 is a geometric sequence :ewpu: )
and made an arithmetic error on #8 so i was off by exactly 10

i dont think a 141 and a 6 are good enough to cut it....sigh
next year, though, next year!


Let $ x,y,z\geq0,$prove

          \[ -{\sum\limits_{cyc}(x^2-y^2)^2}\leq4\sum\limits_{cyc}{(x^3y-xy^3)}\leq\sum\limits_{cyc}(x^2-y^2)^2.\]

[quote="fjwxcsl"]Let $ x,y,z\geq0,$prove
\[ - {\sum\limits_{cyc}(x^2 - y^2)^2}\leq4\sum\limits_{cyc}{(x^3y - xy^3)}\leq\sum\limits_{cyc}(x^2 - y^2)^2.
\]
[/quote]
It suffices to prove that
\[ \sum (x^2-y^2)^2 +4(\sum xy^3 -\sum x^3y) \ge 0\]
\[ \Leftrightarrow \sum (z+x)^2(x-y)(x-z) \ge 0\]
which is true by Vornicu Schur.
By the way, it is equivalent to the following inequality of mine
\[ (x^2+y^2+z^2+xy+yz+zx)^2 \ge 4(x+y+z)(x^2y+y^2z+z^2x)\]


Its here : http://www.mathlinks.ro/Forum/viewtopic.php?p=8854#p8854

we know that $(a^2+d^2)(c^2+b^2) = (ab+cd)^2 + (ac-bd)^2$.
so, we have to prove that $9(ab+cd)^2 \ge 16(ac-bd)^2$,
but his is the same that prove $3ab+3cd \ge 4|ac-bd|$.
we can suppose, WLOG, $ac \ge bd$.
so, we have to prove that $3ab+3cd+4bd \ge 4ac$.
but this is the same that $b(3a+4d) \ge c(4a-3d)$.
we know that $c \leq 2b$, so, we have to prove only that $3a+4d \ge 2(4a-3d)$
bu this is the same that $2d \geq a$  :)


??? Why does it only show half an arrow?

It shows half because it is not justified to be in the center

The forum naturally moves it to the left of the post.

well, can you tell me why you love this quote?

"The intellect is characterized by a fundamental inability to comprehend life."
-Henri Bergson

Is that a Quote?

Any way, one of my favorites is on my signature. But there are a lot of others:

[b]The Optimist Ones[/b]

"If your problem have a solution, what do you worry about? If your problem does not have a solution, what do you worry about? "
Chinease proverb.

"You are allowed to fall, to get up is an obligation"
Russian Proverb.

"As they didn't know that it was impossible, they did it!"
Proverb.

[b]The Funny Ones[/b]
"Money does not give you happiness, but gives you a sensation that similar that it is required a very advanced speciallist to find the difference."
Woody Allen. (He has very good ones, but I suppose I would be censored if I post them here  :D )

"If you have a collegue degree, you can be sure of one thing... That you have a collegue degree."
Graffitti outside the UCR.

"Nobody complains to have what he does not deserve"
Jane Austen

"It is human to make mistakes, but to blame someone else is even more human"
Baltasar Gracián

"There is nothing new under the sun, but how many old things we don't know"
Ambrose Bierce

"Lucky the man that can laugh about himself, he will always have a reason to laugh."
Habib Bourguiba

"In this world, most of the people achive what they deserve, but just the ones who achive success are aware of that"
Georges Simenon

"When your opinions are very similar to the ones of everyone else, it is time to reevaluate and redefine them."
Mark Twain

and of course

"Sometimes turgid and opaque prose shrouds a true statement, and sometimes ignorance is quite nicely expressed."
Evans Harrell.

Best,

My favorite quote

"Gun control means using two hands"
- No clue...so i'll just say Me

    My Second Favorite quote

"You are the dumbest smart person i've ever met"
-Will Smith

    My Third Favorite quote

"Never bring a knife to a gun fight"
-Me

"Dreams are like rainbows, only fools chase them."
                                                                 - Some inspirational poster.

One of my favorite ones...

"640K ought to be enough for anybody."
-Bill Gates

Another favorite...

"The hardest thing in the world to understand is the income tax."
-Albert Einstein

I waited there from 7:30 to 8:00 for like 3 hours!
- friend of mine

elvenchamp777

"When a man says he got rich through hard work, ask him whose"
-Don Marquis

"Abandon all hope all ye who enter here"
-Dante... so appropriate(?) for this forum  :?

