I think that the following is very suggestive
\[ \frac x{x+y+z}+\frac y{x+y+z}+\frac z{z+x+y}=1.  \]

This quation is easy than I think.
 :D This is my solution.
By Holder's we have that
$\sqrt[3]{((a^2)^3+(b^2)^3){((c^2)^3/2)+(a^2)^3/2}^2} \geq (a^2)(c^2)+ (a^2)(b^2)$

It's beautiful ? :D

oh
i think we can replace x^3=a
then it is easier to sovle

I guessed this when I solved the problem:
\[ \frac{a^4}{a^4+\sqrt[3]{\cdots}} \le \frac{a^2}{a^2+b^2+c^2}  \]
But Holder's give solution only, while I tried other ones (because I don't like Holder's). Can... ?

See
http://www.mathlinks.ro/Forum/viewtopic.php?p=490540#490540
In this page,you will see quite general of this problem :)  :)

[quote="detectivehien"]
This is a question of a competition called April 4th Olympiad in VN.
Firstly, I prove this:$(a_1+b_1)(a_2+b_2)(a_3+b_3) \geq(\sqrt[3]{a_1a_2a_3}+\sqrt[3]{b_1b_2b_3})^3$
[/quote]
And this just is Mink's triangle inequality.
Very simple solution :) .


AMC
AIME
quadratics
geometry
MATHCOUNTS
AIME II
algebra

Are we allowed to have scotch tape, scissors, or carbon paper on the AIME?

[quote="davidyko"]Only scratch paper, graph paper, rulers, compasses, and protractors are permitted, unless there was a change of rules this year.[/quote]
It doesn't say clothes are allowed either, better not bring those.

[quote="davidyko"]Fine, "No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted." - 2006 AIME II booklet sitting on my desk.
So unless your clothes are your mathematical aids...(in which case that would be quite strange).[/quote]
Heh, you never know :D
I'm just wondering, I especially think that they wouldn't care about carbon paper

Can anyone clarify?
I don't see what the big deal would be, especially with carbon paper, but if the MAA says no, there must be a good reason.
Also, add construction paper to my list.

Why would you need carbon paper?

I guess you could take it because it wouldn't exactly be um...mathematical aids?

(Unless you scratched formulas into construction paper)

Maybe to help with a geometry problem, but you could probably do it with a compass, protractor, and ruler anyway.

[quote="randomdragoon"][quote="davidyko"]Fine, "No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted." - 2006 AIME II booklet sitting on my desk.
So unless your clothes are your mathematical aids...(in which case that would be quite strange).[/quote]
In my middle school days one team at a regional math meet came in with a team t-shirt that had the quadratic equation written on the back ... that caused quite an uproar.[/quote]
:rotfl:
Last year at state mathcounts this one team had shirts with the distance formula on the back :D

Please, can someone in charge confirm or deny whther these can be used?

Just to add to this, such a discussion interestingly came up last year when scissors turned out to be very useful in solving #10 on last year's AIME I. From what I remember, people were talking as if scissors were allowed, although most people did not think of bringing them.

Read the Teachers Manual: on page 6 is a facsimile of the AIME cover page.  On item 3 it says:
"No aids other than scratch paper, graph paper, ruler, compass and protractor are allowed.  In particular, calculators and computers are not allowed."

The rules say exactly what the rules mean, the rules mean exactly what the rules say.

Steve Dunbar
Director, American Mathematics Competitions


Nice application of log rules and AM-GM (if my solution is correct).

The left expression is equivalent to

$ (p*log_{p}{(p+1)})^{\frac{p}{p+1}} + ((p+1)*log_{p+1}{(p)})^{\frac{p}{p+1}}.$

Since p is a natural number not equal to 1, both of those terms are positive, so we can apply AM-GM.
By AM-GM, this thing is greater than or equal to

$ 2 [p (p+1) log_{p}{(p+1)}*log_{p+1}{(p)}]^{\frac{p}{2p+2}}$

We use the fact that $log_{a}{(b)}*log_{b}{(a)} = 1$ to finally simplify it to

$ 2 [p(p+1)]^{\frac{p}{2p+2}}.$

To show that the two quantities cannot be equal, we use the fact that the equality case of AM-GM is only when all terms are equal. Therefore we would have

$ p*log_{p}{(p+1)} = (p+1)*log_{p+1}{(p)}$

which is equivalent to

$\frac{log_{p}{(p+1)}}{log_{p+1}{(p)}} = \frac{p}{p+1}$

which is further equivalent to

$log_{p}{(p+1)}*log_{p}{(p+1)} = \frac{p}{p+1}$

The left is irrational and the right is rational for natural p,  so the equality case is impossible. So, we are left with

$\sqrt[p+1]{{log_p}^p {\left ( {p+1} \right )}^p} + \sqrt[p+1]{{log_{p+1}}^p {\left ( {p} \right )}^{p+1}} > 2{[p(p+1)]}^{\frac {p}{2p+2}}.$ QED

Well i saw your sollution and it's ok.
Here is my sollution:

$\sqrt[p+1]{{log_p}^p {\left ( {p+1} \right )}^p} + \sqrt[p+1]{{log_{p+1}}^p {\left ( {p} \right )}^{p+1}}$ =
= $ \left ( \sqrt[p+1]{{log_p} {\left ( {p+1} \right )}^p} \right )^p + \left ( \sqrt[p+1]{{log_{p+1}}{\left ( {p} \right )}^{p+1}} \right )^p $$(*)$

Here we use the following inequality : $ \frac {x^n + y^n}{2} \geq \left ( \frac {x+y}{2} \right )^n $ which can be prooved easy by Jensen. So we get:

$ (*) \geq 2 \left ( \frac {{\sqrt[p+1]{log_p (p+1)^p} + {\sqrt[p+1]{log_{p+1}p^{p+1}}}}}{2} \right )^p $ $(**)$ and here
using AM-GM we get: $(**) \geq 2 \left (  \sqrt {\sqrt[p+1]{p(p+1)log_p {(p+1)} log_{(p+1)}p}}   \right )^p = 2 \left [ p(p+1) \right ]^{\frac {p}{2(p+1)}}$ and for prooving that the inequality is strict i prooved like u did.

q.e.d.


Consider all possible sequences of 2004 balls, white, black and red on a line.
I will say transformation to the following process: We choose 200 consecutive balls, the first $100$ balls at the left($in$ $the$ $same$ $order$) are ubicated inmediately on the right of the other $100$ balls, we say that two sequences of balls are $similar$ if one can be obtained from the other with transformations, find the number of sequences of balls which are $not$ $similar$ :read:

[quote="OHO"]find all OR find how many[/quote]
Find how many

nobody$?$, please. Its a very nice and dificult $problem$.


linear algebra
matrix
linear algebra unsolved

Prove that every symmetric matrix has orthogonal eigenvectors.

This is the finite-dimensional [url=http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices]spectral theorem[/url].


$ m$ and $ n$ are positive integers. In a $ 8 \times 8$ chessboard, $ (m,n)$ denotes the number of grids a Horse can jump in a chessboard ($ m$ horizontal $ n$ vertical or $ n$ horizontal $ m$ vertical ). If a $ (m,n) \textbf{Horse}$ starts from one grid, passes every grid once and only once, then we call this kind of Horse jump route a $ \textbf{H Route}$. For example, the $ (1,2) \textbf{Horse}$ has its $ \textbf{H Route}$. Find the smallest positive integer $ t$, such that from any grid of the chessboard, the $ (t,t\plus{}1) \textbf{Horse}$ does not has any $ \textbf{H Route}$.

Can anyone help?
It seems that t=3 can be easily proved
but t=2 seems like a hard case.

If I'm not mistaken, $ t \equal{} 2$ is handled by noticing that only two moves can be made from $ (1,1), (1,7), (7,1), (7,7)$. The moves from these points form a cycle along with $ (3,4), (4,3), (5,4), (4,5)$. We can make similar cycles by looking at translates to $ (1,2), (2,1), (2,2)$. We only have two endpoints on the path, so we can cut at most two of the cycles, not enough.


function
limit
integration
real analysis
real analysis unsolved

Let $ f: [0,1] \rightarrow R$ be a integrable function with the property that for all $ x \in [0,1)$ we have:

    $ \lim_{n\to\infty}{n \cdot \int_{x}^{x\plus{}\frac{1}{n}}{f(t)}dt}\equal{}0.$

Prove that for all $ a,b \in (0,1)$ we have:

   $ \int_{a}^{b}{f(t)}dt\equal{}0.$

let $ F(x) \equal{} \int_{0}^{x}f(t)dt$
then we have $ n[F(x \plus{} \frac {1}{n}) \minus{} F(x)] \equal{} n\int_{x}^{x \plus{} \frac {1}{n}}f(t)\rightarrow F'(x) \equal{} f(x)$

Since we weren't given that $ f$ is continuous, that isn't quite the way to manage it. Close, but we should rearrange the argument slightly.

Define $ F(x)$ as you have.

$ F'(x)\equal{}\lim_{n\to\infty}{n \cdot \int_{x}^{x\plus{}\frac{1}{n}}{f(t)}dt}\equal{}0$ for all $ x.$

Hence (by the Mean Value Theorem) $ F(a)\equal{}F(b)$ for all $ a$ and $ b.$

$ \int_a^bf(t))\,dt\equal{}\int_0^bf(t)\,dt\minus{}\int_0^af(t)\,dt\equal{}F(b)\minus{}F(a)\equal{}0$ and we're done.


inequalities
function
induction
inequalities unsolved

Hello,I have the following problem.
If  $f:R->R$ is a convex function and $a_{k+1}\ge a_{k}$ for $k=1,2,...,n-1$ then prove that
  $a_{1}f(a_{n})+a_{n}f(a_{n-1}))+...+a_{2}f(a_{1})\leq f(a_{1})a_{n}+f(a_{n})a_{n-1}+...+f(a_{2})a_{1}$

Are you sure the function isn't concave instead? Is the sign backwards? Subtracting the RHS from the LHS yields

$(a_1-a_{n-1})f(a_n) + \cdots + (a_2-a_n)f(a_1) \ge 0$

But $a_i > a_j$ for each $i > j$, which means that the coefficients of $f(a_n)$ and $f(a_1)$ are both negative, but $f$ evaluated at the endpoints can be made arbitrarily large without losing the convexity of $f$, making the LHS arbitrarily small.

If $f$ is made concave, the problem looks like it can be solved with smoothing; I'll think about it.

Only take $f(x)=x^2$ abd $x_1=1,x_2=2,x_3=3$ we have:
$x_1f(x_2)+x_2f(x_3)+x_3f(x_1) \ge x_1f(x_3)+x_3f(x_2)+x_2f(x_1)$
And the sign of hardsoul reverse.

You can see a similiar problem in Iran 1999 first round.

Sorry, does anyone know how to  finish the proof by an induction?

Finally I proved it :P .
For $n=3$ it is proved .
Suppose that it's true for n i.e
$x_1\cdot f(x_2)+...+x_n\cdot f(x_1)\ge x_2\cdot f(x_1)+...+x_1f(x_n)$
Now take $x_{n+1}\ge x_n\ge x_{n-1}\ge ...\ge x_1$
Then adding  $x_{n+1}f(x_1)-x_{n}f(x_1)\ge x_1f(x_{n+1})-x_1f(x_n)$ equivalent to  $\frac{f(x_{n+1})-f(x_n)}{x_{n+1}-x_n}\ge \frac{x_{n+1}-f(x_1)}{x_{n+1}-x_1}$ we complete the induction.


Ralph and Waldo each simultaneous hold out some fingers on one hand. Ralph always holds out a prime number of fingers; Waldo always holds out an odd number. What is the probability that the sum of the number of fingers they hold out is even? Express your answer as a common fraction.

WHen prime guy holds out 2 sum is odd, when 3 or 5, even, so it's 2/3

well, since waldo always holds out a odd number, that means to get an even sum, ralph has to hold out an odd number as well

therefore, ralph holding out 3 or 5 fingers would get us an even sum

the probability of it being an even sum would be $ \frac {2 \times 3} {3 \times 3} \equal{} \frac {6} {9} \equal{} \frac {2} {3}$

edit: dang, beaten to it, but mine is more detailed :P


integration
trigonometry
limit
real analysis
real analysis unsolved

Find $\displaystyle\int^\frac{\pi}{2}_0 \sin x\,dx$ only bu using    Riemann  metode.


calculus
integration
arithmetic sequence
number theory unsolved
number theory

2. Let us write every integer $ i$ from $ 1$ to $ 2n$ in the form $ 2^{k_{i}}l_{i}$ where $ i$ is odd and $ k$ is a non-negative integer. Then, because in the set $ \{1,2,...,2n\}$ there are only $ n$ odd integers, then in the set $ \{l_1,l_2,...,l_{2n}\}$ there are only $ n$ distinct integers, therefore by Pigeon-Hole Principle among the $ n+1$ integers $ x_1,x_2,...,x_{n+1}$ selected from the set $ \{1,2,...,2n\}$ we will find distinct integers $ a,b\in\{x_1,x_2,...,x_{n+1}\}$ such that $ l_a=l_b$ and therefore, if ${ a=2^{k_{a}}l_a,b=2^{k_b}}l_b$ and wolog $ k_a<k_b$ (we cannot have $ k_a=k_b$ as then $ a=b$ which is impossible) then ${ b=2^{k_b}}l_b=(2^{k_{a}}l_a)2^{k_b-k_a}=a2^{k_b-k_a}$ so $ a|b$ and we find two integers among the $ n+1$ selected so that one divides the other, QED.

[quote="amcavoy"]1. Show that there exists some integral power of 2 equal to 2002...

2. Given any n+1 integers in [1,2n], show that we can find one element that divides another.

3. How many primes are there among the positive integers (written in the usual base 10) such that the prime begins with 1, ends with 1, and only contains 0's and 1's for digits?

