inequalities geometry trigonometry function calculus AMC USA(J)MO

Prove the inequality for acute triangle abc with area K such that,

27((b^2+c^2-a^2)^2)((a^2+c^2-b^2)^2)((a^2+b^2-c^2)^2) :le: (4K)^6

Gosh, I hope I typed that up right!! Well, have fun with this one.

IW.OAV

[color=cyan]I give you two approaches.

Number 1:  notice that we have an inequality which appears to have an equality case.  Also, equality appears to occur only when a=b=c.  This makes it unlikely that the triangle inequality will prove useful, and instead implies AM-GM.  So, put the area in terms of a, b, and c using Heron's Formula and multiply everything out.  All the way.  Try to make AM-GM work.  An hour or less will tell you whether this really does work out in the end.

Number 2:  b:^2: + c:^2: - a:^2:.  Hmm.  Look familiar?  Think about the Law of Cosines.  K, eh?  How about abSinC/2 ?  Go with it from there.

Notes:  I got bored of number 1 about a third of the way through, and I'm not positive it will work.  And if it doesn't, I'd be really interested in knowing why exactly.  Because, well, it should.  As for number 2, you end up with a trig relationship that I wasn't able to finish off entirely.  But it's definitely very doable that way.[/color]

Sir, I have tried those approaches and they have been futile... can anyone else give me the proof or a link to the proof, it's supposedly a famous one. If there is no proof, well... there's my science fair project!

IW.OAV

Well, you can get someting like:

([(CosA)(CosB)(CosC)]^2)/([(SinA)(SinB)(SinC)]^2) :le: 64/27

I honestly don't know whether that will get you anywhere.

[color=cyan]You should have lost the 64 somewhere, Long.  That expression can be turned around rather nicely into
tan(A)*tan(B)*tan(C) :ge: 3:rt3:.

Clearly, the equality case for this is A=B=C=60 degrees.  This shouldn't be that hard, I don't think.  Perhaps some sort of smoothing argument?  Or replacing tan(A)*tan(B)*tan(C) with tan(A) + tan(B) + tan(C)?

Or use calculus.  That should work.  Since C is an easy funtion of A and B.[/color]

um, yup yup. I should have lost the 64 somewhere, when i canceled out all the powers of 2.

I was thinking of doing the tangents, but I don't where to go after that...

[color=cyan]Join the club :).  Calculus should work, though.  If only I remembered how . . .[/color]

Did you try smoothing? I didn't, but it might work.

[color=cyan]I thought about it, but I haven't tried it.  It certainly looks doable.  Maybe when it isn't 1 AM . . .

I did do it in the case where one of the angles is equal to 60 degrees, although admittedly that doesn't tell you a whole lot . . .[/color]

For a smoothing argument, you might, for example, replace A by 60+x and show that the expression is minimized when x=0.

ahh, I see.

But maybe we're spending lost time. Maybe its not supposed to get (tanAtanBtanC)^2....

[color=blue]I think smoothing works.  First, show it works when one of the angles is 60 degrees.  Then, try smoothing out three angles into a triangle with one 60 degree angle.  I got it to work last night and didn't quite make it work this morning.  But I'm pretty sure it's doable.[/color]

Come on punks, one of yous guys gots ta be able to solve this problem.

Don't think it's insolvable... it's doable. You math hotshots should be able to do this... It's a simple inequality!!! Try using schurs or something, and look at this year's USAMO problem I mean come on, they both have a,b, and c's in em so you should be able to do it... Perhaps this should be an Olympiad problem... IMHO it is of olympiad stature...

Peace

IW.OAV

[hide]Chickens taste good on mondays whereas on other days I enjoy spraying - on my brother cuz it's funny. After that my mom takes us for ice cream and we have lots of fun. I like to eat ice cream because it tastes yummy in my tummy! Shizam![/hide]

[color=cyan](1) why weren't your last 2 messages the same message?
(2) why are you insulting people trying to help you out?
(3) why are you insulting people for not being able to do a problem you can't do either?
(4) why do you care so much, anyhow?
(5) why don't you try and use some of the ideas that we have already given?  The trig thing looks very very promising.[/color]

[quote="Idiot_without_a_village"]Whoa buddy, chill out... I just really wanna solve this problem and methods you guys give aren't really cuttin for me... I don mean to insult anyone[/quote]

If the methods that JBL, ComplexZeta, darkquantum, etc. are giving you aren't working out for you, just say so.  And if YOU want to solve the problem, why are you insulting other people and asking THEM to solve it for you?  If you want to solve it, you should do so yourself.

And if you're kiddin' when you call people "hotshot", make sure you let them know it.  Otherwise they may percieve it as an insult.  It sure does to me.

Is it that hard for you all to take a joke?

People like you make me regret posting the question in the first place.

By the way this is called Ono's inequality or something like that...

And quite frankly, I don't see the need for you to purport on everyone of my posts when all you really do is make other people feel bad. I was just tryin to get peeps to help me with problem in a manner that wasn't so anal.

Thanks.. I think.
IW.OAV

OK, a,b,c are the sides of hte triangle, and let A, B, C be the respective angles.

Note that K=(abSinC/)2=(bcSinA)/2=(acSinB)/2

Thus, (4K)^6= [(abSinC/)2]^2*[(bcSinA)/2]^2*[(acSinB)/2]^2 which is equal to 4^3*[abSinC]^2*[bcSinA]^2*[acSinB]^2

notice that (a^2+b^2-c^2)^2=(2abCosC)^2
                (b^2+c^2-a^2)^2=(2bcCosA)^2
                (a^2+c^2-b^2)^2=(2acCosB)^2

Now, divide the 27 and 4^3 from the left to the right, and the [abSinC]^2*[bcSinA]^2*[acSinB]^2 from the right to the left.

You are left with [(CosA*CosB*CosC)^2]/[(SinA*SinB*Sinc)^2] :le: 1/27

Take the reciprocal and square rooting both sides, we get

tanA*tanB*tanC :ge: 3 root 3

We know that in a triangle, the product of the tangents is equal to the sum of the tangents (this is easily proven by letting the third angle equal 180-(A+B) and then using the tangent sum of angle formulas)

Thus, tanA + tanB + tanC  :ge: 3 root 3

Let tanA + tanB + tanC =x

By the AM-GM inequality,

x/3 :ge: the cube root of tanA*tanB*tanC,

but we know tanA*tanB*tanC=x,

therefore, we get x/3 :ge: the cube root of x

Solving for x by some simple algebraic manipulation, we get that x :ge: 3 root 3.

Thus, tanA*tanB*tanC=tanA + tanB + tanC :ge: 3 root 3.

QED

Hopefully, if that is right, it eliminates the need for painful smoothing arguments or calculus/jensen's theorem.

I guess, Joel, that your not at your optimum at 1am!  :D

[color=cyan]ooh, sexy!  Nice job.

IW.OAV, I am withholding my response to you, except to tell you not to be so rude.[/color]


blogs
limit
calculus
calculus computations

Just to make things readable:

Let $ f \, : (\minus{}\epsilon,\epsilon) \to \mathbb{R}$ be given, $ f(0) \equal{} 0$ and $ f'(0)$ exists.

Show that:

$ \lim_{x \to 0} \frac1x \sum\limits_{n\equal{}1}^k f(\frac{x}{n})  \equal{} f'(0) \sum\limits_{n\equal{}1}^k \frac1n$, where $ k \in \mathbb{N}^*$.

It's sufficient to see that $ \displaystyle \lim_{x\rightarrow 0}\frac{f(x/n)}x\equal{}\lim_{x\rightarrow 0}\frac{f(x/n)\minus{}f(0)}{x/n}\frac 1n\equal{}f'(0)\frac 1n$
 :cool:


vector
linear algebra
linear algebra unsolved

Let H be a subspace of dimension n - 1 of an n-dimensional vector space V. Show that if U is a subspace of V and U $ \nsubseteq$ H, then H + U = V and dim (H$ \cap$U) = dim(U) - 1.

By $ U\nsubseteq H$,then there exist $ x \in U$,but $ x \notin H$.Then it is not hard to prove $ H\plus{}x$ spans $ V$(In fact we can find a basis of $ H$,say it $ e_1,\cdots,e_{n\minus{}1}$,then $ e_1,\cdots,e_{n\minus{}1},x$ is linearly independent,thus it is a basis of $ V$)
For the next,notice:
$ dim(U \cap H)\equal{}dim(U)\plus{}dim(H)\minus{}dim(U\plus{}H)$


Results until now:
IRN1(Kasra): 7 7 - 7 7 -
IRN2(Milad) : 7 7 - 7 7 -
IRN3(Nima) :  7 2 - 7 7 -
IRN4(Mohamad) : 7 7 - 7 7 -
IRN5(Amir) : 7 7 - 7 7 -
IRN6(Mahdi) : 7 7 - 7 - -

wow! that's pretty good up till now. congrats!

As same for others (-) means, I don't know the exact mark  :wink:

The total score will be something like 181 or 182.

congratulation!!! As usual iran do it again and guys proved their ability. i think iran will be 1 or 2 ?   am i right? :lol:

Of course not.

As I know rank of Iranian team will be 6 or 7.

CONGRATULATIONS!!!
I think IRAN will be 3 or 4.

What about Uzbekistan teams result?


trigonometry
limit
calculus
calculus computations

I have 2 problems I got wrong on a quiz and I can't seam to figure them out

lim     [u]xcos(x)[/u]
x->0     cos(x)-1

I know I need to use L'Hospital's rule because its 0/0 but I can't get it to work out

lim      [u] 2[/u]
x->   x^(x-1)

A hint for you. Observe that
\[ \frac{{x\cos x}}{{1 \minus{} \cos x}} \geqslant \frac{1}{x}
\]
provided $ x>0$ small enough.

[hide="Hint for limit 1"]Multiply and divide by (cos x + 1) and use the known limit lim(x->0) (sin x/x) = 1.[/hide]

[hide="Answer to limit 1"]$ \lim_{x \to 0} \frac {x \cos x}{\cos x \minus{} 1} \equal{} \lim_{x \to 0} \frac {x \cos x(\cos x \plus{} 1)}{ \minus{} \sin^2x} \equal{} \lim_{x \to 0} \minus{} \left(\frac {x}{\sin x}\right)^2 \cdot \frac {1}{x} \cdot \cos x(\cos x \plus{} 1)$

$ \equal{} \minus{} 1^2 \cdot \infty \cdot 2 \equal{} \infty$.[/hide]

Regarding your second limit, x tends to what?

X-> 1+ sry for the 2nd one

but for the 1st, we had to solve it using L'Hospitals rule

Using L'Hôpital:

$ \lim_{x \to 0} \frac {x \cos x}{\cos x \minus{} 1} \equal{} \lim_{x \to 0} \frac {\cos x \minus{} x \sin x}{ \minus{} \sin x} \equal{} \frac {1 \minus{} 0}{0} \equal{} \infty$.

[quote="Amandasmith"]X-> 1+ sry for the 2nd one[/quote]

("sry" ?). Then the limit is $ 1^0 \equal{} 1$.


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $f,g:[0,1]\rightarrow\mathbb R$ are integrable and both have integral equal to 1. Prove that there is an interval $[a,b]\subset[0,1]$ such that in that interval both function have integral equal to $\frac12$.

Best,

Here is the interesting reformulation of the problem:

On a blackjack machine you can win or loose one dollar at a time. Show that if two players were playing once every minute in a period, and both of them won 2N dollars, then there was a period during which each of them won exactly N dollars

Source:  Homepage of Vilmos Totik  http://tarski.math.usf.edu/~totik/problems.html


geometry
incenter
circumcircle
perpendicular bisector
angle bisector

i hope this is my own question and no one else has invented it

let ABC be an acute triangle. M,N are on BC,  P,Q are on AC,  X,Y are on AB, such that BM=BA, AC=CN, AP=AB, BC=CQ, AX=AC, BC=BY. (M is towards C, N towards B, P towards C, Q towards A, X towards B, Y towards A) prove that the perpendicular bisectors of MN, PQ, and XY are concurrent.

[hide]It is very easy to prove by suming up segments that the midpoints of MN, PQ, and XY are the tangency points of the inscribed circle of ABC with the sides BC,AC,AB hence these three perpendicular bisectors concur at the incenter I of ABC[/hide]

$BM=AB$ means that the perpendicular bisector of $AM$ is the bisector of $\angle ABC$; similarly, the angle bisector of $\angle ACB$ is the perpendicular bisector of $AN$, hence incenter $I$ of the triangle $\triangle ABC$ is the circumcenter of $\triangle AMN$; similarly for $\triangle BPQ, \triangle CXY$, hence the three perpendicular bisectors pass through $I$.

Best regards,
sunken rock


[b]Problema B4133:[/b]
Fie $ ABCD$ un dreptunghi si punctele $ P\in(BC), \ Q\in(CD)$.
Sa se arate ca in cazul in care $ \triangle{APQ}$ este echilateral, avem: $ S_{AQD}\plus{}S_{ABP}\equal{}S_{CPQ}.$

[b]Problema B4139:[/b]
In triunghiul ascutitunghic $ ABC$ notam cu $ E$ si $ F$ piciarele inaltimilor duse din varfurile $ B$ si $ C$, iar cu $ P$ si cu $ Q$ punctele de intersectie ale dreptei $ EF$ cu cercul circumscris triunghiului.
Aratati ca: $ |AP| \equal{} |AQ|.$

[b]Problema B4140:[/b]
In patrulaterul inscriptibil $ ABCD$ punctele $ E$ si $ F$ sunt mijloacele arcelor mici $ BC$ si respectiv $ AD$, iar cu $ P$ si $ Q$ centrele cercurilor inscrise in triunghiurile $ ABC$ si $ ABD$.
Aratati ca: $ PQ\parallel{}EF$.

