algebra
polynomial
modular arithmetic
algebra unsolved

Let be given an odd prime p and the polynomial :
 $ T(x) = (p-1)x^p-x -1 $ .
 Prove that : there exists an infinite number of possitive integers a such that $ T(a) $ is divisible by $ p^p $

Suppose you've found $a_n$ s.t. $p^n|T(a_n)$. We now look for numbers of the form $a_{n+1}=a_n+tp^n,\ t\in\overline{0,p-1}$ s.t. $p^{n+1}|T(a_{n+1})$.

We have $T(a_n+tp^n)=T(a_n)+tp^nT'(a_n)+p^{n+1}M$. We divide this by $p^n$ to get $\frac{T(a_n+tp^n)}{p^n}=\frac{T(a_n)}{p^n}+tT'(a_n)+pM$, and we want $t\in\overline{0,p-1}$ s.t. the RHS is divisible by $p$. We can clearly choose such a $t$, since $T'(a_n)\equiv -1\pmod p$, so as $t$ goes through $0,1,\ldots,p-1\pmod p$, so does the RHS.

We have shown that for all $n$, there is an integer $a_n$ s.t. $p^n|T(a_n)$, and if there is one such integer, there are infinitely many. In particular, this holds for $n=p$, as desired.


geometry
geometric transformation
vector
analytic geometry
algebra solved
algebra

A region in the plane has area \geq 1. Prove that it contains some points (a, b) and (c, d) such that both a - c and b - d are integers.

I think it is false if you only suppose that the area is  \geq 1. A counter-example is the interior (not the sides) of a square with side 1.

In another hand, the result is true if it is assumed that the area is > 1 (it is known as Blichfeldt's lemma, and uses as a starting point the prove Minkowski's theorem in geometry of numbers) :

Let E be a set with area > 1.
Consider the usual square grid with side 1.

Decompose E as E =  \cup ( E \cap S) where the union is over the squares S of the grid. Let T(S) = t( (E \cap S)' ) where t is the translation such that t(S) = S_0 the square with vertices (0,0),(0,1),(1,1) and (1,0), and (E \cap S)' denotes the set (E \cap S) without the sides of S.
Note that each of this translation has vector with integer coordinates.

We have area[E] = sum area[(E \cap S)'] = sum [T(S)] > 1 = area[S_0] It follows that two of the sets T(S) must have a non empty intersection. Then there exists M(x,y) in the interior of S_0 which belongs to some T(S) and T(S') for some distinct S,S' of the grid. Using the reciprocal translations we deduce that there are two distinct points of E which have coordinates of the form (x+a,y+b) and (x+c,y+d) respectively, where a,b,c,d are integers. And we are done

Pierre.


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

I saw this problem in Engel, USAMO 1980;

$If a,b,c$ are 3 reals that can take any value from 0 to 1 (both inclusive); prove that $(a/(b+c+1))+(b/(c+a+1))+(c/(a+b+1))+(1-a)(1-b)(1-c)$ can take a maximum value of 1.
He has proved it by the following method:

" LHS = f(a,b,c) ; f is strictly convex in each variable, and so is defined on a closed convex cube. "
I understand this. But then he says, "thus, the function assumes extremal values at extremal points". How is this ?

He cites some theorem of Weierstrass. Please help. I need to get this cleared quick. Thanks, in advance.

The values of $a,b,c$ are defined for $[0,1]$

This means that the function takes its values from the unit cube in $\mathbb{R}^{3}$ space.

Now, look at each variable and hold the other two constant. Then the function in Convex in that variable, which you can check by taking second derivities. Since its convex, that means it assumes the maximum on one or both of its endpoints. Since the same is true for the other two variables, you know the whole function takes on its maximum at one of the points

$(0,0,0),(0,0,1),(0,1,1),(1,1,1)$

(We dont have to check all the values since ots symmetric in its variables)
Hope this helps.


If $ x^2\minus{}y^2\equal{}27$ and $ x^3\plus{}y^3\equal{}4941$, find $ 2xy$.

Again, as in the previous thread, is there an "elegant" solution?

If you factor things out, you will get

$ (x\plus{}y)(x\minus{}y)\equal{}27$
$ (x\plus{}y)(x^2\minus{}xy\plus{}y^2)\equal{}(81)(61)$

If we have $ x\plus{}y\equal{}27$ and $ x\minus{}y\equal{}1$ (Sorry, I couldn't really explain anything legit other than guess and check), then solving the system gives $ x\equal{}14$ and $ y\equal{}13$. Now to check this:

$ (27)(14^2\minus{}14*13\plus{}13^2)\equal{}(27)(196\plus{}169\minus{}182)\equal{}(27)(183)\equal{}(81)(61)$

which works.

So $ 2xy\equal{}2(13)(14)\equal{}\boxed{364}$.

we can use Cardano methode to solve  a qubic equation  we will get  y= 13
and x= 14


geometry
rectangle
ratio
perimeter
ceiling function
floor function
number theory

Let $ ABCD$ be a rectangle with $ AB \equal{} m,BC \equal{} n$ and $ m,n\in\mathbb{N},(m,n)\equal{}1$.Rectangle $ ABCD$ is divided into squares of unit length and colored as chess board.
Let $ X$ be the sum of all white pieces,that are obtained by intersecting unit squares by the line $ AC$,similarly $ Y$ is the sum of all black pieces,that are obtained by intersecting unit squares by the line $ AC$.
Find the value $ \frac {X}{Y}$ in terms of $ m,n$.

It's important to specify whether the unit square touching point $ A$ is white or black, right? I'll assume it's white here. Here's a proof with a few sizable holes I don't care to plug.

First, note that if $ m \equal{} da, n \equal{} db$ with $ (a,b) \equal{} 1$ (so $ (m,n) \equal{} d$) then we can subdivide the $ m \times n$ rectangle into $ d$ $ a \times b$ rectangles so that they don't overlap and $ AC$ passes through all of them. $ AC$ passes through the same color squares in each rectangle, so the ratio $ X/Y$ for the $ m \times n$ rectangle is the same as the ratio for the $ a \times b$ rectangle. Therefore it suffices to consider $ (m,n) \equal{} 1$.

Observe now that when $ (m,n) \equal{} 1$ that we can not pass through the vertex of any unit square. So line $ AC$ only passes through edges of unit squares. This means that if we mark the squares line $ AC$ touches we can walk from the square touching $ A$ to the square touching $ C$ through only marked squares. In any such walk (marked squares or not) we always pass through the same number of squares, so if we consider the walk around the perimeter we get $ m \plus{} n \minus{} 1$ squares. This means that line $ AC$ passes through $ m \plus{} n \minus{} 1$ squares.

Since we only pass through the boundaries of the unit squares, $ AC$ always passes through a white, then a black, then a white, and so on. By the assumption in the first paragraph, we touch a white first. If $ m \plus{} n \minus{} 1$ is even, we pass through $ \frac{m \plus{} n \minus{} 1}{2}$ white and black squares both, and the desired ratio is $ 1$. If $ m \plus{} n \minus{} 1$ is odd, we pass through $ \frac{m \plus{} n}{2}$ white squares and $ \frac{m \plus{} n \minus{} 2}{2}$ black squares, and the desired ratio is $ \frac{m\plus{}n}{m\plus{}n\minus{}2}$.

You do not understand the statement,$ X$ is not the number of black unit squares,which are intersected by $ AC$,it is the sum of length of the pieces,which lies on $ AC$ and inside black squares.
For example:
If $ m\equal{}n\equal{}2$,then $ X\equal{}\sqrt{2}$

If $ m \equal{} n \equal{} 2$ then according to the clarification either $ X \equal{} 0$ or $ X$ is undefined  :maybe:

The reduction to relatively prime $ m, n$ is still valid.  If one of $ m, n$ is even then the answer is $ X \equal{} 1$, so the only interesting part is for $ m, n$ odd.  A little experimenting suggests that the answer will always be $ \frac{\left\lceil \frac{m\plus{}n}{2} \right\rceil}{\left\lfloor \frac{m\plus{}n}{2} \right\rfloor}$.  This sounds entirely reasonable to me, but I haven't tried to prove it.

$ X$ doesn't represent the ratio he's asking for, but you are right that $ X/Y \equal{} 0$ or $ X/Y$ is undefined.


I'm working on chapter 7, and was wondering if anyone is kind and could tell me the answers to the problems in the back.  :lol:

oh oops I forgot to say that. volume 1.  :D


How many positive integers $ n$ satisfy \[ (n \plus{} 8)(n \minus{} 3)(n\minus{}12)<0\]

This could only be negative if all three are negative or 1 is negative while the the other two are positive.

Case 1: All 3 terms are negative -

This occurs when n < -8, but clearly, there are no positive solutions for this.

Case 2: One term is negative while the otehr two are positive -

(n+8) and (n-3) are the two biggest terms, so in order for only one of them to be negative, then (n-12) must be it. (n-3) is the second smallest, and this must be positive, so n > 3. However, (n-12) is less than 0, so n < 12. The integer solutions are

$ 4\le n\le 11$

Therefore, there are a total of (11-4+1) = 8 positive integer solutions.


This question looks to me like it more properly belongs in a calculus forum, although perhaps there are elementary ways of solving it.

[hide]
It would follow immediately from the inequality for any positive integer $m$:
\[\sqrt{m}<\frac{2}{3}((m+1)^{3/2}-m^{3/2})\]
But we can rationalize the numerator on the right hand side:
\begin{eqnarray*}\frac{2}{3}((m+1)^{3/2}-m^{3/2})&=&\frac{2}{3}\cdot{(m+1)^{3}-m^{3}}{(m+1)^{3/2}+m^{3/2}\\ &=&\frac{2}{3}\cdot \frac{m^{2}+m(m+1)+(m+1)^{2}}{(m+1)^{3/2}+m^{3/2}}\\ &=&\sqrt{m}\cdot \frac{2}{3}\cdot \frac{1+r^{2}+r^{4}}{1+r^{3}}\\ }\end{eqnarray*}
where $r=\sqrt{\frac{m+1}{m}}>1$. It only remains to show that
\[\frac{1+r^{2}+r^{4}}{1+r^{3}}>\frac{3}{2}\]
But the numerator factors into $(1+r+r^{2})(1-r+r^{2})$ and the denominator is $(1+r)(1-r+r^{2})$ so the left hand side of the above is actually $\frac{1+r+r^{2}}{1+r}$. The fact that this is greater than $\frac{3}{2}$ is a result of
\[2(1+r+r^{2})-3(1+r)=2r^{2}-r-1=(r-1)(2r+1)>0\]
[/hide]


Let,
\[ S\equal{}\sum_{i\equal{}0}^\infty \frac{F_i}{k^{i\plus{}1}}\]
where $ k$ is a nonzero real number.
Suppose the series converges. Then, by Binet's formula,
$ \sum_{i\equal{}0}^\infty \frac{F_i}{k^{i\plus{}1}} \equal{} \frac{1}{\sqrt{5}k}\left[ \sum_{i\equal{}0}^\infty \left( \frac{1\plus{}\sqrt{5}}{2k} \right)^i \minus{} \left( \frac{1\minus{}\sqrt{5}}{2k} \right)^i \right]$
$ \equal{} \frac{1}{\sqrt{5}k}\left[ \frac{1}{1\minus{}(1\plus{}\sqrt{5})/2k} \minus{} \frac{1}{1\minus{}(1\minus{}\sqrt{5})/2k} \right]$
$ \equal{} \frac{1}{\sqrt{5}k}\left( \frac{1}{2k\minus{}1\minus{}\sqrt{5}} \minus{} \frac{1}{2k\minus{}1\plus{}\sqrt{5}} \right)$
Thus,
\[ S\equal{}\frac{1}{k^2\minus{}k\minus{}1}\]

when $ k\equal{}10$, $ \sum_{i\equal{}0}^\infty \frac{F_i}{10^{i\plus{}1}} \equal{} \frac{1}{89}$
Thus,
\[ \sum_{i\equal{}0}^\infty \frac{F_i}{10^i}\equal{}\frac{10}{89}\]

when $ k\equal{}\minus{}10$, $ \sum_{i\equal{}0}^\infty \frac{F_i}{(\minus{}10)^{i\plus{}1}} \equal{} \frac{1}{109}$
Thus,
\[ \sum_{i\equal{}0}^\infty \frac{F_i}{10^i}\equal{}\frac{10}{109}\]


[quote="Toxic_Frog"]We have divided a 1999-sided regular polygon into triangles by drawing its diagonals . Prove that there exists exactly one acute angled triangle .[/quote]

Suspiciously homework-looking again, so a hint only:

How does one infer whether a triangle is acute-angled from the position of its circumcenter?

  darij

thx a lot
[quote]
 Suspiciously homework-looking again
[/quote]

No proposed in a mock olympiad.

Seems to be a VERY simple olympiad problem then.

  Darij


Find all continuos function $f: R\to R$ such that:
$f(x+2f(y))=f(x)+f(y)+y$
I need a nice solution!!