There's no I in TEAM, but there's three in CHILI FRIES
- from a random magazine ad

"that admirable italian gentleman" - Franklin D Roosevelt, in reference to Benito Mussolini

[quote="alan"][quote="Scrambled"]"I do not consider Hitler to be as bad as he is depicted. He is showing an ability that is amazing and seems to be gaining his victories without much bloodshed "-Mahatma Gandhi

i dont really like the quote that much, but the fact that it comes from Gandhi...[/quote]

well, hitler [i]was[/i] nominated for the nobel peace prize... :what?:[/quote]
good point

my favorites
-when the going gets though, give up.
-no, but i did stay at a holiday inn express last night
-and my signature

[quote="aznness"]Err..

I'd like to know who would nominate Hitler for the Nobel Peace Prize. After all, he did throw the world into WWII[/quote]

Who says that he was nominated AFTER WWII. After WWII he was dead (or at least that is the comon belief). Hittler, If I am right, became the head of the german state on 1933 (therefore, he had to be an important figure for at least a few years before that, and do "important" things during that time), therefore, we would have him as an important figure in international politics for around, let's say, 15 years before the start of the second world war. I really don't know much about his story, but it wouldn't be amaizing that he could have some preponderant roll in some important negociations or so, and though been considered as a possible Nober Peace Prize holder.

Fortunatelly he didn't get it. It would have been amusing that one of the most hated figures in history had won such an imporant prize.

Looking around the web, I found [url=http://www.libertyforum.org/showflat.php?Cat=&Board=news_history&Number=293086975&page=0&view=collapsed&sb=5&o=21&part=1&vc=1&t=0]This Thead[/url] in other forum, and also [url=http://www.ihr.org/jhr/v16/v16n5p22_Weber.html]This Article[/url], that paradoxically stablish that Gertrude Stein proposed Hittler as Nobel Prize 1938. I say paradoxically, because Stein was Jewish.

Best,

I read someone's signature that said "It's all fun and games til someone loses an eye"

To add to that

"It's all fun and games til someone loses an eye...
...then it's just a game...FIND THE EYE!"

-Don't know where I heard it from

A couple of my personal favorites:

"It requires wisdom to understand wisdom: the music is nothing if the audience is deaf." - Walter Lippman

"There are none so blind as those who do not wish to see, and none so deaf as those who do not wish to hear."
-Probably some famous guy

[quote="ankur87"]"There are none so blind as those who do not wish to see, and none so deaf as those who do not wish to hear."
-Probably some famous guy[/quote]

I think it comes from the Bible, but last time I read it was 1989, I cannot be 100% sure. I will ask my wife tonight, for sure she can tell us if I am right!

Best,

[quote="A+MATH"]"Hard work pays off later, sleep pays of now"  :D[/quote]

Mathematics is like crime .... It doesn't pay!!!
- Somewhere in the web.

It's difficult to make anything foolproof, for fools are far too ingenious.

-Forget...

[quote="Danbert"][url=http://www.despair.com]Click here[/url].

It would take too long to post all of my favorites  :D[/quote]

wow, that movie is awesome

I am kind of paranoid in reverse. I suspect people of plotting to make me happy. - J. D. Salinger

God is Love,
Love is Blind,
Ray Charles is Blind,
Ray Charles is God.

Nothing is better than God,
Lobsters are better than nothing,
Lobsters are better than God

Lobsters are better than Ray Charles.

"If you can find a better car, buy it!"
that's from a chevy commercial. :D

I really like this quote:

"So fame comes To the men who mean to win it And care about nothing else!"
-from [i]Beowulf[/i]

Was beowulf a good book? It sounds like something Austin or Dickens would write in terms of the language... but is it interesting?

I only read the excerpts of [i]Beowulf[/i] that is in my Literature book.  It's an Anglo Saxon epic poem so it is very different from the usual Austen or Dickens work.  It's more comparable to [i]The Iliad[/i] or [i]The Odyssey[/i].  The excerpts that I read were very interesting and I hope to read the rest of the book someday (if we don't do that in my english class).

an interesting Family Guy quote:

Peter: Oh my god, Brian, there's a message in my Alphabits. It says, 'Oooooo.'
Brian: Peter, those are Cheerios.

I like

"To Cauchy or not to Cauchy, that is the question"

(I used it as a chapter title in a text I wrote on inequalities recently.)

Quote from Family Guy:

"I'd like some ice cream, WITH SPRINKLES! For every sprinkle I find, I WILL KILL YOU!  :mad: "

--Stewey.

i love family guy you have to be this beautiful to ride the quagmire

Don't reach for the stars, you'll burn your hand off.