Semidreptele $ [AO$ si $ [AH$ sunt izogonale in $ \triangle{ABC}$($ O$-fiind centrul cercului circumscris $ \triangle{ABC}$)$ \Rightarrow$ ele sunt izogonale si in $ \triangle{AEF}$.
$ [AH]$ este insa diametru in cercul circumscris patrulaterului inscriptibil $ AEHF\Rightarrow AO\perp{PQ}\Rightarrow$ punctul $ \{M\}\equal{}PQ\cap AO$ este mijlocul corzii $ [PQ]\Rightarrow$ dreapta $ AO$ este mediatoarea segmentului $ [PQ]\Rightarrow |AP|\equal{}|AQ|.$

Pe scurt, cum $ AO$ si $ AH$ sunt izogonale in $ ABC$, iar $ EF$ e antiparalela cu $ BC$, rezulta ca $ AO \perp EF$, de unde concluzia.

These problems are from the current issue of KöMaL; they should not be discussed until the deadline (15 January 2008) expires.

Please do not destroy our work.


blogs
pigeonhole principle
combinatorics proposed
combinatorics

During a certain lecture, each of five mathematicians fell asleep exactly twice.  For each pair of these mathematicians, there was some moment when both were sleeping simultaneously.  Prove that at some moment, some three were sleeping simultaneously.

[hide="from my blog"]
For each professor, there are 2 falling asleep moments, making 10 total. Call this set of moments S. For each pair, there is a moment at which both are asleep, making 10 also. Call this set of moments T. If any two moments in T are the same, we are done. If not, then they are all different.

We can correspond each moment in T with a moment in S because at the beginning of any interval in which two professors are sleeping, one of them must have just fallen asleep. However, the first 2 moments in S have to come before or at the same time as the first moment in T (since two have to fall asleep before two can be sleeping at the same time). Then the remaining 9 moments in T have to correspond with the 8 other moments in S, which means (by Pigeonhole) that 2 moments in T must be the same. Contradiction.

Hence there must be some moment at which three professors are asleep.[/hide]


modular arithmetic
algebra proposed
algebra

$ \frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},...,\frac{1}{2005},\frac{2}{2005},...,\frac{2004}{2005}$ fractions are written by rising sequence. Find next fraction of this $ \frac{5}{41}$ fraction.

[quote="zaya_yc"]$ \frac {1}{2},\frac {1}{3},\frac {2}{3},\frac {1}{4},\frac {2}{4},\frac {3}{4},...,\frac {1}{2005},\frac {2}{2005},...,\frac {2004}{2005}$ fractions are written by rising sequence. Find next fraction of this $ \frac {5}{41}$ fraction.[/quote]
Let us look for the next fraction with bottom part $ q$ : $ \frac{p\minus{}1}q\le \frac 5{41}<\frac pq$ with $ p<q\le 2005$

This is equivalent to $ p\equal{}\left[\frac{5q}{41}\right]\plus{}1$ (and the condition $ p<q$ is trivially verified).

So we just have to find $ \min_{1\le q\le 2005}\frac{\left[\frac{5q}{41}\right]\plus{}1}q$

Let then $ 5q\equal{}41n\plus{}r$ the euclidian division of $ 5q$ by $ 41$ with $ 0\le r<41$ and $ n\plus{}r\equal{}0\pmod 5$

$ \frac{\left[\frac{5q}{41}\right]\plus{}1}q$ $ \equal{}\frac{5n\plus{}5}{41n\plus{}r}$ and so we need to have $ n$ and $ r$ max with $ n\plus{}r\equal{}0\pmod 5$ and $ 41n\plus{}r\le 5\cdot 2005$ and so $ 5$ possible cases:

$ n\equal{}0\pmod 5$ $ \implies$ $ r\equal{}40$ and so $ n\equal{}240$ and $ \frac{240}{1976}\le\frac 5{41}<\frac{241}{1976}$

$ n\equal{}1\pmod 5$ $ \implies$ $ r\equal{}39$ and so $ n\equal{}241$ and $ \frac{241}{1984}\le\frac 5{41}<\frac{242}{1984}$

$ n\equal{}2\pmod 5$ $ \implies$ $ r\equal{}38$ and so $ n\equal{}242$ and $ \frac{242}{1992}\le\frac 5{41}<\frac{243}{1992}$

$ n\equal{}3\pmod 5$ $ \implies$ $ r\equal{}37$ and so $ n\equal{}243$ and $ \frac{243}{2000}\le\frac 5{41}<\frac{244}{2000}$

$ n\equal{}4\pmod 5$ $ \implies$ $ r\equal{}36$ and so $ n\equal{}239$ and $ \frac{239}{1967}\le\frac 5{41}<\frac{240}{1967}$

And the winner is ... $ \boxed{\frac{241}{1976}}$


algebra
polynomial
calculus
integration
group theory
abstract algebra
Galois Theory

Show that no one $n$-th root of a rational (for $n$ a positive integer) can be a root of the polynomial $x^5 - x^4 - 4x^3 + 4x^2 + 2$.

overkill solution: it has galois group $S_5$  :P

Simpler: when there would be some solution of that type, it's minimal polynomial $x^n-a$ would divide the polynomial.
Thus $a$ would be integral (e.g. by Gauß' lemma).
This leaves us with $a \in \{ \pm 1 , \pm 2\}$ and $n \leq 5$, so just some casework (indeed some more stuff shoul exclude even more before casechecking).

Well, the idea is almost that...
This polynomial is irreducible, and we can suppose wlog that n is not greater than 5.
Then the problem is almost immediate :P

Well, how I can study about this "overkill solution" in detail (Galois) on Net? I think it is very interesting!

We can try a generalization:

Let $f ( X ) \in \mathbb{Q}[ X ]$ be an irreducible polynomial over $\mathbb{Q}$, of degree $p > 3$ prime.
If $f$ has precisely two non-real roots in the complex numbers, then no $n$-th root of a rational can be a root of $f$.

The overkill solution (as proposed by ZetaX) would follow from the following Galois theory fact
If $f ( X ) \in \mathbb{Q}[ X ]$ is an irreducible polynomial over $\mathbb{Q}$ and of degree $p$ prime, and $f$ has precisely two non-real roots in the complex numbers, then the Galois group of $f$ is $S_p$.
Together with this classical group theory fact: If $n > 4$, then $S_n$ is not solvable.
I believe this is enough motivation to start learning Galois :)

Johann : Have you got some background in group theory ?

Galois theory gives us that the no radical expression solves it, but when we just look for radicals / $n$-th roots itself it's a lot simpler:
When we already know that it is irreducible and not of type $x^n-a$, then we are done. The reason is that any root (meaning 'nontrivial' roots, not roots of unity) has a minimal polynomial of type $x^m-b$ (these are irreducible by Eisenstein), and irreducibility of the original polynomial solves the rest.
In general, simply factor it into irreducible polynomials (well this isn't that simple to do in reality ;) ).

Also: it's enough that the polynomial has at least one nonreal and two real solutions to have a nonsolvable Galois group.

Edit: some nonsense corrected.

[quote="ZetaX"]
Also: it's enough that the polynomial has at least one nonreal and two real solutions to have a nonsolvable Galois group.
[/quote]

Let $f ( X ) \in \mathbb{Q}[ X ]$ be defined by $f ( X ) = X^4 - 1$.
The galois group $G$ of $f$ is a subgroup of $S_4$
$S_4$ is solvable, therefore $G$ (as a subgroup of a solvable group) is solvable.
Is the mistake on my or your side ?

Sorry, I forgot to mention: it has to be irreducible and prime degree as yours.

[quote="{x}"]

Johann : Have you got some background in group theory ?[/quote]

Well, I have almost no serious background in Groups but I read I.N Herstein's book :). it is an optional study in my faculty course, and I studied by myself.


In any triangle  ABC  prove  that [tex] 18R^3\geq(a^2+b^2+c^2)R+\sqrt{3}abc [/tex]

I think it is very simple. It suffices to see that $ 9R^2\geq a^2+b^2+c^2$ and then the AM-GM. Am I missing something<?


$ p,q,r$and $ a$ are integers so that $ pq \equal{} ra$ ,when $ r$ is prime and $ gcd(p,q) \equal{} 1$.Prove that one of the numbers $ p,q$ is a perfect square.
[edit] -perfect square *

[quote="MellowMelon"]$ p \equal{} r \equal{} 2, q \equal{} a \equal{} 3$?[/quote]
Hmm, I see. Good counter example.
The problem must be rectified....  :maybe:

solve for the case when the variables are $ >3$

[quote="enndb0x"]solve for the case when the variables are $ > 3$[/quote]
p=r=5, q=a=7

it's $ pq\equal{}ra^2$ with $ (p,q)\equal{}1$ and $ r$ prime...:)

[hide="Solution"]
We know that since $ (p,q) \equal{} 1$, one of $ p,q$ must contain $ a^2$ as a factor - two relatively prime factors of a perfect square are of course also perfect squares, so it's impossible for one to just be a factor of $ a^2$. Suppose that $ p$ does. Then in order for $ p$ not to be a perfect square, $ r|p$. But then $ p$ has to equal $ ra^2$ and then $ q$ is $ 1$, a perfect square.[/hide]

no it is $ pq\equal{}ra$ not $ pq\equal{}ra^2$

Except that $ pq \equal{} ra$ means the problem is completely wrong. If you're going to insist on that you need to give a valid correction.

I took this problem from a book , there's a solution but I am not understanding the solution , so here is the solution.
$ p = p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$,$ q = q_1^{s_1}q_2^{s_2}\cdots q_t^{s_t}$ , $ a = a_1^{l_1}a_2^{l_2}\cdots a_u^{l_u}$ are the canonic form of the number $ p$,$ q$ and $ a$ , when $ p$ and $ q$ has no common divisor greater then $ 1$.We have :$ p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m} \cdot q_1^{s_1}q_2^{s_2}\cdots q_t^{s_t} = r \cdot a_1^{l_1}a_2^{l_2}\cdots a_u^{l_u}$ when $ r$ is prime number , so two sides of equation (left side and right side)
are the canonic form of the same integer.If $ r = a_i$ then we can write $ r\cdot a_i^{2l} = a_i^{2l + 1}$. We have :${ p = p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m} = a_1^{{2l_i}_1}a_2^{{2l_1}_2}\cdots a_n^{{2l_i}_n} = ({a_i}_1}^{{l_i}_1} {{a_i}_2}^{{l_i}_2} \cdots {{a_i}_n}^{{l_i}_n})^2$
so one of the numbers $ p$,$ q$ is perfect square.

Don't blame yourself for not understanding the solution; it looks like total garbage. They seemed to have in fact abruptly switched the problem to $ pq \equal{} ra^2$ in the middle, since they act like they said $ r \cdot a_1^{2l_1} a_2 ^ {2l_2} \cdots$ when in the second half. I'm also not sure what they were thinking when they made that last statement about $ p$. It's true if you say WLOG that the exponent of $ a_i$ in $ q$ is odd (has to be odd in one of $ p,q$), but you can't just completely omit that.

They start with $ pq \equal{} ra$ , and in the middle they seem to assume $ pq \equal{} ra^2$  ?? :huh:

Could anyone please help me understand this subtle/absurd solution? :(

From which book did you take that problem?


what is the value of
1- cosx cos2x cos3x cos4x ....cos nx

If $ x \equal{} \frac {(2k\plus{}1)\pi} {2\ell}$ and $ n$ is large enough, the value is $ 1$. Otherwise ...


calculus
integration
trigonometry
calculus computations

How would I find a reduction formula for the integral of $ \frac{1}{(a\plus{}bcos(x))^n}$.
I tried substituting t=tan(x/2) but that doesn't seem to make life any easier.

Went back to this question and found it was not that hard just needed some manipulation:
Assume $ b<>0$.
Then divide out the b term and substitute $ u\equal{}\frac{a}{b}\plus{}\cos x$.
After some work you can express the original $ I_n$ in terms of $ I_{n\minus{}1}$, $ I_{n\plus{}1}$ and a term involving $ \frac{1}{(a\plus{}b\cos x)^n}$.


logarithms
inequalities
triangle inequality

[hide="Sketch of solution"]Let $\log_{3}x = a$. It reduces to $\sqrt{a^{2}+2a+1}+\sqrt{a^{2}-2a+1}= 2$, i.e. $|a+1|+|a-1| = 2$. By the triangle inequality, this is true iff $\textrm{sign}(a+1) = \textrm{sign}(1-a)$, i.e. $-1\leq a \leq 1$. So $\frac{1}{3}\leq x \leq 3$.

Edit: we need $\log_{x}3$ te be defined, so $x\neq 1$. So our solution is $x\in\left[\frac{1}{3},3\right]\setminus\{1\}$.[/hide]

[hide="solution"]
Just slightly different simplifying and factoring than above.

