geometry
algorithm
induction
combinatorics proposed
combinatorics

Two teams, $ A$ and $ B$, fight for a territory limited by a circumference.

$ A$ has $ n$ blue flags and $ B$ has $ n$ white flags ($ n\geq 2$, fixed). They play alternatively and $ A$ begins the game. Each team, in its turn, places one of his flags in a point of the circumference that has not been used in a previous play. Each flag, once placed, cannot be moved.

Once all $ 2n$ flags have been placed, territory is divided between the two teams. A point of the territory belongs to $ A$ if the closest flag to it is blue, and it belongs to $ B$ if the closest flag to it is white. If the closest blue flag to a point is at the same distance than the closest white flag to that point, the point is neutral (not from $ A$ nor from $ B$). A team wins the game is their points cover a greater area that that covered by the points of the other team. There is a draw if both cover equal areas.
Prove that, for every $ n$, team $ B$ has a winning strategy.

Let $ A_{1}, ... , A_{2n}$ the $ 2n$ points around the circunference. Color the A-points with $ A$ and the B-points with $ B$.
Let O the center of circunference and $ B_{1}, ... , B_{2}n$ the midpoints of the arcs $ A_{1}A_{2}, ... , A_{2n}A_{1}$, respectively.
See that if a point $ X$ is in the sector $ A_{i}OB_{i+i}$, this point will belong to the territory with same color of $ A_{i}$ (and if this point is in the sector $ B_{i}OA_{i+1}$, this point will belong to the territory with same color of $ A_{i+1}$).

The problem can be solved using induction.
To n = 2, is easy.
Suppose that for $ n\leq k$, team B has a winning strategy.
To prove that for $ n = k+1$ the proposition is valid, follow the same strategy to win of $ n = k$.
When the team A put your last flag, we have three cases:

1)Team A put the last flag between two A-flags:

Trivial case. A-territory remais the same, and the same for B-territory. So, B-team should only put your (k+1)-flag in any place.

2) Team A put the last flag between two B-flags ($ A_{i}$ and $ A_{i+1}$, wlog):

Let $ S_{a}$ the area of the total territory of $ A$ after the k-flag of $ B$ be used, and $ S_{b}$ the area of B-territory.
Let $ [XOY]$ the area of a sector $ XOY$.
Let M the point with the (k+1)-A-flag, and N the point with the (k+1)-B-flag.
We'll have that the area of A, after your (k+1)-flag used, will be $ S_{a}+\frac{[A_{i}OM]}{2}+\frac{[MOA_{i+1}]}{2}$.
Suppose that, wlog, ${ [A_{i}OM}]\geq [MOA_{i+1}]$.
Now, to win, B-team should put your flag in a point N, sufficiently closed to M and between A_i and M,
because your new area will be $ S_{b}-(\frac{[A_{i}OM]+[MOA_{i+1}]}{2})+([A_{i}OM]-[NOM])$
$ = S_{b}+\frac{([A_{i}OM]-[MOA_{i+1}])}{2}-[NOM]$.
$ [NOM]$ can be too little than we want, and the new area of A-territory will be $ S_{a}-\frac{([A_{i}OM]-[MOA_{i+1}])}{2}+[NOM]$.

3) Team A put your last flag between one A-flag and one B-flag ($ A_{i}$ and $ A_{i+1}$, respectively, wlog):

We'll use the same notations of case 2.
The new area of A will be $ S_{a}+\frac{[A_{i}OM]}{2}$ (it can be easily checked, look a picture).
To win, B-team should only put you (k+1)-flag in any place between $ A_{i}$ and $ M$, because the new areas of A and B will be $ S_{a}$ and $ S_{b}$, respectevely!

i think your solution isn't working.
after the last moves $ S_{a}$ will be more then $ S_{a}\plus{}[A_{i\plus{}1}OM]\cdot1/2$
$ S_{b}$ will be less then $ S_{b}$ and also $ S_{a}\plus{}[A_{i\plus{}1}OM]\cdot1/2$ can be more then $ S_{b}$

Let $ X_{1}$ be the first point where  $ A$ puts his flag. Starting from this point divide the circle with an angle of $ 360/n$ and fix these points $ X_{2},X_{3},...,X_{n}$.
It is obvious that having an $ A$ flag near a $ B$ the territory between them is the same for both. The only territories that count are obtained from $ 2$ nearby identical flags.
The winning strategy for $ B$ is to put his flags in the fixed points $ X_{i}$ until there are no more free of them; if $ A$ puts a flag in a fixed point $ B$ continues putting his flag in the next free fixed point.
The simplest configuration we can have is that $ B$ occupies all the fixed points and $ A$ tries  to neutralize the $ B$ territories by putting an $ A$ flag between $ 2$ nearby $ B$ flags; until now the game is  a draw but there are $ n$ angles(areas) and $ A$ disposes of $ n\minus{}1$ flags (the first one is fixed) so after the "neutralization" $ A$ has only one flag and $ 2$ angles are free(the first one and the last), putting an $ A$ flag near another one identical to it, $ B$ can put his last flag in one of the $ 2$ free angles, closer to the $ X_{1}$ s.t. the new territory of $ B$ is bigger then $ A$`s $ \Rightarrow$ $ B$ wins.
Starting again with this configuration, if $ A$ tries to obtain a territory between $ 2$ nearby $ B$ flags, a $ B$ territory with the maximum angle will be not neutralized and again at the end $ B$ wins.
Now, if $ A$ changes this configuration by putting a flag in the place of a $ B$ fixed point, obtaining so an $ A$ territory, $ B$ will have one more free flag at the end and he can neutralize this $ A$ territory with it.This is valid also in the situation that $ 2$ nearby $ A$ flags are on fixed points; every time when $ A$ tries to obtain a territory, either gives $ B$ one more free flag ,so a chance to neutralize it, or makes free a $ B$ territory with the maximum angle,so bigger then the $ A$ territory.
So, $ B$ has a wining strategy :D .
If smth is not clear or we have some slips, we wait for questions or remarks :) .

$ MihAliza$, please, explain me, your strategy is put B´s flags in the free fix points "in order" am i right?
so, for n=7, if A put A's flags in the fix points $ X_{1}, X_{7}, X_{6}$ and B(in your strategy) put B's flags in $ X_{2},X_{3}$ and $ X_{4}$, then if A put a flag in X5, he is temporally the winner; but when B play, he can put a B flag between and A and B flags or between two As flags, so in the A turn, he is winning yet or the game is in draw, then A can neutralize 2 consecutive Bs and he is winning again, this procces can be repeated. how do you solve this case?

The algorithm is the same. When all the fixed points are occupied, B begins to neutralize A's territories.There are the steps, after $ X1$,$ X7$,$ X6$,$ X5$ are occupied by A and $ X2$,$ X3$,$ X4$ by B.It's B's turn:
$ B$ $ \Rightarrow $ between $ X1$ and $ X7$
$ A$ $ \Rightarrow $ between $ X2$ and $ X3$
$ B$ $ \Rightarrow $ between $ X7$ and $ X6$
$ A$ $ \Rightarrow $ between $ X3$ and $ X4$
$ B$ $ \Rightarrow $ between $ X6$ and $ X5$
And, just how you said, the game until now is a draw.But there is 1 remaining flag for A, and 1 for B, and there are no more teritories to neutralize.But we have 2 free ter. with maximum area: those between $ X1$ and $ X2$, and between $ X4$ and $ X5$, so wherever puts A to save a territory for him (the smartest is to put in one of these, near B's flag), B can put nearer to A's flag in the other. And he wins.
(It's important, that at the end, before the last move of A and B, there are always at least 2 free territories with maximum area, each of an angle, because they put until that moment 2*(n-1) flags, and there are n fix points, and n places between them, the fix points are surely occupied. So there are at least 2 free places,and even if the game until now is a draw,which is the worst case for B, he can save a larger territory then A, with his last step)

Sorry, i can't see where is the mistake in my solution! Are you speaking about the case 3) ?
If yes, you saw how i told to put the last B-flag?

I can be wrong, but i still thinking that my solution work..

Your solution is wrong. Let $ n\equal{}3$: A puts the first flag,B puts symmetrical from the center his flag,A puts near the B-flag and B puts closer to A so that $ S_{b}>S_{a}$; until now we constructed a winning strategy for $ n\equal{}2$. Now if A puts his last flag between the B-flags(in the middle), B has no solution to win. So, the induction does not work in this case. :wink:


Well, nice inequality. But I failed finding the equality and proving it. Give a hint :!:

[quote="Lovasz"][quote="arqady"][quote="Lovasz"]Give a hint :!:[/quote]
Let $c=2c_1.$ :D[/quote]
Excuse me, I don't understand what you meant.[/quote]
If $c=2c_1$ then our inequality equivalent to
$\left(\frac{3a}{a+2b+2c_1}\right)^{\frac{3}{4}}+\left(\frac{3b}{2a+b+2c_1}\right)^{\frac{3}{4}}+\left(\frac{3c_1}{2a+2b+c_1}\right)^{\frac{3}{4}}\geq2$ :)


when a digital watch reads 3:47, the sum of the digits is 14.
after how many minutes will the digits go to 20 for the first time?

[hide]6:59 is the first time, it's 3 hours 12 minutes. [/hide]

[hide="click here"]192 minutes?  at 6:59? [/hide]

[quote="aznmonkey1992"]noooo...
[hide]3:89[/hide][/quote]

lol, i didn't know you came from mars  :rotfl:

yeah. that's why your answer wasn't riht.  :P

[quote="RICHMATH"][hide]29 minutes, 4:16?[/hide][/quote]

Uh, 4+1+6=11  :roll:  they want 20.  ;)

[quote="236factorial"]yes, but can you hide your answer?[/quote][quote="anirudh"]why?[/quote]
because other people may still want to attempt the problem without knowing the answer!

but how can they unless they have a 55 inch monitor.  ;)

not true, I have a 15" or so, and I can see everything around, and also, this spam shall stop  ;)

236! is a good mod  :lol:
[hide]
to answer the question now 6:59 would be the first time it occurs so then that is 182 minutes
[/hide]

[hide]6:59, or 3hrs and 12 min, or 182 min[/hide]

and anirudh, this is not the first time you have spammed. you post an average of 8.56 posts a day.  sure, thats not a whole lot, but lots of those posts are meaningless.  So please stop spamming.

Here's a question:

How many minutes before 3:49 was the total of the digits on the 12-hour clock 20 for the first time?

[hide]
The closest time to 3:47 that the sum of it's digits is 20 is 6:59, so the answer is 192 minutes.[/hide]


when cos(x) and x are variables in one equation, how to solve it?
like these cos(x)+sin(x)+x^2=0, tan(x)=x+c
if cos(x)=x solved by iterative way, and set it as a constant, could others be converted by it?

Hmm... it starts to get pretty ugly. Well, $\sin x+\cos x$, for example, can be written as one sinusoidal function ($it's amplitude is \sqrt{2}$).

They do get pretty ugly, but you can sometimes figure out the bounds for the roots, ie:

For $\cos x+\sin x = x^{2}$ we have $x^{2}= (\sqrt{2}) \cos(x-\frac{\pi}{4}) \leq \sqrt{2}\Rightarrow x^{2}\leq \sqrt{2}$.

For alot of these problems you can use the formula $a \sin x+b \cos x = (\sqrt{a^{2}+b^{2}})\cos(x-\arctan \frac{b}{a})$ (Wikipedia).

I'm not really sure if there's a way to find the exact roots of the others...

uugh...don't waste time on these problems, it is very unlikely that there are nice solutions, furthermore, there is no general solution that is in terms of elementary functions

I may have not written my question clearly.
Thanks, Altheman, you know what I meant. Probably it do not have exact solution.


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations
AMC

Find the number of ordered pairs of integers (m,n) for which mn>=0 and
$m^3+n^3+99mn=33^3$
A)2 B)3 C)33 D)35 E)99

AHSME 1999 #30

[hide]

let . We then get the equation . Subtract the original equation from this new equation to get . Factor  out of the equation to get . So either  or . The second solution only has one solution, , but the first equation has  solutions in non-negative integers , so the total number of solutions is .[/hide]

That's what I did, I just introduced the variable $s$ to make it quicker to write out.

You can set it to anything, it's just that if you set it to $33$ then it cancels out of both sides of the equation and you immediately get $34$ solutions from it.

You just have to kinda notice that...

[quote="1234567890"]My question was [u]why [/u]can you set m+n to 33? Perhaps I'm missing something....[/quote]

Well s=m+n, ignore the s=33 part.

Well you have to see that $s=33$ or else you don't see how you get the second equation.

I know I got this question right on the real thing...

Remember the factorization of a sum or difference of cubes- the three-variable form is also worth knowing.
$x^3+y^3+z^3-3xyz=\frac12\cdot(x+y+z)\cdot((x-y)^2+(y-z)^2+(z-x)^2)$

Setting $x=m,y=n,z=-33$, we have that either $m+n-33=0$ or $m=m=-33$.

Another way to arrive at what jmerry said.  Start with $(m+n)^3)$:

\[ (m+n)^3=m^3+n^3+3mn(m+n)  \]

Subtract the given equation from this:

\begin{eqnarray*} (m+n)^3-33^3 &=& 3mn(m+n)-99mn\\ (m+n-33)((m+n)^2-33(m+n)+33^2) &=& 3mn(m+n-33) \end{eqnarray*}

So either $m+n=33$ or $(m+n)^2-33(m+n)+33^2=3mn$.

