Hi I am new to Mathcounts and how do the problems in the school handbook compare to the actual competition problems for school, chapter, states, and nationals?
Like warmups 1- x are school level etc. etc. . If the problem are even similar at all.

The warmups are somewhat similar to the school round, perhaps a bit harder, and the workouts have problems in the state difficulty.

Of course, some problems vary in difficulty, and workout problems could get kinda hard...maybe about easy national level. The problems do resemble the actual test problems in a way.

It really depends.  This year's handbook seemed a bit easier than prior years.  But I'd say:
Warm ups=School Sprint
Workouts=Chapter Target

Let's say if a student mastered all problems in 2010 handbook.
Will he/she do well in School, Chapter or State round?

How much School/Chapter/State contest problems in a year, say 2008, differ from 2008 handbook problems?

One should do both this year's along with those of previous years to get a firmer grip on the real School/Target Rounds. i think that the 'Warm-Ups' and 'Work-outs' are pretty similar to the real competition.

This varies of course ... but I think it is a good question.  I wonder myself sometimes how to best use the handbooks to simulate competitions (as there are many more handbook problems than competition sets).

I use the later Workouts (6-9) to simulate state-level team rounds, and I think in some cases they are more difficult (depends on the year ... handbooks and competitions vary widely sometimes).  There are national level problems within the later Warm-Ups and Workouts (actually taken from national sets, for example, Workout 7 from this year's handbook contains a previous year's state team round problem).

I would say that Warm-Ups 12-15 are similar to problems 15-25 on a state sprint round, while WU 16-18 are like the middle-pages of a national sprint.

If you are going to simulate a state-level sprint using the handbook, I would say ... take Warm-Ups 6, 12, and 18 ... and try do them all in 40 minutes.  For chapter, take 4, 8, and 12 ... something like that.

I am interested in hearing what students perceive as the difficulty of the competitions versus the handbook problems.


Given two numbers m and n, such that $2n \geq n \geq 0$. Then show that $2^{2n+2} + 2^{m+2} +1$
is a perfect square if and only if $m=n$.

Davron

By hypothesis the given number is an odd square. Say it is (2k+1)^2. Setting these as equals and simplifying yields
 2^(2n)+2^m=k(k+1)
Now if 2n>m We have (2^m)(2^(2n-m)+1)=k(k+1) One of the factors on each side is odd while the other is even. If k is even then k=2^m and k=2^(2n-m) and m=n. If k is odd then k>k+1 (RAA).
If m>2n We have once again k is even, k=2^(2n), and k=2^(m-2n) then k^2=2^m. Thus 4n=m. Then the given square is (2^(2n+1)+1)^2. I think 2n>m may be required for the given result


function
integration
algebra unsolved
algebra

Does there exits a function $f: \ (0;+\infty)\rightarrow (0;+\infty)$ such that $\exists f'(x) \ \forall x>0$ and $f'(x)\ge f(f(x)) \ \forall x>0$

Hope this is not another proof of me being mentally retarded.
Since  $f'(x)\geq f(f(x))>0$, if $f$ exists, it is monotonically increasing. If it was bounded then $f$ would tend (how do you say this in English?) to a limit, $c>0$ as $x$ grows to $\infty$; in addition for fixed $\delta$,   $lim_{x\to \infty}\frac{f(x+\delta)-f(x)}{\delta}=0$,  and since letting  $g(x)=y|[f'(y)=min\{f'(z)| z\in[x,x+\delta]\}]$,  we would have that   $f'(g(x))<\frac{f(x+\delta)-f(x)}{\delta}$, we would get
$0=lim_{x\to \infty}\;f'(g(x))\geq lim_{x\to \infty}f(f(g(x)))=lim_{x\to \infty}f(f(x))=f(c)>0$
which is a non-sense. So it would be unbounded and therefore there would be an $X$ such that $f(f(x))\geq 1+\epsilon \;\; \forall x\geq X$, and so, such that  $f'(x)\geq 1+\epsilon\;\; \forall x\geq X$ for some $\epsilon>0$  (it has no relation with the previous one). This means that   $[f(x)-x]'=f'(x)-1\geq \epsilon>0 \;\;\;\forall x\geq X$,  so the function  $f(x)-x$  would be unbounded; in particular there must exist a $Z$ such that   $f(x)\geq x+1 \;\;\forall x\geq Z$.  But then $\forall x\geq Z$ we would have,
$f(x+1)=f(x)+\int_{x}^{x+1}f'(t)dt\geq f(x)+\int_{x}^{x+1}f(f(t))dt\geq$
$\geq f(x)+\int_{x}^{x+1}f(f(x))dt=f(x)+f(f(x))\geq f(x)+f(x+1)>f(x+1)$,
so the funtion doesn't exist.


in every triangle with a,b,c sides, prove that:
1/a+b-c +1/b+c-a +1/a+c-b>=1/a +1/b +1/c

nice inequality:
put: $ x=\frac{1}{a+b-c}$ and $ y=\frac{1}{a-b+c}$ and $ z=\frac{1}{-a+b+c}$

$ \longleftrightarrow \frac{1}{a}=\frac{2xy}{x+y}$ ;$ \frac{1}{b}=\frac{2xz}{x+z}$ and $ \frac{1}{c}=\frac{2yz}{y+z}$
the inequality is equivalent has:
$ S=\frac{2xy}{x+y}+\frac{2xz}{x+z}+\frac{2yz}{y+z}\leq x+y+z$
we know that: $ (x+y)^2\geq 4xy$
$ \longleftrightarrow \frac{2xy}{x+y}\leq\frac{1}{2}(x+y)$
and $ \longleftrightarrow \frac{2yz}{y+z}\leq\frac{1}{2}(y+z)$
and $ \longleftrightarrow \frac{2zx}{z+x}\leq\frac{1}{2}(z+x)$
by summation:
$ S\leq x+y+z$

[quote="behdad.math.math"]In every triangle with $ a,b,c$ sides, prove that

$ \frac{1}{b+c-a} +\frac{1}{c+a-b}+\frac{1}{a+b-c }\geq\frac{1}{a} +\frac{1}{b} +\frac{1}{c}.$[/quote]$ \sum{\frac{1}{b+c-a}}-\sum{\frac{1}{a}}=\frac{1}{2}\sum{\left(\frac{1}{c+a-b}+\frac{1}{a+b-c }\right)-\sum{\frac{1}{a}}=\sum\frac{(b-c)^2}{a(c+a-b)(a+b-c)}\geq0.}$

note that $ a\plus{}b\minus{}c,b\plus{}c\minus{}a,c\plus{}a\minus{}b>0$

The rest follows. :D


Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?
(A) 45
(B) 48
(C) 50
(D) 55
(E) 58

[hide] $x$ is the distance to his workplace. $y$ is how long it should take for him to be on time.
$\frac{x}{40}=y+3$
$\frac{x}{60}=y-3$
$40y+120=60y-180$
$y=15$
$x=60\cdot12=720$
$\frac{720}{15}=\boxed{B)48}$[/hide]

[hide="maybe simpler"]
since going at 40 mph and 60 mph get him to work the same amount of time late as the other does early, then the desired speed is the harmonic mean of 40 and 60.
harmonic mean is $\frac{2ab}{a+b}$ or in our case, $\frac{2\times60\times40}{60+40}=\fbox{48}$[/hide]

[quote="lotrgreengrapes7926"][hide="Another method"] In this type of problem we can use harmonic mean. $\frac{1}{\frac{\frac{1}{40}+\frac{1}{60}}{2}}=\frac{2}{\frac{3}{120}+\frac{2}{120}}=\frac{2}{\frac{5}{120}}=2\cdot\frac{120}{5}=\boxed{48}$[/hide][/quote]

What's a harmonic mean?

sapphyre571, yours is almost exactly like mine.

The harmonic mean is the reciprocal of the arithmetic mean of the reciprocals of the numbers. For example, $\frac{1}{a}$ and $\frac{1}{b}$ are the 2 reciprocals. The arithmetic mean of those are $\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{\frac{a+b}{ab}}{2}=\frac{a+b}{2ab}$. Then, the reciprocal of that is $\frac{2ab}{a+b}$.

I don't know exactly why it works, but it almost always does in problems involving rates.

Does it always work?
How do you use use it (like know which numbers to average)?

[quote="laughinghead505"]Why would it be the harmonic mean and not some other kind?[/quote]

I don't know.

It is by name it is called harmonic mean.  you can call it whatever you remember it by, and this would help with average speed problems.

[hide] (rate)(time) = (distance)

let $T$ be the time it takes to get to work in minutes.

$40(T+3) = 60(T-3) \Rightarrow T=15$

let $x$ be the desired speed. then $40(18) = 15x$
giving $x = 48$
Making the answer $\boxed{B}$[/hide]

The simplest problem you can use harmonic mean on is "I drove at $x$ mph to point A and drove at $y$ mph on the way back. What was my average speed?" For other problems, you may need to adjust and use it differently.

No, I meant why would it be the harmonic mean instead of, say the arithmetic or geometric?


Is it possible to use Kile on a Windows XP platform?
If possible, how should one install it?

I just downloaded it from sourceforge.
It is an archive of thousands of files, none of them executable.
How do I use it?

It looks like you need KDE (or at least QT) to run Kile. KDE is a desktop environment for Unix-based operating systems. You might be able to get KDE to work in Windows using Cygwin, if you really want to use Kile, but I'd suggest finding a different editor.


inequalities
function
integration
real analysis
real analysis unsolved

Prove that for every continous function $f: [0,1]\to R$ the following inequality holds:
$\frac{1}{4}\int_{0}^{1}f^{2}(x)dx+2\left ( \int_{0}^{1}f(x)dx \right )^{2}\geq 3 \int_{0}^{1}f(x)dx \int_{0}^{1}xf(x)dx$

This really is the typical Cezar Lupu inequality.

Trying some functions we discover that for $f(x) \equiv 3x-1$ we have equality in
\[\int_{0}^{1}f^{2}(x) \, dx+8 \left( \int_{0}^{1}f(x) \, dx \right)^{2}\geq 12 \int_{0}^{1}f(x) \, dx \int_{0}^{1}x f(x) \, dx . \]
So, let's apply an inequality. But what inequality should that be? You've guessed it, it's Cauchy:
\[\int_{0}^{1}f^{2}(x) \, dx \int_{0}^{1}\left( x-\frac13 \right)^{2}\, dx \geq \left( \int_{0}^{1}f(x) \cdot \left( x-\frac13 \right) \, dx \right)^{2}, \]
i.e.
\[\int_{0}^{1}f^{2}(x) \, dx \geq \left( \int_{0}^{1}f(x) \cdot \left( 3x-1 \right) \, dx \right)^{2}. \]
Thus, it suffices to prove that
\[\left( \int_{0}^{1}f(x) \cdot \left( 3x-1 \right) \, dx \right)^{2}\geq 12 \int_{0}^{1}f(x) \, dx \int_{0}^{1}x f(x) \, dx-8 \left( \int_{0}^{1}f(x) \, dx \right)^{2}\]

\[\iff 9 \left( \int_{0}^{1}x f(x) \, dx \right)^{2}-6 \int_{0}^{1}x f(x) \, dx \int_{0}^{1}f(x) \, dx+\left( \int_{0}^{1}f(x) \, dx \right)^{2}\geq \]

\[\geq 12 \int_{0}^{1}f(x) \, dx \int_{0}^{1}x f(x) \, dx-8 \left( \int_{0}^{1}f(x) \, dx \right)^{2}\]

\[\iff 9 \left( \int_{0}^{1}x f(x) \, dx \right)^{2}+9 \left( \int_{0}^{1}f(x) \, dx \right)^{2}\geq 18 \int_{0}^{1}f(x) \, dx \int_{0}^{1}x f(x) \, dx , \]
which is true by $AM-GM$.

I hope I haven't made any stupid mistakes!

I think the solution is correct. Thanx a lot perfect_radio! This bloody problem kept annoying me since the olympiad :) !

Here is a similar but slightly different approach. Note that $1$ and $x-1/2$ are orthogonal as elements of $L^{2}[0,1]$. So, let $f=a+b(x-1/2)+g$ where $g$ is orthogonal to both $1$ and $x-1/2$. Then $\int_{0}^{1}f^{2}\ge a^{2}+\frac{b^{2}}{12}$, $\int_{0}^{1}f=a$ and $\int_{0}^{1}xf=\frac{a}{2}+\frac{b}{12}$, which brings us to the inequality
\[a^{2}+\frac{b^{2}}{12}+8a^{2}\ge 12 a\left(\frac{a}{2}+\frac{b}{12}\right)\]
or, which is the same,
\[3a^{2}+\frac{b^{2}}{12}-ab = 3\left(a-\frac{b}{6}\right)^{2}\ge 0\,..\]
It is clear from here that, to get equality, we must have $b=6a$, $g\equiv 0$, so $f(x)=a(1+6(x-1/2))=a(6x-2)$.