Aim for the stars...

...but first aim for their bodyguards.

The one about ray charles=God and lobsters was in my logic course workbook! :D

Look at my signature. It's one of Murphy's laws


Find the best value of k to this ineq be true for all $ a_1,a_2,....,a_n \geq 0$
$ (a_1^2 \plus{} a_2^2 \plus{} ...a_{n \minus{} 1}^2)(a_2^2 \plus{} a_3^2 \plus{} .. \plus{} a_n^2)(a_3^2 \plus{} a_4^2 \plus{} ... \plus{} a_n^2 \plus{} a_1^2).....(a_n^2 \plus{} a_1^2 \plus{} ... \plus{} a_{n \minus{} 2}^2)\leq k(a_1 \plus{} a_2 \plus{} ..a_n)^{2n}$
There are n-sums,each sum has $ n \minus{} 1$ terms

I think the best value of $ k$ should be $ (\frac{n\minus{}1}{n^2})^{n}$, am I right? :wink:

[quote="Sunkern_sunflora"]I think the best value of $ k$ should be $ (\frac {n \minus{} 1}{n^2})^{n}$, am I right? :wink:[/quote]
Oh ; I think best value of $ k\equal{}\frac {1}{2^{n\plus{}2}}$

[quote="onlylove_math"][quote="Sunkern_sunflora"]I think the best value of $ k$ should be $ (\frac {n \minus{} 1}{n^2})^{n}$, am I right? :wink:[/quote]
Oh ; I think best value of $ k \equal{} \frac {1}{2^{n \plus{} 2}}$[/quote]
yes, it is.Where is your proof onlylovemath

I think if :
+$ n\equal{}1,2,3,5,6,7$. $ k\equal{}\frac{1}{2^{n\plus{}2}}$
+$ n \ge 8$,$ k\equal{} \frac{(n\minus{}2)^n}{n^{n\plus{}2}}$
And I think this problem isn't easy ;)
Now , Why don't we dicuss about an easier problem ?
Find the best constant of $ k$ to this ineq true for all $ a_i \ge 0; a_{n\plus{}1}\equal{}a_1$
$ \prod^{n}_{i\equal{}1}.( a_i^2\plus{}a_{i\plus{}1}^2) \le k(\sum a_i )^2$

[quote="mitdac123"]I think if :
+$ n \equal{} 1,2,3,5,6,7$. $ k \equal{} \frac {1}{2^{n \plus{} 2}}$
+$ n \ge 8$,$ k \equal{} \frac {(n \minus{} 2)^n}{n^{n \plus{} 2}}$
And I think this problem isn't easy ;)
Now , Why don't we dicuss about an easier problem ?
Find the best constant of $ k$ to this ineq true for all $ a_i \ge 0; a_{n \plus{} 1} \equal{} a_1$
$ \prod^{n}_{i \equal{} 1}.( a_i^2 \plus{} a_{i \plus{} 1}^2) \le k(\sum a_i )^2$[/quote]
Are you sure,mitdac123?
I think if we know how to prove my problem,we will prove your problem
It is quite nice,mitdac123


If you are given only a compass and no straight-edge or anything else. Show how you would create a square

I thought I solved this problem before but now I can't do it so I'm starting to doubt I ever solved it.
 :blush:

I think you are asking us to construct the [b]vertices[/b] of a square. Or how can one draw the sides with a compass :?:

EDIT: With a compass can we draw a circle with the center at a given point and touching a given circle?

Yes you are right. Sorry if I was unclear.

You would need to show that those two circles are tangent.

[hide="solution"]First of all trace a circle with center $ O$ and raze $ R$ and pick 2 oposite points $ A$ and $ D$, $ AD$ diametar of $ C(O,R)$.
$ AD \equal{} 2R$.With the sharp claw of the compas into point D, trace the circle $ C_2(D,R)$.Note that the intersection of $ C(O,R)$ with $ C_2(D,R)$ is the point $ C$.
$ \Delta ADC$ is rectangular $ \Rightarrow$ $ AC \equal{} R\sqrt {3}$.With the sharp claw in $ A$ and then in $ D$,trace two circles with the raze $ R\sqrt {3}$.These 2 circles are intersectating in the point $ M$.$ \Delta AMD$ is isosceles therefore $ OM \equal{} R\sqrt {2}$.

