LaTeX
ratio
inequalities
algebra
system of equations
algebra proposed

Dear Yagaron,

I believe that you are surprised of solving this system because it is very nice and beautiful. I am sure this is the prettiest system I created in my life!  :lol: Try to prove it again, just find roots as reals numbers.

[quote="yagaron"]Dear hungkhtn,

Like I said, Your system of 3 equations is not a true system of 3 equations! It is a system of 2 equations!

Is impossible to solve for $a,b,c$ together. has infinitely many solutions.

Or the obviously solution $a,b,c=0$

I didnt see any beauty in this system of 2 equations. Has too big solution and a lot of unecessary numbers.

Sorry, but is not beautiful...  :maybe:[/quote]

We can figure its root. It hasn't infititive root because we can find exactly the ratio $(a: b,b: c,c: a)$. I know this is only 2-equation system, but this fact can not prove that the inequality has infinitive root.

Well, if a=b=c it can be anything...
So you want to find $(a: b,b: c,c: a)$? :maybe:

The answer is $a=b=c$ or $\{a,\ b,\ c\}=\{1,\ \omega ,\ \omega^{2}\}$, where $\omega$ is the cubic root of 1 except 1.

solving for c=1 will give all solutions by multiplying by a factor k, since the equations are homogeneous. As well, the third equation isn't necessary in solving the other two, so we focus on th first and second equations.

2a^3-b^3-1=3a^2b-3b^2
2b^3-a^3-1=3b^2-3a

adding 2 of the first to the second gves:

3a^3-3=6a^2b-3b^2-3a

a^3-1=2a^2b-b^2-a

b^2-(2a^2)b+(a^3+a+1)=0

b=((2a^2)+-sqrt(4a^4-4a^3-4a+4))/2
b=a^2+-sqrt(a^4-a^3+a+1)
b=a^2+-(a-1)sqrt(a^2+a+1)

then, the solutions are given by:

(ak,(a^2+-(a-1)sqrt(a^2+a+1))k,k)

I do believe that this is right, any feedback?

[quote="hungkhtn"]It is not the right answer, Kunny. The problem is to find the root in the set of real numbers only. Come on, guy! before I post my unexpected and very nice solution,[/quote]

Please dont use invisible Ink to write your question.

And I really think that this is just one more system for the infinity set of system, it does not have anithing special..

If You knows that answer you had wrote them. But, i think you dont have it. and neither you knows the meaning of the system.

It's just a symetrial system. just one more

Sorry for someting I sayd, is not for ofense. It's just because does not make sense continue with this discution.

I dont know what you want.

The only Real roots of the system is $a=b=c$
And this is not special...

[quote]
We can figure its root. It hasn't infititive root because we can find exactly the ratio (a: b,b: c,c: a). I know this is only 2-equation system, but this fact can not prove that the inequality has infinitive root.
[/quote]

This is not true...

I Will post some solution in Latex

[hide]

${\left\{a=\frac{1}{2}i \left(i+\sqrt{3}\right) b,c=-\frac{1}{2}i \left(-i+\sqrt{3}\right) b\right\}}$

${\left\{a=\frac{12 b}{\left(3 i+\sqrt{3}\right)^{2}},c=\frac{1}{2}i \left(i+\sqrt{3}\right) b\right\}}$

${\left\{a=-\frac{2 \left(i+3 \sqrt{3}\right) b^{3}}{-2^{2/3}\left(5 i+\sqrt{3}\right) b \sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}+\sqrt[3]{2}\left(5 i+\sqrt{3}\right) \left(\left(i \sqrt{3}+1\right) b^{3}\right)^{2/3}+\left(2 i+6 \sqrt{3}\right) b^{2}},\right.}$
${\left.c=-\frac{\sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}}{\sqrt[3]{2}}-b-\frac{\sqrt[3]{2}b^{2}}{\sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}}\right\}}$

${\left\{a=\frac{16 b^{4}\left(-5 2^{2/3}\left(-3 i+\sqrt{3}\right) b \sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}-5 \sqrt[3]{2}\left(3 i+\sqrt{3}\right) \left(\left(i \sqrt{3}+1\right) b^{3}\right)^{2/3}+12 \left(i+\sqrt{3}\right) b^{2}\right)}{\left(-i+\sqrt{3}\right) \sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}\left(-8 b i \sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}+\left(i+\sqrt{3}\right) \left(\left(2 i \sqrt{3}+2\right) b^{3}\right)^{2/3}+4 (-1)^{5/6}\sqrt[3]{2}b^{2}\right)^{2}},\right.}$
${\left.c=\frac{-(-2)^{2/3}b \sqrt[3]{\left(i \sqrt{3}+1\right) b^{3}}+\sqrt[3]{2}\left(\left(i \sqrt{3}+1\right) b^{3}\right)^{2/3}-2 b^{2}}{2 b}\right\}}$

[/hide]

Like You See, It's impossible for this be real...

Our system is equivalent to
\[\sum (2a^{3}-b^{3}-c^{3}-3a^{2}b+3b^{2}c)^{2}=0\]
or
\[\sum a^{3}(a-b)^{3}+b^{3}(b-c)^{3}+c^{3}(c-a)^{3}=3abc(a-b)(b-c)(c-a)\]
According to the fact that
\[x^{3}+y^{3}+z^{3}=3xyz \Leftrightarrow x=y=z\ or\ x+y+z=0\]
we refer that, except the trivial root $x=y=z$, our system is equivalent to
\[a(a-b)=b(b-c)=c(c-a).\]
This last system is very nice and enough three-equation. That's what I imply in this problem. An unexpected identity.

again how did you get the squared expression in the first place

are there any ways that inspired you to rewrite it that way

again its one of those sos tricks in rewritting them into squares i wish i knew :(

sorry for being so dumb

make a problem with a trick that just you knows is.... humpf

You did not make this problem in this way, you make it in the reverse way.

[quote]This last system is very nice and enough three-equation. That's what I imply in this problem. An unexpected identity.[/quote]

hey dude.. is not 3 equations, is 2... huauha...
You never give up

The $a(a-b)=b(b-c)=c(c-a)$ cant be solved for all variables.

There is reals solutions, but we cannot solve for all variables, because is a 2 equations system
ho

Yet I do not intentionally create this problem in inverse way. If you think that, could you set up a similar one? I believe that it is very complicated, because the identity here is unbelievable, very hard to find, and such identities is very hardly. To be honest, I create this one when I create an inequality, occasionally, I find that identity (and I tried many time but can not find a similar one).

Maybe the problem is hard to find, [b]but the solution is very natural[/b]. Squaring is natural way, moreover, after that, you can figure out the identity easily.

Now I will solve the last system, it is very nice.
\[a(a-b)=b(b-c)=c(c-a). \]
Subtracting side by side, we get
\[(a-b)(a+b)=b(a-c) \]

\[(b-c)(b+c)=c(b-a) \]

\[(c-a)(c+a)=a(c-b) \]
Multiplying all of them, with the assume that $a,b,c$ are distinct (if $a=b$, the system is easy), we get
\[(a+b)(b+c)(c+a)=-abc \]
or
\[(a+b+c)(ab+bc+ca)=0. \]
If $a+b+c=0$ then
\[a(a-b)=b(b-c)=b(b+b+a) \Rightarrow a^{2}-2ab-2b^{2}=0 \Rightarrow \frac{a}{b}=-1+/-\sqrt{3}. \]
Similarly for $\frac{b}{c}$ and $\frac{c}{a}$.

In case $ab+bc+ca=0$, we get $c=-\frac{ab}{a+b}$, therefore
\[a(a-b)=\frac{ab}{a+b}(a+\frac{ab}{a+b}\]
or
\[\frac{a-b}{b}=\frac{a^{2}+ab}{(a+b)^{2}}\]
or
\[\frac{a+b}{b}+\frac{b^{2}}{(a+b)^{2}}=3. \]
We can find out $\frac{a+b}{b}$ and we have done...

Just from two equation, we can find exactly the ratio (in real numbers) of
\[\frac{a}{b},\frac{b}{c}, \frac{c}{a}. \]
This is what I implies in this problem. What a pity, no one can solve it, and it is only unconstructable of yagaron  :(

I can make some inequalities, and construct a System with 3 equations too.

I can make things go harder and harder.

But the system dont have nothing in special.
It's just one more symetrical system.

Creat problems in reverse way just make the final problem harder. because there is many others ways to go to with the problem, but have just one way that go to your ways.

Just Like quantum physics and the theory of multiverse
It's impossible to go to your past, you can go to another, but no to Yours. This is one implication of the theory.

And I really think that there is nothing special in the solution.

Just like Euler's four-square identity
http://en.wikipedia.org/wiki/Euler%27s_four-square_identity

Make this is difficult, but in the reverse way is just one implication


modular arithmetic
algebra
system of equations

Find all integers a,b and c for

$(2a+b)(2b+a)=5^c$

Another one:

Find all integers k such that both k+1 and 16k+1 are square numbers.

Problem:

[color=blue]Find all $k\in\mathbb{Z}$ such that both numbers $k+1$ and $16k+1$ are perfect squares.[/color]

Assume, that $k$ satisfies the conditions.We have:
$k+1=a^2$ and $16k+1=b^2$. We multiply the first equation by $16$ and obtain $16k+16=(4a)^2$.
By subtracking equation $16k+1=b^2$ from $16k+16=(4a)^2$ we have:
$15=(4a+b)(4a-b)$

We can assume, that $a, b$ are positive, so we need to solve two system of equations:

$4a+b=15$
$4a-b=1$

and

$4a+b=5$
$4a-b=3$

We have two solutions: $a=2$, $b=7$ and $a=1$ and $b=1$

In the first case $k=3$, in the second $k=0$. That's all solutions

Thanks, do you know how to solve the first one too?

[hide="1"]This equation has a solution if and only if both factors are powers of 5. So let $2a+b=5^j$ and $2b+a=5^k$ where $j$ and $k$ are nonnegative.

Adding the equations and dividing by 3, $a+b=\frac{5^j+5^k}{3}$ so $a=\frac{2\cdot 5^j-5^k}{3}$ and $b=\frac{2\cdot 5^k-5^j}{3}$. This makes $c=j+k$.

Now $a$ and $b$ need to be integers, so $5^j\equiv -5^k\pmod{3}$. Since $5\equiv -1$ and $5^2\equiv 1$, this works if and only if $j$ and $k$ have opposite parity (one odd, the other even).[/hide]

BTW, for the above,  FWIW, there are no solutions, if one wants  both $a$ and $b$ to be non-negative integers.


calculus
integration
trigonometry
calculus computations

Put $\int \frac{dx}{\sin x \sqrt{\tan x}}$ as a (combintation of) elliptic(s) integral(s)

Thanks.

Merry Xmas.

Check out this site:

http://integrals.wolfram.com/index.jsp

I'm not sure if the answer is expressable in terms of elliptic integrals, but this gives it in terms of a hypergeometric function.

Maple 10 expresses this integral as a rather involved combination of four elliptic integrals. I tried to post the result here, but unsuccessfully.


limit
logarithms
inequalities
real analysis
real analysis unsolved

[color=darkred]Prove that $ \lim_{n\to\infty}n^{\alpha_n}\equal{}e\ ,$ where $ \alpha_n\equal{}\frac {\ln n!}{nH^2_n}$ and  $ H_n\equal{}\frac 11\plus{}\frac 12\plus{}\ldots \plus{}\frac 1n\ .$[/color]

[quote="Virgil Nicula"][color=darkred]Prove that $ \lim_{n\to\infty}n^{\alpha_n} \equal{} e\ ,$ where $ \alpha_n \equal{} \frac {\ln n!}{nH^2_n}$ and  $ H_n \equal{} \frac 11 \plus{} \frac 12 \plus{} \ldots \plus{} \frac 1n\ .$[/color][/quote]
we have
$ \lim_{n \minus{} > \plus{} oo}n^{\alpha_{n}} \equal{} \lim_{n \minus{} > \plus{} oo}e^{\alpha_{n}\ln n}$
let's find asymptotical expression of
$ \alpha_{n}\ln n$ for big $ N$.
we know $ H^{2}_{n}$~$ \ln ^{2}n$
by stirling formula we have
$ n!$~$ \sqrt {2\pi n}(n/e)^{n}$
therefore
$ \ln n!$~$ n\ln n \plus{} o(.)$ I don't consider constant
therefore
$ \frac {\ln n!}{nH_{n}^{2}}$~$ \frac {n \ln n}{n \ln^{2}n} \equal{} \frac {1}{\ln n}$
therefore
$ \lim ... \equal{} e^{\ln n \frac {1}{\ln n} \plus{} 0(.)} \minus{} > 1$
we are done

[quote="Virgil Nicula"][size=123][color=darkred][b]Prove that : $ \lim_{n\to\infty}n^{\alpha_n}=\text{e}$ , where $ \alpha_n=\frac {\ln n!}{nH^2_n}$ and  $ H_n=1+\frac 12+\ldots +\frac 1n$ . [/b][/color][/size][/quote]
[color=darkblue][u][b]Proof.[/b][/u] We will use the following known inequalities : $\{\begin{array}{cccc}
1\blacktriangleright & (\frac n{\text{e}})^n\ <\ n!\ <\ n\cdot(\frac n{\text{e}})^n\ ,\ \forall\ n\ge 7 \\\\
2\blacktriangleright & \ln n\ <\ H_n\ <\ 1+\ln (n+1)\ ,\ \forall\ n\ge 1\end{array}$ .

Thus, from $(1)$ we obtain the inequality : $n\cdot(\ln n-1)\ <\ \ln n!\ <\ n\cdot(\ln n-1)+\ln n$ , which multiplied

by $\frac {\ln n}{n\cdot H_n^2}$ becomes : $\boxed{\ \frac {(\ln n-1)\cdot\ln n}{H_n^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {[n\cdot(\ln n-1)+\ln n]\cdot \ln n}{n\cdot H_n^2}\ }\ ,\ \forall\ n\ge 7$ .

