Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sum \frac {ab}{3 \plus{} bc} \le \frac {3}{4}
\]
Equality holds for $ a \equal{} b \equal{} c \equal{} 1$ and $ a \equal{} b \equal{} \frac {3}{2},c \equal{} 0.$

P.S.

It is equivalent to

$ 3a^2b^2c^2 \plus{} 81 \geq 9abc \plus{}12 \sum a^2b^2 \plus{}4abc\sum a^2b \plus{} 9\sum ab$

Now I am thinking how to solve this one.  :maybe:

Any help is welcome :wink:

Thank you very much. :)

[quote="manlio"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sum \frac {ab}{3 \plus{} bc} \le \frac {3}{4}
\]
Equality holds for $ a \equal{} b \equal{} c \equal{} 1$ and $ a \equal{} b \equal{} \frac {3}{2},c \equal{} 0.$

P.S.

It is equivalent to

$ 3a^2b^2c^2 \plus{} 81 \geq 9abc \plus{} 12 \sum a^2b^2 \plus{} 4abc\sum a^2b \plus{} 9\sum ab$

Now I am thinking how to solve this one.  :maybe:

Any help is welcome :wink:

Thank you very much. :)[/quote]
Well, you should use the known $ \sum a^2b \le 4\minus{}abc$ to transform the inequality into symmetric. It's the main step in my proof.

Well, you should use the known $ \sum a^2b \le 4 \minus{} abc$ to transform the inequality into symmetric. It's the main step in my proof.[/quote]

Is it necessary to expand it?

Dear can_hang2007,

using your hint I don't obtain the inequality case $ a \equal{} b \equal{} 3/2$and $ c \equal{} 0$.

Using your hint we have to prove

$ 7a^2b^2c^2 \plus{}81 \geq 25abc \plus{} 9\sum ab \plus{} 12\sum a^2b^2$

Now put

$ p\equal{}\sum a\equal{}3$, $ q\equal{}\sum ab$ and $ r\equal{}abc$

We have to prove  $ 7r^2 \plus{} 47r \plus{} 81 \geq 12q^2 \plus{} 9q$

Now use shur ineq in the form $ 9 \plus{} 3r \geq 4q$, so

$ \frac{81}{4} \plus{} \frac{27}{4} \geq 9q$  and

$ \frac{243}{4} \plus{} \frac{27}{4}r^2 \plus{} \frac{81}{2} \geq 12q^2$

Adding these last two we have

$ 81 \plus{} \frac{27}{4}r^2 \plus{} \frac{189}{16}r \geq 9q \plus{} 12q^2$

Coming back at our inequality we have

$ 7r^2 \plus{} 47r \plus{} 81 \geq 81 \plus{} \frac{27}{4}r^2 \plus{} \frac{189}{16}r \geq 9q \plus{} 12q^2$

I think it is not right as I don't obtain the equality case I told you.  :(

Where am I wrong?

Can you please post your proof?

Thank you very much.

[quote="manlio"]Dear can_hang2007,

using your hint I don't obtain the inequality case $ a \equal{} b \equal{} 3/2$and $ c \equal{} 0$.

Using your hint we have to prove

$ 7a^2b^2c^2 \plus{} 81 \geq 25abc \plus{} 9\sum ab \plus{} 12\sum a^2b^2$

Now put

$ p \equal{} \sum a \equal{} 3$, $ q \equal{} \sum ab$ and $ r \equal{} abc$

We have to prove  $ 7r^2 \plus{} 47r \plus{} 81 \geq 12q^2 \plus{} 9q$

Now use shur ineq in the form $ 9 \plus{} 3r \geq 4q$, so

$ \frac {81}{4} \plus{} \frac {27}{4} \geq 9q$  and

$ \frac {243}{4} \plus{} \frac {27}{4}r^2 \plus{} \frac {81}{2} \geq 12q^2$

Adding these last two we have

$ 81 \plus{} \frac {27}{4}r^2 \plus{} \frac {189}{16}r \geq 9q \plus{} 12q^2$

Coming back at our inequality we have

$ 7r^2 \plus{} 47r \plus{} 81 \geq 81 \plus{} \frac {27}{4}r^2 \plus{} \frac {189}{16}r \geq 9q \plus{} 12q^2$

I think it is not right as I don't obtain the equality case I told you.  :(

Where am I wrong?

Can you please post your proof?

Thank you very much.[/quote]
Well, the mistake in your proof is using the "not strong enough" inequality.
As your expanding, we have to prove
\[ 7r^2 \plus{} 47r \plus{} 81 \geq 12q^2 \plus{} 9q
\]
This inequality is clear if $ 4q \le 9$, for $ 4q \ge 9$, if you use the third degree Schur inequality $ 3r \ge 4q\minus{}9$, you won't get the result since it's not strong enough, but for the fourth degree Schur $ \sum a^4 \plus{}\sum a^2bc \ge \sum ab(a^2\plus{}b^2)$ or $ r \ge \frac{(4q\minus{}9)(9\minus{}q)}{18} \ge 0$, you will get the result. :)

[quote="can_hang2007"]Well, the mistake in your proof is using the "not strong enough" inequality.
As your expanding, we have to prove
\[ 7r^2 \plus{} 47r \plus{} 81 \geq 12q^2 \plus{} 9q
\]
This inequality is clear if $ 4q \le 9$, for $ 4q \ge 9$, if you use the third degree Schur inequality $ 3r \ge 4q \minus{} 9$, you won't get the result since it's not strong enough, but for the fourth degree Schur $ \sum a^4 \plus{} \sum a^2bc \ge \sum ab(a^2 \plus{} b^2)$ or $ r \ge \frac {(4q \minus{} 9)(9 \minus{} q)}{18} \ge 0$, you will get the result. :)[/quote]

Using your hint and making all computations I obtained

$ 112q^4 \plus{} 2520q^3\minus{}11439q^2\minus{}8046q\plus{}94770 \geq 0$

Is this also your solution?

Thank you very much.

[quote="manlio"][quote="can_hang2007"]Well, the mistake in your proof is using the "not strong enough" inequality.
As your expanding, we have to prove
\[ 7r^2 \plus{} 47r \plus{} 81 \geq 12q^2 \plus{} 9q
\]
This inequality is clear if $ 4q \le 9$, for $ 4q \ge 9$, if you use the third degree Schur inequality $ 3r \ge 4q \minus{} 9$, you won't get the result since it's not strong enough, but for the fourth degree Schur $ \sum a^4 \plus{} \sum a^2bc \ge \sum ab(a^2 \plus{} b^2)$ or $ r \ge \frac {(4q \minus{} 9)(9 \minus{} q)}{18} \ge 0$, you will get the result. :)[/quote]

Using your hint and making all computations I obtained

$ 112q^4 \plus{} 2520q^3 \minus{} 11439q^2 \minus{} 8046q \plus{} 94770 \geq 0$

Is this also your solution?

Thank you very much.[/quote]
I think you should factor the inequality first, then if you apply the Schur, it will be easy for us to factor
\[ 7r^2\plus{}47r\plus{}81\minus{}12q^2\minus{}9q \ge 0
\]
\[ \Leftrightarrow 7r^2\plus{} 47r \plus{}3(9\minus{}4q)(q\plus{}3) \ge 0
\]
If $ 9 \ge 4q$ then it's obvious.
If $ 4q \ge 9$ then,
\[ 7r^2\plus{} 47r \plus{}3(9\minus{}4q)(q\plus{}3) \ge 7( \frac{(4q\minus{}9)(9\minus{}q)}{18})^2 \plus{}47(\frac{(4q\minus{}9)(9\minus{}q)}{18}) \plus{}3(9\minus{}4q)(q\plus{}3)\equal{}\frac{1}{324}(3\minus{}q)(4q\minus{}9)(483q\minus{}28q^2\minus{}135) \ge 0
\]

OK, thank you very much. Now all is clear.  :lol:

You are right, my great mistake was that I had not factorized before applying Shur. :blush:

Many thanks again for your great help.


First, let me apologize in advance if this is the wrong place to post this...I just signed up today, and my problem grew out of a topic we discussed in CS...but I'd like to know...could a moderator please move this to the proper location?

In the game of Nim, we have multiple "stacks" of "coins."  Two players take turns taking as many coins as they please (at least one, though) from a single stack.  The object is to take the last coin.  It turns out that the (time-efficient) solution to a particular configuration deals with the parity of the binary representation of the number of coins in each stack.

I chewed on this problem for a while, and came up with my one problem: consider an m x n lattice of coins.  Two players take turns taking as many coins as they please from a single row OR column.  How can one determine whether an arbitrary board is a "winning" board or a "losing" board?  (The empty board is considered a losing board.)

I've "solved" the problem in Prolog using recursive descent, but that's a ridiculous way to solve a board (solving a 15 x 15 board would take until the Second Coming).  Dynamic programming promises a faster solution, but it uses an excessive amount of memory: 2 ^ (m x n) bits.

Again, I'm sorry if this shouldn't be posted here.

Is there choice not only in the number but also the specific coins removed from a given row or column?
Different choices of which $k$ coins to remove from a given row/column could lead to inequivalent positions.

Yes, we can choose specific coins when taking them out of a row or column.

oooo
oo_o
o_oo
oooo

and

oooo
oo_o
oo_o
oooo

are not equivalent, and both are obtainable from (take 1 coin from row 2) (take 1 coin from row 3).  That's what makes this problem a recursive nightmare...in order to test all possible moves, we not only have to test all possible amounts from each row and column but also all possible arrangements of coins within a row or column.


Will the AIME Problem Series A be held this fall or winter?

The Ad on top of the page reads: "Art of Problem Solving's AIME A Problem Series starts on June 12" so it is offered at summer. I think AIME B is during the fall/winter.


The expression $ \frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $ x=-1$ or $ x=2$, since division by zero is not allowed.  For other values of $ x$:

$\textbf{(A)}\ \text{The expression takes on many different values.} \\
\textbf{(B)}\ \text{The expression has only the value 2.} \\
\textbf{(C)}\ \text{The expression has only the value 1.} \\
\textbf{(D)}\ \text{The expression always has a value between } -1 \text{ and } +2. \\
\textbf{(E)}\ \text{The expression has a value greater than 2 or less than } -1.$

[hide="Click for solution"]
Evaluating, we have:

$ \frac{2x^2\minus{}x\minus{}4\minus{}x}{(x\plus{}1)(x\minus{}2)}\equal{}\frac{2(x\minus{}2)(x\plus{}1)}{(x\plus{}1)(x\minus{}2)}\equal{}2$, or $ \boxed{\textbf{(B)}}$.
[/hide]


calculus
derivative
function
real analysis
real analysis unsolved

Prove that if $f$ is derivatve and semiconvex then $f$ is continuous.

I meant $f: \mathbb{R}\to \mathbb{R}, f(\frac{x+y}{2})\le \frac{f(x)+f(y)}{2}$ and admite a primitive. Prove that $f$ is contiunous.  I have no ideea about how to start. Can anyone give ne a hint (at least  :D )?

What do you mean by [color=blue]admite a primitive[/color] ?

 :)

I mean that it is a derivative function i.e. There exist a function such that $F'=f$.

I found in a book this proof. It seems that it doesn't work, and I have no ideea how to fix it. Any help is welcomed.

[b]Proof[/b]

Let $F: I\to \mathbb{R}$ be the antiderivative of $f$. Consider $g_t(x)=tF(x+(1-t)a)+(1-t)F(x-at)-F(x),x\in I$;
One can see that $g_t(x)\ge 0,\forall x\in I,\forall t\in [0,1]\cap \mathbb{Q}$ and $g_t$is increasing $\forall t\in [0,1]\cap \mathbb{Q}\Rightarrow$ $g_t$ is increasing for any $t\in [0,1]$. This implies that $g_t(x)\ge 0,\forall x\in I,\forall t\in [0,1]$. Hence $f$ is convex $\Rightarrow$ continuous. $\blacksquare$

I think that there is an typo in the expresion of $g_t(x)$.

In what book have you found this proof? :)

This problem was proposed last year at Archimede contest at $12-th$ grade . The solution was in a book with all problem given in contests last year: "Olimpiadele si concursurile de matematica IX-XII " published by RMT.

You could take $g_{t}(x)=tF(x+\alpha(1-t))+(1-t)F(x-\alpha t)-F(x)$ where $\alpha$ is fixed.

Thak you very much dzeta! Nice ideea. The proof seems to work with now (I've edited it).


function
geometry
3D geometry
sphere
topology
real analysis
real analysis unsolved

Hi,

how does one think of limit points? I know that I can check out the definition but how can one visualize them? Any examples available?

An open set can be thought of as a diffuse edge or periphery that never becomes sharp no matter how much you zoom in on it, right? Or can you guys supply me with something better?

I'm trying to get something like an intutive feeling for this subject not only using definitions and memorization. In a problemsolving respect it makes things alot easier, atleast for me.

Great question and absolutely right attitude to the maths.

1. I think of limits as follows. When somebody tells me that $f(x) \rightarrow a$ (let's say, $f: R \rightarrow R$) I think of a thread (the one with which yo mom sews) which is entirely in the square which is the intersection of two bands (2D cylinder constructed on the neigbourhoods of the points). Here tread is the graph of the function and the square is $[x-\epsilon, x+\epsilon] \times [a-\epsilon, a+\epsilon]$ Note that this is not exactly the defenition of a limit because for a limit the value in point $x$ is not important and the neighbourhoods can have different diameter, but these are the details. Btw, for topological questions it's better to think of a continuous function as the one that transforms each open set into an open set, or connected into connected, or compact (=closed interval, a disk including it's circle, etc.) into compact, depending on situation.
2. I think of open sets as if they were just inervals without end points, disks without borders, balls without border sphere, etc. And all the sets topologivcaly equivalent to those, i.e. suppose all these sets are made out of a rubber, then you can stretch them in any direction as u like and by doing this you're getting equivavalent sets.