Simplify, log rules
$\sqrt{\log_{3}x}\cdot \left( \sqrt{ log_{3}x+log_{x}3+2 }+\sqrt{ log_{3}x+log_{x}3-2}\right) = 2$

Since $\log_{3}x$ and $\log_{x}3$ are reciprocles, the terms under the radicals both factor to squared terms. ((x + 1/x)^2 and (x - 1/x)^2)

$\sqrt{\log_{3}x}\cdot \left( \sqrt{ log_{3}x }+\sqrt{ log_{x}3 }+\sqrt{ log_{3}x }-\sqrt{ log_{x}3 }\right) = 2$

Which is $2\log_{3}x = 2$, so $x=3$. (Which means it doesn't matter which term was negative when simplifying the second radical term.)

[/hide]

[quote="Cicatriz"]so $x=3$[/quote]

[quote="Dr Sonnhard Graubner"]hello, i have $x=3$ or $x=\frac{1}{27}$.
Sonnhard.[/quote]

It's true for $x=2$, e.g., I checked it. I think my solution is correct.

I agree with Kurt Godel.

When you take the square roots you have to remember that you get absolute values of the expressions.  Then you have to do casework to see when you get solutions.


A teacher with a class of 7 students has 10 pieces of candy. If each piece of candy is given to a student at random, what is the probability that every student gets a piece of candy?

[quote="usaha"]3/286, Ill post a solution if this is right[/quote]

can you explain your answer so we can deem it correct or incorrect?

edit: your answer isn't logical because there are more pieces of candy than students so there is a better chance than $\frac{3}{286}$ that every student gets at least 1 piece.

[quote="usaha"]No, I think its correct,
If every studnet gets a candy, then there are 3 candies remaining for any studnet, so thats 9C6
The total amount of ways the candy can be distributed is 16C6, and ((9C6)/(16C6) is 3/286[/quote]
Looks solid.

but a piece of candy is randomly given to a random student so the first 2 pieces given can be to the same student...or does that not matter? i suck at probability  :blush:

and by 16C6 do you mean $_{16}\text{C}_{6}$?

[quote="7h3.D3m0n.117"]but a piece of candy is randomly given to a random student so the first 2 pieces given can be to the same student...[/quote]

It "doesn't matter" what order you give out the candies in.  If you treat the candies as being distinguishable then the only thing that matters is the final distribution of candy, so we can assume WLOG that the candy was given out in a way such that $7$ went to all $7$ students first.

[quote="usaha"]No, I think its correct,
If every student gets a candy, then there are 3 candies remaining for any student, so thats 9C6
The total amount of ways the candy can be distributed is 16C6, and ((9C6)/(16C6) is 3/286[/quote]

[hide="For those who do not understand where these numbers came from:"]

[u]Balls and Urns[/u]

Say we draw a diagram depicting this situation, where each candy is a dot (indistinguishable from the others) and each person is an urn, which will be the space between two divider lines. The diagram looks like this:
\[| \cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \ \ \cdot \cdot \ \cdot \]
Those last three dots are the extra candies to be distributed. Now, we need to devise a system to count all of the possible places where we can add these three candies, while not counting anything we don't want, like one person not having any candies. The method I use is to make larger elements out of the balls and walls, so that no matter what I move, each person (urn) will have at least one candy. The elements are boxed:
\[\boxed{\boxed{ | \cdot }}\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\boxed{ | \cdot }\cdot \cdot \ \cdot \boxed{\boxed{|}}\]
Now, we cannot move the double-boxed elements on the left and right, since they determine the outermost urns. Note also that one of the two double-boxed elements needs one ball, so that the left-most or right-most urn always will have at least one ball, along with the others. The single-boxed elements and the extra balls can move around freely; the number of permutations of these is $\frac{ 9!}{6! 3!}= \dbinom{9}{6}$. Now, if do not place the restriction that all children must have candy, the diagram looks thus:
\[\boxed{ \boxed{|}}\cdot | \cdot | \cdot | \cdot | \cdot | \cdot | \cdot \cdot \cdot \cdot \boxed{ \boxed{ | }}\]
The number of permutations in this situation is $\frac{ 16!}{10! 6!}= \dbinom{16}{6}$, and since we want the probability, we divide:
\[\frac{ \dbinom{9}{6}}{ \dbinom{16}{6}}= \boxed{ \frac{3}{286}}\]
[/hide]


Sa se demonstreze ca daca $K_1,K_2,K_3$ sunt subcorpuri ale corpului $K$, diferite de $K$ atunci $K_1\cup K_2\cup K_3$ e diferit de $K$

Este problema 2, ONM, 1987, propusa de C. Nastasescu.
Recomand cartea "Olimpiadele Nationale de Matematica, 1954-2003", care se poate comanda online [url=http://www.librarie.net/carti/39911/Olimpiadele-Nationale-de-Matematica-]aici[/url]. Costa sub 300.000 si are peste 500 pagini. Merită :)

Se poate demonstra ca un corp nu se poate scrie ca reuniune disjuncta de un numar finit de subcorpuri.

O sa schitez o demonstratie prin reducere la absurd:

1) Se trateaza cazul in care corpul $K$ este finit.
2) Daca $K$ e infinit, Consideram cea mai mica multime (in sensul incluziunii) de subcorpuri care au ca reuniune pe $K$. Fie aceasta familie $K_1,K_2,...,K_n$.
3) Se demonstreaza ca intersectia celor $n$ subcorpuri contine un sir infinit de elemente distincte (i.e. este infinita) prin inductie.
4) Se demonstreaza ca intersectia lor are cel mult $n-1$ elemente. Presumpunem ca exista $a_1,...a_n$ apartinand intersectiei si consideram elementele de forma $aa_i+b$ unde $a\in K_1\setminus \cup K_i,b\in K_2\setminus K_1$.

Demonstratia e mai lunga si plina de detalii. Daca e necesar o sa mai completez unde e cazul.

[quote="xirti"]Se poate demonstra ca un corp nu se poate scrie ca reuniune disjuncta de un numar finit de subcorpuri.
[/quote]
Pai mi se pare evident. Daca $0$ e in unul din subcorpuri, fiind disjuncte, n-ar putea fi si in altul, deci clar nu ar mai fi subcorp. Cred ca ar trebui sa definesti mai clar ce intelegi prin disjuncte.

Nu vroiam sa zic nimic prin disjuncta :D  singurele conditii sunt ca subcorpurile sa fie proprii.


Who did you think was going to win nationals.
I thought David was, being the top 7th gradfer last year

I thought it was gonna be mark zhang because he got a 46 on state.  But I don't know if anyone else got a 46..

I thought it would be Neal. Of course.

Actually, I thought it would be Mark Zhang. However, I wasn't very surprised to see Sergei win (my anti-speed, anti-kumon, etc. mind, I consider the written winner the winner). But yeah, Mark Zhang was my bet.

Pardha Ponugoti  and David from IN did really good at state

Were there any other 46s? or  just those 2?

there were no other 46's or 45's besides those three. Sergei got 44

[quote="aznness"]Patricia, cuz shes from CA and so am i[/quote]

same reason, and you should have seen her own at state, i owned aznness over there 3-0 :D, not bragging or anything, and then she came and owned me 3-0, answering before i had even read the question...

[quote="aznness"]Patricia, cuz shes from CA and so am i[/quote]

same reason. Go California!! Go for first place!! Oh wait we didn't win!! NO!!! a;sdlkfjapoeribth aeriobtupnaewrbtonaipoanb!!!!

I thought it was gonna be Mark Zhang cuz he aws really fast at countdown. Btw NoSoupForYou were you the guy who came over to the delaware table at the lunch on Saturday? or was that bob?

My count of AOPSers I met at nats: 2

Don't remember, but here is a way to tell us apart; bob is Asian, I'm not.

AoPSers I met: 10+

Do you mean 10, 11, and 12th written winners?

10) Pardha Ponugoti, Indiana

That's all I know.

10. Pardha
11. Mike Jin- MO
12. KAren Kota- KS

Ahem, we'll ignore that... but they did have a tie for 12th, which is just as good in terms of team score.

[quote="Phelpedo"]CA is always in the top 3, and consistently has more than 1 person in CD.[/quote]

2001, CA was 4th...

CA will own next year because I will be on the team next year!!!
[size=75]I wish[/size]

Yes and when they don't win they're still up there in the top ranks.

[quote="JesusFreak197"][quote="Phelpedo"]TX, IN, and MA all had good years this year. By this I mean an exceptionally strong team. But CA is good nearly every year (although did you know that CA has never had a national champion?)[/quote]

As in, 1st place individual? Didn't TianKai Liu (sorry if I misspelled that) get first place in 2000? Or did he lose in CD?

[b]EDIT:[/b] LynnelleYe... do you have an older sister who goes to North Hollywood High School? If not, it must just be the same last name...[/quote]

He got 2nd after countdown, I think...

[quote="NoSoupForYou"]CA, as a team, won 5 times - more than any other team by far.[/quote]
What about individual winner...........nope :!:  :o  ;)

Nope, no older sister. An irritatingly large number of people have my last name actually...almost every time they announce AMC perfect scorers or USAMO qualifiers or MathCounts top 25% at state or whatever they will have someone with my last name. Unfortunately it's never followed by my first name.

The reason why CA won 5 times as a team is because it has such a large population.  That's one unfair thing about nationals.  Since California has the largest population, it had the greatest chance of selecting the smartest people.

Incidentally, a mixed blessing, because if California were to be on a level playing field with say, Wyoming, then it would be divided into 74 different states, all of whom got to bring 4 people to nations.  Then, many more of your fellow Californians would have been able to make nationals.

The one downside to having such a big population is that students almost never make nationals two years in a row; so, some people from other states can get more experience with the actual national competition because they live in smaller states, and can attend more than one year.

Yeah but TX also has a large population; nowhere near as large as CA but certainly still large. Yet, their win ratio is much lower than their population ratio.


Are you familiar with Balls and Urns? If so, heres what you can do:
$ r$= # of red flowers
$ p$= # of pink flowers
$ u$= # of purple flowers

Since each one must be at least 1, we have:
$ r \plus{} p \plus{} u \equal{} 9$
Which by balls and urns has $ \binom{11}{2} \equal{} \boxed{55}$ solutions.

OR if you dont know balls and urns:
We have 12 flowers. Lets examine this by cases:
CASE 1;1 red flower: We have $ 11$ flowers left. of those, we can have any number from $ 1 \minus{} 10$ of the flowers purple (we cant have 11 otherwise there will be no pink flowers). This has a total of $ 10$ ways.
CASE 2; 2 red flowers: We have $ 10$ flowers left. of those, we can have any number from $ 1 \minus{} 9$ of the flowers purple.This has a total of $ 9$ ways.
CASE 3; 3 red flowers:We have $ 9$ flowers left. of those, we can have any number from $ 1 \minus{} 8$ of the flowers purple.This has a total of $ 8$ ways.
Now we see a pattern. The solution, based on our pattern, is $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} 10 \equal{} \frac {10 \cdot 11}{2}$(which is also $ \binom{11}{2}$)$ \equal{} \boxed{55}$

I should explain the concept first...

SO, here goes nothing.

The formula for balls and urns is $ \binom {n \plus{} k \minus{} 1}{n \minus{} 1}$ which distributes Indistinguishable to Distinguishable.

In your problem John wants to buy a dozen flowers. There are three colors: red, pink, and purple. He wants to have at least one of each color. How many different ways can he choose the dozen of roses?

Since he must have at least one of each color, there are three colors, and so $ 12 \minus{} 3 \equal{} 9$

$ 9 \equal{} n$ in this case.
$ 3 \equal{} k$

Soo, by balls and urns is $ \binom {9 \plus{} 3 \minus{} 1}{3 \minus{} 1}$ $ \equal{}$ $ \binom {11}{2}$ $ \equal{}$ $ \boxed 55$

Did this help?

Thank you so much for the help.  Now I know where to go when I have problems.

Tech.


State an example of a space X and a topology T on X that is induced by no metric of X.  :D

Hope you like the proposal.

Best of luck.

[quote="ZetaX"]Or best, take any non-metric space.[/quote]

ZetaX, could you please tell me what you exactly mean by [i]a non-metric space[/i]? :wink:

A space whose topology is not induced by a metric; i.e. a space that fulfills your requirements  :wink:

Aren't those spaces typically referred to as non-metrizable spaces?

The expression non-metric space doesn't make much sense to me...

Let $ X$ be a set and $ T$ a collection of subsets of $ X$. We say that $ (X,T)$ is $ \emph{a topological space}$ if
1. $ T$ is closed under arbitrary unions.
2. $ T$ is closed under finite intersections.
3. $ \emptyset\in T$ and $ X\in T$.
(Actually 3. follows from 1. and 2. using properties of the empty set)
Topological space $ (X,T)$ is said to be $ \emph{metrizable}$ if there is a metric $ d$ such that $ (X,T) \equal{} (X,T_d)$. If there is no such $ d$ then it is a non-metrizable topological space. For example the sets $ U(a,t) \equal{} \{a\}\cup [t,\infty [,a,t\in\mathbb{R}$ forms a basis of a non-metrizable space. This follows from the fact that $ X$ is not Hausdorff but every metrizable space is Hausdorff.