On the other hand, from the inequalities $(2)$ we can weaken the last chain of inequalities and finally

obtain that : $\boxed{\ \frac {(\ln n-1)\cdot \ln n }{[1+\ln (n+1)]^2}\ <\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}\ <\ \frac {n\cdot(\ln n-1)+\ln n}{n\cdot \ln n}\ }\ \ ,\ \forall\ n\ge 7\ \ (\ast)$ .

Since : $\{\begin{array}{lllll}
\lim\limits_{n\to\infty}\ \frac {(\ln n-1)\cdot \ln n }{[1+\ln (n+1)]^2}=\lim\limits_{n\to\infty}\ \frac {1-\frac {\ln n}n}{[\frac {1+\ln (n+1)}{\ln n}]^2}=1 \\\\\
\lim\limits_{n\to\infty}\ \frac {n\cdot(\ln n-1)+\ln n}{n\cdot \ln n}=\lim\limits_{n\to\infty}\ [1-\frac 1{\ln n}+\frac 1n]=1\end{array}\|\color{white}{.}$ $\stackrel{(\ast)}{\implies}\boxed{\ \lim_{n\to\infty}\ \frac {\ln n\cdot\ln n!}{n\cdot H_n^2}=1\ }$ .

Therefore, $\lim_{n\to\infty}\ n^{\alpha_n}=\lim_{n\to\infty}\ \text{e}^{\alpha_n\ln n}=\lim_{n\to\infty}\ \text{e}^{\frac {\ln n\cdot\ln n!}{n\cdot H_n^2}}=\text{e}$ , so we are done.
[/color]


1, 3, 21, 55 are fibonacci numbers and triangular numbers too.
find all suth numbers.

are there any bigger then 55 ?
is it known or not ?

Yes.
1, 3, 21, and 55 are the only Fibonacci trianglar numbers.

cf.[url=http://www.amazon.com/exec/obidos/ASIN/0471399698/]Fibonacci and Lucas Numbers with Applications[/url] P.12-13

thanks for the information

do you have this book or you have seen it ?

there is a complete proof in it or it is only mentioned that suth theorem exists ?


wait, hooyoung i don't understand ur answer

do u mean 1+log in base 2 of 3?

Yes, that's what he means (I hope, because it's correct).


Auf 100 Affen werden 1600 Kokosnüsse verteilt, wobei einige Affen auch leer ausgehen können.
Man beweise, dass es – ganz gleich, wie die Verteilung erfolgt – stets mindestens vier Affen mit derselben Anzahl von Kokosnüssen gibt.

 [u]Babis[/u]

 Wenn sie eine Loesung haben , bitte schicken sie es mir.

 Danke

Lösungsansatz:
Man versucht mit möglichst wenigen Kokusnüssen, allen Affen so Kokusnüsse zu geben, dass höchstens 3 die selbe Zahl haben.
Dazu gibt man je 3 Affen(Also insgesamt 99) 0,1,...,31,32 Nüsse. (insgesamst 32*33/2=1584 Nüsse).
Dem letzten muss man nun 33 Nüsse geben.
1584+33=1617>1600
=> Selbst bei optimaler Verteilung benötigt man mehr als 1600 Nüsse.


linear algebra
matrix
algebra
polynomial
abstract algebra
linear algebra unsolved

Let $A$ a real matrix such that $A^{3}=-A$.
Prove that $A$ is similar to $B= \begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix}$.

I suppose that $A$ is a $3 \times 3$ matrix :)

The three eigenvalues of $A$ satisfy $\lambda^{3}+\lambda = 0$, i.e. $\lambda \in \left\{ 0, i,-i \right\}$. Since the characteristic polynomial of $f$ is real, we have two cases:

[i]Case 1.[/i] $\lambda_{1,2,3}= 0$, i.e. $A$ is nilpotent. Hence, $A^{3}= 0$, so $A = 0$. Therefore the problem's statement is wrong.

[i]Case 2.[/i] $\lambda_{1}=0, \, \lambda_{2}= i, \, \lambda_{3}=-i$. Thus, $A$ has both non-trivial kernel and non-trivial image. So there are:
- $x \neq 0$ such that $Ax=0$;
- $y \neq 0$ such that there is $t \neq 0$ such that $A t = y$. Hence, $A^{3}t = A^{2}y \, \, \Longleftrightarrow \, \,-At = A^{2}y \, \, \Longleftrightarrow \, \,-y = A^{2}y$. Now let $z = Ay$. That implies $Az =-y$.

Now, it's pretty easy to see that $\left\{ x,y,z \right\}$ is a basis to $\mathbb R^{3}$.
First of all, if one of $y,z$ were in $\ker \, A$, then $y = 0$, contradiction.
So, if $y,z$ are not collinear with the origin, then $A$ has the desired form with respect to the basis $\left\{ x,y,z \right\}$.
Now let's suppose that $y,z=Ay$ are collinear with the origin. Then we'd have that $Ay = ky$, where $k \in \mathbb R$, so $A^{3}y = k^{3}y =-ky =-Ay$. Since $y \neq 0$, that implies $k = 0$. Therefore $y \in \ker \, A$, which is not possible, as mentioned above.

PS: I hope it's correct (it's pretty lengthy though).

Yes, A is $3\times 3$ matrix and $A\neq 0$. :)
Thanks for the quick answer, but our teacher give us a hint : Use the kernel's decomposition. ie: \[\mathbb{ R}^{3}=\ker A \oplus \ker (A^{2}+1)\]


Take $a, b, c > 0$. Maximize and minimize the expression
[b](1)[/b] $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ca}+\frac{c^{2}}{c^{2}+ab}$

Similar: $a, b, c > 0$; maximize and minimize the expressions
[b](2a)[/b] $\frac{a^{2}b}{ab^{2}+c^{3}}+\frac{b^{2}c}{bc^{2}+a^{3}}+\frac{ca^{2}}{c^{2}a+b^{3}}$
[b](2b)[/b] $\frac{a^{3}}{bc(b+c)}+\frac{b^{3}}{ca(c+a)}+\frac{c^{3}}{ab(a+b)}$

edit: by "maximum and minimum" really I mean supremum and infimum, that is, there may be no equality case (for instance if you take a limit to $0$ or $\infty$).

edit2: fixed typo in [b](2b)[/b], the exponent on the $c$.

[b](1)[/b] Minimum is not very interesting.  :oops: We easily see the expression is bigger than 1 by clearing the denominator. Take $a=1, b=c \to 0$ to see that nothing better will work.
Maximum wasn't exactly what I expected: I thought it would be $\frac{3}{2}$ at equality, but actually we can show it at most 2 by clearing the denominators again, and no less than 2 by taking $c \to 0$ while $a, b$ are just anything fixed.

I didn't try [b](2a), (2b)[/b] but based on [b](1)[/b] I guess they'll be similarly boring.  :| However people are still free to post solutions if they wish. :)

Ok I found they are not actually as similar as I thought at first. Well for [b](2a)[/b] take $a=0$. Then we just have $\frac{b}{c}$ which can be as small or large as I want. I guess that wasn't one of my finest creations.

[b](2b)[/b] also looks unbounded above with the given constraints, by taking $b=c=1$ and letting $a$ get large, for example. I guess the min is $\frac{3}{2}$ but I can't prove it 'cause I'm a noob...

of course the minimum of (2b) is $\frac{3}{2}$
from cauchy we have :
$(\sum_{cyc}abc(b+c))\left( \sum_{cyc}\frac{a^{3}}{bc(b+c)}\right)\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{2abc(a+b+c)}\geq \frac{3}{2}$


1. An office supply company manufactures desks and chairs. In a given week, the company can assemble 400 units. They must asemble at least 100 desks and 150 chairs to meet their usual demands. Their profit is $ \$30$ per desk and $ \$20$ per chair. How many of each should they sell to maximize their profit?

2. Maximize $ C\equal{}x\plus{}3y$ with the following constraints:
$ 0<\equal{}x<\equal{}6$
$ 0<\equal{}y<\equal{}6$
$ \minus{}6x\plus{}3y<\equal{}12$
$ 6x\plus{}y>\equal{}4$

Source: Thinkwell Website

Number 1 seems too easy...am I doing something wrong?
[hide="Number 1 Solution"]Since there is more profit from selling desks, we minimize the amount of chairs the company assembles to $ 150$. Then, the company can manufacture $ 400\minus{}150\equal{}250$ desks. In total, its profit is $ 150(\$20)\plus{}250(\$30)\equal{}\boxed{\$10500}$[/hide]

I hate linear programming
[hide="#2 hint"]
Graph them, and find where all the points intersect. Then plug into the original equation.

Alternate method:
Use a simplex matrix which is long and tedious.[/hide]


[quote="lasha"]Find all real valued function f:R--R such, that for any real (x,y) different from 0, we have:
f(x+y)+f(1/x+1/y)=1.
I solved it but I amn't sure in it.[/quote]

$ y=\frac{1}{x}$ we have $ f(x+\frac{1}{x})=\frac12$ if $ x+\frac{1}{x}=t$ then $ f(t)=\frac12,\quad \forall t\in (-\infty,-2]\cup [2,\infty)$.
$ x=-y$ we got $ f(0)=\frac12$ and $ y=-\frac{1}{x}\implies f(x-\frac{1}{x})+f(-x+\frac{1}{x})=1$ so $ t=x-\frac{1}{x}$ then $ f(t)+f(-t)=1\implies f(\frac{1}{t})=1-f(-\frac{1}{t})$. If $ y=x=2t$ we got $ f(4t)+f(\frac{1}{t})=1=f(4t)+1-f(-\frac{1}{t})\implies f(4t)=f(-\frac{1}{t})=\frac12$ so $ f\equiv \frac12$

Put $ x = y = 1$ to get $ 2f(2) = 1$ so $ f(2) = 1/2$. Then let $ y = 2-x$ where $ x \neq 0, x \neq 2$, to get $ f(2)+f(\frac{2}{2x-x^{2}}) = 1$, implying that $ f(\frac{2}{2x-x^{2}}) = 1/2$. This uniquely defines $ f$ for all reals except for those in $ [0,2)$ (as the image of $ \frac{2}{2x-x^{2}}$ is $ \mathbb{R}-[0,2)$).

For $ f(0)$, let $ x =-y$, then $ 2f(0) = 1$ so $ f(0) = 1/2$. For $ f(a), 0 < a < 2$, let $ x = y = a/2$. Then $ f(a)+f(4/a) = 1$. But $ 4/a \geq 2$ if $ 0 < a < 2$, so $ f(4/a) = 1/2$. Thus $ f(a) = 1/2$. This shows that all values of $ f$ are $ 1/2$.

$ f(x) = 1/2$.


category theory
advanced fields
advanced fields unsolved

1. Let $ n \geq 2.$ Show by example that the functor
      \[ \mathbb{Z}/(n) \otimes_\mathbb{Z} \minus{} : \mathbb{Z}\text{\minus{}Mod} \rightarrow \mathbb{Z}\text{\minus{}Mod}
      \] fails to be left exact.

2. Show by example that the functor
      \[ \text{Hom}_\mathbb{Z} (\mathbb{Q}, \minus{}): \mathbb{Z}\text{\minus{}Mod} \rightarrow \mathbb{Z}\text{\minus{}Mod}
      \] fails to be right exact.


[quote="worthawholebean"]In a triangle $ ABC$, take point $ D$ on $ BC$ such that $ DB \equal{} 14, DA \equal{} 13, DC \equal{} 4$, and the circumcircle of $ ADB$ is congruent to the circumcircle of $ ADC$.  What is the area of triangle $ ABC$?[/quote]
[hide]
A=abc/4R
[ADC]=13AC/R
[ADB]=91AB/2R

sinx=sin(180-x)
ADC=180-ADB by def,
So sinADC=sinADB,
Let angle ADC=a
[ADC]=26sina=13AC/R
[ADB]=91sina=91AB/2R
[ADC]/[ADB]=2/7=(13AC/R)/(91AB/2R)=13AC/(91/2)AB=2AC/7AB, so AC=AB.
By Stewart's Theorem,
(169+56)18=18AB^2
AB=15, [ABC]=$ 9\sqrt6$[/hide]
Crap Quatto told me I could've used law of sines...that would've been much simpler for the AC=AB part.

Even simpler would be to use the fact that $\frac{a}{sinA}=2R$, where $R$ denotes circumradius.

Denote the common lengths $\rho$ , $h$ of the circumradii and $A$-altitudes for the triangles $ADB$ , $ADC$ . From the well-known property $bc=2Rh_a$ in a standard triangle $ABC$ obtain $13c=2\rho h=13b$ , i.e. $\boxed{b=c=x}$ . Now in the isosceles triangle $ABC$ exists the relation $AB^2=AD^2+DB\cdot DC$ , i.e. $x^2=13^2+14\cdot 4$ $\implies$ $x=15$ ,  $h=12$ and $\boxed{[ABC]=108}$ . Can calculate and $\rho$ , i.e. $\rho=\frac {65}{8}$ .


The following problem is pretty vauge, what does it want? A closed form? What is the solution?

For positive integers $n$ find
\[\binom{3n}{0}+\binom{3n}{3}+\cdots+\binom{3n}{3n}. \]

$\frac{1}{3}*((\omega+1)^{n}+(\omega^{2}+1)^{n}+2^{n})$ where $\omega^{2}+\omega+1=0$, this is a roots of unity filter

???

How did we go from n_C_k to roots of unity? I only know roots of unity filters in the sense of generating functions, and if that expression is one there must be something obvious I'm missing.

[hide]Consider $f(x) = (1+x)^{3n}$. This is obviously

$f(x) = \sum_{k=0}^{3n}\binom{3n}{k}x^{k}$.

If we plug in $x = 1, \omega, \omega^{2}$, we get

$f(1) = \sum_{k=0}^{3n}\binom{3n}{k}$

$f(\omega) = \sum_{k=0}^{3n}\binom{3n}{k}\omega^{k}$

$f(\omega^{2}) = \sum_{k=0}^{3n}\binom{3n}{k}\omega^{2k}$.