This can be said with more details, fr example, what's the difference between sets topologocally and differentially equivalent to an open interval, etc.


trigonometry
ratio
geometry proposed
geometry

ABCD is a convex quadrilateral for which <ACB + <ADB = 180 degs. If M is the common point of AC and BD, then the following equality holds:
AM*MB = AD*BC + CM*DM.

Best regards,
sunken rock

[color=darkblue]Denote $ \angle{AMB}\equal{}\beta$, $ \angle{BCM}\equal{}\alpha$.

Out relation is equivalent to

$ \dfrac{AM}{AD}\cdot \dfrac{MB}{BC}\equal{}1\plus{}\dfrac{CM}{BC}\cdot \dfrac{DM}{AD}$

$ \dfrac{\sin \alpha}{\sin \beta}\cdot \dfrac{\sin \alpha}{\sin \beta}\equal{}1\plus{}\dfrac{\sin(\alpha\plus{}\beta)}{\sin \beta}\cdot \dfrac{\sin(\alpha\minus{}\beta)}{\sin \beta}$

$ \sin^2 \alpha\equal{}\sin^2 \beta\plus{}(\sin \alpha \cos \beta\plus{} \sin \beta \cos \alpha)(\sin \alpha \cos \beta\minus{} \sin \beta \cos \alpha)$

$ \sin^2 \alpha\equal{}\sin^2 \beta\plus{}\sin^2\alpha \cos^2 \beta\minus{} \sin^2 \beta \cos^2 \alpha$

$ \sin^2 \alpha(1\minus{} \cos^2 \beta)\equal{}\sin^2 \beta(1\minus{} \cos^2 \alpha)$

$ \sin^2 \alpha  \sin^2 \beta\equal{}\sin^2 \alpha  \sin^2 \beta$[/color]

[i] First synthetic solution [/i]
If <ADB = <ACB = 90 degs, the problem is quite trivial from triangles AMD and BMC similar.
Suppose, w.l.o.g. that <ACB > 90 degs and draw BE and AF perpendicular on AC and BD respectively, with E and F on AC and BD respectively, see that triangles BCE and ADF are similar, same for BME and AMF; from their similitude ratio we get:
DF/CE = AF/BE = AD/BC = AM/BM = MF/ME = k ( 1 )
The relation to prove becomes:
AM*MB = AD*BC + (MF + DF)*(ME - CE); substituting AM = k*BM, AD = k*BC and other similar relations from (1) we shall finally get
BM^2 = BE^2 + ME^2, obviously, true.

Best regards,
sunken rock

Take a point $ P$ on the ray $ \overrightarrow{MD}$ such that $ AP\parallel BC.$ Then $ \triangle MPA \sim \triangle MBC,$ $\triangle ADP \sim \triangle MCB$ and let $ k$ denote the similarity coefficient of $ \triangle MCB$ to $ \triangle ADP$

$ \frac {AD}{MC} \equal{} k \Longrightarrow \ \frac {AD}{k} \cdot \frac {MA}{MC} \equal{} MA$ $ \Longrightarrow \frac {AD}{k} \cdot \frac {PM}{MB} \equal{} MA$

$ \Longrightarrow MA \cdot MB \equal{} \frac {AD\cdot PD}{k} \plus{} \frac {AD \cdot DM}{k}$

$ PD \equal{} k \cdot BC \ , \ AD \equal{} k \cdot CM \Longrightarrow MA \cdot MB \equal{} AD \cdot BC \plus{} CM \cdot DM.$

Other metric relations of this quadrilateral $ ABCD$ come from the fact that triangles $ \triangle ADM$ and $ \triangle BCM$ are pseudo-similar $ \Longrightarrow$ $ MA \cdot BC \equal{} MB \cdot AD.$ If $ P \equiv AD \cap BC,$ then $ P,C,M,D$ lie on a circle $ \omega.$ Hence $ (A,B)$ are conjugate points with respect to $ \omega,$ which implies that

$ p(A,\omega) \plus{} p(B,\omega) \equal{} AB^2$ $ \Longrightarrow AM \cdot AC \plus{} BM \cdot BD \equal{} AB^2.$

2nd synthetic solution - idea.
Continuing the work from here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=202546&p=1114247&hilit=finding+a+theorem#p1114247, we shall find another nice solution.

Best regards,
sunken rock


Three players play 4 rounds of a game and end with a tie score of 80. A different player wins each of the final three rounds. When a player wins, s/he doubles his or her points, and the two losers each must subtract the amount the winner gains from their scores. What was the highest score at the end of round 1?

I really didn't get the question..
[hide="Answer"]190 points[/hide]


problem 84:
[hide="answer"]
unless ive made a mistake in cross product, it's $ t^2$
[/hide]
[hide="method:"]
write position vector $ r$ and velocity vector $ v$ as a function of time.
angular momentum $ L \equal{} m(r$ x $ v)$
solve and get time dependence of $ L$ :)[/hide]

Yes da you are right, but I actually gave the problem to illustrate a more elegant method which I saw.

Torque = r x F
           = $ mgvCos \theta t$ (Take only the perpendicular comp)
$ \frac {dL}{dt}$ Proportional to t
=> L Prop to T2.

Nice eh???


Prove that $ 7$ does not divide $ 2^n \plus{}1$ for any positive integer $ n$.

[hide="Hello"]All in Mod 7:

$ 2^1\equal{}2$
$ 2^2\equal{}4$
$ 2^3\equal{}1$
$ 2^4\equal{}2$
repeating...

added 1, is  3 5 2 Mod 7, which is never divisible by 7
[/hide]


function
limit
trigonometry
calculus
calculus computations

Let there be the functions $ f,g : \mathbb{R} \to \mathbb{R}$ such that $ f(a\plus{}b)\equal{}f(a)g(b)\plus{}f(b)g(a)$  for all  $ a,b \in \mathbb{R}$ , and $ \displaystyle\lim_{x\to 0} \frac{f(x)}{x} \equal{}1$.
Find the limit $ \displaystyle\lim_{n\to\infty} \prod_{k\equal{}1}^{n} g \left(\frac{x}{2^{k}}\right)$    .     (where $ x\in \mathbb{R^{*}}$)

Intuition:

$ f(x)$ is $ \sin x$, $ g(x)\equal{}\cos x$. The limit is $ \frac{\sin x}{x}$.

Well the limit is indeed $ \frac{f(x)}{x}$ but finding $ f$ and $ g$ is not essential.You can solve the problem in a simpler mode[hide="like this."]
from the first condition we have : $ f\left(\frac{x}{2}\plus{}\frac{x}{2}\right)\equal{}f(x)\equal{}2f\left(\frac{x}{2}\right)g\left(\frac{x}{2}\right)\equal{}...\equal{}2^{n} f \left(\frac{x}{2^{n}}\right) \prod_{k\equal{}1}^{n} g \left(\frac{x}{2^{k}}\right)$.

So $ \displaystyle\lim_{n\to\infty} \prod_{k\equal{}1}^{n} g \left(\frac{x}{2^{k}}\right)\equal{}\displaystyle\lim_{n\to\infty} \frac{\frac{x}{2^{n}}}{f\left(\frac{x}{2^{n}}\right)}\frac{f(x)}{x}\equal{}\frac{f(x)}{x} \displaystyle\lim_{n\to\infty} \frac{\frac{x}{2^{n}}}{f\left(\frac{x}{2^{n}}\right)}\equal{}\frac{f(x)}{x}.$ [/hide]

[quote]How do you define $ f(0)$ ?[/quote]
Can't see your point.


logarithms
geometry
integration
trigonometry
limit
calculus
complex numbers

Evaluate the sums:
$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(1+n)}$
$\sum_{n=0}^{\infty}\sin(nt)/n$  t in $(0,2\pi)$

really thanks!

in these kind of problems, that is evaluate $\sum a_{n}$, it is often a good idea to study $f(x) = \sum a_{n}x^{n}$ and to use classical theorems on power series ..

What alekk said works for the first one. The second one requires one more idea, pertinent to trigonometric series.

Consider the series $\sum_{n=1}^{\infty}\frac{z^{n}}n$

(Yes, you can identify this function.)

Now, let $z=e^{i\theta}$ be a point on the unit circle. Assuming the series converges (proving that is interesting in its own right), the the power series becomes

$\sum_{n=1}^{\infty}\frac{e^{in\theta}}n$

and the imaginary part of that series is

$\sum_{n=1}^{\infty}\frac{\sin(n\theta)}n$

I very much like this exercise. It involves understanding power series (including such theorems as Abel's Theorem), understanding the complex logarithm, and it finishes with a geometry argument. (Being able to determine all of the angles of an isosceles triangle when you know one will be needed.)

The first series is well known and its sum is $S = ln2$.
For the second one I get $S = \frac{\pi-t}{2}$.
However I don't see where do we have to use a geometry argument and consider an isosceles triangle. Perhaps my answer is wrong. Could someone provide an explanation?

Thanks, Carcul

Another standard way to prove the second one (more real analytic though):

First note that:
$\sum_{k=1}^{\infty}\frac{\sin(kx)}{k}= \int_{\pi}^{x}\sum_{k=1}^{\infty}\cos(kt) \,dt$

Secondly we have:
$\frac{1}{2}+\sum_{k=1}^{n}\cos(kt) = \frac{1}{2}+\sum_{k=1}^{n}\frac{1}{2}(e^{ikt}+e^{-ikt})$
$= \frac{1}{2}\sum_{k=-n}^{n}(e^{it})^{k}= \frac{1}{2}e^{-int}\sum_{k=0}^{2n}(e^{it})^{k}$
$= \frac{1}{2}e^{-int}\frac{e^{(2n+1)it}-1}{e^{it}-1}= \frac{1}{2}\frac{e^{(n+1)it}-e^{-int}}{e^{it}-1}$
$= \frac{1}{2}\frac{e^{(n+\frac{1}{2})it}-e^{-(n+\frac{1}{2})it}}{e^{\frac{i}{2}t}-e^{-\frac{i}{2}t}}= \frac{1}{2}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}$

So I get:
$\sum_{k=1}^{\infty}\frac{\sin(kx)}{k}= \int_{\pi}^{x}\sum_{k=1}^{\infty}\cos(kt) \,dt$
$= \lim_{n \to \infty}\int_{\pi}^{x}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}-\frac{1}{2}\,dt$
$= \lim_{n \to \infty}\int_{\pi}^{x}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}\,dt+\frac{\pi-x}{2}$

The Integral on the left is $0$, because it has the form $\int_{a}^{b}f(x) \sin(kx) \,dx$ for $k \to \infty$ and $f : [a,b] \to \mathbb{R}$ continously differentiable.
So there exists a $K \in \mathbb{R}$ with $f(x) < K$ and $f'(x) < K$, hence:
$\lim_{k \to \infty}\left|\int_{a}^{b}f(x) \sin(kx) \,dx \right| = \lim_{k \to \infty}\left|\left[\frac{-f(x) \cos(kx)}{k}\right]_{a}^{b}+\frac{1}{k}\int_{a}^{b}f'(x) \cos(kx) \,dx \right|$
$\leq \lim_{k \to \infty}\frac{2K}{k}+\frac{1}{k}\int_{a}^{b}K \,dx = 0$


An o-Pod MP3 player stores and plays entire songs. Celeste has 10 songs stored on her o-Pod. The time length of each song is different. When the songs are ordered by length, the shortest song is only 30 seconds long and each subsequent song is 30 seconds longer than the previous song. Her favorite song is 3 minutes, 30 seconds long. The o-Pod will play all the songs in random order before repeating any song. What is the probability that she hears the first 4 minutes, 30 seconds of music - there are no pauses between songs - without hearing every second of her favorite song? Express your answer as a common fraction.

[url=http://www.mnspe.org/mathcounts/08_state_test/2008_solutions.pdf]See #30, Sprint Round[/url]

Quicker way:
Only consider the first 2 songs to play:
so there are a total of 10x9 = 90 different orderings

Now there are 9 cases where the favourite song is first.

Another 2 cases where the 1min or the 30sec song is the first and the favourite song is second so that all of the favourite song is played.

1 - (2+9/90)=79/90[/hide]

1/10 chance that 7th song is first
Also we can choose first or second song and then play the 7th song,
and this is 1/5 * 1/9 = 1/45 probability

add them up and you get 11/90, subtract from 1 to get 79/90

Seems really hard but actually really simple solution


two software gave inf as answer by the way.
But i need the proof. Thanks already


May be you mmisunderstood the problem, they asking you to find the remain of the division.

[b]STOP discussing ISL 04 problems until 10th of June!!![/b]

[quote="Pascual2005"]Hi i was looking to some of the imo shortlist problems from last year and I tried to translate one of them: NT 5:

cho số nguyên $n$, $n>1$ . $P_n$ là tích tất cả các số nguyên dương $x$, $x<n$ and $n|x^2-1$ Tính số dư của $P_n$ khi chia cho $n$.

What i understand is that they are asking us to count odd numbers $x$ less than $n$ such that $n|x^2-1$, for every given $n$. Is this the right formulation? if it is I think the problem is very easy!

Thanks[/quote]
let's see if i can translate it

given integer $n$, $n>1$. $P_n$ is the product of all positive integer $x$, $x<n$ and $n|x^2-1$. Calculate the remainder of $P_n$ when dividing by $n$.

[quote="orl"]analysis, I told you to stop discussing, translating etc. it!!!  :mad:[/quote]
you didn't tell me to stop translating

[quote]STOP discussing ISL 04 problems until 10th of June!!![/quote]
don't you agree discussing and translating are different   :rotfl:

analysis, you are not even known about any problem of ISL 04 until IMO 2005  ;)

[quote="Pascual2005"]Hi i was looking to some of the imo shortlist problems from last year and I tried to translate one of them: NT 5:

cho số nguyên $n$, $n>1$ . $P_n$ là tích tất cả các số nguyên dương $x$, $x<n$ and $n|x^2-1$ Tính số dư của $P_n$ khi chia cho $n$.