In general, when you see an inequality involving a few linear functionals of $f$ and $\int f^{2}$, it may be helpful to use some orthogonal system that is rich enough to allow you to express all the involved linear functionals in terms of the Fourier coefficients of $f$ with respect to this system. Then, instead of an inequality for functions, you'll get one for numbers.


Let $ a,b,c$ be positive numbers such that: $ abc \equal{}1$.Prove that:
$ \frac{\sqrt{a^2\plus{}ab\plus{}2b^2}}{c^3\plus{}3}\plus{}\frac{\sqrt{b^2\plus{}bc\plus{}2c^2}}{a^3\plus{}3}\plus{}\frac{\sqrt{c^2\plus{}ca\plus{}2a^2}}{b^3\plus{}3} \ge \frac{3}{2}$

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ abc \equal{} 1$.Prove that:
$ \frac {\sqrt {a^2 \plus{} ab \plus{} 2b^2}}{c^3 \plus{} 3} \plus{} \frac {\sqrt {b^2 \plus{} bc \plus{} 2c^2}}{a^3 \plus{} 3} \plus{} \frac {\sqrt {c^2 \plus{} ca \plus{} 2a^2}}{b^3 \plus{} 3} \ge \frac {3}{2}$[/quote]
I have an nice solution for this inequality.But with the following inequality:
$ \frac{\sqrt{a^2\plus{}ab\plus{}2b^2}}{c^4\plus{}4}\plus{}\frac{\sqrt{b^2\plus{}bc\plus{}2c^2}}{a^4\plus{}4}\plus{}\frac{\sqrt{c^2\plus{}ca\plus{}2a^2}}{b^4\plus{}4} \ge \frac{6}{5}$
I don't know it 's true or false? :(
 .I hope you can help me.Thanks :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ abc \equal{} 1$.Prove that:
$ \frac {\sqrt {a^2 \plus{} ab \plus{} 2b^2}}{c^3 \plus{} 3} \plus{} \frac {\sqrt {b^2 \plus{} bc \plus{} 2c^2}}{a^3 \plus{} 3} \plus{} \frac {\sqrt {c^2 \plus{} ca \plus{} 2a^2}}{b^3 \plus{} 3} \ge \frac {3}{2}$[/quote]
Dear quykhtn-qa1 !
Can you give some hint with this inequality,thanks.

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ abc \equal{} 1$.Prove that:
$ \frac {\sqrt {a^2 \plus{} ab \plus{} 2b^2}}{c^3 \plus{} 3} \plus{} \frac {\sqrt {b^2 \plus{} bc \plus{} 2c^2}}{a^3 \plus{} 3} \plus{} \frac {\sqrt {c^2 \plus{} ca \plus{} 2a^2}}{b^3 \plus{} 3} \ge \frac {3}{2}$[/quote]
It's easy because $ \sqrt{a^2\plus{}ab\plus{}2b^2}\geq\frac{3a\plus{}5b}{4}$
and $ \sum_{cyc}{\frac {a}{c^3 \plus{} 3}}\geq\frac {3}{4}$
and $ \sum_{cyc}{\frac {b}{c^3 \plus{} 3}}\geq\frac {3}{4}$  (It's old and easy by CS) :P

[quote="zaizai-hoang"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ abc \equal{} 1$.Prove that:
$ \frac {\sqrt {a^2 \plus{} ab \plus{} 2b^2}}{c^3 \plus{} 3} \plus{} \frac {\sqrt {b^2 \plus{} bc \plus{} 2c^2}}{a^3 \plus{} 3} \plus{} \frac {\sqrt {c^2 \plus{} ca \plus{} 2a^2}}{b^3 \plus{} 3} \ge \frac {3}{2}$[/quote]
I have an nice solution for this inequality.But with the following inequality:
$ \frac {\sqrt {a^2 \plus{} ab \plus{} 2b^2}}{c^4 \plus{} 4} \plus{} \frac {\sqrt {b^2 \plus{} bc \plus{} 2c^2}}{a^4 \plus{} 4} \plus{} \frac {\sqrt {c^2 \plus{} ca \plus{} 2a^2}}{b^4 \plus{} 4} \ge \frac {6}{5}$
I don't know it 's true or false? :(
 .I hope you can help me.Thanks :)[/quote]

The second is also correct  :wink:[/quote]
I think,if second true.It's very strong! :)
can you post your solution?thank you!


summer program
MathPath
pigeonhole principle
Duke
college
Stanford

Besides MathPath, what are some of theo ther good MS programs?

For middle school students, I highly recommend attending a program such as

Center for Talented Youth

instead of a program like MathPath.

I don't really think it's good to pigeonhole students at THAT young of an age.

(Disclaimer Yes, I do work for CTY, so there is a bias element..)

the one from Johns Hopkins??

what about Stanford's MS institute?

I think Stanford's one is great as well (JHU, Duke and Stanford all collaborate together).


calculus
derivative
function
induction
algebra
functional equation
real analysis

Find all fuctions $f: \mathbb{R}\to \mathbb{R}$ continuous at $0$ and  satisfying for all real $x,y$ the following equation
\[f(x+y)=f(x)+f(y)+xy(x+y) \]
[hide="the case when there exists $f'(0)$"]Let  $c=f'(0)$, then there exists $\exists f'(x)=x^{2}+c \Longrightarrow f(x)=\frac{x^{3}}{3}+cx$ is the only solution. What if the derivative at 0 does not exist :?:   [/hide]

why kunny?
what;s wrong with my answer?
and by the way how do I have to put the text to be hiden?

Plugging $x=y=0$ into the given functional equation gives the desired answer. This problem is made by the following equality.

$\boxed{(x+y)^{3}=x^{3}+y^{3}+3xy(x+y)}\Longleftrightarrow \frac{1}{3}(x+y)^{3}=\frac{1}{3}x^{3}+\frac{1}{3}y^{3}+xy(x+y).$

\[f(x+y)=\frac13 \cdot (x+y)^{3}+c(x+y)= \frac13 x^{3}+cx+\frac13 y^{3}+cy+xy(x+y)= f(x)+f(y)+xy(x+y) \]
SO $f(x)=\frac13x^{3}+cx$ is solution.

But how do we solve the problem if the [color=red]fuction is continuous in one point 0 and you dont know whether there exists or not[/color] $f'(0)$ :|

Define $g(x)$ so that $f(x)=\frac{x^{3}}3+g(x).$ Then the functional equation reduces to:

$g(x+y)=g(x)+g(y).$

This is a fairly famous problem, and it is known that the only continuous solutions are of the form $g(x)=cx,$ although there do exist discontinuous solutions (the "constuction" of such monsters requires the Axiom of Choice.)

Let $c=g(1).$ By induction, $g(n)=cn.$ By another induction, $g\left(\frac{m}{n}\right)=c\frac{m}{n}.$

Now let $x$ be any real number, and find a sequence $q_{n}$ of rational numbers which tends to $x.$

$g(x)=g(q_{n})+g(x-q_{n})=cq_{n}+g(x-q_{n}).$

Tak the limit. $cq_{n}$ tends to $cx$ while $g(x-q_{n})$ tends to $g(0)=0$ because $g$ is presumed continuous at zero. We get that $g(x)=cx.$

Since $g(x)=cx,$ we have that $f(x)=\frac{x^{3}}3+cx.$

Thank you!
I do had to figure it out from my previous post  :oops:


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Find the equation of the line tangent to $ y\equal{}f(x)$ at $ x\equal{}1$, where $ f(x)\equal{}6e^{5x} \plus{} e^{\minus{}x^2}$?

[hide="what I got"]The derivative of the equation above:
$ \frac{df(x)}{dx}\equal{}30e \minus{} \frac{2}{e}$

Plugging in $ x\equal{}1$ to $ f(x) \rightarrow 6e^4$.

So I set-up point slope form and got
$ y \equal{} (x\minus{}1)(30e \minus{} \frac{2}{e}) \plus{} 6e^4$

I got a feeling that it is wrong...[/hide]

[quote="jeez123"]Plugging in $ x \equal{} 1$ to $ \boxed{f(x) \rightarrow 6e^4}$.

So I set-up point slope form and got
$ y \equal{} (x \minus{} 1)(30e \minus{} \frac {2}{e}) \plus{} \boxed{6e^4}$

I got a feeling that it is wrong...[/hide][/quote]

 :idea:

[quote="jeez123"]Find the equation of the line tangent to $ y \equal{} f(x)$ at $ x \equal{} 1$, where $ f(x) \equal{} 6e^{5x} \plus{} e^{ \minus{} x^2}$?

$ f'(x) \equal{} 30e^{5x} \minus{} 2xe^{\minus{}x^{2}}$

$ f'(1) \equal{} 30e^{5} \minus{} 2e^{\minus{}1}$

Hence: $ m \equal{} 30e^{5} \minus{} 2e^{\minus{}1}$
          $ x_{0} \equal{} 1$
          $ y_{0} \equal{} 6e^{5} \plus{} e^{ \minus{} 1}$
          $ y \minus{} y_{0} \equal{} m(x \minus{} x_{0})$


Let $A_1B_1C_1$ be an acute triangle. $A , B , C$ are respectively belong to segments $B_1C_1 , C_1A_1 , A_1B_1$ such that : $\angle BAC = \angle B_1A_1C_1$ , $\angle ABC = \angle A_1B_1C_1$ , $\angle BCA = \angle B_1C_1A_1$ . Let $(O)$ be the circumcircle of $\triangle A_1B_1C_1$ . $H , H_1$ are respectively the orthocenters of $\triangle ABC$ , $\triangle A_1B_1C_1$ . Prove that : $OH =OH_1$

I found that : $(A_1BC) , (B_1AC)$ and $(C_1AB)$ have a common point namely $H$ . We also have $\R(A_1BC) =\R(AB_1C) =\R(ABC_1) =\R(ABC)$ (it can be prove by applying sin law )
  And it's easy to get $H$ is the circumcenter of $(A_1B_1C_1)$
  Could you please help me to finish it ?


Call a super-integer an infinite sequence of decimal digits: $\ldots d_n \ldots d_2d_1$.

(Formally speaking, it is the sequence $(d_1,d_2d_1,d_3d_2d_1,\ldots)$ )

Given two such super-integers $\ldots c_n \ldots c_2c_1$ and $\ldots d_n \ldots d_2d_1$, their product $\ldots p_n \ldots p_2p_1$ is formed by taking $p_n \ldots p_2p_1$ to be the last n digits of the product $c_n \ldots c_2c_1$ and $d_n \ldots d_2d_1$.
Can we find two non-zero super-integers with zero product?
 (a zero super-integer has all its digits zero)

This problem is probably from some older ISL or something like hat since it was used in a german TST some years ago.
The solution of Rust is incorrect since it doesn't give such sequence (or at least I don't see how it shall work).

The idea behind this is to work in the ring $\mathbb{Z}_2 \times \mathbb{Z}_5$ (being product of two rings of $p$-adic integers).
We are looking for some solution to $xy=0$ other than the trivial ones. Such solution exists, e.g. $(0,1) \cdot (1,0) = (0,0)$ in the ring above.
So taking some $x \equiv 1 \mod 5^n$ , $x \equiv 0 \mod 2^n$ and inversly $y$ will give us the result, and they will fulfill all we want (they are unique determined $\mod 10^n$, thus by highering $n$ the last digits will remain the same).


function induction trigonometry logarithms

If $ f(x)$ is a real valued function of the real variable $ x$, and $ f(x)$ is not identically zero, and for all $ a$ and $ b$ we have $ f(a \plus{} b) \plus{} f(a \minus{} b) \equal{} 2f(a) \plus{} 2f(b)$, then for all $ x$ and $ y$
(A) $ f(0) \equal{} 1$
(B) $ f( \minus{} x) \equal{} \minus{} f(x)$
(C) $ f( \minus{} x) \equal{} f(x)$
(D) $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
(E) There is a positive number $ T$ such that $ f(x \plus{} T) \equal{} f(x)$

Thanks in advance!

[hide="solution"]
$ a\equal{}b\rightarrow f(2b)\plus{}f(0)\equal{}4f(b)$
$ a\equal{}\minus{}b\rightarrow f(0)\plus{}f(\minus{}2b)\equal{} 2f(\minus{}b)\plus{}2f(b)$
$ f(2b)\minus{}f(\minus{}2b)\equal{}2f(b)\minus{}2f(\minus{}b)$
$ f(2b)\minus{}2f(b)\equal{}f(\minus{}2b)\minus{}2f(\minus{}b)$
Thus, $ f(b)\equal{}f(\minus{}b)$, or C.[/hide]

What kind of function would satisfy that relationship?

[hide="Discussion"] First of all, there is a more direct way to get the answer:  substitute $ b \equal{} \minus{}b$ on the LHS and you get $ f(a \minus{} b) \plus{} f(a \plus{} b) \equal{} 2f(a) \plus{} 2f(\minus{}b)$, which we already know is equal to $ 2f(a) \plus{} 2f(b)$.