Sum these up. Notice that $1+\omega+\omega^{2}= 0$ and $\omega^{3}= 1$ so $1^{k}+\omega^{k}+\omega^{2k}= 0$ when $k \equiv 1, 2 \pmod{3}$. When $k \equiv 0 \pmod{3}$, we get $1^{k}+\omega^{k}+\omega^{2k}= 3$. So take one-third of this expression to get the desired sum.

$\frac{1}{3}(f(1)+f(\omega)+f(\omega^{2})) = \frac{1}{3}(2^{3n}+(\omega+1)^{3n}+(\omega^{2}+1)^{3n})$.[/hide]

Which can be simplified further:

[hide]$\omega^{2}+1=-\omega,\omega+1=-\omega^{2}$, hence

$S={1\over 3}\left(2^{3n}+(-1)^{3n}(\omega^{6n}+\omega^{3n})\right)={1\over 3}\left(2^{3n}+2(-1)^{3n}\right)={1\over 3}\left(8^{n}+2(-1)^{n}\right)$[/hide]


Discovered by Ramanujan, but he hadn't proved it,he just writed it down by his intuition.
Is there anyone could prove it.
[img]http://img012.photo.21cn.com/photos/user/24454028/20070402211151/o/8369447.jpg[/img]

Pre-Olympiad forum, maybe?

Continued fractions, especially insane ones like this, are beastly.

On first instinct, seeing the $e$, I would assume that Calculus is needed here, but I may be wrong.

This probably belongs to "Advanced Fields".


calculus
integration
calculus computations

Find the volume of the solid obtained by revolving the region bounded by the graphs of $y=xe^{1-x}$ and $y=x$ around the $x$ axis.


1 rabbit: Wascaly Wabbit
1 dog: Barkely
1 fish: Pythagoras

I had many pets but they all died

3 fish- Wendy, Judy, and........ i forgot the other one

3 baby turtles- Sinnie, Cindy, and Samuel....

yer they all died within a year :(
mmmmhaaaaaaa

Three cats, Pookie, Ditto, and Taz (I only named Taz, lol)
Two guinea pigs, Fluffy and Piggy,
and some fish, Rainbow, Shark, and Stripes.

1 cat: Gusano
I used to had another cat, Phiona -In honor to the golden ratio Phi- but she died.
Between my sisters, my mom and my brother in law, they have 2 cats, 2 dogs, like 4 fish and 5 birds

A hamster Maxime (girl) who was bought in 1993 but died in 1995:

[img]http://users.telenet.be/frederic_bel6/Maxime.JPG[/img]

In 1995 we bought two male guinea pigs : Quickie and Winkie.
Those guys were lazy, nothing like the acrobatic hamster!

Anyway Winkie died in 1999 (didn't reach four years):

[img]
http://users.telenet.be/frederic_bel6/WINKIEzonderdatum.BMP
[/img]

Quickie died in 2003 and reached the age of eight years :

[img]http://users.telenet.be/frederic_bel6/quickie_beerkes.JPG[/img]

[img]http://users.telenet.be/frederic_bel6/dubbelguineasmall.JPG[/img]

[quote="nutz_for2.718281828"]I forgot my pet calculator, computer, my pet trophies, my pet ribbon, my pet certificates and my pet medal :P .[/quote]

my pet pencil died in the sharpener ;)

I had a dog,sally, and my mom gave it away to our cleaner...
Then we had 3 hamsters, one after another...
I forgot the first 2 hamsters name but the 3rd was bono.

i have one kasco, and his name is kakoo.

he has been too wild :!:

because i give him meat every day.

he can speak very well. :D

Ashegh, who is your master? :P

Just kidding!

I also have a pet ipod.

i had hamsters who ate each other

they were very cute

i had dogs who didn't eat each other

they were also very cute

i gave away a dog that bit my sister

he was very cute

i have a dog who was born on 9/11

he is very cute

dude but is your sister cute THAT IS THE MAIN PIECE OF INFORMATION THAT IS CURIOUSLY MISSING FROM YOUR ABOVE POST


inequalities
number theory unsolved
number theory

the integers a and b have the property  that for every nonnegative integer n the number of $ 2^n.a\plus{}b$ is the square of an integer .prove that a=0
thank you very much

[quote="DuongChinh_K41SPHN"]the integers a and b have the property  that for every nonnegative integer n the number of $ 2^n.a \plus{} b$ is the square of an integer .prove that a=0
thank you very much[/quote]

First, suppose $ b\equal{}0$. then, in order to have $ u$ and $ 2u$ perfect squares, we need $ u\equal{}0$ and so $ a\equal{}0$
Then, suppose $ b\neq 0$

Let $ 2^na\plus{}b\equal{}a_n^2$ with $ a_n\geq 0$
Then $ a_{n\plus{}2}^2\equal{}2^{n\plus{}2}a\plus{}b\equal{}4(2^n\plus{}b)\minus{}3b\equal{}4a_n^2\minus{}3b$ and so $ 3b\equal{}(2a_n)^2\minus{}a_{n\plus{}2}^2$

If $ b>0$, we have $ 3b\geq(a_{n\plus{}2}\plus{}1)^2\minus{}a_{n\plus{}2}^2$ and so $ 3b\geq 2a_{n\plus{}2}\plus{}1$
If $ b<0$, we have $ 3|b|\geq(2a_n\plus{}1)^2\minus{}(2a_n)^2$ and so $ 3b\geq 4a_{n}\plus{}1$

If $ a\neq 0$, right parts of these two inequalities grow up to $ \plus{}\infty$ when $ n\rightarrow \plus{}\infty$, which is impossible since left part is fixed.

So $ a\equal{}0$

Just a comment:
You can also find this problem in PEN section,but thw solution given there,is the same.

This is problem from Poland mathematical competition and it was also posted before on this forum. Please visit:
[url]http://www.mathlinks.ro/viewtopic.php?p=150403#150403[/url]


In a bag with 20 marbles, five of them are blue. How many blue marbles must be added to the bag so that the probability of selecting a blue marble at random is $ \frac{1}{2}$?

Let $ x$ be the number of blue marbles we should add.  We are given that
\[ \frac {5 \plus{} x}{20 \plus{} x} \equal{} \frac {1}{2}\]
Cross multiplying yields
\[ 10 \plus{} 2x \equal{} 20 \plus{} x\]
Thus, $ x \equal{} \boxed{10}$

Alternately, you know that when the number of blue marbles and the number of other marbles are equal, there will be a $ \frac{1}{2}$ probability of selecting a blue marble at random.  Since there are $ 5$ blue marbles, there are $ 15$ other marbles.  Thus, we need $ \boxed{10}$ more blue marbles to make them equal.


geometry
parallelogram
trigonometry
Gauss
geometric transformation
reflection
ratio

Let $ ABCD$ be a cyclic quadrilater. Denote $ P\equal{}AD\cap BC$ and $ Q\equal{}AB \cap CD$. Let $ E$ be the fourth vertex of the parallelogram $ ABCE$ and $ F\equal{}CE\cap PQ$. Prove that $ D,E,F$ and $ Q$ lie on the same circle.

i want to show that:$ CD.CQ\equal{}CE.CF$(1)

then i will find these length,according to their relations to each other:

according to a lemma in inversion,if i choose $ P$,the center,and power of $ P$ to circle of the quadrilateral,as the radius$ K$,

i can find the length of $ CD$ this way:

$ CD\equal{}\frac{k.AB}{PA.PB}\equal{}\frac{PB.PC.AB}{PA.PB}\equal{}\frac{PC.PB}{PA}$

$ ABCE$ is paralelogram,then: $ CE\equal{}AB$

we know that:$ QC.QD\equal{}QB.QA$,then:$ QC\equal{}\frac{QB.QA}{QD}$

triangles $ PBQ,PCF$are similar to each other,then:$ \frac{CF}{BQ}\equal{}\frac{PC}{PB}$,

then we conclude:$ CF\equal{}\frac{PC.BQ}{PB}$

now we find the length,lets use them in (1)equality,and check if it is right or not.

$ \frac{PC.AB}{PA}.\frac{QB.QA}{QD}\equal{}AB.\frac{PC.BQ}{PB}$

it concludes that,we should have:$ \frac{QA}{PA.QD}\equal{}\frac{1}{PB}$

$ \frac{PA}{PB}\equal{}\frac{PC}{PD}$

now i should chek that if it is right or not.

ok,it is clear that $ \angle PCQ\equal{}\angle QAD$

$ \sin{PCQ}\equal{}\sin(180\minus{}QAD)$

suppose$ PH,QH'$are the altitudes of triangle $ PDQ$.

we can find the area of$ PDQ$,in two ways:

$ S\equal{}PH.QD\equal{}QH'.PD$

but:$ PH\equal{}PC.\sin{PCQ}$ and $ QH'\equal{}QA.\sin(180\minus{}QAD)$

and we conclude that: $ PC.QD\equal{}QA.PD$and it is proved that:$ \frac{PA}{PB}\equal{}\frac{QA}{QD}$

and every thing is ok,but we should go back these steps from end to first,

and then the proof will be complete :wink:

Let $ K,L,M$ be midpoint of $ BD,CA,PQ$. By Gauss line theorem(See http://www.mathlinks.ro/Forum/viewtopic.php?t=79751) , $ K,L,M$ are collinear. Then reflection of $ B$ in the point $ K,L,M$ are also collinear, i. e, $ D,E,X$ are collinear where $ X$ is reflection of $ B$ in the point $ M$. Since $ XPBQ$ is parallelogram,$ \angle QXP \equal{} \angle PBQ \equal{} \angle CBA \equal{} \angle CDA \equal{} \angle QDP$. Hence $ X,P,D,Q$ are cyclc. From $ XPBQ$ and $ ECBA$ are parallelogram, $ XP\parallel EC$. So $ \angle DEF \equal{} \angle DXP \equal{} \angle DQP \equal{} \angle DQF$. So $ D,E,F,Q$ are cyclic.

[quote="vittasko"]We draw the line through the point $ Q$ and parallel to $ BC,$ which intersects the line segment $ EF$ at one point so be it $ F'.$

It is enough to prove that $ CD\cdot CQ \equal{} CE\cdot CF$ $ \Longrightarrow$ $ CD\cdot CQ \equal{} AB\cdot CF$ $ ,(1)$

From $ E'F\parallel AQ,$ where $ E'\equiv AD\cap EF,$ we have by [b][size=100]Thales theorem[/size][/b], $ \frac {AB}{BQ} \equal{} \frac {CE'}{CF}$ $ \Longrightarrow$ $ AB\cdot CF \equal{} BQ\cdot CE'$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ CD\cdot CQ \equal{} BQ\cdot CE'$ $ ,(3)$ and so, it is enough to prove that $ CD\cdot CQ \equal{} CF'\cdot CE'$ $ ,(4)$

But, the relation $ (4)$ is true, from the cyclic quadrilateral $ DE'QF',$ because of

$ \angle DE'F' \equal{} \angle DAB \equal{} \angle BCQ \equal{} \angle DQF'$ $ \Longrightarrow$ $ \angle DE'F' \equal{} \angle DQF'$ $ ,(5)$

Hence, we conclude that $ CD\cdot CQ \equal{} CE\cdot CF$ and the proof is completed.

Kostas Vittas.[/quote]

Dear Kostas,

 [b]nice solution ! [/b] In the official solution ( very comprehensive )I read the relation :

   $ \frac { CP}{DP} \equal{} \frac {CQ}{BQ}$

 How can we deduce this relation without trigononetry ?

 babis

[quote="vittasko"][quote="stergiu"]....
Dear Kostas,  In the official solution ( very comprehensive )I read the relation :

$ \frac { CP}{DP} = \frac {CQ}{BQ}$

How can we deduce this relation without trigononetry ?

babis[/quote]
Because of the triangles $ ($ schema [b][size=100]t=191681[/size][/b] $ )$ $ \bigtriangleup PDC$ and $ \bigtriangleup QBC,$ have $ \angle PCD = \angle QCB$ and $ \angle PDC + \angle QBC = 180^{o},$

we conclude that $ \frac {CP}{CQ} = \frac {DP}{BQ}$

$ ($ The ratio of their side-segments opposite to their equal angles, is congruent to the ratio of their side-segments opposite to their angles with sum $ 180^{o}$ $ ).$

But, how is the official solution you mentioned? Have you any reference about it?

Best regards, Kostas vittas.[/quote]

 [color=blue]I had found a link in romanian languange, but I have lost it. I will have the official solution next year , when I will buy the book with all math olympiads for year 2008.[/color]
Babis

The official solution is based on some raport equality and one Menelaos.

Let $ AE$ meet $ CD$ at $ K$
The line parallel with $ QC$ through $ P$ meet $ BQ$ at $ M$ $ (1)$
We have $ \angle KCE\equal{}\angle AQD\equal{}\angle PMA$ (because $ CE//BQ$ and (1))
and $ \angle PAM\equal{}\angle DCB\equal{}\angle EKC$ ( because $ A,B,C,D$ are cyclic and $ AE//BC$ )
Thus triangles $ PMA$ and $ ECK$ are similar $ (2)$
We have $ CK/CD\equal{}PA/PD\equal{}MA/MQ$ $ (3)$
From (2) and (3) $ CE/MP\equal{}CK/MA\equal{}CD/MQ$
Thus triangles $ PMQ$ and $ ECD$ are similar
So $ \angle EDC\equal{}\angle PQM\equal{}\angle QFC$
Hence $ Q,D,E,F$ are cyclic $ q.e.d$


modular arithmetic
number theory
least common multiple
floor function
LaTeX
MATHCOUNTS
function

how do u solve mod equations w/ (mod x) at the end?
like $ 57\equiv 2 \pmod{x}$, find all values of x or something.

and there wuz these funny problems:

1)Lisa thought of an integer greater than 10, that when divided by 2, 3, 4, 5, or 6 has a remainder one less than the divisor. When divided by 7 the remainder is one. What is the smallest such integer?

2)If a three-digit number of the form [u]1N1[/u] is idvided by N, the quotient is of the form [u]2N[/u] remainder 5. What is the value of N?