[quote="Hong Quy"][b]lemma[/b]: if $f: R\to R$ satisfies $2f(f(y))=y+f(y) \forall y\in R$ then $f$ must be continous at $x_0$ (for some $x_0$).

 The lemma is very easy, you should prove it by yourself. [/quote]
Really Hong Quy, I'm verry interested to see this very easy proof !   :D  Actually that's impossible !
Have a look ..
Let $B$ be a base of  the  $R$ (real number) vectorial space on the $Q$(rational number) field.
On this vectorial space I define the linear functions $f$ such that $\forall b\in B\ f(b)=\lambda_b*b$, where $\lambda_b\in \{1,-\frac 12\}$
There are $2 ^{card(R)}$ of such functions and they satisfy the relation ship  but only two are continous !!

 :cool:

[quote]Really Hong Quy, I'm verry interested to see this very easy proof !   :D  Actually that's impossible ![/quote]

 You are right. :blush:

I've edit my solution without lemma.In fact,I hadn't seen the condition "$f$ is continuos" before I solved the problem.

Yes,Hong Quy.I got a similar one,after posting.May this problem has an unique solution,as above!!


The product of a set of positive integers is 144. What is the least possible sum of this set of positive integers?

The prime factorization of 144 is 2^4 * 3^2. SInce these are its smallest factors, the answer is (2*4) + (3*2) = 8 + 6 = [u]14[/u].
QED :)

can someone tell me what makes this true?

and will this be true for everything

The closer the 2 factors are, the smaller the sum, so a perfect square would satisfy the need.

Wait, the problem says " a [b]SET[/b] of positive integers", and a set shouldn't have repeated elements. Shouldn't the answer be $2+3+4+6=15$?


[b]Hello[/b]

I would like some help understanding the Chi-Square Test...

This illustration is about rocks - yeah interesting!

I can use the formula to get the results but I would like to know how to interpret

X2 (4) = 9.4 < .05

(I think I have  written this correctly.)
The table of critical values in my book for chi square shows that with four degrees of freedom the critical value of chi square is 9.49 at the 0.05 level of significance.
The example says to reject the null hypothesis and conclude that "on the evidence we have collected the reliance on cereal production as a main source of income is affected by rock type. "

I'm not sure what is significant about the result of this test
 :?:

p.s.
sorry i put the tables up in the wrong order...

Since the calculated p value is greater than the critical value, you reject the null hypothesis, which originally states that there is no statistically significant difference between the results you observed and those expected.  So basically it's saying that your data does not really match your original hypothesis.  Judging from glancing briefly at your tables, your original hypothesis must have been something like:

The rock type does not affect the number of cereal farms.

And since the total was 26 cereal farms, and there were 5 rock types, the results that would have supported that hypothesis would be if each rock type grew approximately 5.2 farms.  However, according to the chi-square test, your results deviate too much to fit that model and therefore rock type does affect the number of cereal farms.

I hope that answers your questions.   :)


I have written the definitions in my works as
\begin{defn}
\label{finiteset}
A set $ A$ is (dollar)\operatorname{finite}(dollar) if there exists $ n\in \mathbb{Z}$ s.t. $ |A| \equal{} n$.
\end{defn}

Well, that was only an example of the form I have written my definitions, of course I won't define such a trivial things. Now, the problem is that I have to write one paper in Finnish and I have to define a term containing the letter ä. How to do that because

\begin{defn}
\label{aarellinen joukko}
Joukko $ A$ on (dollar)\operatorname{äärellinen}(dollar), jos on olemassa $ n\in \mathbb{Z}$ jolle $ |A| \equal{} n$
\end{defn}

outputs only
Joukko $ A$ on rellinen, jos on olemassa $ n\in \mathbb{Z}$ jolle $ |A| \equal{} n$?

Neither
\begin{defn}
\label{aarellinen joukko}
Joukko $ A$ on (dollar)\operatorname{\"a\"arellinen}(dollar), jos on olemassa $ n\in \mathbb{Z}$ jolle $ |A| \equal{} n$
\end{defn}
won't work.

Here is the beginning of the tex-file:

\documentclass[a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage[finnish]{babel}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{stmaryrd}
\usepackage{euscript}
\begin{document}
\newtheorem{lau}{Lause}[section]
\newtheorem{defn}[lau]{M\"a\"aritelm\"a}
\newtheorem{merk}[lau]{Merkint\"a}
\newtheorem{lem}{Lemma}[section]
\newtheorem{cor}[lau]{Korollaari}
\newtheorem{prop}{Propositio}[section]
\newtheorem{teor}{Teoreema}[section]
\begin{titlepage}

The problem is because you want the accent in math-mode. So instead of \"{a} you use \ddot{a} as in
$ \operatorname{\ddot{a}\ddot{a}rellinen}$.
You'll find a list of math-mode symbols on page 56 of [url=http://tug.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url]


Given $ a, b, c > 0$ and $ abc\equal{}1$. Prove that
\[ \frac{1}{(1\plus{}a)^2(b\plus{}c)}\plus{}\frac{1}{(1\plus{}b)^2(c\plus{}a)}\plus{} \frac{1}{(1\plus{}c)^2(a\plus{}b)} \le \frac{3}{8}\]

It exist a nice proof by Am-Gm  :)

[quote="Ji Chen"]$ \frac {3}{8} \minus{} \sum{\frac {1}{(a \plus{} 1)^2(b \plus{} c)}} \equal{} \frac {3}{8} \minus{} \sum{\frac {bc\sqrt [3]{a}}{(\sqrt [3]{a^2} \plus{} \sqrt [3]{bc})^2(b \plus{} c)}}$

$ \equiv\frac {3}{8} \minus{} \sum{\frac {xy^3z^3}{(x^2 \plus{} yz)^2(y^3 \plus{} z^3)}}$

$ \equal{} \sum\frac {x(y \minus{} z)^2}{16(x^2 \plus{} yz)(y^2 \plus{} zx)^2(z^2 \plus{} xy)^2}\bigg[x^3\left(3y^4 \plus{} 2y^3z \plus{} 8y^2z^2 \plus{} 2yz^3 \plus{} 3z^4\right)$

$ \plus{} \frac {2x^2yz}{y^3 \plus{} z^3}\left(3y^6 \plus{} 7y^5z \plus{} 9y^4z^2 \plus{} 12y^3z^3 \plus{} 9y^2z^4 \plus{} 7yz^5 \plus{} 3z^6\right) \plus{} 3xyz\left(y^4 \plus{} z^4\right)$

$ \plus{} \frac {2y^3z^3}{y^3 \plus{} z^3}\left(y^2 \plus{} z^2\right)\left(3y^2 \plus{} 5yz \plus{} 3z^2\right)\bigg]\geq0.$

By the way, if $ a, b, c$ are nonnegative numbers such that $ bc \plus{} ca \plus{} ab \plus{} abc\geq 4,$ then

$ \sum{\frac {1}{(b \plus{} c)[4a^2 \plus{} (\sqrt {97} \minus{} 1)a \plus{} 4]}}\leq \frac {\sqrt {97} \minus{} 7}{32},$

with equality if $ a \equal{} 0,b \equal{} c \equal{} 2.$

See also : http://www.mathlinks.ro/viewtopic.php?t=314506[/quote]
there is no way that this was hand done am i right??
[hide="if it was due to a genius computer program otherwise omit this"]
wat is the fun in inequalities if a computer is gonna solve it instead of you??
do we do inequalities because we like it or because we [b]HAVE[/b] to get solutions with so much desperation as to using these programs???
i am extremely sorry about the above if you have not used a computer programme or any other device other than your brain for this..
[/hide]

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that
\[ \frac {1}{(1 \plus{} a)^2(b \plus{} c)} \plus{} \frac {1}{(1 \plus{} b)^2(c \plus{} a)} \plus{} \frac {1}{(1 \plus{} c)^2(a \plus{} b)} \le \frac {3}{8}\]
It exist a nice proof by Am-Gm  :)[/quote]
see any solution here http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-sa-ng-ta-o/347-la-m-ma-nh-1-ba-i-ca-arqady.html#post1511

[quote="jingjun"]If $ a,b,c > 0$ and $ bc \plus{} ca \plus{} ab \plus{} abc \equal{} 4,$ then

$ \frac {1}{(1 \plus{} a)^2(b \plus{} c)} \plus{} \frac {1}{(1 \plus{} b)^2(c \plus{} a)} \plus{} \frac {1}{(1 \plus{} c)^2(a \plus{} b)}\leq \frac {3}{8}.$

[b]Proof[/b] Substitute $ a \equal{} \frac {2u}{v \plus{} w},b \equal{} \frac {2v}{w \plus{} u},c \equal{} \frac {2w}{u \plus{} v},$ where $ u,v,w > 0.$ Then[/quote]$ (2u \plus{} w \plus{} v)^2(2v \plus{} u \plus{} w)^2(2w \plus{} v \plus{} u)^2\left[3\minus{} \sum{\frac {8}{(1 \plus{} a)^2(b \plus{} c)}}\right]$

$ \equal{} \sum\frac { (v \plus{} w)(v \minus{} w)^2}{u(v \plus{} w) \plus{} v^2 \plus{} w^2}\Big[2u^5 \plus{} 3u^4(v \plus{} w) \plus{} 2u^3(v^2 \plus{} 15vw \plus{} w^2)$

$ \plus{} u^2(v \plus{} w)(9v^2 \plus{} 89vw \plus{} 9w^2) \plus{} u(14v^4 \plus{} 117v^3w \plus{} 198v^2w^2 \plus{} 117vw^3 \plus{} 14w^4)$

$ \plus{} 2(v \plus{} w)(v^2 \plus{} 5vw \plus{} w^2)(3v^2 \plus{} 5vw \plus{} 3w^2)\Big]\geq0.$

[b]Remark[/b] This cool substitution see also : http://www.mathlinks.ro/viewtopic.php?t=169195

By the way, let $ a,b,c$ are nonnegative numbers such that  $ bc \plus{} ca \plus{} ab \plus{} kabc \geq 3 \plus{} k,$ then

$ \frac {1}{(1 \plus{} a)^2(b \plus{} c)} \plus{} \frac {1}{(1 \plus{} b)^2(c \plus{} a)} \plus{} \frac {1}{(1 \plus{} c)^2(a \plus{} b)}\leq \frac {3}{8}$

holds if and only if $ k\geq \frac {\sqrt [3]{332119 \plus{} 5400\sqrt {3945}}}{27} \minus{} \frac {1679}{27\sqrt [3]{332119 \plus{} 5400\sqrt {3945}}} \minus{} \frac {47}{27} \equal{} 0.79199\cdots.$

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that
\[ \frac {1}{(1 \plus{} a)^2(b \plus{} c)} \plus{} \frac {1}{(1 \plus{} b)^2(c \plus{} a)} \plus{} \frac {1}{(1 \plus{} c)^2(a \plus{} b)} \le \frac {3}{8}\]
It exist a nice proof by Am-Gm  :)[/quote]
See here: http://canhang2007.wordpress.com/2009/12/03/inequality-86-t-q-anh/
[quote="jingjun"]$ \sqrt {97}$ is  ugly.But the following inequality is nice:

If $ a,b,c > 0$ and $ bc \plus{} ca \plus{} ab \plus{} abc \equal{} 4,$ then

$ \frac {1}{(b \plus{} c)(1 \plus{} a)^{2}} \plus{} \frac {1}{(a \plus{} c)(1 \plus{} b)^{2}} \plus{} \frac {1}{(a \plus{} b)(1 \plus{} c)^{2}}\leq \frac {3}{8}.$
[/quote]
See here: http://canhang2007.wordpress.com/2009/12/03/inequality-87-unknown-author/


Hi,

I realized that the second Russian team participates in Bulgarian NMO, same for Romania (but 12 grades are excluded). China participates in the Russian NMO. And the fourth German IMO team participates in the China TSTs.  :D  Just kidding. But what else do you know about such traditions to take part in other national competitions ?  :)

Moldavian team participates in Romanian NMO. This year second Bulgarian team participated there too.

This is the first year in a long time when Canadians were allowed to officially write the USAMO.

Yes, the Moldovian team always participates in the Romanian NMO since 1994 (or even earlier). We (the Romanians) used to send a team to Moldovian NMO (that was untill 2000): the first students in Bucharest, that didn't qualify for the National Olympiad would go to the Moldovian NMO.

I belive that in the next years this kind of "exchange" programs between NMOs will become something normal, between most of the leading IMO countries (and not only).

It's indeed a tradition for China and Russia participating at each other's MO.
In CMO2004 the Russian team got 1 gold and 5 silver.

How many golds are usually awarded at CMO2004?
And what was the gold cut-off this year?


I looked @ collegeboard's practice sat 2 math and it seemed pretty easy, as long as you covered the material.

For those people who took the SAT 2s math, on scale of 1-10 and the sat regular math is a 3, how difficult is the sat 2 math?

Are there any other places i can get practice problems?  I searched google for a while and no luck

hmm... i would say a 5 (using your relative scale and guesstimating for you.  for me it was like a 1)  then again, don't trust me since i walked in blind and walked out with an 800.

but seriously don't base everything on my opinion as mine may not correspond to yours.

this may be in a math forum, but you're in the wrong place for opinions on this.

try Yahoo! Answers or something.