For the second, we expand and simplify $m^2+n^2-33m-33n-mn+33^2=0.$  Note that by AM GM $m^2+n^2\ge 2mn$ so sub in $2mn$ and get $0=m^2+n^2-33m-33n-mn+33^2\ge mn-33m-33n+33^2=(m-33)(n-33).$  But for $0\ge (m-33)(n-33)$ we need either $m\ge33$ and $n\le 33$ or $n\ge33$ and $m\le 33.$  However, it is easy to see that if either of $m$ or $n$ is greater than 33, the LHS of the original equation is larger than the RHS (note that since $mn\ge 0$, means we would have to have both $m$ and $n$ positive since each case has one of the variables at least 33).  So we need either $m=33$ or $n=33$...which make $n=0$ and $m=0$ respectively...edit..erm, my work here is a little messed up.  Anyway, we need equality for AM GM, so $m^2=n^2$ and we find $(-33,-33)$ works.

Yeah, jmerry's way is MUCH nicer.  But this way works...dunno if I could have done it during an exam though  :(

Um, elemenop, I'm not sure you can just [i]see[/i] that $m+n=33.$  You're neglecting the other case that I addressed above, aren't you?

Yeah, I immediately saw that $(-33,-33)$, $(0,33)$, and $(33,0)$ worked when looking at the equation. The second two of those roots seem to lend to the fact that $m+n=33$, and plugging it in you can show that the statement is true.

[quote="joml88"][quote="beta"][quote="joml88"]

Um, elemenop, I'm not sure you can just [i]see[/i] that $m+n=33.$  You're neglecting the other case that I addressed above, aren't you?[/quote]

Well elemenop's solution is perfectly fine, what he was pointing to is that he arrived at his solution by intuitively noticing m+n=33 works.[/quote]

Say we were solving $x(x+3)=3(x+3)(x^3+x^2+2)$.  Intuitively $x+3=0$ works, but that doesn't mean there might not be other solutions.  It [i]seems[/i] intuitive...but I don't think you can make the jump that $m+n=33$ so easily.[/quote]

Yeah but the point is that by seeing m+n=33 Elemennop was able to come up with a rigorous solution(the 2nd post) from intuition.

Oh ok.  I didn't read his post well enough.  I read the first line and scanned the rest and, yeah...

jmerry's way is still the nicest ;)

[quote="joml88"]Yeah, jmerry's way is MUCH nicer.  But this way works...dunno if I could have done it during an exam though  :([/quote]

Even though his solution works better/quicker, I think there's a larger chance I'd have done it your way on the exam as opposed to jmerry's way (I wouldn't know/figure out that formula).  Although both chances are slim, and the time might even be worth more spent on other problems or checking my work than trying to get a question like this.  Was the timing different for AHSME than the current AMCs?  Because I think if I got to this question, I wouldn't have enough time (if I didn't notice jmerry's trick).

The timing for the 30 question AHSMEs was 90 minutes (3 minutes per question seems to be the time regardless of questions.

Hmm... I did this test for practice once, and here was my 'solution' :roll:

$(m+n)^3 = m^3+n^3+3mn(m+n)$, $\therefore$ one possibility is $m+n = 33$. This gives us $34$ solutions, and yeah clearly there couldn't be 99 solutions (we could have only missed a case or two) so we can safely assume that $\boxed {35}$ is the answer :D :blush: Hey, it's a #30--you don't really have time to find the case(s) that you missed, especially if there are barely any.


integration
vector
function
calculus
calculus computations

Consider $ y''+\lambda y=0, y(0)=0, y'(L)=0$
Show that if $ \phi_n$ and $ \phi_m$ are eigenfunctions corresponding to the eigenvalues $ \lambda_n$ and $ \lambda_m$ respectively, with $ \lambda_n \not= \lambda_m$ then

$ \int^L_0 \phi_m (x)\phi_n (x)\,dx=0$

I think I know whose course you're in.

While you can do this by explicitly finding the eigenfunctions, there's some possible enlightenment here that can be obtained without doing that.

Write $ \int_0^Lf(x)g(x)\,dx$ as $ \langle f,g\rangle.$ That's an inner product, and you can interpret the result as a statement that the eigenfunctions are orthogonal.

We have that $ \phi_k'' \equal{} \minus{} \lambda_k\phi_k$ for $ k \equal{} m,n.$

$ \lambda_m\langle\phi_m,\phi_n\rangle \equal{} \langle\lambda_m\phi_m,\phi_n\rangle \equal{} \langle \minus{} \phi_m'',\phi_n\rangle \equal{} \minus{} \int_0^L\phi_m''(x)\phi_n(x)\,dx$

Now integrate by parts, twice:

$ \equal{} \minus{} \phi_m'(x)\phi_n(x)\big|_0^L \plus{} \int_0^L\phi_m'(x)\phi_n'(x)\,dx$

$ \equal{} 0 \minus{} 0 \plus{} \phi_m(x)\phi_n'(x)\big|_0^L \minus{} \int_0^L\phi_m(x)\phi_n''(x)\,dx$

$ \equal{} \minus{} \langle\phi_m,\phi_n''\rangle \equal{} \lambda_n\langle\phi_m,\phi_n\rangle$

So $ \lambda_m\langle\phi_m,\phi_n\rangle \equal{} \lambda_n\langle\phi_m,\phi_n\rangle$

From that, either $ \lambda_m \equal{} \lambda_n$ or $ \langle \phi_m,\phi_n\rangle \equal{} 0.$

---

A broader perspective: we have an operator $ T$ defined on the vector space of smooth functions on $ [0,L]$ such that $ y(0) \equal{} y(L) \equal{} 0$ by $ T(y) \equal{} y''.$ The integration by parts argument above shows us that $ \langle T(u),v\rangle \equal{} \langle u,T(v)\rangle.$ The name for that is that $ T$ is a [i]self-adjoint[/i] operator on that inner product space. And self-adjoint operators on a real inner product space are just like real symmetric matrices: the eigenvalues must be real, and eigenvectors for different eigenvalues must be orthogonal.


search
number theory unsolved
number theory

[quote="hieuchuoi@"]Prove that if $(xy+1)(yz+1)(zx+1)$ is a perfect square then $xy+1,yz+1,zx+1$ are all perfect squares, with $x,y,z$ are positive integer numbers.[/quote]
The problem is really difficult,but it was posted long ago.Try to search!!! :wink:


Γνωριζει κανεις παιδια (κυριως στους φοιτητες αναφερομαι...η σε οποιον αλλον γνωριζει...),αν υπαρχει η υπαρχουν διαγωνισμοι μαθηματικων,ειτε εθνικοι ειτε διεθνεις ,για φοιτητες???Αν καποιος ξερει,αν γινετε ας με παραπεμπψει σε καποια ιστοσελιδα για more info....THANKS...

ΡΑΛΛΗΣ :D

File Rakar kai ego gia fititikous mathimatikous diagonismous
epsaxna kai brika ton exis International Mathematics Competition
me site http://www.imc-math.org
Apo oti exo katalabei mporei na simetasxi opios theli arkei
na brisketai sto 1,2,3,4 etos.

για εθνικους διαγωνισμους ..δε νομιζω να υπαρχουν . :!: μονο μεσω ιντερνετ.γιατι αραγε τα ελληνικα πανεπιστημια δεν δραστηριοποιουνται σε αυτο? :huh:

Νομίζω ότι υπάρχει και ο Putnam Ράλλη......
Αλλά δυστυχώς τα ελληνικά πανεπηστήμια από ότι έχω συζητήσει δεν στέλνουν αποστολές,,,,,,, :o

απ'ότι έμαθα μπορεί να πάει κάποιος μόνος του και δεν είναι ανάγκη να σπουδάζει μαθηματικά :clap:
Λένε ότι δεν είναι τοσο αυστηρά όσο στην imo. Έχει γίνει μια συζήτηση στο αγγλικό.
πάντως δεν ξέρω τι γίνεται με τη γλώσσα.


-1 times -1 is equal to 1

a+b=b+a

CD=CD

if CA>CB, then CB<CA

The names? not sure.. i just know them as facts. -1 x -1 is that... that a negative times a negative is a positive? i doubt thats the name.

the a+b=b+a i think is .. the commutative property of addition?

CD=CD - i didnt even know that that had a name. Every Number Equals Itself?

A>B, B<A.. those are the same two inequalities, basically..

[hide="CD=CD"]
reflexive property of equality[/hide]

[hide="a+b=b+a"]
symettric property of equality[/hide]

[hide="a+b=b+a"]
additive property of equality[/hide]
[hide="CD=CD"]
I think that you meant CD=DC which is the multiplicative property of equality,
if I'm not mistaken.[/hide]


What is the difference between the sum of the first $ 2003$ even counting numbers and the sum of the first $ 2003$ odd counting numbers?

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 2003 \qquad
\textbf{(E)}\ 4006$

If you know the formula for the sum of an arithmetic sequence, that would help:
$ S_{n}\equal{}\frac{n(a_{1}\plus{}a_{n})}{2}$
Note that the odd $ n$th odd number is given by $ 2n\minus{}1$ so the 2003rd odd number is $ 4005$.  Plug it into the formula and we get
$ S_{2003}\equal{}\frac{2003(1\plus{}4005)}{2}\equal{}4012009$
For evens, the $ n$th even number is given by $ 2n$ so the 2003rd is $ 4006$.  Into the formula:
$ S_{2003}\equal{}\frac{2003(2\plus{}4006)}{2}\equal{}4014012$
Now subtract and we get a difference of $ \boxed{2003}$.

[hide="fast"]The difference between the nth even and nth odd is 1, so simply 2003.[/hide]

Let an even positive integer be $2n$, since it is positive, there will be an odd integer $2n-1$, and their difference will be $1$. In this situation we have:

$$(2+4+...) - (1+3+...)$$,

Which can be rearranged as:

$$(2-1) + (4-3) + ...$$

Clearly this is just $1 + 1 + ... + 1$, with $2003$ numbers, so $\boxed{2003}$.

The sum of the first $n$ even counting numbers can be written as $n^2+n$ while the sum of the first $n$ odd counting numbers as $n^2.$

Plugging in and subracting gives: $(2003)^2+2003-(2003)^2=\boxed{2003}.$

Can people stop bumping the easiest problems from AMC 10's like ten years ago to give solutions that are exactly the same as those which have been presented? These problems don't really deserve too much discussion...


let $k=x^2-5x+7$, then do cases $k>0$, $k<0$.  i think it's easier to work with the first equation first, and then solve for $k$ in terms of $y$, then substitute that into the second equation to solve.

Whoa.  That is ridiculously messy.  Well, at least we know that for any y which works, so does -y. :D
Also note that we never have the term x alone, we always have $|x^2-5x+7|$.  So let $|x^2-5x+7|=n$, then we have
$n>0$, 0 would make it undefined
$n=|y|^{n|y|}, n^{\frac{2}{n}}+|y|^{\frac{1}{y}}=2$
Just consider all positive values of y, because y is always in absolute value signs, then at the end you just double the result.  Now just graph this system, i suppose, treating n as x.  Hope that helped somewhat...


quadratics
number theory
prime factorization

I posted this in the high school basics forum, but they thought it was slightly above the level of difficulty, so I'm sorry if you have seen this problem 2 times: If $ p$ is a prime greater than or equal to $ 11$, find all possible values of $ k$ if $ p^2$ is $ k$ mod (120) and prove you have found all the values. I did this using a kind of long algebra process. Can someone see a relatively short process.

First I will show my method: Break 120 down to its prime factorization: $ 2^{3}*3*5$. A prime's square must be $ 1 (mod 24)$ or ${ (mod 2^{3}*3}$ because either p-1 or p+1 must be a multiple of 4, and the other must therefore be a multiple of 2, and either p-1 or p+1 must be a multiple of 3, so since $ (p-1)(p+1)+1 (mod 24)=0+1 (mod 24)=1 (mod 24)$. Dealing with the issue of divisibility of 5, we must have 2 cases:

Case A: Either p-1 or p+1 are divisible by 5, so $ p^2$ is $ 1 (mod. 120)$.

Case B: The other alternative is that either p-2 or p+2 are divisible by 5. If p-2 is the one let $ 5k_1=p-2$, or if p+2 is the one, let $ 5k_2=p+2$, therefore, p may be written as either $ 5k_1+2$, or $ 5k_2-2$; therefore in either cases p is $ 4 (mod 5)$. The only result that complies with 4 (mod 5) and 1 (mod 24) (besides 1) is 49 (mod 120).

These produce the desired result, that only 1 and 49 (mod. 120) are possible for the square of any prime greater than or equal to 11. I just wanted to know if there was a shorter process to do this or not. Please help. Thanks!

Probably no. But for divisibility with 5, you can note that quadratic residues mod 5 are 0,1 and 4, but p isn't divisible by 5, so we are left with 1 and 4. Then use CRT to finish the job.

Oh, ok trhanks. So the way I did it was correct, right?


here's the question:

A true or false test with 6 questions has to have at least 2/3 of the answers True. How many possible answer keys can be created?

I have the answer, but I don't know how to do it. Answer:22

Thanks! ;D

[quote="Yongyi781"]3 cases:

1) 4 trues, 2 falses in some order. So we are arranging TTTTFF, and that gives us $ \frac {6!}{4!2!} \equal{} 15$.

2) 5 trues, 1 false in some order. So we are aranging TTTTTF, and that gives us 6 more.

and finally, 3) 6 trues, 0 falses. There is obviously one way we can do this.