3)What is the remainder when 5^207 is divided by 7?

4)What is the remainder when 3^127 is divided by 5?

5)What is the remainder when 10,987,654 is divided by 2,103?

o yeah, for #3 and #4, i no it for one more/less than the number but idk for 2 less/more...

for #1, i tried by finding lcm of 2,3,4,5,6 which is 60, and then -1=59. so $ 59 \plus{} 60k \equiv 1 \pmod{7} \longrightarrow k\equiv 3 \pmod{7}$ wut next?

@math154: wut do u do, u no, on that last time when at the end we had fractions and stuff and it HAD to be an integer, wut if u had to add a number in the 1000's? how would u find?

thx!  :)

I don't quite know what you are asking. Could you clarify?

here we go:

[quote="math154"]He either used guess and check or my method posted above.

@toadoncart:
We have $ x \equal{} 4a \plus{} 2 \equal{} 7b \plus{} 4$ for INTEGERS $ a,b$. Take this $ \bmod4$ to get $ 3b\equiv2\pmod4$, which is essentially what you had. But if we simply said, for all integers $ c$, $ 3b \equal{} 4c \plus{} 2\implies{b \equal{} \frac {4c \plus{} 2}3}$, we wouldn't always have an integer value for $ b$, which we already assumed above. You can't just plug in any value of $ c$...you have to plug in a value of $ c$ such that $ b$ is an integer. So you need to first solve the congruence $ 3b\equiv2\pmod4\implies{3b\equiv6\pmod4}$, and since $ \gcd(3,4) \equal{} 1$, $ b\equiv2\pmod4\implies{b \equal{} 4c \plus{} 2}$ for some integer $ c$ (notice how this holds for ALL values of $ c$). THEN you can plug it back in and solve. I hope this helps![/quote]

I still don't understand.
Do you mean if we had to solve something like:
$ x\equiv500\pmod{1349}$
$ x\equiv600\pmod{1587}$
?
If so, the technique is the same. We let $ x\equal{}1349a\plus{}500$ and substitute it to get $ 1349a\equiv100\pmod{1587}$. Of course, this is very tedious.

For your first question, I think in $ a \equiv b \pmod{n}$, you would have to subtract $ b$ from $ a$.

Then, you would have to find the factors of $ a \minus{} b$, and excluding $ 1$, you have your solution set.

[quote="toadoncart"]how do u solve mod equations w/ (mod x) at the end?
like $ 57\equiv 2 \pmod{x}$, find all values of x or something.

for #1, i tried by finding lcm of 2,3,4,5,6 which is 60, and then -1=59. so $ 59 \plus{} 60k \equiv 1 \pmod{7} \longrightarrow k\equiv 3 \pmod{7}$ wut next?
[/quote]
till 59 its correct, but u messed up after that. we have $ (60k\minus{}1)$ as the number for $ 2,3,4,5,6$ and and it leaves 1 remainder on division by $ 7$. so, $ (60k\minus{}1)\minus{}1\equiv0\pmod7$. now, try and find the answer

[hide="My Way for #3 and #4"]Well, use my way. :P

(I presume that $ a^b\mod{n} \equal{} x$.)

First, find $ a\mod{n}$.  This becomes your new $ a$ value.

$ 5^{207}$

$ 3^{127}$

Then, find $ \lfloor \frac{b}{n} \rfloor \plus{} (b\mod{n})$.  Repeat this step until $ b < n$.

$ 5^{207} \rightarrow 5^{33} \rightarrow 5^9 \rightarrow 5^3$

$ 3^{127} \rightarrow 3^{27} \rightarrow 3^7 \rightarrow 3^3$

Then, simply find the answer.

$ 125\mod{7} \equal{} \boxed{6}$

$ 27\mod{5} \equal{} \boxed{2}$[/hide]

is this memorization stuff, ernie? cuz i like the understanding stuff better, but o well. i might as well just mem it. :)

No, it's just something I learned from 1markkram2.  :oops:

Oh, math154 proved it in a HSB thread.

[quote="ernie"][hide="My Way for #3 and #4"]Well, use my way. :P

(I presume that $ a^b\mod{n} = x$.)

First, find $ a\mod{n}$.  This becomes your new $ a$ value.

$ 5^{207}$

$ 3^{127}$

Then, find $ \lfloor \frac {b}{n} \rfloor + (b\mod{n})$.  Repeat this step until $ b < n$.

$ 5^{207} \rightarrow 5^{33} \rightarrow 5^9 \rightarrow 5^3$

$ 3^{127} \rightarrow 3^{27} \rightarrow 3^7 \rightarrow 3^3$

Then, simply find the answer.

$ 125\mod{7} = \boxed{6}$

$ 27\mod{5} = \boxed{2}$[/hide][/quote]
thats a good trick. if u hav read fermats theorem , u cud hav found even a better one.
[hide="wut u did"]by fermats theorem, we hav $ a^p\equiv a\pmod p$. so, if we want to find $ a^k\pmod p$. then wut u did was split k into several p's and the remainder i.e. $ a^k = a^{p + p + p + p + p\cdots.. + p + r}$, where r is the left power. and then$ a^p\equiv a\pmod p$, we replace and get wut u explained.[/hide]
[hide="better"]$ a^{p - 1}\equiv1\pmod p$, so just divide the power by $ p - 1$ and get the required power.like, here,we hav to find $ 5^{207}\pmod7$. divide207 by (7-1). remainder=3. so, it is equivalent $ 5^3\equiv\pmod7$[/hide]

@anyone new to number theory: my advice is not to follow these shortcut methods. solve from normal congruences as long as u know abt fermat and all and how to prove the trick perfectly. moreover ur forming basics right now, so conc. on normal congruences n concrete y're concepts

Here is a trick for $ a^b \equiv c \pmod{(a \plus{} 1)}$.

If $ b$ is even, $ c$ is $ 1$.

If $ b$ is odd, $ c \equal{} a$.

You need to put dollar signs around the code.

[quote="Sean Fankey"]So with a problem with an unknown mod value, you would just find the largest answer for x and then find all the factors? So in the case of the problem in the first post would it be x = 5, 11, 55?

On a semi-related note, how do you type the mathematical characters? I know it's a type of code called LaTeX, and while I was going to dowload it anyway, do I have to do that to use it directly on AoPS?

It didn't work when i just typed the code (like 57\equiv 2 \pmod{x}).[/quote]

[code]$57\equiv 2 \pmod{x}$[/code]
yields
$ 57\equiv 2 \pmod{x}$

[quote="grn_trtle"][hide="Problem with #1"]
We have a system of congruences:
$ n\equiv{1}\pmod{2}$
$ n\equiv{2}\pmod{3}$
$ n\equiv{3}\pmod{4}$
$ n\equiv{4}\pmod{5}$
$ n\equiv{5}\pmod{6}$
$ n\equiv{1}\pmod{7}$

Chinese Remainder Theorem tells us that there exists a solution for $ n$ which satisfies all the congruences if 2,3,...,7 are pairwise coprime, which they obviously aren't (2,4,6?)
Doesn't this mean there are no solutions? Or can there exist solutions even if they are not pairwise coprime?
[/hide][/quote]

There can be solutions if they are not pairwise relatively prime, but there is not always a solution.

Ex:  $ x\equiv 2\pmod 4$ and $ x\equiv 2\pmod 6$, which is the same thing as $ x\equiv 2\pmod{12}$

Ex 2:  $ x\equiv 2\pmod 4$ and $ x\equiv 1\pmod 2$; the first implies that $ x$ is even, but the second implies that $ x$ is odd.

found an example for the (mod x) problems:
"When 97, 151, and 241 are each divided by a positive integer K, the remainder is the same. What is the largest possible value of K?" ('07-'08 MATHCOUNTS School Handbook)

and for that problem with the "5^207 divided by 7 wut is remainder?" problem, there's a pattern for 5^x=c (mod 7), where c is a constant and x is increasing consecutively right?  :|

$ 97\equal{}ak\plus{}r$
$ 151\equal{}bk\plus{}r$
$ 241\equal{}ck\plus{}r$
Subtracting these two at a time, we have
$ 54\equal{}2\cdot3^3\equal{}(b\minus{}a)k$
$ 90\equal{}2\cdot3^2\cdot5(c\minus{}b)k$
$ 144\equal{}2^4\cdot3^2(c\minus{}a)k$
Thus, $ k$ must be common to all of these numbers, and must be of the form $ 2^x\cdot3^y$, where $ x\leq1$ and $ y\leq2$. Now it's easy to check the possible values: $ 1,2,3,6,9,18$. The answer is $ \boxed{18}$.

[hide="#2"]Subtract 5 (the remainder) from 1N1, which leaves 1(N-1)6.  We know that 2N times N equals a number ending in 6, so the only possibilities for N are 4 and 6.  However, 24*4 is 96, which is not in the form 1(N-1)6.  So the only answer is 6.[/hide]

So, Fermat's theorem says

$ a^{p\minus{}1}\equiv 1 \pmod{p}$

for $ p$ being any prime.

Does it also say that $ a$ must be relatively prime to $ p$?

If it doesn't, than I'm really confused.

Also, let's say the problem was: If you divide $ 5^{45}$ by 12, what's the remainder? So you set it up:

$ 4^{45}\equiv x \pmod{12}$

Using a general form of a mod equation $ a^b\equiv x \pmod{c}$

To use Fermat's theorem you need the mod value to be a prime, so can you just use a prime factor of the $ c$? I guess it would have to be one that is relatively prime to $ a$. And then, would I just do it out, but stop a little early and switch back to the original $ c$, so I don't overshoot?

So in the above problem:

$ 4^{45}\equiv x \pmod{12}$

becomes

$ 4^{45}\equiv x \pmod{3}$

(because you can't use 2 because it's not relatively prime to 4. Also it's useless in these things anyway.)

$ 4^3\equiv x \pmod{3}$

I stopped a little early

$ 4^3\equiv x \pmod{12}$

$ 64\equiv x \pmod{12}$

$ 4\equiv x \pmod{12}$

So when $ 4^{45}$ is divided by 12, it leaves a remainder of 4.

Did I do that right?

[quote="Sean Fankey"]...Does it also say that $ a$ must be relatively prime to $ p$?...
...Also, let's say the problem was: If you divide $ 5^{45}$ by 12, what's the remainder?...[/quote]

1. Yes. But since $ p$ is prime, this simply means that $ a\not\equiv0\pmod p$.
2. You had two different problems, so I will answer both...

a. $ 5^2\equiv1\pmod{12}\implies5^{45}\equiv\boxed5\pmod{12}$

b. $ 4^{45}\equiv1^{45}\equiv1\pmod3$ and $ 4^{45}\equiv0\pmod4$, so $ 4^{45} \equal{} 3a \plus{} 1 \equal{} 4b$.
Taking this $ \bmod3$, we get $ b\equiv1\pmod3\implies b \equal{} 3c \plus{} 1$. Plugging this into the equation above, we get $ 4^{45} \equal{} 12c \plus{} 4\equiv\boxed4\pmod{12}$.

More more general methods, refer to

[url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url]
[url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem]Euler's Totient Theorem[/url]
[url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]Carmichael Function[/url]

Thanks. And sorry, I mean to write $ 4^{45}$ but I accidentally wrote $ 5^{45}$ that first time.

Just randomly, I'm the same as Sean Fankey, but I decided to go with something more mathy.


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

$ \frac {11}{18} \geq \frac {(1 \minus{} a)(2 \minus{} b)}{(3 \minus{} c)^2} \plus{} \frac {(1 \minus{} b)(2 \minus{} c)}{(3 \minus{} a)^2} \plus{} \frac {(1 \minus{} c)(2 \minus{} a)}{(3 \minus{} b)^2} \geq \frac {15}{32}$
 BQ

Why???
Why solve not?
Why ask
Solve = can
ask = no.
Show solution here
Good luck
I know you can
or not
I never not try AM-GM
\[\frac{(1-a)(2-b)}{(3-c)^{2}}+\frac{(1-b)(2-c)}{(3-a)^{2}}+\frac{(1-c)(2-a)}{(3-b)^{2}}\geq\big(\frac{(1-a)(2-b)(1-b)(2-c)(1-c)(2-a)}{(3-c)^{2}(3-a)^{2}(3-b)^{2}}\big)^{1/3} \]
easy
no solve yet
Yet
Lucky
DQ

That is too easy,
See this:

Let $a,b,c>0$,such that $a+b+c=1$,prove that:

$3 \geq \sqrt{a+b^2+c^2}+\sqrt{b+c^2+a^2}+\sqrt{c+a^2+b^2} \geq \sqrt{5}.$
please give a proof.

BQ

[quote="kuing"][quote="minhhoang"]
 $ \le \sqrt {3(a + b + c + 2({a^2} + {b^2} + {c^2}))}$
 $ \le \sqrt {3(a + b + c + {{(a + b + c)}^2})}$
  = 3
[/quote]
the second one I think it should be $\le \sqrt {3(a + b + c + {{2(a + b + c)}^2})}$ ?[/quote]
Oh yes,  thanks kuing, I have a mistake in typing, i will edit it :)


[quote="chessfreak"]a and b are odd integers such that a^2-b^2+1 divides b^2-1. Prove that a^2-b^2+1 is a perfect square.[/quote]

This has been posted 6 times this summer.  But, it is nice to know where the problem comes from, imotc camp 2007 postal problem.  :)  Welcome to the forum  :) and see
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162545
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5935
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161877

i request you all not to post the postal problems or else i will complain to BJV . :mad:

it is the height of unethical behaviour.

the last date  was 25 August,2007 and you are trying to get answers from external help which is strictly prohibited.

undue advantage for people who have access to AOPS.

i hope all of you take my words seriously and refrain from further malpractice. :huh:


don't have any idea with this limit:
compute:
lim n-->oo (lim -->0 ((n! tan(x)*tan(x/2)*..*tan(x/n)-x^n)/x^(n+2))).

the answer should be between:

a)1/18
b)pi
c)pi^2/18
d)pi^2/31
e)pi/18

bye

The answer is pi^2/18, but I don't have a solution yet!!! :(

cheers!