4. Let A be any set of 20 distinct integers chosen from the arithmetic progression 1, 4, 7, ..., 100.  Prove that there must be two distinct integers in A whose sum is 104.[/quote]
[hide="2"]
Group the terms into the n sets
{1, 2, 4, 8, 16, ...}
{3, 6, 12, 24, ...}
{5, 10, 20, ...}
.
.
.
{2n-1}
Every term falls into exactly 1 set, so if we pick n+1 terms, two must be from the same set, meaning that one is a multiple of the other.[/hide]

I don't understand problem 1.
3. It is an open question :
$ n\equal{}\sum_{i\equal{}1}^{m}{e_i}10^{i}$
Where $ e_i\in \{0,1\}$
4.It is quite easy.

1. Followed from: exist n, suth that $ lg2.002\le \{nlg2\}<lg2.003$.
4. Exist at least 2 pairs $ (x,y),(z,t)$, suth that $ x\plus{}y\equal{}104,z\plus{}t\equal{}104$.


[color=cyan]A coin of diameter 2 cm is rolled around the outside of an equilateral triangle of side   5 cm.  Compute, in square centimeters, the total area covered by the coin as it is rolled.[/color]

Is the answer --->[hide]15+pi?????[/hide]

[color=cyan]Well, the form of your answer isn't bad . . . the actual numbers involved are different, though.  Reread the question and take each of your steps very carefully.  Going slowly and not making silly mistakes is the most important part of this problem.[/color]

O man, I can't believe i messed up again.

OK, renewed answer:---->[hide]30+4pi[/hide]

hmm, better be rite...

Ok, if u draw the boundaries of the path of the circle, you see that it is a triangle around the original equilateral triangle, except that at the corners, its rounded (do u see?)

Now, from each of the three vertices, we draw two perpendiculars to both sides of the "triangle", which is really the path of the circle

Wat we are left with is three rectangles, 2x5, and 3 sections of circles, each of which is 120 degrees.

So the area is 3*2*5+3*(1/3)*4pi, which is 30+4pi


function
calculus
integration
calculus computations

In how many ways one can prove that (for example) that $ f(x) \equal{} 2x$ is integrable on $ x\in[0,1]$ ?

Explicitly evaluating the integral from the definition would be one way.  Also, we know $ f$ is bounded on $ [0,1]$ so if you prove that $ \sup\{L(f,P)\}\equal{}\inf\{U(f,P)\}$ where $ P$ is a partition of $ [0,1]$, then that's the definition of being integrable, so there's another way.


calculus
integration
trigonometry
logarithms
complex numbers
calculus computations

Okay friends, here is an exercise:
Show that $ \int\limits_0^\pi {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \equal{} \frac{\pi}{a}$, where $ a$ and $ b$ are real numbers with $ a \neq 0$.

hello, we have $ \int\frac{a\minus{}b\cos(x)}{a^2\plus{}b^2\minus{}2ab\cos(x)}\,dx\equal{}\frac{x\minus{}2\arctan\left(\frac{(a\minus{}b)\cot(\frac{x}{2})}{a\plus{}b}\right)}{2a}\plus{}C$.
Sonnhard.

Can you please show how you obtained that solution?

For $ |r| < 1,$

$ \sum_{k \equal{} 0}^{\infty}r^k\cos kx \equal{} \text{Re}\,\sum_{k \equal{} 0}^{\infty}r^ke^{ikx} \equal{} \text{Re}\,\left(\frac {1 \minus{} re^{ \minus{} ix}}{1 \minus{} 2r\cos x \plus{} r^2}\right) \equal{} \frac {1 \minus{} r\cos x}{1 \minus{} 2r\cos x \plus{} r^2}.$

Case 1: Suppose that $ |b| < |a|.$

$ \sum_{k \equal{} 0}^{\infty}\left(\frac ba\right)^k\cos kx \equal{} \frac {1 \minus{} \frac ba\cos x}{1 \minus{} 2\frac ba\cos x \plus{} \left(\frac ba\right)^2} \equal{} \frac {a(a \minus{} b\cos x)}{a^2 \plus{} b^2 \minus{} 2ab\cos x}$

Then $ \int_0^{\pi}\frac {a \minus{} b\cos x}{a^2 \plus{} b^2 \minus{} 2ab\cos x}\,dx \equal{} \frac1a\int_0^{\pi}\sum_{k \equal{} 0}^{\infty}\left(\frac ba\right)^k\cos kx\,dx$

$ \equal{} \sum_{k \equal{} 0}^{\infty}\frac {b^k}{a^{k \plus{} 1}}\int_0^{\pi}\cos kx\,dx \equal{} \frac {\pi}{a}$

since the integral is zero except for $ k \equal{} 0$ and in that case is the integral of a constant.

Case 2: Suppose that $ |b| > |a|.$

This is a little more problematical.

$ \sum_{k \equal{} 0}^{\infty}\left(\frac ab\right)^k\cos kx \equal{} \frac {b(b \minus{} a\cos x)}{a^2 \plus{} b^2 \minus{} 2ab\cos x}$

With some algebraic maneuvering, we get that

$ \frac{a\minus{}b\cos x}{a^2\plus{}b^2\minus{}2ab\cos x}\equal{}\frac1a\minus{}\frac1a\sum_{k \equal{} 0}^{\infty}\left(\frac ab\right)^k\cos kx.$

Put that in and discover that the integral is zero.

Case 3: $ |a|\equal{}|b|\ne 0$

In this case, this is simply $ \int_0^{\infty}\frac1{2a}\,dx\equal{}\frac{\pi}{2a}.$

=============================

akech's original statement was not quite right. It should have been

$ \int_0^{\pi}\frac{a\minus{}b\cos x}{a^2\plus{}b^2\minus{}2ab\cos x}\,dx\equal{} \begin{cases}\frac{\pi}a,&|a|>|b|\\\frac{\pi}{2a},&|a|\equal{}|b|\ne 0\\0,&|a|<|b|\end{cases}$

with the further note that this is meaningless if $ a\equal{}b\equal{}0.$

$ \int\limits_0^\pi {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \\
\equal{} \int\limits_0^\pi {\frac {1}{2a}\frac {{2a^2 \minus{} 2ab\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \\
\equal{} \int\limits_0^\pi {\frac {1}{2a}\frac {{a^2 \minus{} b^2 \plus{} a^2 \plus{} b^2 \minus{} 2ab\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \\
\equal{} \int\limits_0^\pi {\frac {a^2 \minus{} b^2}{2a}\frac {{1}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \plus{} \pi$
for the remaining part, write
$ a^2 \plus{} b^2 \equal{} c \\
\minus{} 2ab \equal{} d$
Then,
$ \frac {1}{c \plus{} dcos x} \\
\equal{} \frac {sec^2{\frac {x}{2}}}{(c \minus{} d) tan^2{\frac {x}{2}} \plus{} (c \plus{} d)}$
put value of c and d.
corresponding to different cases you will have to break $ ln{|{Denominator}|}$

mathematica:
[code]
if a^3 b != a b^3, ans=(\Pi (1 + Sign[a + b]/Sign[a - b]))/(2 a) [/code]

Thank you, Dr. Merryfield for pointing out my error and also for the complete solution.
My partial solution was for the case $ |a| > |b|$:
Consider the integral:
$ I \equal{} \frac {1}{2} \int \log (a^2 \plus{} b^2 \minus{} 2ab \cos x) dx$
$ \equal{} x\log a \plus{} \frac {1}{2} \int \log \left(1 \plus{} ( \minus{} \frac {b}{a})^2 \plus{} 2( \minus{} \frac {b}{a}) \cos x\right) dx$
$ \equal{} x\log a \plus{} \frac {1}{2} \int \log (1 \plus{} ( \minus{} \frac {b}{a})e^{ix}(1 \plus{} ( \minus{} \frac {b}{a})e^{ \minus{} ix}dx$
$ \equal{} x\log a \minus{} \sum_{n \equal{} 1}^{\infty} (\frac {b}{a})^{n}\frac {\sin xn}{n^2} \plus{} C$.

$ \frac {dI}{da} \equal{} \frac {x}{a} \plus{} \sum_{n \equal{} 1}^{\infty}\frac {b^{n}}{a^{n \plus{} 1}}\frac {\sin xn}{n}$

Also, $ I \equal{} \frac {1}{2} \int \log (a^2 \plus{} b^2 \minus{} 2ab \cos x) dx \implies \frac {dI}{da} \equal{} \int {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx$

Hence $ \int {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \equal{} \frac {x}{a} \plus{} \sum_{n \equal{} 1}^{\infty}\frac {b^{n}}{a^{n \plus{} 1}}\frac {\sin xn}{n} \plus{} C$

Thus, $ \int\limits_0^\pi {\frac {{a \minus{} b\cos x}}{{a^2 \plus{} b^2 \minus{} 2ab\cos x}}} dx \equal{} \frac {\pi}{a}$.

Another perspective: $ a^2\plus{}b^2\minus{}2ab\cos \theta\equal{}(b^2\\minus{}2ab\cos\theta\plus{}a^2\cos^2\theta)\plus{}a^2\sin^2\theta$
$ \equal{}(\minus{}b\plus{}a\cos\theta)^2\plus{}(a\sin\theta)^2\equal{}|ae^{i\theta}\minus{}b|^2$. Introduce complex numbers.

Consider the complex integral $ I\equal{}\int_{|z|\equal{}a}\frac1{z\minus{}b}\,dz$. This can be parametrized as $ \int_0^{2\pi}\frac1{ae^{i\theta}\minus{}b}\cdot iae^{i\theta}\,d\theta$, which becomes in real form
$ I\equal{}\int_0^{2\pi}\frac{i(a\cos\theta\minus{}b\minus{}ia\sin\theta)(a\cos\theta\plus{}ia\sin\theta)}{a^2\plus{}b^2\minus{}2ab\cos \theta}\,d\theta$
$ \equal{}a\int_0^{2\pi}\frac{b\sin\theta\plus{}i(a\minus{}b\cos\theta)}{a^2\plus{}b^2\minus{}2ab\cos \theta}\,d\theta$

The imaginary part of $ I$ is $ a\int_0^{2\pi}\frac{a\minus{}b\cos\theta}{a^2\plus{}b^2\minus{}2ab\cos \theta}\,d\theta$, $ 2a$ times the integral we want to compute. We have a simple pole of residue $ 1$, so $ I\equal{}2\pi i$ if that pole is inside the circle, $ I\equal{}0$ if the pole is outside, and $ I\equal{}\pi i$ as a principal value if it's on the circle.


I am not a spammer, at least, this is the way I use to think about myself, and thus I will not open a new thread for the following problem from today's DeMO exam:

Let Q(n) denote the sum of the digits of a positive integer n. Prove that $Q\left(Q\left(Q\left(2005^{2005}\right)\right)\right)=7$.

[[b]EDIT:[/b] Since this post was split into a new thread, I comment:

The problem is completely analogous to the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 , with the only difference that you have to consider the number $2005^{2005}$ instead of $4444^{4444}$.]

  Darij

But, Darij, won't the method be the same? So we're just left to make some computations

I think Darij knew that when posting  ;)
But the DeMO is going down and down... they are not able to find new problems (or to take them from a little bit more unknown olympiads)


Prove
The sum of the 3rd powers of three consecutive numbers from Z is divisible by 9.

Let´s expand this cubics:

(a-1)^3+a^3+(a+1)^3=a^3-3a^2+3a-1+a^3+a^3+3a^2+3a+1=3a^3+6a=3a(a^2+2). And now we have to prove that 3/a(a^2+2).

1)a=0(mod3) and so is proved
2)a=1(mod3) -> a^2=1(mod3) -> a^2+2=3(mod3) and so is proved
3)a=2(mod3) -> a^2=4=1(mod3) and so is proved too.


Here's a problem from my math homework that my math teacher couldn't answer (just parts d and e) So if anyone knows how to answer d or e, it would be greatly appreciated

26. The ten cards, ace to ten in hearts are removed from a deck of cards, shuffled, and placed in a row. The ace to ten of clubs are also removed, shuffled, and placed in a row just below the first cards.

a) How many different pairs (of ten cards) can be made? (answer: (10!)2)
b) How many of the pairings in 26a have the same card in the third position? (answer: (10!)(9!) )
c) How many of the pairings in 26a have the same card in the fourth position? (answer: (10!)(9!) )
d) How many of the pairings in 26a have exactly one of the pairs the same? (any position) (answer: ???)
e) How many of the pairings in 26a have at least of of the pairs being the same? (answer: ???)

Answers and explainations for parts d and e would be appreciated.

What do you mean by part a?  I would say that there are C(20, 2) pairs of cards, 10 pairs by the poker definition.  What exactly do you mean?  How many orderings?  I think you need to explain to me what you are doing better, because I'm not really sure what we are counting.

Okay -- and we consider a heart and club to be paired when they are directly above (below) each other?

For part d, you have several things to do:

First, [hide]you have to pick the pair that will stay the same.[/hide]

Second, (and this is the hard part),

[hide]you have to make sure none of the other pairs is still together.[/hide]  Now, I have a question:  If we were playing with only 4 cards, {A, 2, 3, 4}, and I had

A 2 3 4

A 2 3 4

as one arrangement and

A 2 4 3

A 3 4 2

as another arrangement, do the 44 pairs count as the same pair?  I assume that is what "any position" means, but I just wanted to check.

This second bit is rather hard.  The problem is known as "derangements," and you could try a search for that word on this forum in order to find out about it.  I personally would start by looking at how many derangements of 1 card, 2 cards, 3 cards, and so on, there are, and trying to find a pattern.

For part e, try using the principle of inclusion-exclusion.

I still feel like the question isn't well-posed, though -- it leaves me with too many "what exactly do they mean?" feelings.