$ 15 \plus{} 6 \plus{} 1 \equal{} \boxed {22}$, as desired.[/quote]

Wow, thank you so much! :D


Let $ S \equal{} \{ 1,2,3,4,\cdots ,10 \}$. We wish to color all of the elements of $ S$ such that no two elements $ a$ and $ b$, with $ a \minus{} b$ odd, have the same color. How many distinct colorings of $ S$ are there?

Surely there was something about the number of available colors?  Or some notion of what "distinct" means for colorings?

clarifications: There are three colors. Colorings are distinct if at least 1 pair of corresponding elements is non-monochromatic.

[hide]From the condition: if a number of one parity is colored, no numbers of the opposite parity can have that color.

Case I: Use only two colors.
Then you color all the odd numbers 1 color and the even numbers another color. There are $ 3\cdot{2}\equal{}6$ ways to do this.

Case II: Use three colors.
Then one of the parities must be monochromatic. Choose a parity and a color to color it; there are $ 2\cdot{3}\equal{}6$ ways to do this.

Of the $ 5$ numbers of the other parity, color each one as either one of the remaining colors, such that the five numbers are not monochromatic. There are $ 2^5\minus{}2\equal{}30$ ways to do this.

In total, there are $ 6\plus{}6\cdot{30}\equal{}186$ colorings.[/hide]


superior algebra
superior algebra unsolved

(i) Let $ G$ be a group and $ x\in G$. Suppose that there is an integer $ n\geq 1$ s.t. $ x^n\equal{}e$. Show that there is an integer $ m\geq 1$ s.t. $ x^{\minus{}1}\equal{}x^m$.

(ii) Let $ G$ be a finite group. Show that given $ x\in G$, there exists an integer $ n\geq 1$ s.t. $ x^n\equal{}e$.

(i) $ x\cdot x^{n\minus{}1}\equal{} x^{n\minus{}1} \cdot x \equal{}e$ thus $ m\equal{}n\minus{}1$.
(ii) The set $ \{ x,x^2,x^3,...\}$ can not contain infinitely many distinct values because $ G$ is finite thus there exist $ k\ne l \in \mathbb N$ such that $ x^k\equal{}x^l$ so $ x^{|k\minus{}l|}\equal{}e$.


logarithms
number theory proposed
number theory

(i) Fix a prime $p$ and integers $a_{1},\ldots,a_{k}\in \mathbb{Z}$. Prove that if $x_{1},\ldots,x_{k}\in \mathbb{Z}$ are any integers not divisible by $p$, and $n_{1},\ldots,n_{k}$ any positive integers,  then either $a_{1}x_{1}^{n_{1}}+\cdots+a_{k}x_{k}^{n_{k}}= 0$; or else the exponent of $p$ dividing $a_{1}x_{1}^{n_{1}}+\cdots+a_{k}x_{k}^{n_{k}}$ is less than $2\log{k}+2kp \sum_{1}^{k}\log{|n_{i}a_{i}x_{i}|}$.

(ii) Use this to deduce the following result: fix integers $a_{1},\ldots,a_{k}$ and rational numbers $y_{1},\ldots,y_{k}\in \mathbb{Q}\setminus \mathbb{Z}$ (with no $y_{i}$ an integer). Then there are only finitely many $n$ such that $a_{1}y_{1}^{n}+\cdots+a_{k}y_{k}^{n}$ is a *nonzero* integer.

--Vesselin

Wow, Vess, this is a terrible result. Note that it immediately implies and old conjecture of Nathanson (proved in 1979) saying that if $3^{n}|5^{n}-2$ then $n=1$. Indeed, your estimation yields  $v_{3}(5^{n}-2)\leq log (4(10n)^{12})$, which gives a very good upper bound for $n$. Also, note that this estimation (a)) yields a trivial solution for the problem with binary digits. This makes me believe that a proof of $a$ should either be extremely inventive or use a very high arsenal.

Ok, this seems to be unbelievably difficult, so let's start with some small steps. Has anyone ANY idea about the case $k=2$?


modular arithmetic
number theory unsolved
number theory

a) Prove that each element of the set  ${n!+2,...n!+n}$  has a prime divisor that does not divide any other element of the set

b) Prove that for every positive integer  k there exist a set of n consecutive numbers such that each element of the set has k distinct prime divisors that do not divide any other element of the set .

 the first part i already  solved , but for the second i  havent found a solution  :blush:

a) Assume that the statement is false. Suppose that, for some positive integer $r$ such that $2 \leq r \leq n$, a randomly chosen prime divisor $p$ of $n!+r$ also divides $n!+s$ for some $s$ between $2$ and $n$. Then $p$ divides $r-s$. It follows that $p \leq |r-s|$, then $p \leq |n-2|$. Suppose this happens for all prime $p$ which divide $n!+r$; that is, all its prime divisors are at most $n-2$. Then, for all such $p$, since $p<n$, we have $p \mid n!$ and thus $p \mid r$.

Then all prime divisors of $n!+r$ must also divide $r$. Suppose $p \mid r$ and $r=pk$ with $k>1$; also $p^a \mid r$ but $p^{a+1}$ doesn't divide $r$. Then $n!=p \times r \times N$ for some $N$, since $p<r<n$. Then $p^{a+1}$ divides $n!$ but not $r$; so $p^{a+1}$ doesn't divide $n!+r$. This can't happen for all prime divisors of $r$, since $n!+r>>r$. So, for some $p$, we must have $a=k=1, r=p$. Then our number is $n!+p$, whose only prime divisor less than $n$ can be $p$. Then $n!+p=p^b$. Since $p$ must divide some other element of the set, $2p \leq n$. But then $p, 2p$ both divide $n!$, thus $p^2$ divides $n!$ but not $p$; then $p^2$ doesn't divide $n!+p$; then $b=1, n!=0$, absurd.

b) We do it by induction. For $k=1$, we have proved it above. Now suppose it is true for $k$; that is, we have numbers $A+1, A+2, \ldots, A+n$ satisfying the condition (with $k$ prime divisors). Let $P=(A+n)!$, and $B=P+A$. Then consider the numbers $B+1, B+2, \ldots, B+n$. For each $i$, $A+i \mid B+i$. Also, if $p \mid A+i$, and $p$ is one of those $k$ prime divisors, then $p \mid P$, so $p \mid B+j$ iff $p \mid A+j$ iff $i=j$, so these $k$ prime divisors still work. We must find another one for each $B+i$.

Suppose that, for some $i$, $B+i$ doesn't have any prime divisor which divides no other $B+j$ (other than the aforementioned $k$). It is clear that it must then have no prime divisors $>A+n$, since the original $k$ primes were $\leq A+i$ (since they divided $A+i$), and any prime dividing $B+i, B+j$ must divide $i-j$ and then be $<n$. Now, suppose $p$ is one of the failing primes, that is, $p \mid B+i, p \mid B+j$. Then $p<n<A+n$. Since $(A+n)!=P$, and $p \mid (A+n)!$, we have $p \mid A+i, A+j$. It follows that all failing primes for $B+i$ also divide $A+i$. It would then be enough to find a prime $p$ which divides $B+i$ but not $A+i$. Suppose there's no such $p$. Then, as above, there must be a prime $p$ dividing $A+i$ such that $p(A+i) \mid B+i$ (since $B+i>>A+i$). However, if $p \neq A+i$, we'll have $p(A+i) \mid P$, so that $p(A+i) \mid A+i$, absurd. Then $A+i=p$. This implies $k=1$. If our numbers $A+1, \ldots, A+n$ for $k=1$ were $(n+2)!+2, \ldots, (n+2)!+n+2$, then they were all composite, which makes $A+i=p$ impossible.

Another proof of part (b)
================

Take $k\times n$ distinct primes $p_{11}, p_{12}, \dots, p_{1k}, p_{21}, \dots, p_{nk}$ greater or equal to $n$. Now, for the chinese remainder theorem, there exists a number $A$ such that
\[ A\equiv 0\pmod{p_{11}}\quad A\equiv 0\pmod{p_{12}}\quad\cdots\quad A\equiv 0\pmod{p_{1k}}  \]
\[ A\equiv -1\pmod{p_{21}}\quad A\equiv -1\pmod{p_{22}}\quad\cdots\quad A\equiv -1\pmod{p_{2k}}  \]
\[ \cdots  \]
\[ A\equiv -n+1\pmod{p_{n1}}\quad A\equiv -n+1\pmod{p_{n2}}\quad\cdots\quad A\equiv -n+1\pmod{p_{nk}}  \]
Now $A, A+1,\dots, A+n-1$ are the $n$ consecutive integers we are looking for.


[color=blue]There are two bags. In bag A there are three black and two white marbles. In Bag B, there are two white and one black marbles.  One marble is randomly selected from bag A and placed into bag B. One marble is then randomly selected from bag B. What is the probability that the marble selected from bag B is black? Express your answer as a common fraction.[/color]

[hide]Two cases:
1. marble from bag A is white
    2/5

2. Marble from bag A is black
    3/5

$2/5\times1/4+3/5\times2/4=2/5$[/hide]

[size=75][color=darkred]Hidden.
-nebula42[/color][/size]

[quote="espark52"](lol, just ignore this post, i just put it there to get it up to the top)[/quote]
Please [b]do not[/b] post something like this. Posts should be at least somewhat related to the topic at hand and should be useful to other users. Posts with very little content, posts to move the topic to the top of the list, and posts just to increase post count should be avoided.


$x,y,z$ are positive real numbers such that $xy+yz+zx=3xyz$ prove that :

$\frac{yz}{x^{2}}+\frac{yx}{z^{2}}+\frac{zx}{y^{2}}+\frac{3}{2}\geq \frac{9xyz}{2}$

the given inequality is equivalent to proving

$2\sum_{cyc}\frac{1}{x^{3}}$ $+$ $\frac{3}{xyz}$ $\geq$ $9$

let $\frac{1}{x}=a$,$\frac{1}{y}=b$,$\frac{1}{z}=c$

so the given condition and inequality become

$a+b+c$ $=$ $3$

$2\sum_{cyc}(a^{3}+\frac{abc}{2})$ $\geq$ $9$

by schur' inequality

$\sum_{cyc}(a^{3}+abc)$ $\geq$ $\sum_{sym}a^{2}b$

but,  $\sum_{sym}a^{2}b$ $=$ $\sum_{cyc}[a^{2}(3-a)]$ $\geq$ $\sum_{cyc}(3-a^{3})$

where the last step follows from jensen's inequality
this gives us the required result

With substitution $ x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c} $ demonstrated inequality is rewritten as follows:
$ (a, b, c >0, a+b+c=3)\Rightarrow 2\sum{a^3}+3abc\geq 9 $
Noting $ bc+ca+ab=3(1-q^2),0\leq q<1 $ must show that $ abc\geq 2(bc+ca+ab)-5 $
which is equivalent to $ abc\geq 1-6q^2 $
But according to the method UVW, we double inequality:
 $ (1+q)^2(1-2q)\leq abc\leq (1-q)^2(1+2q)\Leftrightarrow 1-3q-2q^3\leq abc\leq 1-3q+2q^3 $
Is enough to show that:
$ 1-3q-2q^3\geq 1-6q^2\Leftrightarrow q^2(2q-3)\leq 0\Leftrightarrow 2q\leq 3 $
But $ q<1\Rightarrow 2q<2<3 $, and demonstration ended!
_________
sandu Marin


trigonometry
real analysis
real analysis unsolved

a) Prove that the series $\sum_{n\geq 1}{\frac{1}{n^3\cdot sin(n\cdot\pi\cdot\sqrt{2})}}$ converges absolutely.
    b) Compute this sum!!!! :idea:

No, I say the second question is very hard. The first one is only a joke.

[quote="jmerry"]You say that's hard? The series still converges absolutely if we replace $n^3$ with $n^2$.

On the other hand, I have no idea how to approach computing the exact sum.[/quote]
It is also very interesting and not an easy(particularly for me) problem, that $n^2$ could even be replaced by $n^a$,for every $a>1$ i mean, prove that the series $\sum \frac{1}{n^a sin( n \pi \sqrt2)}$ converges absolutely for every $a>1$.

Harazi,

From wich ens paper did you find this problem ?

It is not an examination, I saw it in a kind of article, but I don't have references  since I only noted the result on one of my papers with exercices.

I don't remember exactly the anwser, all I know is that it involved 360 and pi, but not zeta(3) But maybe you want to show us your computations?  ;)

I edited now the formula for $\frac{1}{\sin(\pi x)}$ is correct


A sum is formed by alternatingly adding and subtracting consecutive odd integers starting with 1 and ending with 1997 as indicated. What is the sum?

$1-3+5-7+9-11+...+1993-1995+1997$

[quote="lotrgreengrapes7926"][quote="math92"][hide]what i would do is match up every number combined with a minus sign so for example
1-3
5-7
7-9
11-13
13-15
and so on
each of these makes -2
1+1997/2 [is that what you do to get number of pairs?]
so there are 999 pairs including 1997
so
998(-2)+1997=x
-1996+1997=x
1=x
[/hide][/quote]

Was your answer 1? I agree with pianoforte. There are $499$ pairs of $-2$ plus a $1997$, not $999$ pairs. So the answer is $499*-2+1997=1997-998=\boxed{999}$ :D[/quote]I agree with pianoforte as well.

i just didn't know the formula or how to get the number of pairs.


What is the longest you have run at one time...without stopping? I am just curious to see who among us isa runner. Also, WHY did you run that distance?