[quote="Lagrangia"]okay I got a solution, but just gonna post a hint:

use lim y-->0 (tany-y)/y^3.

cheers!! :D[/quote]

 i have a solution Lagrangia, i follows your hint

when x tends to 0

n! tgx.tg(x/2)....tg(x/n)

= x^n (1 + x/3 + o(x^))(1+x/3.2+o(x))....(1+ x/3.n + o(x))

= x^n + x^(n+2)/3.(\sum_{k=1 to n}1/k + o(x^(n+2))

thus when x-->0 and n-->+oo the limite is

dzeta(2)/3 =  \pi/18

There is another problem simmilar to this one with sin:

Compute lim (x-->0) [n!x^n-sinx*sin2x*..*sinnx]/x^(n+2)

The result is: n*(2n+1)*(n+1)!/36.

cheers! :D :D

[quote="Lagrangia"]

Compute lim (x-->0) [n!x^n-sinx*sin2x*..*sinnx]/x^(n+2)

The result is: n*(2n+1)*(n+1)!/36.
[/quote]
This one is cool for sunday midnight,
sin u = u - u^3/6 +o(u^3) when u tends to 0
n!x^n-sinx*sin2x*..*sinnx = n!x^n(1 - (1-x/6+o(x))...(1-n^2x/6+o(x)) = n!x^n((1+2+...+n)/6x+o(x))
the limit of the problem is
(1+2+...+n)/6 = n(n+1)(2n+1)/36
chers
Moubinol


There are infinitely many piles of stones. Some pile $ X$ has $ 1$ stone, the pile after that has $ 2$ stones, the next $ 3$ stones ...  a pile $ Y$ has $ n$ stones. The pile next to $ Y$ has $ n\minus{}1$ stones, the next one $ n\minus{}2$ stones ... a pile $ Z$ has $ 1$ stone. To the left of pile $ X$, all piles are empty; to the right of pile $ Z$, all piles are empty. (So the piles have stones as follows:$ ..0, 1, 2, 3.. n, n\minus{}1, n\minus{}2 ... 2, 1, 0, 0 ...$)
In one move, we can :
- remove one stone and move it to the pile $ 3$ to the left.
- remove one stone each from two adjacent piles, and place one stone, in the pile to the right of the two adjacent piles; also, we can do the reverse of this: take one stone from one pile, and place one stone in each of the two piles left of it.
For which $ n$ can be we end up with just one stone?

The first move lets us transform 2 piles into a single one with all the stones of the two if the piles' distance is divisible by three.(The "left" part is not important.To move some stone three places right, we can move all oter stone three places left).

The second move combined with the first lets us remove two stones from two piles and put one into a third if none of the three piles' (pairwise) distance is divisible by three(And the converse also).

So let us transform the whole arrangement into three consecutive piles(with the first move) and forget all others. Now the new single move is--- remove 2 stones from two piles and put one into [b]the[/b] third(And the converse also).

Now we can obtain that one last stone can not be achieved iff the sizes of the piles(number of stones in it) are pairwise of the same parity.(same mod 2)

The number of stones in each three piles can be derived in terms of n, and then it will be easy to show that [u][b]n must not be divisible by 3.[/b][/u]


A jar contains only nickels, dimes and quarters. There is at least one of each type of coin in the jar. If the total value of the coins in the jar equals 60 cents, how many quarters are in the jar?

wat does that have to do with the actual question?

There can clearly only be 1 or 2 quarters.  If there are 2 quarters, then the total value of the coins must be at least $ 2(25)\plus{}10\plus{}5\equal{}65$ cents, which is too large.  Thus, there is $ \boxed{1}$ quarter


integration
trigonometry
complex analysis
complex analysis unsolved

[hide="Hint"]$ \int_{ \minus{} \infty}^{ \plus{} \infty}\frac {\cos 2x}{1 \plus{} x^4}dx \equal{} \Re\left(\int_{ \minus{} \infty}^{ \plus{} \infty}\frac {e^{i2x}}{1 \plus{} x^4}dx\right)$[/hide]

hello, show that $ \int_{\minus{}\infty}^{\infty}\frac{\cos(2x)}{1\plus{}x^4}\,dx\equal{}\frac{e^{\minus{}{\sqrt{2}}}\pi(\cos(\sqrt{2})\plus{}\sin(\sqrt{2}))}{\sqrt{2}}$.
Sonnhard.


algebra
polynomial
number theory
relatively prime
Galois Theory
superior algebra
superior algebra unsolved

Its not so hard but its nice.
Let $ n,m$ be relatively prime positive integers.
Let $ E$ be the cyclotomic extension $ x^{n}\minus{}1$ over $ \mathbb{Q}$.
Let $ F$ be the cyclotomic extension $ x^{m}\minus{}1$ over $ \mathbb{Q}$.
Let $ K$ be the cyclotomic extension $ x^{nm}\minus{}1$ over $ \mathbb{Q}$.
Then, prove that,
$ \mbox{Gal}(K/\mathbb{Q})\simeq\mbox{Gal}(E/\mathbb{Q})\times\mbox{Gal}(F/\mathbb{Q})$.

Let $ L,M$ be Galois extensions over $ K$ such that $ L\cap M \equal{} K$. Then $ \mathrm{Gal}(LM|K)\cong\mathrm{Gal}(L|K)\times\mathrm{Gal}(M|K)$.
Or check it directly as we know that the $ n$-th cyclotomic extension has Galois group $ (\mathbb Z / n\mathbb Z)^{*}$.

Suppose we replace the 2 polynomials by
x^k -2 and x^j -2.( k,j odd and co-prime). Don't we have a similar result?


How many lattice points are there in the interior of a circle with radius 4.5 units centered at the origin on the Cartesian plane?

There are $ 4\times 4 \plus{}1 \equal{} 17$ lattice points along the axes inside the circle $ x^{2} \plus{}y^{2} \equal{} \frac {81}{4}$.

Then, considering th first quadrant, $ (4, 1), (4, 2), (3, 3), (2, 4)$ and $ (1, 4)$ are the point furthest from the origin and inside the circle. This gives a further $ 13$ points in the first quadrant inside the circle and so $ 4 \times 13 \equal{} 52$ point in all four quadrants. So the total number of lattice points is $ 17 \plus{} 52 \equal{} 69$.


linear algebra
matrix
induction
superior algebra
superior algebra solved

Let A, B be square matrices of order n, AB=BA, A^1977=B^1978=E. Show that A+B+E is invertible.

Am I missing smth or is the problem false?

take A=B=E=On and we have AB=BA and also A^1977=B^1987=E, but X=A+B+E is not invertible!

Sorry, I forgot to say that E = Indentity matrix

Namdung

Lets denote X=A+B+E then if det(X)=O then the ecuation: XY=0 has different solutions from On. So we have: B+E=-AY.
It's easy to prove using induction that (B+E)^p=(-1)^pA^pY.
Because A^1977=E then (B+E)^1977*Y=Y so: ((B+E)^1977-E)*Y=On= (B^1978-E)Y.

Let p[x]=(x+1)^1977-1 and q[x]=x^1978-1 be two polinomials.
We have that p[B]Y=q[B]Y=0.
we have that (p,q)=1 and also (q[x]-1)/(x-1) is the minimal polinomial.
So there exists polinomials f[x] and g[x] so that pf-gq=1 we have also that Y=(f[B]-g[B])p[B]Y=0, which is impossible.

so we have X<>On.

hope it's okay!

I think the generalization works just fine!

Let A, B be square matrices of order n, AB=BA, A^k=B^(k+1)=E. Show that A+B+E is invertible.

Am I right?

Namdung why do you denote the identity mtrix as E and not I (eventually In if we are talking about matrices nxn) ? :) is that the custom in Vietnam?

this problem is hard and I have seen it in a lot of contest, of course with different values of k besides 1977 :)

[quote="Lagrangia"]I think the generalization works just fine!
Let A, B be square matrices of order n, AB=BA, A^k=B^(k+1)=E. Show that A+B+E is invertible.
[/quote]

We can see A,B matrices in Mn(C) (R is included in C)

Let  q(x)=x^k-1, q(A)=0, q plit with simple roots in C => A is diagonalisable.

The same idea with r(x)=x^(k+1) -1, we get B is diagonalisable.

A,B commute and are diagonalisable so there exist a common basis
in wich A,B are simultanously diagonalisable.

Eigenvalues a of A are in the set of the roots of q
Eigenvalues b of B are in the set of the roots of r

In the common basis A+B+I is similar to a
diag(a_1+b_1+1;...;a_n+b_n+1) where
a_m \in {exp(2i t\pi /k) where 0 \leq t \leq k}
b_m \in { exp(2it \pi/(k+1)) where 0 \leq t \leq k+1 }

any a_m+b_m+1 =/=0 so det(A+B+I)=/=0
A+B+I is invertible.

[quote="Valentin Vornicu"]Namdung why do you denote the identity mtrix as E and not I (eventually In if we are talking about matrices nxn) ? :) is that the custom in Vietnam? [/quote]

I'm russian-spoken man and E taken from "единица". Ofcourse in Vietnam, most of people use I_n, but sometimes someone (as me) use E.

Namdung


Let $x, a, b$ be positive integers and \[ x^{a+b} = a^bb \] Prove that $a=x$ and $b=x^x$

It's from Iran: http://www.mathlinks.ro/Forum/viewtopic.php?t=56766
Posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=56767 and answered in http://www.mathlinks.ro/Forum/viewtopic.php?t=36341 .


I'm sure you can do some case based analysis and figure out the probabilities etc. but I'm almost positive pocket aces is better than a queen-king suited.

[quote="bigboy82892"]If you're playing Texas Hold'em, what is a better hand to be given: two aces, or a queen and a king of the same suit? Please explain why.

Well, I said that AA would be better because the probablility of winning with the other ones are less.

Any help?[/quote]
uh yea..if this is real texas hold em you would have to analyze so many cases because the five board cards could turn your original hand into anything (straight, two pair, full house, flush...), but naturally speaking ,two aces is always the best hand (altughou I challenge someone to waste time to actually caculate all the possibilities...) -
oh yea theres a differnece if there were two people and one got AA and the other got suited KQ in [i]the same hand[/i], because then we know that the two aces and king and queen are already taken frmo the board, whereas you are comparing only AA to the rest of the field or KQ to the rest of the filed individually, not both are dealt in the same hand

well, we're assuming you're the only player ;] (I know, it's lame...)

heres a website with all the probabilities of something happening.  it shows the math too!!

wikipedia is awesome :D

[url=http://en.wikipedia.org/wiki/Poker_probability]poker[/url]

well, that doesn't really answer the question...  :|

The answer is definitely two aces; you can get a 2-pair, 3-of-a-kind, Full house, or four-of-a-kind out of it and have a pair anyway if none of those come up.  With Q-K, you need a straight.

[quote="miyomiyo"]  With Q-K, you need a straight.[/quote]

Not necessarily.  Since the Q and K are suited the chances of a flush are much higher than the chances of a straight.  The other player has two aces, which are obviously not suited.  If the player with QK gets the flush the player with AA has a low chance of beating it since the only way to win is with
1. A higher flush or a straight flush
2. four of a kind
3. Full house

Yes, but for the two hands needed with Q-K, you need 5 cards and have nothing if you don't get the 3 cards you want.


trigonometry
calculus
calculus computations

Find the [b]points[/b] for which the tangent line to the polar curve $ r\equal{}1\plus{}\sin \theta$ is horizontal or vertical.

Note: When solving a trigonometric equation, express in general solution formed, i.e. $ \theta \equal{} \pi/6\plus{}2k\pi, k \in \mathbb{Z}$, not just $ \theta \equal{} \pi/6$. Also, consider the case when $ dy/d\theta \equal{} dx/d\theta \equal{}0$. Use L'Hopital?

I am stuck. It is my first time doing such problems so please help out so that I can solve similar problems.

Let $ x\equal{}r\cos\theta\equal{}(1\plus{}\sin\theta)\cos\theta$ and $ y\equal{}r\sin\theta\equal{}(1\plus{}\sin\theta)\sin\theta.$

Find $ \frac{dx}{d\theta}$ and $ \frac{dy}{d\theta}.$

Any point for which $ \frac{dx}{d\theta}\equal{}0$ while $ \frac{dy}{d\theta}\ne 0$ is a vertical tangent.

Any point for which $ \frac{dy}{d\theta}\equal{}0$ while $ \frac{dx}{d\theta}\ne 0$ is a horizontal tangent.

Any point for which $ \frac{dx}{d\theta}\equal{}0$ and $ \frac{dy}{d\theta}\equal{} 0$ both needs to be investigated further; there are many possibilities including no tangent line at all. I'll leave those details to you.


What number results when 42 is increased by $ 120\%$?


LaTeX
inequalities unsolved
inequalities

If $ a_1$, $ a_2$, ..., $ a_n$ are positive reals satisfying $ \sum_{i \equal{} 1}^n \frac {1}{1 \plus{} a_{i}} \equal{} 1$, then prove that
$ \sum_{i \equal{} 1}^n \sqrt{a_{i}}\geq \left(n \minus{} 1\right)\sum_{i \equal{} 1}^n\frac {1}{\sqrt {a_{i}}}$.

Here is a nice solution found in my notes:

1. If $x_i\geq{x_j}$ then $ \frac{\sqrt{x_i}}{1+x_i}\leq\frac{\sqrt{x_j}}{1+x_j}$.

2. From Chebishev's theorem and 1:

    $1\geq\frac{1}{n}\cdot{\sum\frac{1}{\sqrt{x_i}}\cdot{\sum\frac{\sqrt{x_i}}{1+x_i}}}$.
3. From Cauchy:

    $\sum{\frac{\sqrt{x_i}}{1+x_i}\cdot{\sum{\frac{1+x_i}{\sqrt{x_i}}}\geq{n^2}}}$.