[quote="picounts"]Sorry, I just copied it from the book word-for-word.
You will end up with two rows of ten cards each, with the second row right below the first. The first row only contains hearts (Ace to ten), and the second only clubs (Ace to ten). There would be 10! ways to arrange each row, for a total of (10!)2 permutations.[/quote]

Thats kinda weird anyway...

Lets simplify this to hearts and clubs with 2h, 3h, 4h being 3 hearts and 2c,3c,4c being 4 clubs.... then if you lay it out like this:

2h 3h 4h
2c  3c  4c

It's different from

2h 4h 3h
2c 4c 3c

By how you've defined it. This might be what you meant, but its kind of odd.

Yes, I agree -- this definition of pairings is very strange.  Are you really convinced your answer to the first question is correct?  A much more "usual" answer to a question like this would be 10!.

Yeah, pretty sure. For example, with 3 cards instead of 10 we woud have (3!)2:

Ad 2d 3d
Ac 2c 3c

Ad 2d 3d
Ac 3c 2c

Ad 2d 3d
2c Ac 3c

Ad 2d 3d
2c 3c Ac

Ad 2d 3d
3c Ac 2c

Ad 2d 3d
3c 2c Ac

Ad 3d 2d
Ac 2c 3c

Ad 3d 2d
Ac 3c 2c

Ad 3d 2d
2c Ac 3c

Ad 3d 2d
2c 3c Ac

Ad 3d 2d
3c Ac 2c

Ad 3d 2d
3c 2c Ac

And so on

I understand what you are saying -- however, the conventional definition of pairing doesn't rely on what order your pairs are in.  Thus,

A  2
2  A

2  A
A  2

would normally be considered to be the same.  It makes the later questions much nicer, and it makes them make more sense.

Wow, I feel slow today....

So, to get the number of ways of counting we want, can't we just multiply the "nicer" way of counting by 10P10=10! ?


look it in the dicktionary

try wikipedia.com

[quote="Tecnogram888"]What are combinatorics?  like, the definition of them and what r they classified as and stuff.[/quote]


AMC
AIME
probability
pigeonhole principle

9. A fair coin is to be tossed ten times. Let $i/j$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i + j$

consider the string of H's and T's, say that there are $h$ H's, then there are $10-h$ T's, and $11-h$ spots for the H's to go for a good configuration, so we choose $h$ of those $11-h$ spots to place the H's, and giving all the possible numbers of H's in a summation:

$\frac{ \sum_{h=0}^{5}\binom{11-h}{h} }{2^{10}}=\frac{9}{64} \implies m+n=73$

[hide]5 heads: We must distribute these 5 heads among 6 slots, so there are 6 possibilities (ie, HTHTHTHTHT, HTHTHTHTTH, HTHTHTTHTH, HTHTTHTHTH, HTTHTHTHTH, THTHTHTHTH).

4 heads: 4 heads are to be placed into 7 slots.  So there are $\binom74=35$ ways.

3 heads: $\binom83=56$ ways

2 heads: $\binom92=36$ ways

1 head: 10 ways

No heads: 1 way

So in total, we have \[ \binom65+\binom74+\binom83+\binom92+\binom{10}1+\binom{11}0=6+35+56+36+10+1=144 \]  good possibilities.  There are $2^{10}$ total ways, so the probability in question is \[ \frac{144}{2^{10}}=\frac{9}{64}\implies m+n=9+64=\boxed{73} \][/hide]

[quote="Tecnogram888"]athlet, whered you get the 11 from?  i thought it was 10 tosses?
k8, where did you get those 6 possibilities? why are they the only ones?
how come you guys didnt take into account the probability of getting a TTTTTTTTTT, HTTTTTTTTT, THTTTTTTTT, etc.  2 heads, 3 heads, etc.?[/quote]

When he counts 11 slots, he's basically counting the possible strings of heads and tails.  For example,

_T_ T_ T_ T_ T_  and you can place 5 heads in one of 6 slots.  There are 6, not 5 slots (endpoints count! :D )

The rest follows.  For example, if there are 5 heads, then we calculate (6 c 5).  And if there are 4 heads, there would be 6 tails, so 7 spots to put them (7 c 4) etc. all the way down to no heads.

The reason why we can't have 6 heads, is because then we'll have 4 tails, and by Pigeonhole, 2 of these heads must be beside each other...an impossibility.  So we consider when there are 5 or fewer heads.

And as for the last possibility...they did take into account the probability of getting all tails etc.  All tails is represented by (11 c 0) = 1.  etc.  And as for 1 head, 2 heads, etc, both have considered the possible strings containing those numbers of heads, combinatorially of course.

Hope this helps.

[hide="Another Solution"]After 2 coins are chosen, the possibilities are HT, TH, and TT.  The only significant characteristic of each possibility is the outcome of the last flip, because it determines our possibilities for the outcomes of the next flip.  So after two coins are chosen, we can write the number of possibilities as 2T, 1H, where the letter denotes the outcome of the last flip.

After 3 coins are chosen the number of possibilities ending in T is equal to the total number of possibilites after 2 coins are chosen, because we can add a T to any possibility without breaking any restrictions.  The number of possibilities ending in H is equal to the number of possibilites ending in T after 2 coins are chosen, because we can only add an H to possibilities ending in T.

2:   2T, 1H
3:   3T, 2H
4:   5T, 3H
5:   8T, 5H
6:   13T, 8H
7:   21T, 13H
8:   34T, 21H
9:   55T, 34H
10:   89T, 55H

If you notice that these are all Fibonacci numbers, then you can save some time.

So the probability is $\frac{89+55}{2^{10}}=\frac{9}{64}\Rightarrow\frac{i}{j}=73$[/hide]


How many ways can $ 5$ distinct paperback books and $ 1$ hardcover book be arranged on a shelf if the hardcover must be the rightmost book on the shelf?

Since the placement of the hardcover book is set, we need to find the ways to arrange $ 5$ books in a row, which is $ 5!\equal{}\boxed{120}$.


I actually planned to go to todays math jam, but couldn't get thru after trying for about 40 minutes :( Suppose the classroom site was too busy for a lowly dialup connection to get through to :)

I've been trying continuously for the last hour. About 10 minutes ago, I actually got through to the page, but now the Java applet won't load.

Weird - has anyone else had this problem?  Maybe you need to update your Java?

I got reapetedly kicked out and finally gave up.

What sort of internet connection do you have?

That explains it.  Many wireless connections are fragile - lots of very brief disconnections.  This is no problem if you are surfing the net, in which you don't have to stay connected continuously, but our classroom requires you stay connected contininuously (without interruptions, no matter how small).  Your wireless has lots of tiny interruptions - you lose contact with the classroom each time.

oh, now i see why i get kicked out once or twice an hour,

Also check ur CPU processing percentage in the task manager. I rmr once not being able to get in just b/c I was running at 100% and didn't know it.

Nah my java thingy is fine.. and I have been in once before (though I remember it took a while then too). It must just be the speed of my modem (my friend gets in fine)

well if you have a pentium 2 or lower java is very slow i know because i have one. and it takes like a century to use maple 9 on them.

Na, I don't have one of those. I tried disabling my firewall, tried downloading a different internet browser, and updating java. Nothing works. It just says it can't load the applet.. if anyone knows anything about Java , this is what it said.. (a tiny bit got cut off, cos I copied it into messenger and then later into here)

[quote="Java thing"]
load: class com.diginet.digichat.client.DigiChatApplet not found. java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet at sun.applet.AppletClassLoader.findClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.applet.AppletClassLoader.loadClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.applet.AppletClassLoader.lo

Caused by: java.io.IOException: open HTTP connection failed. at sun.applet.AppletClassLoader.getBytes(Unknown Source) at sun.applet.AppletClassLoader.access$100(Unknown Source) at sun.applet.AppletClassLoader$1.run(Unknown Source) at java.security.AccessController.doPrivileged(Native Method) ... 10 more

Exception: java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet 29/02/2004 13:19:05 sun.plugin.util.PluginLogger log INFO: Exception: java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet Finding information ... 29/02/2004 13:19:05 sun.plugin.util.PluginLogger log
[/quote]

I asked the folks at Digichat and they asked that you run a tracert.  Here's how:

[quote]Open up a command line (for windows, start > run > cmd), then type "tracert <host>", where host is replaced with the actual host.  This will tell us if they can establish a connection to our servers.[/quote]

Let me know what that yields and we'll see if we can get you connected.

Cool! So, uh.. this is probably obvious, but whats the host? :D

i believe the host would be artofproblemsolving.com, doing a trace route just shows you what servers/junction points you pass when trying to connect to a certain host, and how long it takes.

[quote="TripleM"]Oh.. well here we go... [/quote]

Hi, TripleM, I for one would love to see you participating in math jams, so I hope your technical problems get resolved. It does seem that some of those hops are rather long (more than a second, in one instance I noted). I used to live in Taiwan (till July 2001), and in those days trans-Pacific connections in the north Pacific were sometimes painfully slow. I had a lot of problem in Taiwan, too, with local modem pool latency.

Now back in North America, usually connecting to North American sites through a local Internet service provider with a local broadband connection to a good Internet backbone, I usually enjoy high speed of connection, but I do notice that when I type messages here on AoPS (as I am doing right now) that sometimes the characters don't show up on the screen right away. I DON'T seem to have that issue on other sites at which I type to reply to PHPBB boards. So I'm still investigating what speed is workable for connecting to AoPS. We do get the math jams here, and stay on them successfully.

I suppose/hope that these days there is a good fiber optic cable between New Zealand and North America. But perhaps sometimes just the extra router connections along the way may be adding latency to your connection. Best wishes to you for getting connected and participating in the math jams.

check out http://www.edgehill.ac.uk/mltr/lessons/viewfaq.htm it has some instructions on fixing Java problems, i have a feeling that your problem is cause 2 or 3 at that site, but just go through it and see if anything helps.

My bad - I should have read what I posted more closely.  Try host5.digichat.com for the host.

Unfortunately, it appears that one of the routers between you and the digichat servers is bad.  Sadly, this is not on either your end or ours, so we're pretty powerless to fix it.  Hopefully the problem will be resolved in the near future and you'll be able to join our Math Jams in the future.


modular arithmetic
number theory unsolved
number theory

[quote="pluricomplex"][quote="pavel25"]Prove that if $p$ is a prime and $p^2+2$ is a prime then $p^3+2$ is a prime![/quote]

because $p$ must be $3$. Indeed, if $p,p^2+2$ are both primes then if $(p,3)=1$ then $p^2+2\equiv0\pmod{3}$ which cotract! So $p=3$ and $p^3+2=29$ is prime![/quote]
thanks pluricomplex! and i have to add that $p=1$ too!


Find the value of the sum:
\[ \sum_{i=1}^{2001} [\sqrt{i}] \]
where $[ {x} ]$ denotes the greatest integer which does not exceed $x$.

[hide]
I have an another approach.We can consider on cartesian coordinates.The value $\sum_{i=1}^{2001} [\sqrt i]$ denotes the number of lattice point $(x>0,y>0)$ below $y=\sqrt x$.We can calculate the number of lattice point by considering the number of lattice point $(y>0,x>0)$ below $x=y^2$.Namely, $44\cdot 2001-\sum_{y=1}^{44} (y^2-1)=\boxed{58718}$.


Let $f(x)=2x-sinx$. Prove that : there exists a real $b$ and functions $h(x)$and  $g(x)$ satisfies the coditions
1. $g(x)=bx+h(x)$
2. $h(x)$is periodic
3. $f(g(x))=x$ for all x

Obviosly b=1/2 and
(1) $h(2x-sinx)=\frac{sin x}{2}$.
Because $2x-sinx$ increase in $(0,2\pi )$, then exist unique function$h: (0,4\pi)\to (-\frac{1}{2},\frac{1}{2})$, suth that (1). Obviosly$h(x)$ had period $4\pi.$


superior algebra
superior algebra solved

for any set $X$, let $P(X)=\{G|G \subseteq X\}$ be the power set of $X$.  For any $G$ and $H$ in $P(X)$ define.
$G+H=\{x|x \in G \cup H, x \notin G \cap H \}$  and $G\cdot H= G \cap H$

under these two operations how can I show $P(X)$ is a ring with unity?

Nice, I never saw that before  :)

zero is the empty set, unity is whole $X$
to prove the other necessary things: draw some graphic!

It's just a construction of a Boolean ring from a Boolean algebra. Another way to look at it is as being the direct product $\mathbb Z_2^X$.


geometry
similar triangles
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14821[/img]

 :(

This problem is coming from the past.

$ \bullet$ When the triangles $ \bigtriangleup AO_{1}B$ and $ \bigtriangleup A'O_{1}B'$ are directly similar, we have that $ \frac {O_{1}A}{O_{1}A'} \equal{} \frac {AB}{A'B'}$ $ ,(1)$ and $ \frac {O_{1}B}{O_{1}B'} \equal{} \frac {AB}{A'B'}$ $ ,(2)$

From $ (1),$ we conclude that the point $ O_{1},$ lies on the [b][size=100]Appolonius circle[/size][/b] $ (C_{1}),$ with diameter the segment $ MM',$

where $ M,\ M',$ defined points on the line segment $ AA',$ such that $ \frac {MA}{MA'} \equal{} \frac {M'A}{M'A'} \equal{} \frac {AB}{A'B'}$ $ ($ $ M,$ between $ A$ and $ A'$ $ ).$

Similarly from $ (2),$ we conclude that the point $ O_{1},$ lies also on the [b][size=100]Appolonius circle[/size][/b] $ (C_{2}),$ with diameter the segment $ NN',$

where $ N,\ N',$ defined points on the line segment $ BB',$ such that $ \frac {NB}{NB'} \equal{} \frac {N'B}{N'B'} \equal{} \frac {AB}{A'B'}$ $ ($ $ N,$ between $ B$ and $ B'$ $ ).$

There are two points $ O_{1}$ and $ O'_{1},$ with the property as the problem states, as the intersection points of the circles $ (C_{1})$ and $ (C_{2}).$

$ \bullet$ By the same way, in the case of inversely similar triangles $ \bigtriangleup AO_{2}B$ and $ \bigtriangleup B'O_{2}A',$

when we have that $ \frac {O_{2}A}{O_{2}B'} \equal{} \frac {AB}{A'B'}$ $ ,(3)$ and $ \frac {O_{2}B}{O_{2}A'} \equal{} \frac {AB}{A'B'}$ $ ,(4)$

the point $ O_{2},$ lies on the [b][size=100]Appolonius circles[/size][/b] $ (C'_{1}),\ (C'_{2}),$ with respect to the segments $ AB',\ BA'$ respectively.