[quote="mathgeniuse^ln(x)"]In a class election, one candidate received more than $94\%$ (but less than $100\%$) of the votes cast.  What is the least possible number of votes cast?[/quote][hide] $\frac{1}{x}<.06=\frac{3}{50}\Longrightarrow x>\frac{50}{3}$. $x=17$. [/hide]

[hide]$\frac{16}{17}$ is the smallest fraction whose decimal representation is bigger than 0.94. so 17[/hide]


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry unsolved

$ D_aM_a$ is the perpendicular bisector of the side $ BC.$ Perpendicular to $ D_aM_a$ at $ M_a$ is the sideline $ a.$ $ N \in a$ is foot of perpendicular from $ S$ to $ a.$ $ S$ is the Fermat point $ \Longrightarrow$ $ \angle BSC \equal{} 120^\circ.$ Inversion with center $ S$ and power $ SN^2$ takes the line $ a$ into a circle $ \mathcal A'$ with diameter $ SN.$ Unknown rays $ SB, SC$ through the inversion center $ S$ go to themselves, intersecting the known circle $ \mathcal A'$ at unknown $ B', C'.$ Circumcircle $ \mathcal P$ of $ \triangle SBC$ passing through the inversion center $ S$ goes to the line $ p' \equiv B'C'.$ This unknown line is tangent to a circle $ \mathcal Q'$ concentric with the circle $ \mathcal A'$ and diameter $ \frac {_{SN}}{^2},$ because of the angle $ \angle B'SC' \equal{} \angle BSC \equal{} 120^\circ.$ This circle can be constructed. Let $ T$ be reflection of $ S$ in the line $ D_aM_a$ and $ T'$ inversion image of $ T'.$ Since the circle $ \mathcal P$ goes through $ T,$ the line $ p'$ goes through $ T'.$ Construct tangents from $ T'$ to $ \mathcal Q'$ and pick one that separates $ S$ and the common center of $ \mathcal A', \mathcal Q'.$ This tangent intersects the circle $ \mathcal A'$ at $ B', C',$ so that $ \angle B'SC' \equal{} 120^\circ,$ it is the line $ p'.$ Rays $ SB', SC'$ cut the line $ a$ at $ B, C.$ Circumcircle $ \mathcal O$ of $ \triangle D_aBC$ is also circumcircle of $ \triangle ABC.$ Internal bisector line of the angle $ \angle BSC$ cuts $ \mathcal O$ at $ A,$ such that $ \angle CSA \equal{} \angle ASB \equal{} \angle BSC \equal{} 120^\circ.$


The followig problem is from Bulgarian Winter Mathematical Competition this year for 8th grade. No one of the contestants solved the problem.
Find all integer numbers "a" and "b" such that $ a^{2} > \frac {2.4.6\cdots2008}{1.3.5\cdots2007} > b^{2}$.
I know some solutions.

And of course "a" and "b" must be consecutive numbers!

It remains me Canadian Math Olympiad 1997 problem.

Is it the best selection for a and b? or we just find a, b with the suitable coefficients.

There are only two pairs of consecutive numbers "a" and "b" (depends on if the numbers are positive ot negative). We need "a" and "b" such that they are [b]consecutive[/b] numbers.

Good work! The solution I have is almost the same. Finally (7,8) and (-7,-8) are the numbers we wanted to find.


logarithms
function
parameterization
calculus
analytic geometry
graphing lines
slope

The number of solutions of the equation $ 2^x(4\minus{}x) \equal{} 2x\plus{}4$ is ______.

This is my first time doing a pre-olympiad problem, so my answer might be wrong.
[hide="My Solution"]
$ 2^x(4 - x) = 2x + 4 \\
2^x(4 - x) = 2(x + 2) \\
\log{(2^x)(4 - x)} = \log{(2)(x + 4)} \\
\log{2^x} + \log{(4 - x)} = \log{2} + \log{(x + 4)}$
Now we have two log functions being applied on both side to the same base (10) so we can equate the parameters (I think that's what they're called).
$ 1.) 2^x = 2 \\
x = 1%Error. "bigspace" is a bad command.
\\
2.)2^x = x + 2\\
no\,solutions? \\
3.)4 - x = 2 \\
- x = - 2 \\
x = 2%Error. "bigspace" is a bad command.
\\
4.)4 - x = x + 2 \\
2x = 2 \\
x = 1$

$ x = 0,1,2$ all work when you plug them back in. As for Case #2, I am not sure how you would find the value of $ x$ for such equations. At most there will be 4 solutions, but if Case#2 has no solutions, then there are 3.
[/hide]

[quote="sunehra"]
$ 2^x(4 \minus{} x) \equal{} 2x \plus{} 4 \\
2^x(4 \minus{} x) \equal{} 2(x \plus{} 4)$
[/quote]

i think it's rather

$ 2^x(4 \minus{} x) \equal{} 2(x \plus{} 2)$

no??

Yea it is XD. I just checked. It appears the solutions still work though for some odd reason.

[quote="sunehra"][hide="My Solution"]
Now we have two log functions being applied on both side to the same base (10) so we can equate the parameters (I think that's what they're called).[/hide][/quote]

Um... what? It's not true that $ \log a \plus{} \log b \equal{} \log c \plus{} \log d$ necessarily implies that $ a \equal{} c$ or $ a \equal{} d$. You don't even need to take logs.

I'm not sure how to solve this problem, but here's my partial

[hide="Progress"]Notice that if $ x > 4$, then $ 2^{x} (4 \minus{} x) < 0$, so $ 2x \plus{} 4 < 0$. But this is impossible if $ x > 4$.

If $ x < \minus{} 2$, then $ 2^{x}(4 \minus{} x) > 0$, so $ 2x \plus{} 4 > 0$. But this is impossible for $ x < \minus{} 2$.

So we have $ \minus{} 2\le x \le 4$. The solutions are easy to verify: $ x\equal{}0,1,2$, but I don't know how to get farther than that.[/hide]

This problem is like trying to solve $ 2^{x} \equal{} 3x$. It's pretty annoying (to get a closed form of $ x$), but we can make a guess on how many solutions there are...

[hide="Calculus outline"]
LHS has exactly one inflection point, so use an argument with concavity and slopes of tangent lines to argue that a line can only intersect at most three times, giving the three solutions.[/hide]

hello, write your equation in the following form:
$ 2^x(4\minus{}x)\equal{}2(x\plus{}2)$
$ 2^{x\minus{}1}(4\minus{}x)\equal{}x\plus{}2$
$ 2^{x\plus{}1}\minus{}x\cdot2^{x\minus{}1}\equal{}x\plus{}2$
$ 2^{x\plus{}1}\minus{}2\equal{}x(2^{x\minus{}1}\plus{}1)$
$ x\equal{}\frac{2^{x\plus{}1}\minus{}2}{2^{x\minus{}1}\plus{}1}$
Now you can consider the function
$ f(x)\equal{}x\minus{}\frac{2^{x\plus{}1}\minus{}2}{2^{x\minus{}1}\plus{}1}$.
Sonnhard.

a simple study of office and you see that this function is zero 3 points


Let u be a fixed positive integer. Prove that the equation $n! = u^{\alpha} - u^{\beta}$ has a finite number of solutions $(n, \alpha, \beta).$

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

This is not a formal proof (maybe someone would like to formalize it): problem is equivalent to proving that $n!=u^{\alpha}(u^{\beta}-1)$ has finite number of solutions. By a well-known lemma if $p^k || n!$, and $p \perp u$ and $p \perp u-1$ then $p^k||\beta$. That actually finishes the problem because $u^{\beta}-1$ is growing much faster than $n!$ by above observation (that shouldn't be hard to prove).

What lemma¿ I see problems, e.g. $n=4$, $p=3$, $u=2$.

Well, please forget about this nonsense above  :blush: I have realized later after I had posted it that it was completely wrong.

OK lets try to kill it (the idea is more or less that one of Megus)...

Lemma: when $ p$ is a odd prime and $ p \nmid a$, then the order of $ 1 + ap^k \mod p^s$ is $ p^{s - k}$ for all $ s > k$.
(But I will not post the proof as long as noone wants to see it: it's boring calculation and induction)

Now to the problem:
We want that $ v_p(n!) = v_p(u^b - 1)$ for all primes $ p$ not dividing $ u$.
Look at any odd prime $ p$ dividing $ u - 1$ (if there is none, so when $ u = 2^k + 1$, just consider any odd prime dividing $ u^s - 1$ for some $ s$).
Let $ p^k \parallel{} u^s - 1$, so $ u^s = ap^k + 1$, $ p \nmid a$, and by the above $ u^s$ has order $ p^{t - k}$ seen $ \mod p^t$.
When we also choose $ s$ to be minimal with $ p|u^s - 1$, we get that the order of $ u \mod p^n$ is $ s \cdot p^{t - k}$.

Now lets compute what the order has to be for given $ n$:
There are $ \left\lfloor \frac {n}{p}\right\rfloor > \frac {n}{p} - 1$ multiples of $ p$ in $ n!$, thus $ v_p(n!) > \frac {n}{p} - 1$.
We need that $ u^b \equiv 1 \mod p^{v_p(n!)}$, thus $ b$ being a multiple of $ p^{v_p(n!) - k}$.

So we have (finally!):
$ n^n \geq n! = u^a(u^b - 1) > u^{p^{v_p(n!) - k}} - 1 > u^{p^{\frac {n}{p} - k - 1}} - 1$
giving:
$ n^n \geq u^{p^{\frac {n}{p(p - 1)} - k - 1}}$
which by taking logarithms yields:
$ n^2 \geq n \cdot \log(n) \geq \log(u) \cdot p^{\frac {n}{p(p - 1)} - k - 1}$
or equivalently:
$ \log(u)^{ - 1} n^2 + \cdot (k + 1) \geq n \cdot \log(n) \geq \left( p^{\frac {1}{p}\right)^n}$
But this can only happen for finetely $ n$ since one the left is a polynomial, on the right is an exponential function.

So the problem is dead, dead, dead, any questions ;)

Edit: simplified it a bit.
Edit number two: changed the error mentioned below.

I suppose it should be $p^{s-k}$ instead of $p^{s-k-1}$ in the lemma (check the case for $k=1$: $(1+ap)^{p^{s-2}}\equiv 1+ap^{s-1} \mod p^s$) though it doesn't influence the solution which is nice  :)

Thanks, I corrected it.
But it's very similar to what you mentioned above  :)

can i ask something about these lemmas you people keep coming with?
do you make them to suit the needs for this specific problem or have
you read them some where and recall them  as you notice they would
be usefull for these problems?

The one I used above is another way of expressing something some people call "Hensels Lemma" (see the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Formulary[/url]).
In general, some lemmas reappear from time to time. Others more rarely, and there are also some that you just make up for this special problem.
Thus I would say that you can't classify them.

Here is an alternative solution based on cyclotomic polynomial :)
To ZetaX:do you have a nice proof for the lemma below?

Here's a new solution with LTE.

First, note that for any fixed $n$ there are only finitely many solutions because $\alpha$ is bounded by considering a prime divisor of $u$ and then $\beta$ is fixed when $\alpha$ and $n$ are known. It suffices to show that only finitely many $n$ have solutions, then. We will prove that $n$ is bounded from above.

Rewrite the equation as
        \[ n! = u^{\beta} (u^{m}-1 ) \]
        where $m=\alpha-\beta.$ Now, consider an odd prime $p$ not dividing $u$ and denote the order of $u$ modulo $p$ by $d$. For sufficiently large $n$, $p$ must divide $n!$ so $d$ must also divide $m$. By the Lifting the Exponent Lemma,
        \[ \nu_p(u^m-1) = \nu_p(u^d-1) + \nu_p( \frac{m}{d} ) \leq \nu_p( u^d - 1 ) + \log_p \left(\frac{m}{d} \right) .\]
        Letting $c=\nu_p(u^d-1) - \log_p(d)$, it becomes clear that $\nu_p(u^n-1) \leq \log_p(m) + c$ for a fixed constant $c$. However by Legendre,
        \[\nu_p(u^m-1) = \nu_p(n!) = \sum_{k=1}^{\infty} \left\rfloor\frac{n}{p^k} \right\rfloor \geq \frac{n}{p} - 1. \]
        From these two inequalities, we deduce that
        \[ {m}  \geq p^{\frac{n}{p} - c} \]
        for a new constant $c$. Then, a chain of inequalities reveals that
        \[ n^n \geq n! = u^{\beta} (u^m - 1 ) \geq u^{\log_p{m} + c} \geq u^{p^{\frac{n}{p} - c} } . \]
        Taking logs of each side gives that
        \[ n \ln{n} \geq p^{\frac{n}{p} - c} \ln{u} , \]
        which can only hold for finitely many $n$ since the left hand side is less than a polynomial of $n$ and the right hand is an exponential in $n$.

Not my solution with Zsigmondy's.

Rewrite the equation as $n!=u^a(u^b-1)$. Then
$$a=v_u(n!)=\sum_{k=1}^{\infty} \Biggl\lfloor\frac{n}{u^k}\Biggr\rfloor < \sum_{k=1}^{\infty}\frac{n}{u^k}\le n$$
so $a\le n-1$. Moreover, by Zsigmondy's theorem there exists a prime number $p$ such that $p|u^b-1$ and $p\nmid u^m-1$ for any $m<b$. Hence $ord_p(u)=b$, so because from Fermat's Little Theorem we have $p|u^{p-1}-1$, it follows that $b|p-1$, so $b\le p-1$. But since $p|n!$, we have $p\le n$, thus $b \le p-1 \le n-1$. Therefore
$$ n!=u^a(u^b-1)\le u^{n-1}(u^{n-1}-1) $$
which clearly has a finite number of solutions.