If you remember concepts well, then math SAT Subject Test should be a breeze

The trickiest stuff on there are like... sums of geometric series.
To the average math class student, that type of formula is forgotten quickly, but if you remember things like that, then you'll do fine.

just make sure you watch the clock

If you look at the average range of SAT scores the average (well reputed) college accepts, the SAT II math score average is generally significantly higher than the regular SAT math score average.  I know many people who took the SAT II and got an 800, but got less than that on the regular SAT math.  With this in mind, I would say that the SAT II is much easier.  Note that the problems might not be that much easier, but scoring is more lenient.


vector
linear algebra
matrix
algebra
system of equations
linear algebra unsolved

1. Can any of you tell me what an [i]affine space[/i] is? A formal definition is nowhere to be found on the Internet...

2. Is the solution set of an $ m \times n$ non-homogeneous system of equations an examples of an affine space? What would its dimension be?

I look forward to hearing from you soon.

Best Regards.

See the [url=http://en.wikipedia.org/wiki/Affine_space]Wikipedia article[/url].

Thanks for taking the time to reply, comrades.

From [b]LkNsngth[/b]'s post I deduce that the first question in 2 has a negative answer. The [b]wikipedia[/b] article makes me think that the answer must be afirmative, though... What's going on here?

Also, I'm still doubtful about the dimension-[i]thingy[/i]. How could I go about calculating it?

Salutations.

i accidentally posted my answer to this in a new topic, here is the link to it http://www.mathlinks.ro/viewtopic.php?t=233136 (sorry).

The solution to a set of linear equations is a translate of a vector space (the nullspace of the corresponding matrix) and so can be given the structure of an affine space.  Its dimension is the dimension of the corresponding nullspace, or the nullity of the corresponding matrix.

[quote="t0rajir0u"] Its dimension is the dimension of the corresponding nullspace, or the nullity of the corresponding matrix.[/quote]

Even if the system is non-homogeneous?

Thanks again for taking the time to reply.

Yes.  If the system $ \mathbf{Ax} \equal{} \mathbf{b}$ has a solution $ \mathbf{x} \equal{} \mathbf{x}_0$, then the set of all solutions takes the form $ \mathbf{x} \equal{} \mathbf{x}_0 \plus{} \mathbf{v}$ where $ \mathbf{v}$ is in the nullspace of $ \mathbf{A}$ - like I said, the solution set is a translate of the nullspace.  There are two senses in which a translate of a vector space can be given a meaningful dimension:

- as an affine space (it has the same dimension as the vector space, which I believe is by definition), or

- as an [url=http://en.wikipedia.org/wiki/Algebraic_variety]algebraic variety[/url].  The notion of dimension here is much more subtle, but the two concepts agree in this case.

Things are much clearer now, esteemed t0rajir0u.

Thanks for caring, man! :lol:


logarithms
real analysis
real analysis solved

No, $n x_n$ can diverge.  For example, define $x_n = \frac{\ln n}{n}$.

[quote="harazi"]Let $x_n$ be a sequence that converges to 0 and such that $| n(n+1)x_{n+1}-n^2 x_n|$ is bounded. Does it follow that $nx_n$ converges?[/quote]

It says $|u_{n+1} - u_n| = O(1/n)$ where $u_n = nx_n$, so that $u_n = O( \log n )$ and any asymptotic behavior of $u_n$ consistent with this estimate can be realized.


[quote="Stun"]Given 2 real numbers $x\ge 0$ and $\alpha\ge 1$
Define the sequence  $\{a_{n}\}$ as follow : $a_{1}=x \ ; \ a_{n+1}=1+\frac{a_{n}^{2}}{n^{\alpha}}\ \forall n\ge 1$
1/ Prove that $\exists \delta=\delta(\alpha)$ such that
    if $x<\delta$ then $\lim_{n\rightarrow+\infty}a_{n}=1$
    if $x\ge \delta$ then $\lim_{n\rightarrow+\infty}a_{n}=+\infty$
2/ Prove that $\delta(\alpha)$ is increasing[/quote]

What does the backwards E notation mean?

Look at the LaTEx code, it meas exists. :wink:


trigonometry geometry AMC AIME USA(J)MO USAMO MATHCOUNTS

Is trig helpful for AMC [b]10[/b]?

thank you

AMC 10 typically does not cover trigonometry, but AMC 12 does.

I asked this a bit ago.  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=234122

A Question:

Is trig needed for AMC 12?

If so then what specific areas should you be familiar with?

Yes, it's asked in the AMC 12.
I don't quite get what you mean by which areas. Just be familiar with them all I guess, although I personally don't know too much the part that deals with graphs.

Perhaps you should look at the amc faq?

Trig could be helpful for some of the harder geo problems in AMC 10, I believe.
(Such as the area of a triangle formula that uses sin)

So would a thorough read of the trig in Volume 2 be sufficient?

Probably...but I think the basic trig in Intro to Geo and AoPS Volume 1 would help just enough, but if you want more clarification on trig, or better methods, AoPS Volume 2 surely could help.

i would suggest getting Plane Trigonometry by S.L. Loney. Its a classic and explains concepts very well.

Trigonometry in AMC-12 is such a broad topic. It can mean simple computation of sine, cosine, and tangent of triangles to complex numbers and use of binomial theorem with DeMoivre's. So technically, anything that is beyond algebra II on AoPS volume II and is not part of NT can be considered as "area" of trigonometry.

For AMc-10 and trigonometry, I think it helps depending on the person. When I was taking AMCs, I hated geometry and solved almost all geometry with trigonometry (with calculator too but that's not for you folks anymore!  :) ). General idea can even help to give you what kind of construction is needed so definitely, keep in mind but you don't have to use it.

When my academics settle down bit, I need to combine the advices and tips for studying AMCs because this type of question arises every year about the same time.

[quote="cognos599"]no problem in AMC 10 will REQUIRE trig. but trig. can be helpful on some problems.(2008 AMC 10 #25, which is also 2008 AMC 12 #22).[/quote]

Yeah, the trig solution was easier, but the other didn't require it.

[quote="diophantient"]Knowing anything is helpful in doing anything.[/quote]
This is not true.  I suggest to you, diophantient, not to make such claims before carefully evaluating them.
Using the word "anything" twice, namely, is the mistake in the statement.
Counterexample: If you are asked to evaluate $ 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 10$, then will knowing the fact that $ \frac {a \plus{} b}{2}\ge \sqrt {ab}$ for positive integers $ a,b$ be helpful?  I'm sure that some crazy solution can be found using that fact, but it is certainly not "helpful" to use it.

I think what you are trying to say is that one should be familiar with any topic in mathematics in order to get an extra edge on contests such as the AMCs.  That is wrong, too.  I'm sure that knowing something like Fermat's Last Theorem will not be helpful to you on the AMCs (I'm not talking about AIME, USAMO, and beyond, just AMC).

What is correct is to say that it can be helpful to know certain topics in order to solve contest problems such as those on the AMCs, even if the topics aren't necessary to know in order to solve the problems.  This is certainly true for trigonometry: trigonometry is extremely usable on AMC and even MATHCOUNTS problems (and helpful, in that it provides a faster solution than non-trigonometry methods).
Example: Find the area of a triangle with two side lengths of $ 12$ and $ 15$ and the included angle of those two sides having an angle measure of $ 120^\circ$.

This type of problem actually does show up on MATHCOUNTS (and maybe AMC, although I'm not quite sure) problems, and it is extremely easy to solve with the formula $ A \equal{} \frac {1}{2}ab\sin C$, where $ A$ is the area of a triangle with two side lengths of $ a$ and $ b$ and the included angle of those two sides having an angle measure of $ C$ radians (or $ C\cdot\frac {180}{\pi}$ degrees).
Other problems (for example, the one mentioned by cognos599) on the AMC 10 can be solved quicker with trigonometry.

So, to make a long answer short, yes, trigonometry is helpful for the AMC 10.


complex analysis
function
limit
complex numbers
topology
real analysis
algebra

What are the reasons why normal proving techniques have failed when it comes to the riemann hypothesis? Please post specific attempts/failed techniques if you can.

[quote="ZetaX"]I have no clue what a "normal" proving technique is.
The problem is difficult, that's all.[/quote]

I would consider any technique that has been used in the past to prove something, normal.

The most obvious approach is to prove that the reciprocal $ 1/\zeta(s)$ is holomorphic in the half-plane $ \re s > \frac12$. There is a [i]formal [/i]Dirichlet series for it; all that's left to prove is that the series converges.  :P

The book "Scattering theory for automorphic functions" by Lax and Phillips has a section "How not to prove the Riemann Hypothesis".

This will be a silly question. However, could anybody tell me what exactly is Riemann's Hypothesis asking? I checked Wiki and only find unexplained nouns as zeta-function.

For some $ s$ (namely those complex $ s$ with real part $ >1$) the sum $ \zeta(s) : \equal{} \sum_{n\equal{}1}^{\infty} \frac{1}{n^s}$ converges. A bit more checking gives, that for those $ s$ it is complex differentiable (so $ \lim_{h \to 0} \frac{\zeta(s\plus{}h)\minus{}\zeta(s)}{h}$ exists, where $ h$ can be complex).
Now there is a fundamental theorem from complex analysis: Let $ U \subset \mathbb C$ have an accumalation point (a point where infinitely many elements of $ U$ are; e.g. if $ U$ is a nonempty open set); If two complex differentiable functions $ f,g$ are defined on connected sets $ X,Y \subset \mathbb C$ and agree on $ U \subset X,Y$, then $ f$ and $ g$ are the same on all of $ X\cap Y$. As a result, there is at most one complex differentiable function that continues a given such function to a connected set $ X$ (note that this is essentially different from real analysis where e.g. a function on an interval can be extended to the reals in many ways).
Now there exists a complex differentiable function $ \zeta$ that is defined on $ \mathbb C \backslash \{1\}$ and agrees with that function mentioned above. And by that theorem, this function is the only such. We call it the Zeta-function.
It has some properties (e.g. a functional equation relating $ \zeta(s)$ to $ \zeta(1\minus{}s)$) and has some "trivial" zeros at negative even integers (and no other real zeros). The Riemann hypothesis says: all other zeros have real part $ \frac 12$.

One can avoid the discussion of analytic continuation by considering $ (1\minus{}2^{1\minus{}s})\zeta(s)\equal{}\sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\minus{}1}}{n^s}$. This series converges in the half-plane $ \mathrm{Re}\,s>0$, and the hypothesis is that its sum is not equal to zero unless $ \mathrm{Re}\,s\equal{}\frac12$.


(1) If $ \frac {a!}{b!}$ is a multiple of $ 4$ but not a multiple of $ 8$, then what is the maximum value of $ a \minus{} b$?

(2) Less than $ 50$ people are at a party.  Each person shakes everyone else's hand.  If there is an odd number of total handshakes at the party, what is the largest number of people that could be at the party?

Thanks a lot!

[hide="1"] We assume $ a\ge b$ (or else $ \frac{a!}{b!}\not\in\mathbb{Z}$).  So, $ \frac{a!}{b!}\equal{}a(a\minus{}1)(a\minus{}2)\cdots(b\plus{}1)$.  Notice that there are $ a\minus{}b$ terms in this product, so what we are maximizing in the number of terms.  We claim that the maximum number of terms is $ 3$, which is achieved when $ a\equal{}4t\plus{}1,b\equal{}4t\minus{}2$.  Suppose there was a greater number of terms that satisfies the conditions.  Notice that $ b\plus{}1,b\plus{}2,...,a\minus{}1,a$ are a sequence of $ a\minus{}b$ terms where $ a\minus{}b\ge4$.  Thus, there is a multiple of $ 4$ among them.  This multiple of $ 4$ lies in one of the halves of the sequence, each of which has at least $ \lfloor\frac{a\minus{}b}{2}\rfloor\ge2$.  Thus, in the other half there are still left at least 2 consecutive integers.  Thus one of them must be a multiple of $ 2$ and so the total product is divisible by $ 8$.
[/hide]

[hide="2"] The total number of handshakes is $ \binom{n}{2}$ where $ n$ is the number of people.  This is equal to $ \frac{n(n\minus{}1)}{2}$.  For this to be odd, we must have $ n$ or $ n\minus{}1$ leave a remainder of $ 2$ when divided by $ 4$.  Thus, the greatest amount of people is $ 47$.[/hide]

I don't get how this first statement is made:

[quote]We assume $ a\ge b$ (or else $ \frac {a!}{b!}\not\in\mathbb{Z}$).  So, $ \frac {a!}{b!} \equal{} a(a \minus{} 1)(a \minus{} 2)\cdots(b \plus{} 1)$.[/quote]


What is (are) they? They look nice. I want to use them  :) .

I know this is an old thread, but I'd like to answer my own question in case anyone else wants to know. The AoPS books use Palatino (at least Volume 1 & 2 do, I don't have the other books), as well as the matching math font.