What i understand is that they are asking us to count odd numbers $x$ less than $n$ such that $n|x^2-1$, for every given $n$. Is this the right formulation? if it is I think the problem is very easy!

Thanks[/quote]

 Hi ! you are so clever ! CAn you teach me ? i'm just 15 . And i haven't studied math so well . i want to learnt more from you !


analytic geometry
graphing lines
slope
function
limit
combinatorics solved
combinatorics

This is very very difficult problem.

On the plane with lattice points,there is a frog.
First it is on the point(0,0).At each second if it is on point (x,y) it goes to point (x+1,y) or (x,y+1).
Prove that for each n there are n collinear points on the frog's path.

If nobody don't answer me I sit here and begin crying. :(  :(  :(

[quote="Omid Hatami"]Prove that for each n there are n collinear points on the frog's path.[/quote]

Sorry I just don't get the wording of it?

You mean after n moves there are n collinear points? that's not true...

:? a little explaining could do miracles here :)

He said for each $n$ there are $n$ collinear points on the frog's path. So an arbitrary number of collinear points can be found in the frog's path?!? That's what I thought. :?

[quote="grobber"]It looks pretty clear to me: the path of the frog consists of infinitely many lattice points (a set of lattice points). Show that in this set we can find $n$ collinear points, no matter how large $n$ is.[/quote]
Why is it so clear ???... You say that for each infinit set of lattice points we can find
$n$ collinear points no matter how large $n$ is. That's not true take for example $f(x)=x^2$. Its graph contains infinitely many lattice points but no 3 collinear. Furthermore the frog can alternate its steps. That is $s1 s1 s2 s1 s1 s1 s2 s2 s1 s2... $ where $s1 : (x,y)->(x+1,y) s2 : (x,y)->(x,y+1)$ It's really not that easy to find n collinear points. I didn't give it much thought but it doesn't look so trivial to me.

Ok, I got it but we will prove the result on the following form :
"The frog's path is supposed to be infinite. Then for each positive integer $n$, it contains at least $n$ of collinear points."

Sorry, but we are gone for another long post...

Let's give any path of infinite length $O(0,0) = M_0,M_1,...,M_k,...$.
For each $k \geq 0$ let :
$h(n)$ (resp. $v(k)$) be the number of horizontal (resp. vertical) jumps made by the frog from $M_0$ to $M_k$.
Thus $h(k),v(k)$ are nonegative integers, $h(k)+v(k) = k$, and $h(k+1) - h(k) = 0$ or $1$.
Moreover $M_k(h(k),v(k)).$
Let $n$ be a given integer.

Lemma :
If there exist integers $x,y,z$ with $(x,y)$ =/= $(0,0)$, such that
$xh(k)+yv(k) = z$ for at least $n$ distinct integers $k$ then there are at least $n$ collinear points on the frog's path.

Proof :
Let $k_1,k_2,...$ be an infinite increasing sequence of positive integers such that $xh(k_i)+yv(k_i) = z$ for each $i$.
Then $x(h(k_{i+1}) - h(k_i)) = -y(v(k_{i+1}) - v(k_i))$ for each $i$.
If $y = 0$ then all the $M_{k_i}$ have the same $x$-coordinate so that they are all collinear.
Otherwise, we have $ \frac {v(k_{i+1}) - v(k_i)} {h(k_{i+1}) - h(k_i)} = \frac {-x} {y}$, so that the lines ($M_{i+1}M_i$) have all the same slope, and then the $M_i$ are all collinear.

From above, we have $0 \leq \frac {h(k)} {k} \leq 1$. The sequence $\{ \frac {h(k)} {k} \}$ is then bounded so that it has a convergent subsequence. Let $L$ be the limit of some such subsequence.

It is a well-known result that we may find two integers $a,b$ such that $| L - \frac{a} {b}| < \frac {1} {2b^2}$ and $b > n$. (note that it is not assumed that $a$ and $b$ are coprimes, so that if $L = \frac {p} {q}$ is rational then choose $a = p(n+1)^2$ and $b = q(n+1)^2$, and if $L$ is irrational then choose a convergent of $L$, in the sense of continued fractions)
Let $N = (2m+1)(n-1)$ where $m$ is an arbitrary positive integer.
Let $f(k) = bh(k) - ak$.

Then :
- Either each of the $N+1$ values of $f(k)$, for $k = 0,1,...,N$, are in the set $\{-m,...,m\}$ : In that case, from the pigeon-hole principle, at least $n$ of them are equal. It follows that there exist an integer $z$ such that for at least $n$ distinct integers $k$, we have $x = f(k) = (b-a)h(k) - av(k)$.
From the lemma, it follows that there are at least $n$ collinear points on the frog's path, and we are done in that case.

- Or, there exist at least one integer $k \in \{0,...,N \}$ such that $f(k) \notin \{-m,...,m \}.$
Note that ${ \frac {N} {m+1} = 2(n-1) - \frac {1} {m+1} < 2n}$ thus $ \frac {m+1} {N} > \frac {1} {2n}.$
For such a $k$, we have $ | \frac {f(k)} {k} | \geq \frac {m+1} {N} > \frac {1} {2n}.$
It follows that $| \frac {h(k)} {k} - \frac {a} {b}| > \frac {1} {2nb} > \frac {1} {2b^2}.$
Since $m$ is arbitrary, choosing $m$ bigger and bigger, we then find an infinite number of integers $k$ such that $| \frac {h(k)} {k} - L| > \frac {1} {2nb} - \frac {1} {2b^2} = \epsilon > 0.$

It foolws that there are an infinite number of $k$ such that $  \frac {h(k)} {k} - L > \epsilon$ or such that $ \frac {h(k)} {k} - L < - \epsilon.$
Suppose that we are in the first case and choose a rational number $r$ such that $L + \frac { \epsilon} {2} < r < L + \epsilon$ (in the second case, the reasoning is the same as below with $r$ such that $L - \epsilon < r < L - \frac { \epsilon} {2}$).
Let $r = \frac {p} {q}.$
Then, for an infinite number of $k$, we have $ \frac {h(k)} {k} > L + \epsilon > r$, which gives $h(k) > rk$ for an infinite number of $k$.
In another hand, since $L$ is the limit of a subsequence of $\{ \frac {h(k)} {k} \}$, it follows that there an an infinite number of $k$ such that $  \frac {h(k)} {k} < L + \frac { \epsilon} {2} <$ so that $h(k) < rk$ for an infinite number of $k$.

But $h(k) - rk = \frac {qh(k) - pk} {k}$ then the function $g(k) = qh(k) - pk$ changes its sign an infinite number of times.
Moreover, for each $k$, we have $g(k+1) - g(k) = q(h(k+1) - h(k) - p$ thus $-p \leq g(k+1) - g(k) \leq q-p$, from which we deduce that the sequence $\{ g(k+1) - g(k) \}$ is bounded. Since the function $g(k)$ takes only integer values, changes its sign an infinite number of times, it then follows that the function $g$ takes the same value an infinite number of times. That is : there exists an integer $c$ such that, for an infinite number of distinct values of $k$, we have $g(k) = c$, and then $(q-p)h(k) - pv(k) = c$.
From the lemma, it follows that the frog's path contains an infinite number of collinear points, and we are done.

Pierre.

Another problem.
This came to my mind but I couldn't solve it.
Suppose $ P$ is subset of lattice points in plane.$\alpha$ is a positive real number.
We now that in every $n*n$ square (square has sides parallel to the x-axis and y-axis) with $(0,0)$ at center we have at least $\alpha .n$ points of $ P$ in this square.
For which $\alpha$ the following sentence is correct.
For every $k$ there are $k$ colinear points in $ P$

At first sight, I would say that no $\alpha$ has this property, in the sense where I think it could be done to construct such a set $P$ with few collinear points. But from now, I have no idea to achieve such a construction.
Note that the initial problem gave a stronger hypothesis than the density of $P$.

A famous result, known as Szemeredi's lemma states that if a set $P$ of the integers has a positive upper density (that is $\lim_n \sup \frac {|P \cap [1,n] |} n  > 0$ ) then $P$ contains arbitrary long arithmetic progressions.

Clearly, I agree that the converse is not true, since this has recently been shown that the set of primes contains arithmetical progressions of arbitrary finite lengths (wow!).

In this point of view, for the two-dimensional case you are asking for, I think you probably have to ask for a set with at least $\alpha \cdot n^2$. In that case, I think the result follows from the well-known Sperner's lemma and the Hales-Jewett's theorem.

Pierre.

Please explain more.Also tell me your proof for $\alpha n^2$ with Heils Jewet theorem (Also tell me the Heils Jewet theorem).I would be thankful if you could prove that  $\alpha.n$ is not correct.

Excuse me but I must go.


A kite is inscribed inside a rectangle. Prove that the diagonals of the kite are the side lenghts of the rectangle. ( I encountered this problem during homework and I was like woah is there a formal way of proving this?)

[hide]The vertices of the kite are on the sides of the rectangle.

Since the diagonals of a kite are perpendicular to each other, the sides of the kite are parallel to the sides of the rectangle.

Therefore, the diagonals of the kite are congruent to the sides of the rectangle. [Try to prove this using congruent triangles.] [/hide]


[color=purple] With any $ M$ in plane and three positie real numbers $ x,y,z$ prove that :
   $ x\frac {MB.MC}{a} \plus{} y\frac {MC.MA}{b} \plus{} z\frac {MA.MB}{c}\ge\frac {\sqrt {xy \plus{} yz \plus{} zx}(R^2 \minus{} OM^2)}{R}$
  Have fun ! [/color]

Use two lemma:
Lemma 1. Triangle $ ABC$, area $ S$. We have:
$ x.bc\plus{}y.ac\plus{}z.ab\ge \sqrt{xy\plus{}yz\plus{}zx}.S$
Lemma 2. Triangle $ ABC$, $ M$ is any point inside it. $ A', B', C'$b such that $ MA'\perp BC, MB'\perp AC, MC'\perp AB$ and $ A', B', C'\in BC, CA, AB)$ then:
$ S[A'B'C']\equal{}\frac{1}{4}\left(1\minus{}\frac{OM^2}{R^2}\right).S[ABC]$


geometry
modular arithmetic
geometric transformation
homothety
combinatorial geometry

To [i]clip[/i] a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by the three segments $AM, MN$, and $NC$, where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$.  In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon.  A regular hexagon ${\cal P}_6$ of area 1 is clipped to obtain a heptagon ${\cal P}_7$.  Then ${\cal P}_7$ is clipped (in one of the seven possible ways) to obtain an octagon ${\cal P}_8$, and so on.  Prove that no matter how the clippings are done, the area of ${\cal P}_n$ is greater than $\frac 13$, for all $n \geq 6$.

Nice problem!

Let the hexagon be $ A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$. Note that $ P_{n}$ is convex (this is easy with induction). On each side $ A_{i}A_{i\plus{}1}$, where $ A_{i}\equal{}A_{i\plus{}6}$, there must exist two points, say $ X_{i}$ and $ Y_{i}$, where $ X_{i}$ is between $ Y_{i}$ and $ A_{i}$, that are vertices of $ P_{n}$. By the convexity of $ P_{n}$, $ Y_{i}X_{i\plus{}1}$ is inside $ P_{n}$ for $ i\equal{}1,2,\ldots,6$. Furthermore, $ X_{i}$ and $ Y_{i}$ are adjacent vertices, so we have $ P_{n}>[X_{1}Y_{1}X_{2}Y_{2}\ldots X_{6}Y_{6}]$.

Let $ A_{i}A_{i\plus{}2}\cap A_{i\plus{}1}A_{i\plus{}3}\equal{}B_{i}$. It follows from the convexity of $ \Delta A_{i}A_{i\plus{}1}A_{i\plus{}2}$ that $ B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}$ is completely contained within $ X_{1}Y_{1}\ldots X_{6}Y_{6}$, so it is contained in $ P_{n}$. But it's a straightforward calculation to find that $ [B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}]\equal{}\dfrac{1}{3}$, so we're done.

It seems like this bound can be improved significantly...I'm thinking it's true for 1/2.

Sorry about the 5.5 year revive darn. Also, I pleased the above poster by getting a bound of 1/2. For a $n$-gon $A_1A_2 \ldots A_n$ define its [i]outsum[/i] to be $\sum_{i = 1}^n[A_{i}A_{i+1}A_{i+2}]$ (indices $\pmod{n}$ ofc).

WLOG we clip the triangle $A_{n-1}A_nA_1$, and our polygon becomes $A_1A_2 \ldots A_{n-1}A_n'A_{n+1}$. What the the outsum of our new polygon?

Well, we add $\frac{1}{4}[A_{n-1}A_nA_1]$ to our total clipped area. But note that the outsum decreases by at least $\frac{1}{2}[A_{n-1}A_nA_1]$. This is because the areas $[A_{n-2}A_{n-1}A_n], [A_nA_1A_2]$ decrease to $[A_{n-2}A_{n-1}A_n'], [A_n'A_1A_2]$ (to be exact, half as much area), while $[A_{n-1}A_nA_1]$ is replaced $[A_{n-1}A_n'A_{n+1}]+[A_n'A_{n+1}A_1] = \frac{1}{2}[A_{n-1}A_nA_{n+1}]$. But the outsum is at least positive, and the clipped area decreases by half as much as the outsum, so its at most 1/2. $%Error. "blackbox" is a bad command.
$

Here is another solution obtaining a half, copy pasted from an email (hence no LaTeX).

Define one cut X to be a child of another Y if one of the endpoints on X lies on the cut formed by Y.

Note that at most 6 cuts are not children because of cutting the vertices/corners off.

Now, the crucial lemma is that a cut's child cuts off at most 1/4 of the area the parent cuts off and this is really not hard to prove (I don't have a diagram, but basically if the parent cuts XZ where vertex Y is between, and you have YZ>ZA with A on YZ the new cut cuts off the image of XA upon a homothety with scale factor 1/2 about Z).