4. From 2 and 3 we get:

   $\sum{\frac{1+x_i}{\sqrt{x_i}}}=\sum{\sqrt{x_i}}+\sum{\frac{1}{\sqrt{x_i}}}\geq{n}\cdot{\sum{\frac{1}{\sqrt{x_i}}}}$.

5. From 4:

   $\sum{\sqrt{x_i}}\geq(n-1)\cdot{\sum\frac{1}{\sqrt{x_i}}}$ and we are done.

Yes,it's righ.
My proof:
[tex]\sum\frac{a_{i}}{1+a_{i}}=n-1[/tex]
It becomes:
[tex]\sum\sqrt{a_{i}}.\sum\frac{1}{1+a_{i}}\geq\ \sum\frac{a_{i}}{1+a_{i}}\sum\frac{1}{\sqrt{a_{i}}}[/tex].
Or
[tex]\sum\frac{(\sqrt{a_{i}}+\sqrt{a_{j}})(\sqrt{a_{i}}-\sqrt{a_{j}})^2(\sqrt{a_{i}a_{j}}-1}}{(1+a_{i})(1+a_{j})\sqrt{a_{i.}a_{j}}\geq\ 0[/tex]
But:
[tex]\frac{1}{1+a_{i}}+\frac{1}{1+a_{j}}\leq\ 1[/tex]
Hence:
[tex]\sqrt{a_{i}a_{j}}\geq\ 1[/tex]
sorry,i don't know LATEX well


counting
distinguishability
combinatorics unsolved
combinatorics

In a group of n people, each person has some set of coins with positive integer dollar values, and each person has at most one of each type of coin. Each person then either does not put any money into a pot or puts exactly one coin in that pot, and the total monetary value of the pot is counted. Given that for each integer k with $ 0 \le k \le 999$, there is exactly one way of achieving a total pot value of $ \$k$, and that there is no way of achieving a total pot value greater than $ \$999$, what is the greatest possible value of n? (Note that two ways of putting money into the pot are distinguishable iff at least one person puts a different value into the pot between the two ways.)

[hide="one person suffices..."]ummm, n=1, where he has coins of value 1,2,3,...,999? Maybe you mean every person has at most one coin?[/hide]

That'd be the minimal number of people ;)

If the various people have $ c_1, c_2, \ldots$ coins, then the total number of values that can be achieved is $ c_1 \cdot c_2 \cdots \equal{} 1000$.  Thus, as $ 1000 \equal{} 2^3 \cdot 5^3$, the theoretical maximum would be six people, three with one coin and three with four coins.  Then we just need to construct this.  In fact, it's straightforward to do so: $ \{1, 2, 3, 4\}$, $ \{5, 10, 15, 20\}$, $ \{25, 50, 75, 100\}$, $ \{125\}$, $ \{250\}$, and $ \{500\}$ does the trick.  I think it's more or less clear how to generalize this with any number in place of 999.


inequalities
geometry
inequalities proposed

Given a triangle $ ABC$ with three sides: $ BC\equal{}a,CA\equal{}b,AB\equal{}c$ such that:$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}a^2b^2c^2$.Let $ r_{a},r_{b},r_{c}$ denote the $ A$ -exradius,$ B$ -exradius, $ C$- exradius respectively;$ r$ be the the radius.Prove that:
$ r_{a}\plus{}r_{b}\plus{}r_{c} \ge \frac{9}{4r}$

[quote="Dr Sonnhard Graubner"]hello, is $ r$ here the radius of the incircle of the given triangle?
Sonnhard.[/quote]
Yes,you are right :)


The positive integers are arranged in a row in some order, each occuring exactly once. Does there always exist an adjacent block of at least two numbers somewhere in this row such that the sum of the numbers in the block is a prime number?

do u mean that all the positive integers from 1 to n for arbitrary n are listed, or are all pos. integers(infiinite) arranged?

"The positive integers," without qualification, refers to ... the positive integers.

I copied this verbatim from Tournament of Towns, and I take it to mean ALL positive integers.. one to infinity. And I don't know how to begin this problem =P


geometry
geometric transformation
homothety
ARML
ratio

Points $P$ and $Q$ are on circle $O$, and chord $\overline{PQ}$ is drawn. A second circle is drawn with diameter $OP$, crossing $PQ$ at $S$. If $OP = 7$ and $PQ=12$, compute $PS$.

Can't make homothecy any simpler...
[hide]
$P$ is the point of homethecy, and the ratio is $\frac 12$ since the radius becomes the diameter of the smaller circle.  Then, by homethecy, $PS = \frac 12 \cdot PQ = 6$
[/hide]

homothety is really not necessary for this problem...it is just a pair of similar triangles...
let OP hit the large circle at T, then

$\angle PSO=90=\angle PQT$ since they subtend a semicircle, then $\angle OPS=\angle TPQ$ by reflexivity, so $\triangle PSO\sim\triangle PQT$ by AA

furthermore, note that this is how that we prove that circles are homothetic when they are tangent internally


ok...so here's how it goes....im an oldie (second page in the member list by date! woot woot!)..and I have been a tad busy for AoPS...but Ive decided to come back to it.  At least once a week.  To start off...a question that has been perturbing me for some time now...

family guy or simpsons...

you cant quite compare just first 3 seasons...or entire shows....but...which is better

[s]the good old days when people didn't necropost[/s] [color=#eee]because everything was new[/color]

[quote=mathleticguyyy]The good old days when "W T F" wasn't banned[/quote]

The good old days when I did all the 14 year bumps -_-

[quote=mathleticguyyy]The good old days when "W T F" wasn't banned[/quote]

that was a 14 year bump, and @above was a 2 year bump :wow:


Anyone can help me with this proof?
f(x)=(x+a)q(x) + f (-a)?

It's hard to understand your question as stated.

What is $ f$, a function?  A special type of function?  Some other type of object which may not be a function at all?  What is $ q(x)$?  What is $ a$?

I'm going to assume that $ f(x)$ is a polynomial with coefficients in $ \mathbb{R}$, and that $ a \in \mathbb{R}$.  The division algorithm for polynomials tells us that there exist polynomials $ q(x)$ and $ r(x) \in \mathbb{R}[x]$ such that $ f(x) \equal{} (x \plus{} a)q(x) \plus{} r(x)$ and the degree of $ r(x)$ is less than the degree of $ x\plus{}a$, or $ r(x) \equal{} 0$.  Since $ x\plus{}a$ has degree one, $ r(x)$ must be a constant.  Now plug in $ \minus{}a$ on both sides of the equation and you are done.

That's basically a ‘special’ case of the remainder theorem: when the divisor is $ x \plus{} a$, or $ x \minus{} (\minus{}a)$.

Suppose $ \frac {f(x)}{x \plus{} a} \equal{} q(x) \plus{} \frac {r}{x \plus{} a}$, where $ q(x)$ is the quotient and $ r$ is the remainder. Multiplying by $ x \plus{} a$, we end up with $ f(x) \equal{} (x \plus{} a)\cdot q(x) \plus{} r$. Now, if we evaluate the function $ f$ when $ x \equal{} \minus{}a$, we get that $ f(\minus{}a) \equal{} [(\minus{}a \plus{} a)\cdot q(\minus{}a) \plus{} r] \equal{} r$; therefore, we can say that the remainder $ r$ when a polynomial $ f(x)$ is divided by $ x \plus{} a$ is equal to the value of $ f(\minus{}a)$.


find all integers $x$ such us $1+x+{x}^{2}+{x}^{3}+{x}^{4}$ is a perfect square.

Not sure of the ultimate source, but it appeared on New Zealand team training in 2004.

If $40x+55>0$, then note that
\[ (8x^2+4x+3)^2+40x+55=64(x^4+x^3+x^2+x+1)=(8x^2+4x+4)^2-16(x-3)(x+1). \]
For $x>3$ or $x<-1$, $x^2-2x-3>0$, so $64(x^4+x^3+x^2+x+1)$ lies strictly between two perfect squares and hence cannot be one itself.  Checking $-1\leq x\leq 3$, we find the solutions $x=-1$, $x=0$, $x=3$.

If $40x+55<0$, then note that
\[ (8x^2+4x+2)^2+4(4x^2+12x+15)=64(x^4+x^3+x^2+x+1)=(8x^2+4x+3)^2+40x+55. \]
In this range, $4x^2+12x+15$ is positive always and $40x+55$ is negative, so $64(x^4+x^3+x^2+x+1)$ lies strictly between two perfect squares and hence cannot be one itself.

thanks :clap2: thats correct the only solutions are $-1,0$ and $3$


integration
trigonometry
function
real analysis
real analysis unsolved

there (in this forum ) was posted problem by didilica
prove that $ \int _{0}^{\plus{}oo}f(x) \frac{\sin x}{x}dx \equal{} \int_{0}^{\pi/2}f(x)dx$ if $ f(x)$ is periodical continious function with period $ \pi$
How many solution do you know ofthisproblem? I am not sure but I think some hint about fourier expansion of $ f(x)$ or substitution $ x\equal{}ty$ will help but..

also please say what is the name of this theorem?


Given $ \lim_{n\to \infty} (a_{n} \minus{} a_{n \minus{} 2}) \equal{} 0$ Prove that $ \lim_{n \to \infty} \frac {a_{n} \minus{} a_{n \minus{} 1}}{n} \equal{} 0$

It suffices to show that $ \lim_{n\to\infty} \frac{a_{2n}}{n} \equal{} 0$ and $ \lim_{n\to\infty} \frac{a_{2n\minus{}1}}{n} \equal{} 0$, respectively. But this easily follows from the basic fact about Cesaro mean that if $ \lim_{n\to\infty} a_n \equal{} A$, then $ \lim_{n\to\infty} \frac{a_1 \plus{} \cdots \plus{} a_n}{n} \equal{} A$.


AMC USA(J)MO USAMO trigonometry MATHCOUNTS AIME ARML

[quote="logic812"]Currently, in Korea, I am in this Korean math class where everyone of the 15 person class has gotten a 23+ on AMC 8, gotten a 132+ on AMC 10, gotten at least a 6 on AIME, and there is one American/Korean, that got a twenty-something on the USAMO.  :ewpu: OMG! (He's 13 years old!)
Scary....
Jealousy overpowers me when I see these people, but I think I can handle it.
Anyone else seen a few very very very (emphasis on very) nerdy people?[/quote]Me.  He has an AoPS account   [hide="I am pretty sure it is"]cognos599[/hide] Heh, just kidding.  More seriously, there is David Benjamin. :)

Wait, so why are you in a math class during vacation?

idk, vacation.(parents)
No seriously, these korean kids are DEMENTED about just about everything school-related.
Allen Yuan got a perfect of the indiv. in ARML, but he didn't seem so smart.
(partly because of cognos's explanation of him?)

wait what? i never said anything about Allen Yuan! (besides that he is smart)

He is smart (compared to basically everyone in our school)
You said something like he didn't even make blue MOP...that gave me the impression
that he wasn't SO smart.

first of all, I NEVER SAID THAT, and second of all, he's a freshman, and it is nearly impossible to make it to blue mop in 9th grade or below. so stop saying lies that could possibly offend other people!  :)

[quote="undefined117"]unimpossible,

hmm idk about david, but nathan definitely isn't nerdy.. neither is cognos.

But wth is the point of this topic?

CCY SHOULD START MODDING ALSKDJFIOSDMFK[/quote][url]http://www.artofproblemsolving.com/Forum/memberlist.php?level=2[/url] that's the list of mods i think

I have no clue how ccy posted here.  He might have modded north america forum or something at the time.

[quote="the future"][quote="cognos599"] and it is nearly impossible to make it to blue mop in 9th grade or below.  :)[/quote]

*cough* lyndon ji *cough* in 2009[/quote]

Um actually if anyone's gonna make blue mop next year its gonna be Anaxerzia...

or pythag or whatever his name is who got a 15 as a sixth grader

[quote="the future"]Dude no.
Lyndon > other people[/quote]
Yeah, Lyndon is really good with the whole AMC competitions (and the increasingly harder ones that follow it).  Although, he may be unlucky next year as he was with Mathcounts Nationals this year.  You can hardly expect someone who got Lyndon's place in Nats to have made USAMO, or even better, done well at it.
The great part about this whole forum is that Lyndon pretends to be stupid ("I'm going to get 227th at Nationals"), so you can make fun of him and call him dumb all you want, and he won't deny it.

But yeah, Lyndon probably will make MOP next year.

We should make an AHL club... all hail Lyndon!!!!

Dude its only because y'all are Indiana pplz

And he will no doubt make MOP next year since he made USAMO this year and got the highest score in Indiana

I need to pull a not_trig next year :P

wow...

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!

[quote="undefined117"]what was lyndon's USAMO score again?

of course, this is alllll boosting someone's ego. a LOT :D or maybe not.[/quote]

9

Any estimations on Lyndon's 2009 USAMO score?
I'm thinking around a 18, do I sound like I'm off by a lot?

He'll get a 27 xD :D

I don't care if I don't belong in the Indiana forum... xD

the Hoosier forum has infiltrators. SHUN THE INFILTRATORS XD jk.

crap who else thinks I'm beginning to turn into the catalyst of spam?

This is quickly turning into spam. (Hint, hint)... :)

I think I may need to go help Missouri beat BA in spam volume :D

[quote="serialk11r"]uh yea that just proved my point[/quote]Thanks, I try. :)

YO GUYZ

LETS ALL MAKE OHIO THE SPAMMIEST FORUM EVER xD


For $ a,b,c\ge0$ and $ k\ge2$, prove
\[ \sum\frac{ab\plus{} \frac1{2k} (a\minus{}b)^2}{a^2\plus{}b^2\plus{}kc^2}\ge \sum\frac{c^2}{ a^2\plus{}b^2\plus{}kc^2}\ge \sum\frac{ab}{ a^2\plus{}b^2\plus{}kc^2}.\] ;)

After using $ ab \leq \frac{a^2\plus{}b^2}{2}$ the second inequality is almost in SOS form and its easy to finish. I dont think you need $ k\geq 2$ for this.