[code]\usepackage{mathpazo}[/code]

In my opinion, Palatino looks better than the default Computer Modern font. (The latter places heavy emphasis on the vertical strokes (imitates a horizontal calligraphic pen); the former has the thickest lines on a slightly offset angle (imitates a tilted calligraphic pen).)


$\alpha=\frac{\pi}{7}$とする。このとき、次の等式を証明せよ。
\[ \frac{1}{\sin{\alpha}}= \frac{1}{\sin{2\alpha}} +\frac{1}{\sin{3\alpha}} \]


Let $ABC$ be an isosceles triangle at angle $A$ and $\hat{A}= 36^{o}$.Prove that the ratio $\frac{BA}{BC}$ is irrational.

The triangle can be seen in a regular pentagon.

Remark the golden ratio, we have ${\frac{BA}{BC}=\frac{\sqrt{5}+1}{2}}.$


In a party of $n$ persons, every set of $k$ persons have exaclty one common friend. What is the number of friends of the person with most friends.

PLEASE A MODERATOR DELETE THIS POST. THIS PROBLEM IS ALREADY POSTED IN THE COMBINATORICS SECTION.. THANKS


linear algebra
matrix
vector
linear algebra unsolved

Diagonalize the matrix B =

$ \left(\begin{array}{cc}{acc}
1 & 2 \
3 & 4 \
\end{array}\right)$

by a change of basis.
~~~

I don't know how to diagonalize "by a change of basis." How do you do that?

I tried to find the determinant of $ \lambda I - B$. I got real, but ugly, eigenvalues of $ \frac{5 \pm \sqrt{33}}{2}$. I tried to find the eigenvector for $ \frac{5 + \sqrt{33}}{2}$, but I wasn't successful. (I got the zero vector.)

Let's try the eigenvalue $ \lambda=\frac{5+\sqrt{33}}2.$ We want to find the null space of

$ B-\lambda I=\begin{pmatrix}\frac{-3-\sqrt{33}}2&2\\
3&\frac{3-\sqrt{33}}2\end{pmatrix}.$

Now, I'm not going to spend any time row reducing this. If we've done things right, it is singular, and after subtracting the appropriate multiple of the the top row, the second row should be zero. Now find some vector such that the top row times it is zero. We can just rearrange what we see to get

$ x=\begin{pmatrix}2\\ \frac{3+\sqrt{33}}2\end{pmatrix}.$

At this point, I will do one check:

$ \begin{pmatrix}3&\frac{3-\sqrt{33}}2\end{pmatrix}
\begin{pmatrix}2\\ \frac{3+\sqrt{33}}2\end{pmatrix}=
6+\frac{9-33}4=0.$

It checks. This is our eigenvector.

For the other eigenvector, we should just be able to just reverse the sign of the square root.

Now, let $ P=\begin{pmatrix}2&2\\ \frac{3+\sqrt{33}}2&\frac{3-\sqrt{33}}2\end{pmatrix}.$

We explicitly compute that $ P^{-1}=-\frac1{2\sqrt{33}}{\begin{pmatrix}\frac{3-\sqrt{33}}2&-2\\ -\frac{3+\sqrt{33}}2&2\end{pmatrix}}$

Then $ P^{-1}BP=\begin{pmatrix}\frac{5+\sqrt{33}}2&0\\ 0&\frac{5-\sqrt{33}}2\end{pmatrix}.$

There were a couple of arbitrary choices in that process; it would be possible for a correct answer to look a little different.


function
real analysis
real analysis unsolved

Let $ f: [a,b]\minus{}> R$  a function continuos on $ [a,b]$ and derivable on $ (a,b)$  with $ f(a)\equal{}f(b)$ .
Prove that  exist $ c_{1} , c_{2} \in (a,b)$ s . t .
$ af^{\prime} (c_{1}) \plus{}bf^{\prime} (c_{2}) \equal{}0$

What's wrong with taking $ c_1\equal{}c_2\equal{}c$ where $ f'(c)\equal{}0$ (such $ c$ exists according to the standard Rolle's theorem). I'm pretty sure you meant something else, but what? :?


Gabriel A. Moens, from Brigham Young University Law Review once stated “the idea that church and state have different roles in society and that respectful separation between them should be maintained is essentially a good idea.? Because I agree with those words, I must negate the resolution: Democracy is best served by a strict separation of church and state.

Point of Clarification: This resolution must first be established in the United States before we go beyond its borders. If we try to worry about every democracy, we interfere with laws set up by separate countries.

V: Prosperity, in that people are more prosperous when religion is able to be valued by a governments citizens.
VC: Democracy, because when we value support religious beliefs, we help the success of our democratic governments. This makes them more prosperous.

If a man can survive and prosper without something, then that object isn’t that important. This leads me to my first contention.

Contention I: Democracies can function and flourish without a strict separation of church and state. Great Britain, a democratic country has civil rights, an effective economic system and working parliament. But it also has a state church, which is the Anglican Church. Yet England is a great example of a supreme democratic nation. France, Spain, and England were all religious nations when they abolished slavery, which was before the United States did. India, another example, is a well functioning democracy. It has a majority of Hindus. India’s state-church interaction benefits the lower caste population that’s being discriminated by the Hindu religion by granting them money. Indonesia, another successful democratic country with a predominant religion of Muslim had a female prime minister. Even though under Islamic Laws, women are placed inferior to men, a woman led an Islamic nation before. Social Welfare, greatly valued in democracies can be hindered by a strict separation of church and state. Religious organizations best supports social welfare, especially in the area o f addiction treatment. Faiths-based organizations often have a higher success rate then secular treatments. If a strict separation were in place, it would have denied public access to such organizations.

Contention II: A strict separation of church and state doesn’t guarantee and can restrict freedoms. In some cases, Native Americans were allowed to continue to use a hallucinogenic drug called peyote. Had a strict separation been in place, the Native Americans would not have been granted this right. The same pertains to the nuns working for free in hospitals instead of being paid minimal wage. Exceptions are often made for religious reasons. A strict separation also hinders an individual’s rights to expression. In France, public schools banned the wearing of any religious symbols. Headscarves are a predominant symbol in Muslim faith, but the schools thought by banning them; they were giving the Muslim girls more freedom from their religion. However, the Muslim girls protested, declaring that by banning a harmless religious symbol, they were being discriminated. Once again, we must realize that by a strict separation of church and state, anything somewhat religious in any public school and government official building would be banned. This brings us to my third contention.

Contention III: This resolution is severely detrimental to the education process.
As stated in the definition provided, strict separation is having no correlation. Schools are one form of a governmental institution, hence the federal funding, state funding, et cetera. By such a regulation as the resolution proposes, classes in public schools would be harmed.
Sub point A: Classes such as World History would be harmed, because much of what is taught comes from religious based events, such as Egyptian study, Roman Empire, the Greek civilization, and England's empire.
Sub point B: Many other classes would be harmed as well, including Speech and Debate. Under this resolution, we could not even discuss this type of topic, because it brings religious views into a governmental institution, and would be violating the resolution by affirming it.

A strict separation of church and state not only hinders individual rights, it is also not necessary to best serve democracy. When a strict separation is in place, the government is unable to support any religious groups, thus denying charities from helping the less fortunate. For these reasons and more, I urge you to vote negative.

Good Case. Comments:

I don't agree with the observation. The resolution is not limited to the United States, and without proving why it's important to establish it here, it's not an observation but rather a contention - not to mention a contention which shouldn't belong.

I don't think you should run democracy as a criterion to judge prosperity as that's contendable. Democracy has not always been the most prosperous type of government in history. Use it as a value if anything, and you must specify what aspects of democracy you are refering to.

I'd rephrase your first contention: Democracies flourish under non-strict separation of church and state.

You must be aggressive and use attacking arguments. Don't go with the neutral statement.

 ;)  Where did you get your second contention from? Honestly.

As an AFF, my definition would be a separation of church and state as a separation of religious and state institutions, as defined by Wikepedia and many other dictionaries. The resolution does not specifically refer to religion itself, only that religion and state and made separate entities which do not directly impact each other.

You prove that strict separation is bad, but what if the AFF goes on the offensive? You haven't clearly stated why non-strict is necessarily better.

Otherwise, very good case.

I believe it's a fairly good case.  :P

But even so, there's a few problems I see. The greatest is probably the fact that most of what you're doing is presenting examples instead of offering arguments. Debate is about your ability to argue, not find facts. I must say though, it is a very interesting approach. I generally see people with not enough "tangible proof" in LD, instead of too much. It's also kind of short. I'm not sure what time limits you compete under, but you should always try to hit within 10 seconds of your full time on the first speech. In my non-expert opinion, that just shows that you actually do have a lot to say and back up your case with. Having 8 minutes and making a 2 minute speech would definitely not be something I want to do too often, unless of course I'm on NEG and the AFF case that just got presented was absolutely horrendous.

Oh yeah...contention III doesn't need to be there. I see absolutely no way in which it actually benefits your case, and heck...I'd even argue it's kind of off-topic.

You could argue your contention three, but not on the grounds your presenting.

Also, it's a NEG MithsApprentice, so I think it is of decent length.

He must reserve at least 3 minutes for rebuttal.

;) You post your case up here for us to critique it so I'm going to be honest.

  With this case, you will not qualify into whatever you are trying for. Is this for states? I know it's not for nationals because I can tell you are a fairly new LD debater just from your case. I can tell you now, you have huge wholes in your case that need to be patched up.
  Here in Hawaii, we just had N.I.T on another island, Kauai, and the day I left I heard a fellow LDer going to the same debate as I was. Her case was very similar to yours and like I predicted, she didn't even make it through the 5th round. (If you haven't figured, I am a high school LD Debater from Kahuku High School in Hawaii on the island of Oahu. I am currently a senior.)
  Lets help you patch up you case. First of all, you do know that the U.S is not a Democracy...don't you? We are a Republic. Why? We have representatives. "...to the Republic for which it stands" an example for you from our own Pledge of Allegiance.
  Secondly, your case is not really leading to the point of you value and value criterion. Besides, Democracy as a VC? Democracy is a very weak bases and usually used as a value. How about the Value "Societal Welfare" and VC "Status Quo"? It is easily linked to EVERYTHING in Neg and also you can make arguments (which you really need because I see no arguments whatsoever).