Now, every guy's children actually sum to at most 1/2 his area.  Once again this is trivial, because the heights are bounded by things like d(A,XZ)<=YZ.

So create a directed graph, edges directed from parents to children.  Delete edges to remove all cycles.  Now we have a binary tree.  The total area clipped off is at most 6*(1/24)*(1+1/2+1/4+...)=(1/4)*2=1/2 with equality never, so 1/2 is the answer.

Problem 3 of the 2010 Cono Sur Olympiad (http://www.obm.org.br/export/sites/default/como_se_preparar/provas/Cone_sul/docs/Dia1-portugues.pdf ) asks for a bound of $2/3$.

This can be achieved by noting that the centroids of triangles $A_i A_{i+1} A_{i+2}$ are never removed, as well as some point in each of the sides. The convex hull of the 12 points has area at least 2/3.

Another way to get $\frac{2}{3}$ is to note that $f_n = 4\sqrt{3} \cdot \text{Area of } P_n - \sum_{l \text{ is a side of }P_n} l^2$ increases over time. This is because when we clip two sides with lengths $a, b$ and angle $\theta$ in $P_n$ to get $P_{n+1}$, the area drops by $\frac{ab \sin{\theta}}{8}$ while the sum of squared lengths drops by $a^2 + b^2 - \frac{a^2}{4} - \frac{b^2}{4} - \frac{a^2+b^2 - 2ab\cos{\theta}}{4} = \frac{a^2+b^2+ab \cos{\theta}}{2}$. The preceding claim then reduces to $ab(-\cos{\theta} + \sqrt{3} \sin{\theta}) \leq a^2 + b^2$, which is true. Note that equality holds here if and only if $a = b$ and $\theta = \frac{2\pi}{3}$, corresponding to the initial situation -- in fact, equality can hold for the first 6 clip operations, after which the angle becomes larger and this estimate is no longer tight.

Going back to the original problem, suppose initially each side length is $1$, then the area of the regular hexagon is $\frac{3\sqrt{3}}{2}$. Thus the initial value of the semi-invariant is $12$, which means we have $\text{Area of } P_n \geq \sqrt{3} \geq \frac{2}{3} = \text{Area of } P_6$ for all $n$.

Let the vertices be $A_1$, $A_2$...$A_6$. Connect each $A_i$ to $A_{i+2}$, where indices are taken modulo six. We get the following diagram.
[asy]
size(6cm, 6cm);
pair A=(0,2), B=(sqrt(3),1), C=(sqrt(3),-1), D=(0,-2), E=(-sqrt(3),-1), F=(-sqrt(3),1);
draw (A--B--C--D--E--F--cycle, cyan+linewidth(2));
draw (A--C--E--cycle, heavycyan+dashed+linewidth(2));
draw (B--D--F--cycle, heavycyan+dashed+linewidth(2));
dot (A, red);
dot (B, red);
dot (C, red);
dot (D, red);
dot (E, red);
dot (F, red);
[/asy]
It is not hard to see that no cut can penetrate the interior of the inner hexagon. One can compute that the inner hexagon is $\frac{1}{3}$ of the whole hexagon. Hence, the area of $P_n$ will always be greater than $\frac{1}{3}$.

Here's just the proof of $1/3$ (from [i]Twitch Solves ISL[/i]). Call the original hexagon $ABCDEF$. We show the area common to triangles $ACE$ and $BDF$ is in every $\mathcal{P}_n$; this solves the problem since the area is $1/3$.
For every side of a clipped polygon, we define its [i]foundation[/i] recursively as follows: [list] 	[*]$AB$, $BC$, $CD$, $DE$, $EF$, $FA$ 	are each their own foundation 	(we also call these [i]original sides[/i]). 	[*]When a new clipped edge is added, 	its foundation is the union of the foundations 	of the two edges it touches. [/list] Hence, any foundations are nonempty subsets of original sides.

[b][color=red]Claim:[/color][/b]  	All foundations are in fact at most 	two-element sets of adjacent original sides.
[i]Proof.[/i]  	It's immediate by induction 	that any two adjacent sides have at most two elements 	in the union of their foundations, and if there are two, 	they are two adjacent original sides. $\blacksquare$
Now, if a side has foundation contained in $\{AB,BC\}$, say, then the side should be contained within triangle $ABC$. Hence the side does not touch $AC$. This proves the problem.

([url]https://artofproblemsolving.com/community/q1h2722725p23691094[/url])

here's another solution for $\frac{1}{2}:$

Scale up the figure of the hexagon such that it has a side length of $1.$ The hexagon now has an area of $\frac{3}{2}\sqrt{3}.$

Notice that every time $P_n$ is clipped, no side of the figure completely disappears, so the figure must touch all sides of the original regular hexagon. In addition, since the figure always remains convex, we can take the intersections of the figure and each side of the original regular hexagon, and the polygon formed will have a lesser area than the clipped figure. It suffices to find the minimum area of a hexagon that has one vertex on each side, which can be done by finding the maximum area of the six triangles that are formed by the area outside the inner hexagon but inside the outer hexagon. This is clearly equal to the maximum of $\frac{1}{2}\sin{120^{\circ}}(a(1-b)+b(1-c)+c(1-d)+e(1-f)+f(1-a))=\frac{3\sqrt{3}}{4}(a+b+c+d+e+f-ab-bc-cd-de-ef-fa)=\frac{3\sqrt{3}}{4}\left( \frac{1}{4}-\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right) + \frac{1}{4}-\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right) + \frac{1}{4}-\left(c-\frac{1}{2}\right)\left(d-\frac{1}{2}\right) + \frac{1}{4}-\left(d-\frac{1}{2}\right)\left(e-\frac{1}{2}\right) + \frac{1}{4}-\left(e-\frac{1}{2}\right)\left(f-\frac{1}{2}\right) + \frac{1}{4}-\left(f-\frac{1}{2}\right)\left(a-\frac{1}{2}\right) \right).$ Clearly we want to minimize $\sum_{cyc}\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right).$ We can minimize each term by setting one factor equal to $1-\frac{1}{2}=\frac{1}{2},$ and the other factor equal to $0-\frac{1}{2}=-\frac{1}{2}.$ This is achievable with $(0,1,0,1,0,1),$ which corresponds to three pairs of concurrent points at the vertices of the hexagon, which form a triangle. This triangle clearly has half the area of the hexagon, and scaling back down to a hexagon with area $1,$ the figure $P_n$ cannot have a lesser area than the triangle, which has area $\frac{1}{2}.$

[i]A slightly different solution:[/i]
[rule]
Let $A_1A_2 \cdots A_6 = \mathcal P_6$ be the regular hexagon. For any $1 \le i \le 6$, define $\mathcal R_i$ as the region of triangle $A_{i-1}A_i A_{i+1}$.

[b][color=#f00]Claim: [/color][/b] Fix any $j \ge 6$. Then any three vertices of $\mathcal P_j$ must be contained in some $\mathcal R_i$ and conversely, for any $\mathcal R_i$ ($1 \le i \le 6$), some three consecutive vertices of $\mathcal P_j$ will be contained in $\mathcal R_i$.

[b][color=#00f]Proof: [/color][/b] Of course, the assertion is true for $\mathcal P_6$. Now if we assume the assertion is true for $\mathcal P_k$, then it is not hard to see that it is true for $\mathcal P_{k+1}$. $\square$

Now fix any $j \ge 6$. Let $P$ be the regular hexagon formed by lines $P_{i-1} P_{i+1}$ for $1 \le i \le 6$. Our claim also gives us that any side of $\mathcal P_j$ has to intersect with the interior of $P$, and it cannot be completely contained in some side of $P$. Moreover, each of the regions $\mathcal R_i$ contains at least one vertex of $\mathcal P_j$. Now I am pretty sure that this is enough to imply that area of $\mathcal P_j$ is more than the area of $P$ (though, I can't get how to prove this rigorously). Now it is not hard to prove that area of $P$ is $\frac{1}{3}$, so we are done. $\blacksquare$

Let the hexagon be $ABCDEF$. Consider the hexagon in the intersection of $\triangle ACE$ and $\triangle BDF$, which has exactly $\frac13$ of the desired area. We show that it lies in all $\mathcal{P}_n$.
Now, consider the vertices of the hexagon as a cyclic set of vectors. Clipping is equivalent to replacing adjacent vectors $a, b, c$ with $a, \frac{a+b}{2}, \frac{b+c}{2}, c$.

[b][color=pink]Claim:[/color][/b] Suppose we have adjacent vectors $a, b, c, d$. Then at no point does clipping give us a vector which is a weighted average containing both $a$ and $d$.
[i]Proof.[/i] There are always two vectors not containing either components of $a$ or $d$ between the two under the operation of clipping. $\blacksquare$
Noteably, if $v$ and $u$ are two adjacent vertices in $\mathcal{P_n}$, then the subset of $\{a, b, \dots, f\}$ contained in one of them has at most three elements.
As such, all edges of $\mathcal{P}_n$ lie in some triangle formed by three adjacent vertices of the original hexagon.


Hey Manuel,

Puerto Rico is quite a nice place it seems.  :D

[url=http://www.spiegel.de/fotostrecke/0,5538,PB64-SUQ9MTUxNjcmbnI9MQ_3_3,00.html]Here[/url] you have 172 pics of those 86 woman participating in the competition, each featured woman got two images. One close shot and another one further away. Winner is the 18 years old Zuleyka Rivera Mendoza from Puerto Rico. Second was the Japanese girl Kurana Chibana, third was Swiss girl Lauriane Gillieron, fourth Lourdes Arevalos from Paraquay and fifth Tara Conner from the USA. Hot babes, dude.  :10:

Warum die Seite ist in Deutsch?
¿Y por qué pusieron ese topic aquí?

Capaz que porque la chica esta habla castellano. Igual, me pareció sin sentido, y sobre todo luego de que a un topic en el foro de la IMO se le hallan borrado algunas fotos porque la moayoría de los usuarios no lo encontraban de buen gusto

No creo que sea sin sentido, de otra forma lo hubiera borrado! Orl es amigo personal de Manuel y ha hablado mucho conmigo expresandome siempre su gusto por las muejres latinas, es por eso que cuando Miss Universo (que aunque para algunos no sea interesante (como para mi) pero que hay que respetar) es ganado por una latina, Orl simplemente quiere hacer que Manuel se entere de esto y dar un link...

De todas maneras, mientras esto se haga en las discusiones generales no hay problema


geometry calculus integration quadratics

A rancher plans to use 600 ft of fencing to enclose a rextanular corral and to divide it into two parts by a fence parallel to the shorter sides of the corral.  Find the dimensions of the corral if its area is 8000ft^2.

I've been workin on this problem and can't figure it out...i know to use substitution with 3x+2y=600 and xy=8000 but can't make it work.  I dont know if im forgetting something or what.
Please and Thanks!!  :)

I'm going to assume the dimensions have to be integral...

[hide]
Add your two equations to get $xy+3x+2y = 8600$. Add $6$ to both sides and factor the LHS to get $(x+2)(y+3) = 8606$. Factor 8606 to get $2 \times 13 \times 331$. The only way to divide this into two numbers and get an answer that makes sense is to divide it into $26 \times 331$. Since x is shorter, $x+2 = 26$ and $y+3$ = 331. The dimensions of the corral are $24$ by $328$.[/hide]

Ubemaya, there are a few issues with that

1. As the writer pointed out, you can't assume the dimentions are integral
2. [quote]The only way to divide this into two numbers and get an answer that makes sense is to divide it into $26 \times 331.$[/quote]

There are many more ways, and none of them work in the first place.

[hide="solution"]
Given that you wanted to use a substitution, what you should do is solve one equation for one variable and substitute it into the other equation.  So,

$3x+2y = 600 \Longrightarrow y =-\frac32x+300$

which means that
$xy = x(-\frac32x+300) = 8000$

This leaves you with

$-\frac32x^{2}+300x-8000 = 0$
or
$3x^{2}-600x+16000 = 0$.

Now it should be easy to solve this quadratic and then you can calculate $y$ using the first equation.
[/hide]


abstract algebra
induction
vector
algorithm
number theory proposed
number theory

Problem 1)
a) For a prime number $ p$, find all permutations $ a_1,a_2,\cdots,a_p$ of $ 1,2,\cdots,p$ with $ i\plus{}a_i | j\plus{}a_j$ for all $ 1\leq i<j \leq p$
b) (infinite case) Is there a permutation $ a_1,a_2,\cdots$ of all natural numbers (i.e., $ \mathbb{N}$) with with $ i\plus{}a_i | j\plus{}a_j$ for all $ i<j$?