There also two points $ O_{2}$ and $ O'_{2},$ with the property as the problem states and the solution is completed.

Kostas vittas.


[quote="iamhe"]If a*b=22 and a+b=12,
what is absolute value of b-a?
A)2 sqrt 66
B)2 sqrt 14
C)5 sqrt 66
D)2 sqrt 10
E)5 sqrt 14[/quote]

[hide="generic"]
$ a\plus{}b\equal{}12$
$ (a\plus{}b)^2\equal{}144$
$ a^2\plus{}2ab\plus{}b^2\minus{}4ab\equal{}144\minus{}88$
$ (a\minus{}b)^2\equal{}56$
$ |a\minus{}b|\equal{}|b\minus{}a|\equal{}2\sqrt{14}$

[/hide]


Let $ H$ be the orthocenter of an acute triangle $ ABC$ whose edgesare $ BC\equal{}a;CA\equal{}b;AB\equal{}c$
Prove that
$ \frac{HA^2\plus{}HB^2\plus{}HC^2}{a^2\plus{}b^2\plus{}c^2} \leq (cotA.cotB)^2\plus{}(cotB.cotC)^2\plus{}(cotCcotA)^2$


The value of $ \tan 10^{\circ}$ is denoted by $ p$. Find, in terms of $ p$, the value of

(i) $ \tan 55^{\circ}$,

(ii) $ \tan 5^{\circ}$,

(iii) $ \tan \alpha$, where $ \theta$ satisfies the equation $ 3 \sin(\theta \plus{} 10^{\circ}) \equal{} 7 \cos (\theta \minus{} 10^{\circ})$.
.

[hide="i"]$ \tan 55^{\circ}\equal{}\tan(10\plus{}45)^{\circ}\equal{}\frac{\tan 10\plus{}\tan 45}{1\minus{}\tan 45\tan 10}\equal{}\frac{p\plus{}1}{1\minus{}p}$[/hide]

[hide="ii"]$ \frac{2\tan 5}{1\minus{}\tan^2 5}\equal{}p\Rightarrow 2\tan 5\equal{}p\minus{}p\tan^2 5$
$ \Rightarrow \tan 5\equal{}\frac{\minus{}2\plus{}\sqrt{4\plus{}4p^2}}{2p}\equal{}\frac{\minus{}1}{p}\plus{}\frac{\sqrt{1\plus{}p^2}}{p}$[/hide]


How many positive factors does the number 900 have?

Since $ 900\equal{}2^2\cdot3^2\cdot5^2$, there are $ (2\plus{}1)^3\equal{}\boxed{27}$ factors.

[hide="To find the number of factors of a number"]
You find the prime factorization of that number and multiply the number of exponents + 1 for each prime number. Yeah, I know it sounds confusing, but lemme give you an example. To find the number of factors for the number $20$, you factors it with primes, so $20=2^2\cdot5^1$. Any factor(s) of $2^{2}5^{1}$ must be in the form $2^{x}5^{y}$, where $2\ge{x}\ge0$ and $1\ge{y}\ge0$, and they must be integers. Therefore, $x$ could be $0,1,2$ and $y$ could be $0,1$. We multiply them to get $\boxed{6}$ factors. This is how we determine that $20$ has $6$ factors.
[/hide]

Now, we know that $900=2^{2}3^{2}5^{2}$, so we multiply by $3\cdot3\cdot3=3^{3}=\boxed{27}$.

Hope I helped!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[color=transparent]Please rate, like, comment, or subscribe to my YouTube channel at youtube.com/user/DAMENAOCL12[/color]

[size=50]teehee[/size]

[quote=Greenacid]Thanks! Forgot how to find the number of factors but I know now! I know you probably won't see this because it has been 7 years though[/quote]

Lol


geometry
incenter
analytic geometry
linear algebra
matrix
trigonometry
function

I, G, H are incenter, gravity center (medicenter) and orthocenter of triangle ABC respectively (there are no two equal sides).
a) Is it posible IH to be parallel to some of the sides of a triangle?
b) Is it possible triangle IGH to be equaliteral?

$ \text{1). In barycentric coordinate I(a; b; c) and G(1; 1; 1)} \Longrightarrow \\
\text{IG line equation: }\begin{equation*} \begin{vmatrix} x & y & z \\
a & b & c \\
1 & 1 & 1 \end{vmatrix} \end{equation*} = 0 \Longleftrightarrow (b - c).x + (c - a)y + (a - b).z = 0 \text{ and BC line equation: x = 0.}$
$ \text{ 2). }\boxed{{IG}\|{BC}}\Longleftrightarrow$
$ \Longleftrightarrow \begin{cases} \text{(b - c).x + (c - a)y + (a - b).z = 0} \\
x = 0 & \text{have solution: }y^2 + z^2\ne{0} \\
x + y + z = 0 \end{cases} \end{equation*} \Longleftrightarrow$
$ \Longleftrightarrow \begin{equation*} \begin{cases} \text{(c - a)y + (a - b).z = 0} \\
y + z = 0 \end{cases} \end{equation*} \text{have solution: } y^2 + z^2\ne{0} \Longleftrightarrow$ $ \begin{equation*} \begin{vmatrix} c - a & a - b \\
1 & 1 \end{vmatrix} \end{equation*} = 0$
$ \Longleftrightarrow \boxed{ b + c = 2.a.}$
$ \text{3). Have: H(tgA; tgB; tgC), and barycentric equation the line IH: }$
$ \begin{equation*} \begin{vmatrix} x & y & z \\
a & b & c \\
tgA & tgB & tgC \end{vmatrix} \end{equation*} = 0 \Longleftrightarrow \\
\Longleftrightarrow(b.tgC - c.tgB).x + (c.tgA - a.tgC)y + (a.tgB - b.tgA).z = 0\text{ and: } \\
\boxed{{HI}\|{BC}}\Leftrightarrow{c.tgA - a.tgC = a.tgB - b.tgA}\Longleftrightarrow\boxed{(c + b).tgA = a.(tgB + tgC).}$

Thank you a lot, but the condition of a) were wrong. I edited it. It should be IH to be parallel to some of the sides.

For b) I've an idea. If it is possible give an answer may we from:
$ AI^{2} \plus{} BI^{2} \plus{} CI^{2} \equal{} AH^{2} \plus{} BH^{2} \plus{} CH^{2}$
To get ABC is equaliteral? If yes it is impossible, because it is impossible IG=GH for non equaliteral triangle. Please if you can - verify this hypothesis.

[b][color=darkblue]Prove easily that $ IH\parallel BC$ $ \Longrightarrow$ the projection of $ A$ on $ BC$ belongs to $ (BC)\ !$
Thus, $ \boxed {IH\parallel BC}$ $ \Longleftrightarrow$ $ \frac {2p}{a} \equal{} \tan B\tan C$ $ \Longleftrightarrow$ $ a\sin B\sin C \equal{} 2p\cos B\cos C$ $ \Longleftrightarrow$ $ abc \equal{} 8R^2p\cos B\cos C$ $ \Longleftrightarrow$ $ 4Rpr \equal{} 8R^2p\cos B\cos C$ $ \Longleftrightarrow$ $ \frac rR \equal{} 2\cos B\cos C$ $ \Longleftrightarrow$ $ \frac rR \equal{} \cos (B \minus{} C) \minus{} \cos A$ $ \Longleftrightarrow$ $ 2\cos A \plus{} \cos B \plus{} \cos C \equal{}$ $ 1 \plus{} \cos (B \minus{} C)$ $ \Longleftrightarrow$ $ \cos^2\frac A2 \minus{} \sin^2\frac A2 \plus{} \cos\frac {B \plus{} C}{2}\cos\frac {B \minus{} C}{2} \equal{}$ $ \cos^2\frac {B \minus{} C}{2}$ $ \Longleftrightarrow$ $ \sin^2\frac {B \plus{} C}{2} \minus{} \cos^2\frac {B \plus{} C}{2} \equal{}$ $ 2\sin\frac B2\sin\frac C2\cos\frac {B \minus{} C}{2}$ $ \Longleftrightarrow$ $ \left(\tan\frac B2 \plus{} \tan\frac C2\right)^2 \minus{} \left(1 \minus{} \tan\frac B2\tan\frac C2\right)^2 \equal{}$ $ 2\tan \frac B2\tan\frac C2\left(1 \plus{} \tan\frac B2\tan\frac C2\right)$ $ \Longleftrightarrow$ $ \boxed {\left(\tan\frac B2 \plus{} \tan\frac C2\right)^2 \equal{} 1 \plus{} 3\tan^2\frac B2\tan^2\frac C2}\ .$ Denote $ \tan\frac B2 \equal{} x$, $ \tan\frac C2 \equal{} y\ .$ Thus, $ IH\parallel BC$ $ \Longleftrightarrow$ exist $ x > 0$, $ y > 0$ so that $ (x \plus{} y)^2 \equal{} 1 \plus{} 3x^2y^2$, i.e. exists $ x > 0$ so that the equation (in $ y$) $ (3x^2 \minus{} 1)\cdot y^2 \minus{} 2x\cdot y \plus{} 1 \minus{} x^2 \equal{} 0$ has at least a  positive root. It is possibly, for example $ \frac {\sqrt 3}{3} < x < 1\ !$

I hope that there are no any mistakes in the upper proof. Here are two remarks :

$ \boxed {IG\parallel BC}$ $ \Longleftrightarrow 3r \equal{} h_a$ $ \Longleftrightarrow$ $ 3ar \equal{} ah_a$ $ \Longleftrightarrow$ $ 3ar \equal{} 2pr$ $ \Longleftrightarrow$ $ 2p \equal{} 3a$ $ \Longleftrightarrow$ $ \boxed {b \plus{} c \equal{} 2a}\ .$[/color]

[color=darkred][u]Exercise.[/u]$ \boxed {IG\perp BC\Longleftrightarrow b \equal{} c\ \ \vee\ \ b \plus{} c \equal{} 3a}$ (it is interestingly).[/color][/b]

I've only one stupid question. How you get the condition IH||BC?

$ IH\parallel BC\Longleftrightarrow$ $ [BHC]\equal{}[BIC]$ (the areas).

Very good idea. I think your proof is very good (if i don't have a mistake, too). And I also think there are infinitely many triangles with required property.

[b][color=darkblue]Here is the particular case $ \|\begin{array}{c} \tan\frac B2 = \frac {r}{DB} = x = \frac 23\in(\frac {\sqrt 3}{3},1) \\
 \\
\tan\frac C2 = \frac {r}{DC} = y = \frac {6 - \sqrt {21}}{3} < 1\end{array}$, where $ D\in BC\cap C(I,r)\ .$[/color][/b]

[b][color=darkred][u]Can someone to draw[/u] a triangle $ ABC$ for which $ \frac {r}{10} = \frac {DB}{15} = \frac {DC}{12 + 2\sqrt {21}}\ ?$[/color]

[color=darkblue]Please answer me if you"ll obtain $ IH\parallel BC\ !$ Thank you a lot ![/color][/b]

I don't understood you in depth. But I think:
By your conditions we may find $ sin \alpha$ and $ sin \gamma$
Do you want more details? I think this triangle is constructible. And the construction is easy.

$ \boxed{IH\|BC}\Longleftrightarrow (b \plus{} c).tgA \equal{} (tgB \plus{} tgC).a\Longleftrightarrow \\
\Longleftrightarrow (b \plus{} c).\frac {tgB \plus{} tgC}{1 \minus{} tgB.tgC} \plus{} (tgB \plus{} tgC).a \equal{} 0\Longleftrightarrow tgB.tgC \equal{} \frac {a \plus{} b \plus{} c}{a} \equal{} \frac {h_a}{r}.$
$ \text{Similar of Virgil Nicula assertion!}$

Thank you Virgil Nicula and mihai miculita! I think you are right. In original statement of the problem were written to find dimension of the angles of the triangles for a) and b). But I think it is not a good idea, because in my opinion for a) we may have infinitely many triangles with this property. Am I right?

$ \text{Using the sin and cos laws, have: }\\
(b\plus{}c).tgA\equal{}a.(tgB\plus{}tgC)\Longleftrightarrow (b\plus{}c).\frac{sinA}{cosA}\equal{}a.(\frac{sinB}{cosB}\plus{}\frac{sinC}{cosC})\Longleftrightarrow \\
\Longleftrightarrow \dots \Longleftrightarrow \frac{b\plus{}c}{b^2\plus{}c^2\minus{}a^2}\equal{}a.(\frac{1}{a^2\plus{}c^2\minus{}b^2}\plus{}\frac{1}{a^2\plus{}b^2\minus{}c^2})\Longleftrightarrow \\
\Longleftrightarrow \boxed{2.a^3.(b^2\plus{}c^2\minus{}a^2)\equal{}(b\plus{}c).(a^2\plus{}c^2\minus{}b^2).(a^2\plus{}b^2\minus{}c^2)}.$

[color=darkblue][b]The relations $ \left\|\begin{array}{c} 9\cdot GI^2 \equal{} p^2 \plus{} 5r^2 \minus{} 16Rr \\
 \\
HI^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2 \\
 \\
9\cdot GH^2 \equal{} 36\cdot R^2 \minus{} 8\cdot \left(p^2 \minus{} r^2 \minus{} 4Rr\right)\end{array}\right\|$ are well-known.