One solution is $ f(x) \equal{} cx^2$.  Are there others? [/hide]

t0rajir0u, to add on to what you said, substituting $ b \equal{} a$ yields $ f(0) \plus{} f(2a) \equal{} 4 f(a)$. Substituting $ a \equal{} 0$ yields $ 2f(0) \equal{} 4 f(0)$, implying that $ f(0) \equal{} 0$, so $ f(2a) \equal{} 4 f(a)$, so $ f(a) \equal{} f(2a) / 4$.

If we let $ f(1) \equal{} c$, it follows from induction that $ f(2^k) \equal{} 4^k c$ for all integers $ k$ (positive and negative.)

Not sure where else we can go from there... other than it's even.

I wouldn't spend too much time. I'm almost positive that the same kind of noncontinuous solutions that the Cauchy equation has will show up here. But I think you can get $ f(x) \equal{} cx^2$ if the function is defined on rationals without too much trouble.

[quote="TZF"]Rephrase that as, if $ x$ is a power of $ 2$, then $ f(x) \equal{} cx^2$. It is then natural to see if this relationship also holds true if $ x$ is not necessarily a power of $ 2$.[/quote]
This is not at all the case, and you don't even need Cauchy-type solutions to see this.  Consider the function $ f(x) \equal{} c \left( \cos (2 \pi \log_2 |x| ) \right)^2$.


calculus
integration
logarithms
calculus computations

Observe that $ x^{4n}+1 = (x^{2n}-\sqrt{2}x^{n}+1)(x^{2n}+\sqrt{2}x^{n}+1)$.

The rest is easy.

The result I obtained is:

\[  \frac{\sqrt{2}}{4n}\left( \ln(x^{2n}-\sqrt{2}x^{n}+1)-\ln(x^{2n}+\sqrt{2}x^{n}+1) \right)\]

show your work even though the answer is right

can someone work this fully work this out?

[quote="mdk"]Ascertain the indefinite integral $ I=\int\frac{x^{3n-1}-x^{n-1}}{x^{4n}+1}\ dx$[/quote]
[color=darkblue][b]Proof.[/b] Observe that $ I=\int\frac{x^{n-1}\left(x^{2n}-1\right)}{x^{4n}+1}\ dx=\frac{1}{n}\cdot\int\frac{\left(x^{n}\right)^{2}-1}{\left(x^{n}\right)^{4}+1}\cdot\left(x^{n}\right)'\ dx$. Using the substitution $ y=x^{n}$ obtain $ I=\frac{1}{n}\cdot K\left(x^{n}\right)+\mathcal C$, where
$ K=\int\frac{y^{2}-1}{y^{4}+1}\ dy=\int\frac{1-\frac{1}{y^{2}}}{y^{2}+\frac{1}{y^{2}}}\ dy=\int\frac{\left(y+\frac{1}{y}\right)'}{\left(y+\frac{1}{y}\right)^{2}-2}\ dy$. Using the substitution $ z=y+\frac{1}{y}$ obtain $ K=L\left(y+\frac{1}{y}\right)+\mathcal C$, where
$ L=\int\frac{1}{z^{2}-2}\ dz=\frac{1}{2\sqrt 2}\ln\left|\frac{z-\sqrt 2}{z+\sqrt 2}\right|+\mathcal C$. Therefore, $ K=\frac{\sqrt 2}{4}\cdot \ln\frac{y^{2}-y\sqrt 2+1}{y^{2}+y\sqrt 2+1}+\mathcal C$ and $ \boxed{I=\frac{\sqrt 2}{4n}\cdot\ln\frac{x^{2n}-x^{n}\sqrt 2+1}{x^{2n}+x^{n}\sqrt 2+1}+\mathcal C}$.

[b]Remark.[/b] Can apply directly the substitution $ t=x^{n}+\frac{1}{x^{n}}$ and obtain $ I=\frac{1}{n}\cdot\int \frac{1}{\left(x^{n}+\frac{1}{x^{n}}\right)^{2}-2}\cdot\left(x^{n}+\frac{1}{x^{n}}\right)'\ dx$, i.e. $ I=\frac{\sqrt 2}{4n}\cdot\ln\frac{x^{2n}-x^{n}\sqrt 2+1}{x^{2n}+x^{n}\sqrt 2+1}+\mathcal C$.[/color]

[quote="Virgil Nicula"]
[b]Proof.[/b] Observe that $ I=\int\frac{x^{n-1}\left(x^{2n}-1\right)}{x^{4n}+1}\ dx=\frac{1}{n}\cdot\int\frac{\left(x^{n}\right)^{2}-1}{\left(x^{n}\right)^{4}+1}\cdot\left(x^{n}\right)'\ dx$. Using the substitution $ y=x^{n}$ obtain $ I=\frac{1}{n}\cdot K\left(x^{n}\right)+\mathcal C_{1}$, where
$ K=\int\frac{y^{2}-1}{y^{4}+1}\ dy=\int\frac{1-\frac{1}{y^{2}}}{y^{2}+\frac{1}{y^{2}}}\ dy=\int\frac{\left(y+\frac{1}{y}\right)'}{\left(y+\frac{1}{y}\right)^{2}-2}\ dy$. Using the substitution $ z=y+\frac{1}{y}$ obtain $ K=L\left(y+\frac{1}{y}\right)+\mathcal C_{2}$, where
$ L=\int\frac{1}{z^{2}-2}\ dz=\frac{1}{2\sqrt 2}\ln\left|\frac{z-\sqrt 2}{z+\sqrt 2}\right|+\mathcal C_{3}$. Therefore, $ K=\frac{\sqrt 2}{4}\cdot \ln\frac{y^{2}-y\sqrt 2+1}{y^{2}+y\sqrt 2+1}+\mathcal C_{2}+\mathcal C_{3}$ and $ \boxed{I=\frac{\sqrt 2}{4n}\cdot\ln\frac{x^{2n}-x^{n}\sqrt 2+1}{x^{2n}+x^{n}\sqrt 2+1}+\mathcal C_{1}+\mathcal C_{2}+\mathcal C_{3}}$.
[/quote]

nice !  :coolspeak:


I really want to know how to prove Holder ineqality……This maybe the most simple one:
有实数列$a_1,a_2,a_3...a_n$与$b_1,b_2,b_3...b_n$以及$\frac{1}{p}+\frac{1}{q}=1$
求证:$({a_1}^p+{a_2}^p+...+{a_n}^p)^{\frac{1}{p}}({b_1}^q+{b_2}^q+...+{b_n}^q)^{\frac{1}{q}} \geq |a_1b_1+a_2b_2+...+a_nb_n|$

One approach is to use Young's inequality.

As mentioned above, Young's inequality can be used.
If p,q are positive such that $\frac{1}{p}+\frac{1}{q}=1$ and a,b are positive, then $\frac{a^p}{p}+\frac{b^q}{q} \ge ab$
Proof: This is directly true by using the fact that ln is concave.

Now, without loss of generality, assume $\sum a^p = 1$ and $\sum b^q=1$, then by young's inequality,
$\sum a_1b_1 \le \sum \frac{a_1^p}{p} + \frac{b_1^q}{q} = \frac{1}{p}+\frac{1}{q} = 1$


when you get a problem wrong/right it will show u a solution that you get to rate. are there multiple DIFFERENT solutions for each problem?

There are a lot of ways to solve most probability problems, and the solution usually states the easiest method for the specific problem.

If you find an easier solution, you can post it on the forum too!  :)

[quote="007math"]I don't think there are different solutions that they show, but most problems could be solved a unique way than the solutions provided.[/quote]

yeah they could
different solutions would be easier to understand for different ppl

If you believe that there is a better solution or a solution that can be better understood, you could always suggest it on the forum, and maybe an administrator would consider it as an alternate solution.

We have multiple solutions on the "to be added" list for Alcumus. I would love to see this as well since not only may one solution be easier than another, it's helpful to understand how you can solve problems in different ways.


AMC
USA(J)MO
USAMO
modular arithmetic
algebra
polynomial
limit

It is a nice and famous problem (given at 1995 USAMO and, I believe, some Russian olympiad before this): let $f : \mathbb{N} \to \mathbb{Z}$ be such that $f(x+m) \equiv f(x) \pmod{m}$ for all $x,m \in \mathbb{N}$. If $f(n)$ is bounded by a polynomial (that is, there are constants $C,k$ such that $|f(n)| < Cn^k$ for all $n$), then $f(n)$ [b] is[/b] a polynomial (that is, there exists a polynomial $Q \in \mathbb{Q}[x]$ such that $f(n) = Q(n)$ for all $n$).

Now, I propose a stronger result: it suffices that $f$ is bounded [i] exponentially[/i]!

[b]Claim: [/b] Let $f : \mathbb{N} \to \mathbb{Z}$ be such that $f(x+m) \equiv f(x) \pmod{m}$ for all $x,m \in \mathbb{N}$. Suppose that there exists a constant $A > 0$ such that $|f(n)| < A^n$ for all $n$. Then, $f(n)$ must be a polynomial in $n$.

--Vesselin

I think this is true if and only if $A<e$. Let $l_n=lcm(1,2,\ldots,n)$

Firstly, we prove the following lemma:

Lemma 1: $\lim_{n \rightarrow \infty} log_n l_n=e$.

Proof: $l_n=\prod_{p<n, p^k\leq n<p^{k+1}} p^k$. As there are $\sim \frac{n}{\ln n}$ primes not exceeding $n$, and each $p^k$ is at most $n$, we get $l_n\leq e^n$.
From the other sides, $p_n|l_n$, where $p_n$ is the product of all primes not exceeding $n$. take now a $\beta<1$. There are $\sim \frac{n}{\ln n}-\frac{n^{\beta}}{\ln {n^{\beta}}}\sim \frac{n}{\ln n}$  prime numbers between $n$ and $n^{\beta}$.
Therefore $p_n\geq n^{\beta \cdot \frac{n}{\ln n}}=e^{\beta n}$. Taking $\beta \rightarrow 1$ we can deduce the claim.

I even guess that we can prove that $2l_n<e^{n}$, which would disprove the claim for $A=e$, but I don't know yet whether this is true.

This allows us to disprove the claim for $A>e$ immediately: construct $a_n$ recurrently: take $a_{n+1}=b_{n+1}+l_n<A^n$ where $b_{n+1}$ is the smallest integer to satisfy $b_{n+1}\equiv a_k \pmod{n+1-k}$.

Obviosly $a_n$ can't be a polynomial, as it is asimptotically exponential.

However, for $A<e$ the claim is true:

Lemma 2: For $p\leq n$, $\sum_{i\equiv k \pmod{p}} (-1)^k \binom {n}{k}\equiv 0 \pmod{p}$($p$ an odd prime).

This is proven easily by induction using Pascal Recurrence, as for $n=p$ it's obviously true.

Consider the interpolation polynomial $p$ which gives $p(0)=a_0, \ldots,p(n)=a_n$. It's easy to check that $p(n+1)=\sum_{i=0}^n \binom {n+1}{k} (-1)^{k-1} a_k$.

Now applying the lemma $2$, we deduce that $p(n+1)\equiv a_{n+1-p} \pmod{p}$(because for each residue modulo $p$ the congruent numbers modulo $p$ appear a number of times divisible by $p$) for $p<n$.
 Hence $a_{n+1}-p(n+1)$ is divisible by $p(n)$. Moreover, $p(n+1)<A^{n+1}$, which is easily proven.
Now $a_{n+1}<A^{n+1}$ hence $|a_{n+1}-p(n+1)|<A^{n+1}$. But it's divisible by $p_n$, and in the process of proving Lemma 1 we've shown that $p_n>A^n$ for all $A<e$, $n$ sufficiently big.
Hence $|a_{n+1}-p(n+1)|=0$ so $a_{n+1}=p(n+1)$. Repeating this argument for $n+1$ we get $a_{n+2}=p(n+2)$ etc.

Happy Easter!

Happy Easter, Iura!

You are right. My approach was exactly the same, but my mistake was in assuming a congruence of the form $\sum_{k \equiv i \pmod{q}} \binom{n}{k}(-1)^k \equiv 0 \pmod{p^{\lfloor (n-1)/(q-1) \rfloor}}$ for prime powers $q=p^a$, but actually this only holds for $q=p$ ... Sorry about that.

Nevermind, here is another problem :D:

For an odd prime $p$ and any residue $k_0 \pmod{p}$, show that
\[ \sum_{k \equiv k_0 \pmod{p}} \binom{n}{k}(-1)^k \equiv 0 \pmod{p^{\lfloor \frac{n-1}{p-1} \rfloor}}. \]

I remembered this thread, and believe we had made an error here: iura's proof only works for $ A < e \minus{} 1$, and not for any $ A < e$. The elementary result proven in this thread is thus: [i] Let $ A < e\minus{}1$. If $ a_n, n \geq 1$ is a sequence of integers such that $ |a_n| < A^n$ for $ n \gg 0$, and $ a_{n\plus{}p} \equiv a_n \pmod{p}$ for all primes $ p$ and all $ n$, THEN $ a_n$ is a polynomial in $ n$. [/i]

But the stronger question remains unsettled: if $ a_n$ is a sequence with the mentioned periodicity-mod-$ p$ property which is NOT a polynomial in $ n$, must $ \lim\sup \frac{\log{|a_n|}}{n} \geq 1$?

