It is rather easy.
Denote H be the ortho projection of A on BC. $ \angle OAD\equal{}\angle HAD$, so $ \angle ADP\equal{} \angle ADB$, hence result.

[quote="anonymous111"][quote="mr.danh"]It is rather easy.
Denote H be the ortho projection of A on BC. $ \angle OAD \equal{} \angle HAD$, so $ \angle ADP \equal{} \angle ADB$, hence result.[/quote]
Why $ \angle OAD \equal{} \angle HAD$   :blush:
thanks[/quote]
$ \angle BAH \equal{} 90^0 \minus{} \angle ABC \equal{} \frac {180^0 \minus{} 2\angle ABC }{2} \equal{} \angle CAO$

$ \Leftrightarrow \angle HAD \equal{} \angle BAD \minus{} \angle BAH \equal{} \angle DAC \minus{} \angle CAO \equal{} \angle OAD.$

[quote="apollo"]An acute-angled triangle $ ABC$ is inscribed in a circle with center $ O$.
Let $ D$ be the intersection of the bisector $ A$ with $ BC$ and suppose that the perpendicular to $ AO$ though $ D$ meets the line $ AC$ in a point $ P$ interior to the segment $ AC$.show that triangle $ ABP$ is isosceles.[/quote]

 I think I have seen this problem in an Italian olympiad  with a solution in som issue in Crux. Tryin to solve the problem I had found the same solution as mr. Danh .The solution in Crux was a little different .The problem is really easy , but basic.It uses the fact that $ AO, AD$ are isogonal wrt $ \angle A$, where $ AD$ is an altitude.

[b] Babis[/b]

Dear Mathlinkers,
I present a proof based on well-knowed theorems.
Let 1 the incircle, A', B' the points of contact of 1 wrt BC, CA.
1. It is knowed that DP is tangent to 1.
2. According to Newton's theorem applied to the tangential quadrilateral ABDP, AD, BP and A'B' are concurrent.
3. Let U this point of concurrence.
4. According to "An unlikely concurrence" (Honsberger R., Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, New Mathematical Library, 1995 p. 31), AU is perpendicular to BU.
5. In the triangle ABP, the A-bissector being also the A-altitude, ABP is A-isoceles.
Sincerely
Jean-Louis


sole e besiyar sade, fekr konam male IMO taiwan bashe.

hameye zoj hayer moratab e $(a,b)$ az aadad e sahih va mosbat ra biyabid, ke $a^2b+a+b$ bar $ab^2+b+7$ bakshpazir bashad.

$ab^{2}+b+7\mid a^2b+a+b\Rightarrow ab^{2}+b+7\mid a^2b^2+ab+b^2 , ab^{2}+b+7\mid a^2b^2+ab+7a\Rightarrow ab^{2}+b+7\mid (a^2b^2+ab+b^2)-(a^2b^2+ab+7a)\Rightarrow ab^{2}+b+7\mid b^2-7a$
حالا حالت بندی می کنیم:
$1)b^2-7a> 0\Rightarrow b^2< ab^2+b+7< b^2-7a<b^2\Rightarrow pas-dar-in-halat-javabi-nadarad$
$2)b^2-7a< 0\Rightarrow ab^2< ab^2+b+7< 7a-b^2<7a\Rightarrow b^2<7\Rightarrow b=1,2$$IF(b=1)\Rightarrow a=11,49$

$3)b^2=7a\Rightarrow \left\{\begin{matrix}
a & = & 7 & k^2\\
 b& = & 7 & k
\end{matrix}\right.$


inequalities
combinatorics unsolved
combinatorics

Let $A$ be a non-empty subset of the set of all positive integers $N^*$. If any sufficient big positive integer can be expressed as the sum of $2$ elements in $A$(The two integers do not have to be different), then we call that $A$ is a divalent radical. For $x \geq 1$, let $A(x)$ be the set of all elements in $A$ that do not exceed $x$, prove that there exist a divalent radical $A$ and a constant number $C$ so that for every $x \geq 1$, there is always $\left| A(x) \right| \leq C \sqrt{x}$.

Actually the inequality should be reversed - the general result is if $2$ is changed for $k$ then $|A(x)| \geq C x^{\frac{1}{k}}$ - unfortunately I cannot prove it now.

Consider the set consisting of all number having just digits $0,1$ in base $4$ or just digits $0,2$ in base $4$.
We can see that every number $0\leq k\leq 3$ can be uniquely writteln as $a+b$ where $a \in \{0;1\}$ and $b \in \{0;2\}$, so this set is divalent radical.

Now we can see that $|A\bigcap \{1,2,\ldots,x\}|<4\sqrt x$

Nice, I didn't know it could be estimated from below by this. As I said above reversed inequality is also true (i.e. there exists another constant  $B>0$ such that $|A(x)|\geq B\sqrt{x}$  - it is an exercise in Narkiewicz's "Teoria Liczb


How do you find the center of intersection of the two lines each of the form ax+by+c=0.

Eg: the center of intersection of the lines $ 3x \plus{} \frac{7}{2}y \plus{} \frac{5}{2} \equal{} 0$ and $ \frac{7}{2}x \plus{} 2y \plus{} \frac{5}{2} \equal{} 0$ is $ (\minus{}\frac{3}{5},\minus{}\frac{1}{5})$.

But how to obtain that coordinate ,,, ?