I ran 8 Monday.  The most that I have run is about 10.  I do it because I run cross country. We do 7-10 miles on long run days, but most other days we normally do about 5-7.  I want to do a half marathon(13.1 miles) but I don't really have much time between track and cross country.  When I do have time there normally aren't many road races so I ll just wait til later in life to do a longer distance.

i haven't run that much all summer! i just realized that!
as to the topic, i've run 5-6, but usually i only run about 2-3 since most of my friends live near by. we started a habit of just turning up at each other's doorsteps and not having a reason to be there. but running just for running's sake is fun too. i don't run competitively, just for fun.

longest...hmm probally about 10.  average about 3.  I am also a cross country runner though I didn't make my schools team this year because the stupid distances they made us run for tryouts was like 1 1/2-2 miles.  Gosh darn don't they know distance runner start out slow a lot but speed up later in races???  Anyone else had that kinda problem with a new coach or something?

At my school we don't cut people from cross country or track.  In fact the coach wants the team to be as large as possible.  We have no speed requirements also.  Our coach has gotten a lot better gradually increasing beginners mileages and having the older kids run farther than say the middle schoolers.  This doesn't mean that don't work hard and do good at meets.  We work extremely hard and are one of the best teams in the state.  Our girls team has won the state championship 2 out of the last 3 years and are in really good shape to win again.  The boys were state runners up two years ago but unfortunately we finished a horrendous 8 th(worst in like the history of our school) last year.  We look to improve greatly this year.

I can usually run two miles, sometimes three each day. I run to stay in good shape. It's a great workout and relaxing for me...I don't have to think about my problems when I run; it's a good escape.

I can run about 15 feet before I collapse.

Since I'm still in middle school, we still have P.E. classes. Anyway, during one of our run days, I did about 12-13 laps around the track. (4 laps=1 mile). But usually I don't run that far.

[quote="yif man12"]i wonder if anyone here can go the marathon distance. my friend tried to enter into the hartford half-marathon, but they didn't let him because he was "too young" and he was forced down to just running a single mile[/quote]

I doubt it.  But then again you never know.  To run a marathon it takes a lot of training.  I'm sure that a lot of ppl could do a marathon if they trained.  It would probably take close to a year to fully prepare for a marathon for most ppl but anyone can run a marathon if they put their mind to it.   What really is crazy is ultrarunners.  They consider a marathon short.  I know that there are races that go up to 100 miles! maybe even more!  I think the craziest thing is those people who set records on treadmills.  I believe the very recently set record for miles run on a treadmill in a 24 hour period is 153.76 miles if I remember correctly!!! I mean running 100 miles outside is sort of understandable.  But I can not imagine running in the same room for 150+ miles.

Hey, I'm a cross country runner as well. Maybe math geniuses are also great runners, no? I see a correlation.... The farthest I've ever run is probably 10K, but I'm preparing for a half marathon over the summer...

i know a grl that ran & biked 18 miles in one day... really scary...

My math teacher from the 2003-2004 year participated in the ironman triathlon,
it's domething like:
~2 mile swim
~120 mile bike ride
~26 mile run

Well, the most Ive ever run at one time was 4-5 miles...hehe.... :lol:

Did anyone here ever run more than 20 miles??????? thats impossible for me.....

impossible for me to run more than 2 miles (well, sorta)

i bet you could run more than 2 miles. just start slowly, keep going and try not to stop. all it takes is a little patience and determination. what i've found out with a lot of people is that if they can run 1 mile, they usually can run 2.

As my friends who are in cross country say to me, "our sport is your sport's punishment."  My dad's obsessed with running, so i can manage two miles or so.  But I hate to run.  I'd gladly hike 26 miles but marathoners have to be insane.

I like that saying.  :P  I think that runners are a little bit insane.  But then again who isn't?

the longest i've run in the past few months is 1.5 miles.  simply because i'm a sprinter who cares only about leg muscle mass o_O.  longest distance i compete in is 400m.

however, i have run up to 6 miles when i was doing crosscountry.  turns out it was a horrible mistake >_<

I can't run very far. I have superlow endurance!

like 5.

i've also run really really hilly races..


Mafia calculus LaTeX blogs factorial logarithms geometry

A while ago, I (as r15s11z55y89w21) started the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266455]AoPS Hybrid[/url] game. ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299414]Here's dragon96's revival thread[/url].) Now, I wish to bring it back, with a loaded twist. Now, you write an equation! And don't limit yourself just to users!

The next person who posts after you posts what they evaluated the equation as, and posts a new one!

And of course, be courteous. Refrain from disparaging innuendos and the such.

Here's the first!

(isabella2296)^3-15(AIME15)+james4l

A mod with water splashed all over them, but isn't that smart.

AIME15 + 1

one more than a mod

$ \text{musicnmath^\inf}$

(musicnmathnmusic)^{4/3}

$ \frac {1}{\text{dragon96}} \plus{} \frac {1}{96}$

PhireKalkrszyw * 1.15*10^11

james4l + pythag011

(dragon96)! = $ Meewhee009^{42}$

$ Meewhee009^{(dragon96)(PhireKaLk6781)}$

musicnmath*[i]i[/i]+ $ 96^2/3$ + $ levans^dragon$
Monkeygirl13 * Maybach

ERROR:  Division by 0.

(Games+Mafia!)^2-Fun Factory

A mystery of a guy shooting.

$D^2=awesome=CHUCK NORRIS=?$

MORE AWESOME NESS

${\sqrt[3]{levans^\pi}}$ + ${\sqrt[42]{REAPISAWASTEOFTIME}}$

A beast admin that is the dad of the awesome reap/sniper :D.

New problem.
halowarfare-$asf

$\sqrt[3]{El_{Ectric}}$

Problem:

TwinDracula  + $\sqrt[Jongy]{LaTeX^{MCQ}}$

Error:  Two awesome.

Reap+is+a+waste+of+AOPS

ReapIsAWasteOfTime in AoPS without time.

Dsquared + PhireKaLk6781 =

And I want to make 2 equations, closely related:

Dsquared's posts + PhireKaLk6781 =

[quote="fprosk"]5.

prezcoin+my avvie=?[/quote]

[quote="halowarfare17"]Oqwardness

${\sqrt[spam]{\pi}}$ - $\sqrt[D^2 Avatar]{My Avatar_{MyFirstAvatar}}$[/quote]

The answer is actually [b]A revolution in Panem.[/b]

Anyone ever notice that this has been getting more and more spammy?

This is Fun Factory. It's ok to be kind of spammy.

No, it's not. Pretty much anything that is not what the thread is for is not okay.

Can we continue

$\frac{dojo}{dragon96}\sqrt{exmath}$


algebra
polynomial
Rational Root Theorem

Let $ p(x) \equal{} a_nx^n \plus{}...\plus{} a_1x \plus{} a_0$ be a polynomial with integer coeffi-
cients. If $ r$ is a rational root of $ p(x)$, show that the numbers $ a_nr$, $ a_nr^2\plus{}
a_{n\minus{}1}r, . . . , a_nr^n \plus{} a_{n\minus{}1}r^{n\minus{}1} \plus{} ...\plus{}a_1r$ are all integers.

Since $ r$ is a rational root, what can you say about it?  In particular, what can you say about its relationship with the coefficients?

Well, using the rational root theorem, we get
$ r$ is a factor of  $ \frac {a_0}{a_n}$
Clearly $ r*{a_n}$ is a whole number multiple of of $ {a_0}$
but I cant find a pattern to this:
$ a_{n}r^{2}\plus{}a_{n\minus{}1}r, . . . , a_{n}r^{n}\plus{}a_{n\minus{}1}r^{n\minus{}1}\plus{}...\plus{}a_{1}r$

Isnt that the Newtonian Relations? I read it in a book somewhere, but never bothered to think about it  :blush:


geometry
circumcircle
trigonometry
ratio
parallelogram
geometric transformation
reflection

$ BA{'}{\parallel{}}SP$ and $ CA{'}{\parallel{}}PT$ then $ \angle{A{'}SP}\equal{}\angle{SA{'}B}$ analogously $ \angle{A{'}TP}\equal{}\angle{TA{'}B}$

$ BS\equal{}\cos{C}(BP)$ and $ BA{'}\equal{}\cos{C}(2R)$ then, $ \frac{BS}{BA{'}}\equal{}\frac{BP}{2R}$ analogously $ \frac{CT}{CA{'}}\equal{}\frac{BP}{2R}$ then $ BA{'}S$ is similar to $ CA{'}T$ because $ \angle{SBA{'}}\equal{}\angle{TCA{'}}\equal{}\frac{\pi}{2}$ (and the ratios) then $ \angle{BA{'}S}\equal{}\angle{CA{'}T}$ and therefore $ <A{'}SP\equal{}<A{'}TP$

Nice solution...

I will present another synthetic solution (without the use of ratios etc.)

Consider points Q,R such that $ BQ//\equal{}A'R \perp AA'$ and $ BQ\equal{}A'R\equal{}BP(\equal{}CP)$ then $ BQRA'$ is parallelogram so $ QR \perp BA$ and if $ QR \cap BA\equal{}G$ then $ AGA'R$ is cyclic so $ \angle RAA'\equal{} \angle RGA'\equal{} \angle GA'B \equal{}\angle BA'S$ (since $ BQ$ is the reflection of $ BP$ across $ BC$). Similarly, $ \angle CA'T\equal{}\angle RAA'$ so $ \angle BA'S\equal{}\angle CA'T$.

I posted this exercise here, not because it is difficult or something but because it is useful for a simple solution(without polar theory) to another exercise (which was posted by Silouan and was given to the Bulgarian team):

Let $ M,N$ be two interior points in a circle $ C(O,R)$ such that $ O$ is the mid point of $ MN$. If $ S$ is a point that lies on the circle and $ E,F$ the intersection points of the lines $ SM, SN$ with the circle then prove that the if the tangents at $ E,F$ meet at $ I$ then the perpendicular bisector of $ MN$ goes through the mid point of $ SI$.

[quote="staretak"]Let ABC be a triangle and let AA' be a diameter of its circumcircle (O). The tangents at B,C of (O) intersect at P. If PS, PT are perpendicular to AB, AC respectively, show that $ \angle{A'SP} \equal{} \angle{A'TP}$[/quote]
Nice problem!
[hide="Alternate Solution"]
Let $ N$ be the midpoint of $ A'B$ and $ Q$ be that of $ A'C$.  Thus, $ \angle NOB \equal{} \frac {\angle BOA'}{2} \equal{} \angle BAA'$.  Also, $ \angle A'BA \equal{} 90 \equal{} \angle ONB$ since $ N$ is the midpoint of $ A'B$, so $ \triangle ONB\sim \triangle ABA'$.  Also, $ \angle PBS \equal{} 180 \minus{} \angle ABC \minus{} \angle PBC \equal{} 180 \minus{} \angle ABC \minus{} \angle BAC \equal{} \angle ACB \equal{} \angle AA'B$.  Since $ \angle PSB \equal{} 90 \equal{} \angle ABA'$, we see that $ \triangle PSB\sim \triangle ABA'\sim \triangle ONB\implies \triangle BSN\sim \triangle BPO$.  Similarly, $ \triangle CQT\sim \triangle COP\cong \triangle BPO\sim \triangle BSN$.  Since $ N$ and $ Q$ are the midpoints of $ A'B$ and $ A'C$, we see that \[ \triangle BA'S\sim \triangle CA'T\implies \angle CTA' \equal{} \angle BSA'\implies 90 \minus{} \angle BSA' \equal{} 90 \minus{} \angle CTA'\implies \angle A'SP \equal{} \angle A'TP\]  [/hide]

[quote="staretak"][color=darkred]Let $ ABC$ be a triangle and let $ AA'$ be a diameter of its circumcircle $ w = C(O)$ . The tangents at $ B$ , $ C$ of $ w$ intersect at $ P$ . If $ PS$ , $ PT$ are perpendicular to $ AB$ , $ AC$ respectively show that $ \widehat {A'SP}\equiv\widehat {A'TP}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Prove easily that the following problem ("any similarity of the notations is absolutely random") :[/color]

[quote][color=darkred]Let $ A'$ be the orthocenter of the triangle $ AUV$ . Denote $ \{\begin{array}{c} B\in AU\cap VA' \\
\ C\in AV\cap UA'\end{array}$ and the middlepoints $ P$ , $ S$ , $ T$ of the segments $ [UV]$ , $ [UB]$ , $ [VC]$ respectively. Then $ \{\begin{array}{c} PS\perp AB \\
\ PT\perp AC\end{array}$ , the lines $ PB$ , $ PC$ are the tangents from the point $ P$ to the circumcircle of the triangle $ ABC$ and $ \widehat {A'SP}\equiv\widehat {A'TP}$ .[/color][/quote]

[color=darkblue]If we"ll [b][i]erase[/i][/b] the points $ U$ , $ V$ , then we"ll obtain the proposed problem ![/color]


When written as a decimal, the fraction $\frac{m}{n}$ (with $m < n$) contains the consecutive digits 2, 5, 1 (in that order). Find the smallest possible n.

Well-known AIME problem.  I forget the year.  It's been solved several times.

Here's the outline:
[hide]
Assume $\frac{m}{n}= \text{blah}.251\ldots$. Otherwise multiply by a sufficient power of $10$, which doesn't change $n$. Then just subtract away the blah part so $\frac{m}{n}= .251\ldots$.

So $251n \le 1000m < 252n$. Let $n = 4a+b$. The inequality becomes $1004a+251b \le 1000m < 1008a+252b$. Since we want to minimize $a$ first, we clearly take $b = 3$ to get as close as possible to a multiple of $1000$ (a more rigorous argument can be applied here). Then we require that

$4a+753 < 1000$ and $8a+756 > 1000$.