Let $(n,b,c)$ be a solution.
Replace $(u,\alpha,\beta)$ with $(a,b,c)$ for convinence.
By LTE,$\nu_p(a^b-a^c)=\nu_p(a^{b-c}-1) \le \nu_p(a^{(b-c)(p-1)}-1)=\nu_p(a^{p-1}-1)+\nu_p(b-c)\ge \nu_p(a^b-a^c)=\nu_p(n!)>\frac{n}{p}-1$
Now let $\nu_p(b-c)\ge \frac{n}{p}-k$ for some constant $k$ independent of $n$ and hence $n^n>n!=a^b-a^c \ge a^{p^{\frac{n}{p}-k}}$ and by taking logarithms we will be done.

[quote=ryan17]Not my solution with Zsigmondy's.

Rewrite the equation as $n!=u^a(u^b-1)$. Then
$$a=v_u(n!)=\sum_{k=1}^{\infty} \Biggl\lfloor\frac{n}{u^k}\Biggr\rfloor < \sum_{k=1}^{\infty}\frac{n}{u^k}\le n$$
[/quote]

Isn't it true for prime u?

[quote=SerdarBozdag][quote=ryan17]
$$a=v_u(n!)=\sum_{k=1}^{\infty} \Biggl\lfloor\frac{n}{u^k}\Biggr\rfloor < \sum_{k=1}^{\infty}\frac{n}{u^k}\le n$$
[/quote]

Isn't it true for prime u?[/quote]

You're right; Legendre's formula works only when $u$ is a prime. So @ryan17 takes a wrong step here.

Suppose that $n, b, c$ are positive integers such that $n! = a^{b} - a^{c}$ for a given $a$.
Thus by Exponent Lifting, we have that for all primes $p$, \[v_p(a^b - a^c) = v_p(a^c) + v_p(a^{b-c} - 1) = cv_p(a) + v_p(a^{b-c}-1) = v_p(n!) \]

Note that $a$ and $a^{b-c} - 1$ are relatively prime so let $p_1, p_2, ... ,p_u$ be the primes which divide $a$.\\

Now consider $p \in \text{\textbraceleft} p_1, p_2, ... , p_u \text{\textbraceright}.$
Then we know that $cv_p(a) = v_p(n!)$, so it must be true that
\[c \hspace{1mm}|\hspace{1mm} \text{gcd}( v_{p_1}(n!), v_{p_2}(n!), ... , v_{p_u}(n!)). \]
so $c$ can take on only finitely many values for each $n$. Now let $r$ be the least prime which is greater than every prime which divides $a$ or $a^{b-c}-1$, so $n < r$ and thus $n$ is bounded. For each choice of $n$ and $c$, there is at most one value of $b$ which satisfies $n! = a^{b} - a^{c}$, so we are done.

@above how do we know that there are finitely many primes which divide $a^{b-c} - 1$? Note that $a$ is fixed, but $b-c$ is not.

[hide=Solution] The equation rewrites as $n!=u^{\beta}(u^{d}-1)$ for $d=\alpha-\beta$. Suppose that $u \neq 1$ and take a prime $p \mid u$; we have $p \nmid u^d-1$, so $\beta \leq \beta\nu_p(u)=\nu_p(n!)<\frac{n}{p-1}\leq n$, so $\beta<n$.

Notice that for a fixed $d$, there are finitely many solutions for $n$: $n!=u^{\beta} (u^d-1)<u^n(u^d-1):=M \cdot u^n$ for a constant $M$, but $n!$ grows faster than an exponential function, so $n$ is bounded, giving finitely many solutions for a fixed $d$.

Unless $d=1,2,6$ (but for each of these cases there are finitely many solutions, so we ignore them), we have that by Zsigmondy $u^d-1$ has a primitive prime divisor $q$, so $d=\mathrm{ord}_q(u) \mid q-1$, so $q>d$ divides $n!$, which yields that $d<n$. Hence, we obtain $n!=u^{\beta}(u^d-1)<(u^2)^n$, but again since $n!$ grows faster than an exponential function, we obtain than $n$ is bounded, hence there are finitely many solutions in this case as well.


When I tried to post something in #H34N1's blog, I got

[quote]SQL requests not achieved

DEBUG MODE

SQL Request: SELECT t.topic_last_post_id, t.topic_title, t.topic_last_poster, t.topic_last_username, t.topic_last_time, u.username FROM phpbb_topics t LEFT JOIN phpbb_users u ON u.user_id = t.topic_last_poster WHERE forum_id = 419 AND topic_moved_id = 0 ORDER BY topic_last_post_id DESC LIMIT 1
Line : 379
File : CH_functions_post.php[/quote]

It actually still posts it, though.

I got the same error on leetspeak's blog.


Let $k\ge 14$ be an integer, and let $p_k$ be the largest prime number which is strictly less than $k$. You may assume that $p_k\ge \tfrac{3k}{4}$. Let $n$ be a composite integer. Prove that [list=a] [*] if $n=2p_k$, then $n$ does not divide $(n-k)!$, [*]  if $n>2p_k$, then $n$ divides $(n-k)!$. [/list]

The [url=http://www.kalva.demon.co.uk/apmo/asoln/asol033.html]solution on Kalva[/url] seems incomplete to me... at least I see no justification why we'd only have those 3 cases, e.g. why we cannot have $ q<r$. Can anyone explain me why it is obvious?

[quote="Peter"]You may assume that $ p_{k}\ge\tfrac{3k}{4}$. [/quote]And my question is if the statement above is always true :) I mean I know how to prove that $ p_{k}\geq\frac{k}{2}\plus{}3$ but that's about it ...

There are many variants on that theorem. :) For sufficiently high $ a$ we have a prime above $ \frac{b\minus{}1}{b}a$, I believe.

The link solutions on kalva appears not to work . Anyway I balieve that the assumtion quoted by valentin vornicu could be a wrong assution if we take into account large enough prime gaps or lagoons startin at some  big enough n! +2.

Phill: Use [url=http://web.archive.org/web/%2A/http://www.kalva.demon.co.uk]the Wayback Machine[/url] to access Kalva (if you live in GB, use a proxy like TOR to access the Wayback Machine...). And the bounds for $ p_k$ are correct, they are a rather known generalization of the Bertrand postulate.

  darij


geometry
inradius
ratio
geometry proposed

Given triangle $ ABC$ with sidelengths $ a,b,c$. Tangents to incircle of $ ABC$ that parallel with triangle's sides form three small triangle (each small triangle has 1 vertex of $ ABC$). Prove that the sum of area of incircles of these three small triangles and the area of incircle of triangle $ ABC$ is equal to
$ \frac{\pi (a^{2}\plus{}b^{2}\plus{}c^{2})(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}{(a\plus{}b\plus{}c)^{3}}$

(hmm,, looks familiar, isn't it? :wink: )

let me post my solution:
let $ P , Q$ the intersection of the paralle with $ AB$ and the sides $ AC$ and $ AB$ respectively  , let $ x = AP,y = BQ, PQ = z$,the quadrilateral $ APQB$ has an incircle so $ c + z = x + y$ and by thales $ \frac {b - x}{b} = \frac {a - y}{y} = \frac {z}{c}$ , solving this system we find easily :
$ PQ = z = \frac {c(a + b - c)}{a + b + c} ,CP = b - x = \frac {b(a + b - c)}{a + b + c}$, and $ CQ = \frac {a(a + b - c)}{a + b + c}$. let $ m = \frac {(a + b - c)}{a + b + c}$.
the inradius of triangle $ CPQ$ is $ \sqrt {\frac {(ma + mb - mc)(ma + mc - mb)(mb + mc - ma)}{4(ma + mb + mc)}} = mr$ ,where $ r$ is the inradius of triangle $ ABC$. and with the same way for the other triangles .
the sum of areas is $ \pi(m^2 + n^2 + p^2 + 1)r^2$ where $ p = \frac {(a - b + c)}{a + b + c}$ and $ q = \frac {( - a + b + c)}{a + b + c}$.
which equal to:
${ \pi([\frac {[a + b - c}{a + b + c}]^2 + [\frac {a - b + c}{a + b + c}]^2 + [\frac { - a + b + c}{a + b + c}]^2 + 1)\frac {(a + b - c)(a + c - b)(b + c - a)}{4(a + b + c}}$
$ = \pi[\frac {(a^2 + b^2 + c^2)(a + b - c)(a + c - b)(b + c - a)}{(a + b + c)^3}$.

Well, actually, I think this is a routine problem. I'm using similiarities, and the ratio between the base and the height of 1 small triangle and 1 big triangle.

[quote="dyd"]Given the triangle $ ABC$ with sidelengths $ a,b,c$ . Tangents to incircle  of $ ABC$ which are parallel with triangle's sides form three small triangles ( each small triangle has one vertex of $ ABC$ ). Prove that the sum of area of incircles of these three small triangles and the area of incircle of triangle $ ABC$ is equal to $ \frac {\pi (a^{2} \plus{} b^{2} \plus{} c^{2})(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}{(a \plus{} b \plus{} c)^{3}}$ [/quote]

[b][u]Proof.[/u][/b] Denote the incircle $ w\equal{}C(r)$ of $ \triangle ABC$ , the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ where the circle $ w$ touches the triangle $ ABC$ , the points $ M\in (AB)$ , $ N\in (AC)$ for which $ MN\parallel BC$ and $ MN$ is tangent to $ w$ , the incircle $ w_1\equal{}C(r_1)$ and the semiperimeter $ p_1$ of $ \triangle AMN$ . Observe that $ \triangle AMN\sim\triangle ABC$ and $ p_1\equal{}p\minus{}a$ and $ \frac {r_1}{r}\equal{}\frac {p_1}{p}$ $ \implies$ $ r_1\equal{}\frac {r(p\minus{}a)}{p}$ . In conclusion, $ r^2\plus{}\sum r_1^2\equal{}r^2\plus{}\left(\frac rp\right)^2\cdot\sum (p\minus{}a)^2\equal{}$ $ \frac {r^2}{p^2}\cdot\left[p^2\plus{}\sum (p\minus{}a)^2\right]\equal{}$ $ \frac {r^2}{p^2}\cdot\sum a^2\equal{}$ $ \frac {S^2}{p^4}\cdot\sum a^2\equal{}$ $ \frac {(p\minus{}a)(p\minus{}b)(p\minus{}c)}{p^3}\cdot\sum a^2\equal{}$ $ \frac {(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}{(a\plus{}b\plus{}c)^3}\cdot\left(a^2\plus{}b^2\plus{}c^2\right)$ .


I need advice about what colleges I should apply to. Initially, I was thinking about applying only to colleges in California. But now I think that it would be better if I apply to some other colleges too. My grade 10 GPA is 3.0 and the grade 11 GPA is a bit better. I never had access to honor classes or anything like that. My SaT score is, in my opinion, good (2250) but I have to compensate for a bad GPA. I have (just once) cleared the INMO which is the indian equivalent (well, more or less so) of the USAMO. I might also get a good recommendation from a professor at the local mathematics univesity (well, at least he is reputed in India for his work). I could send a couple of articles, too, to places like Caltech. (I can only wish they were difficult enough to be called research papers.) Do you have suggestions about what colleges care for the few qualifications I have? I certainly know that the GPA is very important and so, I have no hopes for MIT or Caltech.

How did you calculate your GPA?  I believe that the Indian system of grading is quite different from that of the US.

About how many students are in your grade (batch) in your school?  Do you know approximately what is your rank in the grade?

Someone told me that my GPA was 3.0. I do not whether it's true.
I found an article about GPA conversion in India on wikipedia, but I doubt its accuracy. Or it may be for universities in India instead of high schools.

Purdue?
 The message is too small. Please make the message longer before submitting.

I think for this.. the [b]best[/b] thing to do  is to go to the College Confidential forum: [url]http://talk.collegeconfidential.com/[/url]

They let you fill out your stats profile thing


modular arithmetic
function
number theory
least common multiple

1) Find number $ \overline{ab}$ such that $ 2^{999}\equiv \overline{ab}$ (mod 100)

2) Prove $ 2^{2^{5}}\plus{}1\equiv 0$(mod 641)

[hide="1"]$ \lambda(100)\equal{}20$ and $ 999\equiv 19\bmod 20$, so $ 2^{999}\equiv 2^{19}\bmod 100$

I happen to know that $ 2^{12}\equiv 96\equiv \minus{}4\bmod 100$, so we can reduce this to $ \minus{}4\cdot 2^7\equiv \minus{}12\equiv \boxed{88}\bmod 100$[/hide]

[hide="2"]
This is a pretty lame solution.
$ 2^{2^5}\equal{}2^{32}\equiv 256^{4}\equiv 154^2\equiv 640\bmod 641$, so $ 2^{32}\plus{}1\equiv 0\bmod 641$
[/hide]

[quote="Dr Sonnhard Graubner"]hello, or
$ 2^{2^5} \plus{} 1 \equal{} 4294967297 \equal{} 641\cdot 6700417$.
Sonnhard.[/quote]

Right, because everyone knows the factors of $ 2^{32}\plus{}1$ off the top of their heads.

[quote="sondhtn"][quote]This is a pretty lame solution.
$ 2^{2^5} \equal{} 2^{32}\equiv 256^{4}\equiv 154^2\equiv 640\bmod 641$, so $ 2^{32} \plus{} 1\equiv 0\bmod 641$[/quote]

You can show that $ 154^2\equiv 640\bmod 641$(use modunlo theory)[/quote]

And "modunlo theory" would be? In any case, it's not that hard to square it out and take it $ \mod 641$, just tedious.