Therefore, $ GI \equal{} HI \equal{} GH$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} 5p^2 \equal{} 18R^2 \plus{} 26Rr \plus{} 11r^2 \\
 \\
3p^2 \equal{} 12R^2 \plus{} r^2 \plus{} 16Rr\end{array}\right\|$ $ \Longleftrightarrow$

$ \left\|\begin{array}{c} p^2 \equal{} 3\cdot\left(2R^2 \plus{} 2Rr \minus{} 3r^2\right) \\
 \\
3R^2 \plus{} Rr \minus{} 14r^2 \equal{} 0\end{array}\right\|$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} R \equal{} 2r \\
 \\
p \equal{} 3r\sqrt 3\end{array}\right\|$ $ \Longleftrightarrow$ $ a \equal{} b \equal{} c\ .$

In conclusion, $ \triangle\ GHI$ is equilateral iff $ \triangle\ ABC$ is equilateral, i.e. $ G\equiv H\equiv I\ .$[/b][/color]

Excellent job, Virgil Nicula! Very good.

Virgil Nicula, I've an idea - you use very interesting equalities for the three distances. How do you get them? May you find similar relations for the sides of the triangle by its angle bisectors?

I ask you, because I've posted a problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=172431 and I'm interested is there a solution in elementary functions and if there is a solution how to find it and if there is no a solution - how to prove it.

P.S. Just to know the triangle cannot be equaliteral because GIH>90 for every triangle. I found it after I found this problem on another forum and you had solved it.


Dear Mathlinkers,
Let ABC be a triangle, Na the Nagel's point of ABC, (1) the incircle of ABC, I the center of (1), DEF the contact triangle of ABC, H' the orthocenter of DEF, and U, V, W the antipoles of D, E, F wrt (1).
Prouve : the isogonal of H' wrt the triangle UVW is the Nagel's point Na of ABC.
Sincerely
Jean-Louis

[b]Lemma:[/b] "$ AU, BV, CL$ concur at Nagel point of triangle $ ABC$". It's well-known so I leave proof for the reader  :oops:
Applying this lemma above, we will prove that $ AU$ and $ UH'$ are the isogonal lines wrt $ \angle VUW. (*)$
$ \Leftrightarrow \angle AUW\plus{}\angle VUH'\equal{}180^o$
But $ \angle AUW\equal{}\angle AUE\plus{}\angle EUW\equal{}\angle AUE\plus{}\angle IFE$
And $ \angle VUH'\equal{}\angle FUH'\minus{}\angle FUV\equal{}\angle FUH'\minus{}\angle IEF$
Thus we need to show $ \angle AUE\plus{}\angle FUH'\equal{}180^o$
On the other side, $ U$ is the antipole of $ D$ wrt $ (I)$ then $ U$ and $ H'$ are symmetric through the midpoint of $ EF$.
Denote $ L$ the second intersection of $ AU$ and $ (I)$.
The well-known result about harmonic quadrilateral: $ UH'$ and $ UL$ are isogonal lines wrt $ \angle FUE$
Which follow that $ \angle AUE\plus{}\angle FUH'\equal{}\angle AUE\plus{}\angle EUL\equal{}180^o$
Therefore $ (*)$ is true. Similarly for $ BV, CL$ we completed the proof.


limit
integration
trigonometry
calculus
real analysis
function
absolute value

Suppose $f$ is bounded on $[0,\infty)$ and

$\lim_{a\to\infty}\int_0^af(x)dx=L$ exists.

Prove that $\lim_{t\to0^+}\int_0^{\infty}e^{-tx}f(x)dx=L.$

Application: once you have proved this, use it in a computation of

$\int_0^{\infty}\frac{\sin x}xdx.$

nice one.
also known as laplace transform

Tr: saying that this is related to the Laplace transform doesn't solve it. There's work to do.

If $\int_0^{\infty}|f(x)|dx$ is finite, the question isn't very interesting. For the theorem to gain you anything, you might as well assume the indefinite integral is conditionally convergent (as in the application.)

Also, the hypothesis "$f$ is bounded" is much stronger than we really need; it would be sufficient for $f$ to be locally (Lebesgue) integrable. That is, $\int_a^b|f(x)|dx$ is finite for each finite $a$ and $b$ - that's enough.

$F(u)=-\int_{u}^{+\infty}f(t)dt$  tend to $0$ when $u$  tends to $+\infty$

$\epsilon>0, \exists A>0, \forall u\geq A, |F(u)|\leq \epsilon$
For $x\geq 0$ and $A\leq \alpha < \beta$
Integration by parts gives
$\int_{\alpha}^{\beta}e^{-xt}f(t)dt=[e^{-tx}F(t)]_{\alpha}^{\beta}+x\int_{\alpha}^{\beta}e^{-xt}F(t)dt$
This implies
(1)  $|\int_{\alpha}^{\beta}e^{-xt}f(t)dt|\leq 3\epsilon$

(2)  ${| \int_{0}^{+\infty}(e^{-tx}-1)f(t)dt|\leq |\int_{0}^A}(e^{-tx}-1)f(t)dt|+|\int_{A}^{+\infty}e^{-xt}f(t)dt|+|\int_{A}^{+\infty}f(t)dt|$

the first integral is bounded by $||f||_{\infty}.A.(1-e^{-Ax})$ tends to $0$ when
x tends to $0$
the second integral is less than $3\epsilon$
the third one is less than $\epsilon$

The LHS of (2) tends to $0$ when x tends to $0$

Mathlinks great place to be for practice the Art of Problem Solving
thank you Valentin Vornicu for this forum.

[quote="Moubinool"]the first integral is bounded by $ ||f||_{\infty}.A.(1-e^{-Ax}) $ tends to $ 0 $ when x tends to $ 0 $ [/quote]
I made the comment that we don't really need $f$ to be bounded, just locally integrable, so I'd like not to use $||f||_{\infty}$ if possible. Since $A$ has already been chosen above, this integral is bounded by $(1-e^{-Ax})\int_0^A|f(t)|dt,$ and that does go to $0$ as $x\to0.$

Thank you Moubinool, that works nicely. We don't need $f$ to be bounded; this works just fine for, say, $f(x)=x\cos(x^3).$

(Note that Moubinool reversed the usage of the letters $t$ and $x$.)

Now, what about $\int_0^{\infty}\frac{\sin x}xdx$?

Suggestion: can you differentiate $G(t)=\int_0^{\infty}e^{-tx}\frac{\sin x}xdx$?

[quote="Kent Merryfield"]
Suggestion: can you differentiate $G(t)=\int_0^{\infty}e^{-tx}\frac{\sin x}xdx$?[/quote]

This one is well known for Oral Examination X-ENS in France
I let it for harazi, grobber, alekk good training

Hints: Show that G is derivable for $t \geq 0$, find the expression of $G'$ without
integral symbol, take a primitive and use the continuity of $G$ at $0$, also limit
of G when $t$ tends to $+\infty$

I think that $G'(t)=-\frac{1}{t^2+1}$. Anyway, I won't post the proof I found, since it is too long and ugly. There must be a nice solution for this monster.

Yes, for $t>0, G'(t)=-\frac1{t^2+1}.$ As for the justification of that:

$\frac{G(t+h)-G(t)}h=\int_0^{\infty}\frac{e^{-(t+h)x}-e^{-tx}}h\cdot\frac{\sin x}xdx$

We now want to take the limit as $h\to0$ and bring that limit inside the integral sign. By the Mean Value Theorem, $\frac{e^{-(t+h)x}-e^{-tx}}h=e^{-sx}$ for some $s$ between $t$ and $t+h$. If we restrict $h$ to $|h|<\frac t2$ then this is bounded in absolute value by $e^{-tx/2}.$ Using also the estimate $\left|\frac{\sin x}x\right|\le 1$ we get that the integrand for the difference quotient is dominated by the integrable function $e^{-tx/2}.$ Then the interchange of limit and integral is justified by the Dominated Convergence Theorem. Once you've done that, the actual computation of $G'(t)$ is a routine calculus exercise.

One thing to note: because we're going to use the "continuous version of Abel's Theorem" proved above, we only need to do these calculations for $t>0$, not worrying about what happens when $t=0.$


The equation $ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$ has three real roots.Prove that:
                   $ 2(ab\plus{}c)\le\ a^3\plus{}4\sqrt{2(a^2\minus{}b)}$


Hello,

Thank you, that solution works. But how did you come up with it in the first place?
Also, how do I show that the whole expression indeed evaluates to $LHS-RHS$ without expanding everything and brute-force algebra? Is there a trick that I'm missing?


Cells of certain organisms have organelles that produce ATPs and use in the synthesis of organic substances from carbon dioxide. These organelles are:

a) the lysosomes
b) the mitochondria
c) the chloroplasts
d) the Golgi system
e) the nucleoli

Here's one way to go about the problem: which of the organelles definitely DON'T do what you're asking?  Lysosomes are similar to vacuoles, which aid in the digestion of macromolecules.  The Golgi apparatus aid in the processing and packaging of macromolecules.  The nucleoli works with the transcription of ribosomal DNA.

That leaves you down to the mitochondria and the chloroplasts, both of which deal with ATP.  Your goal, is to figure out which one fits.  The key is that the question description gives you all the information you need-one of those two organelles should scream out at you, as its function is carry out the process which converts carbon dioxide into organic compounds.

I don't know  :blush:

I only copy and cut this question from my book. Do you know the answer ?

 :D

Yes, I know the answer.  Just look at the last line of my first post.  If you know what does that you've got the answer (and I already narrowed it down to mitochondria and chloroplast for you, so you don't have much work to do).

Well ... then ... the answer is C ? chloroplasts ?

Thanks. I think this is the answer. It's correct ?

yeah correct!!!

though both mit and chl prod energy,org matter in synth in chl only (i.e. frm CO2)

[quote="eli140"]This is kinda related but i forgot what the fold inside the mitochondria was called, can anyone remind me?[/quote]

Cristae (sing. crista). Just look at Wiki or one of the diagrams in your bio book (Campbell or whatnot).


The digits from 1 to 6 are arranged to form a six-digit multiple of 5. What is the probability that the number is greater than 500,000? Express your answer as a common fraction.

Because it is a multiple of 5, the last digit must be 5. There are 5 other digits besides 5. The number is only greater than 500000 if the first digit is 6. Because there are 5 possible digits, and only 1 results in the number being greater than 500000, the answer is $ \frac {1}{5}$.

hmm...
wait
if the first digit has to be a $ 6$, and the last digit has to be a $ 5$, then the remaining four digits can be arranged in $ 24$ ways
there are $ 6!$ ways to arrange every digit, so that's $ 720$
so, the answer is $ \frac {24} {720} \equal{} \frac {1} {30}$

No... because we are already GIVEN that the last digit is 5. It's conditional probability.

Don't divide by 6 because it already says its divisible by 5

In conditional probability, the probability of an event A given an event B is $ \frac {P(A \cap B)}{P(B)}$. We are looking for the probability that the number is greater than 500000 GIVEN that it is a multiple of 5. This is obviously different from the probability that the number is greater than 500000 AND is a multiple of 5, which is what you found, vallon22.


algorithm
number theory
Euclidean algorithm

find the minimum integer value of $ n$ for which $ 3n \minus{} 233$ and $ 8n \plus{} 201$ will have a common positive divisor, other than $ 1$.

By the Euclidean Algorithm,
$ \gcd(3n\minus{}233,8n\plus{}201)$
$ \equal{}\gcd(3n\minus{}233,2n\plus{}667)$
$ \equal{}\gcd(n\minus{}900,2n\plus{}667)$
$ \equal{}\gcd(n\minus{}900,2467)$
$ 2467$ is prime, so the smallest possible value of $ n$ is $ 900\plus{}2467\equal{}\boxed{3367}$.

[b]@ OP:[/b] Remember to write "positive" integer next time :)


probability
group theory
expected value
combinatorics unsolved
combinatorics

In a simple graph on $ n$ points, the points have degrees $ 0<d_1\leq d_2\leq ...\leq d_n$. Prove that it is possible to select at least $ \sum_{i\equal{}1}^n\frac{2}{d_i\plus{}1}$ points, such that the subgraph formed by these points contains no cycle.

I had no clue how to proceed until I accidentally came across the the Sixth Proof of Turan's Theorem in
[url=http://mathdl.maa.org/images/upload_library/22/Ford/Aigner808-816.pdf]Martin Aigner, [i]Tur\'{a}n's Graph Theorem[/i], American Mathematical Monthly 102 (1995), pp. 808-816[/url].

Slightly restated (replace the graph by its complementary graph), this proof shows the following strengthening of Turan's theorem:

[color=blue][b]Theorem 1.[/b] Let $ n$ be a positive integer, and let $ G$ be a simple graph. Assume that $ V\left( G\right) = \left\{ 1,2,...,n\right\}$. Then, there exists a subset $ I$ of $ V\left( G\right)$ such that $ \left\vert I\right\vert \geq\sum\limits_{i = 1}^{n}\frac {1}{\deg i + 1}$ and such that no two vertices in $ I$ are connected.[/color]

Here (and in the following), we use the following [i]notations[/i]:

For any graph $ G$, we denote by $ V\left( G\right)$ the set of the vertices of $ G$. Two vertices $ a$ and $ b$ of $ G$ are said to be [i]connected[/i] if and only if the graph $ G$ has an edge joining $ a$ and $ b$. For every $ v\in V\left( G\right)$, we denote by $ N\left( v\right)$ the set of neighbours of $ v$ in $ G$, and by $ \deg v$ the degree of $ v$ (so that $ \deg v = \left\vert N\left( v\right) \right\vert$).

A cycle of a graph is always supposed to be simple and contain more than two vertices.