[color=darkblue][b]The main construction of the figure[/b][/color]
Pick up into the compas claws the raze and trace the circle.We have discovered the measure of $ R\sqrt {3}$,that we take into the claws and trace a circle,and therefore we obtain $ R\sqrt {2}$ and we trace the circle with it and in this way we obtain the square[/hide]

How do you choose the points A and D so that you know it is a diameter?

I don't follow what you said at the end either.

How do you choose a diameter of the circle? How do you get points A and D such that AOD is a straight line?

Based on balan razvan's solution, I created this picture with a little modification.

Construction:

[$ (X,YZ)$ means circle with center in point $ X$ and with radius length same as length of $ YZ$. New points are not defined to be intersection of previous circles. $ <12>$ means intersection of the 1st and 2nd circle.]

Take points $ P_1$, $ P_2$ in the distance of the square's side length. We can do that.

$ (1).$ $ (P_1,P_1P_2)$
$ (2).$ $ (P_2,P_2P_1)$
$ P_3=<12>$
$ (3).$ $ (P_3,P_3P_1)$
$ P_4=<13>$
$ (4).$ $ (P_4,P_4P_1)$
$ P_5=<14>$
$ (5).$ $ (P_5,P_5P_3)$
$ (6).$ $ (P_2,P_2P_4)$
$ P_6=<56>$
$ (7).$ $ (P_6,P_1P_2)$
$ P_7,P_8=<17>$

$ P_1P_7\&P_6P_8$ is the intended square.

Proof:

Since $ P_1P_7\&P_6P_8$ is a rhombus I'll prove the angle $ P_6P_7P_1$ is a right angle.
Note that $ P_6P_1,P_5P_3,P_2P_4$ are respectively perpendicular to $ P_2P_5,P_2P_3,P_5P_4$ because $ P_5P_1P_2$ collinear( Sum of angle at $ P_5P_1P_2=P_2P_1P_3+P_3P_1P_4+P_4P_1P_5=60^0+60^0+60^0=180^0$)$ ,P_5P_1=P_2P_1,P_2P_6=P_5P_6$}

$ P_1P_6^2=P_6P_2^2-P_2P_1^2=P_2P_4^2-P_2P_1^2=P_2P_5^2-P_4P_5^2-P_1P_2^2=4P_1P_2^2-P_4P_5^2-P_1P_2^2=2P_1P_2^2=P_1P_7^2+P_6P_7^2$

Proved.


inequalities Vieta quadratics algebra polynomial vector trigonometry

Can someone explain the AM-GM and Cauchy-Schwarz Inequalities? I don't really understand them that well. (or how to use them)

Also, can anyone explain Vieta's Formulas? are they only related to the quadratic or what?

AM-GM : If $ a_1, a_2, ... , a_k$ are nonnegative reals, then:

$ \frac {a_1\plus{}a_2\plus{}...\plus{}a_k}{k} \ge \sqrt[k]{a_1a_2....a_k}$

Cauchy-Schwarz: If $ x_1,x_2,...,x_n,y_1,y_2,...,y_n$ are reals, then:

$ (\sum^n_{i\equal{}1} x_iy_i)^2 \le (\sum^n_{i\equal{}1} x_i^2)(\sum^n_{i\equal{}1} y_i^2)$

I really can't help you about when using them.

Viete's formulas : It's boring for me to write.

Links:

AM-GM: http://en.wikipedia.org/wiki/AM-GM_inequality
CS: http://en.wikipedia.org/wiki/Cauchy-Schwartz_inequality
Viete's formulas: http://en.wikipedia.org/wiki/Vieta%27s_formulas

The AM-GM inequality is short for the Arithmetic Mean-Geometric Mean inequiality.

It just says the average of a set of numbers is greater than or equal to the geometric average.
So to use the positive integers for an easy example.. $ \frac {1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n}{n} \geq \sqrt [n]{n!}$

The Cauchy-Schwarz is the magnitude of of two multiplying numbers is less than or equal to the magnitude of each number multiplying.
That is $ |abcd...n| \leq |a|\cdot|b|\cdot|c|\cdot|d|\cdot...\cdot|n|$

This is similar to the triangle inequality which states $ |a \plus{} b \plus{} c \plus{} ... \plus{} n| \leq |a| \plus{} |b| \plus{} |c| \plus{} ... \plus{} |n|$.

I don't really know Vieta's, so http://en.wikipedia.org/wiki/Vieta%27s_formulas.