  Moving on to your contentions. Awesome, Awesome Contention 3, but know EVERYTHING with each country you mention or else your opponent can severely rip your case apart. Contention 2 gives your opponent the chance to take the win. I say scratch the whole C2. No matter what religion or culture says peyote is okay, it is still  an illegal drug and NO ONE can get away with it. Native Americans should be able to smoke peyote cause its apart of  their religion but other people who smoke it for personal needs cannot? That is the biggest whole in your case. No matter what, whether there is a strict separation or not, what is illegal is illegal! No bends, no turns, no leaning for ANYONE. I must admit though, the muslim girls and headscarves, very good arguements. Maybe even faith-based programs would be good. For example, Alcoholics Anonymous or Noacrotics Anonymous. Maybe for the criterion I gave you, a good contention would something like "Religious organizations can assist the government's achievement of societal welfare". Just a suggestion. Okay now your Contention 3. WoW! You either need to drop it or you need to work on it and go deeper into it. Tell WHY it is important and HOW it would be dtrimental in the EXACT way.Education process, not a very good agrument topic on you part. Very good for your opponent to poke holes in. How about something like "The resolution is severely detrimental to society". You can say that 'by strictly separating Church and state,your opponent is advocating no intervention by government. By allowing acts such as; Allowing people to torturine themselves or others at their consent as a religious rite to show dedication to their God, every criminal will be commiting crimes in the name of their God. The federal government will not have the power to interfere. Thus putting society at risk'. Just another suggestion.
  You do have a really awesome case. It just that you have many big topics that have a strong Affirmative side to it that could kill your case IF your opponent was smart and knew what they were doing.
  You have an awesome quote. I'm Mormon also, in fact I am on a B.Y.U-Hawaii computer right now. (My Grandfather is a professor here). Keep up the good work and stick with LD. Just be confident and stay cool, calm, and collected. Good luck! If you need any help or there is anything I can do for you, just e-mail me. Waugh@hawaii.edu

If you want to take a look at my case, you can e-mail me and i'll e-mail it back to you. You can take ideas from it but I would advise you not to copy it straight because their might be things that you dont really know about and will cause you to lose. But you can always use it as just an outline.  :lol:  Okay. Aloha.[/quote]


Firstly if $f(x)=f(y)$ then $f(a^2+f(x))=f(a^2+f(y))$
From here $x+f(a)^2=y+f(a)^2 \Longrightarrow x=y$
We get that $f(x)$ is injective.
Taking $x=0$ we get $f(f(y))=y+f(0)^2$ $(1)$
So there exists $t\in R$ such that $f(t)=0$
Taking $x=y=t$ we get $f(t^2)=t \Longrightarrow f(f(t^2))=f(t) \Longrightarrow t^2+f(0)^2=0$
So $t=f(0)=0$
Taking $y=0$ we get $f(x^2)=f(x)^2=f(-x)^2$
From injectivity of $f$ we get $f(-x)=-f(x)$
Taking $y=0$ we get $f(x^2)=f(x)^2$.So $f(x)\ge 0$ if and only if $x\ge 0$  $(2)$
Taking $y=-x^2$ we have $f(x^2+f(-x^2))=-x^2+f(x)^2$
$f(x^2-f(x^2))=f(x)^2-x^2$
From $(2)$ we get that $f(x)=x$
That's all ;)


If [i]a [/i]and [i]b [/i]are distince, odd primes, then how many distinct positive factors does [i]4a(squared)b(cubed)[/i] have.

Thanks

[hide]Ok since 4=2^2. Then you have 2^2*a^2*b^2. Using this you add one to each exponent  and you can find 3,3,3. You multiply the exponents so it would be 3*3*3. The answer is 27.[/hide]

[hide]The number is $2^2a^2b^3$.   Each factor can be divided by $2^0,\ 2^1,\ 2^2$, it can be divided by $a^0,\ a^1,\ a^2$ and so on.  Thus there are $3\times 3\times 4=36$ factors because every factor is equal to $2^xa^yb^z$ where $x$ can be 0, 1, or 2, and so on.[/hide]

I would recommend learning and using LaTex (like above).


Determine if $ x^2\plus{}xy\plus{}y^2\equal{}2$ has a solution in $ \mathbb{Q}$.

Solution.
[hide]
We want to find solutions of $ m^2 \plus{} mn \plus{} n^2 \equal{} 2k^2$ in integers (and $ k\neq 0$).

Let $ (m,n,k)$ - solution with minimum value of $ |k|$ ($ k\neq 0$).
If $ m,n$ both odd, then left side is odd.

If one of $ m,n$ is even (for example, $ m$), then $ m^2 \plus{} mn$ is divisible by 2, so $ n^2$ is divisible by 2. So, we have $ m,n$ - even. Thus, $ m^2 \plus{} mn \plus{} n^2$ is divisible by 4. It means that $ 2k^2$ is divisible by 4. So, $ k$ is even.

$ m,n,k$ - even. So, $ (\frac {m}{2},\frac {n}{2},\frac {k}{2})$ - solution too. But we chose solution $ (m,n,k)$ with a minimum value $ |k|$. Contradiction.
[/hide]


[quote="Limp Bizkit"]Let $S_{XYZ}(l)$ be the set of all $T$-s that $TX^2+TY^2+TZ^2=l$, where $l$ is a constant (of course it is big enough).
Now imagine a triangle $ABC$ and points $A'$ and $C'$ inside the $\angle ABC$, which satisfy to following condidtions: $AA'=CC'$ and $AA'||CC'$. Prove that for every (big enough) $m$, we can find out a $n$(this is big enough too), which satisfies this: $S_{ABC}(m)=S_{A'BC'}(n)$[/quote]
If $M$ is the medians intersection point of triangle $XYZ$
We know the following formula
$PM^2=\frac{1}{9}(3(PA^2+PB^2+PC^2)-(AB^2+AC^2+BC^2))$
So the set $S_{XYZ}(l)$ is a circle with centre in $M$.
If $S_{ABC}(m)=S_{A'BC'}(n)$=>the medians intersections points of triangles $ABC$ and $A'BC'$ are same, BUT this condition is wrong, therefore  the problem is wrong too :rotfl: .
I think in the correct version of this problem asking to prove that for every (big enough) $m$, we can find out a $n$( big enough too), which satisfies this: $S_{ABC'}(m)=S_{A'BC}(n)$,now anything is true,becaus how we can see the triangles $ABC'$ and $A'BC$ have same medians intersection point ;),we can choose $n$ such that the radius of circles are equal,and the problem is done.
 [hide="Only for Limp Bizkit"]Ara Neso xntirt sxalaaaaaaaaaaaaa :rotfl:  :rotfl: [/hide]


Princeton
college
MIT
inequalities
Putnam
real analysis
Gauss

[quote="new"]:?

The youngest perfect scorer was the former GDR participant Pavel Kroeger.
When he attended 7th grade, he managed this performance.
A student from Australia was the youngest participant.
If I am not mistaken he was 11 years during his first IMO.
In the next days I  could provide more detailed records.[/quote]

terry tao (australia) was that australian student. he got gold when he was 13, after which he went to uni i think. if he had continued, he would have easily got about 5 Gold ...

now did that tao guy had a "gold childhood" or what? :P he didn't score max. points though, so I guess he gaved up pretty quickly :)

um, tao didnt do the IMO after he was 13 for some reason  :? . (anyway, he was the one that proved we can find an arbitrarily long sequence of primes in a.p.).

Terry went off to uni afterwards, making himself ineligible.. (might have been 1 year in between or something though..)
Oh yeah. Vinoth, get back to doing problems :-P

Terry Tao is the same one involved in the recent proof that there are primes in arithmetic progressions of length k for any natural number k.

I believe he was the youngest to receive a Gold Medal at IMO. He did so in 1988 at (just barely) the age of 13 - (his 13th birthday was during the week of the competition). He also recieved a Bronze Medal in 1986 and a Silver Medal in 1987.

Right after his third time at IMO he went to Flinders University of South Australia where he got his Bachelors and Masters degrees at the age of 16. He then went to Princeton University where he got his PhD in 1996 at the age of 20. He now teaches and works at the University of California, Los Angeles.

If he had competed in IMO until he was 19 years old it's safe to say that he could have won as many as 7 Gold Medals and 9 Medals in total.

What I find funny about [url=http://www.math.ucla.edu/~tao/]Terrence Tao[/url] is that he gives the [url=http://www.math.ucla.edu/~tao/putnam/]Canadian Mathematical Olympiad 1999[/url] as regular course work.  :lol:

This course was designed as preparation for the Putnam contest.

Well,

Bombieri could prove the Prime Number Theorem at age 11.
Selberg was reading Ramanujan's collected works in his early/mid teen years.
Deligne read Bourbaki commutative algebra in high school.
Don Zagier started MIT at age 12 or so, and won the Putnam competition
during one of his years there.  He went to do his PhD with Atiyah in his
late teens.
Kazhdan started working with Gelfand around age 13.
Beilinson was learning Grothendieck-Riemann-Roch in his mid teens.
Grothendieck independently rediscovered Lebesgue measure theory
at age 15.

One of the basic nonlinear binary codes (Kerdock? Nordstrom? I forgot)
is named after a 14-year-old British student who attended a lecture about
coding theory and worked out the example.

John Conway made some discoveries in knot theory during high school.
Simon Norton got 3 gold medals at IMO and as a child-prodigy type,
may have been fairly young for the first one.

Bau Chau Ngo who works on geometric Langlands' program and recently
proved the Fundamental Lemma (25 yrs notorious open problem) with
his advisor Laumon, got a perfect score at IMO at a fairly young age,
14 or 15.

Noam Elkies was also 14-15 for his first IMO gold, a few years before
disproving Euler's conjecture on sums of fourth powers.

Terry Tao was young but not particularly spectacular at the IMO.
He became spectacular later, during and after graduate school.

[quote="fleeting_guest"]Don Zagier started MIT at age 12 or so, and won the Putnam competition
during one of his years there.  He went to do his PhD with Atiyah in his
late teens. [/quote]

Funny story about Don Zagier, actually.  He graduated high school speaking something like 7 languages when he was 12, I think.  Maybe 13.  He applied to Oxford, but they told him that they don't accept anyone under age 16 as an undergraduate.  So he went to MIT for 2 years, graduated, and at 15 was accepted as an Oxford Ph.D. student since they didn't have age restrictions on who could seek higher degrees.

[quote="orl"]What I find funny about [url=http://www.math.ucla.edu/~tao/]Terrence Tao[/url] is that he gives the [url=http://www.math.ucla.edu/~tao/putnam/]Canadian Mathematical Olympiad 1999[/url] as regular course work.  :lol:
[/quote]

Q5 is actually somewhat difficult, IMHO.  I have not yet seen a solution that streamline with typical inequality solutions

I'd forgotten the Zagier story about Oxford.  I think Gross tells it
occasionally in connection with the Gross-Zagier formula.

Also, I think Drinfeld's IMO 24k was at a young age, 14-15.

Kontsevich would have been a natural for the IMO, and posed problems in
Kvant, but as far as I know was not involved in the competition.
He started university in Moscow several years early (15 or so) and
I think also Gelfand's seminars, so he may have had better things to do
at the relevant age.

[quote="fleeting_guest"]I'd forgotten the Zagier story about Oxford.  I think Gross tells it
occasionally in connection with the Gross-Zagier formula. [/quote]

That's where I heard it :)

[quote="Tokyo"]r these guys geniuses or just their efforts?[/quote]I think they are more of a genius than a hard worker (but they are certainly both).  :roll:

And I disagree with Valentine. I think that they are actually more a result of having trained themselves to be [i]efficient[/i] thinkers rather than being "natural" geniuses or concientious hard workers.

As one of my math teachers once said -- its better to work efficiently than to work hard.

Effective thinkers are people who've somehow, either consciously or subconsciously, learned to synthesize information in an unusual way. How they learn to do this is probably a combination of random circumstances and deliberate conditioning.

Once they've begun to think efficiently, and have had positive experiences from doing so, it then becomes a positive feedback system. They can continually use that knowledge to learn and sysnthesize even more knowledge and information! So its sort of an exponential learning curve based on their random initial conditions and the initial conditions of their childhood experiences.  I think the element of random circumstances is what makes it impossible to recreate such individuals at will.

Either way, I think it's an oversimplification to attribute such achievements to either unexplainable "genius" or simply to self-willed "hard work."

There's probably nothing particularly different about them genetically than from any other normal bright person, and they probably don't work "harder" than even some students who fight to make good scores on even much simplier math exams.

By looking at each of these people individually, I bet you could find things in each of their personalities or childhoods which created an initial condition that fertilized their ability to think very efficiently.