Problem 2)
For a prime number $ p$, prove that there is a permutation $ a_1,a_2,\cdots,a_{p\minus{}1}$ of $ 1,2,\cdots,p\minus{}1$ with partial sums $ \sum_{i\equal{}1}^ja_i$ ($ j\equal{}1,2,\cdots,p\minus{}1$) producing pairwise distinct remainders module $ p$. What about the case when $ p$ is replaced with $ p^2$ (I couldn't solve this part).

coud you give a hint(just a small one :) ) for part $ a$?
because i got many answers and here are two of them which are somehow non-trivial:
for example let $ t$ be a large number and also $ l$ be another large number for which $ \gcd(l\plus{}1,t\minus{}1)\equal{}1$ then we'll have :
$ \gcd(t(l\plus{}1),tl\plus{}1)\equal{}1$
so by Dirichlet's theorem we have infinitely many primes of the form $ t(l\plus{}1)k\minus{}(tl\plus{}1)$ for some $ k\in N$
let $ p$ be one of those primes then put $ tl\plus{}1\equal{}i$ which implies $ t|t\plus{}(i\minus{}1)$ and also $ t\plus{}i\minus{}1|p\plus{}i$
now one can easily see that the permutation :
$ <a_p,...,a_i,a_{i\minus{}1},a_{i\minus{}2},...,a_{t},a_{t\minus{}1},...,a_{1}>\equal{}<i,...,p,t,t\plus{}1,...,i\minus{}1,1,...,t\minus{}1>$
does the work.
or for example if we take a large number like $ s$ for which $ 4(s\minus{}1)\minus{}s\equal{}p$ where $ p$ is a prime then the permutation :
$ < s,s\plus{}1,...,p,s\minus{}1,1,...,s\minus{}2>\equal{}< a_{p},a_{p\minus{}1},...,a_{s},a_{s\minus{}1},a_{s\minus{}2},...,a_{1}>$
does the work.
also one can easily see that there is a prime $ p$ which fulfills both of the above conditions(let $ t\equal{}3f$) and so there exist two permutations like the two above.there exists also some other permutations which are easy to get...
thx for your answer in advance... :)

Sorry, ... in the first problem part a), it is $ p\minus{}1$ instead of $ p$! my mistake!. Now you will see it is not very difficult and based on your comment I think you got the idea. Anyway the original problem in part a) actually deals with finding such permutations for any natural number. these permutations don't have a good-looking representation of course, but they introduce some interesting situations which can be used to pose many other nice problems.

so we suppose that $ p$ is an integer not necessarily prime.
at first if one of the integers $ i \plus{} a_i$ is divisible by $ p$ then directly we have $ a_p \equal{} p$ and we had $ p \le a_{p \minus{} 1} \plus{} p \minus{} 1 |2p$ which implies that $ a_{p \minus{} 1} \equal{} 1$ and then we get $ a_{p \minus{} 2} \ge 2, a_{p \minus{} 2} \plus{} p \minus{} 2|p$ which implies that $ a_{p \minus{} 2} \equal{} 2$ and so we'll get the unique permutation $ < p,1,...,p \minus{} 1 >$.
note that from now ,none of the integers $ i \plus{} a_i$ is divisble by $ p$.suppose that $ i$ is the smallest number for which $ i \plus{} a_i > p$ then obviously we'll have $ i \plus{} a_i \equal{} ... \equal{} p \plus{} a_p$ .
put $ a_i \equal{} t$ then we get $ a_{i \plus{} 1} \equal{} t \minus{} 1,...,a_p \equal{} t \plus{} i \minus{} p$
we know that if for an integer $ k$ we have $ a_k \equal{} p$ then for some $ 0 \le l \le p \minus{} i$ we have $ t \minus{} l \equal{} a_k \equal{} p$ but we know that $ t \le p$ which implies that $ t \equal{} a_i \equal{} p$
so we have $ 1 \plus{} a_1 | ... | (i \minus{} 1) \plus{} a_{i \minus{} 1} | i \plus{} p \equal{} (i \plus{} 1) \plus{} (p \minus{} 1) \equal{} ... \equal{} p \plus{} (p \minus{} (p \minus{} i))$
so we get  $ \{a_i,...,a_{p}\} \equal{} \{p,p \minus{} 1,...,i\}$ which gives $ \{a_1,...,a_{i \minus{} 1}\} \equal{} \{1,2,...,i \minus{} 1\}$.
now we have $ 1 \plus{} a_1 | ... | (i \minus{} 1) \plus{} a_{i \minus{} 1}$ with $ \{a_1,...,a_{i \minus{} 1}\} \equal{} \{1,2,...,i \minus{} 1\}$ which is in fact the same problem but this time for the case $ p\equal{}i\minus{}1$.i wanted to use induction when i saw this but unfortunately :(  i didn't succeed...
so between the permutations for the case $ p\equal{}i\minus{}1$ we must choose those who has this property $ a_{i\minus{}1}\plus{}i\minus{}1 | p\plus{}i$.
if we do the same thing for the case $ p\equal{}i\minus{}1$ we obtain two cases:
1-the permutation $ <a_{i\minus{}1},a_{i\minus{}2},...,a_1>\equal{}<i\minus{}1,1,2,...,i\minus{}2>$ which gives us the permutation:
$ < i,i \plus{} 1,...,p,i \minus{} 1,1,...,i\minus{}2>\equal{}< a_p,a_{p \minus{} 1},...,a_{i},a_{i \minus{} 1},a_{i \minus{} 2},...,a_{1}>$
with $ 2(i\minus{}1)|p\plus{}i$.(if such $ i$ exists)
2-if $ j$ is the smallest integer for which we have $ a_j\plus{}j>i\minus{}1$ where $ 1 \le j <i\minus{}1$ then we'll have $ a_{j}\equal{}i\minus{}1$ (by the same reasoning for $ p$) and so we have $ a_{i\minus{}1}\equal{}j,...a_j\equal{}i\minus{}1$.
but we must also consider the condition $ i\minus{}1\plus{}j | p\plus{}i$
now we have $ \{a_{j\minus{}1},...,a_1\}\equal{}\{j\minus{}1,...,1\}$ and also $ 1\plus{}a_1 |...|j\minus{}1\plus{}a_{j\minus{}1}$
and so we again repeat the whole story ...
so, finally ,we arrive at these two permutations:
1-
$ <a_p,...,a_{i_1},a_{i_1\minus{}1},...,a_{i_2},a_{i_2\minus{}1},...,a_{i_3},...,a_{i_t},a_{i_t\minus{}1},...,a_{i_{t\plus{}1}}\equal{}a_1>
\equal{}<i_1,...,p,i_2,...,i_1\minus{}1,i_3,...,i_2\minus{}1,...,i_{t\minus{}1}\minus{}1,i_{t\plus{}1}\equal{}1,...,i_t\minus{}1>$
where $ i_l \ (1 \le l \le t)$ are $ \plus{}$ integers and we have $ i_{l\plus{}1}\plus{}(i_l\minus{}1)|i_{l}\plus{}(i_{l\minus{}1}\minus{}1)$ and also $ p\plus{}1\equal{}i_0>i_1>i_2>...>i_{t\plus{}1}$
(if such integers exist)
2-
$ <a_p,...,a_{i_1},a_{i_1\minus{}1},...,a_{i_2},...,a_{i_t},a_{i_t\minus{}1},...,a_1>\equal{}$
$ <i_1,...,p,i_2,...,i_1\minus{}1,...,i_{t\minus{}1}\minus{}1,i_t\minus{}1,1,...,i_t\minus{}2>$
where $ i_l \ (1 \le l \le t\minus{}1)$ are $ \plus{}$  integers such that $ i_{l\plus{}1}\plus{}(i_{l}\minus{}1) | i_{l}\plus{}(i_{l\minus{}1}\minus{}1)$ and also $ 2(i_t\minus{}1)|i_{t\minus{}1}\minus{}1\plus{}i_t$ and also $ p\plus{}1\equal{}i_0>i_1>i_2>...>i_t$
(if such integers exist).
but i don't think that this is the real solution and i'm sure that there should be sth about $ i_l$(some important properties maybe...)
could u post your solution for this part or at least give me a small hint :) ?
thx in advance for your reply  :)

I think shoki almost solved the problem. Now I shall explain my proof for the general case where $ a_1,a_2,\cdots,a_n$ is a permutation of numbers $ 1,2,\cdots,n$ ($ n$ could be $ \infty$ to cover the part(b)) with $ i\plus{}a_i | j\plus{}a_j$ for any $ 1\leq i<j \leq n$. My approach is the same as shoki's last post, but there is small difference: shoki's approach is top-down while mine is bottom-up!. In shoki's approach we start from the last block of numbers (namely $ \{a_i,...,a_{p}\}$) and determine the numbers in that block (i.e., proving that $ \{a_i,...,a_{p}\} \equal{} \{p,p \minus{} 1,...,i\}$) and then we move over to the remaining block ($ \{a_1,...,a_{i \minus{} 1}\} \equal{} \{1,2,...,i \minus{} 1\}$) and repeating the argument over this set for $ n\equal{}i\minus{}1$. That sounds like a recursive. The only problem is, for the numbers in this set, we get an extra condition (in addition to $ i\plus{}a_i | j\plus{}a_j$ ) of $ a_{i\minus{}1}\plus{}i\minus{}1 | p\plus{}i$ which somehow breaks the recursion. Usually in these cases, where the recursion is broken due to dependency (like $ a_{i\minus{}1}\plus{}i\minus{}1 | p\plus{}i$) of the smaller problem ($ n\equal{}i\minus{}1$ in this case) to the main problem, we have two choices: add this dependency to the original problem and apply the recursion to this new stronger problem (i.e., we can add the extra condition of $ n\plus{}a_n|M$ to the main problem where $ M$ is an arbitrary integer), or we can take the bottom up approach instead. That's what I did, however it gives a similar sequence of divisibility as shoki stated. Lets start from the very beginning, namely $ a_1$ (instead of the last number $ a_n$). Visualize the axis of integer numbers with $ n$ points $ 1,2,\cdots,n$ (arrange from left to right) it. In the beginning we are on the point $ a_1$ and in each step we go from the current number $ a_i$ to the next number $ a_{i\plus{}1}$ and draw an arrow from the former point to the latter point to indicate the movement. We do these movements until we arrive at the point $ a_n$ which is the end. There is an interesting thing about these steps: let $ b_i\equal{}i\plus{}a_i$ ($ i\equal{}1,2\cdots,n$), then suppose our current number is $ a_i$ and we are about to jump to $ a_{i\plus{}1}$. Since we have $ b_i | b_{i\plus{}1}$, we get two possibilities:
1) $ b_{i\plus{}1} \geq 2b_i$: in this case $ a_{i\plus{}1} > a_i$, i.e., on the axis we have a jump to the right of $ a_i$. Call this a big-jump.
2) $ b_{i\plus{}1}\equal{}b_i$: in this case $ a_{i\plus{}1}\equal{}a_i\minus{}1$. On the axis it means moving to the left neighbor point. Call this a left-jump.

Fig 1 shows a general scheme of how this movements may look like on the integer axis. I partition the movements to some blocks of successive left-jumps (a block may happen to be empty), separated by a right-jump from each other. the blocks of left-jumps are shown in Fig 1. for example the block 1 and block 2 are separated by the big-jump from the point $ a'$ to $ a_{j_1}$ and block 3 is an empty block ...

Since $ a_1,\cdots,a_n$ is a permutation of $ 1,2,\cdots,n$, so any number from $ 1,2,\cdots,n$ must be covered exactly once during our movements. In the beginning we are over the point $ a_1$. I claim that we have to take left-jump all the way down to the point $ 1$, before we can take a big-jump. Suppose in contrary, after some (maybe zero) left-jumps starting from $ a_1$, we take a big-jump from the point $ 1<a\leq a_1$ to another point $ a'$ which must be to the right of $ a_1$ . Then the point $ 1$ shall never be covered in the future movements, unless we cover the point $ a_1$ again which is contrary to the assumption of permutation. The reason is that our backward movements (i.e., to the left) is one unit, and so any series of movements which takes us from the point $ a'$ (in the right of $ a_1$) to $ 1$ (in the left of $ a_1$) has to step over $ a_1$ (why?). So the first block of left-jumps shall be $ a_1,a_1\minus{}1,\cdots,1$. Now suppose from point $ 1$ we take a big-jump to the point $ a_{j_1}$. The same argument for the point $ a_{j_1}$ instead of $ a_1$ shows: a series of successive left-jump starting from this point has to be done all the way down to the point $ a_1\plus{}1$ before we do the next big-jump (why?). In general, let $ a_{j_t}$ be the landing point of the $ t$-th big-jump (for convenience set $ j_0\equal{}1$). Then we must left-jump all the way down to $ a_{j_{t\minus{}1}}\plus{}1$ before we do the next (if any) big-jump. Fig 2 shows the general scheme when we have four big jumps. This figure essentially tell anything about the structure of such permutation $ a_i$. All we need to know (to determine the structure uniquely) is the values of $ a_{j_t}$, which I denote by $ k_t$, for $ t\equal{}0,1,2\cdots,m$ where $ n\geq m \geq 0$ is the number of big-jumps. Fix a value for $ m$, for $ t\equal{}0,1,2\cdots,m$ let $ c_t\equal{}b_{j_t}$. $ c_t$s are kind of free variables, however there are two restrictions for them: $ 1< c_0 < c_1 < \cdots < c_m$ and $ c_0 | c_1 | \cdots | c_m$. Call such a vector $ (c_0,c_1,\cdots,c_t)$ of integers with these two restrictions a division-vector. Now one can see it is easy to express $ k_t$s in terms of $ c_t$s. Based on Fig 2 we shall get $ k_0\equal{}c_0\minus{}1, k_1\equal{}c_1\minus{}c_0$ and $ k_t\equal{}k_{t\minus{}2}\plus{}c_t\minus{}c_{t\minus{}1}$, so one easily gets: $ k_t \plus{} \epsilon(t) \equal{} \sum_{q\equal{}0}^t (\minus{}1)^{q}c_{t\minus{}q}$ for $ t\equal{}0,1,\cdots,m$ where $ \epsilon(t)$ is equal to $ 1$ if $ t$ is even and $ 0$ otherwise. In other words, the term $ k_t \plus{} \epsilon(t)$ equals to the "alternative summation" of the division vector $ (c_0,\cdots,c_t)$. For the infinite case (when $ n\equal{}\infty$), any infinite vector $ (c_0,c_1,\cdots)$ will uniquely determines the values of $ k_t$ which in turn, determines the whole permutation uniquely. For the finite case, we have one more restriction of $ k_m\equal{}n$ which grantees the size $ n$ for our permutation. So the problem has come down to identify the division vectors $ (c_0,\cdots,c_m)$ with their alternative summation equal to $ N\equal{}n\plus{}\epsilon(m)$, or

$ N\equal{}n \plus{} \epsilon(m) \equal{} \sum_{q\equal{}0}^m (\minus{}1)^{q}c_{m\minus{}q} (1)$

Well at least we can find an easy way to generate all possible values of the division vector $ c_t$s. Thanks to the fact that $ c_i | c_j$ for $ i<j$, we conclude that $ c_0\geq 2$ is a divisor of $ N\equal{}n\plus{}\epsilon(m)$. By dividing both sides of (1) by $ c_0$ we get a similar expression with $ m\minus{}1$ summands (instead of $ m$):