This was a conjecture by Ruzsa; I don't know if it has been settled. (Anyone?)

Here is the gap in iura's proof which needs the stronger condition $ A < e\minus{}1$ (instead of just $ A < e$):

[quote="iura"]
Consider the interpolation polynomial $ p$ which gives $ p(0) \equal{} a_0, \ldots,p(n) \equal{} a_n$. It's easy to check that $ p(n \plus{} 1) \equal{} \sum_{i \equal{} 0}^n \binom {n \plus{} 1}{k} ( \minus{} 1)^{k \minus{} 1} a_k$.

Now applying the lemma $ 2$, we deduce that $ p(n \plus{} 1)\equiv a_{n \plus{} 1 \minus{} p} \pmod{p}$(because for each residue modulo $ p$ the congruent numbers modulo $ p$ appear a number of times divisible by $ p$) for $ p < n$.
 Hence $ a_{n \plus{} 1} \minus{} p(n \plus{} 1)$ is divisible by [edit: $ \prod_{\substack{p < n \\ prime}} p$ -- Vess]. Moreover, $ p(n \plus{} 1) < A^{n \plus{} 1}$ [!!! -- Vess], which is easily proven.
[/quote]

Now, note the [!!!]. It is not true that $ |p(n\plus{}1)| < A^{n\plus{}1}$, from $ p(n\plus{}1) \equal{} \sum_{i\equal{}0}^n \binom{n\plus{}1}{k} (\minus{}1)^{k\minus{}1}a_k$ and $ |a_k| < A^k$! Instead, these inequalities imply $ |p(n\plus{}1)| < \sum_{i\equal{}0}^n \binom{n\plus{}1}{k} |a_k| < \sum_i \binom{n\plus{}1}{k} A^k \equal{} (A\plus{}1)^{n\plus{}1}$; and to ensure that this quantity is smaller than $ \prod_{\substack{p < n} \\ \text{prime}} p \sim e^n$ for $ n \gg 0$, we need to have $ A < e\minus{}1$.

The result does not necessarly hold when $ limsup |a_n|^{\frac 1 n} \leq e$. This appears in [1].
Another reference about this kind of problem is [2].

[1] A.M.M. pb 10185 (1994, p.85) and 10553 (1999, p.68-69)
[2] D. Callan, "When does $ m\minus{}n$ divides $ f(m)\minus{}f(n)$? A look at column-finite matrices", Fibonacci Quaterly, 1997, p.290-299.

Pierre.


geometry
rectangle
integration
calculus
calculus computations

Consider the intersection of the line $ \ell_k : y\equal{}k$ and the curve $ C: y\equal{}x^2\plus{}1$. Revolve the rectangle with vertices at the origin and the intersection of $ \ell_k$ and $ C$ about the $ x$ axis. Find all values of $ k$ for which the volume generated by the region $ A$ is equal to the volume generated by the region $ B$.

Here's my work.  Let me know if there are errors in it.

$ l_k$ and $ C$ intersect at $ x \equal{} \sqrt {k \minus{} 1}$, and the given problem implies the condition $ k > 1$ for the region $ A$ to exist.

The volume of the cylinder formed when the total rectangular region ($ A \plus{} B$) is revolved about the $ x$-axis is $ \pi k^2 \sqrt(k \minus{} 1)$.

Thus, the sume of the volume generated by $ A$, $ V_A$, and the volume generated by $ B$, $ V_B$, is

$ V_A \plus{} V_B \equal{} \pi k^2 \sqrt {k \minus{} 1}$

Thus, when $ V_A \equal{} V_B$ (which the problem asks for), $ V_B \equal{} \frac {\pi}{2} k^2 \sqrt {k \minus{} 1}$.

$ V_B$ can be found using the disk method:

$ V_B \equal{} \pi \int_0^{\sqrt {k \minus{} 1}} (x^2 \plus{} 1)^2 \ dx$

This can be evaluated to get

$ V_B \equal{} \frac {\pi}{15} \sqrt {k \minus{} 1} \ (3k^2 \plus{} 4k \plus{} 8)$

Using the two expressions for $ V_B$, solve for $ k$.

$ \frac {\pi}{15} \sqrt {k \minus{} 1} \ (3k^2 \plus{} 4k \plus{} 8) \equal{} \frac {\pi}{2} k^2 \sqrt {k \minus{} 1}$

Solving this yields a few extraneous solutions, leaving a final answer of

$ k \equal{} \frac {4}{9} \plus{} \frac {4 \sqrt {10}}{9}$


inequalities trigonometry geometry circumcircle inradius logarithms function

Let [tex]A,B,C[/tex] be the angles of a triangle, Prove that

[tex]\sin{A/2}\sin{B/2}\sin{C/2}\leq1/8[/tex]

(This is from p.6 of Kiran Kedlaya's MOP notes on inequalities)

Note: This inequality implies that [tex]r\leq{1/2R}[/tex], where [tex]R[/tex] is the circumradius of [tex]ABC[/tex] and [tex]r[/tex] is the inradius of [tex]ABC[/tex]. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=86151#86151]here[/url] for a proof of the identity [tex]\sin{A/2}\sin{B/2}\sin{C/2}=r/4R[/tex]

[hide]Taking ln of both sides and using a property of logs we get [tex]\ln\sin A/2 + \ln\sin B/2 + \ln\sin C/2 \leq 3(\ln 1/2)[/tex]. Note that the function [tex]f(x) = \ln\sin(x) [/tex] has second derivative [tex] -1/\sin(x)^2[/tex] which is always negative, so that the function is concave. Then we apply Jensen's Inequality to the function to get [tex] \ln\sin A/2 + \ln\sin B/2 + \ln\sin C/2 \leq 3( \ln(A+B+C)/6) = 3(\ln 30) = 3(\ln 1/2)[/tex], which is what we wanted to show.[/hide]

Nice proof. There is an elementary proof which doesn't require knowing the concavity of the function [tex]\ln\sin x[/tex] (which requires calculus to prove). Actually, [tex]1-8\sin A/2\sin B/2\sin C/2[/tex] can be written as a sum of squares.

Dont post the second proof yet, Im going to try it. Does the concavity of ln(sin(x)) really require calc? Of course its very easy with calc. Is there another method?

[quote="tetrahedr0n"]Does the concavity of ln(sin(x)) really require calc? Of course its very easy with calc. Is there another method?[/quote]

I have no idea.

Concavity is essentially a property of the derivatives -- the only other way to define concavity that I know of is in terms of Jensen's inequality, which obviously isn't helpful in this case.

There are other ways to prove concavity. In fact you can prove concavity for things that aren't even differentiable (though concavity implies continuity).

The basic definition of concavity is a function for which t*f(a) + (1-t)*f(b) :le: f(t*a + (1-t)*b) for any a and b and any t betwen 0 and 1. [It turns out that this stronger statement is equivalent to just letting t be any fixed number between 0 and 1 (say 1/2 for instance)].

For most functions, proving that it is concave without using calculus is not easy. But it is possible.

Gauss202 -- Can you give an example of a function for which is it not sufficient to let t=1/2? (That is, can you come up with a function that would appear to be concave if you were only to check t=1/2 that it not actually concave?)

Yes. If f is not continuous then you can define it to be non-concave on the span of some basis element of R over Q, while keeping it concave on the span of the other bases. So I guess I should say - if f is continuous then you can take t to be some fixed number between 0 and 1 (such as 1/2). Good point!


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Show that $ \bar{\mathbb{Q}}$ is not a finite extension of $ \mathbb{Q}$.

Should I prove that the roots are linearly independent over $ \mathbb{Q}$?

They aren't; $ \sqrt{2} \plus{} (\minus{}\sqrt{2}) \equal{} 0$.

What you want to do is just add one of the roots and find the dimension.

I actually meant the positive roots, my mistake.I think those are independent.

But I see how it's easier just to adjoin a root.

I have another question, not exactly on the same subject but close enough.

Say you have $ F$ a field and $ E_1$, $ E_2$ finite extensions of that field. Will one extension be necessarily a subfield of the other or at least contain a subfield isomorphic to the other? I think the answer is yes but I cannot prove it.

$ F \equal{} \mathbb{Q}, E_1 \equal{} \mathbb{Q}[ \sqrt {2} ], E_2 \equal{} \mathbb{Q}[ \sqrt {3} ]$.  Generally, field extensions form a [url=http://en.wikipedia.org/wiki/Lattice_(order)]lattice[/url], not a chain.

1.Would this mean that if extensions $ E_1$ and $ E_2$ would have different dimensions over $ F$ then one would be the subfield of the other?
2. Actually this question arose from the following exercise: Argue that every finite field extension $ E$ of $ \mathbb{R}$ is either $ \mathbb{R}$ or is isomorphic to $ \mathbb{C}$.

I solved the problem using uniqueness of algebraic closure but I still can't answer the above question which in this case would be: If I assume that $ [E: \mathbb{R}]\equal{}3$ does this imply that $ E\geq \mathbb{C}$?

$ E_2 \equal{} \mathbb{Q}[x]/(x^3 \minus{} x \minus{} 1)$.  Lattices don't behave like that, either.  The standard example to keep in mind is the lattice of subsets of a given set:  two sets can have different cardinalities without one being the subset of the other.  If you're more number-theoretically inclined, think of the lattice of divisors of a given positive integer.

The given property of $ \mathbb{R}$ is a consequence of the fundamental theorem of algebra.  Every finite field extension of $ \mathbb{R}$ is of the form $ \mathbb{R}[x]/(P(x))$ for some polynomial $ P$ which is irreducible over $ \mathbb{R}$ if and only if $ P$ is a quadratic with negative discriminant or linear; in other words, $ \mathbb{R}$ can't have any finite field extensions of degree greater than $ 2$.  Now show that the extensions of degree $ 2$ are all isomorphic to $ \mathbb{C}$.

In any case, this is a very special property of $ \mathbb{R}$:  the lattice of finite field extensions of $ \mathbb{Q}$ is very complicated indeed.


function
algebra
domain
induction
limit
algebra unsolved

Who is n, and who is de domain and image of the function?

I think $f: N\to N$. In this case it is very easy $f(n)=n$.
Consider n=1 $f(f(f(n)))+f(f(n))+f(n)\ge 3$ if  is equality then f(1)=1. If f(2)=a>2, then f(a) or f(f(a))=1. It give $ffff(a)+fff(a)+ff(a)<3f(a)$ contradition. By induction $f(n)=n \ \forall n\in N$.

I have prove that if the condition is $f: R^{+}\to R^{+}$then the solution is also $f(x)=x$
first,$f(x)<3x$
then $f(x)+ff(x)+fff(x)<(1+3+9)f(x)$then $f(x)>\frac{1}{13}x$
$\dots \dots \dots$
we can find two sequence $a_{n},b_{n}$in this way such that $a_{n}x\leq f(x)<b_{n}x$ and have the property :$\lim_{n\to \inf}{b_{n}}=\lim_{n\to \inf}{a_{n}}=1$

[quote="Hawk Tiger"]I have prove that if the condition is $f: R^{+}\to R^{+}$then the solution is also $f(x)=x$
first,$f(x)<3x$
then $f(x)+ff(x)+fff(x)<(1+3+9)f(x)$then $f(x)>\frac{1}{13}x$
[/quote]
I think it is not true.

Thank you for pointing out my mistake.  :oops:
But  I think the solution can be modified :
first,$f(x)<3x$ $ff(x)<3x$
then $f(x)+ff(x)+fff(x)<(1+3+3)f(x)$then $f(x)>\frac{1}{7}x$
then$3x=f(x)+ff(x)+fff(x)>\frac17x+\frac87ff(x)$ $ff(x)<\frac52x$
$\dots \dots \dots$
we can find two sequence $a_{n},b_{n},c_{n}.d_{n}$in this way such that $a_{n}x\leq f(x)<b_{n}x$$c_{n}x<ff(x)<d_{n}x$ and have the property :$\lim_{n\to \inf}{b_{n}}=\lim_{n\to \inf}{a_{n}}=1$
$\lim_{n\to \inf}{c_{n}}=\lim_{n\to \inf}{d_{n}}=1$
I don't have enough time to post the detail now,I 'll post a complete solution soon.
I hope this method is correct now :)

To rust,thank you for the complete solution. :)
I guess if the condition is $f: R\to R$ and $f(x)$is continuous,there is also exactly one solution$f(x)=x$,(there is a another similar well-known problem if $f: R\to R$,and f is continuous,and $ff(x)+f(x)=2x$then f(x)=x)
But I can't solve this supposition,Would you like to have a try?

[quote="Hawk Tiger"] if $f: R\to R$,and f is continuous,and $ff(x)+f(x)=2x$then f(x)=x)
[/quote]
It is not true. For example f(x)=-2x is solution.