[quote="gauss202"]And I disagree. I think that they're actually more a result of having trained themselves to be [i]efficient[/i] thinkers rather than them being natural geniuses or necessarily hard workers.
[/quote]I was not talking only about their high-school acomplishments, but rather more about their college ones ;)

[quote="fleeting_guest"]Terry Tao was young but not particularly spectacular at the IMO.
He became spectacular later, during and after graduate school.[/quote]

I wouldn't exactly say that. Scoring Bronze, Silver, and Gold at IMO at the ages of 11-13 is pretty spectacular in itself. It is really the age factor that is the most impressive thing about all of the IMO participants, because the 19 year age limit is rather arbitrary.

If you allowed people to participate until they were 25 or 30 or older, it would be a completely different competition. It is the fact that such young people can answer such difficult questions that is impressive. So the younger, the even more impressive to me. It is clear that if some of those who went to college at an early age had stayed to compete for longer they would have won many more Gold Medals and probably come close to some perfect scores.

Hmmm, This is an nice intellectual disquisition in here; but can anyone elaborate/elucidate the explanation presented by gauss and valentine?

There was a legend in my high school, now a well known mathematician, who graduated a  few years before I attended who just might fit the bill, Noam Elkies.

From Wiki: born August 25, 1966 and

http://www.imo-official.org/participant_r.aspx?id=10464 shows a perfect score in 1981 which would have made him 14 at the time.

[quote="davidstone"][quote]Hello. Please post here names of youngest perfect scorers or gold medallist at IMO. Comments would be welcomed . THANKS[/quote]
Terence Tao must surely have been the youngest gold medallist. But he didn't get a perfect score, and nor did Pawel Kroger.
[url=http://tinyurl.com/ye5nvzb]Kroger[/url] got 40 and 27 in 1972-73.
[url=http://tinyurl.com/y8ueg5t]Tao[/url] got 19, 40, 34 in 1986-88.

So who [b]does[/b] hold the record as the youngest ever perfect scorer?

David[/quote]

Greetings to Ireland. [url=http://sites.google.com/site/pawelkroeger/]Pawel Kröger[/url] got indeed a full score of 40/40 at 14th IMO 1972 in Poland. Please check page 32/96 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=87&t=295730]here.[/url] It is written that he has been in grade 7 at this time which means he has been around age 13 at this time.


A particle of mass $M$ and energy $E$ collides elastically with a particle of mass $m$ which was resting before the collision. Find the energy of the first particle after the collision. Assume that both particles move only in one dimension.

well djole is $M$ the inertial mass or the recorded mass of the moving body

There is a convention that when you say that a relativistic particle has a mass of $m$, it is understood that this is the rest mass. I'm a follower of this rule, so...

well thanks . i  just wanted to clear my doubt because i follow a convention that says $M_{0}$ as the particles inertial  mass and $M$ as it's relativistic mass

this problem can be dun by normal method + [i]patience[/i]

Can you do it? :P It's not enough to say that it can be done ;)

well u r right the thing is it can be done and i know how to do it .the only thing is that i wrote it out of loss of patience .ok let me give the sol^n soon

I'm waiting :) Take your time, it's quite tricky...

hmmm... let me post my approach and see
 let's solve the problem by using $four-vectors$

for simplicity i take the inertial mass of 1st body as $M_{0}$ ana inertial mass of 2nd body as $m_{0}$

so before collision
$\vec P_{1}$  $=$  $M_{0}$ $\gamma_{1}$ $(c,v_{1},0,0)$
$\vec P_{2}$ $=$  $m_{0}$ $\gamma_{2}$ $(c,0,0,0,)$

after collision,

$\vec P_{1}^{'}$ $=$ $M_{0}$ $\gamma_{1}^{'}(c,v_{1}^{'},0,0)$

$\vec P_{2}^{'}$ $=$ $m_{0}$ $\gamma_{2}^{'}(c,v_{2}^{'},0,0,)$

but by the conservation laws
$\vec P_{1}$ $+$ $\vec P_{2}$ $=$ $\vec P_{1}^{'}$ $+$ $\vec P_{2}^{'}$

this implies, $|\vec P_{1}^{'}|^{2}$ $=$ $( \vec P_{1}^{'}$ $+$ $\vec P_{2}^{'}-\vec P_{2})^{2}$

also since this is an elastic collision

we have ,
$M_{0}$ $( \gamma_{1}-1 )$ $+$ $m_{0}$ $( \gamma_{2}-1 )$ $=$ $M_{0}$ $( \gamma_{1}^{'}-1 )$ $+$ $m_{0}$ $( \gamma_{2}^{'}-1 )$

also we can equate the corresponding parts of the momentom conservation equation using

$|\vec P_{1}|^{2}$ $=$ $-M_{0}^{2}c^{2}$
similarly we get
$|\vec P_{2}|^{2}$ $=$ $-m_{0}^{2}c^{2}$
$|\vec P_{1}^{'}|^{2}$ $=$ $-M_{0}^{2}c^{2}$
$|\vec P_{2}^{'}|^{2}$ $=$ $-m_{0}^{2}c^{2}$

also, $\vec P_{1}.\vec P_{2}$ $=$  $-M_{0}\gamma_{1}m_{0}\gamma_{2}c^{2}$
similarly get the other dot products..

also $\gamma_{2}$ $=$ $1$
further
$\gamma_{1}$ $=$ $\frac{E}{M_{0}c^{2}}$

using this we tend to get
$\gamma_{2}^{'}$ ehich gives us
$E^{'}$ $=$ $M_{0}$ $\gamma_{2}^{'}$ $c^{2}$

this is one approach  the other is  to use the centre of mass frame of refrence for analysis after collision which is an inertial frame of refrence (since our initial frame of refrence was inertial though there doesn't exist something called inertial)

well abnyways we use this fundamental invariant property of he centre of mass frame of refrence which states that

in the centre of mass frame $\sum \vec P_{i}$ $=$ $(\frac{E_{tot}}{c},0,0,0)$
 hence equate both sides by the previous obtained equations.....

The general procedure is correct. I hope you'll at least find the time to write down the expression for the final result. I have two questions:

- why is  there a minus in $|\vec P|^{2}=-M^{2}c^{2}$?

- I think you need to find $\gamma'_{1}$, not $\gamma'_{2}$, but that's probably a typo. You're supposed to find the energy of $M$ after the collision.

surely i will write down the final expression for i will be a stisfied man then

the second one is surely a typo. :)

but i have no clue as to why i did the first one as there was modulus on the vector... :o


Express $ (0.5 \plus{} \frac23)^2$ as a common fraction.

$ \frac{1}{2} \plus{} \frac{2}{3} \equal{} \frac{7}{6}$, and that squared is $ \boxed{\frac{49}{36}}$.


geometry
circumcircle
blogs
geometric transformation
homothety
cyclic quadrilateral
power of a point

There is cyclic quadrilateral ABCD(center O)

And intersection of AC,BD is E.

There is a point P is ABCD and X,Y,Z,W is circumcenter of ABP,BCP,CDP,DAP

Then,

XZ,  YW,  OE    are concurrent!!

This is the Generalization of the China TST 2006 geometry problem,though it is not hard,maybe even easy.But anyway this is nice application of the pole-polar theory.
Proof:
Let $ \omega(O),\omega_1(O_1),\omega_2(O_2),\omega_3(O_3),\omega_4(O_4)$ be the circumcircles of the $ ABCD,\triangle APB,\triangle BPC,\triangle CPD,\triangle DPA$,respectively.Denote $ \omega_1\cap\omega_3 \equal{}P,N$ and $ \omega_2\cap \omega_4 \equal{}P,M$.Then
The point of intersection of $ O_1O_3$ and $ O_2O_4$ is the circumcenter of $ \triangle PNM$.
Consider circumcircle of $ \triangle PNM$,denote it as $ \zeta(I)$.Now we must prove that $ I,O,E$ are collinear.
Denote $ AD\cap BC\equal{} F$,$ AB\cap CD\equal{}G$.Then it is well known that $ OE\perp FG$($ FG$ is the polar of $ E$ w.r.t $ \omega$).Hence
it is sufficiently to prove that $ OI\perp EG$.
It is well known that radical axes of the 3 circles passes through one point,hence
$ PM,AD,BC$ are concurrent.
$ PN,AB,CD$ are concurrent.
Since $ PM$ is the radical axe of $ \zeta$ and $ \omega_2$,and $ AD$ is radical axe of $ \omega$ and $ \omega_2$,thus radical axe
of $ \zeta$ and $ \omega$ passes through $ F\equal{}PM\cap AD$.Similarly the radical axe of $ \zeta$ and $ \omega$ passes through $ G$,hence
$ FG$ is radical axe of $ \zeta$ and $ \omega$,and we are done.

[quote="Erken"]This is the Generalization of the China TST 2006 geometry problem,though it is not hard,maybe even easy.But anyway this is nice application of the pole-polar theory.
Proof:
Let $ \omega(O),\omega_1(O_1),\omega_2(O_2),\omega_3(O_3),\omega_4(O_4)$ be the circumcircles of the $ ABCD,\triangle APB,\triangle BPC,\triangle CPD,\triangle DPA$,respectively.Denote $ \omega_1\cap\omega_3 \equal{} P,N$ and $ \omega_2\cap \omega_4 \equal{} P,M$.Then
The point of intersection of $ O_1O_3$ and $ O_2O_4$ is the circumcenter of $ \triangle PNM$.
Consider circumcircle of $ \triangle PNM$,denote it as $ \zeta(I)$.Now we must prove that $ I,O,E$ are collinear.
Denote $ AD\cap BC \equal{} F$,$ AB\cap CD \equal{} G$.Then it is well known that $ OE\perp FG$($ FG$ is the polar of $ E$ w.r.t $ \omega$).Hence
it is sufficiently to prove that $ OI\perp EG$.
It is well known that radical axes of the 3 circles passes through one point,hence
$ PM,AD,BC$ are concurrent.
$ PN,AB,CD$ are concurrent.
Since $ PM$ is the radical axe of $ \zeta$ and $ \omega_2$,and $ AD$ is radical axe of $ \omega$ and $ \omega_2$,thus radical axe
of $ \zeta$ and $ \omega$ passes through $ F \equal{} PM\cap AD$.Similarly the radical axe of $ \zeta$ and $ \omega$ passes through $ G$,hence
$ FG$ is radical axe of $ \zeta$ and $ \omega$,and we are done.[/quote]

Oh, thank you.

But I don't know why FG is polar of E.....

You need to know some properties of harmonic fours and polar theory.If you know them,see here:
http://www.mathlinks.ro/viewtopic.php?t=168866

See my blog for the proof of the fact above :)

Very nice solution, Erken  :)  I have an another solution base on polar theory, but it is worse than yours  :)
Let $ S \in XZ \cap YW$
Denote $ u$ is the polar of the point $ U$ wrt $ (O)$, and $ (mn)$ is the pole of the line passes through $ M,N$ wrt $ (O)$
Let $ A_1 \in WX \cap AA$, and $ B_1, C_1, D_1$ are defined similarly.
We have $ A_1,B_1, C_1, D_1$ are collinear, because all of them lie on the radical axis of the circle $ (O)$ and the point-circle $ P$
Then $ A(wx) = a_1, B(xy) = b_1, C(yz) = c_1, D(zw) = d_1$ are concurrent at a point, called $ Q$
From this, and not that $ (wx)(xy) = x \parallel AB, (xy)(yz) = y %Error. "parralel" is a bad command.
BC, (yz)(zw) = z \parallel CD, (zw)(wx) = w \parallel DA$, we have there exists a homothety $ f$, centered at $ Q$ such that: $ f(A) = (wx), f(B) = (xy) , f(C) = (yz), f(D) = zw$
Now let $ AB \cap CD = F, AD \cap BC = G$, we have $ f(F) = x \cap z = (xz), f(G) = (yw)$
So $ FG \parallel (xz)(yw) = s$, then $ E$ - the pole of $ FG$, $ S$ - the pole of $ s$ and $ O$ are collinear.


calculus
integration
floor function
calculus computations

{x}
∫           14 + [x] dx       where {.} denotes fractional part and [.] denotes integral part
0

The expression you've written is meaningless: $ x$ gets either to be the variable of integration or involved in the bounds, not both.  The typical way to make something like this sensible is to guess that you meant $ \int_0^{\{t\}} 14 \plus{} \lfloor x \rfloor \, dx$, a not very interesting integral (since the integrand is constant on the interval over which we're integrating).

How about
[x]
∫     2x – [x] dx  where [.] denotes greatest integer
0

Please re-read the first sentence of my previous post and think about its content.  If you come to understand its content, edit your post appropriately.  If you don't understand its content, try to form a precise statement of what you don't understand and I or someone else will help explain it to you.

why can't one variable used for both purposes?

How the hell can you integrate something over a variable interval ?

Right: the notation $ \int_a^b f(x)\, dx$ means that you fix an interval $ [a, b]$ and then do the process known as definite integration over this interval.  The $ x$ in $ f(x)dx$ is a placeholder -- it doesn't exist outside the integral.  This is very similar to the fact that $ \sum_{i \equal{} 1}^{5} i \equal{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5$ -- the index of summation doesn't have any meaning outside the sum.

But your answer seems to be wrong

0<{x}<1, then in (0,1), [x] = 0

the answer thus should be 14{x}

I didn't give an answer, so I'm not sure how you think that it could be wrong.  :wink:


The Wildcats scored an average of 60 points per game during the 1998 season. They averaged 72 points for each of their 13 home games. How many points did they average for each of their 12 away games?

Fromt he 72 point games, the team scored 13*12 points over 60.