[quote="grn_trtle"]Right, because everyone knows the factors of $ 2^{32} \plus{} 1$ off the top of their heads.[/quote]

It is the fifth Fermat prime, which is rather famous.

Well, since its asking you to prove its 0 mod 641, you could expand it and then do some long division, espeically if you know 2^16 off the top of your head.

Well, $ 641 \equal{} 5^4 \plus{} 2^4 \equal{} 5\cdot2^7 \plus{} 1$, so
\[ 2^{32}\equiv(2^7)^4(2^4)\equiv\left( \minus{} \frac15\right)^4( \minus{} 5^4)\equiv \minus{} 1\pmod{641}.\]But if you are me then you have memorized which primes less than a thousand $ 154$ is a primitive fourth root of unity in. :roll:

[quote="grn_trtle"]$ \lambda(100) \equal{} 20$[/quote]

Please forgive my ignorance, but what is this function?

Don't worry about asking questions; there's nothing wrong with it. That function is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] which gives you the smallest integer $ k$ such that $ a^k \equiv 1 \pmod{n}$.

Wow, thanks. This is gonna be incredibly useful. O_O

This may sound incredingbly dumb, but... for the Carmichael Theorem to be true, doesn't n and a have to be coprime? If that's the case, 2 is not coprime with 100, so this wouldn't work.

Yes, you are right, since obviously $ 2^{20} \not\equiv 1 \pmod{100}$; grn_trtle probably forgot to write out CRT. On the other hand, his solution still holds true; can you figure out why? (it has to do with $ 4|2^{19}$)

Oh, nice.
[hide]
We have to prove that $ 2^{19}\equiv 2^{999}$ (mod 100).
That is because $ 2^{19}\equiv 2^{999} \equiv 0$ (mod 4) and $ 2^{\lambda (25)}\equiv 1$ (mod 25), since 2 and 25 are coprime. Then, we can see that $ \lambda (25) \equal{} 20$.
$ 2^{999}\equiv (2^{20})^{49}\cdot 2^{19} \equiv 2^{19}$ (mod 25).
So $ 2^{19}\equiv 2^{999}$ (mod 4) and $ 2^{19}\equiv 2^{999}$ (mod 25).
Now, we know that if $ n_1, n_2, n_3 ,...,n_r$ are possitive integers. If for every $ i \equal{} 1, ..., r$, $ a \equiv b$ (mod $ n_i$), then: $ a \equiv b$ (mod lcm[$ n_1, ..., n_r$]).
Using that, $ 2^{19} \equiv 2^{999}$ (mod lcm[4, 25]), and lcm[4, 25] = 100.
Thus, $ 2^{19} \equiv 2^{999}$ (mod 100).
[/hide]

BTW, I almost forgot to ask, what is CRT?

Chinese Remainder Theorem. It's kind of what you used when you went from $ 2^{19} \equiv 2^{999} \pmod{4}$ and $ 2^{19} \equiv 2^{999} \pmod{25}$ to $ 2^{19} \equiv 2^{999} \pmod{100}$.

Ah, yes, I know what it is. I didn't realize that it was shortened like that. Thanks.

Is this a corect way to solve the second subdivision?
$ 2^{640} \equiv 1 mod  641$
Taking the 20th root,
$ 2^{32} \equiv 1 mod 641$

OR

$ 2^{32} \equiv \minus{}1 mod 641$

Then i proved that 641 cannot divide $ 2^{32} \minus{} 1$ by expanding it

So, 641 divides $ 2^{32} \plus{} 1$

[quote="grn_trtle"][quote="Differ"]It is the fifth Fermat prime, which is rather famous.[/quote]

Oh cool, didn't know that. It's just that that poster is notorious for sticking things into Mathematica and pasting them here to increase his post count or something.[/quote]
yup. and calls himself a doctor.
[quote="grn_trtle"][hide="1"]$ \lambda(100) \equal{} 20$ and $ 999\equiv 19\bmod 20$, so $ 2^{999}\equiv 2^{19}\bmod 100$

I happen to know that $ 2^{12}\equiv 96\equiv \minus{} 4\bmod 100$, so we can reduce this to $ \minus{} 4\cdot 2^7\equiv \minus{} 12\equiv \boxed{88}\bmod 100$[/hide]

[hide="2"]
This is a pretty lame solution.
$ 2^{2^5} \equal{} 2^{32}\equiv 256^{4}\equiv 154^2\equiv 640\bmod 641$, so $ 2^{32} \plus{} 1\equiv 0\bmod 641$
[/hide][/quote]
how do you know that $ \lambda(100) \equal{} 20$? (This would have been pretty useful on this year's 24 on the 10A)

My question has not been answered yet..................

[quote="gokussj3"]Is this a corect way to solve the second subdivision?
$ 2^{640} \equiv 1 mod 641$
Taking the 20th root,
$ 2^{32} \equiv 1 mod 641$

OR

$ 2^{32} \equiv \minus{} 1 mod 641$

Then i proved that 641 cannot divide $ 2^{32} \minus{} 1$ by expanding it

So, 641 divides $ 2^{32} \plus{} 1$[/quote]

Sorry but I dont think taking the 20th root or something always works  , say for example  , we know that

$ 2^{4} \equiv 1 \bmod{5}$

But you take something like a 4th root ,you get that ,

$ 2 \equiv 2 \bmod{5}$ ,it is neither $ 1$ nor $ \minus{} 1$ .

I think at the max if $ p$ is an odd prime ,by euler's criterion you can say that , $ a^{\frac {p \minus{} 1}{2}} \equiv \pm1\bmod{p}$

sorry if i am wrong .


Consider a subset $A$ of $84$ elements of the set $\{1,\,2,\,\dots,\,169\}$ such that no two elements in the set add up to $169$. Show that $A$ contains a perfect square.

[quote="April"]Consider a subset $ A$ of 84 elements of the set $ \{1,\,2,\,\dots,\,169\}$ such that no two elements in the set add up to 169. Show that $ A$ contains a perfect square.[/quote]
I think it is obvious at all from $ 169\equal{}5^2\plus{}12^2$ and $ 169\equal{}2.84\plus{}1$
If the set doesn't  contain 25 or 144 it will contain a pairs of $ (k,169\minus{}k)$ ,it gives contradiction .

In other words, consider the family of $84$ doubletons $D_k = \{k, 169-k\}$, for $1\leq k \leq 84$. If $169 \in A$, then we are done, since $169 = 13^2$. If not, since then $A \subset \{1,2,\ldots,168\} = \bigcup_{k=1}^{84} D_k$, and we need have $|A \cap D_k| \leq 1$ for all $1\leq k \leq 84$, it follows  $|A \cap D_k| = 1$ for all $1\leq k \leq 84$. But $D_{25} = \{25, 144\} = \{5^2,12^2\}$, and so $A$ will contain a perfect square.


best team ever with all ohio people

-me
-tjhance
-xpmath
-tong
-hs_potter

Hey, wait a minute... :o  (jk  :lol:  I know how sweet they are at math)

meh I'm probably going to defect and join anaxerzia

Let me confirm with Anaxerzia...and if there seems to be problems, I'll take Tong's place, since he's like virtually uncontactable =o

I can't, I'm already on an all soon to be 8th grade team:P. Sorry.


trigonometry
function
algebra unsolved
algebra

It is easy to prove that

$\sin a+\sin 2a +\sin3a+...+\sin na =\frac{\sin{\frac{(n+1)a}{2}}\sin{\frac{na}{2}}}{\sin {\frac{a}{2}}}$

but it is very strange to notice that cancelling the symbol $\sin$ from this formula you can obtain the well known formula

$a+2a+3a+...+na = \frac{(\frac{(n+1)a}{2})(\frac{na}{2})} {\frac{a}{2}}$

Are there other functions which satisfy this strange property?

Dear Manlio, the reason of this strange consideration is very simple: while x ~ 0 then sin(x) ~ x. So, the other examples we can find among such functions that equivalent to x near 0, like $e^{x}-1,tg(x), ln(1+x)...$

Maybe this is one of the reasons that zero is called 'trivial'  :D

PS: Namdung!!! It's been a long time since I see a post from you. Happy New Year.

Thank you 3X.lich. Happy new year to you and all Forum's members. BTW, your permutations are very nice.


topology
geometry
3D geometry
sphere
real analysis
real analysis unsolved

suppose that $ A,B\subset \mathbb{R}^2$.
[b]a)[/b] if $ A$ and $ B$ are homeomorphic, are their complements homeomorphic ?
[b]b)[/b] what if $ A$ and $ B$ are compact?
[b]c)[/b] what if $ A$ and $ B$ are compact and connected?

for a) take $\mathbb{R}$ and $S^1\setminus\{(1,0)\}$.
Inversion proves that they are homeomorphic. However $\mathbb{R}^2\setminus\mathbb{R}$ is not connected, while $\mathbb{R}^2\setminus (S^1\setminus\{(1,0)\})$ is.

For (b), take two circles external to each other (the complement has three connected components, two of which are simply connected) and two circles, one of which is internal to the other one (the complement has three connected components, only one of which is simply connected).

for (c) make grobber's circles in A and B touch at one point. I'm not sure this example works

[quote="sam-n"][b]c)[/b] what if $ A$ and $ B$ are compact and connected :? ?[/quote]

In this last case the answer is positive. I'll just sketch a proof.

First, notice that if $ K$ is a compact connected subset of the plane, every bounded connected component of $ \mathbb R^{2}\setminus K$ is simply connected. If we view the whole thing as taking place on the sphere $ \mathbb S^{2}$, the unbounded component becomes just a component of $ \mathbb S^{2}\setminus K$ minus a point. All simply connected open sets of the plane are homeomorphic, so all we need is to prove that for any two embeddings of a compact set $ K$ in $ \mathbb S^{2}$, the number $ r$ of components of $ \mathbb S^{2}\setminus K$ is the same (here number means any finite number or $ \aleph_{0}$). This follows from [url=http://en.wikipedia.org/wiki/Alexander_duality]Alexander Duality[/url] (more precisely, the case of the theorem needed here says that the reduced homology group $ \tilde H_{0}(\mathbb S^{2}\setminus K)$, which is free of rank $ r \minus{} 1$, is isomorphic to the Cech cohomology group $ H^{1}(K)$, so in particular it does not depend on the embedding we choose for $ K$).


trigonometry
quadratics
algebra
quadratic formula
trig identities
Law of Cosines

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

[hide="Solution"]Drawing a diagram will help you get the most out of this solution.

Since we are given two $60^\circ$ angles, we are tempted to think $30,60,90$ right triangles. It turns out that this is exactly the case!

Draw perpendiculars from points $D$ and $C$ to $AB$. Let the points of intersection be $X$ and $Y$, respectively. Now that we have two convenient triangles, we know that $AX=5$, $DX=5 \sqrt{3}$, $BY= 4$, and $CY= 4 \sqrt{3}$.

However, we are still missing a segment of $AB$; $XY$. We solve this problem by dropping yet another perpendicular from $C$ to $DX$, giving us another right triangle. We also realize that $ED=DX-CY= \sqrt{3}$. This gives us $CZ=XY= \sqrt{141}$ by the Pythagorean Theorem.

Therefore, $AB= 5+ \sqrt{141} +4=9+ \sqrt{141}$. We find that our answer is $141+9=150$ [/hide]

[hide="Answer"]Drawing the perpendiculars $CX$ and $DY$ to $AB$, we have two 30-60-90 triangles, so $AY=5$ and $BX=4$, and $DY=5\sqrt{3}$ and $CX=4\sqrt{3}$. Drawing the perpendicular $CZ$ to $DY$, we have $DZ=\sqrt{3}$, so $CZ=\sqrt{141}$ by Pythagorean Theorem. Therefore, $AB=9+\sqrt{141}\Rightarrow \boxed{150}$.[/hide]

[hide]Since $m\angle{A}= m\angle{B}= 60^\circ$, extend $AD$ and $BC$ to a point, call it $E$, to form equilateral $\triangle ABE$.

Inside $\triangle ABE$, there is $\triangle DCE$. It is given that $DC = 12$. Let $x =$ the length of $AB$. Since $\triangle ABE$ is equilateral, $AE = BE = x$, and it is given that $AD = 10$ and $BC = 8$, so $DE = x-10$ and $CE = x-8$.

Since $\triangle ABE$ is equilateral, $\angle E = 60^\circ$. Applying the law of cosines to $\triangle DCE$ at $\angle E$, $12^{2}= (x-10)^{2}+(x-8)^{2}-2(x-10)(x-8)\cos{60^\circ}$. Simplify to get $x^{2}-18x-60 = 0$.

Apply the quadratic formula to get $x = 9 \pm \sqrt{141}$

$9+141 = \boxed{150}$.[/hide]


AMC
AIME
analytic geometry
rotation
complex numbers
geometry
similar triangles

There is a complex number $ z$ with imaginary part $ 164$ and a positive integer $ n$ such that
\[ \frac {z}{z \plus{} n} \equal{} 4i.
\]Find $ n$.

[hide="Answer"]
Let $ z\equal{}a\plus{}bi$, where $ b\equal{}164$. Then we hvae $ \dfrac{a\plus{}164i}{a\plus{}n\plus{}164i}\equal{}4i\implies a\plus{}164i\equal{}4i(a\plus{}n)\minus{}656$. Equating real and imaginary parts, $ a\equal{}\minus{}656$ and then with some computation $ n\equal{}697$.[/hide]

hello,let $ z\equal{}a\plus{}bi$, then we have from the problem $ \frac{a\plus{}bi}{a\plus{}bi\plus{}n}\equal{}4i$ with $ b\equal{}164$, multiplying by the denominator we get the equation $ a\plus{}656\plus{}(164\minus{}4a)i\equal{}4in$, from here we have $ a\equal{}\minus{}656$ and we get $ n$ from $ 164\plus{}4\cdot656\equal{}4n \Leftrightarrow n\equal{}697$.
Sonnhard.