$ N/c_0\minus{}(\minus{}1)^m \equal{} \sum_{q\equal{}0}^{m\minus{}1}(\minus{}1)^qc_{m\minus{}q}/c_0$

In fact the above expression induces the length $ m\minus{}1$ division vector $ (c_1/c_0,\cdots,c_m/c_0)$ with its alternative sum equal to $ N/c_0\minus{}(\minus{}1)^m$. Based on this recursive relation we can come up with the following approach to generate all possible vectors $ (d_0,d_1,\cdots,d_m)$ where $ d_i\equal{}c_i/c_{i\minus{}1}$ for $ m\geq i > 0$ and $ d_0\equal{}c_0$ for a division vector $ (c_0,\cdots,c_m)$ with the alternative summation equal to $ N$: First decide whether $ m$ should be odd or even, so we get a fixed value $ \epsilon$ for $ \epsilon(m)$. Then in step $ i\equal{}0,1,\cdots$, if $ N\equal{}0$ then stop here. Otherwise let $ d_i\geq 2$ be an arbitrary divisor of $ N$. Replace $ N$ with the new value of $ N/d\plus{}(\minus{}1)^{i\plus{}\epsilon}$, if this new value is greater than $ 1$. Then goto step $ i\plus{}1$.
At the end of the above algorithm we get a vector $ (d_0,\cdots,d_m)$ for some $ m$ and we accept it only if $ m$ has the same parity as we decided at the beginning of the algorithm. By choosing all possible divisors in each step (and all possible parities for $ m$), we can produce all possible vectors $ d_i$ which determine the vectors $ c_i$ which lead to all desired permutations.
Anyway, to make this answer more mathematics-friendly, we may pose some restrictions on $ n$ which limits the choices of division-vectors with their alternative sum equal to $ n$. That's an interesting problem on its own and it leads to some in-depth study of the division vectors. For example, it is interesting know that which integers show up in some division-vector with its alternative sum equal to $ n$. many other questions can be asked, I'll write more about the problems related to it.


A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is 13 less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is 27 more than the mean of $\mathcal{S}.$  Find the mean of $\mathcal{S}.$

[hide="Solution"]
Denote the sum of the members of $\mathcal{S}$ by $s$.  Denote the number of members of $\mathcal{S}$ by $n$.  Then we have $\frac{s+1}{n+1} = \frac{s}{n} - 13$ and $\frac{s+2001}{n+1} = \frac{s}{n} + 27$.

Multiplying both equations by $n(n+1)$ gives us $n(s+1) = s(n+1) - 13n(n+1)$ and $n(s+2001) = s(n+1) + 27n(n+1)$.  Subtracting the equations gives $2000n = 40n(n+1) \implies 50 = n+1 \implies n = 49$.  Plugging back in to the first equation, we have $49s + 49 = 50s - 13 \times 49 \times 50 \implies s = 49 + 13 \times 49 \times 50$.  This gives us $\frac{s}{n} = 1 + 13 \times 50 = \boxed{651}$.
[/hide]

[hide="Answer"]We have $x$ distinct numbers in $S$. Therefore, $\frac{S+1}{x+1}+13=\frac{S}{x}\Rightarrow S=x(13x+14)$ and $\frac{S+2001}{x+1}-27=\frac{S}{x}$, so $\frac{S+1}{x+1}+13=\frac{S+2001}{x+1}-27\Rightarrow x=49$. From there, we have $\frac{S+1}{50}+13=\frac{S}{49}\Rightarrow S=31899\Rightarrow \frac{S}{x}=\boxed{651}$.[/hide]

[quote=carlos8]Sorry for the 12 yr bump lol, but why in this problem don't we have to consider the case where S contains 1 and/or 2001?[/quote]

Because then $S \bigcup \{1\}$ would just be $S$, so the mean couldn't be $13$ less than the mean of $S$.

[hide=Solution]Let $\mathcal{S}$ have sum of elements $a$ and amount of elements $b$. Then $\frac{a+1}{b+1}=\frac{a}{b}-13$ and $\frac{a+2001}{b+1}=\frac{a}{b}+27$. Subtracting gives $\frac{2000}{b+1}=40$, so $b=49$. This means $\frac{a+1}{50}=\frac{a}{49}-13$, or $49(a+1)=50a-13 \cdot 2450 \implies a = 31899$. Dividing gives $\boxed{651}$.[/hide]


probability
Alcumus
conditional probability

[quote="jonathanchou711"]Two boxes are each filled with colored socks. Box A has 2 red socks and 4 blue socks. Box B has 4 red socks and 4 blue socks. A box is chosen at random, and a sock is drawn from the box. Given that the sock chosen is blue, what is the probability that it came from box A?[/quote]
There is a $ \frac {1}{2}$ probability that box A is chosen. From there, there's a $ \frac {4}{6} \equal{} \frac {2}{3}$ chance that a blue sock is picked from box A. Also, there's a $ \frac {1}{2}$ chance of picking Box B, and of picking a blue sock from Box B. Total chance for picking a blue sock=$ \frac {1}{4} \plus{} \frac {1}{3} \equal{} \frac {7}{12}$.

Probability for picking a blue sock from box A=$ \frac {\frac {1}{3}}{\frac {7}{12}} \equal{} \boxed{\frac {4}{7}}$.

I'm not sure if I'm right though...

[quote="xpmath"]I belive that is correct.

This problem is an example of [url=http://en.wikipedia.org/wiki/Conditional_probability]Conditional Probability[/url].

For this problem, you're attempting to find the probability that you drew a blue ball from box A, knowing that you drew a blue ball.

In general, for conditional probability problems, you wish to find the probability that an event A occurs knowing that event B has already happened. This is written as $ P(A|B)$ and is found by using the well known formula, $ P(A|B) \equal{} \dfrac{P(A \text{ and } B)}{P(B)}$. In other words, you find the probability that both events occur and divide by the probability that the second event occurs.

Sorry for the awful explanation; conditional probability is somewhat hard to understand.[/quote]
It's ok, I got it though. Intermediate C and P explains this pretty well though...

Thanks guys!! I got 4 points on Alcumus from this!!!! w00t!!


$ (1,1)$ is obvious solution, so product is $ 1$>

$ x\plus{}y\equal{}\frac{1}{x}\plus{}\frac{1}{y}$

$ x^2y\plus{}xy^2\equal{}x\plus{}y$

$ x^2y\minus{}x\plus{}xy^2\minus{}y\equal{}0$

$ x(xy\minus{}1)\plus{}y(xy\minus{}1)\equal{}0$

$ (x\plus{}y)(xy\minus{}1)\equal{}0$

Thus either $ x\equal{}\minus{}y$ or $ xy\equal{}1$. As x is positive, y must be negative for the first case to work. This is a contradiction. Thus $ xy\equal{}1$

You forgot to mention that x=-y is OK with x=y=0 but then you would divide by 0 which you can't. Also, is that your method of doing FTW problems in such a way that takes 30 seconds, and not 1.5 or is it higher post count-mania?

AIME15's solution works, but I was showing why it always works.

Also, whether or not one can divide by 0 is irrelevant. It says the numbers must both be positive.

[quote="GameBot"]The sum of two positive numbers is equal to the sum of the reciprocals of the same two numbers. What is the product of these two numbers?[/quote]

Easier Solution: $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x+y}{xy} = x+y \implies xy = 1$.


inequalities
Euler
inequalities unsolved

Let $ a$, $ b$, $ c$ be positive reals. Prove the inequality
$ \left(a \plus{} \frac {1}{b} \minus{} 1\right)\left(b \plus{} \frac {1}{c} \minus{} 1\right) \plus{} \left(b \plus{} \frac {1}{c} \minus{} 1\right)\left(c \plus{} \frac {1}{a} \minus{} 1\right)$
$ \plus{} \left(c \plus{} \frac {1}{a} \minus{} 1\right)\left(a \plus{} \frac {1}{b} \minus{} 1\right)\ge 3$.

Your puzzle is amazing. It looks like one of the IMO test but I think it is more difficult than it. Well, frankly, I don't believe yours is right and I am looking for a counterexample. Why?
Because :
 I suppose that $ c \geq b \geq a $
$ abcLHS-abcRHS=(a^2b^2c^2+2ab^2c-a^2b^2c+2a^2bc+2ab-a^2b-a^2bc^2-7abc+2abc^2+2bc+2ac+1-a-ac^2-c-ab^2c^2-b^2c-b); $
(Hope I caculate right)
$ L(a,b,c)=(a^2b^2c^2+2ab^2c-a^2b^2c+2a^2bc+2ab-a^2b-a^2bc^2-7abc+2abc^2+2bc+2ac+1-a-ac^2-c-ab^2c^2-b^2c-b) $
$ L(a,b,c) $ looks like a monster and is very weird . So I use the trick like $ c=a+x+y $, $ b=a+x $
and the expand $ L(a,b,c) $. Oh my godness! It looks more weird and seem can be less than zero. Here it is $ L(a,a+x,a+x+y)=1-ax^2y^2-3a-2x-y+8ax+2x^2+4ay+2xy+8a^3y+14a^2x^2+2a^2y^2+4ax^3+4a^5x+16a^3x+2a^5y+6a^4x^2+a^4y^2+4a^3x^3+a^2x^4-20a^2x-10a^2y-11ax^2-x^2y-10a^4x-5a^4y-12a^3x^2-6a^2x^3-ay^2-2a^3y^2+14a^2xy+6ax^2y+2axy^2+6a^4xy+6a^3x^2y+2a^3xy^2+2a^2x^3y+a^2x^2y^2-11axy+6a^4+6a^2+a^6-10a^3-x^3-3a^5-12a^3xy-9a^2x^2y-3a^2xy^2-2ax^3y-ax^4 $
(Very brutal caculation,isn't it ?). And I am hopeless at that point.
So... I think it is maybe wrong in the puzzle itself. Or am I wrong?)

[quote="euler_vn"]Let $a,b,c>0$ .Prove that:
$\(a+\frac{1}{b}-1)(b+\frac{1}{c}-1)+(b+\frac{1}{c}-1)(c+\frac{1}{a}-1)+(c+\frac{1}{a}-1)(a+\frac{1}{b}-1)\ge 3$[/quote]
I saw a solution in http://diendantoanhoc.net/forum
Euler has post this problem there.
This is :
Two of $\{(a+\frac{1}{b}),(b+\frac{1}{c}),(c+\frac{1}{a})\}$ is lager or smaller than 2.

If $a+\frac{1}{b}$ and  $b+\frac{1}{c}$

So $\{a+\frac{1}{b}-2)(b+\frac{1}{c}-2) \geq 0$

And $(b+\frac{1}{c})(c+\frac{1}{a})+(c+\frac{1}{a})(a+\frac{1}{b}) \ge (c+\frac{1}{a})+4$

The problem is correct. See http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 (rather indirectly - see post #7 and the subsequent ones of this topic).

  darij

This problem is proposed by Vasc and there are 2 solution in Old and New ineq  ;)


Determine all values of $x$ in the interval $0 \leq x \leq 2\pi$ which satisfy the inequality \[ 2 \cos{x} \leq \sqrt{1+\sin{2x}}-\sqrt{1-\sin{2x}} \leq \sqrt{2}.  \]

Note that $ \sqrt{1+sin2x}-\sqrt{1-sin2x}\leq\sqrt{2}$ is always true for all real x since $ \sqrt{1-sin2x}\geq\ 0$ and $ \sqrt{1+sin2z}\leq\sqrt{2}$
From $ 2cosx\leq\sqrt{2}$
$ \frac{\pi}{4}\leq\ x\leq\frac{7\pi}{4}$
Now consider $ 2cosx\leq\sqrt{1+sin2x}-\sqrt{1-sin2x}$
$ 4cos^{2}x\leq\ (1+sin2x)+(1-sin2x)-2\sqrt{1-sin^{2}2x}$
$ 4cos^{2}x\leq\ 2-2\sqrt{1-sin^{2}2x}$
$ \sqrt{1-sin^{2}2x}\leq\ 1-2cos^{2}x$
$ 1-sin^{2}2x\leq\ 1-4cos^{2}x+4cos^{4}x$
$ 1-4sin^{2}xcos^{2}x\leq\ 1-4cos^{2}x+4cos^{4}x$
$ 4cos^{4}x+4sin^{2}xcos^{2}x-4cos^{2}x\geq\ 0$
$ 4cos^{2}x(cos^{2}x+sin^{2}x-1)\geq\ 0$
But $ cos^{2}x+sin^{2}x-1=0$, so,
$ 0\geq\ 0$
So it means that all x such that $ \frac{\pi}{4}\leq\ x\leq\frac{7\pi}{4}$ works

From $2\cos x\leq\sqrt{2}\iff\cos x\leq\tfrac{\sqrt{2}}{2}$, we find that $x\in[\tfrac{\pi}{4},\tfrac{7\pi}{4}]$ is necessary, and $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|\leq\sqrt{2}$ holds for all $x$, since the absolute difference between $\sqrt{1+\sin2x}$ and $\sqrt{1-\sin2x}$ attains its max. when $|\sin2x|$ attains its maximum of $1$, which yields $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|=\sqrt{2}$. Henceforth disregard the upper bound of $\sqrt{2}$.