It follows that $a = 31$ is the smallest value, which means $n = 4(31)+3 = 127$.[/hide]


An infinite sequence of light bulbs and a infinite sequence of switches are both numbered by the positive integers. Ecah switch has a finite number of positions. Whether a bulb is on or off depnds only on the positions of a finite number of switches. In any setting of the switches, at least one bulb is on. Prove that thre exists a finite set of bulbs such that for any setting of the switches at least one of them is on.

Pierre.

I hope "an infinite sequence" means that we have countable number of bulbs? I invented proof for this case.

Yes, as far as I know, sequence always means countable, and Pierre mentioned that they are numbered with the positive integers.

We may assume, for simplicity, that light bulb $b_i$ depends on switches $s_1,s_2,\ldots,s_i$. Let's assume we can put out $b_1,b_2,\ldots,b_n$, for any $n$. In the sequence of such positions, $s_1$ has a certain position $s_1^{i_1}$ which appears infinitely many times (because it has only finitely many positions), so choose this position for $s_1$. In the sequence of arrangements which put out $b_1,b_2,\ldots,b_n$ and which have position $s_1^{i_1}$ for the first switch, there is a position $s_2^{i_2}$ which appears infinitely many ntimes for th second switch, so choose this position for $s_2$. We can repeat the process, and find $s_k^{i_k}$ for switch $s_k$. It's easy to see that these positions will put out all the light bulbs, a contradiction.

That means that there is some $n$ for which we can't put out all the $b_1,b_2,\ldots,b_n$.


Given two planes

$ \Pi_{1} : a_{1}x \plus{} b_{1}y \plus{} c_{1}z \plus{} d_{1} \equal{} 0 \\
\Pi_{2} : a_{2}x \plus{} b_{2}y \plus{} c_{2}z \plus{} d_{2} \equal{} 0$,

the following equation is true:

$ \cos \theta \equal{} \frac {a_{1}a_{2} \plus{} b_{1}b_{2} \plus{} c_{1}c_{2}}{\sqrt {a_{1}^{2} \plus{} b_{1}^{2} \plus{} c_{1}^{2}}\sqrt {a_{2}^{2} \plus{} b_{2}^{2} \plus{} c_{2}^{2}}}$

($ \theta$, in this case, is known as the [url=http://mathworld.wolfram.com/DihedralAngle.html]dihedral angle[/url]). Plugging in our values, we find that

$ \cos \theta \equal{} \frac {39}{\sqrt {1554}} < 1$,

therefore the two planes intersect (note that the planes would be parallel iff $ \cos \theta \equal{} 1$).

[quote="sharkman"]Determmine whether the planes are parallel.

4x - y + 2z =5 and 7x - 3y + 4z =8[/quote]

[hide="Vector solution."]
-----
$ 4x\minus{}y\plus{}2z\equal{}5\implies 4(x\minus{}1)\minus{}(y\minus{}1)\plus{}2(z\minus{}1)\equal{}0 \implies 4x'\minus{}y'\plus{}2z'\equal{}0 \implies [4,\minus{}1,2]\cdot[x',y',z']\equal{}0$
$ 7x\minus{}3y\plus{}4z\equal{}8\implies 7(x\minus{}1)\minus{}3(y\minus{}1)\plus{}4(z\minus{}1)\equal{}0 \implies 7x'\minus{}3y'\plus{}4z'\equal{}0\implies [7,\minus{}3,4]\cdot[x',y',z']\equal{}0$

The two planes thus have normal vectors $ [4,\minus{}1,2]$ and $ [7,\minus{}3,4]$, respectively, which are not parallel, and thus the planes are not parallel.
-----
[/hide]


invariant
monovariant
function
inequalities
combinatorics proposed
combinatorics

Some people are in a building with several rooms. Each minute a person leaves a room and enters one with at least as many people. Show that eventually all the people are in one room.

[hide]I solved this using induction. Can you solve it using invariant or monovariant?[/hide]

I think that the sum of the squares of the inverses of the number of peoples is monovariant.(i.e, only increases or decreases)

[quote="t0rajir0u"]You can take any convex function of the number of people in each room; for example, the sum of the squares of the number of people in each room is also a monovariant (strictly increasing).[/quote]Wow, that's a new insight for me. Thank you so much!

The key is that at every point the sequence of numbers of people in each room [url=http://en.wikipedia.org/wiki/Majorization]majorizes[/url] the previous sequence.  Schur-convexity and [url=http://en.wikipedia.org/wiki/Karamata%27s_inequality]Karamata's inequality[/url] are relevant here.

I think that this has appeared on Leningrad MO.


Given $x,y,z \geq 0$, prove:
\[{xy \over \sqrt{xy+2z^{2}}}+{yz \over \sqrt{yz+2x^{2}}}+{zx \over \sqrt{zx+2y^{2}}}\geq \sqrt{xy+yz+zx}. \]

Weighted Jensen on convex function $x \rightarrow \frac{1}{\sqrt{x}}$:
\[{xy \over \sqrt{xy+2z^{2}}}+{yz \over \sqrt{yz+2x^{2}}}+{zx \over \sqrt{zx+2y^{2}}}\geq \frac{xy+yz+xz}{\sqrt{\frac{\sum_{cyc}{x^{2}y^{2}}+2(xyz^{2}+xy^{2}z+x^{2}yz)}{xy+yz+xz}}}\]
Now, we see that
\[\sum_{cyc}{x^{2}y^{2}}+2(xyz^{2}+xy^{2}z+x^{2}yz)=(xy+yz+xz)^{2}\]
so the inequality is proven.

I'm a bit confused...
Do you mean that I didn't explain my solution well ?
Or do you mean that you perfectly understand my solution, but that you think it is too complicated because there exists a very easy solution?

If you mean I didn't explain well, I must admit that you're right.  :)
Sorry, here's a better explanation:

Because $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $(0,+\infty)$ (because $f''(x)= \frac{3}{4\sqrt{x^{5}}}$) we can apply the Weighted Jensen's inequality:
\[\frac{\frac{xy}{\sqrt{xy+2z^{2}}}+\frac{xz}{\sqrt{xz+2y^{2}}}+\frac{yz}{\sqrt{zy+2x^{2}}}}{xy+yz+xz}=\frac{xy.f(xy+2z^{2})+xz.f(xz+2y^{2})+yz.f(zy+2x^{2})}{xy+yz+xz}\]
\[\geq f(\frac{xy^{2}+2xyz^{2}+y^{2}z^{2}+2x^{2}yz+x^{2}z^{2}+2xy^{2}z}{xy+yz+xz}) = f(\frac{(xy+yz+xz)^{2}}{xy+xz+yz})\]
\[= \frac{1}{\sqrt{xy+xz+yz}}\]


Strange!

  $a,b,c$ are sides of triagle then $a+b >c$ so $\frac{a+b}{c}> 1$
you know what I mean?

Oops, small problem, made a + instead of a -, I change it now.

Thank you.

Left inequalty is equivalent with $\prod_{cyclic}{(a+b-c)}>0$ which is is true because $a,b,c$ form a triangle.

Right inequality is just Schur, so no need for triangle condition.

[quote="SBHD"]If a,b,c are the sites of a triangle, prove that
$2 < \frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}-\frac{a^{3}+b^{3}+c^{3}}{abc}\leq 3$[/quote] \[(*) \iff \sum_{cyc}(a-b)^{2}{a+b-c \over 2abc}\ge 0.\] Anyway, it was posted, http://www.mathlinks.ro/Forum/viewtopic.php?t=73396

I think this inequality is extremel easy!The second inequality is Schur and you can prove it easily by using S.O.S(i think we don't nedd to use it!)


Whats the point of getting Mathematica again?
Ive never gotten it but do they send you a CD to install it?
Since it student edition does that mean you can only use it as on you are a student?
If so how would they make it so that you can only have it as long as you are a student?
What is the lowest grading level you can get (gold, silver, bronze) to get mathematica from what you have seen in the years you've participated in USAMTS.

Thanks :DDD

Mathematica is a very powerful calculator. That's the reason. :)

They send you a CD, i believe... I didn't get my mathematica from USAMTS, so someone verify.

Student edition doesn't mean you can only keep it for a certain amount of time. You can keep it for as long as you want.

You get to choose prizes, starting with gold winners, and then silver gets the prizes left over, etc. Im not sure how long Mathematica usually lasts though...

It lasted to the beginning of bronze last year, but if you were bronze you had to get up at 3 AM (EST) or stay up until midnight PST to get it.

[quote="pianoforte"]//If it lasts long enough, could you get Mathematica as an honorable mention?//

Seems tautological.. if you can't possibly get Mathematica, it probably didn't last long enough  :P .

I suggest stop worrying about prizes. Go do the problems and try to get the best score that you can. You can worry about prizes at the end, when you know what score you got and your placing, and when the prizes have been determined. The point of USAMTS is to work on cool problems, not whether you can get Mathematica.[/quote]
No, I meant are you eligible for Mathematica as an HM? or is it only for gold, silver, bronze?

I agree that the point of USAMTS is to get better at math and solving cool problems, but a powerful calculator like Mathematica is a great (secondary) incentive  :)


A = (-3, -6) B = (-2, 4)
How do I determine the slop equation from line passing A to B?

y = mx + b


We define two sequences $ \{a_n\}_{n \equal{} 1}^{\infty},\{b_n\}_{n \equal{} 1}^{\infty}$ as
$ a_{n} \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$, $ a_1 \equal{} 1,a_2 \equal{} i$, and $ b_n \equal{} \frac {a_{n \plus{} 1}}{a_n}$.(where $ i$ is the imaginary unit)

Show that all $ b_1,b_2,b_3,\cdots ,b_n,\cdots$ are on the circumference of the same circle in the complex plane.

[hide="outline"]Find $ b_1,b_2$, and $ b_3$ and find their circumcircle, and prove that the rest lie on that circle.[/hide]


geometry
function
Triangle
area
IMO Shortlist

Three points $A,B,C$ are such that $B \in ]AC[$. On the side of $AC$ we draw the three semicircles with diameters $[AB], [BC]$ and $[AC]$. The common interior tangent at $B$ to the first two semi-circles meets the third circle in $E$. Let $U$ and $V$ be the points of contact of the common exterior tangent to the first two semi-circles. Denote the area of the triangle $ABC$ as $S(ABC)$. Evaluate the ratio $R=\frac{S(EUV)}{S(EAC)}$ as a function of $r_1 = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$.

$X=EA\cap (AB),\ Y=EC\cap (BC)$, where, in general, $(PQ)$ is the semicircle with diameter $(AB)$. It's obvious tht $X,\ Y$ are the projections of $B$ on $EA,\ EB$ respectively. At the same time, $E$ is on the radical axis of $(AB)$ and $(BC)$, so $EX*EA=EY*EC$, so $\triangle EYX \sim \triangle EAC$. This means that $\angle BYX=\angle EXY=\angle ECA=\angle BCY$, so $XY$ is tangent to $(BC)$. In the same way we show that $XY$ is tangent to $(AB)$. This means that $X=U,\ Y=V$.

$\frac{S(EUV)}{S(EAC)}=(\frac{UV}{AC})^2=(\frac{EB}{AC})^2=\frac{AB*BC}{(AB+BC)^2}=\frac{r_1*r_2}{(r_1+r_2)^2}$ (because triangles $EVU$ and $EAB$ are similar and $EUBV$ is a rectangle).


Every positive integer can be represented as a sum of one or more consecutive positive integers. For each $n$ , find the number of such  represententation of $n$.

[quote="cuopbientcd"]n=1  is easy.
n=2k+1 then n=k+(k+1)
If n is even number then n must has odd factor(why?) if  can express
 I let $n=2^{i}.(2k+1)$
- If $2^{i}>k$then  I write $n=(2^{i}-k)+(2^{i}-k+1)+...+(2^{i}+k)$
- If $2^{i}<k$ the I write$n=(k-2^{i}+1)+..+(k+2^{i})$[/quote]

You NOT understand problem  :D

NO I DON'T THINK THIS .
IF  n=1 or n odd or n=2^i(2k+1) then n only express as sum of two or more consecutive positive integers . and  orther situation then  can't express. ARE YOU OK?

Are you sure? :D
[quote="N.T.TUAN"]Every positive integer can be represented as a sum of [b]one or more [/b]consecutive positive integers. For each $n$ ,[b] find the number of such  represententation of $n$[/b].[/quote]

[hide="Answer without proof"]The answer is the number of divisors of the odd part of $n$.  Each such factor corresponds in a reasonably natural way to one sum of consecutive integers.[/hide]

[b]My Solution[/b]

Let us call a set of consecutive integers which sum to $n$ a [i]Sum[/i]. Let us call a sum a [i]Positive Sum[/i], if all of the integers in the sum are positive, and let us call a sum a [i]Negative Sum[/i] if at least one of the integers of the sum are non-positive. For example, the sum $(1+2+3+4)$ is a Positive Sum and the sums $(0+1+2), (-2+-1+0+1+2+3)$ are Negative Sums.

Now we show that for a fixed positive value of $n$, the number of Negative Sums equals the number of Positive Sums. First we show that given a Positive Sum, we can construct a unique Negative Sum from it. Let $n = a+1~+~a+2~+~a+3~+~a+4~\dots +~ a+k$, where $a$ is a whole number. Now consider the Negative sum $n = -a~+~-(a-1)~+~-(a-2)~+~\dots +~0~+~1~+~2~+~\dots a~+~a+1~+~a+2~+~\dots +~a+k$. This Negative sum can be constructed from the Positive Sum. For instance, the Negative Sum corresponding to $2+3+4$ is $-1 + 0 + 1 + 2 + 3 + 4$, and the negative sum corresponding to $1+2+3$ is $0+1+2+3$. Thus, the number of Positive Sums $\le$ number of Negative Sums. Now we see that we can construct a unique positive sum from a negative sum, in the exact reverse of the proof of our previous claim. So, we conclude that the number of Negative Sums $\le$ number of Positive Sums. Hence, number of Positive Sums equals the number of Negative Sums.