Now, knowing the proof of Theorem 1, it is easy to extend it so that it applies to the original problem:

[color=blue][b]Theorem 2.[/b] Let $ n$ be a positive integer, and let $ G$ be a simple graph with no isolated vertex. Assume that $ V\left( G\right) = \left\{ 1,2,...,n\right\}$. Prove that there exists a subset $ I$ of $ G$ such that $ \left\vert I\right\vert \geq\sum\limits_{i = 1}^{n}\frac {2}{\deg i + 1}$ and such that every cycle of the graph $ G$ contains at least one vertex from $ V\left( G\right) \setminus I$.[/color]

[i]Proof of Theorem 2 (sketched).[/i] For any permutation $ \pi\in S_{n}$ (where $S_n$ denotes the $n$-th symmetric group), define a subset $ J\left( \pi\right)$ of $ V\left( G\right)$ by
$ J\left( \pi\right) = \left\{\pi\left( k\right) \mid k\in V\left( G\right) \text{ and }\left\vert N\left( \pi\left( k\right) \right) \cap\left\{\pi\left( 1\right) ,\pi\left( 2\right) ,...,\pi\left(k - 1\right) \right\} \right\vert \leq1\right\}$.

(In other words, this subset $ J\left( \pi\right)$ is the subset we obtain if we start with an empty set and traverse the vertices of $ G$ in the order $ \left( \pi\left( 1\right) ,\pi\left( 2\right) ,...,\pi\left( n\right) \right)$, putting a vertex into the set unless the set already contains two or more neighbours of this vertex at the time it is being traversed. But I guess the formula above is easier to understand than this, uhm, explanation.)

Now, for every $\pi\in S_n$, let us show that

[color=green][b](1)[/b][/color] every cycle of the graph $ G$ contains at least one vertex from $ V\left( G\right) \setminus J\left( \pi\right)$.

[i]Proof of [color=green][b](1)[/b][/color].[/i] Let $c$ be a cycle of the graph $G$. Take the greatest $ k\in V\left( G\right)$ such that $ \pi\left( k\right)$ is a vertex of the cycle $c$. Then both neighbours of $\pi\left(k\right)$ in the cycle $c$ lie in the set $ N\left( \pi\left( k\right) \right) \cap\left\{\pi\left( 1\right) ,\pi\left( 2\right) ,...,\pi\left( k - 1\right) \right\}$. Therefore, $ \left\vert N\left( \pi\left( k\right) \right) \cap\left\{\pi\left( 1\right) ,\pi\left( 2\right) ,...,\pi\left( k - 1\right) \right\} \right\vert \leq 1$ cannot hold. Thus, $\pi\left(k\right)\neq J\left(\pi\right)$ (by the definition of $J\left(\pi\right)$), so that $\pi\left(k\right)$ lies in  $ V\left( G\right) \setminus J\left( \pi\right)$. Since $\pi\left(k\right)$ is a vertex of the cycle $c$, it follows that $c$ contains at least one vertex from $ V\left( G\right) \setminus J\left( \pi\right)$. Since we have proven this for every cycle $c$, we have thus proven [color=green][b](1)[/b][/color].

We thus know that [color=green][b](1)[/b][/color] holds for every $\pi\in S_n$. Hence, in order to prove Theorem 2, we can take $ I = J\left( \pi\right)$, provided we find some $ \pi\in S_{n}$ satisfying $ \left\vert J\left( \pi\right) \right\vert \geq\sum\limits_{i = 1}^{n}\frac {2}{\deg i + 1}$.

Now, fix some $ i\in\left\{ 1,2,...,n\right\}$, and randomly choose a permutation $ \pi\in S_{n}$ (in a way that all permutations are equally likely, i. e., by rolling a fair $ n!$-sided dice). Then, the probability $ \mathcal{P}\left( i\in J\left( \pi\right) \right)$ is easily seen to be $ \frac {2}{\deg i + 1}$ (in fact, $ \mathcal{P}\left( i\in J\left( \pi\right) \right)$ is the probability that $ \pi^{ - 1}\left( i\right)$ is one of the two smallest elements of the set $ \left\{\pi^{ - 1}\left( j\right) \mid j\in\left\{ i\right\} \cup N\left( i\right) \right\}$; now, this set has $ \deg i + 1$ elements, and all orderings of its elements are equally likely, so the probability that a given element is one of its two smallest elements is $ \frac {2}{\deg i + 1}$). Thus, the expected value of $ \left\vert J\left( \pi\right) \right\vert$ is $ \sum\limits_{i = 1}^{n}\mathcal{P}\left( i\in J\left( \pi\right) \right) = \sum \limits_{i = 1}^{n}\frac {2}{\deg i + 1}$. Consequently, there exists some $ \pi\in S_{n}$ such that $ \left\vert J\left( \pi\right) \right\vert \geq \sum\limits_{i = 1}^{n}\frac {2}{\deg i + 1}$. This ends our proof of Theorem 2.

Note that Theorem 1 also appeared in
Noga Alon, Joel H. Spencer, [i]The Probabilistic Method[/i], 2nd edition,
The Probabilistic Lens: Turán's Theorem.
I have no idea whether Theorem 2 also appears in this book.

[b]EDIT:[/b] mathmanman just pointed out that this was also discussed at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4300 .

  Darij


number theory
prime numbers
number theory proposed

Find all prime numbers ( not necessary to discriminate them from each other) such that product of them equal to 10 multiply their sum.

Let $ Pr_n\equal{}\prod_{i\equal{}1}^np_i\equal{}10\sum_{i\equal{}1}^np_i\equal{}10S_n,p_1\le p_2<...\le p_n$. Obviosly $ 2\le n<6$, because $ \frac{2^6}{6}>10$.
$ p_n\equal{}\frac{10S_{n\minus{}1}}{Pr_{n\minus{}1}\minus{}10}$. If $ n\equal{}2$ we get $ p_2\equal{}\frac{10p_1}{p_1\minus{}10}\to p_1>10\to p_2$ is not prime.
If $ p_1>2$, then $ 10S_{n\minus{}1}$ is even and $ Pr_{n\minus{}1}\minus{}10$ odd, therefore $ 2|p_3, p_n>2$ give $ p_n$ is not prime.
Consider $ n\equal{}3$. If $ p_1\equal{}2$, then $ p_3\equal{}\frac{10p_2\plus{}20}{2p_2\minus{}10}\equal{}\frac{5(p_2\plus{}2)}{p_2\minus{}5}$ give $ p_3$ is not prime.
Therefore $ 4\le n\le 5$. Let $ Pr\equal{}p_2*...p_{n\minus{}1},S\equal{}p_2\plus{}...\plus{}p_{n\minus{}1}$, then $ p_n\equal{}\frac{5(S\plus{}2)}{Pr\minus{}5}$.
If $ 5\not |Pr$, then must be $ p_n\equal{}5, Pr\equal{}S\plus{}7$ and $ p_i,i\equal{}2,..n\minus{}1$ equal 2 or 3. It is easy to chek, that it had nod solution.
Therefore $ 5|Pr$ or one of primes $ p_2,..,p_{n\minus{}1}\equal{}5$. Let $ a\equal{}\frac{Pr}{5}\equal{}\frac{p_2...p_{n\minus{}1}}{5}, b\equal{}S\minus{}5\equal{}p_2\plus{}..p_{n\minus{}1}\minus{}5$, then $ p_n\equal{}\frac{b\plus{}7}{a\minus{}1}$.
If $ n\equal{}4$, then must be $ a\equal{}b\equal{}p$ and $ p\equal{}3$ or $ p\equal{}5$.
It give solution $ (2,3,5,5)$.
If $ n\equal{}5$ $ p_n\equal{}\frac{p\plus{}q\plus{}7}{pq\minus{}1}\ge 5$ have not solution.

in the question Does the product of primes means just a.b(i.e.product of two primes) OR a.b.c.d.....(product of any number of primes)

sorry if my english is not good :(

I think $ 2\le n<6$ is not obviosly  :roll: The reason is $ 2^6\ge 10\times 6$. But I think it is not logic  :blush:
\[ 10(a_1\plus{}a_2\plus{}...\plus{}a_n)\equal{}a_1a_2...a_n,\forall a_n\ge ...\ge a_2\ge a_1\ge 2\]
With $ a_1,a_2,...,a_n$ are prime numbers. I think it is not easy to give $ n<6$
I misunderstood something in your post. For expamle: why $ p_1>10 \to p_2$ is not prime  :maybe: Can you explain clearly? Thank you so much  :wink:

Denote the numbers by $ p_{1},....,p_{n}$. We have $ \prod p_{i} \equal{} 10 \sum p_{i}$, hence, WLOG, $ p_{1}\equal{}2, p_{2}\equal{}5$. Now we have:
$ \prod_{i\equal{}3}^{n} p_{i} \equal{} \sum_{i\equal{}3} p_{i} \plus{}7$
Obviously (I sound like one of the most active members on this forum...  :P ) n>3.
If $ n\equal{}4: xy\equal{}x\plus{}y\plus{}7 \implies (x\minus{}1)(y\minus{}1)\equal{}8 \implies x\equal{}2^{k}\plus{}1, y\equal{}2^{3\minus{}k}\plus{}1$. The only possible solutions are $ p_{3}\equal{}3, p_{4}\equal{}5$ (and of course vice versa).
If n>4:
None of the primes is bigger than 3. If WLOG $ p_{3} \ge 5$, we have: $ (p_{3}\minus{}1)(p_{4}\minus{}1) \ge 4$, or: $ p_{3}p_{4} \ge p_{3}\plus{}p_{4}\plus{}3$, and than: $ p_{3}p_{4}p_{5} \ge (p_{3}\plus{}p_{4}\plus{}3)(p_{5}) \ge p_{3}\plus{}p_{4}\plus{}3\plus{}p_{5}\plus{}8$ (because: $ (x\minus{}1)(y\minus{}1) \ge 9, x\equal{}p_{3}\plus{}p_{4}\plus{}3 \ge 10, y\equal{}p_{5} \ge 2$), a contradiction (If we use $ p_{i}p_{j} \ge p_{i}p_{j}$ for the rest of the primes).
Hence, the only primes are 2 and 3. We need to solve $ 2^{k}3^{t}\equal{}2k\plus{}3t\plus{}7$. If k=0 or t=0, we have an easy contradiction.
I will show that t=0:
If $ p_{3} \ge 3$, in the same way as above: $ p_{3}p_{4}p_{5}\ge p_{3}\plus{}p_{4}\plus{}p_{5}\plus{}5$
If n=5: we need to solve $ 3yz\equal{}3\plus{}y\plus{}z\plus{}7, z,y\equal{}2,3$. It's simple to check that there are no solutions.
If n>5: $ p_{3}p_{4}p_{5}p_{6}\ge (p_{3}\plus{}p_{4}\plus{}p_{5}\plus{}5)(p_{6}) \ge p_{3}\plus{}p_{4}\plus{}p_{5}\plus{}5\plus{}p_{6}\plus{}10$, a contradiction. Hence, t=0, another contradiction.

Hence, the only solution is {2,3,5,5}


search geometry calculus derivative videos

We are having this debate in our (suburban) public elementary school. Should all children in the class be taught at the same level or should they be divided into, say, three levels and taught different material? I run a math club and it is quite apparent that the top 1/3 is just coasting with the current material. The counterpoint is that having all kids do the same material really pulls the kids not doing well by providing them with role-models.

 Are there any authoritative study of this? What does your school do?

[quote="rsinha"] The counterpoint is that having all kids do the same material really pulls the kids not doing well by providing them with role-models.
[/quote]
Just one small remark: I doubt that somebody who does virtually no work in class but still gets excellent grades can really serve as a good "role model". Other kids may envy him but what is there to learn from looking at such a student? He would serve as a good role model if he had to struggle with the material and still came a winner but this would require the classroom material be ajusted according to the abilities of the top students in the class, which is something very few teachers would dare to do and which may be not the best idea for several other reasons. So, I am inclined to disagree with this particular objection. Of course there may be other arguments against separation and there are many alternative techniques of giving different workloads to different students.

[quote="rsinha"]We are having this debate in our (suburban) public elementary school. Should all children in the class be taught at the same level or should they be divided into, say, three levels and taught different material? [/quote]

Well, I can see where you are going. Our school are divided into different classes in subjects. For example, for math, we usually have GT in one class and Non-GT in the other. Then my teacher sits with the struggling GT people (usually the same people) and the higher, more, advanced GT people are doing their own work.

A search for "ability grouping" will turn up a bunch of links like [url=http://www.ericdigests.org/pre-927/grouping.htm]this one[/url]. But why bother reading them when you can just say "Research shows..." ?  :)

I would advise* against ability grouping. Chances are that it will correlate with race/ethnicity, which will sooner or later make a story in your local newspaper.  :| More math-intensive extracurricular activities could be a better way to go.

(*) having no knowledge or experience in the field :ninja:

Yeah in my old elementary days everyone learned practically the same thing. Even in the GT program, you just learn the same thing everyday...it was kinda boring  :D

Give students the option of choosing a normal class (easy and good for their moral but might be boring), challenging GT class (might drop morale). Make sure their parents know about it. There's nothing worse then a young student discouraged about math.

[quote="randomdragoon"]When I was in elementary school, once a week, a bunch of us would go to a separate room with a math specialist to talk about some more advanced material for about an hour or something. Then we'd get independent study work for the rest of the week so we wouldn't be slacking off while the rest of the class learned material that we already knew. I thought the system worked pretty well.[/quote]

I don't like the once a week thing. That means if you have a question you would have to ask it in 6 days, which could be unhelpful (other than posting on AoPS that is). The people who take geometry at my middle school have the once a week thing and they just hang out in a math class of Algebra II and do nothing, which is kinda wasting time. Preferablly, Id either go to a class specifically about geometry at a high school or something rather than learn with that much space between lessons and lectures.