[quote="spanferkel"]For $ a,b,c\ge0$ and $ k\ge2$, prove
\[ \sum\frac {ab \plus{} \frac1{2k} (a \minus{} b)^2}{a^2 \plus{} b^2 \plus{} kc^2}\ge \sum\frac {c^2}{ a^2 \plus{} b^2 \plus{} kc^2}\ge \sum\frac {ab}{ a^2 \plus{} b^2 \plus{} kc^2}.\]
;)[/quote]

 $ \sum\frac {c^2}{ a^2 \plus{} b^2 \plus{} kc^2}\ge \sum\frac {ab}{ a^2 \plus{} b^2 \plus{} kc^2}$

The inequality does not hold

try $ k\equal{}3,a\equal{}1,b\equal{}\frac{1}{2},c\equal{}1$

Okay, I should be more careful  :oops: . Anyway, what i got is \[ \sum \frac{2c^2\minus{}2ab}{a^2\plus{}b^2\plus{}kc^2} \geq \sum \frac{(c^2\minus{}a^2) \plus{} (c^2\minus{}b^2)}{a^2\plus{}b^2\plus{}kc^2} \equal{}\]  \[ \sum (a^2\minus{}b^2) \left( \frac{1}{b^2\plus{}c^2\plus{}ka^2} \minus{} \frac{1}{c^2\plus{}a^2\plus{}kb^2} \right) \equal{}\] \[ \sum (a^2\minus{}b^2)^2 \frac{1\minus{}k}{(b^2\plus{}c^2\plus{}ka^2)(c^2\plus{}a^2\plus{}kb^2)}\] which proves the inequailty only for $ k \leq 1$.

[quote="theSA"]if I am not wrong, for right part we can assume $ a \ge b \ge c$ which gives the desired.[/quote]

try $ k\equal{}3,a\equal{}2,b\equal{}1,c\equal{}\frac{1}{2}$


calculus
integration
logarithms
function
real analysis

Calculate
$ \int_0^1\ln(x)\ln(1\minus{}x)\,dx$.

I have absolutely no idea how to do this and would like to find out.

[hide=" idea?"]
Since $ x$ is in the radius of convergence of $ \ln$, we can write
\[ \ln(1 \minus{} x) \equal{} \minus{} \sum_{k \equal{} 1}^{\infty} \frac {x^k}{k}
\]
So we want
\[ \int_{0}^{1} \ln(x) \sum_{k \equal{} 1}^{\infty} \minus{} \frac {x^k}{k} \ dx
\]
Switching the order of summation, we get
\[ \sum_{k \equal{} 1}^{\infty} \minus{} \frac {1}{k} \int_{0}^1 x^k \ln(x) \ dx
\]
Now we integrate by parts (setting $ u \equal{} \ln(x), dv \equal{} x^k dx$)
\[ \sum_{k \equal{} 1}^{\infty} \minus{} \frac {1}{k} \left[ \frac {1}{k}x^{k \plus{} 1}\ln(x) \minus{} \int_{0}^1 \frac {1}{k}x^k \ dx \right]
\]

\[ \sum_{k \equal{} 1}^{\infty} \minus{} \frac {1}{k^2} \left[ x^{k \plus{} 1} \ln x \minus{} \frac {1}{k} x^{k \plus{} 1} \right]_{0}^1 \equal{} \sum_{k \equal{} 1}^{\infty} \frac {1}{k^3}
\]
Notice how this last expression does not actually have a nice closed form, which is an excellent indication that the above solution is grievously wrong.
[/hide]

I seem to remember seeing something somewhere that you can't always switch the order of a sum and an integral? I'm not exactly sure. Mathematica can compute this integral exactly, but I see no way to do it by hand so far.

$ \zeta(3)$ isn't a particularly nasty answer to a question like this; many such questions (where the indefinite integral is not elementary but a particular definite integral can be described) have similar answers.

I'm afraid I don't know necessary and sufficient conditions so that a sum and integral can be switched (a sufficient condition is that the entire summation, considered as a Riemann sum, converges absolutely), but solutions along those lines are common for problems like these.

The answer (according to mathematica), should come out to $ 2\minus{}\frac{\pi^2}{6}$, so the solution given above is wrong somewhere, but the form does suggest that the zeta function comes into play at some point.

Has anyone actually checked that the steps are performed correctly? Perhaps the method is right and I just made some mistake somewhere...

[quote="Phelpedo"]Now we integrate by parts (setting $ u = \ln(x), dv = x^k dx$)
\[ \sum_{k = 1}^{\infty} - \frac {1}{k} \left[ \frac {1}{k}x^{k + 1}\ln(x) - \int_{0}^1 \frac {1}{k}x^k \ dx \right]
\]
[/quote]

The indefinite integral should be $ \frac {1}{k + 1} x^{k + 1} \ln x - \int \frac {1}{k + 1} x^k \, dx = \frac {x^{k + 1} \ln x}{k + 1} - \frac {x^{k + 1}}{(k + 1)^2}$.

Edit:  This gives the answer given by Mathematica.  In particular, after evaluating the definite integral the remaining summation is

$ \begin{eqnarray*} \sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} &=& \sum_{k=1}^{\infty} \frac{(k+1) - k}{k(k+1)^2} \\
 &=& \sum_{k=1}^{\infty} \left( \frac{1}{k(k+1)} - \frac{1}{(k+1)^2} \right) \\
 &=& 1 - (\zeta(2) - 1) \\
 &=& 2 - \frac{\pi^2}{6} \\ \end{eqnarray*}.$

I didn't actually take the test, but...

Integrating by parts with $ \ln(1 \minus{} x)dx \equal{} du$ and $ \ln(x) \equal{} v$ gets me to a different integral: $ \int\frac {(x \minus{} 1)(1 \minus{} \ln(1 \minus{} x))}{u}du$. If you expand it out the only one you can't do immediately is $ \int\frac {\ln(1 \minus{} x)}{x}$... but I probably screwed up somewhere.

Edit: Obviously I'm being sloppy with my definite and indefinite integrals, but it's late. I'll try again in the morning.

I want to say you can do something like ln(1/x)*ln(1/1-x) and this gives you a better chance of actually getting somewhere with long chains of integration by parts, but i havent had much success with this problem.

[quote="mathgeniuse^ln(x)"]I guess we can always use the Taylor series for ln(x+1), and plug those in, and tehn integration by parts

This gets us the answer we are looking for.[/quote]

But then we get a negative "x" in ln(x+1)... man ln really sucks... stupid discontinuity. Argh.

EDIT: yay, it has been explained to me.

Er... that's exactly Phelpedo's solution, and with my fix it gives the correct answer.

This is such an awesome problem!

But I always find it strange to write the sum of series in terms of $ \pi$...since that in itself is just a sum of a series. So we are basically write one sum in terms of another...

See here  https://artofproblemsolving.com/community/q1h1532768p9227579


I've noticed that there will be an olympiad geometry class in the spring.  Will there ever be a beginning, intermediate geometry course, or one less challenging than the olympiad class?

Mr.Ocax

yep, an intro geometry course will be in the summer

and you know how?... since i didn't see anything on the online classes board

Late spring or summer for the intro to geometry class.  I think Syntax probably picked this up from my comments in one of the other geometry class threads.


induction
combinatorics unsolved
combinatorics

Each member of a parliament has no more than three enemies among other MPs. Prove that the parliament can be split into two houses such that each MP has no more than one enemy in the same house.

use induction
for $n=4$ it's clear
suppose there are $n$ members in two houses and this is a good partition, the last member has no more than $3$ enemies so in one house he has no more than one enemy, he goes there and we are done

[quote="Cmmmrc"]Sure?

I mean: suppose Mr. Last Member joins party A. There could be Mr. Last's Enemy in A, with only an enemy in A. Adding Mr. L.Member does not violate the limit of three enemies, but then Mr. Enemy has two enemies in A: his former one and Mr. Last Member.[/quote]

It doesn't matter. Because if $E$ is the number of enemy pairs there are counting in both parties and $F$ the number of friendly bonds then in the process of transfering Mr LM from party $B$ to party $A$ you're reducing the number of enemyness by 2 units and increasing it by one, thus leading to an overall reduction of 1 unit of enemyness. In other words, what you should do in your case is simply transfer Mr Enemy to party B and continue this process. This can't go forever so you'll end up with parties $A$ and $B$ in both of which every person has only one enemy, (if there were two then you're just not finished).

Love the word I just invented enemyness

Daniel

Ok. But this is a completely different solution.

My point is that Zorro's solution by induction does not work, not that the thesis is false.

M.

we will prove it for $n$ using induction
suppose there is an even cycle $v_1 \leftarrow v_2 \leftarrow ... \leftarrow v_{2k} \leftarrow v_1$
let the first house be A and the second B. take to A every vertex from that cycle with odd number and the rest to B. there are no two enemies in one house. becouse there are $n-2k$ vertices left we can split them correctly
if there is no even cycle then there must be odd cycle (for tree it's trivial)
$v_1 \leftarrow v_2 \leftarrow ... \leftarrow v_{2k+1} \leftarrow v_1$
take vertices with odd numbers to A and with even to B. In B we do not have enemies and in A we have one such pair - $v_1,v_{2k+1}$.
now such method will work: take enemies of $v_1,v_{2k+1}$ to the B and their enemies take to A and their to B and so on. it's good becouse in every step we do not have a pair of enemies in one house (except $v_1,v_{2k+1}$)
otherwise we would have an even cycle. when we finish every vertex from A and B has at least 2 enemies in one house except those who does not have 2 enemies in the parliament
there are less then $n$ vertices left so we can use assumption


This time, let's start at 20, because there are less girls than boys.

Same rules as the last revival, girls subtract the number of stars they have, boys add.

Good luck all! :)
[b]
20[/b]

Wow girls outnumbered the guys 3-1 and we started at 20...hmm quite fair right?

haha lol Good job angel!!!

Nice try, Blazer!

Congrats me, izzy, and angel!! :D w00t!

@blazer: last time the girls were outnumbered, but next time I will certainly start at 25.. :wink:

yeah rofl that's wat i was telling jjfun but watever we will won =P

Ok, guys.. I made a round three. Blazer, angel, izzy, please go there. Good job for this round guys! :D


Sorry to post this random post, but could anyone tell me how can I write LaTeX code in this post. I mean, I know how to write in LaTeX but, how can I make it active. Whenever I type a piece of code and hit the preview button, I always get back the same code I wrote!

-Swapnil

put it between dollars: $x^2$ will result in what you want. Topic closed now since we have a latex forum. ;)


inequalities
number theory proposed
number theory

Let $p$ be an odd prime and $a,b$positive integers satisfying$(p+1)^a-p^b=1$.Show that $a=b=1$

It's a bit of a heavy handed approach, but you could use Catalan's Conjecture/Theorem to show $a=1$ or $b=1$, and then some simple inequalities will finish it off.

In my opinion (I am not sure) you can try solving for (p+1)^a = 1+p^b in base p.

If (b > 1) then sure enough RHS has a represenation in the form of 100..01 in base p (where we have b+1 digits in total).

And what I cannot prove (but rely on someone to prove) is that LHS cannot have such a representation.

-Thank You

Let $n$ be any positive integer which is odd. I will prove that \[(n + 1)^a - n^b = 1 \Rightarrow a = b = 1.\]

Modulo $n + 1$ we get $(-1)^b \equiv -1$, so $b$ must be odd.

Hence \[0 \equiv (n + 1)^{a - 1} = \frac{n^b + 1}{n + 1} = n^{b - 1} - n^{b - 2} + \dots + 1 \equiv b.\]

So, $n + 1$ divides $b$ if $a > 1$, which is impossible (since $n + 1$ is odd).

So, we must have $a = 1$ and hence $b = 1$.

Someone please tell me, was I onto anything??? Can we proceed using the base conversion idea???

-Thanks in advance


group theory
abstract algebra
algorithm
superior algebra
superior algebra unsolved

In group $G$ with condition: $S^{6}=T^{2}=(ST)^{2}$ construct characteristyc table.
The first step: you must find order of a group $G$.
...
I I have not  mistake im my calculation then the order is equal to $26$

[quote="owk"]By Lagrange's theorem, the order of each element divides the order of the group. Hence, $\#G$ cannot equal 26. owk[/quote]
No.let $S=T=e$

So if you are not talking about the group with generators $S,T$ and relations as given above, what other group do you have in mind? owk

I agree with you. OK I firts I said: If I have not mistake in my valculation then Order of a group is equal to $26$. I have mistake. Please solve the problem.  I calculate again and show that order of a group is equal to $24$

[quote="jmerry"]That's a nonsense answer, Extremal- we can certainly find elements satisfying these relations in any subgroup. The real question: what is the largest possible group generated by $S$ and $T$ with these relations? It's certainly not a group of order 26.
Since $STST=e$, $STS=S^{-1}=TST=T^{-1}ST$, and the subgroup generated by $S$ is normal. This is the defining relation of a dihedral group, and $G$ is the dihedral group $D_{6}$ of order 12.[/quote]
I don't agree with you. I don't said: $(ST)^{2}=e$

[quote="owk"]I suspected another case of sloppy writing. In this case, 24 is correct. From $T^{2}=(ST)^{2}$ we get $T=STS$. Since $S^{-6}T\cdot T=1$, we must also have $1=T\cdot S^{-6}T=S^{6}(S^{-6}TS^{-6})T=S^{6}T^{2}=S^{12}$. On [url=http://web.usna.navy.mil/~wdj/gap/small_groups.html]this page[/url], the group is called [i]dicyclic group of order 24[/i] and denoted by $Q_{12}$. owk[/quote]
Good work.
I agree with you.
And alse : you can use Algoritm Kokster-Todda to find order of this group.

and also Note that $D_{12}$ has a same characteristic table


geometry
analytic geometry
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, P a point so that the pedal triangle A'B'C' of P is also a cevian triangle, A"B"C" the antimedian triangle of ABC, L the orthocenter of A"B"C", A* the intersection of the A"-altitude of A"B"C" with AA' and cyclically B*, C*.
Prouve : A*B* // A'B'.
Sincerely
Jean-Louis

Dear Mathlinkers,
any ideas concerning L (de Longchmps's point)?
Sincerely.
Jean-Louis

We shall show that P, its isogonal conjugate and the De Longchamps point of ABC are collinear iff the pedal triangle of P WRT ABC is perspective with ABC. Since I don't have a synthetic proof of this fact, I'll introduce this lemma apart

[b]Lemma:[/b] Let $P$ be a point of the plane of $\triangle ABC$ and $P^{-1}$ be its isogonal conjugate. $L \equiv X_{20}$ is the De Longchamps point of $\triangle ABC$ (orthocenter of its antimedial triangle). Then, $L,P,P^{-1}$ are collinear iff the pedal triangle of $P$ WRT $\triangle ABC$ is perspective with $\triangle ABC,$ i.e. if $P$ moves on the Darboux cubic of $\triangle ABC.$

Using barycentric coordinates WRT $\triangle ABC,$ the equation of the line $\ell$ passing through $P(u:v:w)$ and its isogonal conjugate $P^{-1} \left (\frac{a^2}{u}:\frac{b^2}{v}:\frac{c^2}{w} \right)$ is defined by

$\ell \equiv u(c^2v^2-b^2w^2)x+v(a^2w^2-c^2u^2)y+w(b^2u^2-a^2v^2)z=0$

If $L \equiv X_{20}$ lies on $\ell$ $\Longrightarrow \  \sum_{\text{cyclic}} (S_AS_B+S_AS_C-S_BS_C)(c^2v^2-b^2w^2)x=0$

Then comparing this latter expression with the equation of the Darboux cubic:

$\sum_{\text{cyclic}} (S_AS_B+S_AS_C-S_BS_C)(c^2y^2-b^2z^2)x=0$

We conclude that $L,P,P^{-1}$ are collinear and the proof of the lemma is completed.