Also there are QM and HM:
Power mean:

$ M_p(x_1,x_2,...,x_n)\equal{} ( \frac {\sum^n_{i\equal{}1} x_i^m}{n})^\frac{1}{m}$

$ QM: m\equal{}2$
$ AM: m\equal{}1$
$ GM: m\equal{}>0$
$ HM: m\equal{}\minus{}1$

For $ p<q$ holds $ M_p \le M_q$ for numbers $ x_1,x_2,...,x_n$

so $ QM \ge AM \ge GM \ge HM$

The equality holds for $ x_1\equal{}x_2\equal{}...\equal{}x_n$

Check out: http://en.wikipedia.org/wiki/Power_mean

Vieta's formulas apply to all polynomials, but let's look at only monic polynomials (leading coefficient is 1) for now.  The basic idea is that if a monic polynomial $ P(x)$ has degree $ n$, then we can write $ P(x)\equal{}(x\minus{}r_{1})(x\minus{}r_{2})\cdots (x\minus{}r_{n})$.  Now consider expanding the right side.  You can think of this expansion like this: each term upon expanding the whole thing comes from choosing either the $ x$ or the $ \minus{}r_{i}$ from each term, and multiplying all your choices together.  Then we get a $ x^{n\minus{}k}$ term by choosing $ n\minus{}k$ of the $ x$'s and $ k$ of the $ \minus{}r_{i}$'s.  When we combine all these terms, we get the sum of all possible products of $ k$ $ r_{i}$'s times $ x^{n\minus{}k}$, and we get a negative sign if $ k$ is odd and no negative sign if $ k$ is even.

Therefore, if our polynomial is $ P(x)\equal{}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{0}$, then $ a_{n\minus{}1}\equal{}\minus{}(r_{1}\plus{}r_{2}\plus{}\cdots \plus{}r_{n})$.  Also, $ a_{n\minus{}2}\equal{}r_{1}r_{2}\plus{}r_{1}r_{3}\plus{}\cdots \plus{}r_{n\minus{}1}r_{n}$, the sum of the roots taken two at a time.  Vieta's gives us a relation like this for every coefficient, but the most useful ones are the first one, which gives you information about the sum of the roots, and the last one, $ a_{0}\equal{}(\minus{}1)^{n}r_{1}r_{2}\cdots r_{n}$, which gives you information about the product of the roots.

To deal with polynomials that are not monic, we just have to divide through by the leading coefficient first.  This means, for example, that instead of $ a_{n\minus{}1}\equal{}\minus{}(r_{1}\plus{}r_{2}\plus{}\cdots \plus{}r_{n})$, the formula we want is $ a_{n\minus{}1}/a_{n}\equal{}\minus{}(r_{1}\plus{}r_{2}\plus{}\cdots \plus{}r_{n})$.  This formula now works for all polynomials, not just monic ones.  The other formulas would be $ a_{n\minus{}2}/a_{n}\equal{}r_{1}r_{2}\plus{}r_{1}r_{3}\plus{}\cdots \plus{}r_{n\minus{}1}r_{n}$, and so on.

Vieta's formulas are useful primarily in problems involving polynomials (duh).  Specifically if you're given certain coefficients of a polynomial but not all the coefficients, then you may not be able to work with the polynomial directly, but you have valuable information about the roots.  Another application is this: if you have a system of equations like $ a\plus{}b\plus{}c\equal{}6$, $ ab\plus{}bc\plus{}ca\equal{}11$, $ abc\equal{}6$, then $ a,b,c$ are the roots of the polynomial $ x^{3}\minus{}6x^{2}\plus{}11x\minus{}6$, which is much easier to deal with than the original system of equations.

[quote="Fermat's Little Turtle"]The Cauchy-Schwarz is the magnitude of of two multiplying numbers is less than or equal to the magnitude of each number multiplying.
That is $ |abcd...n| \leq |a|\cdot|b|\cdot|c|\cdot|d|\cdot...\cdot|n|$[/quote]

what?  as stated, the above is actually an equality, and is definitely not equivalent to cauchy-schwarz.

$ QM$ is quadratic mean:

for $ n$ numbers: $ \sqrt{\frac{a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2}{n}}$

Also is used: $ RMS$ which means Root Mean Square.