[quote="davidyko"]Hmm, in middle school, I, along with a few other people, had an "independent study" from the second semester of 7th grade through the end of 8th grade. We never went to class at all, we worked from AoPS Vol. I, with a teacher there to help us if we needed it. I don't know if this is practical or not at your school, but that's what we did to remedy the problem.[/quote]

Aww lucky! I wanted to take some AoPS courses... :D

i think that dividing the class into more and less advanced sections should be one of the top priorities of the school
its completely stupid to discourage the top students from learning, even if it means taking away attention from some of the people who need help
this is one of the main reasons america is now falling way behind asia in math and science :(

personally, i get to take an AoPS class and work on the material during class
i hadn't been able to do so before, and it has made me a happier, and smarter person.  a few years ago, i was acting up in class cuz i had nothing to do :P
[quote="mlok"]
A search for "ability grouping" will turn up a bunch of links like this one. But why bother reading them when you can just say "Research shows..." ? Smile

I would advise* against ability grouping. Chances are that it will correlate with race/ethnicity, which will sooner or later make a story in your local newspaper. Neutral More math-intensive extracurricular activities could be a better way to go.

(*) having no knowledge or experience in the field [/quote]
i have to say that i reeeealy disagree with you
if they write a story about it in the newspaper, just say, "well i guess these people work harder cuz they got into the better class we didn't have any bias"
and not doing so will just discourage the top students
if i had not lived in an area with lots of good math resources, i probably would have became dissillusioned with math, thinking it was too easy and  boring.  the only reason that i like math is that i was introduced to AoPS and harder math

[quote="daermon"]i think that dividing the class into more and less advanced sections should be one of the top priorities of the school
its completely stupid to discourage the top students from learning, even if it means taking away attention from some of the people who need help
this is one of the main reasons america is now falling way behind asia in math and science :( [/quote]

I agree but I believe the reason why elementary schools do not seperate students is the lack of available resources. No extra rooms, no extra textbooks, no extra teachers, etc. cause advanced students to be holed up with for lack of better term, lesser students. Also, there can be a minimal amount of students say 3, that would require advanced classes but the school would not provide the extra work and money into providing the students their needs. Thats why in middle school, where there is a higher chance of advanced students, that there is the extra funding for advanced classes. Of course in high school then the classes seperate in to 4 different skill levels (academic, honors, GT, and AP)

Everyone at my school doesn't even know what Art of Problem Solving is so there is next to no chance of us getting taught from their books...

[quote="Danbert"]This is interesting for me largely because I grew up with an rather different system.  At the elementary school I went to in 1st - 4th grade (moved for 5th & 6th), I was in an 'accelerated' / 'gifted' program within a normal elementary school.  In 1st and 2nd grade, all the students worked on the same material at the same time, but once we got to 3rd grade we could go at our own pace.  This meant we read the book on our own, did the homework our teacher assigned us, asked for help if we didn't understand something, and took the chapter tests when we were ready.  If you got a poor grade (C or below) you had to redo the chapter.  This had the end result of letting me finish "7th grade math" in 4th grade (of course, I had to repeat it when I moved because no one believed me...).  In my opinion, this is an excellent system - it allows students to work at their own level of ability, provides 'role models' (i.e. the students farther ahead), allows talented people to work ahead and actually enjoy themselves in math class.   The 'redo-on-failure' part also ensures that people actually understand the material once they're done, unlike a traditional classroom setting where poor-performing students are just dragged along with the rest of the class because the teacher can't spare the class time to go back over material.
Naturally, I'm a bit biased in favor of this system...  :D[/quote]

I wished more schools offered these kind of policies. I guess an alternative to enhancing a youngster's education in elementary school is to enroll him/her into extracurricular programs, which I've seen in a few places. And of course, a number of problems arise, which include:
1) Risk of overworking the student. Of course this would be a lot less if the student wasn't forced (like in most places in Asia. So there is a trade-off involved)
2) Since the material taught in elementary school is not very broad, there will bound to be overlapping material --> waste of time for the student in his math class
3) Sort of an extension to 2, many schools (especially elementary, at least it applies for the one I went to) are simply too rigid to let you skip grades in a specific course, let alone let you take an apititude test. And depending on the environment and the concerned student, being surrounded by a class of upper years (aka "bigger kids") may be intimidating, especially if intellect is not regarded highly in the particular student body.

[quote="rsinha"]I have a clear bias in this debate :) I think the kids should be given different material. However the administrators in charge of making this decision are convinced that this will adversely affect the overall class atmosphere and they are backed by a professor of math education whom they consulted for curriculum design. They see it as a policy decision of individual good vs. societal good.

 They do start separating the kids in higher grade but I work with these kids in my math club and I see first hand how much more they can learn and how excited they are with more challenging material. Given all the debate about US kids falling behind their peers, I find it crazy to hold these kids back.  I can pass myself as a mathematician but this doesn't carry the same weight as a professor of math [i]education[/i]. So I am hoping for an authoritative study that I can cite in my argument.

Sorry if I sound frustrated.[/quote]
personally, I find the situation infuriating.  The amount of potential wasted through poor, inadequate, counter-productive, and generally faulty teaching and teaching methods is tragic.  I've become very interested in education lately.

I personally attended a very small Montessori school from age 4-10, and despite some issues with the teachers and some obvious improvements which could have been made, that time was one of the best educational experiences of my life.  I know personally that, perhaps as a result of the Montessori program- although if it is, I would call it a benefitial result- I am only motivated to learn by my own curiosity and desire for knowledge.  This has made my more conventional schooling extremely frustrating at times, since the authoritarian structure of middle school and high school demand that one does an excessive amount of things only because they are required.... and why is there this host of requirements?  the answer seems to differ depending on the situation, but the most common answer I have been given is that students need to learn how to do stuff they don't want to do... how to do what they're told, and how to submit to authority.   :mad:  :mad:  :mad:  :mad:  :mad:  :mad:

Oh it makes me mad.

Kids don't like school.  This seems to have been accepted as fact.  But I really think it is more symptomatic of the failings of our system.  I've suffered a baffling amount of worthless classes and work, sadistic or idiotic authority figures, and other bad things in my education, and the sad thing is, I consider myself extremely fortunate despite all this when I compare my education to others'.

anyhow, as evidenced by my post, I'm thinking about eduction... I certainly don't have it worked out yet.  But I think a key component of an improved educational system is  differentiation between levels of abilities and PLACEMENT STUDENTS AMONG ACADEMIC PEERS, something it seems our society is loath to do for several reasons (a desire for normalcy being top-most).

[quote="plucesiar"]
3) Sort of an extension to 2, many schools (especially elementary, at least it applies for the one I went to) are simply too rigid to let you skip grades in a specific course, let alone let you take an apititude test. And depending on the environment and the concerned student, being surrounded by a class of upper years (aka "bigger kids") may be intimidating, especially if intellect is not regarded highly in the particular student body.[/quote]

skipping grades doesn't do much to alleviate the underlying issue.  The speed and depth of a course is much more important than the level of the material, IMO.

Our middle school deals with this in a fairly good manner.

Out teams are divided based on math level (Pre-Algebra all the way to Pre-Calculus).

Below Pre-Algebra kids are placed on the "Jaguar Team" (lower than normal).
Pre-Algebra kids to Algebra kids are placed on the "Panther Team" (normal).
Beyond Algebra kids are placed on the "Tiger Team" (advance).

This is for 8th grade.

I think the school absolutely must allow a student to learn what they're ready to learn, especially at the elementary level.  I was extremely lucky to have an extremely cooperative elementary school that pretty much satisfied whatever needs I had.  I really can't say what might have happened to my enthusiasm for math if that hadn't been the case.  Certainly, there are students who would have been turned off to math at that point.  And I really don't buy the tired old argument about enhancing the atmosphere of the classroom.  It's irresponsible to hold a student back from their full potential just because the other students haven't reached that level.  And on top of that, honestly, what good does it do to add in the kids who are bored and already know what they're doing to the class?  Maybe in high school, they can help the other kids master it, but (a) most of them won't want to, and (b) elementary school kids are not math tutors.  I was completely incapable of helping other students understand math the way I did back then.  Instead of restricting the more talented students, invest in good teachers who can inspire enthusiasm in students.

Now the disclaimer: I really don't have the authority to speak on this, as I'm not any kind of educator or anything.  This is just my personal opinion, which is not based on any solid facts except my own (limited) experience.

[quote="plucesiar"]
And depending on the environment and the concerned student, being surrounded by a class of upper years (aka "bigger kids") may be intimidating, especially if intellect is not regarded highly in the particular student body.[/quote]

When I was in 4th grade, I skipped up to an 8th grade algebra class; since my elementary school was adjacent to a junior high, I could take it in a normal classroom, mixed in with the most advanced 8th grade kids.  I didn't find it at all intimidating, actually.  All the kids were really nice to me, and I think they thought it was pretty cool to have a 4th grader in their class.

Coincidentally, that class was where I was first introduced to the concept of a "math team".

At my elementary school there wasn't a GT program, but in my opinion, if a students clearly isn't challenged at all, there should be some effort so they can stay interested. For example, when I was in fourth grade, my teacher had me go to a 5th grade room for math and I definitely learned a bunch that year (more than before at least). Next year, I (read: my parents) became more involved and I studied more at home.

In middle school and higher, there should definitely be classes offered that are ahead of grade level since there are so many more students, which means that more are going to be ahead of their grade level (as in enough to fill a classroom at least).

Oh, and while we're on the topic of elementary school math, has anyone heard of the Every Day Mathematics books? Well there are a few videos on youtube about them, this one has been particularly active:
[url=http://youtube.com/watch?v=Tr1qee-bTZI]Youtube video[/url]

I think it's nice to have different levels, as that would encourage students to try harder, but the schools would be wary of trying to have different levels because that would take more time. Not all teachers would want to do that, though. My brother is in elementary school, and he is coasting through math very easily. If there was a 'advanced' level, I'm pretty sure it would encourage him to do more than he is right now and think more than he is. (I do know that several of the boys, including him, try to finish their math work in the fastest amount of time.)

In my old elementary school, we had an enrichment program. Once or twice each week, we'd be taken out of class to discuss something that extended the curriculm.

I think my school did the Every Day Mathematics books. It's hard to remember.


given n is positive integer ,
 consider sequence {x_n } :0<x_1<x_2<...<x_n
Find min P

P is clearly bigger than 1/n. Also, if we take x_1=a_1, x_2=a_1+a_2,...,x_n=a_1+...+a_n such that a_1,...,a_n-1 tend to zero and a_n tends to infinity, the expression tends to 1/n. So, if I understood corectly the problem, then the answer is 1/n and is an infimum rather than a minimum.

You're right!
Problem doesn't have min
and 1/n is infimum  :D


Solve for $ x$. Express your answer as a common fraction.
\[ \displaystyle \frac{1}{3} \plus{} \left(\frac{2}{3}\right)^2 \equal{} \left(\frac{1}{3}\right)^2 \plus{} x\]

$ \frac {1}{3} \plus{} (\frac {2}{3})^2 \equal{} (\frac {1}{3})^2 \plus{} x$

$ \frac {3}{9} \plus{} \frac {4}{9} \equal{} \frac {1}{9} \plus{} x$

$ \frac {6}{9} \equal{} x$

$ x \equal{} \frac {2}{3}$

that's interesting
does $ a \plus{} b^2$ always equal $ a^2 \plus{} b$?

Obviously no, take 2 and 1 as examples. Probably just a coincidence....


let $ C$ an inscriptible circle of $ ABC$.
$ E \in$ [$ AB$] and $ D \in$ [$ AC$] such that $ (DE)$ is parallel with $ (BC)$  and $ (DE)$ is tangent in $ (C)$.
if $ p$ is the perimetre of $ ABC$, what is it the maximal value of $ DE$.


limit
probability
linear algebra
matrix
vector
probability and stats

Fix an integer $ k\geq 2$.  Let $ X_1,X_2,\ldots$ be i.i.d with common distribution $ X$, where $ P(X \equal{} i) \equal{} 1/k$ for $ 1\leq i\leq k$.

For $ n\in\mathbb{N}$, set $ \tau_{n,k} \equal{} \inf_t\{S_t\in\{n,\ldots,n \plus{} k \minus{} 1\}\}$

Prove for $ 1\leq j\leq k$,
\[ \lim_{n\to\infty} P(S_{\tau_{n,k}} \equal{} n \plus{} j \minus{} 1) \equal{} \frac {2(k \minus{} j \plus{} 1)}{k(k \plus{} 1)} \ \left( \equal{} \frac {k \minus{} j \plus{} 1}{1 \plus{} 2 \plus{} \ldots \plus{} k}\right)
\]
This is inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241891]this[/url] problem.

here is an idea, though I have not checked all the computations:
Fix $ k$ and call $ p_{n,i}(k)$ the probability that the random walk finishes at $ n\plus{}i$, knowing that the state $ k$ has been visited. Then one knows that for all $ 0 \leq i \leq k\minus{}1$:
\[ p_{n,i}(n\minus{}1) \equal{} \frac{1}{k} \]
and if defines for $ 0 \leq j \leq k\minus{}1$:
\[ p_{n,i}(n\plus{}j)\equal{}0 \quad \textrm{if} \quad j \neq i\]
\[ p_{n,i}(n\plus{}i)\equal{}1 \]
then it is clear that:
\[ p_{n,i}(m) \equal{} \frac{1}{k} \sum_{j\equal{}1}^k p_{n,i}(m\plus{}j) \]
What we are looking for is $ P(i)\equal{}\lim_{n \to \infty} p_{n,i}(0)$.