Hah, this is one of the two that I made dumb dumb dumb mistakes on.. leading to a $ 10$ AIME  :mad:

Let $ A$ be the affix of $ z$ on the complex plane, and let $ B$ be the affix of $ z\plus{}n$, and let $ O$ be the origin. The initial conditions tell us that $ A,B$ lie on the line $ y\equal{}164$, and that $ AO \perp BO, AO\equal{}4BO$. We are looking for the length $ AB$.
Let $ P$ be the foot of $ O$ onto segment $ AB$. We see that $ OP\equal{}164$. Then, by similar triangles, $ \frac{AO}{BO}\equal{}4\equal{}\frac{PA}{OP}\equal{}\frac{OP}{PB}$, implying that $ PA\equal{}4\cdot 164, PB\equal{}\frac{164}{4}$. From here, we find that $ AB\equal{}AP\plus{}PB\equal{}16\cdot41\plus{}41\equal{}\boxed{697}$.

[quote="Brut3Forc3"]Let $ A$ be the affix of $ z$ on the complex plane, and let $ B$ be the affix of $ z \plus{} n$, and let $ O$ be the origin. The initial conditions tell us that $ A,B$ lie on the line $ y \equal{} 164$, and that $ AO \perp BO, AO \equal{} 4BO$. We are looking for the length $ AB$.
Let $ P$ be the foot of $ O$ onto segment $ AB$. We see that $ OP \equal{} 164$. Then, by similar triangles, $ \frac {AO}{BO} \equal{} 4 \equal{} \frac {PA}{OP} \equal{} \frac {OP}{PB}$, implying that $ PA \equal{} 4\cdot 164, PB \equal{} \frac {164}{4}$. From here, we find that $ AB \equal{} AP \plus{} PB \equal{} 16\cdot41 \plus{} 41 \equal{} \boxed{697}$.[/quote]
hmm, did you really use this geometric solution on the actual test... I just brute-forced it using thoughtless algebra which I thought might be quicker...
(by the way, what's affix?)

An affix is an association of a complex number with a point on a coordinate plane.

OK, but maybe I'm being dumb today... how do you know $ z$ and $ z\plus{}n$ are perpendicular?

Multiplying by 4i rotates the affix by 90 degrees and scales it by 4. (polar form)

omg how could i not realize that $ i\equal{}cis(\pi)$ but thanks anyways.... definitely i need more sleep now (only gotta sleep 2 hrs b/c i got a terrible insomnia  :( )

I always thought $ i\equal{}cis(\pi /2)$... :ewpu:

yeah sorry stupid typo...  :mad:  :blush:

[quote="Brut3Forc3"]Let $ A$ be the affix of $ z$ on the complex plane, and let $ B$ be the affix of $ z \plus{} n$, and let $ O$ be the origin. The initial conditions tell us that $ A,B$ lie on the line $ y \equal{} 164$, and that $ AO \perp BO, AO \equal{} 4BO$. We are looking for the length $ AB$.
Let $ P$ be the foot of $ O$ onto segment $ AB$. We see that $ OP \equal{} 164$. Then, by similar triangles, $ \frac {AO}{BO} \equal{} 4 \equal{} \frac {PA}{OP} \equal{} \frac {OP}{PB}$, implying that $ PA \equal{} 4\cdot 164, PB \equal{} \frac {164}{4}$. From here, we find that $ AB \equal{} AP \plus{} PB \equal{} 16\cdot41 \plus{} 41 \equal{} \boxed{697}$.[/quote]

You're just showing off.

[hide="Faster Solution"]
\[\frac{z}{z+ n} = 4 i  \implies \frac{z+n}{z} = - \frac{1}{4} i \implies 1 + \frac{n}z = - \frac 14 i\implies \frac nz = -1 - \frac 14 i\implies \frac{z}{n} = - \frac{16}{17} + \frac{4}{17} i .\]
Equating imaginary parts give us
\[ \frac{164}{n} = \frac{4}{17} \implies \boxed{n = 697} .\][/hide]

When i did this as practice, i just set them equal and "bashed". It only took around 2 minutes

[hide=Quick sol!]We have z=4iz+4in, hence we have $z=\frac{4i}{1-4i}n=\frac{4i-16}{17}n$, hence $\frac{4}{17}n=164$, hence $n=164\frac{17}{4}=\boxed{697}$[/hide]

It's easy we just need to less then two minutes to find that n=697

MY FIRST COMPLEX NUMBER AIME PROBLEM!  :winner_first: In my birth year :love:

Let $z = a + 164i$, replacing that into the equation yields:
$$\frac{a + 164i}{n + a + 164i} = 4i$$
Cross multiplying gets us:
$$a + 164i = (4(a + n))i - 656$$
Rearrange.
$$a + 656 = (4(a + n))i - 164i$$
$$a + 656 = (4(a + n) - 164)i$$
This implies that $(4(a + n) - 164)i$ is real, so $4(a + n) - 164 = 0$.
$$\implies a = -656$$
$$\implies n = \boxed{697}$$

[hide=Solution]
We let $z=a+164i$.

We can substitute $z=a+164i$ into $\frac{z}{z+n}=4i$ to get
\begin{align*}
\frac{a+164i}{a+n+164i} &= 4i \\
a+164i &= -656+4ai+4ni.
\end{align*}

We see that both $a$ and $n$ are real so we can equate real and imaginary parts to get
\begin{align*}
a &=-656 \\
164 &= 4a+4n \\
41 &= a+n \\
41 &= -656+n \\
\boxed{697} &= n. \\
\end{align*}
[/hide]

[hide = Solution]
So we can name $z = a + 164i$ and sub that into the equation. We get..
$\frac{a + 164i}{a+ n+ 164i} = 4i.$ If we expand we get.. $a + 164i = 4i(a+n) -656.$
We can equate the real and imaginary parts to create $2$ equations. We get $a = -656$ and $164i = 4i(a+n)$. since $a = -656$ we can sub that into the second equation and we get $n$ to be...
$164i = 4i(a+n) \Rightarrow 41 = -656 + n  \Rightarrow n = \boxed{697}$
[/hide]

Easy, but complex (see what I did there)
-----
Let $z=a+bi$. Our equation is just $$\frac{a+164i}{a+164i+n}=4i \rightarrow a=-656$$ from equating coefficients. Resubstituting and equating $\Im(z)$ gives $4n-2624=164$ from where we simply get $n=\boxed{697}$.


The angle bisectors of triangle ABC intersect sides BC, CA and AB at points A1, B1 and C1, respectively. On line A1B1, denote by F the perpendicular foot point of C1. Prove that line FC1 bisects angle AFB.

Theorem 2 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20781#20781]http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 post #2[/url], applied to P = incenter of triangle ABC.

  darij

$B_1$, $C_1$ and the intersection of $BC$ with the external bisector of $\widehat{BAC}$ are collinear, hence $F$ lies on the Applollonius circle of $\triangle ABC$, done.

Best regards,
sunken rock

My solution:
Let $P$,$Q$, and $R$ be the feet of the perpendiculars from $A$, $B$, and $C$ to $A_1B_1$, respectively. Then since $AP \parallel C_1F \parallel BQ$,
$\frac{PF}{FQ}=\frac{AC_1}{C_1B}$
Also, since $\triangle APB_1 \sim \triangle CRB_1$,
$\frac{AP}{AB_1}=\frac{CR}{CB_1}$
Similarly,
$\frac{BQ}{BA_1}=\frac{CR}{CA_1}$
Dividing the two,
$\frac{AP}{BQ}= \frac{AB_1}{CB_1} \cdot \frac{CA_1}{BA_1}=\frac{AC_1}{C_1B}$
By Ceva's theorem.
Thus, since $\frac{PF}{FQ}=\frac{AP}{QB}$ and $\angle APF= \angle BQF =90$, $\triangle APF \sim \triangle BQF$. Thus, $\angle PAF=\angle QBF$. But $\angle PAF= \angle AFC_1$ and $\angle QBF= \angle BFC_1$, thus $FC_1$ bisects $\angle AFB$
Note that for this proof all we need is that $AA_1$, $BB_1$ and $CC_1$ concur, as Darij mentioned, but they do not need to be angle bisectors.


linear algebra
matrix
vector
inequalities

[quote="AngelaS"]A is a hermitian matrix and has orthonormal eigenvectors and real
eigenvalues ordered as: lambda1<=lambda2<=lambda3<=....<=lambdan
Prove that for any unit vector x, lambda1<= <x,Ax><=lambdan[/quote]

No help?  :(

Diagonalize it: $U$ is a unitary matrix, and $U^{*}AU$ is the matrix $\begin{pmatrix}\lambda_{1}&0&\cdots&0\\ 0&\lambda_{2}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_{n}\end{pmatrix}$
Now $x=Uv$, where $v$ is the unit vector $(c_{1},c_{2},\cdots,c_{n})^{T}$, and $\langle x,Ax\rangle=(Uv)^{*}A(Uv)=v^{*}(U^{*}AU)v= \lambda_{1}|c_{1}|^{2}+\lambda_{2}|c_{2}|^{2}+\cdots+\lambda_{n}|c_{n}|^{2}$. By elementary inequalities, this is between $\lambda_{1}|v|^{2}=\lambda_{1}$ and $\lambda_{n}|v|^{2}=\lambda_{n}$. That includes the fact that this inner product is real.


So the question is can anyone survive a succesfully detonated bomb in a plane flying 10,000m?
As it turns out the answer is yes. Check out http://en.wikipedia.org/wiki/Vesna_Vulovi%C4%87 ! :)

Yes, well. Without a parachute that...wow. But I'd say most people could survive a 30,000 feet drop with a parachute on. A big parachute.

Wouldnt you reach maximum velocity at a certain point due to drag?

Yeah but even at that velocity you'd die on impact usually... that's quite impressive if I say so myself.

it was lucky she landed in snow

shed be dead otherwise

Yes the snow was lucky. It would be good to land in snow or water or fluff or something.

Still to survive

1) a detonating terrorist bomb a couple of feet away
2) plane broke in half -> no oxygen
3) plane crash from 10,000 meters
and 4) to have no permanent injury (after 1 year) ...

I think cats have a nonlethal terminal velocity.

Are you saying that a cat can survive something like this? :)  :?

That's interesting. So a cat can survive a fall from 30,000 feet?

It would still hurt though.  :|

[quote]Are you saying that a cat can survive something like this?[/quote]
It is well known that this is the case  if you tape a buttered toast on the back of the cat.  There are two laws:
[quote]Law 1 - A buttered toast always land on the buttered side.
and  Law 2 -A  Cat always land on its feat. [/quote]
voila  - Obviously , only solution is -  Cat will never touch the ground.  :)

Seriously - Terminal velocity (among other factors) depends on the size of the object.  A small mammal  (say size of a mouse) would reach terminal velocity and most likelywill  survive if it lands on  a soft ground.  ( I have known hamsters surviving a fall of 50+ meter)  Anything smaller (say a roach) the chances are excellent. For a cat,  if the cat orients itself well, the terminal velocity is "non-fatal"   There was a study I have seen people quoting   from the Journal of the American Veterinary Medical Association. [quote]Two vets examined 132 cases of cats that had fallen out of high-rise windows and were brought to the Animal Medical Center, a New York veterinary hospital, for treatment. On average the cats fell 5.5 stories (BTW most will reach terminal velocity from this height) , yet 90 percent survived. (Many did suffer serious injuries.)[/quote]  Any mammal larger than a cat and terminal velocity iwill be high enough to be fatal ..

About the person surviving after a fall from the plane - I seem to recall the original story from the newspapers  - what was incredible  that  the person was lucky that some  hikers were actually close by who were able to find her and  rescue her.

I hope Valentin doesn't mind me bumping his old topic back up.
However, I find this one much more interesting than "I have a laptop!" and other spamtopics that can be found in Round Table nowadays.

I once did a parachute jump when I was fifteen (a tandem jump).  For the first seconds it was very confusing and then I regained my notion of up and down.  I'm pretty sure those first seconds are when you are still gaining speed.  An average human reaches a terminal velocity of about 60 meters per second, which is theoretically attained after about six seconds (and about 183 meters).
There is certainly no difference between 1 kilometer and 2 kilometers.
....Except the cold and low air pressure of course.  At about four kilometers there wasn't a noticable difference in air pressure, but the altitude combined with the high speed was already enough to make it ice cold!  There is no such thing as free fall when you have attained the terminal velocity.  You need glasses to protect your eyes, your hands freeze and the wind blows into your face that hard that it becomes hard to breathe.
I was told that unlike jumps from about 4 kilometers, a jump from about 10 kilometers requires a much more advanced suit.

But seriously, I had never heard about this attack  :o .  As I understand it, "Ustace" were Croatian catholics, and in the second WWII they fought for the Nazis and against Tito's partizans.  What were they trying to accomplish (a flight from Copenhagen that lands in Zagreb, there could be quite a few Croatians on board too right?)

On a tangent, if you jumped out of a plane, at say 10,000 feet, would it make any difference at all if you took off your shirt and attempted to use it as a parachute?