Oh,I am wrong again. :D ,it should be $ff(x)+x=2f(x)$ then $f(x)=x$
Sorry.


trigonometry
geometry
perimeter
trig identities
Law of Sines

Consider triangle $ ABC$. If $ AB\equal{}5$ and $ AC\equal{}8$ and $ sin C\equal{}0.5$, then what is the value of $ cos B$, given that triangle $ ABC$ is obtuse?
$ A. \minus{}\frac{3}{5}$
$ B. \minus{}\frac{4}{5}$
$ C. \frac{4}{5}$
$ D. \frac{3}{5}$
[hide="Why doesnt this work?"]
Using law of sines, we have
$ \frac{sin B}{8}\equal{}\frac{sin C}{5}$.
$ 5sinB\equal{}4$ ==>  $ sin B\equal{}\frac{4}{5}$, so $ cos B\equal{}\frac{3}{5}$. Now because the problem gave that the triangle is obtuse, one would think to make this answer negative, since cos in the 2nd quadrant is negative. However we do not know the length of the third side, and since there can be only one obtuse angle in a triangle, we don't know if that's longer then 8. So using law of cosines...
$ 25\equal{}64\plus{}BC^{2}\minus{}2(8)(BC)(cos30)$
$ 25\equal{}64\plus{}BC^{2}\minus{}8\sqrt{3}BC$
$ BC^{2}\minus{}8\sqrt{3}BC\plus{}39\equal{}0$
$ \frac{8\sqrt{3}\pm\sqrt{192\minus{}4(1)(39)}}{2}$
$ \frac{8\sqrt{3}\pm\sqrt{36}}{2}$
$ 4\sqrt{3}\plus{}3$, which is indeed greater than 8 so the obtuse angle doesn't belong to 8, meaning the angle B is acute, so cosine would be positive.[/hide]

whats wrong with ur problem? it just shows that angle B is acute=36.87, so leaving angle A the obtuse angle

You don't need to use law of cosines.
Simply, if $ \sin B \equal{} \frac {4}{5}$, just use the famous identity $ \sin^2 B \plus{} \cos^2 B \equal{} 1$.

$ \sin^2 B \plus{} \cos^2 B \equal{} 1$
$ \Leftrightarrow \left( \frac {4}{5}\right)^2 \plus{} \cos^2 B \equal{} 1$
$ \Leftrightarrow \cos^2 B \plus{} \frac {16}{25} \minus{} 1 \equal{} 0$
$ \Leftrightarrow \cos^2 B \minus{} \frac {9}{25} \equal{} 0$
By factoring,
$ \Leftrightarrow \left( \cos B \minus{} \frac {3}{5}\right) \left(\cos B \plus{} \frac {3}{5} \right) \equal{} 0$
So, the value of $ \cos B$ is $ \frac {3}{5}$ or $ \minus{} \frac {3}{5}$. But since $ \angle{B}$ is obtuse, it can be concluded that $ \cos B \equal{} \minus{} \frac {3}{5}$. Finally, $ A$ is the answer.

I have another problem about triangle.

The perimeter of triangle $ PQR$ is $ 15$ centimeters. $ PQ$ is $ 2$ centimeters longer than $ QR$. If $ \angle{Q}\equal{}\frac{2\pi}{3}$, find the length of each side of this triangle.

You know , you could just have drawn the triangle and all of it would have been clear. The length of the 3rd side is
3+ 4*3^0.5.
And so because B=37degrees. cos37 is 3/5.


The greedy algorithim can be helpful.  I don't remember it off the top of my head though.

I researched the Greedy Algorithm and that led me to Egyptian fractions, which is what I think we are dealing with. But I am still not understanding the steps to how to solve these equations.

Thanks for the lead, though.

If the numbers don't have to be distinct than this is trivial since
$\frac{1}{x} + \frac{1}{y} = \frac{1}{N}$
has a solution where $x = y = 2N$ since we get
$\frac{1}{x} + \frac{1}{x} = \frac{2}{x} = \frac{1}{N}$

Did you want the solution for any number of variables instead??

[quote="vinny919"]Did you want the solution for any number of variables instead??[/quote]

Yes.

Assuming that each number has to be distinct, you can use the greedy algorithm. While the algorithm might not give you an unique solution, it should give you one. Also, you cannot really control how many fractions are in the end result using this algorithm (as far as I know), although you might be able to split another fraction up a bit more.

In your example:

$\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$

The greatest fractions smaller than $\frac{1}{7}$ that is of the form $\frac{1}{n}$ where $n\geq1$ is $\frac{1}{8}$. Subtracting $\frac{1}{8}$ from $\frac{1}{7}$ gives us the other fraction as $\frac{1}{56}$.

$\frac {1}{x} + \frac{1}{y} = \frac 17$

$\frac{x+y}{xy} = \frac 17$

$7x + 7y = xy$

can't you do that?

A [hide="General Solution"]$\frac 1x+\frac1y=\frac1N$

$\frac yy \cdot \frac 1x+\frac xx \cdot \frac1y=\frac 1N$

$\frac{y}{xy}+\frac{x}{xy}=\frac 1N$

$\frac{y+x}{xy}=\frac 1N$

$N(x+y)=xy$

$Nx+Ny=xy$

$Nx=xy-Ny$

$Nx=(x-N)y$

$y=\frac{Nx}{N-x}$[/hide]

We can rewrite $\frac{7}{m}+\frac{7}{n}=1$, we are to imagine the form of intercept to equation of the line.

[quote="ankur87"]I do not know what they are called but something like this:

Find integers $m$ and $n$ such that $\frac{1}{m}+\frac{1}{n} = \frac{1}{7}$

My answer is that m and n both equal 14. That seems to easy and I don't know if they have to be separate integers.
I recall reading that, for similar problems, there is a way to solve these by starting with $\frac{1}{2}$ or something like that.

If anyone knows what I'm talking about (the name or the method of solving), I'd appreciate a little help

Thanks[/quote]

[/hide=Solution]
1/m + 1/n = 1/7

7m + 7n = mn
mn - 7m - 7n = 0
mn - 7m - 7n + 49 = 49
(m - 7)(n - 7) = 49

Yeah.
[/hide]
[/hide]

[quote="Treething"][quote="ankur87"][hide]I do not know what they are called but something like this:

Find integers $m$ and $n$ such that $\frac{1}{m}+\frac{1}{n} = \frac{1}{7}$

My answer is that m and n both equal 14. That seems to easy and I don't know if they have to be separate integers.
I recall reading that, for similar problems, there is a way to solve these by starting with $\frac{1}{2}$ or something like that.

If anyone knows what I'm talking about (the name or the method of solving), I'd appreciate a little help

Thanks[/hide][/quote]

[hide="Solution"]
1/m + 1/n = 1/7

7m + 7n = mn
mn - 7m - 7n = 0
mn - 7m - 7n + 49 = 49
(m - 7)(n - 7) = 49

Yeah.
[/hide]
[/quote]In general the equation $\frac1x+\frac1y=\frac1n$ for a given $n$ and positive integer unknowns $x,y$ has as many different solutions as factors has $n$ (up to permutation of the terms). The solution given by Treething is easely generalizable $\frac1x+\frac1y=\frac1n\Rightarrow n(x+y)=xy \Rightarrow xy-(x+y)+n^2=(x-n)(y-n)=n^2$. If we assume that $x\leq y$, then $(x-n)$ is a factor of $n$.

[quote="ankur87"]I do not know what they are called but something like this:

Find integers $m$ and $n$ such that $\frac{1}{m}+\frac{1}{n} = \frac{1}{7}$
[/quote]

I think we call such a fraction as Egyptian Fraction.

By the procedure explained by some guys have explained, the answer is $(m,n)=(8,56),(14,14),(56,8),(6,-42),(-42,6)$


[quote="silouan"]If $x_i>0$  and $x_iy_i-z_i^2>0$  for $i<0$ then
$\frac{n^3}{(\sum_{i=1}^nx_i)(\sum_{i=1}^ny_i)-(\sum_{i=1}^nz_i)^2}\leq\sum{\frac{1}{x_iy_i-z_i^2}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59017


So I was browsing through a Bleach fansite forum one day and stumbled upon this interesting exchange: (May contain spoilers regarding the actual manga)

http://forums.bleachportal.net/showpost.php?p=3618592&postcount=278
http://forums.bleachportal.net/showpost.php?p=3618628&postcount=279
http://forums.bleachportal.net/showpost.php?p=3620430&postcount=357
http://forums.bleachportal.net/showpost.php?p=3620642&postcount=360
http://forums.bleachportal.net/showpost.php?p=3620880&postcount=369 <--- Interesting tidbit
http://forums.bleachportal.net/showpost.php?p=3620899&postcount=371
http://forums.bleachportal.net/showpost.php?p=3620914&postcount=372 ...and the rest of them shuts up after this.

Post other stupidity you found on the internet here! No Bash/QDB/FML or similar unless it's work-safe.

Thank you for linking us to this very interesting discussion on your favorite anime message board.  I, for one, had no idea that the internet was full of such ill-informed persons!

You should see the flame wars that erupt about whether 0.9999999999...=1

http://www.casiocalc.org/?showtopic=1757&st=0

I randomly googled this and got the above...

Feeling an urge to split my hair as going through their discussion...

Why bother? Anybody who isn't curious enough to look it up themselves obviously isn't going to be doing anything in life that requires knowing that 0.9999... = 1. So if your goal is to improve people's general understanding of mathematics, then trying to explain it to these folks isn't a good use of your time.

If that's not your goal, then stop bashing your head over it -- as a matter of fact, stop thinking about it. Go do something more useful (or at least something fun).


If the lengths of a kite's diagonals are 6 cm and 12 cm, can one of the kite's sides be twice as long as another? Prove your answer.

If the sides are $ a$ and $ 2a$, and if, for example, the diagonals bisect each other at right angles into two equal line segments of length $ 3$ and two line segments of lengths $ b$ and $ 12 \minus{} b$, then by Pythagoras,

$ a^{2} \equal{} b^{2} \plus{} 9$ and $ 4a^{2} \equal{} (12 \minus{} b)^2 \plus{} 9$.

Multiplying both sides of the former equation by $ 4$ and substituting into the latter, we get (I think)

$ b^{2} \plus{} 8b \minus{} 39 \equal{} 0$ which has a real positive solution.

On the other hand, if the diagonal are divided into segments of lengths $ 6 \plus{} 6$ and $ b \plus{} (6\minus{}b)$, then we get the equation $ b^{2} \plus{} 4b \plus{}24 \equal{}0$ (I think) which does not have real solutions, so there is only one possible kite under the given conditions.


[hide]First, $ 2222\equiv3\pmod{7}$, and $ 5555\equiv4\pmod{7}$. So it becomes $ 3^{5555}+4^{2222}$ in modulo 7. From Fermat's, we have $ 3^6$ and $ 5^6$ are both 1 (mod 7), so it reduces to $ 3^5+4^2\pmod{7}\rightarrow{259\pmod{7}\equiv0\pmod{7}}$.[/hide]

This problem has already been posted. See
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180357 .
 :)


Let ABC be a triangle such each interior angle is $ \leq \frac {\pi}{2} $.

The points of the triangle sides are colored with $ red $ and $ blue $ .

Prove that there are 3 points M,N,P colored with the same color, such that the triangle MNP is right .

 :cool:   Do you know the source of this problem ?!  :cool:

Let A'B'C'  the inscribed triangle in ABC s.t. A'B' is ortogonal to BC, B'C' is ortogonal to AC and A'C' is ortogonal to AB. (*)

Two of tre points in {A',B',C'} must have the same colour X. Let's suppose those are A' and B'. If B or C are of the same colour of A' we fisish the proof. If B and C have different colour wrt those of B' and A' we considere the ortogonal projection of C' on BC and call it C". If C' has colour Y=/=X then if C" is coloured X then B'A'C"  is the triangle MNP; if C" is Y then C'C"B  is MNP.

If instead C' has colour X too and A has colour Y (the case A Y is obvious) consider the projection A" of A on BC and analise the cases.

PS
Sorry for my pour english. I hope at least the idea is clear.

(*) could be a simple but nice exercise to construct such a triangle.


This is some FIITJEE test sum, I think his answer is wrong and as usual, I can't understand head or tail of the solution.

Find the number of tuples (x1,x2,x3,x4,x5) such that $ x_1x_2x_3x_4x_5 \equal{} 2310$

.

[hide]I think the answer given was $ 16.5^5$[/hide]

P.S: Check this out. Another FIITJEE question. http://www.mathlinks.ro/viewtopic.php?p=1095007#1095007

Its correct only. What didnt you understand?  :maybe:

If it helps, i did that fiitjee derangements sum by classical fraud :D
Obviously, the total number of cases (which is the denominator in the probability) is $ \frac {11!}{2!2!2!}$
In the choices given, all the denominators were $ 11!$ only. So, obviously, the numerator should be a multiple of 8, which was true only in one case.

Q.E.D.  :D


Prove that $\mathop{\lim }_{n \to+\infty }\left|{y_{n}}\right| = \left| L \right|$ when you know that $\mathop{\lim }_{n \to+\infty }y_{n}= L$. Why the inverse sentence is not a theorem, for $L \ne 0$?

I believe the forward statement just follows from the fact that $| |L|-|y_{n}| | \le |L-y_{n}|$ by the triangle inequality.