Thus, to make up for it, the team must score 13*12 points below 60 in the 12 days, or an average of 13 points under 60, which is 47.

Or:

They score a total of 60*25=1500 pointz
72*13 in home games.which is 936
Thus in away games they scored 1500-936=564
divide by 12

47

Or we could use weighted averages. Letting x be the number of points below 60, 12x=13x12 so x=13, so 60-13=47. Pwnt.

EDIT: 1200th post! w00t!  :D

Or:

13 home games: 13*12 pts up
12 away games: 12*13 pts down => 13 pts below average => 47 pts

@mewto and james: thats the same solution I put.


geometry
incenter
trigonometry
angle
midpoint
IMO Shortlist

[quote="orl"]The incenter of the triangle $ ABC$ is $ K.$ The midpoint of $ AB$ is $ C_1$ and that of $ AC$ is $ B_1.$ The lines $ C_1K$ and $ AC$ meet at $ B_2,$ the lines $ B_1K$ and $ AB$ at $ C_2.$ If the areas of the triangles $ AB_2C_2$ and $ ABC$ are equal, what is the measure of angle $ \angle CAB?$[/quote]
[hide="Solution"]
Let the tangency points between the incircle with $ AB$ and $ AC$ respectively be $ D$ and $ E$.  Let the midpoints of $ CD$ and $ BE$ be $ Y$ and $ X$ respectively.  Furthermore, let the semiperimeter of $ \triangle ABC$ be $ s$, $ AB \equal{} c$, $ BC \equal{} a$, and $ AC \equal{} b$.  It is well-known that $ B_1$, $ X$, and $ K$ are collinear as are $ C_1$, $ Y$, and $ K$.  Notice that\[ 1\equal{}\frac {[AB_2C_2]}{[ABC]} \equal{} \frac {\frac {AC_2\cdot AB_2\sin \angle BAC}{2}}{\frac {AB\cdot AC\sin \angle BAC}{2}} \equal{} \frac {AC_2}{AB}\cdot \frac {AB_2}{AC}\]Apply Menelaus to $ \triangle AEB$ with transversal $ C_2XB_1$, we have that $ \frac {AC_2}{C_2B}\cdot \frac {BX}{XE}\cdot \frac {EB_1}{AB_1} \equal{} 1$.  We have that $ BX \equal{} XE$, so\[ \frac {AC_2}{C_2B} \equal{} \frac {AB_1}{EB_1} \equal{} \frac {\frac {b}{2}}{\frac {b}{2} \minus{} (s \minus{} a)} \equal{} \frac {b}{b \minus{} (a \plus{} b \plus{} c \minus{} 2a)} \equal{} \frac {b}{a \minus{} c}\]which means that $ \frac {C_2B}{AC_2} \equal{} \frac {a \minus{} c}{b}$, so $ \frac {AB}{AC_2} \equal{} 1 \plus{} \frac {C_2B}{AC_2} \equal{} \frac {a \plus{} b \minus{} c}{b}$.  Similarly, $ \frac {AC}{AB_2} \equal{} \frac {a \plus{} c \minus{} b}{c}$, so\[ 1 \equal{} \frac {AC}{AB_2}\cdot \frac {AB}{AC_2} \equal{} \frac {a^2 \minus{} (b \minus{} c)^2}{bc} \equal{} \frac {a^2 \minus{} b^2 \minus{} c^2 \plus{} 2bc}{bc}\]\[ bc \equal{} a^2 \minus{} b^2 \minus{} c^2 \plus{} 2bc\]\[ a^2 \equal{} b^2 \plus{} c^2 \minus{} bc\]\[ \cos \angle BAC \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc} \equal{} \frac {1}{2} \equal{} \cos 60\]from which it follows that $ \angle BAC \equal{} 60$. [/hide]

One can use [b]Cristea's theorem ( the Theorem of Transversal )[/b] to find $ AB_2$ and $ AC_2$, getting the same value for the ratio $ \frac{AB \cdot AC}{AB_2 \cdot AC_2}$ as to [i]The QuattoMaster 6000[/i].

Best regards,
sunken rock

My Method：

Let $BK, CK$ intersect $AC, AB$ at $D, E$ respectively，$CC_3 \parallel B_1K, BB_3 \parallel C_1K,CC_3 \cap BB_3=F,FJ \parallel AB, FL \parallel AC$
Then $C_2, B_2$ are the midpoints of $AC_3, AB_3$ respectively，$K$ is the common midpoint of $AF, EJ, DL$ . $\triangle BEK \cong \triangle BGK$
$\frac{AC}{AC_3}=\frac{FC}{FC_3}=\frac{CJ}{2KJ}$，similarly，$\frac{AB}{AB_3}=\frac{BL}{2KL}$，$S(AB_3C_3)=4S(AB_2C_2), BK$ bisector $EG,GL \parallel BK$

So on the basis of this, we can get：

$S(ABC)=S(AB_2C_2) \Longleftrightarrow 4S(ABC)=S(AB_3C_3) \Longleftrightarrow \frac{CJ}{KJ}=\frac{KL}{BL}$

$\Longleftrightarrow \frac{CG}{BG}=\frac{KL}{BL} \Longleftrightarrow GL \parallel CK \Longleftrightarrow CK$ bisector $DG \Longleftrightarrow \triangle CDK \cong \triangle CGK$

$\Longleftrightarrow \angle KEB+\angle KDC=\angle KGB+\angle KGC=180^{\circ} \Longleftrightarrow A, D, K, E$ are on a circle

$\Longleftrightarrow \angle BAC+\angle BKC=180^{\circ} \Longleftrightarrow \angle BAC=60^{\circ}$ . [b]Q. E. D.[/b]


function
algebra
polynomial
integration
complex analysis
Functional Analysis

Consider the Banach space of all continuous functions on the closed unit disc with the maximum norm.

Show that the set of polynomial functions is not dense in this Banach space.

More interesting question: what is the closure of the polynomials?

A proof that $ f(z)\equal{}\overline{z}$ is not in the closure of the polynomials:

Let $ P(z)$ be any polynomial. Let $ C$ be the unit circle, with its standard parameterization.

$ \int_C(f(z)\minus{}P(z))\,dz\equal{}\int_Cf(z)\,dz$ by Cauchy's Theorem.

$ \int_Cf(z)\,dz\equal{}\int_C\overline{z}\,dz\equal{} i\int_0^{2\pi}e^{\minus{}i\theta}e^{i\theta}\,d\theta\equal{}2\pi i.$

Hence,

$ 2\pi\equal{}\left|\int_C(f(z)\minus{}P(z))\,dz\right|\le\int_C|f(z)\minus{}P(z)|\,|dz| \le2\pi\|f\minus{}P\|_{\infty}$

where $ \|f\minus{}P\|_{\infty}\equal{}\sup_{z\in\overline{D}}|f(z)\minus{}P(z)|$ is the uniform norm.

Thus $ \|f\minus{}P\|_{\infty}\ge 1$ and $ f$ cannot be approximated by polynomials.

-----

The closure of the polynomials in $ C(\overline{D})$ is usually called $ A.$ It is the set of functions continuous on the closed unit disk and holomorphic on the interior of the disk. For such a function, the Cesaro means of the partial sums of the power series centered at zero should give a sequence of polynomials converging uniformly to the function.

On the other hand, the set of polynomials in both $ z$ and $ \overline{z}$ is dense in $ C(\overline{D}),$ by a version of the Weierstrass approximation theorem.


OK, it has become clear that this forum has been overrun by spam.  This is going to stop now.

[b]The Games & Fun Forum will be locked for 24 hours[/b], so that we can clean out all the spam, and so those of you who have been abusing this forum can cool off.  When the forum is reopened tomorrow, users who spam will be subject to banning without warning.

Spam is unacceptable.  All posts should have some [b]meaningful[/b] content, even here in Games & Fun.  Spam not only interferes with users who wish to have a coherent discussion, it also unnecessarily taxes our database and our web server.

In the future, there will be a zero-tolerance policy on "3"'ing, pointless ziggurats, and other egregious spam.

Let me remind you of our user conduct policy (which is in full at http://www.artofproblemsolving.com/Legal/AoPS_L_Comm_Terms.php )

[quote]a. not to harm minors or cause minors to be harmed in any way;

b. not to forge headers of postings, e-mails, or otherwise manipulate identifiers in order to disguise the origin of any Content transmitted through the Services;

c. not to impersonate any person or entity, including, but not limited to, an AoPS employee, forum moderator, guide or guest, or falsely state or otherwise misrepresent your affiliation with any person or entity;

d. not to upload, post, e-mail, transmit or otherwise make available any Content that is unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable;

e. not to upload, post, e-mail, transmit or otherwise make available any Content that you do not have a right to make available under any law or under contractual or fiduciary relationships (such as inside information, proprietary and confidential information learned or disclosed as part of employment relationships or under nondisclosure or confidential disclosure agreements);

f. not to upload, post, e-mail, transmit or otherwise make available any information that you consider confidential in nature, which posting may cause you or any third party to lose rights to the intellectual property associated with the information;

g. not to upload, post, e-mail, transmit or otherwise make available any Content that infringes any patent, trademark, trade secret, copyright or other proprietary and lawful rights of any party;

h. not to upload, post, e-mail, transmit or otherwise make available any unsolicited or unauthorized advertising, promotional materials, "junk mail," "spam," "chain letters," "pyramid schemes," or any other form of solicitation. You understand that such postings may violate laws of various jurisdictions, and that you may be subject to prosecution or to civil suit;

i. not to intentionally or unintentionally violate any local, state, national or international law;

j. not to "stalk" or otherwise harass another; or collect or store personal data about other users;

k. to upload, post, e-mail, transmit or otherwise make available only material which the User has created and to which the User owns the copyrights; and

l. to report any violations of the above immediately to AoPS.
[/quote]Parts (d) and (h) of the above are particularly significant, and are what have resulted in this 24-hour "cooling-off" period.

We provide this forum -- for free -- for your enjoyment.  Don't abuse the privilege.

I do not see how (d) or (h) is violated in G&FF.

I still do not see how (d) or (h) is violated in G&FF.

[quote="Klebian"][quote="akalra1"]some forums (like slot machine) are locked

can you unlock them?

and if you can't, is it ok if we start a new thread?

thanks[/quote]

Obviously not, otherwise they wouldn't have been locked.[/quote]

what was wrong with slot machine

was that too spammy?

can we start a new "clean" slot machine thread?

i liked it

For now, do not introduce a new slot machine thread.

WHAT is the MAX KB for an upload of an unharmful not spamming picture?

What was wrong with that game Duck, Duck, Goose?

jesus guys.

stop asking questions and use common sense, if you type up a post and it looks like something i'd say, it's spam. VIEWER DISCRETION IS ADVISED.


Determine, with proof, all integers $ x$ for which $ x(x\plus{}1)(x\plus{}7)(x\plus{}8)$ is a perfect square.

Let $ u\equal{}x\plus{}4$. This transforms the product into $ (u\plus{}3)(u\minus{}3)(u\plus{}4)(u\minus{}4)\equal{}(u^2\minus{}9)(u^2\minus{}16)$.
Now let $ v\equal{}u^2\minus{}12$. Then we have $ (u^2\minus{}9)(u^2\minus{}16)\equal{}(v\minus{}4)(v\plus{}4\minus{}1)\equal{}(v^2\minus{}16)\minus{}(v\minus{}4)\equal{}v^2\minus{}v\minus{}12$. This will be between $ (v\minus{}1)^2$ and $ v^2$ if $ v> 13$. We have $ v\equal{}u^2\minus{}12\equal{}(x\plus{}4)^2\minus{}12$ so $ v>13$ if $ x>1$ or $ x<\minus{}9$. The cases of x in between are dealt with by brute force so $ x\equal{}1,0,\minus{}1,\minus{}4,\minus{}7,\minus{}8,\minus{}9$


A square has all of its vertices on lattice points. The length of the square is $\sqrt{p}$ where $p$ is prime. Find the number of lattice points that are contained in the square.

i) To prove don't have other lattice point in each segment trought 2 lattice points with size $\sqrt{p}$, for any prime $p$, is easy. Suppose wlog that one of the points is $A=(0,0)$, and the other $B=(a,b)$. If one point $C=(c,d)$ are in AB, we will have that $c = ka$ and $d = kb$, for any positive rational $k$. Suppose that $k = \frac{x}{y}$, where $gcd(x,y)=1$ and $y<x<\sqrt{p}$.
We will have that $AB^{2}=AC^{2}k^{2}\Longrightarrow y^{2}p = x^{2}.AC^{2}\Longrightarrow x^{2}|p$, impossible.

ii) Now, you can use Pick's theorem. The area of our square is $p$, and no lattice points are in your boundary.

So, there are $p+1$ points in your interior.

I assume you mean $p-1$ points, because there are 4 points on the boundary, namely the vertices themselves :)
I didn't know about Pick's theorem, so I solved it by drawing a bounding box and subtracting the lattice points of 4 smaller triangles from the lattice points in the bigger square.


There was no diagram accompanying this problem. I'm not visualizing what my book's answer key calls the "correct" answer. How can one reason through to the approved answer?

Question:

Points P, Q, R, S, T, and U are all different points lying in the same plane. Points P, Q, and U lie on the same line. The line through points P and Q is perpendicular to the line through points R and S. The line through points R and S is perpendicular to the line through points T and U. Which of the following sets contain points that must lie on the same line?

(a) {P, Q, R}

(b) {Q, R, S}

(c) {Q, R, T}

(d) {Q, T, U}

(e) {R, T, U}

That's the whole problem description. I have seen an answer identified as the "right" answer, but I'm not sure I believe it. How would you work it out?

I just drew a picture and it turned out to be pretty clear. P, Q, and U are already said to be on the same line; same thing for R and S - but neither one necessarily has to be on the line of P, Q, and U. So that already eliminates all of the answer choices except for (d).