Now that's what you call problem solving  :P

I don't know penguin,but i s'pose a cape will make a fine one indeed.But i rather not go out in the public without clothes so a cape would be fine for me.

Why don't you try it;then you'll solve a really really difficult problem for skydivers. :harhar:

Why did you revive a 9 year old topic? ._.

It is actually 10 years old, then a year after it died it was revived, then 9 YEARS LATER it was revived AGAIN by Robot7620. Wow.

1. Please read [url=http://www.artofproblemsolving.com/community/c170h1125888_reviving_old_threads_with_contentless_posts]this[/url] before posting (a few years ago, this would have been common sense, but the times change?).

2. Quit replying to spam posts (hint: also spam); just report them and a moderator will take care of it.


linear algebra
matrix
vector
linear algebra unsolved

A an $n$x$n$ permutation matrix is defined as a matrix with columns $e_{1}$,$e_{2},\ldots,e_{n}$ (the standard unit vectors) in any particular order.

Prove that the product of two permutation matrices is again a permutation matrix.

This is inmediately clear when you realize that for a vector x, Ax permutes the entries of vector x.
So for permutation matrices A en B ABx is a product of two permutations, which is again a permutation.

Another more formal (boring) way to see it is:
If $\sigma$ is the permuation and $A_{\sigma}$ is the matrix.
Denote by $B=A_{\sigma}A_{\sigma'}$, $b_{i,j}=\sum_{k=1}^{n}\delta_{i,\sigma(k)}\delta_{k,\sigma(j)}=\delta_{i,\sigma \circ \sigma '(j)}$.
So $A_{\sigma}A_{\sigma'}=A_{\sigma \circ \sigma '}$


Does anyone know the proof for the theorem to solve infinate geometric sequences?

s= t1/ (1- r)  with t1 being term 1 and r being the ratio of change


group theory
superior algebra
superior algebra solved

Need help,
show that if n is odd then the set of all n-cycle consists of two conjugacy classes of equal size in An

Thx

Take $(1 2 ... n)$, his conjugacy class contains elements $(f(1) f(2) ... f(n))$ with $f$ in $A_n$, so this class is cardinal $\frac{n!}{2}\times \frac{1}{n}$ ($\frac{1}{n}$ since $(12...n)=(n12...)=...$, $n$ representations for one cycle).
Now take $(g(1)g(2)...g(n))$ where $g$ is a transposition, clearly he is not in the first class then you have your second class with same cardinal $\frac{n!}{2}\times \frac{1}{n}$. Union of both classes is cardinal $(n-1)!$ we have all our $n$-cycles ...
I hope I've not done mistakes

I'm not sure that what you wrote completely proves that the cardinal of the conjucacy class of an $n$-cycle is $\frac{(n-1)!}2$. An (another) idea is to check that the commutator of an $n$-cycle is the cyclic group it generates which has $n$ elements, hence the cardinal of its conjucacy class is $(n-1)!/2$.


probability
rotation
geometry
geometric transformation
reflection
pigeonhole principle
conditional probability

Robert deals 9 cards in a square from a normal 52 card deck.  He then pretends that it is a tic-tac-toe board (black cards against red cards).  If the probability that black won is $\frac{n}{1000}$, compute $n$ to the nearest integer.

Well what happens if both win  :maybe: like if you have

XXX
OOO
XOX

or something like that?

Since black and red have the same chance of winning, wouldn't it just be $\frac{500}{1000}$?

What if neither of them win?  Or, as [b]A M Y[/b] pointed out, both of them win?

And what about if say...black won twice and red won once?

BBB
RRR
BBB

or

BRB
BRB
BRB

[quote="Gyan"][quote="miyomiyo"]A M Y:  That counts as a tie.
tOrajirOu:  Ties do not count as black winning.
samath:  There is the case where they both win or both lose.[/quote]

In order that  one can make ANY sense out of this problem, [b]you may at least try to define what do you mean by "black won".
[/b]
[hide]An  ordinary tic-tac-toe is played by altering two types (say X and O's).. to see which one (if any) can get three of the same type in a line. "Random" 9 cards from a deck need not even have equal (or with a difference of 1) cards of  alternate color...
[/hide][/quote]

Umm... do you know how to play tic-tac-toe?

[quote="Hamster1800"]
And what about if say...black won twice and red won once?
[/quote]

Did you read my response when A M Y asked the same exact question?

[quote="samath"]
Oh ...
I read the problem to mean, assuming one person won, what is the probability that black won.
[/quote]

Yeah, I figured that was the problem.

[quote="chesspro"]
Perhaps the question should be reworded to:

What is the probability that exactly three black cards lie in the same row, column, or diagonal, and that no three of the red cards lie in the same row, column, or diagonal?
[/quote]

That is a PERFECT rewording of the question.

Approximate Answer:

I estimate a 50% black and red probability per square. There are 2^9=512 possible combinations.

Black has 2 Diagonals: There are 2^4=16 ways this could happen.
Black has 1 Diagonal:There are 2(2*2^4+2^4)=96 was this could happen
SF: 112. We consider cases without Black diagonals:
Black has center and 2 adjacent rows: There are 4*2^2=16 ways this could happen.
Black has center and 1 row: There are 4*2^2=16 ways this could happen (Red having no row crops up first here)
SF: 144
We consider cases with a Red center:
Black has 4 rows: 1 way.
Black has 3 rows: 4 ways.
Black has 2 rows: 7 ways
Black has 1 row: 4*2*2=16 ways.
Total: 172

172/512=43/128. N=336.

1) Yes I did read your response. The question was not EXACTLY the same like you said. The prior question did not account for black winning "more" than red. Mine did.

2) When you ask if someone knows how to play Tic-Tac-Toe, it sounds very demeaning. He was pointing out that it's possible for all 9 cards to be the same color, where as in an "ordinary" game of Tic-Tac-Toe, it's not possible for anything other than a 5-4 split.

3) Yes, I agree, the rewording is great. Try to be as clear as you can when stating a problem and define any corner cases that you may have contained.

[hide="attempt"]
Suppose that someone wins. There's a 50% chance of black winning and red winning. So we just count the cases where nobody wins and subtract the probability of hitting one of those.

1) Both have a row:
6 ways to choose the two rows, 8 ways for the last row: 6*8=48
But we overcounted! We counted the cases I described in my last post twice. This is 2*(3 choose 2) = 6 cases. Therefore, there are only 42 of these cases.

2) both have a column. Same as above: 42

3) neither win.
Let's count manually. This will be exactly the same as cat's games in normal tic-tac-toe. If one player gets 6, he/she can't lose.
BBR
RBB
BRR

RBR
BBR
RRB

These can be rotated and reflected, so there are 8 ways for the top one to occur and 4 for the bottom one, so we remove 24 more.

The total number of cases we're removing is 108

So I get $\frac{2^{9}-108}{2^{1}0}= \frac{512-108}{1024}= \frac{404}{1024}= \frac{202}{512}= \frac{101}{256}$.

Unfortunately, I just realised that my answer is wrong becacuse as more and more black cards are dealt, red cards become more and more likely. Anyone have a solution that incorporates this? (Mine should work for randomly assigning colors, but not for cards. Also, mine's a bit larger than Bill The Maniac's. Anyone have a third opinion?
[/hide]

There is a 30-point difference between yours and mine. That means that you counted 30 that I didn't.

I can't find any errors in yours.
I can't find any errors in mine.

Ouch.

[quote="miyomiyo"][quote="Gyan"][quote="miyomiyo"]A M Y:  That counts as a tie.
tOrajirOu:  Ties do not count as black winning.
samath:  There is the case where they both win or both lose.[/quote]

In order that  one can make ANY sense out of this problem, [b]you may at least try to define what do you mean by "black won".
[/b]
[hide]An  ordinary tic-tac-toe is played by altering two types (say X and O's).. to see which one (if any) can get three of the same type in a line. "Random" 9 cards from a deck need not even have equal (or with a difference of 1) cards of  alternate color...
[/hide][/quote]

Umm... do you know how to play tic-tac-toe?

[/quote]

Yes, I  believe, I do know the rules of tic-tac-toe! But I [b]don't[/b] believe it is played with  "randomly putting 9 cards from a card deck !"
What's next: "problem" about "pretend poker with X's and O's " and  asking the probability of  full house in your next game of tic-tac-toe?

Don't want to be harsh or insulting, but  one has to be a  a little precise to pose  a mathematical  "problem" .. the problem as originally posted, IMO, is so vague that it sounds silly and  nonsensical.  (An obvious example:  you can draw nine red cards from a deck, but  you are not going to see nine "X's in a standard tic-tac-toe game)

Hamster1800, Bill The Mack: I did not look very carefully,  but the considering  your version  of the problem statement, the answer is not that easy (or fun), it just requires brute-force counting.

You first have to calculate the probability that you will have $n$ read cards out of 9 (Binomial type formula)  and then counting in how many ways one can win. (Brute force - more of less)

Based on chesspro's use of the word "exactly", I will assume that situations with redundant wins will not count.

[hide="Casewise"]Suppose:

3 black cards are dealt. There are only two ways to win:
XOO          OOX
OXO  and   OXO
OOX          XOO

4 black cards are dealt. Same as before, except one of the six reds is replaced with a black in each case, for a total of 12 ways.

5 black cards are dealt. We may either use diagonals or select a row or column.

>> For diagonals, there are 2 diagonals and we must replace two of the red cards so as not to form a row or column. There are two ways to do this keeping them on the same side of the diagonal and 2 ways if they are on different sides, for a total of 4 ways per diagonal, or 8 ways.

>> For rows/columns, there are 6 to pick from, and of the remaining 6 spaces, we must choose 2 to cover with black cards so as not to form a second row/column OR a diagonal. There are 3 rows and columns to be formed, and 2 diagonals, which we must subtract from our original 6*6=9 possibilities per original row/column, to produce 4 ways for each of the 6 rows and columns, or 24 ways.

For 6 cards, we may either use diagonals or rows/columns.

>> For diagonals, there are only two ways to color the board:
XOX
OXX
XXO as well as its reflection across the diagonal >> 4 ways.

>> For the central row and column, there are only 2 ways to color each >> 4 ways.

>> For a side row or column, there are only two ways to color >> 8 ways.

>> For 7 cards or more cards, we cannot create a diagonal without redundancy.

Also, if we use a row or column, we must block off each of the other two strips of spaces to prevent a red win. However, with 7 cards, we must use 4, and thus by pigeonhole, at least 2 will line up and create a redundant win. Hence, we are done finding possibilities.

Ways to win with 3 black cards: 2
Ways to win with 4 black cards: 12
Ways to win with 5 black cards: 32
Ways to win with 6 black cards: 18

If three black cards are dealt, then there are $C_{3}^{9}=84$ ways for the cards to be dealt, so the conditional probability of winning given 3 black cards is $P_{3}=2/84=1/42$

$P_{4}=\frac{12}{C_{4}^{9}}=12/126=2/21$
$P_{5}=\frac{32}{C_{5}^{9}}=32/126=16/63$
$P_{6}=\frac{18}{C_{6}^{9}}=18/126=1/7$[/hide]

Someone can calculate the probabilities of each number of cards being drawn for me and finish the estimation =) I'm going to bed

[hide="extending my solution"]
I actually just realised that counting the ties still works. Whatever the probability that the 9 cards split 4-5, you multiply that by the ways to tie a 4 v 5 game and subtract from 1, then halve it. But like others said, it's still brute force (mine doesn't require casework, though) and not a pretty problem.
[/hide]

On a side note, who here is 100% invincible at tic-tac-toe? (i.e., knows a perfect strategy for going first and one for going second) I'm guessing everyone, because we're all pretty smart here.

First, there is no way to win everytime, your opponent can stop you everytime, they just have to know your strategy.  Second, you can be delt any number of cards, not just 3, 4, 5, and 6.  You didn't include 7, 8, or 9.  If you get that many, you obviously have won, but you can get 7, 8, or 9 cards and need to include that.


$ a > 1$,  compute  the  sum

$ \sum_{i \equal{} 0}^{\infty} \frac {2^n}{a^{2^n} \plus{} 1}$


Se considera doua triunghiuri asemenea $A_{1}B_{1}C_{1}$ si $ABC$
 iar primul este in interiorul celui de-al doilea. Demonstrati ca exista un punct ce se corespunde lui insusi prin asemanare.

Ce faci, Cezare ? Ai pus problema asta si la avansati.

O demonstratie foarte rapida ar fi asta:

Asemanarea care duce triunghiul mare in triunghiul mic e o aplicatie continua de la trinughiul mare in el insusi. Din teorema de punct fix a lui Brouwer are un punct fix, si gata :D.

Bai, Alex, bravo, exact la asta m-am gandit si eu. Sau mai poti sa faci si cu contractii, dar o solutie asa mai scolareasca, nu exista? :)

eu n`am inteles enuntul . ce punct cautam si ce "restrictii" sunt asupra lui ?

Cautam un punct fix. Cezar n-a mai mentionat altceva.

Cred ca asemanarile directe sunt date de $z \mapsto \alpha z+\beta$, iar cele inverse de $z \mapsto \alpha \overline z+\beta$. E destul de usor de dedus ca punctul fix exista daca si numai daca $\alpha \neq 1$ sau ($\alpha = 1$ si $\beta = 0$). In problema de fata, $|\alpha| \neq 1$ ($|\alpha|$ = "coeficientul de asemanare"), pentru ca un triunghi e in interiorul celuilalt.