Let $ A$ be a $ n\times n$ matrix of numbers and $ \mathbf x$ a $ n\times1$ matrix of vectors. ($ \mathbf x\equal{}[\mathbf v_1\,\cdots\,\mathbf v_n]^T$)
Does $ A\mathbf x\equal{}\mathbf x$ imply $ A\equal{}I$?

No. All $ A\mathbf{x}\equal{}\mathbf{x}$ for a single nonzero vector $ \mathbf{x}$ implies is that $ 1$ is an eigenvalue of $ A.$

Simple example: $ A\equal{}\begin{bmatrix}1&\minus{}4\\0&7\end{bmatrix},\quad
\mathbf{x}\equal{}\begin{bmatrix}1\\0\end{bmatrix}.$

Here I mean $ \mathbf x\equal{}\begin{bmatrix}
\mathbf v_1 \\
\vdots\\
\mathbf v_n\\
\end{bmatrix},$ where $ \mathbf v_1, \cdots, \mathbf v_n$ are nonzero vectors, not numbers.

All right, $ A\equal{}\begin{bmatrix}1&\minus{}4\\0&7\end{bmatrix},\quad \mathbf{x}\equal{}\begin{bmatrix}1&1\\0&0\end{bmatrix}.$

If $ X$ is invertible, then of course $ AX\equal{}X$ would imply $ A\equal{}I.$ If $ X$ is not invertible, then no.

[quote="ifai"]Here I mean $ \mathbf x \equal{} \begin{bmatrix} \mathbf v_1 \\
\vdots \\
\mathbf v_n \\
\end{bmatrix},$ where $ \mathbf v_1, \cdots, \mathbf v_n$ are nonzero vectors, not numbers.[/quote]
OK. And how about if $ \mathbf v_1, \cdots, \mathbf v_n$ are vectors in general vector space, or they are column vectors?

Don't go there- you're going to have trouble even defining the product. As written, it only makes sense if the $ v_i$ are row vectors.

The key condition is linear dependence/independence. If the $ v_i$ are linearly independent, $ A\equal{}I$ is the only solution. If the $ v_i$ are dependent, there are other solutions.


geometry
algorithm
abstract algebra
number theory
superior algebra
superior algebra theorems

Through studying Topics in Algebra by Herstein, some online sources, and occasionally this forum, I've figured out most of the basics in abstract algebra, and I like the subject. I was wondering if anyone could recommend a book/textbook with which I could continue learning about it and preferably doesn't waste too much time on the very basics. Some specific topics I have interest in learning are algebraic geometry, category theory, and algebraic number theory.

Thanks.

A great introduction to some of the ideas in algebraic geometry is [url=http://www.amazon.com/Ideals-Varieties-Algorithms-Computational-Undergraduate/dp/0387946802]Cox, Little, and O'Shea[/url].  The book is written very accessibly and puts a great deal of emphasis on motivation and avoiding "dryness" but still covers its subject thoroughly.  (It doesn't even assume knowledge of abstract algebra - nevertheless, there is a great deal to learn from it.)

I would recommend [i]Introduction to Commutative Algebra[/i] by Atiyah and Macdonald. That book has great problems, just like Herstein, and it continues right where Herstein left off. The material is commutative ring theory, so it is very applicable to algebraic number theory and algebraic geometry. I also like the homological algebra in the modules section, which is very useful for later math. (Another benefit is that, only from what I've heard, if you do all of the problems in the book, you'll have an understanding of schemes, which are an important but complicated idea in algebraic geometry.)

Unfortunately, the book doesn't have tons of geometric intuition, so if that's what you're looking for, you might want to try [i]Commutative Algebra[/i] by Miles Reid.


function
integration
limit
calculus
real analysis
advanced fields
advanced fields unsolved

I want to prove the following statement:

Let $ u$ be harmonic in $ \mathbb{D}$ and suppose that there exists a $ C > 0$ such that
$ \int_0^{2\pi} |u(re^{it})| \frac {dt}{2\pi} \leq C$
for all $ 0 < r < 1$. Then there exists a function of bounded variation $ \mu$ defined on $ [0,2\pi]$ whose total variation is less than $ C$ such that
$ u(z) \equal{} \int_0^{2\pi} \frac {1 \minus{} |z|^2}{|e^{it} \minus{} z|^2} \, d\mu$

There is a hint to use Helly's selection theorem, but I don't get it. I am unfamiliar with the theorem, and I don't see how to apply it.

I am supposing the proof should run like this (please note I am guessing wildly on things, so please don't take my steps for true):
If for $ 0 < r < 1$ we let $ u_r(z) \equal{} u(rz)$ then $ u_r$ are harmonic in $ |z| < \frac {1}{r}$. So we know that for $ z \in \mathbb{D}$
$ u_r(z) \equal{} \frac {1}{2\pi}\int_0^{2\pi} \frac {1 \minus{} |z|^2}{|e^{it} \minus{} z|^2}u_r(e^{it}) \, dt$
and for fixed $ z$,  $ \lim_{r\to 1} u_r(z) \equal{} u(z)$

If we let $ \mu_r(t) \equal{} \frac {1}{2\pi}\int_0^{t} u(re^{is}) \, ds$ then they satisfy the criterions for Helly's theorem, so there is a $ \mu$ of bounded variation and a sequence $ (r_n) \in (0,1)$ such that $ r_n \to 1$ and for continuous functionals $ T \in \mathbb{C}^*$ we have $ T(\mu_{r_n}(t)) \to T(\mu(t))$ as $ n \to \infty$

Based on a proof of a similar statement for $ p > 1$ where we can apply reflexivity, I am guessing that this should imply that
$ \lim_{n\to \infty} \int_0^{2\pi} v \, d\mu_{r_n} \equal{} \int_0^{2\pi} v \, d\mu$ (*)
At least for $ v$ the Poisson kernel. I don't seem to quite get this, since there is a $ t$ involved in the weak convergence statement above, while for a continuous $ v$,  I would like to look at the functional $ T(\mu) \equal{} \int_0^{2\pi} v \, d\mu$.

Since my $ \mu_r$ are actually absolutely continuous, we have
$ \frac {1}{2\pi}\int_0^{2\pi} v \, u_r(e^{it}) \, dt \equal{} \int_0^{2\pi} v \, d\mu_{r}$
and this would complete the proof, modulo that I can't show (*)

Sorry for the mess, but when I started this post I didn't know what I was doing at all. I think I am close now. Please someone help me fix this. Meanwhile I will try to understand it better, and probably edit my post a million more times.

You are on the right track. All you need to do is to integrate by parts and express $ \int P(z,t)\,d\mu_r(t)$ as $ \int Q(z,t)\mu_r(t)\,dt\plus{}c\mu_r(2\pi)$, thus showing that pointwise convergence of $ \mu_r$ to $ \mu$ implies the convergence of the integrals. Then you can integrate by parts back to return to the integral with respect to $ d\mu$. :)

As a side remark, I find the language using integrals with respect to $ d\mu$ where $ \mu$ is a function of bounded variation rather archaic. The language of signed measures, weak convergence, Riesz theorem and weak compactness of the unit ball in the dual space seems to be much more convenient. Of course, this remark applies only if you are familiar with Lebesgue integration.

Thanks a lot. I was a little suspicious of the integration by parts first, but combining some theorems on Lebesgue-Stieltjes integrals and some on Riemann-Stieltjes integrals I have meticulously checked that yes, it is ok, although in one step you have a Lebesgue integral.

How come integration by parts is such a powerful tool also in "abstract" mathematics (as opposed to what americans refer to as calculus)? It's often used to get integrands with better convergence properties, although I am not fully used to such techniques, and am often mesmerized by their application. One example that comes to mind is when you have a periodic function $ f$ and look at integrals of $ f(nx)$ - I don't remember the exact formulation. I had such a question on a take-home exam and I provided a whopping four pages of approximation by simple functions and use of some strong convergence theorems, and then I was shown a three-line solution with integration by parts.

One more question: In the statement of Helly's theorem, there is a weak convergence statement, that $ T(f_k(t)) \to T(f(t))$ for continuous functionals. In this example, the relevant property was pointwise convergence; but in fact, isn't pointwise convergence [b]equivalent[/b] to the weak convergence? So what is the point of weak convergence?  :maybe:

[quote="Kalle"]In this example, the relevant property was pointwise convergence; but in fact, isn't pointwise convergence [b]equivalent[/b] to the weak convergence? So what is the point of weak convergence?  :maybe:
[/quote]
Try to do everything in $ \mathbb R^3$ instead of $ \mathbb R^2$ where the integral is taken over the 2-dimensional sphere. Where will be the Stiltjes integral, integration by parts, and Helly's theorem now? The result is exactly the same nevertheless ;).


please help us everyone!!!!!! we must have the solution before Monday (19/10) :wallbash: My teacher will check our exercise on Monday!


The Randomest Question ever. What type of gum do you like the most? YAY

I don't normally chew gum, but I'd have to say that the one my parents seem to prefer is Excel.

Personally, I like Trident Splash.

Gum makes my tongue super-sensitive.

Lemons FTW.

I hate gum.

darn, you must be pretty beast if you chew lemons
anyways, 5 cobalt if you don't count price, trident otherwise

I don't really chew gum that much (braces), but I like Orbit Bubblemint. I also like Wrigley's Fruit chewed at the same time with those little Wonka Nerds

[quote="shelly32494"]trident tropical twist!!! ;]][/quote] me too!!! :D  :D  :D

[quote="jonathanchou711"]five solstice.[/quote]

makes me feel sick... but five itself isnt bad

Everyone's choices- taste or long-lasting?

I would prefer healthy, then long lasting, then taste, in that order.

You know that kind of new 5 gum zing?

It tastes weird.. It's kind of like those doctor's gum where they give you gum that tastes minty and after a while it becomes all crumbly and you can swallow it! That's the same with zing.  It becomes swallowable and dry.

[quote="ra5249"]that's what she said[/quote]

-___-;;;

@ OP's title: gum in Spanish is chicle
chiste = joke

I assume that's what you were going for, since I searched for and couldn't find any brand of gum called "Chiste"

:o and :rotfl: to ra's comment on "swallowable and dry"

and yes that's espanol for gum, i taKE espanol. and i was working on the hw. But not now cuz i have SUMMER! and no, im not hispanic.........


Find first
(1) decimal digit of 2^987654321
(2) two decimal digits of 2^987654321
(3) three decimal digits of 2^987654321
(4) n decimal digits of 2^98765321 (represent as function of n)

Do not post the same post in different forums. This is already in the intermediate forum. Mods, please delete this.

Yes, he is right. There is no need to double post. If you want to participate on the discussion of this thread, go to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36661]the thread on intermediate forum[/url]. This is locked... and to be deleted.


Divisibility
modular arithmetic
number theory
IMO Shortlist

Let $b,n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_k$ such that $b - a^n_k$ is divisible by $k$. Prove that $b = A^n$ for some integer $A$.

[i]Author: Dan Brown, Canada[/i]

[quote="orl"]Let $ b,n > 1$ be integers. Suppose that for each $ k > 1$ there exists an integer $ a_k$ such that $ b \minus{} a^n_k$ is divisible by $ k.$ Prove that $ b \equal{} A^n$ for some integer $ A.$

[i]Author: unknown author, Canada[/i][/quote]
[hide="Solution"]
Assume that $ b$ has a prime factor, $ p$, so that $ p^{xn \plus{} r}\parallel{}b$ with $ 0 < r < n$.  Then, we can let $ k \equal{} p^{xn \plus{} n}$.  It follows that $ b\equiv a_k^n\bmod p^{xn \plus{} n}$.  Since $ p^{xn \plus{} r}\parallel{}b$, we see that $ p^{xn \plus{} r}\parallel{}a_k^n$.  Then, $ p^{x \plus{} \frac {r}{n}}\parallel{}a_k^n$, which is a contradiction since $ \frac {r}{n}$ is not an integer.  Thus, we have a contradiction, so $ r \equal{} 0$, which means that only $ n$th powers of primes fully divide $ b$, so $ b$ is an $ n$th power. [/hide]

It may be interesting to know that if $ 8 \nmid n$, then it suffices to consider primes $ k$ only.
More generally (more or less the Grunwald-Wang-theorem):

Let $ n,b$ be a integers, $ n > 0$. If for all primes $ p$ there is an $ a_p$ such that $ p|b \minus{} a_p^n$, then the following holds:
a) If $ 8 \nmid n$, then $ b \equal{} a^n$ for some integer $ a$.
b) If $ 8|n$ then $ b \equal{} a^{\frac n2}$ for some integer $ a$.

This was posted and partially solved before.

For those knowing algebraic number theory, this can be generalised even more (those don't knowing algebraic number theory, please stop here :D ):

Let $ K$ be a number field, $ b \in K$, $ n \equal{} m \cdot 2^s > 0$ an integer ($ m$ odd).
If $ b$ is a $ n$-th power in all but finitely many completions $ K_v$ (where $ v$ runs through the primes of $ K$), then:
a) If $ K[\zeta_{2^s}]|K$ is cyclic, then $ b \equal{} a^n$ for some $ a \in K$.
b) Otherwise, $ b \equal{} a^{\frac n2}$ for some $ a \in K$.