What's the whole purpose of using Quadratic Mean or Root Mean Square? :maybe:

Here's one problem and a solution to it:

If $ a,b$ and $ c$ are lenghths of sides of the triangle, then the following inequality holds:

$ \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{c\plus{}a\minus{}b}\plus{}\sqrt{b\plus{}c\minus{}a}\le\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

[hide="Solution(using inequality QM-AM)"]

$ \frac{\sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{c\plus{}a\minus{}b}}{2}\le\sqrt{\frac{\sqrt{a\plus{}b\minus{}c}^2\plus{}\sqrt{c\plus{}a\minus{}b}^2}{2}} <\equal{}>$
$ \frac{\sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{c\plus{}a\minus{}b}}{2}\le\sqrt{a}$
Do the same for each two square roots on the LHS and you'll get the inequality![/hide]

This is one of many problems where Quadratic mean is used.

The root mean square or quadratic mean is just another way to take the "average" of a list of positive reals.  As you've seen there are a lot of other ways to take averages: the arithmetic mean, the geometric mean, the harmonic mean, the quadratic mean.  The quadratic mean of $ a_{1},a_{2},\ldots ,a_{n}$ is defined as $ \sqrt{\frac{a_{1}^{2}\plus{}a_{2}^{2}\plus{}\cdots \plus{}a_{n}^{2}}{n}}$.  The reason you don't see quadratic mean in an inequality is because there's a lot of inequalities involving it - the quadratic mean of a list of positive reals is always greater than or equal to the arithmetic mean, the geometric mean, and the harmonic mean.  There is a chain of inequalities you should know, $ QM\ge AM\ge GM\ge HM$.

If you want to go even further, then there is a more general type of mean called the "power mean" that Bugi talked about in his post earlier.  Basically, all of the common means (QM, AM, GM, HM) are types of power means, but there are infinitely many other power means, and they all fall into a chain of inequalities analogous to $ QM\ge AM\ge GM\ge HM$ (this is actually a sub-inequality of this infinite chain of inequalities).

[b]CAUCHY-SCHWARZ INEQUALITY[/b]

[b]Background/Vector Definition:[/b] Cauchy-Schwarz says that for any two vectors $ x$ and $ y$, $ | x \cdot y | \leq \| x \| \| y \|$.

The dot product of two vectors can be defined as $ x \cdot y \equal{} \| x \| \| y \| \cos \theta$, where $ \theta$ is the angle between the two vectors. So intuitively, the inequality is equivalent to $ | \cos \theta | \leq 1$, a true statement.

Note that equality happens only when $ \cos \theta \equal{} \pm 1 \implies \theta \equal{} n \pi$; or in other words, when $ x$ and $ y$ are in the same direction.

[b]Algebraic Form:[/b] This is the form you will usually see the inequality being used in.

This is a result of a certain way of computing the [url=http://en.wikipedia.org/wiki/Dot_product#Definition]dot product[/url] of two vectors; that is, for vectors $ x \equal{} (x_1, x_2, \ldots x_n)$ and $ y \equal{} (y_1, y_2 , \ldots y_n)$, we have that:

$ x \cdot y \equal{} x_1 y_1 \plus{} x_2 y_2 \plus{} \ldots \plus{} x_n y_n$

Now, let's go back and square the original vector equation: $ ( x \cdot y )^2 \leq \| x \| ^2 \| y \| ^2$

We know from the distance formula that $ \| x \| ^2 \equal{} x_1 ^2 \plus{} x_2 ^2 \plus{} \ldots \plus{} x_n ^2$, and similarly for $ y$. We also know an expression for $ x \cdot y$ in terms of $ x_i, \; y_i$. So let's go ahead and fill these in:

$ (x_1 y_1 \plus{} x_2 y_2 \plus{} \ldots \plus{} x_n y_n )^2 \leq (x_1 ^2 \plus{} x_2 ^2 \plus{} \ldots \plus{} x_n ^2)(y_1 ^2 \plus{} y_2 ^2 \plus{} \ldots \plus{} y_n ^2)$

Or, in summation notation, $ \boxed{\left(\sum_{i \equal{} 1}^n x_i y_i\right)^2\leq \left(\sum_{i \equal{} 1}^n x_i^2\right) \left(\sum_{i \equal{} 1}^n y_i^2\right)}$

Since the vector form works for any two real-valued vectors, $ x_i, \; y_i$ can be any real numbers (not necessarily positive). Equality happens when $ x$ and $ y$ are in the same direction, or in coordinate terms, when $ x_i/y_i$ is constant for all $ i$.

Another Lemma used sometimes in proving Cauchy-Scharz that can be very very useful:

\[ \frac{a^2}{x} \plus{} \frac{b^2}{y} \ge \frac{(a\plus{}b)^2}{x\plus{}y}\]

Note that this could be generalized to n variable by applying this multiple times.

what is a problem that can be solved by Cauchy Schwartz?
What is HM?