We claim that the number of sums is twice the number of divisors of the odd part of $n$. Let $n = a+1~+~a+2~+~a+3~+~a+4~\dots +~ a+k$, where $a$ is an integer. We know that $a+1~+~a+2~+~a+3~+~a+4~\dots +~ a+k = \frac{k}2 \cdot (2a + k + 1)$. So, we have $n = \frac{k}2 \cdot (2a + k + 1)$. So, we conclude that $k\mid 2n$. Now we have $2a + k + 1 = \frac{2n}{k}$, and so, $2a = \frac{2n}{k} -k -1$. As $a$ is an integer, we conclude that $\frac{2n}{k}, k$ have different parity. Note that at least one of $\frac{2n}{k}, k$ is even. Both cannot be even for integral value of $a$. So, either $k$ is odd or $\upsilon_2(k)=\upsilon_2(2n)$. Now assume that $n = 2^t (2l+1)$, where $t,l$ are non negative integers. So, $2n = 2^{t+1} (2l+1)$. When $k$ is odd, from $k\mid 2n$, $k\mid 2l+1$. If $k$ is even, we have $2^{t+1}\mid k$ and $\mid 2^{t+1}(2l+1)$. So, the number of even $k$ equals the number of odd $k$. As the number of odd $k$ is equal to $\tau (2l+1)$ from $k\mid 2l+1$, we have the total number of Sums is $2\cdot \tau (2l+1)$. We need the number of positive sums, which is $\tau (2l+1)$, the number of divisors of the odd part of $n$.      [b]Q E D[/b]


What is your predicted score/

How hard do you think its gonna be

is it getting a lot harder each year?...cuz i just did the 2001 one and it was pretty easy...i mean...it wasnt so hard...lol

Some people (including me) believe that odd years are harder than even years. So, it might be hard...

LOL bryant what's ur answer for #2?
i got 65 but i am not sure if it's correct
someone in the discussion post got 56 :huh:

65...lol...if ur not sure just put it into the equation [N/3]=[N/5]+[N/7]-[N/35]... :P

It'll probably be same difficulty as last year's. It's made by same country...

[quote="Tomaths"]coments on your score???[/quote]
dis is me first year...dun kno...probubly around 20...

woah maybe a bit over confident (no offence)

So, the contest is tomorrow.
Good luck to everybody writing APMO! :)

[b]Please do not discuss the contest till Wednesday because the people in Asia are writing the contest on Tuesday. NOTE: Last yr, it said on the contest paper that you can't discuss the problems via internet till the problems are posted on the APMO website.[/b]

Man, I am gonna bomb this contest. I was sick for the past week with a fever and I just got a little better now and probably my parents won't even allow me to write the olympiad. Life sucks. :(

Sports psychology... As corny as it sounds, you're more likely to bomb it if you say you're going to bomb it. A gold medalist never has a doubt in his/her mind that they were going to win. If you end up writing it, you're talented, you'll destroy those questions.

Good luck to everyone!

[quote="bryant124"]lol...i just skimmed through the 2002 competition...the last problem seems to be the easiest...and the first one is soooo hard(at least to me...)i dont get it...dont they put the problems in the order of increasing difficulty?cuz if the first problem is really hard then i might get really nervous and wont be able to solve any of the problems...i think generally i can solve 2-4 problems...[/quote]
2002 is much easier relative to 2006, don't know why~maybe i'm just more used to the problems on 2002?
and i agree, 1st question on 2002 is the hardest(4me) and 4th question is the easiest tho solved within a few steps.

ya Good Luck! to all APMO writers tomorrow...

P.S. rem u'll probably pwn!!!!

Not yet. The problems must first appear on the official website:
http://www.kms.or.kr/competitions/apmo/

btw...when r the results gonna come out?

They usually E-mail the results( your score and rank) around the beginning of May.

....end of May?? wow they reli spend a loooong time on marking....
i think i got 1, 3 and a part of 5.....but 3 seems too easy (!), so maybe i did wrong....and 5...if i was not wrong...it was a very surprising answer

daaaaaaaaaaaaaaaamn!!!!!!! i cudnt sleep last night and got up and proved #4 in 10 minutes...and i was just 1 step away from solving it in the actual test, but then my head was messed up and  i just thought it was impossible and gave up.................................. :ewpu:  :wallbash:  :wallbash_red: now im soooooo mad i can cut myself....oh nvm...i just did...

Bah, I got #1 (with a really crappy solution that was about 2 pages long >.<), #2 (with a proof, but not a really good one >.<), and #5 without proof. I'm going to get at most 16, but even that's wishful thinking.

It beats what I did last year, though, I only got 12 last year (though the 2006 #5 was a joke...)

the australian results are out, and it was rare for people to have scored 7s on question 1, in fact noone did, for q2, a mark was lost for not using the condition that AC>AB or something, which meant that there was only one possibly diagram, so most ppl got 6 for q2, lol everyone got 0 on q4, several ppl got 2 marks on q3 and 1 person solved q5.

Wow...Australia is fast!

Yeah, for #1, I'm told you lost 1 point for not mentioning the integers could be negative.  I missed this, and it seems alot of other people did too.

lol...i mentioned the integers could be negative in #1, and i also used the condition that AB>AC(or maybe its the other way, i dont remember) in #2...damn...i should have believed in myself and not give up so fast...#4 was totally easy...

$\longrightarrow$ bryant: Well you still have two more years so..dont worry i guess...

how does australia have their results so fast?

lol...yea...2 years is a reeeeeeally long time...but i dont know...my parents want to move to US... :( ...but theyr really slow in making decisions :P ...

errrrrrr...they still havent posted the problems...i wanna see their solution to #4...mine was just geometrically innovative... :rotfl:

edit: the australians only got 0 or 2 on #3...so i guess there is a lot more to it than we thought...

I'd hope I got that god dang number 4, and well, maybe I'm missing a few steps in the middle, but I'd believe I got parts of it. For number 2, grrr, only after carpo explained to me for 10 min i sorta got it.

[quote="sen"]hey bryant, are you into kobe?
man i love him.[/quote]
lol...yea...huge fan...btw...we can discuss apmo now...i feel my solution to #4 is better than the official solution(which was just tooooo long and boring to read...)...btw i didnt prove it until recently...

more easy than the other years.. i solved the first 4 problems in one hour.

(i'm not contestant, Brazil aren't participant in APMO  :( )


I just wanted to know the ratio of males to female on aops. :D  I'm hoping no one will vote other... I put it just in case... :) Post your votes here if you want to! (Yes! 100% female so far!  :D I'm a girl.)

Now, what exactly would “other“ mean?  :)  :?:

Ummmmmmmm... I don't know :lol: Maybe I should edit it and make it a both? :D

Not at all, darling, not at all. You are obviously well acquinted with people around here  :D

I voted female but i'm actually a guy :ninja:

stop it! it's not lame and i think i kind of gave her the idea.

you can ignore moogra...he always talks nonsense(especially on aim)

Hey! You guys have to post truthfully! Or else it defeats the whole purpose! :mad:  :lol: (It's NOT lame and Funcia did give me the idea =D)

The poll has ALREADY BEEN MADE...at least revive that one instead of making another...

hehe i don't even remember this poll but it seems i wrote alot

SMACK THAT
[color=white]ALL ON THE FLOOR[/color]
SMACK THAT
[color=white]GIVE ME SOME MORE[/color]
SMACK THAT
[color=white]'TIL YOU GET SORE[/color]
SMACK THAT

[quote="i_like_pie"][quote="kstan013"]i is alien from venus[/quote]
And you speak English? :roll:[/quote]
yes, course of. all we alien speak engulish.
we is vry smart.

Why why did did you you repeat repeat each each option option twice twice? :D

Male or female? Yes :harhar:

[quote="i_like_pie"][quote="kstan013"]i is alien from venus[/quote]
And you speak English? :roll:[/quote]
Babelfish. Babelfish, my friend.

[quote="mathnerd314"]Why why did did you you repeat repeat each each option option twice twice? :D[/quote]
[quote="laughinghead505"]Heeeeeeeeeey! Where did it go?  :(  I'll show the poll not to disappear :mad:

EDIT: Not fair! It messed up! Please vote for the first ones only.[/quote]
Third post in the topic...


function
complex analysis
complex analysis unsolved

if f(z) is an entire function of genus h, then how large and how small can the genus of f(z^2) be?

thx


inequalities
function
inequalities unsolved

$a,b,c,d$  are real numbers. Prove that:

$(a+b+c+d)^2\leq 3(a^2+b^2+c^2+d^2)+6ab$

Let $x=a+b,y=c,z=d$ and just use Jensen's inequality for the convex function $f(x)=x^2$

Pierre.

Well, using Cauchy-Schwarz when all the terms in a sum are equals is exactly the same thing than Jensen's inequality and the convexity of $f(x)$ :D  :D

Pierre.

Clearly all of these solutions are equivalent. One must be blind not to observe that

$3\left( a^{2}+b^{2}+c^{2}+d^{2}\right) +6ab=3\left( \left( a+b\right) ^{2}+c^{2}+d^{2}\right)$

and

$\left( a+b+c+d\right) ^{2}=\left( \left( a+b\right) +c+d\right) ^{2}$,

so the inequality just reduces to $3\left( x^{2}+y^{2}+z^{2}\right) \geq \left( x+y+z\right) ^{2}$, what is rather well-known.

  Darij


analytic geometry
algebra
function
domain
topology
calculus
geometry

I didn't know which forum to put this in.  Hope I guessed correctly.

In then pursuit of defining hypervolume in 4+ dimensions, my friend and I tried to create a definition of inside and outside.  He proposed that from any path that can be extended indefinately to reach the end of the coordinate axis will always intersect a figure if it is inside the figure.  I proposed the counterexample:
[tex]\displaystyle{r=2-2^{\frac{-\theta}{2*\pi}}}[/tex]
for the domain [0, :inf: )

That's supposed to be a -:theta:/(2* :pi: ) in the exponent.  Graph it on your calculator, it'll probably help.  It's a spiral.  He said it's closed.  Is it?

It would be great if someone posted a picture of it on these forums...

Well, it doesn't really make sense to talk about the inside and outside of a region unless it is defined by a simple closed curve. Yours isn't a simple closed curve; it's just a curve.

(A simple closed curve is a curve defined parametrically by

[tex]\displaystyle \begin{cases} x=f(t) \\ y=g(t),\end{cases}[/tex]

where the curve is defined on [tex]t\in[0,1][/tex] so that f(0)=f(1) and g(0)=g(1) but apart from that, if f(a)=f(b) and g(a)=g(b), then a=b.)

Your nomenclature is a bit confusing, because "open" and "closed" regions mean something different in math than how you're using them. What I think you mean is bounded and unbounded regions.

But what you're talking about is addressed by the Jordan Curve Theorem. The Jordan Curve Theorem says that any rectifiable simple closed curve* divides the plane into two regions - an inside and an outside. While the theorem seems intuitively obvious, it's actually pretty hard to prove. You need some topology to do it. This theorem turns out to be very important in multivariable calculus and complex analysis though.

*A curve is defined to be a continuous mapping from a finite interval into the plane. Rectifiable means that it has finite arc length, simple means that it's not self-intersecting, and closed means that it maps the endpoints of the interval to each other. The simplest example of a rectifiable simple closed curve is just a circle.

Here is an interesting site with some information about the JCT - http://www.math.ohio-state.edu/~fiedorow/math655/Jordan.html

Hmmmm.  Ok, well, does my figure have area?  The limit of the figure is a circle with radius 2, so in theory it should have area of 4 :pi: .  But is that possible?  If so it's certainly a fun paradox: infinite arc length but finite area.


Can we talk about those problems in Day1 in the evening??

Is there any reason for this? Because everyone has to take it at the same time, right? Then again, they always restrict discussion for an extra day or so, so that would make sense, too.

But everyone does have to take it at the same time.  And we're all in the US or Canada (I think).  And the problems are available on the internet too.

I was going to request that discussion be halted on Tuesday, April  19, from 12:00 noon EDT until Thursday, April 21, at 12:00 EDT.
Since the issue has now been raised, I'll request it now, and request it again on Monday.

Although the vast majority are taking the USAMO simultaneously, in the past couple of years there are some distant timezones in the Pacific Rim and Europe where we allow students to start or end the USAMO within the test-taking period but not finish simultaneously.  (That way, they can take the USAMO in something like reasonable daylight hours.)  True, all students are supposed to be sequestered from each other by this mechanism, but I want to seal all avenues that I can to maintain the highest possible integrity.

There is another reason:  We now ask that all the answers be immediately faxed to us, but practical experience shows that it takes a couple of hours before all the papers have been faxed to us.  If discussion is permitted following the end of the Day 1 or Day 2 USAMO period, given the speed with which discussion actually takes place, there is a possibility someone could read a solution or partial solution, quickly add something to their paper and then fax it in.  Again, I agree the possibility is remote and extreme, but I want  to maintain the utmost level of integrity, so I ask for a moratorium on discussion to forestall even the remote and extreme.