Proving this is some work in (pre)global class field theory (the main step being to show that if $ L|K$ has all but finitely many primes totally split, then $ K \equal{} L$).

Note that a) indeed requires $ 8\nmid n$:
Let $ b \equal{} 16$, $ n \equal{} 8$. We have $ x^8 \minus{} 16 \equal{} (x^2 \minus{} 2)(x^2 \plus{} 2)(x^2 \minus{} 2x \plus{} 2)(x^2 \plus{} 2x \plus{} 2)$. Of course one of the numbers $ \minus{} 1, \minus{} 2,2$ is a quadratic residue $ \mod p$. It means that for every prime $ p$ one of the equations $ (x \plus{} 1)^2 \plus{} 1 \equal{} 0$, $ x^2 \plus{} 2 \equal{} 0$, $ x^2 \minus{} 2 \equal{} 0$ has a solution $ \mod p$ and so does $ x^8 \minus{} 16 \equal{} 0$. And $ 16\neq A^8$.

[Moderator edit: See also http://www.mathlinks.ro/viewtopic.php?t=4874 for this counterexample.]

I also didn't see it at first, but $ p$ is not necessarily prime. Using this, it gets really easy (using valuations).
But you will have a lot of fun to show that if we require that $ 8 \nmid n$, then we need only to look at primes $ p$.

[b]Solution[/b]

Assume that there is no $A$ such that $b=A^n$. Then there must exist a prime number $p$ such that, if $p^a \| b$, then $n \not | a$. Assume now that $mn < a < (m+1)n$ for some $m \in \mathbb{N}$. Taking $k=p^{(m+1)n}$ yields that

\[b \equiv a_{k}^n \pmod{p^{(m+1)n}}\]
Which implies that $p^{a} |a_{k}^n$ and, since $a_{k}^n$ is a perfect $n$th power, that $p^{(m+1)n} | a_{k}^n$. Hence $b \equiv 0 \pmod{p^{(m+1)n}}$ and $n|a$ which is a contradiction. $\blacksquare$

Does anyone have a solition with looking at a prime divisor of $n$?

[quote=ali666]let $ k\equal{}b^2$:
$ b^2|b\minus{}a^n_k \Rightarrow a^n_k\equal{}b(bx\plus{}1)$ but $ gcd(b , bx\plus{}1)\equal{}1$ therefore $ b\equal{}A^n$ for some integer $ A$. :D[/quote]

Can you explain more detailed?

hehe my solution if exist p: Vp(b)=r.n+s ( 0<s<n) then choose k=p^(r(n+1)) then Vp(ak) > r then Vp(ak) >= r+1 then Vp(b-ak^n) = rn+s < r(n+1)

in this problem b and n are not fixed right? they can vary, just like k?

Let $b=p^{an+c}l$ where p is prime and $(p,l)=1$ and $1\leq c\leq n-1$. Let $k=p^{an+n}$, $k|b-a_k^n,=>v_p(k)\leq v_p(b-a_k^n)<=>an+n\leq v_p(b-a_k^n)=min\left\{v_p(b),v_p(a_k^n)\right\}\leq an+c$ . Which is plainly a contradiction, thus $b=m^n$

[quote=vjdjmathaddict]A different [u]Solution[/u]
write $b$ as $b=B^n \times j$ where $gcd(B,j)=1$ now if $j=1$ then we are done.
Suppose $j$ is not equal to $1$ then consider a prime which divides $j$ and consider $k=p^n$
now since $p|j$ then $p|a_k$ this implies $p^n|{a_k}^n$ implies $p^n|j$ but then it follows that $p^{n}|B$ by definition of B.
So this a contradiction and hence we are infact done.[/quote]
 Wrong: If you take B to be the max n-th power that divides b, then not necessarily GCD(B,j)=1, but if you take by construction that GCD(B,j) , B and j are clearly unique, but not necessarily all n-th power that divides b go to B. Take for instanceb=(2^n)*(3^(n+1)), then B=2^n and j=3^(n+1) are the only air (B,j), but 3^n divides j, and therefore no contradiction

Nice problem. Hope, my solution is not the same as anyone else's :blush:

$\rightarrow$. Suppose there exists a prime $\text{p}$ such that $b = p^{\alpha n + e_0}x , 1 \le e_0 < n$.
Where, $\text{gcd} ( x,p)=1$ and $\text{v}_p(b)$ is not congruent to $0 \pmod{n}$.

Now, let $k= p^{n (\alpha +1)}$. Now, $\text{v}_p(b-a_k^n)$ must be greater than or equal to $\text{v}_p(k) = n(\alpha + 1)$.

$\blacksquare$ [b]Case 1.[/b]: $\text{v}_p(a_k) \le \alpha$.
$\implies \text{v}_p(b)$$ \ge \text{v}_p(a_k^n)$$ \implies \text{v}_p(b-a_k^n) =\text{v}_p(a_k^n) \le \alpha $
[i]contradiction.[/i]

$\blacksquare$ [b]Case 2.[/b]: $\text{v}_p(a_k) \ge \alpha +1$.
$\implies \text{v}_p(a_k^n) \ge \text{v}_p(b) \implies \text{v}_p(b-a_k^n) = \text{v}_p(b) < n(\alpha + 1)$
[i]contradiction.[/i]

Henceforth, we conclude that $e_0 \equiv$$ 0 \pmod {n} \implies b=m^n$.

Assume that $n \nmid v_p(b)$ for some prime $p$. Then let $v_p(b)=na+r$ with $0 < r < n$ and $\frac{b}{p^{v_p(b)}}=c$. Then taking $k=p^{n(a+1)}$ and $a_k=m$ for simplicity,  we get that $$p^{n(a+1)} \mid p^{na+r}c-m^n.$$ This obviously implies that $p^a \mid m$. Let $\frac{m}{p^a}=o$. We then get that $$p^n \mid p^rc-o^n.$$ Since $r>0$, we get that $p \mid o$. Thus $$p^n \mid p^rc-p^n(\frac{o}{p})^n \implies p^n \mid p^rc,$$ contradiction since $r<n$. Thus $n \mid v_p(b)$ for all primes $p$ and $b$ is a perfect $n$th power.

$ k | b - a^n_k$ so $tk = b - a^n_k$ or $b - tk = a^n_k$. the idea is to choose $k$ such that LHS is in from $bx$ and $ gcd(b , x)=1$ so then $b = A^n$. Let $k = b^2$ then we have $b(1-bt) = a^n_k$ and we have $ gcd (b , 1-bt) = 1$.
we're Done.

NT Skills are peaking. Took a solid 40 minutes to finish after plugging $k=b^2$. :(

[i]Solution:[/i] We need to choose an optimal $k$ for this. Let $k=b^2$ and call $a_{b^2}=X$.
\[b^2 \mid b-X^n \implies b-X^n=b^2w\]
for some integer $w$. Rearranging we get $b(1-bw)=X^n$. As $\gcd(b,1-bw)=1$, both the factors must be powers of $n$. This is sufficient to conclude that $b=A^n$. $\blacksquare$

Let $p$ be a prime. Let $k\in \mathbb{Z}$ such that $kn>v_p(b)$.

Then, we have
$$a_{p^{kn}}^n\equiv b\pmod{p^{kn}}$$

Since $v_p(b)<kn$, we have
$$v_p(a_{p^{kn}}^n)=v_p(b)$$
So, $n|v_p(b)$. Doing this for every prime completes the proof.

Not that bad.

[hide=solution]Let $p$ be an arbitrary prime that divides $b$ and let $v_p(b)=m.$ It suffices to show that $n\mid m.$ Take $k=p^{m+1}.$ Then, $$a_k^n\equiv b\pmod {p^{m+1}}.$$ However, note that $b$ is a multiple of $p^m$ but not a multiple of $p^{m+1}$ because $v_p(b)=m.$ This implies that $a_k^n$ is also a multiple of $p^m$ but not a multiple of $p^{m+1}.$ Hence $v_p(a_k^n)=m.$ However, $v_p(a_k^n)=n\cdot v_p(a_k),$ so $v_p(a_k^n)$ is a multiple of $n,$ implying that $n\mid m,$ as desired.[/hide]

hi channing from NT3

The statement is equivalent to the following:

[b]Restatement.[/b] For every positive integer $k$, there exists nonnegative integers $x, r$ such that $$x^n =b+rk$$ for this choice of $b$.

Pick some $p \mid b$, and choose $k = p^{mn}$ for some large positive integer $m$ (in particular, we can have $\nu_p(b) < mn$.) Then notice that $$n\nu_p(x) = \nu_p(b+rk) = \text{min}(\nu_p(b), \nu_p(rk)) = \nu_p(b)$$ because the two terms in the minimum are unequal by our choice. This implies $n \mid \nu_p(b)$ for all $p \mid b$, which finishes.

@below I was thinking that $n \mid \nu_p(k)$ for a while which was not required, and I was too lazy to get rid of the extra restriction.

Wait why did we have to choose $k=p^{mn}$ rather than just $p^m$ @above

also hi

[hide=Oneliner]Pick $k=b^2$ then we have $b-a^n=cb^2$ for some $a,c$ then $a^n=b(1-cb).$ Note that for $p\mid b$ $n\mid \nu_p(b(1-cb))=\nu_p(b)$ so we're done.[/hide]

For each prime $p$ take $k=p^e$ such that $e>v_p(b)$. Then we obtain $n\mid v_p(b)$. $\blacksquare$ (same solution as #50)

Suppose not. Then there exists $p$ such that $n\nmid \nu_p(b)$. Let $\nu_p(b) = t$. Setting $k = p^{t+1}$ gives $\nu_p(c^n) = n\nu_p(c) = \nu_p(b)$ for some $c$, which is impossible.

Take $k=b^2$ so that $a_{b^2}^n \equiv b \pmod{b^2}$. Then, $a_{b^2}^n = b(bj+1)$ for some nonnegative integer $j$, but $(b, bj+1)=1$, so $b=A^n$ for some integer $A$, as desired.

The main idea of the problem is the assertion $k=b^2$.
Then for every prime $p$ dividing $b$, we clearly need $n\mid\nu_p(b)$. $\blacksquare$

$k=b^2$ implies that $a_{b^2}^n = b(1-cb)$ for some $c$. However, for all $p \mid b$, $\nu_p(b(1-cb)) = \nu_p(b)$ and we're done since $n \mid \nu_p(b(1-cb))$. $\square$

Another one liner?

The assertion $$k=b^2\implies a_{b^2}^n = b(1-kb)\implies n\cdot v_pa_{b^2}=v_pa_{b^2}^n=v_p(b(1-kb))=v_pb\forall p\mid b\implies n\mid v_pb,$$ which is just the problem rephrased since each exponent of a prime in b must be a multiple of n. $\blacksquare$

Consider a $k$ such that for every prime $p$ dividing $b, \nu_p(k)>\nu_p(b)$ (here $k=b^2$ works easily), note that $k\mid b-a_k^n$, if $\nu_p(b)=n\nu_p(a_k)$ we are done, so assume the contrary then
\[\nu_p\left(b-a_k^n\right)=\min\{\nu_p(b),n\nu_p(a_k)\}>\nu_p(k)\]
Which is impossible as $\nu_p(b)<\nu_p(k)$


This year the in  a college has members from all 4 classes - 9 freshmen, 11
sophomores, 18 juniors and 22 seniors.
a. The photographer wants to line up the students in one line for a photograph so that all of the students in each class will be together. How many ways can this be done? Can anyone solve this problem?

treat each "class group" as a block. It really doesn't matter how many students are in each class, if I'm reading the question correctly. It's basically finding the number of permutations of 4 distinct "objects".

Here we are not considering the order. We are concerned with just lining them up . I thought this is a combination problem not permutation. Try once again and solve it.


Let $t_j=\frac{j(j+1)}{2}$, and $S_n=\displaystyle\sum_{j=1}^{n} t_j$. Find the smalles possible value of $n+k$ if $S_{n+k}-S_n>2007$

this uses a classic result, first notice that $t_j=\binom{j+1}{2}$

it can be shown through pascal's identity, i believe that the specific name is the vandermonde identity, but anyway, you get

$S_n=\binom{n+2}{3}$

clearly, we want to minimize $n$ since we are subtracting that, so the last binomial coefficient is defined when $n=1$, then

$\binom{k+3}{3}>2008$

checking a few values, $k=24$, so $n+k=\boxed{25}$

Ok, I see.  Your solution seems to be right. My solution had an error.
I made this up for an exam without a calculator. Do you think the last step is too much computation (3mins/problem avg.).


Find all prime numbers p and q such that p divides q+6 and q divides p+7

bnick, you need better titles for your posts.  (LaTeX is nice, too.)

[quote="bnick17"]Find all prime numbers $p$ and $q$ such that $p$ divides $q+6$ and $q$ divides $p+7$[/quote]
[hide="Boring solution"]
Note that if $q = 2$ then $p \mid 8$ and so $p = 2$, but $2 \not\mid 9$, a contradiction.  Thus $q$ is odd and so $p \mid q+6$ is odd as well.  Thus $p+7$ is even so in fact $q$ divides $\frac{p+7}{2}$.