As for $2\cos x\leq|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$, assume that $2\cos x$ is nonnegative, because otherwise the inequality clearly holds since $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ is nonnegative. By this assumption, both sides are nonnegative, so we can square both sides of our inequality to obtain
$$4\cos^2x\leq(1+\sin2x)+(1-\sin2x)-2\sqrt{(1+\sin2x)(1-\sin2x)}=(1+1)-2\sqrt{1-\sin^22x}=2-2\cos2x=2-2(2\cos^2x-1)=4-4\cos^2x$$
by the double-angle formula. Rearranging, $8\cos^2x\leq4\iff\cos x\leq\tfrac{\sqrt{2}}{2}$ (and $\geq0$). Combining this with the $\cos x<0$ solutions yields $\boxed{x\in[\tfrac{\pi}{4},\tfrac{7\pi}{4}]}$, which is precisely the interval from the $2\cos x\leq\sqrt{2}$ condition. Hence, this is the requested answer. $\blacksquare$

My work: $E=\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}=\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}-\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x}=$
$=\left| \sin x+\cos x \right|-\left| \sin x-\cos x \right|=\sqrt{2}\left( \left| \cos \left( \frac{\pi }{4}-x \right) \right|-\left| \sin \left( \frac{\pi }{4}-x \right) \right| \right)$
So ${{E}^{2}}=2\left( 1-\left| \sin \left( \frac{\pi }{4}-x \right)\cos \left( \frac{\pi }{4}-x \right) \right| \right)$ thus ${{E}^{2}}=2-\left| \sin \left( \frac{\pi }{2}-2x \right) \right|=2-\left| \cos 2x \right|\le 2\Rightarrow E\le \sqrt{2},\forall x\in \left[ 0,2\pi  \right]$
On the other hand, we have $4{{\cos }^{2}}x\le 2-\left| \cos 2x \right|\Leftrightarrow 2\cos 2x\le -\left| \cos 2x \right|\Rightarrow \cos 2x<0$

Algebrical solution:
Squaring all part, we get $4{{\cos }^{2}}x\le 2-2\sqrt{1-{{\sin }^{2}}2x}\le 2$ or $4{{\cos }^{2}}x\le 2-2\left| \cos 2x \right|\le 2$ and the rest is easy


function
search
arithmetic series
algebra
system of equations

[1] In an arithmetic series, the terms of the series are equally spread out. For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?

[2] If the function f satisfies the equation f(x + y) = f(x) + f(y) for every pair of real numbers x and y, what are the possible values of f(0)?

[hide="Number 1"]
Let the next term be $ 3 \plus{} k$, and so on. Then $ 3 \plus{} nk \equal{} 136\implies nk \equal{} 133$.

$ \sum_{j \equal{} 0}^{n}3 \plus{} kj \equal{} 3(n \plus{} 1) \plus{} \frac {kn(n \plus{} 1)}{2} \equal{} 1390$.

Solving the system of equations, we find that $ k \equal{} 7$, so the first three terms are $ \boxed{3, 10, 17}$.
[/hide]

[hide="Number 2"]
Put $ y \equal{} 0$ to get $ f(x) \equal{} f(x) \plus{} f(0)\implies \boxed{f(0) \equal{} 0}$.
[/hide]

Edit: Sorry, I meant $ 17$, fixed it.

[hide="#1"]The sum of an arithmetic series is $ \frac{n(a_1 \plus{} a_n)}{2}$, where there are $ n$ terms, and $ a_1$ is the first term and $ a_n$ is the last. Knowing this,

$ \frac{n(3 \plus{} 136)}{2} \equal{} 1390$

$ n \equal{} 20$

We also know that $ a_n \equal{} a_1 \plus{} (n\minus{}1)d$, so

$ a_{20} \equal{} 3 \plus{} 19d \equal{} 136$

$ d \equal{} 7$

Therefore, the first 3 terms are $ \boxed{3, 10, 17}$.[/hide]

[hide="#2"]Plug in $ x\equal{}y\equal{}0$.

$ f(0 \plus{} 0) \equal{} f(0) \plus{} f(0)$

$ f(0) \equal{} 2f(0)$

$ \boxed{f(0) \equal{} 0}$[/hide]

At the part f(0)=2f(0), how did you know to subtract instead of divide?
Whenever subtraction and division are a possibility do you always subtract instead?
Dividing by f(0) we get 1=2.
Is this because this is like dividing by 0?

You have $ f(0) \equal{} 2f(0)$.  You now have two cases:

Case 1: $ f(0) \neq 0$.  In this case, we can divide by $ f(0)$ to get $ 1 \equal{} 2$, which is absurd.  Thus, this case cannot happen.

Case 2: $ f(0) \equal{} 0$.  In this case, all conditions are satisfied, so this is okay.

(We still have to verify that there actually is a function $ f$ that satisfies the given equation such that $ f(0) \equal{} 0$ -- alternatively, there could be no such functions that satisfy the equation (but that does not happen in this case).)

In other words, before you divide, you must check that the thing you're dividing by isn't 0.

Yes, you must subtract in case that value is 0 because dividing by 0 will make the equation undefined, like in this problem.

[quote="MrMagician"]Thank you guys very much.

For problems like number 2 I have a great amount of trouble of even understanding where to start, and even comprehending what that question is asking. For example having the idea to let x=0, or let x and y both = 0 is not at all obvious to me. "For every pair of real number x and y..." Questions like this seem like a great amount of work is going to be involved, but the solution is so quick and simple. Do you know where I could get practice problems like these (any websites) or do you guys have any advice? Thanks.[/quote]

Search the forums for "functional equations" and look on Google. AoPS vol 2 has a chapter on them as well.

The idea here was that if the equation holds for all real $ x$ and $ y$, you can freely substitute in values for $ x$ and $ y$. We were looking for $ f(0)$, so putting one of $ x$ or $ y$ as $ 0$ seemed like a good idea.

(also, I meant $ 17$ in the earlier post, not $ 13$. Typo.)


inequalities
function
integration
symmetry
real analysis
real analysis unsolved

Show that for all continuous differentiable functions $ f \in C^1([\minus{}\pi,\pi])$ we have

$ \left|\int\limits_{\minus{}\pi}^\pi (\cos(x)f(x) \minus{} \sin(x)f'(x))\, dx \right| \leq \sqrt{2\pi}\left(\int\limits_{\minus{}\pi}^\pi (|f(x)|^2 \plus{} |f'(x)|^2)\, dx\right)^{\frac12}$.

Claim: $ \langle f,g\rangle \equal{} \int_{\minus{}\pi}^{\pi} (f(x)\overline{g(x)}\plus{}f'(x) \overline{g'(x)}) \, dx$ is an inner product on $ C^1[\minus{}\pi, \pi]$

Proof: Symmetry and linearity are clear. Positive definiteness follows since $ \langle f,f \rangle \equal{} \int_{\minus{}\pi}^{\pi} (|f(x)|^2\plus{}|f'(x)|^2) \, dx \geq \int_{\minus{}\pi}^{\pi} |f(x)|^2 \, dx \geq 0$ iff $ f \equal{} 0$

The result then follows by Cauchy-Schwarz


inequalities
logarithms
inequalities proposed

For positive reals $ a_{i}$'s and $ b_{i}$'s, let $ a_{1}\geq\cdots\geq a_{n}$ and $ b_{1}\geq\cdots\geq b_{n}$. And let $ \{ b_{1},\cdots,b_{n}\}\equal{}\{ c_{1},\cdots,c_{n}\}$. Prove that
\[ (a_{1}\plus{}c_{1})\cdots(a_{n}\plus{}c_{n})\geq (a_{1}\plus{}b_{1})\cdots(a_{n}\plus{}b_{n}).\]

We have that

$ (a_{1}\plus{}b_{1},a_{2}\plus{}b_{2},\ldots,a_{n}\plus{}b_{n})\succ (a_{1}\plus{}c_{1},a_{2}\plus{}c_{2},\ldots,a_{n}\plus{}c_{n})$,

thus, applying Karamata inequality to $ f(x)\equal{}\ln x$, we obtain that

$ (a_{1}\plus{}b_{1})\cdots (a_{n}\plus{}b_{n})\leq (a_{1}\plus{}c_{1})\cdots (a_{n}\plus{}c_{n})$.

Exactly. Moreover, this holds for more bunch of variables:
$ (a_{11}\plus{}a_{21}\plus{}\cdots\plus{}a_{n1})\cdots(a_{1m}\plus{}\cdots\plus{}a_{nm})$
$ \leq (a_{1 w_{1}(1)}\plus{}\cdots\plus{}a_{n w_{n}(1)})\cdots (a_{1 w_{1}(m)}\plus{}\cdots\plus{}a_{n w_{n}(m)})$. and like Chevishev's inequality, if we add all combinations, we get another inequality.


trigonometry
MATHCOUNTS
calculus
geometry
summer program
MathPath
abstract algebra

My school's starts sometime in October...

I think our school round is in December/January.

There's too much competition at my school, so I probably won't get into the team.  ~4 people are 2 years ahead of the regular curriculum(trig), and 5 (edit:6) other people and I skip one grade are in Geom/Alg2...

Obviously the people in Trig are going to dominate at Mathcounts.

Edit: When did this split from the MC thread?

My school started last week or something, and I don't know when the competitions are.

Trig in middle school? No fair. I'm only on Alg 2. :wallbash: :blush: ;)

Hope to see some of you at Nationals... (i.e. I hope to get to Nationals and if I don't I'll cry)

[quote="ccy"]4 people are 2 years ahead of the regular curriculum(trig)

Obviously the people in Trig are going to dominate at Mathcounts.[/quote]

Wow I'm only at Alg I
I have no clue when my school's mathcounts season starts.

[quote="star99"]Trig in middle school?! Wow[/quote]

That's not that surprising. Many people at my school have already learned trig on their own. Although many people are good at math, they don't care about it. (I'm not one of those people  :P )

A surprising thing is that nr1337 learned Calculus BC in [b]7th grade[/b]  :o

My school's mathcounts season started on 9/14. The highest course that my school has is geometry, though it's possible to take Alg2 if you took alg 1 in 6th grade. But the geometry students don't necessarily dominate the team. Last year, the best four people were 2 geometry students, 1 Alg2 student, and 1 Alg1 student.

[quote="Drunken_Math"]A surprising thing is that nr1337 learned Calculus BC in [b]7th grade[/b]  :o[/quote]

That's extremely rare... I knew two guys who did that, and I was in a very advanced program. They had both done Statistics by 8th grade, and one guy was done with all the math that a highly gifted program in high school had to offer to him by 11th grade. The other guy went to an early-entrance college program after 8th grade. O_o

I did Calc AB in 8th grade, but didn't really do much of anything, so I did it again in 9th. :oops:

Trig is nice; I learned it a few years ago, but it didn't really give me a huge advantage.

Speaking of people who know calc, I believe I remember Treething liking calculus in the 7th grade too.

Wow. I'm scared and impressed. I'm pretty good in trig (learned by self) but I never comprehended calculus.

Everyone else from Mathpath seems so smart... and I feel dumb now. :blush:

Don't feel dumb. Most kids do Algebra 1 in 9th grade. Also, note that going so far ahead doesn't really help in MC. Sirorange was at the normal math level and got 14th in the country!

Oh yeah, and to answer the question, we start MC about January (explains my 187th place at nats :P )

eh...
My school isn't very math-loving.
As in, we only started doing mathcounts last year.
And,we only did chapter and state competitions, with only 4 participants (no individuals)
This year, we might do minicomps (our school isn't hosting one).
If we [u]do[/u] do them, then our first one will be October 1st.

And, just to go along with the conversation, I'm doing Geometry in 8th.
 :|

[quote="biffanddoc"]Don't feel dumb. Most kids do Algebra 1 in 9th grade. [/quote]

That sounds like what goes on here, but i feel like going on to trig now.  :P

Believe me, you don't need to be in a higher math level to succeed in MATHCOUNTS. Don't worry about it at all.

I never had a class in anything above Geometry before 8th grade, and I think I did pretty well in MathCounts.

You can worry about catching up in high school, believe me. In spite of not having had anything above Geometry before then, in my three years of high school I was able to master Trig, Calc, Multivariable Calc, Diff Eq, Linear Algebra, and learn fair amounts of Abstract Algebra, Analysis, and Topology. There's plenty of time.

I wasn't moping about Mathcounts. All you need in Mathcounts is Alg 1 and Geometry.

I didn't even make it to nationals. :wallbash:

To sum up this whole topic: It helps to know things beyond Algebra 1, and beyond basic geometry, but it's not necessary.

Don't forget that Pre-Algebra does too according to Jesusfreak197...even though I know very few people who haven't skipped it...and I have no idea what you learn in Pre-Algebra... :lol:

----------
You have reached the offices of the Supreme Lord Voldemort. His Exquisite Evilness is in a meeting with his Death Eaters, crafting deviously sinister plans for the demise of one Harold Potter. Please try again after the All Powerful Dark Master has conquered the Earth. Thank you and have a nice day.

I am in 8th grade, and currently taking Algebra II/trig. Last year I skipped Geometry and most of Algebra I and learned Algebra II on my own . Hopefully I'll be able to test out of Alg II soon, and if I do I will take Math Analysis = Pre-Calc.

[quote="JesusFreak197"]You need to know Pre-Algebra to learn Algebra. It's basically simple solving for variables, like $x+12=19$, so $x=?$.[/quote]

:|
Oh.
Thank you.
I always thought that was part of like third grade math...or maybe that waas back in china.
btw you are not a geek because Ivan can solve the rubix cube in 15 seconds, which is faster, like I commented on you xanga.

[quote="biffanddoc"][quote="JesusFreak197"]You need to know Pre-Algebra to learn Algebra. It's basically simple solving for variables, like $x+12=19$, so $x=?$.[/quote]
For our pre-alg it was a little of basic algebra and geometry and counting/probability[/quote]

Which part is Geometry? :D

[quote="biffanddoc"]We learned basic 2-d and 3-d geometry such as finding the area/surface area of a cone[/quote]
Oh I see..so you took Pre-Algebra?  And how far in MC are you?