Cauchy-Schwarz problem:
Prove $ (x^2\plus{}1)(y^2\plus{}1)\geq(x\plus{}y)^2$ for real $ x,y$.
HM is another way to find an average; for $ a_1, a_2, \ldots, a_n$, the harmonic mean is
$ \frac{n}{\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots\plus{}\frac{1}{a_n}}$.
It can be used as a problem for certain weighted rate averages.
For example, the average speed paradox problem of traveling $ x$ miles at 30 mph, then $ x$ miles at 60 mph does not have an average of 45 mph. Rather, the answer turns out to be the harmonic mean of 30 and 60.

Here are my efforts on the cauchy-swarz problem:
Expanding and rearranging we have $ x^2y^2 \plus{} 1 > 2xy \implies (xy)^2 \minus{}2xy \plus{} 1 > 0 \implies (xy\minus{}1)^2 > 0$ which is true by the trivial inequality. Im still not so clear on cauchy schwarz, how do you tell which$ x_1y_1$   to multiply in the equation:
$ (x_1 y_1 \plus{} x_2 y_2 \plus{} \ldots \plus{} x_n y_n )^2 \leq (x_1 ^2 \plus{} x_2 ^2 \plus{} \ldots \plus{} x_n ^2)(y_1 ^2 \plus{} y_2 ^2 \plus{} \ldots \plus{} y_n ^2)$

Can someone state Cauchy in words/solve the inequality brutforc3 gave using cauchy?
Thanks  :)

Brut3Forc3's problem:

$ x_1\equal{}x, x_2\equal{}1, y_1\equal{}1, y_2\equal{}y$
Now use Cauchy-Schwarz...

Brut3Forc3 posted a problem which can be solved by Cauchy-Schwarz:
$ (x^2\plus{}1)(y^2\plus{}1)\ge(x\plus{}y)^2$
From the inequality: $ ({x_1}^2\plus{}{x_2}^2)({y_1}^2\plus{}{y_2}^2)\ge(x_1y_1\plus{}x_2y_2)^2$-Cauchy-Schwarz(one case).
If we put $ x_1\equal{}x,x_2\equal{}1,y_1\equal{}1,y_2\equal{}y$ we get the Brut3Forc3's inequality.

[quote="mathemonster"]So typically in a proof of cauchy u have to say let x_1 = 1 and so on?[/quote]
Not really. This inequality is well-known enough that you can simply do something like:
Using Cauchy, we have $ (x^2\plus{}1)(1\plus{}y^2)\geq(x\plus{}y)^2$.
In this case, I rearranged the terms of $ y^2\plus{}1$ to make it more clear what corresponded to what.
Here is another Cauchy problem: Let $ a_1,a_2,\ldots,a_n$ be positive reals. Prove that $ (a_1\plus{}a_2\plus{}\ldots\plus{}a_n)(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\ldots\plus{}\frac{1}{a_n})\geq n^2$.
[quote="mathemonster"]Expanding and rearranging we have $ x^2y^2 \plus{} 1 > 2xy \implies (xy)^2 \minus{}2xy \plus{} 1 > 0 \implies (xy\minus{}1)^2 > 0$ which is true by the trivial inequality. Im still not so clear on cauchy schwarz, how do you tell which to multiply in the equation:[/quote]
Comments on your solution: The inequalities can also be equal. :P
Also, I noticed you were able to use the trivial inequality to solve it. From $ x^2y^2 \plus{} 1 \geq 2xy$, we can also use AM-GM. In fact, for 2 variables, AM-GM and the trivial inequality are [i]very[/i] related. Cauchy is also very related to AM-GM in 2 variables.
I propose these 2 problems:
Prove AM-GM in 2 variables with
a) the trivial inequality
b) Cauchy-Schwarz inequality

[u][b]1. Problem:[/b][/u]
$ (a_1\plus{}a_2\plus{}...\plus{}a_n)(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n})\ge(\sqrt{a_1}\times\sqrt{\frac{1}{a_1}}\plus{}\sqrt{a_2}\times\sqrt{\frac{1}{a_2}}\plus{}...\plus{}\sqrt{a_n}\times\sqrt{\frac{1}{a_n}})^2\equal{}n^2$ This can be proven also by $ AM\minus{}HM$ inequality.
[u][b]2. Problem[/b][/u]
a)Let's start with true inequality: $ (\sqrt{a}\minus{}\sqrt{b})^2\ge0$ Now use binomial formula on LHS, then add $ 2\sqrt{ab}$ on both side, and at the end divide both sides by $ 2$. That's it!!!
b)$ (a\plus{}b)(b\plus{}a)\ge(\sqrt{ab}\plus{}\sqrt{ab})^2\equal{}4ab$ so we get $ (a\plus{}b)^2\ge4ab$. As both sides are positive, $ a\plus{}b\ge2\sqrt{ab}$ also holds. Now divide both sides by 2 and it's over!!!