Case 1: $p = q+6$.  Then $q$ divides $p+7 = q+13$ so $q = 13$, $p = 19$.

Case 2: $p \neq q+6$, so in particular $p \leq \frac{q+6}{2}$.  Then $q \leq \frac{\frac{q+6}{2}+7}{2}= \frac{q+20}{4}$ so $q \leq 5$ and $p \leq 5$.  There are only 4 cases to check, and none of them work.[/hide]

Anyone have anything cleverer?

[hide="this also requires some casework"]
Clearly $p\neq q$ (otherwise $p$ divides both $p+7$ and $p+6$), so $p$ and $q$ are relatively prime. Now observe that
\[p|(6p+7q+42),\ q|(6p+7q+42)\Rightarrow pq|(6p+7q+42)\]
Let $6p+7q+42=kpq$. If $k$ is even, then $q=2$. This means $p|8$ so $p=2$, but this fails because $2\not| 9$. Similarly, if $k$ is divisible by 3, then $q=3$. This means $p|9$ so $p=3$, but this fails because $3\not| 10$. Therefore either $k=1$ or $k\geq 5$.

If $k=1$, then $6p+7q+42=pq$ implies
\[(p-7)(q-6)=84\]
Both factors are positive (if both negative, product less than 84). Then $q>6$ and $p>7$, so $q-6$ can't be divisible by 2 or 3, $p-7$ can't be divisible by 7. The only case that remains is $p-7=12,q-6=7$ which leads to $(19,13)$.

If $k\geq 5$, then $6p+7q+42\geq 5pq$ implies
\[(5p-7)(5q-6)\leq 252\]
As shown before, $q=2$ and $q=3$ are bad, so $q\geq 5$. This means $p\leq \frac{252}{19}<18\Rightarrow p\leq 3$. In either case, $p|6$ so $p=q$, but this is a contradiction!

Therefore $\boxed{(19,13)}$ only solution.
[/hide]


Hurray! Another COP. [b][i][u]I'm in[/u][/i][/b]. [b][i][u]Yes[/u][/i][/b].

Wow, endless problem supply. I'm in.  :P

I'll do it. Can you have one or more of tokenadult, rcv, or frost13 check the problems for clarity so we don't have to ask questions?

I'll try to do it, if I can remember that is.

I'm in at least through the proof-free rounds...

[quote="Silverfalcon"]If you're interested in participating in COP 4th, please sign up here by putting "I'm in" or "Yes!" or et cetera.

Thanks.[/quote]

w/e i'll do it

To RC-7th:

I'll try to do it.. But often times, I do not want the problems to be out while it's in process so I rarely do that. Of course, making problems and realizing the mistakes make me a better problem writer though. :)  I remember writing the my former competition and had huge mistakes while the mistakes in here are usually minor.

To shobber:

Please read the stickied thread "About COP (Contest of Opportunity)" on this forum.

Thank you.

Is there going to be a COPSD? If there is, i'll be joining that instead.

Done.

It's on the Basics Forum now. :)  :D

I didn't sumbit the last one.................. I gotta do this one to make up for the stupidity.  :lol:

[quote="solafidefarms"]Can you do both the COP and COPSD? If so, I'm in, else I'll just do the COP.[/quote]

If you're middle schooler, yes.

If you're graduated from middle schooler, sorry no. :lol:

what's the difference between

COPSD and COP?/

and I'm in for both ...just notify me
....

I'm kinda busy these days, so I'm not on every often times

[quote="Silverfalcon"]If you're interested in participating in COP 4th, please sign up here by putting "I'm in" or "Yes!" or et cetera.

Thanks.[/quote]

Et cetera.

[quote="shinwoo"]what's the difference between

COPSD and COP?/
[/quote]

I'm pretty sure COPSD is for middle schoolers only

I 'm in  :lol:
(this is the first post for me)
What is middle schooler's age? (I'm 14  :D )

You just have to still be in middle school :)

Or your country's equivalent, if applicable.

I'm proud to announce that the fourth COP will start in few weeks.

Be ready!

I'll post a thread when problems are ready.

P.S. This is same for COPSD.


Hi, I was wondering if anyone knows of any (possibly online) math contests that don't run during the school year?  Just trying to plan out my summer a little, see if there's anything of the sort to do...

Mu Alpha Theta National Convention, but the registration is already over. and it is pretty expensive ( I think 500 dollars for registration and sleeping and stuff) but it goes for a week and it is pretty well-organized and worked out. This year it is in Hawaii(which is why iam going despite hating the competition itself).

Other than that, I dunno.


function
algebra
functional equation
algebra unsolved

[quote="orl"]Find all functions $f: \mathbb{Q} \mapsto \mathbb{C}$ satisfying

(i) For any $x_1, x_2, \ldots, x_{1988} \in \mathbb{Q}$, $f(x_{1} + x_{2} + \ldots + x_{1988}) = f(x_1)f(x_2) \ldots f(x_{1988})$.

(ii) $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ for all $x \in \mathbb{Q}$.[/quote]

at the beginning, we suppose $f(x)\neq0$ for all $x \in \mathbb{Q}$.

by $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ we find $\frac{f(x)}{\overline{f(x)}}=\frac{f(1998)}{\overline{f(1998)}}\longrightarrow\frac{f(x)}{\overline{f(x)}}=\frac{f(y)}{\overline{f(y)}}=$costant for all x,y.
This means that, if $f(x)=p(x)e^{i\theta(x)}$, $\frac{p(x)e^{i\theta(x)}}{p(x)e^{-i\theta(x)}}=e^{2i\theta(x)}=$costant, so $2i\theta(x)=$costant $(mod$ $2\pi)$, $\theta(x)=$costant $(mod$ $\pi)$: if we make $p(x)$ vary in the positive and negative reals, that means that $\theta(x)$ is costant.

now consider the first relation: if we put $x_1=x_2=\ldots=x_{1998}=0$, we have $f(0)^{1998}=f(0)$, so f(0) is one of the 1997 complex 1997-roots of the unity, $f(0)=e^{\frac{2ki\pi}{1997}}$ (and this is $\theta(x)$ for all x)

now let's put $x_1=x_2=\ldots=x_{1998}=a$, and then we have $f(a)^{1998}=f(1998a)$
again, let's put $x_3=x_4=\ldots=x_{1998}=a$, $x_1=a+b$, $x_2=a-b$ , from which we have $f(a)^{1996}f(a+b)f(a-b)=f(1998a)$.
From this two relations we have $f(a)^2=f(a+b)f(a-b)$, changing
$f(\frac{x+y}{2})^2=f(x)f(y)$,for all x,y.
This means also that $f(\frac{x+y}{2})^2=f(x+y)f(0)$, and then
$f(x+y)f(0)=f(x)f(y)$.

With the substitution $f(x)=p(x)e^{i\theta(x)}=p(x)e^{\frac{2ki\pi}{1997}}$ we have
$p(x)e^{\frac{2ki\pi}{1997}}p(y)e^{\frac{2ki\pi}{1997}}=p(x+y)e^{\frac{2ki\pi}{1997}}e^{\frac{2ki\pi}{1997}}$
$p(x)p(y)=p(x+y)$, with $p(x)$ function $p: \mathbb{Q} \mapsto \mathbb{R}$ and $p(0)=1$.

This functional equation is trivial: the solutions are $p(x)=a^x$ for $a>0$, from which $f(x)=a^xe^{\frac{2ki\pi}{1997}}$ ($k\in\mathbb{Z}/_{1997}$).
It can be easily seen that these functions satisfy the two text relations.

Now suppose that there is an y such that $f(y)=0$.
in the first relation, if we put $x_1=y$, $x_2=z-y$, $x_3=x_4=\dots=x_{1998}=0$, we see that
$f(z)=f(y)f(z-y)f(0)^{1996}=0$ for all $z\in\mathbb{Q}$.
so $f(x)=0$ is another solution. this ends the demonstration.

hi everyone

EDIT: exscuse me, i treated 1988 as 1998....  :D   :D


It is trivial to dissect a triangle into $ n$ triangles for any $ n\ge 2$.
Same is true for convex quadrilaterals. It is not hard to show that every
convex pentagon can be dissected (partitioned) into $ 6$ convex pentagons.
What about the general case?

[b]Problem.[/b] Is it true that every convex $ n$-gon can be dissected into
finitely many convex $ n$-gons, for every $ n\ge 3$?

The problem is probably not new, but one of my students asked me the question
and I do not know the answer. Any reference is welcome.

Dan

nice question, i will be back with a solution if i figured this out  :)


Apparently, KoMaL for this year starts in September. At least that's what the web site said.

So, how does turning in solutions work? I signed up in July, but this is the first time I've done it.

You have to contact them directly. I have no idea what their regulation are.


Kung Fu Madness, does anyone play this game?

Click [url=http://www.kungfumadness.com/PlayerCreate.cfm?Friend=2583]Here[/url] to create an account.

I'm a black belt BTW, my account = sin4pi.

I made an account. My name is DarkKnight. Orange belt. :P

wtf????? I didnt get an email with my password. I think i made like 5 accounts in like 5 different email accounts but still havnt gotten any password

I'd like to start a dojo if  enough AOPSers sign up.

But I need at least 6 or so more first.

[quote="elvenchamp777"]you can count me in.  I'm a purple belt and my name is elvenchamp.  what do dojo do though?

elvenchamp777[/quote]

I have three acounts so that makes 5. And i know a guy that would be willing to join. I think hes a second dan. My accounts are brown belt, brown belt and second dan. working up one of my brown belts and using the other to get some money because i don't care about that one's record. To find my accounts search DarkKnight.

[quote="elvenchamp777"]I'm black now :D  and is there some other way to get money cause I'M SOOOO POOR lol  and how do you transfer money from one account to another?

elvenchamp777[/quote]

To transfer money to another account, go to the gold ledger and click send gold to friend. I made a dojo with three people i've fought and 2 accounts that are me. Please message me to join. It's called Arcane Vault. I've sent a message to you and lucky707 asking you to join my dojo but you will have to leave the one you are currently in. Remember to check your mail. ;)

Could ya invite WindSlicer to the clan? Thanks!

I think I'm Green Belt  workin' my way up :P

[quote="miraculouspostmaster"]Guiseppe - Black Belt[/quote]

Why did you leave our dojo after you joined?


Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of

$ P\equal{}\sum\frac{m_a}{r_a}$

[quote="Ligouras"]Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of

$ P \equal{} \sum\frac {m_a}{r_a}$[/quote]
A computer program shows that $ P_{min}\equal{}3.$

[quote="Ligouras"]Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of

$ P \equal{} \sum\frac {m_a}{r_a}$[/quote]

$ \frac{h_a}{r_a}\equal{} \frac{2(s\minus{}a)}{a}$ , $ s\equal{}\frac{1}{2}(a\plus{}b\plus{}c)$

$ \sum\frac {m_a}{r_a} \ge \sum\frac {h_a}{r_a}\equal{}2\sum\frac {s}{a}\minus{}6 \ge 3$


Am I reading correctly:
$ e^{1\minus{}x}\equal{}17.5$ and $ 3\minus{}2e^{x\minus{}1}\equal{}0$
I'm assuming these are two separate problems so, for the first one, we can decompose the exponent into $ e\cdot\frac{1}{e^{x}}\equal{}17.5$.  We can multiply both sides by $ e^{x}$ and divide by 17.5 to obtain $ \frac{e}{17.5}\equal{}e^{x}$.  Take the natural log of both sides to get $ x\equal{}\ln\frac{e}{17.5}$.  Using properties of logs, we can rewrite $ x\equal{}\ln{e}\minus{}\ln{17.5}$.  Since the first natural log is 1, our final answer is $ x\equal{}1\minus{}\ln{17.5}$.

For number 2, it's basically the same deal.  Move the 3 over and divide both sides by 2 to obtain $ e^{x\minus{}1}\equal{}\frac{3}{2}$.  From there, we follow the same procedure as the first problem.

1. $ e \minus{} x \equal{} 17.5$?
2. $ 3 \minus{} 2e^{x \minus{} 1} \equal{} 0$
These the questions? I probably misread so no need to look at my corrections, but
[b]Please[/b] refer to the link provided in the hint.
[hide="Hint"]
For the first question, just subtract $ e$ from both sides and divide by $ \minus{} 1$.
For the second one, subtract both sides by $ 3$, divide by $ \minus{} 2$, then use logarithms. (If you're having trouble with logs, refer to this link,[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175427[/url], and look how stimnc did it).[/hide]

[quote="Magic Man"][size=150]solve to 3 s.f [/size]

e^1-x = 17.5
3-2e^x-1 = 0

if someone could show me how to do them that would very helpful
thank you
MAGIC MAN[/quote]
1. 1-x=ln 17.5, so x=1-ln 17.5
2. 2e^(x-1)=3, so e^(x-1)=3/2 and x-1=ln 1.5, so x=1+ln 1.5