One can see that if one defines $ p_{n,i} \equal{} p_{n,i}(0)$ then
\[ p_{n,i} \equal{} \frac{1}{k} \sum_{j\equal{}1}^k p_{n\plus{}j,i} \]
Hence, if $ V_{n}\equal{}(p_{n,i}, p_{n\minus{}1,i}, \ldots, p_{n\minus{}k\plus{}1,i})$ then $ V_{n} \equal{} AV_{n\minus{}1}$ where
\[ P \equal{} \left( \begin{array}{cccc}
\frac1k & \frac1k & \ldots & \frac1k  \\
1 & 0 & \ldots & 0 \\
0 & 1 & \ldots & 0 \end{array} \right)
\]
is a stochastic matrix. General theory of Markov chains says that $ P^k$ converges towards a matrix $ M$ whose row are all equal to a fixed vector $ (\pi_1, \ldots, \pi_n)$ which is the normalized $ 1$-eigenvector of $ P^t$ . Then one can check that $ P(i)\equal{}\pi_i$


$ \textbf{A}$ is a matrix, $ \textbf{x}$, $ \textbf{y}$ are vectors.

Prove that $ (\textbf{Ax},\textbf{y})\equal{}(\textbf{x},\textbf{A}^{*}\textbf{y})$

I know this should be easy but for some reason, I can't seem to get there.  I might be overthinking or missing something.

The "right" way to do this is to express the inner product itself by using matrix multiplication: $ \langle\mathbf{x},\mathbf{y}\rangle \equal{} \mathbf{y}^*\mathbf{x}.$

Then the result you're looking for is simply a matter of the associativity of matrix multiplication:

$ \langle\mathbf{Ax},\mathbf{y}\rangle \equal{} \mathbf{y}^*(\mathbf{Ax}) \equal{} (\mathbf{y}^*\mathbf{A})\mathbf{x} \equal{} (\mathbf{A}^*\mathbf{y})^*\mathbf{x} \equal{} \langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle$

---

Added: this is how to argue it for vectors in $ \mathbb{R}^n$ or $ \mathbb{C}^n,$ and $ n\times n$ matrices. Since you explicitly used the word "matrix" in your post, I assumed this is what you meant.

If the setting is a general inner product space, then this result is not a theorem at all but rather the very definition of the adjoint of an operator.


Two small circles are tangent to a large circle and pass through the center of the large circle, as shown. The region interior to the large circle and exterior to the small circles is shaded. If the diameter of the large circle is 4 inches, then what is the area of the shaded region in square inches?

The answer that I got is 8pi, but the actual answer is 2pi. Can anyone explain how could this be.

You probably mixed the diameter and radius up. To find the area of a circle, you use $ \pi r^2$. I think you ended up using $ \pi d^2$.

[hide]The large circle has a radius of 2, and the smaller circles each have a radius of 1. So, $ 4\pi\minus{}2(\pi)\equal{}2\pi$[/hide]

Yeah if you do pi d^2 you get 16pi-8pi=8pi, which is 4 times too much.

It will end up as $ 4$ times too much because the diameter is twice the radius, and $ 2^2$ is obviously $ 4$.


calculus
integration
logarithms
trigonometry
calculus computations

Given that:

$\int_{0}^{\infty}\;\frac{\ln{(1+x^{2})}}{1+x^{2}}\;dx\;=\;\pi\ln{2}$

find a similar expression for:

$\int_{0}^{\infty}\;\frac{\ln{(1+x^{2})}\;\cdot\;tan^{-1}\;{x}}{1+x^{2}}\; dx$

[ PS: I have a feeling [u]didilica[/u] will answer this one first... :D ]

Let $I=\int_{0}^{\infty}\frac{\ln(1+x^{2})}{1+x^{2}}\arctan x dx$.

Use the substitution $x=\tan t$ and you get that

$I=-2\int_{0}^{\pi/2}t\ln \cos t dt$.

Now use the following relation:

$\ln \cos x =-\ln 2+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\cos 2nx$, $-\pi/2<x<\pi/2$, and you should get that

$I=\frac{\pi^{2}}{4}\ln 2+\frac{7}{8}\xi(3)$.


I got the following error:

Tex capacity exceeded, sorry [input stack size = 5000]

So....I assume that I typed too much....How do you break into multiple files but still keep the page number/reference/table of contents thingies right???

That error message is a bit misleading. It just means you've made an error typing somewhere, for example a missing } which has caused TeX to go into a loop. TeX is quite capable of coping with massive books so you're very unlikely to have typed too much.

If you can't find the error (in TeXnicCenter click on the ! on the toolbar or press F9 to find the errors) then post your code here.

I also have the same problem. My code is here:

\documentclass[12pt]{article}
\newtheorem{theorem}{Theorem}[theorem]
\begin{document}
a
\begin{theorem}
a
\end{theorem}
\end{document}

The second line should be
\newtheorem{theorem}{Theorem}
or, if you want the numbering reset every chapter,
\newtheorem{theorem}{Theorem}[chapter]
or similarly for sections etc.
What you asked for, is for the theorem counter to reference the theorem counter giving an infinite loop hence exceeding TeX capacity.

See also [url=http://en.wikibooks.org/wiki/LaTeX/Theorems]LaTeX/Theorems[/url].


Let $ \frac {m}{n}$, in lowest terms, be the probability that a randomly chosen positive divisor of $ 10^{99}$ is an integer multiple of $ 10^{88}$ . Find $ m \plus{} n$.

Source: $ 1988$ AIME #$ 5$

[u]EXTRA CHALLENGE:[/u] Determine the number of $ 0$'s in the decimal form of $ 10^{99}$.

$ 10^{99} \equal{} 2^{99} \cdot 5^{99}$, so there are $ 100^2$ positive divisors to choose from. For a chosen divisor to be a integer multiple of $ 10^{88} \equal{} 2^{88} \cdot 5^{88}$, we must have the divisor be in the form $ 2^a \cdot 5^b$, where $ 88 \le a, b \le 99$. Thus, there are $ 12^2$ ways to choose the divisors. The probability is
\[ \frac {m}{n} \equal{} \frac {12^2}{100^2} \equal{} \frac {3^2}{25^2} \equal{} \frac {9}{625},
\]
so $ m \plus{} n \equal{} 9 \plus{} 625 \equal{} \boxed{634}$.

Is that actually a challenge? In $ 10^{99}$, there's $ \boxed{99}$ zeros.


Τι μπορείτε να συμπεράνετε για τα παρακάτω σχήματα; Πώς είναι δυνατόν να συμβαίνει κάτι τέτοιο;

[u]Διευκρίνιση[/u]: Η ορθογραφία στο κείμενο της εικόνας δεν είναι δική μου  :P

[img]http://users.otenet.gr/~mostel/exer.jpg[/img]

[quote="RAKAR"]:P ...to xo ksanadi..
oreo mostel... :)[/quote]

Έχεις κατά νου κάποια απόδειξη γιατί όντως γίνεται αυτό;

Εγώ για να τη βρω μου 'σπασε λίγο το κεφάλι  :P

Χε χε ...
Πολύ καλό. Το ήξερα

[color=blue][u]Hint:[/u][/color]
Δες αν όντως το αρχικό με το τελικό τρίγωνο είναι ίσα

[hide="Απάντηση"]δεν είναι! :rotfl: [/hide]

Ωραίο!!!Και εγώ όμως το είχα ξαναδεί....

Εγώ την είχα λύσει με εφαπτομένες πάντως  :P

PS: Το μόνο που χρειάζεται στα παραπάνω δύο σχήματα είναι μεγάλη καλλιτεχνία στα σχήματα, έτσι ώστε να φαίνεται αυτή η οφθαλμαπάτη..

φαινεται οτι μονο εγω δεν το ξερω, ομως δεν καταλαβαινω γιατι πρεπει να να αποδηξω οτι συμβαινει αυτο το αποτελεσμα.μπορει να εχουμε το ιδιο και με αλλα πολυγωνα στο επιπεδο, :maybe:

[quote="mostel"]Εγώ την είχα λύσει με εφαπτομένες πάντως  :P
[/quote]
Α, ωραία τότε.
Βασικά εννοούσα, όχι να δει κάποιος αν είναι ίσα τα τρίγωνα, αλλά αν [u]είναι[/u] τρίγωνα

@mostel
Δε νομίζω να θέλει καλλιτεχνία. Το δοκίμασα σε χαρτί και είναι δύσκολο με "γυμνό μάτι" να φανεί η ατέλεια

@kostas
Άσε την απόδειξη. Βρες που είναι το λάθος. Δες και όσα γράφω παραπάνω

Ακριβώς, έτσι έδειξα ότι δεν είναι τρίγωνα  :)


limit
parameterization
probability
inequalities
real analysis
real analysis unsolved

Calculate:for $x>1$
\[\lim_{n \to+\infty}e^{-nx}\sum_{k=1}^{n-1}\frac{(nx)^{k}}{k!}\]

I find the answer is $0$,when $x>1$;is $1$,when $0<x<1$.But I don't know what will happen when $x=1$.And what about $x<0$ can somebody offer helps?

Here is a suggestion, it might be a mess.

$e^{-nx}= \sum_{k=0}^{\infty}\frac{(-nx)^{k}}{k!}$

$\sum_{k=0}^{n-1}\frac{(nx)^{k}}{k!}= \sum_{k=0}^{\infty}a_{k}$ where $a_{k}= \{ \begin{array}{c}(nx)^{k}/k! \mbox{ if }k\geq n-1\\ 0 \mbox{ if }k>n-1 \end{array}$

Then you can multiply,
$\sum_{k=0}^{\infty}a_{k}\cdot \sum_{n=0}^{\infty}\frac{(-nx)^{k}}{k!}= \sum_{k=0}^{\infty}b_{k}x^{k}$

Where $b_{k}= \sum_{i=0}^{k}a_{k-i}\cdot \frac{(-n)^{i}}{i!}$

Hey,thank you,but I have try to do with that approach when I post the problem but get nothing :(

Consider a Poisson-distributed random variable $u$ with parameter $nx$; $\frac{(nx)^{k}}{k!}$ is the probability that $u$ is exactly $k$, so this sum is the probability that $u$ is between $1$ and $n-1$.

If $x=0$, the sum is zero for all $n$.
If $0<x<1$, the mean and variance are both $nx$. By the elementary estimate known as Chebyshev's inequality, the probability that $u$ is at least $n$ is at most $\frac1{t^{2}}$, where $t=\frac{n-nx}{\sqrt{nx}}=\sqrt{n}\frac{1-x}{\sqrt{x}}$. This goes to zero as $n\to\infty$, so the limit is 1 (the $k=0$ term tends to zero and is irrelevant).
If $x>1$, the probability that $u$ is less than $n$ is at most $\frac1{t^{2}}$, with $t=\frac{nx-n}{\sqrt{nx}}=\sqrt{n}\frac{x-1}{\sqrt{x}}$. This goes to zero as $n\to\infty$, so the limit is 0.
If $x=1$, we need stronger estimates. In this case, that's the central limit theorem or some variant. A Poisson distribution with large parameter looks approximately normal- and the normal is symmetric around its mean. The limit is $\frac12$ in that case.

This approach says nothing about $x<0$.

The $x=1$ case has certainly appeared on the forum before, and it hasn't been long.

[quote="jmerry"]
The $x=1$ case has certainly appeared on the forum before, and it hasn't been long.[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146401

I think there is an older thread where myth solved it using non-probabilistic methods.


inequalities
function
ratio
real analysis
real analysis unsolved

In the following  $a,b, A,B$ [b] are positive numbers[/b].  [b]  Let $I: =[0,\infty)$ ,   $f: I \to {\mathbb R}$ be  such that  $|f(x)|\le Ae^{Bx}\; ,\; \ ; \forall x \in I .$
Verify that the series $\sum\limits_{k=0}^{\infty}\frac{(ab)^k}{k!}f\left(\frac{k}{a}\right)$ is convergent.  [/b]
[color=green][b]
If $f$ is assumed to be convex on $I$ prove  $\begin{array}{|c|} \hline \\ \displaystyle \sum\limits_{k=0}^{\infty}\frac{\left((a+1)b\right)^k}{k!}f\left(\frac{k}{a+1}\right) \le \displaystyle e^b \sum\limits_{k=0}^{\infty}\frac{(ab)^k}{k!} f\left(\frac{k}{a}\right) \\ \\ \hline \end{array}\; .$
Show that the equality is attained for any linear function.
[/b][/color]

It seems to me RHS $e$ should be replace by $e^b$
Could you to check this professor LUPAS ?

[quote="Moubinool"]It seems to me RHS $e$ should be replace by $e^b$  [/quote]
Right ! Thanks,  I make the necessary correction !

$|\sum \frac{(ab)^k}{k!}f(k/a)|\leq A \sum \frac{(ab)^k}{k!}e^{Bk/a}$
the serie at the RHS is convergent since the ratio test is verify.
-----------------------------------------------

For the inequality write $p/(a+1)=\sum_{q=0}^{p}w(q)\frac{p-q}{a}$ with
$w(q)=\frac{C_p^q}{(a+1)^p}$
$w(0)+...+w(p)=1$

Use convexity of f and arrange the RHS to get a

(Le Barron Louis Augustin Cauchy (Polytechnicien) ) product

$\sum \frac{((a+1)b)^p}{p!}f(p/(a+1)) \leq \sum_{p\geq 0}\sum_{q=0}^{p}\frac{(ab)^{p-q}}{(p-q)!}f((p-q)/a)\frac{b^q}{q!} = e^b\sum \frac{(ab)^k}{k!}f(k/a)$

[quote="Moubinool"] ......(Le Barron Louis Augustin Cauchy (Polytechnicien) ) product........[/quote]
Nice !


geometry
3D geometry
algebra proposed
algebra

Can we always find integer solutions to

$a^n+b^n+c^n=d^n$ for $n \neq 0$  in  $\mathbb{N}$

you've already solved this? wow, i'd like to see a solution plz... :D

Not quite no. I was hoping someone else might have a good idea. I know its true for squares, cubes and forth powers, but beyond that, it gets tougher!