[quote="biffanddoc"]I made nats lats year (from Delaware means nothing) and was a poor 184th there. So that wasn't really good (did you beat 24 at states last year? That's how bad Delaware is :( )

Im doing ALG II almost done then i do trig :)[/quote]

I didn't get to nats.
I am from the Golden State.
*hint hint Arnold Schwarzenegger-whateever-his-name-is*
I THINK I was 29th at State last year.
I don't really know though.
Me is doing Alg 2 too.
Trig seems hard. :D
Maybe that's because I am stupid. :ninja:

You're not stupid. 29th in the state of Garibaldis and dog-face butterflies  :) is better than 4th written in DE (although i was first countdown :) )

[quote="biffanddoc"][quote="Altariel"]What's a Garibaldi?
I have never seen a dog-faced butterfly before.
So if I went to Delaware I would be smart?[/quote]
Garibaldi is Calfornia's state marine animal
California dog faced butterfly is California's state insect
If you came to Delaware you would be really smart[/quote]

But what is a Garibaldi?  Ok I am going off topic again.
But I don't know anything about California and I live there.
So I couldn't possibly be smarter than you if I lived in Delaware.
All I know is that it's the first state.
At least...I think it is.
Seems like no one has Yahoo Messenger nowadays.


If $ M\in \mathcal{M}_3(\mathbb{C})$, we have $ \chi_M(X)\equal{}X^3\minus{}\mbox{tr}(M)X^2\plus{}\frac{1}{2}(\mbox{tr}(M)^2\minus{}\mbox{tr}(M^2))X\minus{}\det(M)I_3$, where $ \chi_M$ is the characteristic polynomial.
Hence, by Cayley-Hamilton we have $ (AB)^3\equal{}\det(AB)I_3\equal{}\det(BA)I_3\equal{}(BA)^3$.


is alcumus just intro to counting and probability right now?

yes
the only topic currently in Alcumus is Intro to Counting and Probability
I believe the next topic to be added is Intro to Algebra

ya. Well, that's cause intro to prob..class has alcumus subjects, that we have to complete


Who here is taking the introductory course to counting and probability?  I am.

Thanks for letting us know, math92, 4everwise, and ... 4everwise !?


probability
combinatorics unsolved
combinatorics

Could any of you guys help me with this problem:

A box of 20 tubes contains 16 good and 4 defective tubes. What's the probability that the third tube is defective if one of the other two was found to be good and the other was found to be defective?

-Swapnil

18 tubes are remaining out of which 3 are defective.

So probablity of the third tube being defective is 1/6


floor function
combinatorics proposed
combinatorics

The positive integers from 1 to $n^2$ are placed arbitrarily on the $n^2$ squares of a $n\times n$ chessboard. Two squares are called [i]adjacent[/i] if they have a common side. Show that two opposite corner squares can be joined by a path of $2n-1$ adjacent squares so that the sum of the numbers placed on them is at least $\left\lfloor \frac{n^3} 2 \right\rfloor + n^2 - n + 1$.

[i]Radu Gologan[/i]

Consider the row $r$ having the largest sum of numbers written in its squares. Consider the path going from the NW corner - down on column $1$ until row $r$, then right on row $r$ until last column, then down on column $n$ until the SE corner, and the path going from the NE corner - down on column $n$ until row $r$, then left on row $r$ until first column, then down on column $1$ until the SW corner. The largest sum of numbers written on one of these paths will obey the given inequality.


limit
real analysis
real analysis unsolved

Here is what I have:

Let $x_{n}=\frac{f_{n}}{f_{n+1}}$. Then the two conditions read:

$x_{n}\left(1+\frac{x_{n}-x_{n-1}}{x_{n+1}-x_{n}}\right)\rightarrow a+b$,

and

$x_{n+1}x_{n}\frac{x_{n}-x_{n-1}}{x_{n+1}-x_{n}}\rightarrow ab$.

Let $l=\lim x_{n}$, assuming that the limit exists. We get, using the second limit, that:

$x_{n+1}\left(x_{n}\frac{x_{n}-x_{n-1}}{x_{n+1}-x_{n}}+x_{n}\right)-x_{n}x_{n+1}\rightarrow ab$, from which it follows that

$l(a+b)-l^{2}=ab$, in other words $l=a$ or $l=b$.


solve in positive integers
\[\left(\sqrt[3]{a}+\sqrt[3]{b}-1\right)^{2}\ =\ 49+20\sqrt[3]{6}\]

Posted and solved: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120693[/url]


calculus
integration
calculus computations

Hey, for one who has down the integral, do you think there is need to memorized the table of integrals?

What do you mean "down the integral"?  I wouldn't memorize a table of integrals if I could derive the integral instead, or if I had the table handy.


conics
projective geometry
geometry theorems
geometry

[b]Theorem.[/b] Let $ABCD$ be a cyclic (and not necessarily convex) quadrilateral (or, more generally, a quadrilateral inscribed in a conic). The line passing through $AB \cap CD$ and $AD \cap BC$ is the polar of $AC \cap BD$.
[b]Proof.[/b] Let $\omega$ be the circle or conic in the problem statement, and let $\ell$ be the line passing through $AB \cap CD$ and $AD \cap BC$. Let the tangents at $A$ and $C$ to $\omega$ intersect at $P$ and let the tangents at $B$ and $D$ to $\omega$ intersect at $Q$. One can easily see that $PQ$ is the polar of $AC \cap BD$ with respect to $\omega$. Now, by applying Pascal's theorem to the degenerate hexagon $AABCCD$ (where, for example, the line $AA$ is the tangent at $A$ to $\omega$) we get that $P \in \ell$. Similarly, applying Pascal's theorem for the degenerate hexagon $ABBCDD$ we get that $Q \in \ell$. Hence $PQ = \ell$, and we are done.

[quote="Arne"][b]Theorem.[/b] Let $ABCD$ be a cyclic (and not necessarily convex) quadrilateral (or, more generally, a quadrilateral inscribed in a conic). The line passing through $AB \cap CD$ and $AD \cap BC$ is the polar of $AC \cap BD$. [/quote]

The statement can be summarized as "the polar triangle is self-dual", and I think you don't need explicit reference to Pascal's theorem to prove it.  It is just a direct consequence of the fact that polarization gives a duality between points and lines.    It is also essentially the same as the construction with straightedge only (no compass) of the tangent lines from an external point to a conic.


induction
number theory
number theory unsolved

This problem is Rosen Elementary Number Theory 5th ed 3.6 92.
Using pi(2n)-pi(n)<nlog(4)/log(n) (where n>1), prove that pi(2n)=<nlog(64)/log(n).
pi(x) is the number of primes less than or equal to x.

It is straightforward induction with the case distinction $ n\equal{}4m$ and $ n\equal{}4m\plus{}2$


Hi all,

I noticed on the NYC math team website http://www.nycmathteam.com/ that results are out for the first two Senior A contests (as well as some of the B division, junior & soph-forsh scores -- use the sidebar on the right to download the .pdfs) and I was wondering how everyone likes the questions so far, if there were any of particular interest/difficulty/annoyance, etc.

well though i'm a big boy now and i don't take these contests anymore (;)), i get to see some of these questions when my AMC trainees ask me questions about them. aaand... i must say that i really liked the problem with the set of 6 elements where you're asked to pick triples and not repeat pairs among them. it was a damn quaint problem for a NYCIML contest :)

I'm not sure what to think of the adjective "quaint," but I'm glad you liked the problem :lol:.  I thought it had a pretty nice idea wrapped up in it.

For those who haven't seen it, the full question is F07A23 (contest 4, question 5, or at least that's what it was numbered in the last revision I saw):
$ S$ is a set with 6 elements. Gauss chooses subsets of $ S$ with 3 elements so that no two of the chosen subsets have more than 1 element in common. Compute the greatest number of such subsets that Gauss can choose.

The score sheets (updated for nearly every team; Bronx Science and a couple of private schools seem not to have submitted scores yes) indicate that just under half of participants got that one right.

well, quaint because it admits the solution below. my student asked me for a solid explanation that didn't require listing sets feverishly, so i told him this...

the hypothesis means that no pairs are repeated across different picked sets, and of course they're not repeated across the same set, so right off the bat it's clear we can pick at most $ \frac{\binom{6}{2}}{\binom{3}{2}} \equal{} 5$ sets, with equality iff every pair appears exactly once. but each element appears in 5 pairs, which means that each of the 5 sets (if we could pick that many) would contain each of the 6 elements--obviously absurd. so we can pick at most 4. now to the construction, which i think is even more elegant than the previous argument... notice that $ \binom{4}{2}\equal{}6$, and let your prospective sets be $ A,B,C,D$. then to each pair of these sets, assign a different one of your 6 elements and put that element in both of them. each of $ A,B,C,D$ is in 3 pairs, so each will get exactly 3 elements. that's your construction. :)

heh, takes longer to explain than to think out.

joel, did you write this problem? if you did, hats off to you

Yes, I did write it :D.  Actually, I wrote all of the Senior A this semester, hence my interest in it :).  Next semester, I think the Senior A is going to be part my work and part David Hankin's, and some of my questions may end up appearing on Junior or other contests.

[quote="pleurestique"]... which means that each of the 5 sets (if we could pick that many) would contain each of the 6 elements--obviously absurd. ...[/quote]
This isn't quite right -- the sets are triples, not pairs, so the contradiction you can derive is a little different than what you've actually written.  That's a very nice description of the construction, though.  My "official" solution was less nice than this.  It's definitely true that it takes more space to write down than most #5s do -- I was a little worried that meant it was harder than it should be, but it seems to have worked out ok.

you're right. the contradiction comes from the fact that each element will occur in an even number of pairs among the triples that we've chosen (2 for each triple it appears in), so there's no way for an element to occur in 5 pairs.

I have two questions to ask. The first one is: during the second junior meet a question was given concerning multiplicity of roots of a quadratic. The question seemed to be too ambiguous and Mr. Coccoros (spelling?) told us that the question wouldn't count toward the final tally of scores. This was okay at the time but later on when i checked the scores online someone seemed to have been given a perfect score for the contest, bizzare? My second qusetion is if you could supply a solution for one of the questions on the senior A contest which involved three correct answers the concept of drawing balls out of a container (sorry i'm not aware of the exact wording of the question).

1) Perhaps they gave all participants 1 instead of 0 on that question?  (This is just a guess, of course.)

2) Cocoros.

3) The Senior A question was contest 2, question 6:
F07A12 A jar holds $ g$ green and 17 blue marbles for some integer $ g$. Pascal reaches into the jar and selects three marbles at random, without replacement. Compute all three possible values of $ g$ so that the probability that Pascal chooses three green marbles is the same as the probability that Pascal chooses two green and one blue marble.
This was the hardest question on a relatively easy contest -- 28% of respondants answered it correctly.  Maybe we'll wait and see if anyone else wants to offer a solution before I do it?

Well, that didn't work.  Here's the "official" solution:
[hide]There are $ g(g \minus{} 1)(g \minus{} 2)$ ways that Pascal can choose three green marbles and $ 3 \cdot 17g(g \minus{} 1)$ ways Pascal can choose one blue and two green marbles.  The probabilities of these events are the same if and only if the number of ways to perform them is equal, so we must solve $ g(g \minus{} 1)(g \minus{} 2) \equal{} 51g(g \minus{} 1)$.  Clearly $ g \equal{} 0$ and $ g \equal{} 1$ are roots, leaving us with the equation $ g \minus{} 2 \equal{} 51$ or $ g \equal{} 53$.  (Note that if $ g \equal{} 0$ or $ g \equal{} 1$, the probabilities of the two events are both 0.)[/hide]


real analysis
function
real analysis unsolved

E is a subset of a set X.
"A MAXIMAL element of E need not be an UPPER BOUND for E"
//can someone give me an example??????

limsup En = {x: x in En for infinitely many n}
liminf En = {x: x in En for all but finitely many n}
//I'm so confused on these descriptions....actually, I've problem understanding the concept of limsup and liminf for sets... can someone help me with these?

Thanks alot!

1) Consider all pairs of real numbers $(x,y)$ ordered as follows: $(x,y)\le (x',y')$ if and only if $x\le x'$ and $y\le y'$. Let $E=\{(1,3),(2,1),(2,2)\}$. The elements $(1,3)$ and $(2,2)$ are maximal in this set (an element is maximal when there is no other one greater than it). However, neither of them is an upper bound for $E$. An upper bound would be an element that is greater than (or equal to) each element of $E$. For instance, $(4,4)$ and $(2,3)$ are upper bounds for $E$. (Neither belongs to $E$, of course). In fact, $(2,3)$ is the [b]least[/b] upper bound.

2) There is another way to put this definition: \[\limsup E_{n}= \bigcap_{m=1}^{\infty}\bigcup_{n=m}^{\infty}E_{n}\]  \[\liminf E_{n}= \bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}E_{n}\] I don't know if this is helpful, though. Let's consider the characteristic functions of these sets: \[\chi_{E}(x)=\begin{cases}1,\quad{\rm if }x\in E\\ 0,\quad{\rm if }x\notin E\end{cases}\] The characteristic function of $\limsup E_{n}$ is precisely $\limsup_{n\to\infty}\chi_{E_{n}}$, where the $\limsup$ is understood in the usual way. Same for $\liminf$.


inequalities
algebra
polynomial
trigonometry
real analysis
real analysis unsolved

let $ P$ be a monic poly, $ \deg P>1$. show that $ |P(x)|\ge 2$ for some $ x\in(\minus{}2,2)$.

The extremes here are scaled Chebyshev polynomials: $ P_n(x) \equal{} 2\cos\left(n\cos^{ \minus{} 1}\frac x2\right)$.


i know these inequalities for turning the prod of the variables to a sum or at least sum of their G_means , and these are usefull for cases that means are not usefull for inequalities that have prode $\geq$ sum. does anyone know anything else  :?:

$1.(a_1+b_1)(a_2+b_2)...(a_n+b_n) \geq (\sqrt[n]{a_1a_2...a_n}+\sqrt[n]{b_1b_2...b_n})^n$

$2$.for $,a_1,a_2,...,a_n \geq 1$ we have
$(1+a_1)(1+a_2)...(1+a_n) \geq \frac{2^n}{n+1}(1+a_1+a_2+...+a_n) $

hey guys , dont you even know one inequality?