I trust that you will not find a 48-hour moratorium on discussion to be a hardship.  Remember that only in the last two years has the USAMO been visible on the internet as means of disseminating the contest.  And this forum has only been in existence about the same amount of time.  Your slightly older colleagues can remember when discussion about USAMO problems among themselves had to wait weeks.  You should be able to wait a couple of days.

Steve Dunbar
AMC Director

Another good reason not to discuss the tests right after Day 1 is that you need to rest for Day 2.  After the test on Day 1, you should mentally take the rest of the day off.

Also, keep it slient will help each one of you to regroup -- do not let unnecessary wrong info. flying around to hurt your performence the next day. We do that in the IMO all the time.

Best of luck!

Zuming

After staring at those problems for 4.5 hours, I could wait a while to talk about them.  A break is nice once in a while.

Can't wait for college so I don't have to worry about these days anymore.

[quote="DPopov"]Definitely! No reason to rush and figure out how many you got right, you are only likely to panic for the next day and not perform your best![/quote]

it's not how many you got right. It is more like how MUCH you got right.


Steve Dunbar and Zuming Feng will be hosting a Math Olympiad Summer Program (MOsP) Math Jam at 7PM EST/4PM PST on June 2nd.  Students who will be attending MOsP are encouraged to attend as the AMC Director and IMO Team Leader will discuss MOsP with the students.

Students who are not attending MOsP are encouraged to attend, though out of politeness to students attending the summer program I would ask that those students refrain from asking questions.

The Math Jam schedule will be updated tomorrow with the Math Jam originally scheduled for the 2nd moved to the 9th.

[quote="MithsApprentice"]AHHH :-(

*mutters about ARML team leaving town on June 2nd*

What's gonna be discussed?[/quote]

The Math Jam will be an information session about MOsP.  Nothing vital will be discussed, but it should help orient students toward the program.  You can always take a look at the transcript later.


$I$
Prove that if
$(1+a_1)(1+a_2)...(1+a_n)=2^n$
then $1\geq a_1a_2a_3...a_n$

$II$
If $a+b+c=2$ and $a,b,c<1$
Prove that
$\frac{abc}{(1-a)(1-b)(1-c)}\geq 8$

They are trivial but hard for me :(  :(  :(  PLZ help me

for the first one,using chebycheff $n-1$ times,we obtain
$(1+a_1)(1+a_2) \geq 2(1+a_1a_2)$;$2(1+a_1a_2)(1+a_3) \geq 4(1+a_1a_2a_3)$...
until we have $2^{n-2}(1+a_1a_2...a_{n-1}) \geq 2^{n-1}(1+a_1a_2...a_n)$.So,the left member is $\geq$ the quantity in the left of the last inequality.
We have $2^n \geq 2^{n-1}(1+a_1...a_n)$,so $2 \geq (1+a_1...a_n)$,and $1 \geq a_1...a_n$,Q.E.D.

$a_i + 1 \geq 2\sqrt{a_i}$, so $2^n \geq 2^n\prod{\sqrt{a_i}}$, so $\prod{a_i} \leq 1$.

can somebody tell me where I am going wrong here....
we know that,
$a+b+c=2$,
$\frac{abc}{1-a-b-c+ab+ac+cb-abc}$
$\frac{abc}{1-2+ab+bc+cb-abc}$ - $\boxed1$
we know that $a^2+b^2+c^2<=a+b+c$, so
$a^2+b^2+c^2+2ab+2ac+2bc=4$
$a+b+c+2(ab+ac+bc)<=4$
Solving,$ab+bc+ca<=1$
so, equating in $\boxed1$
$\frac{abc}{-abc}$
$=-1$
wher am I going wrong? :?  :(

[quote="riddler"]wher am I going wrong? :?  :([/quote]
Well, $a^2+b^2+c^2\le a+b+c$ is not quite true.
Although it works for $0\le a,b,c\le 1$, it doesn't work for negatives...

have you got an alternate solution? :?  :(  I hate inequalities

This may.. or may not work.

For 2: Put in $1-a = x$, $1-b = y$, $1-c=z$. Then $x,y,z$ are positive numbers. From the restriction we had, we get that $x+y+z=1$. We need to show

$(1-x)(1-y)(1-z) \geq 8xyz$ or

$(x+y)(x+z)(y+z) \geq 8xyz$.

But $x+y \geq 2\sqrt{xy}$ and so on, multiply the three together and we get the result.

For the first one,just use Bernoulli and AM-GM and the conclusion follows imediately. ;)

Pls correct me if I'm wrong.
[hide]
[tex](a+b+c=2)\ \wedge\ (a,b,c<1)\ \Rightarrow\ 0<a,b,c[/tex]

[tex](a-b)^2\ge 0\ \Leftrightarrow\ a^2+b^2\ge 2ab\ \ \ (1)[/tex]
[tex](b-c)^2\ge 0\ \Leftrightarrow\ b^2+c^2\ge 2bc\ \ \ \ (2)[/tex]
[tex](c-a)^2\ge 0\ \Leftrightarrow\ a^2+c^2\ge 2ac\ \ \ (3)[/tex]

[tex]\frac{(1)+(2)+(3)}{2}\ \Leftrightarrow\ a^2+b^2+c^2\ge ab+bc+ac[/tex]

[tex]4=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac\ge 3ab+3bc+3ac[/tex]

[tex]\Longleftrightarrow\ ab+bc+ac\le \frac{4}{3}[/tex]

[tex]\frac{abc}{(1-a)(1-b)(1-c)}=[/tex]
[tex]=\frac{abc}{-1+ab+bc+ac-abc}\ge\frac{abc}{-1+\frac{4}{3}-abc}=\frac{3abc}{1-3abc}[/tex]

[tex]\frac{3abc}{1-3abc}\ge 8\ \Leftrightarrow\ 27abc\ge 8[/tex]

[tex]abc\ge \left (\frac{2}{3}\right )^3[/tex]

[tex]\sqrt[3]{abc}\le \frac{a+b+c}{3}[/tex][/hide]

Btw for (2) $a,b,c \geq 0$ must hold. If we let $a<0$ then we have $b+c=2+|a|$ and $b,c>1$ which contradicts the fact that $a,b,c<1$. Also, if two or more variables are $<0$ a similar argument holds. Thus we can improve the bound in (2) to $0 \leq a,b,c <1$

With this said, I do not think riddler is going wrong there in his argument and tetrahedron's solution is very nice.

[quote="siuhochung"]the sign of abc is wrong[/quote]

Please explain.

[quote="DPopov"]Thus we can improve the bound in (2) to [tex]0\le a,b,c<1[/tex][/quote]

Why couldn't we improve it a little more? Letting [tex]a=0[/tex] and plugging into [tex]a+b+c=2[/tex] we get either [tex](b<1)\ \wedge\ (c>1)[/tex] - contradiction or [tex](c<1)\ \wedge\ (b>1)[/tex] - contradiction or [tex]b=c=1[/tex] - contradiction.

So we see that [tex]0<a,b,c<1[/tex] must hold.

[quote="aappaa"]
You need
[tex]abc\ge \left (\frac{2}{3}\right )^3[/tex]
to be true. But you only have
[tex]\sqrt[3]{abc}\le \frac{a+b+c}{3}[/tex]
which is of reversed sign. That means the inequalities you used are too weak.[/quote]

Of course you're right. I must've been blind.  :( Thanks.

[quote="Gohan"]$II$
If $a+b+c=2$ and $a,b,c<1$
Prove that
$\frac{abc}{(1-a)(1-b)(1-c)}\geq 8$[/quote]

$\frac{abc}{(1-a)(1-b)(1-c)}\geq 8\Longleftrightarrow\frac{8abc}{(2-2a)(2-2b)(2-2c)}\geq 8$

$\Longleftrightarrow\frac{abc}{(b+c-a)(c+a-b)(a+b-c)}\geq 1\ (\because a+b+c=2)\cdots[\#]$

Let $b+c-a=x,c+a-b=y,a+b-c=z$, by $a+b+c=2,\ a,b,c<1$, we have $b+c-a=2(1-a)>0\Longleftrightarrow x>0$,

similarly $y>0,z>0$.Thus $a,b,c$ can be written as $a=\frac{x+y}{2},b=\frac{y+z}{2},c=\frac{x+y}{2}$.

Then $[\#]\Longleftrightarrow \frac{(x+y)(y+z)(z+x)}{xyz}\geq 8$, which can be proved by using $A.M.\geq  G.M.$

kunny

Of course you are correct apaa that $0 <a,b,c <1$, I just didn't write that since most inequalities hold for non-negatives so I thought it makes little difference. The big tip off is that you get

\[abc \geq 8(1-a)(1-b)(1-c)\]

so you know that you need to use some form of AM-GM.

Put $a= \frac{2x}{x+y+z}, b=\frac{2y}{x+y+z},c=\frac{2z}{x+y+z}$. Then it's enough to prove that
 $xyz \geq (x+y-z)(x+z-y)(y+z-x)$
 You can prove that    $ x^2\geq (x+y-z)(x+z-y) ,y^2\geq (y+x-z)(y+z-x),z^2\geq (z+x-y)(z+y-x)$
or  $(xyz)^2 \geq \left( (x+y-z)(x+z-y)(y+z-x)\right)^2$


calculus
integration
function
geometry
real analysis
real analysis solved

hi

show that the integral of
x*(x-1)*....(x-p) with p an integer >=1   over [0,p]

equals zero if and only if p is even

if p is even it is obvious

however when p is odd it is difficult

my apologies if this does belong in the calculus section

It is actually closer to analysis than to algebra. Anyway, here is the solution.

Note that, for odd $p$, the function $f(x)=x(x-1)\dots(x-p)$ looks on $[0,p]$ like a symmetric sequence of humps with alternating signs, the first and the last hump being negative. We shall show that the area of the first and the last humps taken together is greater than the area of all positive humps, which is enough to conclude that the integral is not $0$. To this end, observe, that for any integer $m<p/2$ and any $t\in(0,1)$ such that $m+t<p/2$, we have
\[
|f(m+t)|=|f(t)|\frac {1+t}{p-t}\frac{2+t}{p-1-t}\dots\frac{m+t}{p-(m-1)-t}\leqslant \frac 2{p-1}|f(t)|
\]
(the first factor is, clearly, not greater than $\frac 2{p-1}$, and all other factors do not exceed $1$). Therefore, the area of each positive hump lying to the left of the middle is less than $\frac 2{p-1}$ times the area of the first hump (if the central hump is positive, we consider only its left half) Also, we have only $\frac {p-1}2$ positive humps. Therefore, the total area of the positive humps lying to the left of $p/2$ (including the left half of the central hump) is strictly less that the area of the first negative hump. The rest is clear.

If $p=2k$ is even the graph of $f(x)=x(x-1)\dots(x-2k)$ has a the point $A(k;0)$ for symetric center $[0,p]$

If $p=2k+1$ is odd the line $x=\frac{2k+1}{2}$ is a symetric axis for the graph of
$f$


[quote="Vihang"]Find all functions f from the set of positive reals to the set of positive reals satisfying

           f(f(x)) + f(x) = 6x

Can this be generalised? (replace 6 by any number k :huh: )[/quote]

Yes I can :D $f(af(bx))+ cf(dx) = ex$ , where $a,b,c,d,e > 0$
:cool:  But can you solve it ?? :cool:


Dont know if this is the right section to post this but, oh well.In a equation what does     P(x) mean?

For example:
$P(x)\sum_{i=1}^9 (-x){}^2{}^0{}^-{}^i(x+1)^i$

[quote="Sir.Aaron"]
$P(x)\sum_{i=1}^9 (-x){}^2{}^0{}^-{}^i(x+1)^i$[/quote]

I think this should be

$P(x) = \sum_{i=1}^9 (-x)^{20-i} (x +1)^i$.

$P(x)$ is simply a polynomial function.

Edit:  A far simpler example, consider

$P(x) = x^2 + 2$

This expression is to be read "P of x is equal to x squared plus two."  It is a function that has specific values for specific $x$.  For example,

$P(1) = 1^2 + 2 = 3$
$P(2) = 2^2 + 2 = 6$
$P(3) = 3^2 + 2 = 11$

etc.


LaTeX
logarithms
algebra
polynomial
irrational number

Hey there ;

Prove or disprove :

if a and b are irrational number then a^b is also irrational number. :?:

Good luck

$ a\equal{}\sqrt 2, b\equal{}2log_2 3$ are irrational, but $ a^b\equal{}3\in Z$.

[hide="The standard example"] $ \left( \sqrt {2}^{\sqrt {2}} \right)^{\sqrt {2} } \equal{} 2$.  :) [/hide]
What if we add the condition that $ a, b$ are algebraic?

LaTeX suggestions for Rust:

\log gives $ \log$
\bf Z gives $ \bf Z$; \mathbb{Z} gives $ \mathbb{Z}$.

[b]

Hello :

what do you mean a,b are algebraic

could you write the def. to think by it.

thanks[/b]

$ a$ is algebraic $ \Leftrightarrow \exists P(x) \in \mathbb{Z}[x] : P(a) \equal{} 0$.

So $ \sqrt{2}$ is algebraic because $ x^2 \minus{} 2 \equal{} 0$ is such a polynomial but $ 2 \log_2 3$ is not (well, as far as I know).

[b]Hey
t0rajir0u

I expect it's true.

I've just tried to prove it by many tries , but all of them failed . :(

So, tell is it true or not ?

And please write the proof. :blush:

IF not true, give me the counter example.

 :)
Thanks.
[/b]

See http://en.wikipedia.org/wiki/Gelfond_theorem or http://mathworld.wolfram.com/GelfondsTheorem.html for the full theorem.