geometry
circumcircle
perpendicular bisector
geometry unsolved

In $ \triangle ABC$, $ BX$ is the altitude. $ D,E$ are midpoints of $ AB,BC$ respectively. The circumcircle of triangle $ DXC$ has $ O$ as its circumcenter. Prove that $ OE$ is perpendicular to $ AC$.

[hide]This follows immediately from the fact that the line through $ E$ perpendicular to $ AC$ is the perpendicular bisector of $ CX$...[/hide]


induction
combinatorics unsolved
combinatorics

All the grids of a $m\times n$ chess board ($m,n\geq 3$), are colored either with red or with blue. Two adjacent grids (having a common side) are called a "good couple" if they have different colors. Suppose there are $S$ "good  couples". Explain how to determine whether $S$ is odd or even. Is it prescribed by
some specific color grids? Justify your answers.

[quote]All the grids ...[/quote]
grids = squares?
If yes, then for given $m$ and $n$ it ispossible that $S$ is odd and it is possible $S$ is even.
Or I am missing something.  :maybe:

The problem asked to find the conditions that determined S was odd or even. That is to say, S may be odd or even, which is determined by some conditions.

I look through the Chinese version, the question is:
whether S is odd or even is determined by the color of which grids? On what conditions  is S odd? And on what conditions is S even?

I mean we can't obtain parity of $S$ from numbers $n$ and $m$. So what kind of conditions do they want?

The official solution put the chess board into three parts:
Part I: four grids in corners;
Part II: grids on four sides, not including four corners;
Part III: other grids, which in the chess board, not on the sides.
The official solution said that parity of S is determined by the coloring style of Part II. Here m and n are only demanded >= 3.

Do you mean part II ?
Anyway my opinion is :(  :(  :(

It seems that it is one part of the problem, and not the easiest one, to understand what the question is.... :(
I agree : I hate this kind of statements.

Pierre.

This problem, in this case, is almost trivial:
Consider the red squares. If we consider a connex figure formed of red squares, and it is not near the extremities of the board, it is easy to see it has an even number of $good\ couples$(proof by induction on number of squares in the figure). Now surround the $m \times n$ table with blue squares. Apply the property for all the red squares, then you see that only the ones on the sides and not on the corners have contributed an odd number of times to the number of $good\ couples$.

[quote="DusT"]This problem, in this case, is almost trivial[/quote]
That is why no one write solution  ;)

Classify all grids into three parts: the grids at the four corners (first part), the grids along the borderlines (not including the four corners, the second part) , and the other grids (the third part).

We claim that the parity of $S$, or in other words whether or not it is even or odd, is determined by the colors of the grids in the second part. More specifically, when there are odd blue grids among the grids in the second part , $S$ is odd. Otherwise $S$ is even.

[b]Proof:[/b]
We fill all red grids with label number $1$ and all blue grids with label number $-1$. Denote the label numbers filled in the grids in the first part by $a , b , c, d$, and those in the second part by $x_1, x_2,\dots, x_{2m+2n-8}$, and in the third part by $y_1 , y_2 ,\dots ,y_{(m-2)(n-2)}$.

For any two adjacent grids we write a label number denoting the product of two label numbers of the two grids on their common edge. Let $E$ be the product of all label numbers on common edges. There are $2$ adjacent grids for every grid in the first part, so its label number appears twice in $E$. There are $3$ adjacent grids for every grid in the second part, thus its label number appears three times in $E$. There are $4$ adjacent grids for every grid in the third part, thus its label number appears four times in $E$. Therefore,

$E=(abcd)^2(x_1x_2\dots x_{2m+2n-8})^2(y_1y_2\dots y_{(m-2)(n-2)}^4=(x_1x_2\dots x_{2m+2n-8})^3$

If $x_1x_2\dots x_{2m+2n-8}=1$, then $E=1$, and in this case there are an even number of good couples.
If $x_2x_2\dots x_{2m+2n-8}=-1$, then $E =- 1$, and in this case there are an odd number of good couples.  $\square$

I believe this is also an official solution (there were two solutions presented).