Let us restate the proposed problem as follows:

[b]Problem:[/b] $P$ is a point on the plane of $\triangle ABC$ such that its pedal triangle $\triangle A'B'C'$ is perspective with $\triangle ABC.$ Let $\triangle A_0B_0C_0$ be the antimedial triangle of $\triangle ABC$ and $L$ be its orthocenter, i.e $X_{20}$ of $\triangle ABC.$ Lines $AA',BB',CC'$ cut $LA_0,LB_0,LC_0$ at $X,Y,Z,$ respectively. Then $\triangle XYZ$ and $\triangle A'B'C'$ are homothetic.

$\triangle ABC$ and $\triangle A'B'C'$ are perspective through perspector $Q \equiv AA' \cap BB' \cap CC'$ and orthologic through the orthology centers $P$ and its isogonal conjugate $P^{-1}.$ Hence, by Sondat theorem $Q,P,P^{-1}$ are collinear. From the above lemma, we conclude that $Q,L,P,P^{-1}$ are collinear. From $\triangle QA'P \sim \triangle QXL,$  $\triangle QB'P \sim \triangle QYL,$  $\triangle QC'P \sim \triangle QZL$ we get

$\frac{\overline{QX}}{\overline{QA'}}=\frac{\overline{QY}}{\overline{QB'}}=\frac{\overline{QZ}}{\overline{QC'}}=\frac{\overline{QL}}{\overline{QP}}$

Thus, $\triangle XYZ$ and $\triangle A'B'C'$ are centrally similar through center $Q$ and similarity coefficient $\frac{_{\overline{QL}}}{^{\overline{QP}}}$ $\Longrightarrow$ $B'C' \parallel YZ,$ $C'A' \parallel ZX$ and $A'B' \parallel XY. \square$

Dear Luis and Mathlinkers,
I come back with the lemma...for which I haven't also a synthetic proof.
We know that the de Longchamps point L is a PC-point (L being the symmetric of the PC-point H (orthocenter) wrt O (center of the circumcircle)).
But how can we prove synthetically that L is the Ceva point of a PC-point?
Do some one know a construction of this PC-point?
Sincerely
Jean-Louis

Dear Mathlinkers,
an article (last of three) concerning the first synthetic proof of the PC-lines and the de Longchamps point has been put on my website
http://perso.orange.fr/jl.ayme vol. 6 Pedal-cevian lines go through the de Longchamps's point, p.33
Sincerely
Jean-Louis


Find the smallest positive integer, $ N$, such that the product of 135 and $ N$ is a perfect square.

Let's prime-factor $ 135$ and see what we can find...

...the prime factorization happens to be $ 3^3\times5$. We notice that one number is cubed ($ 3^3$) and one is just itself ($ 5$). In order for a number to be a perfect square, it's prime factorization must be completely made up of squares or other even powers. Thus, in order to make the smallest product that's a perfect square, we multiply first by $ 3$ to raise the $ 3^3$ to an even power, then by $ 5$, to make $ 5$ squared. The total amount we multiply by is $ 3\times5\equal{}\fbox{15}$.

[quote="gunchu"]I don't get this problem.
Please Help.[/quote]

It means find the smallest positive integer such that $135N$ is a perfect square.
Is it clear now ?

What you need to do is prime factorize $135$ into $3^3 \times 5.$ Then both powers need to be equal for a perfect square. If we multiply by $15$ we get $3^4 \times 5^2$. I hope this helps.


The median of the set of numbers $ \{$12, 38, 45, $ x$, 14$ \}$ is five less than the mean. If $ x$ is a negative integer, what is the value of $ x$?

If x is negative, it is the smallest and thus 14 is the median.
Now setup an equation:

$ 14 \plus{} 5 \equal{} \frac{12 \plus{} 38 \plus{} 45 \plus{} 14 \plus{} x}{5}$

$ 95 \equal{} 109 \plus{} x$

$ x \equal{} \minus{}14$


[quote="pkerichang"]What do you guys think???(I hope it's low because of number 5...(bunch of people brute force)).[/quote]
I wouldn't worry too much about whether you qualify for the AIME through the USAMTS - the AMC is a more important way to qualify due to a better score getting you a better USAMO index.

[quote="zanttrang"][quote="K81o7"]Do you think people would get points off for a brute force sort of method on #5?[/quote]
No, not if it's complete an correct - however, it certainly won't get a commendation, and if it's not well-organized, the grader may take a point off just because it's too hard to verify.[/quote]

That's what I'm worrying...In order to go for completeness, I wrote 4 pages with some of brute force work shown (I spend my first page explaining notations I use in the proof).  I hope the grader doesn't get frustrated...


The company PyramidoCo sold pyramids of various sizes. They shipped them in rectangular boxes just big enough to hold a single pyramid. Their best-selling model (Square Pyramid 1A) fit perfectly in a 11 x 11 x 15 inch box. What was the volume of a Square Pyramid 1A?

[quote="10000th User"][quote="math92"]can you tell me the formula for the area of a pyramid? I forgot it... :?[/quote]
Hi math92, Pyramid Volume $V_\text{pyramid}=\frac{1}{3}\times B\times h$, $B$ is area of base, $h$ is altitude[/quote]

Thank you for the formula!!!

[hide="Solution"]

Base: 11x11.
Height: 15.

The above is not "the way it is." The base could be 15x11 too. :)

$V_\text{pyramid}=\frac{1}{3}\times B\times h$

$ = \frac {1}{3} \times 121\times 15$

$ = \frac {1}{3} \times 1815$

$ = 605  sq.  inches$

[/hide]


Let $a,b,c$ are positive number. Prove that:
$2 \sqrt{ \frac{ab+bc+ca}{a^2+b^2+c^2 }} + \sqrt{ \frac{a+b}{c}+ \frac{b+c}{a}+ \frac{c+a}{b}-2} \ge 4$
What do you think?? :D

[quote="Lovasz"][quote="zaizai-hoang"]Let $a,b,c$ are positive number. Prove that:
$2 \sqrt{ \frac{ab+bc+ca}{a^2+b^2+c^2 }} + \sqrt{ \frac{a+b}{c}+ \frac{b+c}{a}+ \frac{c+a}{b}-2} \ge 4(*)$
What do you think?? :D[/quote]
This is SOS, but I wonder if it is a true inequality.
$(*) \iff \sum_{\text{cyclic}} (a-b)^2 \cdot \left(\frac{1}{ab(\sqrt{\sum \frac{a+b}{c}-2})+2}-\frac{1}{a^2+b^2+c^2+\sqrt{(a^2+b^2+c^2)(ab+bc+ca)}}\right) \ge 0$[/quote]

Your solution isn't nice. I sure  solution of problem very nice. :D
Use a lemma :P

[quote="Lovasz"][quote="zaizai-hoang"]Use a lemma :P[/quote]

What's lemma?[/quote]
Yes, my lemma very interesting. :D
[color=green]Lemma[/color] Find the best  $k$ such that inequality true:
$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge \frac{k(a^2+b^2+c^2)}{ab+bc+ca}+6-k$

Now, I think you can prove that... :D

[quote="stancioiu sorin"]Zaizai-Hoang, I think it must be    $\frac{a^2+b^2+c^2}{ab+bc+ca}$. :| If is that, then the inequality is very easy, using that $\frac{a}{b}+\frac{b}{a}$>2.[/quote]
I misunderstand ... :( .
My solution use this lemma and AM-GM inequality.
But I have a solution, too. Use p,R,r menthod :D

I find the best $k=4$.
Then use AM-GM :D


geometry
3D geometry
sphere
circumcircle
geometry unsolved

Given a solid sphere, construct its diameter using compass and straightedge.

I encountered this at a Junior Math Battle which was supposed to be for Grade 10's and under.  But I don't think anyone managed to solve this.

This is a classical construction. Assume we are given a solid sphere $ \mathcal{E}$ and a plane $ \pi$ for ruler-compass constructions. Draw a circumference $ \omega$ with arbitrary radius and  center $ P$ on $ \mathcal{E}$ and take three points $ A,B,C$ on $ \omega.$ Construct the rectilinear triangle $ \triangle ABC$ on $ \pi$ by transporting the chords $ BC,CA,AB$ with the compass. Let $ O'$ be the circumcenter of $ \triangle ABC.$ Then its circumradius $ O'A \equal{} \varrho$ is the orthogonal projection of the spherical radius $ PA$ on the plane $ ABC.$ Therefore, if $ P'$ is the antipode of $ P$ on $ \mathcal{E},$ the right $ \triangle APP',$ given its leg $ AP$ and altitude $ AO' \equal{} \varrho$ on the hypotenuse $ PP',$ is constructible. This produces the diameter $ PP'$ of the spherical surface $ \mathcal{E}.$


function
limit
real analysis
real analysis solved

Given continious function $f:[0,\infty){\longrightarrow}R$,such that $\lim_{n{\rightarrow}\infty}{f(x+n)}=0,\forall x\in{[0,\infty)}$.Does that follow that $\lim_{x{\rightarrow}\infty}{f(x)}=0.$Justify your answer.

Take $f(x)=k(x-k)$ for $x\in[k,k+\frac{1}{k}]$, $f(x)=k(k+\frac{2}{k}-x)$ for $x\in [k+\frac{1}{k},k+\frac{2}{k}]$ and $f(x)=0$ for $x\in[k+\frac{2}{k},k+1]$, $k\in\mathbb{N}$. Function $f(x)$ satisfies all conditions and $\lim_{x\to\infty}f(x)$ doesn't exist.

Let $g_k(x)=f(x+k)$ for $x\in[0,1].$ The condition in the problem, that $f(x+n)\to 0$ for each $x$, can be rephrased as $g_k(x)$ converges to zero [b]pointwise[/b]. That condition is not sufficient. A condition that would be both necessary and sufficient: $g_k$ converges to the zero function [b]uniformly[/b]. Looking at things in that light reveals Myth's example for what it is (and what Myth certainly understood it to be): a fairly standard example of a sequence of functions that converges pointwise but not uniformly. The functions are piecewise linear, and are easier to understand from a simple drawing than from  the formulas.

Perfect.Then guys you can easily notice that if we add the condition $f$ to be uniformly continious then $\lim_{x{\rightarrow}\+\infty}{f(x)}=0$

Actually, this folows from what Kent said: take $g_n:[0,1],\ g_n(x)=f(x+n)$. With this extra condition (uniform continuity of $f$), we know that the sequence of functions $g_n$ tends to $g=0$, because $g_n(x)\to g(x)=0$ uniformly.


when working with LaTeX => PDF, I can't use the ntheorem package, because I'd need to put LaTeX => PS => PDF, but if I use this profile, I can't put a image with the PNG extension.

what I want is to mix these things, so that I can put PNG images and use the ntheorem package.

You can use ntherorem provided you don't use pstricks commands which start with \ps. If you need these then you will have to convert PNG images to EPS before you compile. OpenOffice Draw will do this as will lots of other programs.

[quote="stevem"]You can use ntherorem provided you don't use pstricks commands which start with \ps.[/quote]
ohhh, how can you do that then?

for example, can you produce

[img]http://img518.imageshack.us/img518/8382/asdasdar5.png[/img]

by compiling with LaTeX => PDF?

The shading s done using PSTRicks and hence a \ps command. As [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/ntheorem/ntheorem.pdf]the manual[/url] says:
[quote]By default, the box is given as a \psframebox (see pstricks package) with shadecolor
as linecolor and fillcolor. All these parameters can be changed by setting
\def\theoremframecommand{hany box commandi}
before declaring the theorem type (for examples, the user is referred to section 4).[/quote]so use the framed package rather than PSTricks.

Okay, I almost got what I want, I have the framed package and I put:

[code]\theoremclass{Theorem}
\def\theoremframecommand{%
\shadowbox}
\newshadedtheorem{moreImportantTheorem}{\color{white}}
\begin{moreImportantTheorem}
\end{moreImportantTheorem}[/code]

That will produce a white box with a black edge, I want to change the background color. (For a moment I thought that I should change \newshadedtheorem{moreImportantTheorem}{\color{white}}, where white is there, but that doesn't work.)

yes! that works!!

thanks stevem!!

one more question: can we increase the size of the black line?