And to see that (d) actually is correct, you see that T does have to be on the line with P, Q, and U in order for T-U to be perpendicular to R-S.

[quote="gauss202"]And to see that (d) actually is correct, you see that T does have to be on the line with P, Q, and U in order for T-U to be perpendicular to R-S.[/quote]

THANK YOU! I see that the young learner I was working with made a crucial error in labeling his diagram, and that was what threw him off. Then I was thrown off because I didn't notice the diagramming error at first. I got further confused because at first I transcribed my answer key incorrectly, getting a different answer from the correct answer you found. The answer I was disbelieving was actually a WRONG answer.  :oops: The key point is that point U is mentioned twice, with sufficiently constraining conditions that it must be on the same line as three of the other five named points.

For those of you following along at home, this question comes from the May 1997 Saturday SAT I math section, as reprinted on page 471 of 10 Real SATs, Third Edition (2003).


There are 200 people in a room.
160 speak English, 180 speak Spanish, and 175 speak Mandarin.

What is the least amount of people that speak all 3 languages?

[color=red]So all we need to find out is the maximum amount of people that don't speak one of the languages. [/color] [hide] There can be 20 people who don't speak Spanish, 40 who don't speak English, and 25 who don't speak Mandarin. Since we are solving for the LEAST amount of people who speak all three, we find tht there are 85 people who don't speak a language, giving us 115 people, minimum, that speak all three.[/hide] OR [hide] There are 515 total languages known, out of 600 possible. To minimize the amount of three language speakers, we must maximize the amount of two language speakers. We now have the equation:
2x+3y=515
x+y=200
It solves as x= 85, y= 115, so there are 115 people who speak all three languages.[/hide]
____________________
Flying is learning how to throw yourself at the ground and miss. :)

Usually it is looked down upon when someone just posts an answer.


In $\triangle{ABC}$, $AD$, $BE$, $CF$ are the three interior angle bisectors with $D,E,F$ on the sides of the triangle. Prove: $\triangle{DEF}\leq\frac{1}{4}\triangle{ABC}$.

Suppose $BC=a,AC=b,AB=c$,then $\frac{BD}{BC}=\frac{c}{b+c},\frac{BF}{BA}=\frac{a}{a+b}$
$\frac{\triangle BDF}{\triangle ABC}=\frac{ac}{(a+b)(b+c)}$,
Similar,$\frac{\triangle AEF}{\triangle ABC}=\frac{bc}{(a+b)(a+c)}$,
$\frac{\triangle BDF}{\triangle ABC}=\frac{ab}{(a+c)(b+c)}$,
Just needto prove $\sum\frac{bc}{(a+b)(a+c)}\geq\frac{3}{4}$
Rewrite it like this:$\sum bc(b+c)\geq\frac{3}{4}(a+b)(a+c)(b+c)$
$4\sum bc(b+c)\geq3[2abc+\sum bc(b+c)]$
$\sum bc(b+c)\geq6abc$
By AM-GM,easy to sea it's correct.


Russell rolls a pair of dice with his eyes closed.  His sister tells him that the sum of the dice is greater than or equal to nine.  Knowing this information, Russell calculates the probability that he rolled doubles.  What is that probability expressed as a common fraction?

All posibilities: $ 6\minus{}3$ ,$ 6\minus{}4$ ,$ 6\minus{}5$, $ 6\minus{}6$ ,$ 4\minus{}5$ ,$ 5\minus{}5$
Doubles: $ 5\minus{}5$ ,$ 6\minus{}6$
Probability: $ \frac{1}{3}$


trigonometry
complex numbers
algebra proposed
algebra

Prove that
a) $ sin^2\frac{\pi}{7}\plus{}sin^2\frac{2\pi}{7}\plus{}sin^2\frac{3\pi}{7}\equal{}\frac{7}{4}$
b) $ tan30\plus{}tan40\plus{}tan50\plus{}tan60\equal{}\frac{8\sqrt{3}}{3}cos20$

What level of math are you in? This can be done relatively easy if you have a good grasp of complex numbers and their relation to trigonometric functions. If you do not, your only alternative forseeable to me is to break down each of the $ \sin$'s into dealable components


Let $ f: S\rightarrow T$ be a function. Prove that the following statements are equivalent
(a) $ f$ is one-to-one on $ S$.
(b) $ f/(A\cap B)= f(A)\cap f(B)$ $ \forall$subsets $ A,B$ of $ S$.
(c) $ f^{-1}/[f/(A)]=A$ $ \forall$subsets $ A$ of $ S$.
(d) For all disjoint subsets $ A$ and $ B$ of $ S$ $ f/(A) \cap f(B)=\emptyset$.
(e) For all subsets $ A$ and $ B$ of $ S$ with $ B\subseteq A$, we have $ f/(A $ \ $  B) = f(A)$ \ $ f/(B)$.

(a) $ \Rightarrow$ (b)

Suppose $ f$ is one-to-one. Let $ y\in f^\to (A\cap B)$, then $ \exists x\in A\cap B$ such that $ f(x)=y$, so we have $ f^\to (A\cap B)\subset f^\to(A) \cap f^\to (B)$.

Let $ y\in f^\to (A) \cap f^\to (B)$, then $ \exists x_{1}\in A$ and $ \exists x_{2}\in B$ such that $ f(x_{1})=y$ and $ f(x_{2})=y$, but by $ 1-1$ we have $ x_{1}=x_{2}=z$. So $ z\in A\cap B$ with $ f(z)=y$, therefore $ y\in f^\to (A\cap B)$ and thus $ f^\to (A) \cap f^\to (B) \subset f^\to (A\cap B)$. Therefore $ f^\to (A\cap B)=f^\to(B\cap A)$.

$ (b) \Longrightarrow (c)$

[hide]It is easy to show that $ A \subseteq f^{-1}(f(A))$. So assume (b) and suppose there exists $ a\in S$ such that $ a \notin A$ but $ f(a) \in f(A)$. Then there exists $ b \in A$ such that $ f(b) = f(a) \in f(A)$. Now let $ A = \{ a \}$ and $ B = \{ b \}$. Then a contradiction easily follows since $ \emptyset = f(\emptyset) = f(A%Error. "capB" is a bad command.
) = f(A)\cap f(B) = \{ f(a) \}\neq \emptyset$.[/hide]

$ (c) \Longrightarrow (d)$

[hide]Assume (c) and let $ A, B$ be disjoint subsets of $ S$. As $ f(A)\cap f(B)$ is a subset of $ f(S)$, if it is nonempty, then its inverse image must be nonempty too. But since $ f^{-1}(f(A)\cap f(B)) = f^{-1}(f(A)) \cap f^{-1}(f(B)) = A \cap B = \emptyset$, $ f(A)\cap f(B) = \emptyset$ and (d) is proved.[/hide]

$ (d) \Longrightarrow (e)$

[hide]Assume (d) and let $ B \subseteq A \subseteq S$. Also, let $ C = A\backslash B$. Then $ B$ and $ C$ are disjoint and $ f(B)\cap f(C) = \emptyset$ by (d). Also $ f(A) = f(B\cup C) = f(B) \cup f(C)$, so $ f(A)$ is a disjoint union of $ f(B)$ and $ f(C)$. This implies that $ f(A)\backslash f(B) = f(C) = f(A\backslash B)$, as desired.[/hide]

$ (e) \Longrightarrow (a)$

[hide]Assume (e) and suppose $ f$ is not injective. Then there exists $ a, b \in S$ such that $ a \neq b$ but $ f(a) = f(b)$. Now let $ A = \{ a, b \}$ and $ B = \{ b \}$. Then $ B \subseteq A \subseteq S$ but $ \{ f(a) \}= f(A\backslash B) = f(A)\backslash f(B) = \emptyset$, a contradiction. So $ f$ must be injective.[/hide]


[quote="Silverfalcon"]The triangle $ ABC$ has vertices in such manner that $ AB \equal{} 3, BC \equal{} 4,$ and $ AC \equal{} 5$. The inscribed circle is tangent to $ AB$ in $ C'$, $ BC$ in $ A'$ and $ AC$ in $ B'.$ What is the ratio between the area of the triangles $ A'B'C'$ and $ ABC$?

A. 1/4
B. 1/5
C. 2/9
D. 4/21
E. 5/24[/quote]

[hide]$ BC'\equal{}BA'\equal{}x$, $ CA'\equal{}CB'\equal{}y$, and $ AB'\equal{}AC'\equal{}z$. We have $ x\plus{}z\equal{}3$, $ x\plus{}y\equal{}4$, and $ y\plus{}z\equal{}5$ so that $ x\equal{}1$, $ y\equal{}3$, $ z\equal{}2$. Then using $ [ABC]\equal{}\frac{1}{2}ab\sin C$ a bunch of times, we get $ [AB'C']\equal{}\frac{8}{5}$, $ [BC'A']\equal{}\frac{1}{2}$, and $ [CB'A']\equal{}\frac{27}{10}$, and also $ [ABC]\equal{}6$. Then subtract the areas to get $ [A'B'C']\equal{}\frac{6}{5}$ and I get B. 1/5.[/hide]


algebra
polynomial
quadratics
Rational Root Theorem
number theory
relatively prime
factoring quadratics

What is it?

I see this:

Identify all of the real roots of $ 4x^4\minus{}21x^3\plus{}18x^2\plus{}19x\minus{}6\equal{}0$.
Step 1: Use the Rational Root Theorem to identify possible rational roots.
$ \frac{{ \pm 1, \pm 2, \pm 3, \pm 6}}{{ \pm 1, \pm 2, \pm 4}} \equal{}  \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{4}, \pm \frac{1}{4}, \pm \frac{3}{4}$
So $ p\equal{}\minus{}6$ and $ q\equal{}4$.

From where the whole Step 1 came from  :huh:

hello, i think here is something for you
http://www.trans4mind.com/personal_development/mathematics/polynomials/rationalRootTheorem.htm
Sonnhard.

Imagine the polynomial:
$ P(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_{1}x^{1}\plus{}a_{0}$.
Notice that, if you know that there exists a rational root, which we can express as the simplified fraction $ \frac{p}{q}$, then, when we plug this value in for $ x$, we get that:
$ a_{n}(\frac{p}{q})^{n}\plus{}a_{n\minus{}1}(\frac{p}{q})^{n\minus{}1}\plus{}...\plus{}a_{0}\equal{}0$.
When we multiply by $ q^{n}$, we then get that"
$ a_{n}p^{n}\plus{}a_{n\minus{}1}p^{n\minus{}1}q\plus{}...\plus{}a_{0}q^{n}\equal{}0$.
Notice that every single term on the left hand side is divisible by $ q$ except $ a_{n}p^{n}$.  But $ 0$ is divisible by $ q$ too! So, that must mean that $ a_{n}p^{n}$ has to be divisible by $ q$.  We know that $ p^{n}$ can't, since $ p$ and $ q$ are relatively prime from our definiton of the simplified fraction $ \frac{p}{q}$, so that must mean $ a_{n}$ is divisible by $ q$, or, in other words, $ q$ is a factor of the leading coefficient. In similar fashion, we can easily prove that $ p$ is a factor of $ a_{0}$.
So, when we know that a polynomial has a rational root, it must be of the form $ \frac{a}{b}$, where $ a$ and $ b$ are factors (possibly negative) of the constant and leading coefficient, respectively.
This is called the [u]rational root theorem[/u].

Thanks a lot  :lol:
I'm in the process of understanding now  :rotfl:

Basically,

Given a polynomial $ P(x)$, $ (ax \minus{} b)$ is a factor of $ P(x)$ if and only if $ P(\frac {b}{a}) \equal{} 0$.

When finding possible rational roots of $ P(x)$,

Take the constant term of $ P(x)$ and take all of its factors, both positive and negative.
Divide each factor of the constant term by both the positive and negative factors of the coefficient of the leading term.

For example

$ P(x)\equal{}6x^4\plus{}2x^3\minus{}17x\plus{}8$
The possible roots of $ P(x)$ are then
$ \frac{\pm1, \pm2, \pm4, \pm8}{\pm1, \pm2, \pm3, \pm6} \equal{} \pm1, \pm2, \pm4, \pm8, \pm\frac{2}{3}, \pm\frac{4}{3}, \pm\frac{8}{3}, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{4}, \pm\frac{1}{6}$

It's like a method of factoring, just expanded into higher degrees of exponents. Remember how at first when factoring quadratics, you looked for factors of the last term, considering both positive and negative choices?


Gillis 2000:
prove that for any n>=0 there is no real solution to the equation
x^2000/2001^n + x^1999/2000^n + ... +x^2/3^n + x/2^n + 1 = 0.

Gillis 1984:
if x + y + z = 2pi and cos x cos y cos z = 1/4
prove 4cos 3x cos 3y cos 3z - 12cos 2x cos 2y cos 2z = 1.

actually I have two nice solutions to the first one but only brute force solution to the second.

by reccurence for n=0 the equation :
$ 1\plus{}x\plus{}x^2\plus{}...\plus{}x^{2000}\equal{}0$ has no real solution
we suppose that the equation $ x^{2000}/2001^n \plus{} x^{1999}/2000^n \plus{} ... \plus{}x^2/3^n \plus{} x/2^n \plus{} 1 \equal{} 0$ have no real solution ,
suppose for exemple that $ x^{2000}/2001^n \plus{} x^{1999}/2000^n \plus{} ... \plus{}x^2/3^n \plus{} x/2^n \plus{} 1>0$ for all real number.
we considere the polynomial $ P(x)\equal{}x\plus{}x^2/2^{n\plus{}1}\plus{}...\plus{}x^{2001}/2001^{n\plus{}1}$
we have $ P'(x)\equal{}x^{2000}/2001^n \plus{} x^{1999}/2000^n \plus{} ... \plus{}x^2/3^n \plus{} x/2^n \plus{} 1>0$ so P(x) is bijective ,then the equation $ P(x)\equal{}0$ have one solution which is x=0 ,so conclusion the equation $ 1\plus{}x/2^{n\plus{}1}\plus{}...x^{2000}/2^{n\plus{}1}\equal{}0$ have no real solution,  we are done  :)


Let $\mathcal S$ denote the set of points inside or on the border of a triangle $ABC$, without a fixed point $T$ inside the triangle. Show that $\mathcal S$ can be partitioned into disjoint closed segemnts.

[i]Yugoslavia[/i]

Suppose the point $T$ is strictly inside the triangle, and not on the borders (the other cases will follow from this construction). Let $\ell_a,\ell_b,\ell_c$ be the lines parallel to $BC,CA,AB$ respectively which pass through $T$. Our set of segments will consist of:

(a) those segments parallel to $BC$, with endpoints on $AB,AC$, lying between $\ell_a$ and $BC$ (but not on $\ell_a$);

(b) the segments parallel to $CA$, with endpoints on $AB,\ell_a$, lying between $\ell_b$ and $CA$ (not on $\ell_b$);

(c) the segments parallel to $AB$, with endpoints on $\ell_a,\ell_b$, lying between $\ell_c$ and $AB$.