[b]Thanks ZataX .

It's wonderful theorem[/b]

If you have access to this rare but unique book, it gives a proof of Gelfond-Schneider's theorem:

Niven, I., [i]Irrational Numbers[/i], The Carus Mathematical Monographs No 11, 1963, MAA.

I'm so thankful.

I'll try to find that book.

Thanks again. :D


Will there be any questions on the contest that require knowledge of Calculus?

[quote="Yasha"]Although knowing calculus might help with some of the problems, all of the problems will have a way to solve them that doesn't require calculus.[/quote]

I think taht. Because, calculus is the development of basic knowledge. So, we do not need to be serious to think about calculus, but only to master the basic knowledge and develop it.


If $ a,b,c$ are [b]real[/b] numbers such that $ abc(a \plus{} b \plus{} c)\geq 0,$ then

$ a^4\plus{}b^4\plus{}c^4\plus{}abc(a\plus{}b\plus{}c)\geq 2\left(b^2c^2\plus{}c^2a^2\plus{}a^2b^2\right),$

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

[quote="Ji Chen"]If $ a,b,c$ are [b]real[/b] numbers such that $ abc(a \plus{} b \plus{} c)\geq 0,$ then

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c)\geq 2\left(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2\right),$

with equality if $ a \equal{} b \plus{} c \equal{} 0.$[/quote]
It is here
http://www.mathlinks.ro/viewtopic.php?t=229280


A merchant placed on display some dresses, each with a marked price.  He then posted a sign “$ \frac{1}{3}$ off on these dresses.”  The cost of the dresses was $ \frac{3}{4}$ of the price at which he actually sold them.  Then the ratio of the cost to the marked price was:

$ \textbf{(A)}\ \frac{1}{2} \qquad
\textbf{(B)}\ \frac{1}{3} \qquad
\textbf{(C)}\ \frac{1}{4} \qquad
\textbf{(D)}\ \frac{2}{3} \qquad
\textbf{(E)}\  \frac{3}{4}$

[hide="Click for solution"]
Let $ S$ be the selling price, $ M$ the marked price, and $ C$ the cost.  Then $ S\equal{}M\minus{}\frac{1}{3}M\equal{}\frac{2}{3}M$, $ C\equal{}\frac{3}{4}S\equal{}\frac{3}{4} \cdot \frac{2}{3}M\equal{}\frac{1}{2}M$, so $ \frac{C}{M}\equal{}\frac{1}{2}$ or $ \boxed{\textbf{(A)}}$.
[/hide]


[quote="byrdgyrl"][hide]Wouldn't it be 1:2 since the coins are fair?  [/hide][/quote]

How do you arrive at that answer?

[hide]I got 3/4. You can get HT, TH, HH, TT. [/hide]

[hide]the possibility that exactly one is head is 1/2
the possibility that 2 is heads is 1/4
you add them up
and up get 3/4
right?[/hide]

[quote="mathprincess09"][hide]wouldnt it be 3/4 coz there can be a probability of having
TH HT TT HH
so only 3 out of 4 contain heads so its 3/4![[/hide][/quote]

I get confused very easily if i have to write out the head/tail thing that you did if it's a very large number.... that's why I almost always miss the head/tail questions...  :blush:

[quote="MCrawford"]Two fair coins are flipped.  What is the probability that at least one coin is heads?  Express your answer as a common fraction.[/quote]

[hide]flipping one head is 1/2

flipping two heads is 1/4

1/2+1/4=3/4[/hide]

[hide]
u get it by writing out the possible solutions
they are: HH HT TH TT
3 out of the 4 contain heads
so... the answer is 3/4
 :D  :D  :D  :D [/hide]

[hide]
Also $\frac34$
Four different resultes and three of them we have at least one head.[/hide]


I have trouble understand this:

Use our geometric approach to show that for any obtuse angle
$\theta$
$\sin\theta=\sin(\pi - \theta)$

Solution: In our figure we have
$AB=\sin \angle AOB$
from right trangle AOB (with AO=1). Since A is in the second quadrant, we have
$AB = \sin \angle AOC=sin \theta.$
Equating these expressions for AB, we find
$\sin\theta=\sin\angle AOB=\sin(\pi-\theta)$
as desired.

AoPS II pg. 15

just imagine a unit circle. $O$ is the origin, $A$ is in the second quadrant on the circle, $C$ is the pt. on the circle with coordinate $x=1$, $\triangle AOB$ is in the second quadrant, and $\theta=\angle AOC$. $B$ is directly below $A$ so that $AB$ is perpendicular to the $x$-axis. Since $A0=1$, you know $\sin\angle AOB=\frac{AB}{AO}=AB$. Also, from the definition of sine, $\sin\angle AOC=\sin\theta=AB$. Since $\angle AOB$ and $\theta$ are supplementary, $\angle AOB=\pi-\theta$. So $\sin\theta=AB=\sin\angle AOB=\sin(\pi-\theta)$


Prove that $ \sqrt{2\sqrt{3\sqrt{4\sqrt{...\sqrt{(m\minus{}1)\sqrt{m}}}}}}<3$ for all integers $ m\geq2$

[quote="dnkywin"]Prove that $ \sqrt {2\sqrt {3\sqrt {4\sqrt {...\sqrt {(m \minus{} 1)\sqrt {m}}}}}} < 3$ for all integers $ m\geq2$[/quote]
[hide="hint"]Induction Principle![/hide]


Suppose that in a convex hexagon, each of the three lines connecting the midpoints of two opposite sides divides the hexagon into two parts of equal area. Prove that these three lines intersect in a point.

This problem is not very difficult.
 Let $ABCDEF$ is a  convex hexagon and $M, N, P, Q, R, S$ are respectively the midpoints of $AB, BC, CD, DE, EF, FA$. We will show that if $MQ, NR, PS$ divide  the hexagon into two parts of equal area, then $MQ, NR, PS$ are concurrent.
 Suppose $X, Y, Z$ are the intersections of $MQ$ and $NR$, $NR$ and $PS$, $PS$ and $MQ$.
 If $X$ lies on $PS$ then obviously it must coincide with $Y$ and $Z$, and thus $MQ, NR, PS$ are concurrent.
 WLG, we can assume that $X$ inside the convex pentagon $PDEFS$. Hence $S, Z, Y, P$ are collinear in that order. We have
  $S_{AMZS} = S_{DPZQ}$
  $S_{CNYP} = S_{SYRF}$
  $S_{BMZYN} = S_{BMXN}-S_{XYZ} = S_{EQXR}-S_{XYZ}$.
  Thus
  $S_{DEFSP}= S_{ABCPS} = S_{AMZS} + S_{CNYP} + S_{BMZYN}=S_{DPZQ}+S_{SYRF}+S_{EQXR}-S_{XYZ}= S_{DEFSP}+S_{XYZ}$
   Hence, $S_{XYZ} = 0$ (contradiction)
  Therefore $MQ, NR, PS$ are concurrent.

I think we can also generalize this problem because in my proof (if it is 100% correct) I don't use the facts that $M, N, P, Q, R, S$ are midpoints.
 Let $ABCDEF$ be a convex hexagon. Suppose there exist $M, N, P, Q, R, S$ lying on the sides $AB, BC, CD, DE, EF, FA$ such that $MQ, NR, PS$ divide the hexagon into two parts of equal areas. Then $MQ, NR, PS$ are concurrent.


geometry
3D geometry
tetrahedron
analytic geometry

MPSI Class Entrance Test 2007
Test time : 4 hours
English Version
The following exercises can be solved independently and done in any order. The questions are listed from easiest
to most difficult. Calculators are not permitted. Solutions should be written in French or in English.
We note N the set of positive integers and N* = N \ {0} the set of strictly positive integers.
1. Let n and m be in N* and let E be a set of nm distinct elements. A n-grouping of E is a partition of E
into m-elements subsets. For example if n = 2, m = 3 and E = {1,2,3,4,5,6} the partition
{{1,3,5},{2,4,6}} is a 2-grouping. Find the number of n-grouping of E.
2. Let f :]0,+/[1]0,+/[ be the map defined by f(x) =
x
x + 2
for all x > 0.
We consider the following equations : (E1) : f(x) = x ; (E2) : f(f(x)) = x ; (E3) : f(f(f(x))) = x and so on :
(En) : f(f(f(…f(x)))) = x for all n Î N*.
Solve the equation (En).
3. Let n Î N*. What is the maximum value of
=
n
k
k
k
a
1
where a1 ,…,an are real numbers such that :

=
=
n
k
k
a
1
0 and 1
1
 =
=
n
k
k
a ?
4. Two ferryboats leave at the same time the opposite banks of a river. One is faster than the other and
they meet at 720m from the nearest bank. Once arrived, each ferryboat stops for 10 minutes and goes
back. On the return the ferryboats meet again but at 400m from the nearest bank. How wide is the
river ?
5. If the sum of the lengths of the six edges of a trirectangular tetrahedron TABC is s, determine it’s
maximum volume.
6. For n Î N*, we note En the set of all n-digits numbers whose digits are only 1 and 2. For example
E2 = {11,12,21,22}. Prove that there is one, and only one, integer in En that can be divided by 2n.
7. For n Î N*, we note f(n) the product of all positive divisors of n and t(n) the number of such divisors.
For example if n = 10 we have f(10) = 1 × 2 × 5 × 10 = 100 and t(10) = 4.
(a) Let n Î N*. Prove that f(n)2 = nt(n) .
(b) Let n and m be in N*. Prove that f(n) = f(m) then n = m.
8. Let T : N4 1N4 be the map given by T(a) = (|a1 – a2|,| a2 – a3|,| a3 – a4|,| a4 – a1|) for all quadruple
a = (a1 , a2, a3, a4). For n Î N* we not T n = T  T      T (n times).
(a) Prove that for all a Î N4 the coordinates of T 4(a) are all even.
(b) Prove that for all a Î N4, we can find n Î N* such that T n(a) = 0.
* * *
A
T
B
C
Figure 1: the angles ATB,BTC,ATC are right

[quote="happyme"]Copying and pasting web page math is not good.[/quote]

Exactly what I was thinking. Number 2 is already screwed up.
I found the link:
http://nickshome.myweb.hinet.net/MPSI_2007.pdf

What's MPSI? Meggy-Meg Pasta S** Institute?

Who would want pasta s**?
That's nasty :p[/i]

I wrote the first three words that came to mind (the first and third go hand in hand (or * in *), and I was hungry at the time).


linear algebra
combinatorics unsolved
combinatorics

$ A_1,A_2,...,A_{n\plus{}1}$ are subsets of $ \{1,2,...,n \}$
prove that there exist two disjoint subsets of $ \{1,2,...,n\plus{}1 \}$ X and Y such that
$ \bigcup_{i \in X}{}A_i\equal{}\bigcup_{i \in Y}{}A_i$

If I am not mistaken, this problem appeared in AMM and was solved using Linear algebra by Lindstorm.

Pohoata gave an inductive proof.
http://www.cpohoata.com/2008/10/02/440/

I don't understand this part: [quote]Thus after taking away Ai, the number of elements in G does not reduce and the relation we have just written above remains an equality.[/quote] (in case a) on the 3rd page)
could someone explain? :blush:

no offense to pohoatza's proof which is nice too, but the linear algebra proof is I think the most elegant you can get.


algebra
polynomial
function
calculus
calculus computations

Let p(x) be a real polynomial such that for all x, p(x)+p'(x)>= 0. Does it follow that for
all x, p(x)>=0?

$ p(x)+p'(x) \geq 0 \rightarrow e^{x}p(x)+e^{x}p'(x) \geq 0 \rightarrow (e^{x}p(x))' \geq 0$ for all $ x \in \mathbb{R}$

So $ e^{x}p(x)$ is an increasing function on the real line.

Suppose there is an $ x_{0}$ such that $ p(x_{0}) <0$.  Then we must have $ p(x)<0$ for all $ x \leq x_{0}$.

This implies that $ p(x)$ is a polynomial of odd degree.  But this would contradict the fact that $ p(x)+p'(x) \geq 0$ for all $ x \in \mathbb{R}$.

[quote="edoo"]This implies that $ p(x)$ is a polynomial of odd degree.[/quote]

Actually not -- it could be a polynomial of even degree with negative leading coefficient.  In that case, it's also easy to see that the condition $ p(x)+p'(x) \geq 0$ is violated for some value of $ x$.

[quote]Actually not -- it could be a polynomial of even degree with negative leading coefficient. In that case, it's also easy to see that the condition $ p(x)+p'(x) \geq 0$ is violated for some value of $ x$.[/quote]

     I mentally ruled out that possibility since it would fail the condition $ p(x)+p'(x) \geq 0$ in limit.  But I did not write that, so thanks for making the argument complete.  :)


Show that if $ a$ is an integer such that $ 2$ does not divide $ a$ and $ 3$ does not divide $ a$, then $ 24 \mid (a^{2}\minus{}1)$.

hello, since $ 3$ not divided $ a$ we get only two cases:
$ a\equiv 1 \mod 3$ this implies $ a\minus{}1\equiv 0 \mod 3$
$ a\equiv 2 \mod 3$ from here we get $ a\plus{}1\equiv 0 \mod 3$
Since $ a$ is odd (because $ 2$ not divided $ a$) we can write $ a\equal{}2m\minus{}1$ with some $ m\in \mathbb{N}$,this implies
$ (a\minus{}1)(a\plus{}1)\equal{}4m(m\minus{}1)$ hence $ 8|4m(m\minus{}1)$.
Sonnhard.