I think I have learnt it before ,,, but I can't seem to recall ,,,

Your guide will be appreciated ,,, !

hello, multiplying the second equation by $ - \frac {7}{4}$ we get
\begin{eqnarray*}
3x + \frac {7}{2}y + \frac {5}{2} &=& 0\\
- \frac {49}{8}x - \frac {7}{2}y - \frac {35}{8}& =& 0
\end{eqnarray*}
adding both we have $ 3x - \frac {49}{8}x + \frac {5}{2} - \frac {35}{8} = 0$, from here we get $ x = - \frac {3}{5}$
plugging this in the first equation and solving this for $ y$ we get $ y = - \frac {1}{5}$.
Sonnhard.

oh yea ,,, ! now i remember ,,, it's solving by elimination ,,, !
gosh ,,, i sometimes get these small problems confused from somewhere deep in some complicated examples ,,, !

Thank anyway ,,, !

[size=75]-Topic closed-[/size]

Just a formula I made up:

If the lines $ ax\plus{}by\plus{}c\equal{}0$ and $ dx\plus{}ey\plus{}f\equal{}0$ meet at point $ (x,y)$, then:

$ x\equal{}\frac{fb\minus{}ce}{ae\minus{}db}$

$ y\equal{}\frac{\frac{\minus{}afb\plus{}ace}{ae\minus{}db}\minus{}c}{b}$

That is rather long though.

[quote="PowerOfPi"]Just a formula I made up:

If the lines $ ax \plus{} by \plus{} c \equal{} 0$ and $ dx \plus{} ey \plus{} f \equal{} 0$ meet at point $ (x,y)$, then:

$ x \equal{} \frac {fb \minus{} ce}{ae \minus{} db}$

$ y \equal{} \frac {\frac { \minus{} afb \plus{} ace}{ae \minus{} db} \minus{} c}{b}$

That is rather long though.[/quote]

You can get $ y\equal{}\frac{cd \minus{} fa}{ae \minus{} bd}$ by replacing the corresponding coefficients in your first equation with the y- coeffiecient. Another way is to use the matrices, and calculate the determinants, and divide to get x and y. It's in Art Of Problem Solving Volume 1: The basics, and I think Introduction to algebra, and probably a lot more other books(they're all awesome).

lolx ,,, no need for that formula ,,, !

as you know our mind have only the limited cells to remember all maths formulas ,,, !  :P

such a formula, though welly constructed, I believe, has no use.

Elimination is just as fast, and I would prefer not to plug in so many terms...

No offense, of course :lol:


Joe multiplies a number by 4, adds 1, and then divides by 3, getting a result of 7. Sue divides the same original number by 3, adds 1, and multiplies by 4. What result does she get? Express your answer as a common fraction.

Let the number be x.
We know that 4x+1=21, so x=5.
Plug it in, and (5/3)+1=8/3.
8/3*4=32/3.[/b]


Does the amount of money we said we can pay on the scholarship application have any role in whether or not one gets admitted? Like, if someone said that they can only pay X dollars, and Mathcamp can't provide that much, is it possible that the applicant would be denied on those grounds?

Just curious.


A hacker is attacking a computer network. Each computer on the network is connected to various other computers. The hacker releases a worm on a source computer. When a worm infects a computer, that computer can propagate a copy of the worm to a connected computer at a rate of once per second. Suppose the hacker knows the layout of the network; i.e., he knows which computers are connected to which. How should the worm proceed to infect the whole network as quickly as possible?

Also, if possible, describe an algorithm for computing the minimum time till total network infection. (Admittedly a computer science background will help here, but it is not absolutely necessary.)

Common sense tells me to attack the computer that is connected to the most other computers (i.e. the server or root of the network), because the server can then infect lots of computers.

We really can't do this without more information, I think.

suppose the number of computers is $n$

the number of people is $p$

the gravity is $g$

the velocity is $v$

etc

if you dont have more information, creat this assuming logical things
in physics we do it a lot

"can we approximate the variable velocity to a constant velocity?"

if the result does not change a lot the result of the result, so, we can do this

and, this problem can be found on the "Difficult Problems and DIfficult People" amazing magazine, edition 24


prove that all of integer numbers  such that $y^{2}=x^{3}+1$
are (-1,0), (0,1), (2,3)

the equation is quitefamous, and  the solution uses elliptic curves of course.  THis is a special case of Catalan's Conjecture.

(haha, the identical problem was discussed in SIMUW 2005 Elliptic curves class by Prof. Robert Pollak)


function
induction
limit
algebra proposed
algebra

Let $\mathcal{F}$ be the set of all functions $f : (0,\infty)\to (0,\infty)$ such that $f(3x) \geq f( f(2x) )+x$ for all $x$. Find the largest $A$ such that $f(x) \geq A x$ for all $f\in\mathcal{F}$ and all $x$.

Let $\mathcal{G}$ be the set of all $\alpha\in (0,\infty)$ with $f(x)\geq \alpha x$ for all $f\in\mathcal{F}$ and all $x\in (0,\infty)$.
Then $f(x) \geq f( f({2x\over 3}) )+{x\over 3}>{x\over 3}$ for all $f\in\mathcal{F}$ and all $x\in (0,\infty)$, so ${1\over 3}\in\mathcal{G}$ and $\mathcal{G}$ is not empty.
Also $(0,\infty)\to (0,\infty),x\mapsto{x\over 2}$ is obviously in $\mathcal{F}$. So ${1\over 2}$ is an upper bound for $\mathcal{G}$.
Therefore $A: =\sup \mathcal{G}\leq{1\over 2}$ exists and again $A\in\mathcal{G}$.
Then for all $f\in\mathcal{F}$ and all $x\in (0,\infty)$ we have
$f(x) \geq f( f({2x\over 3}) )+{x\over 3}\geq A f({2x\over 3})+{x\over 3}\geq A^{2}{2x\over 3}+{x\over 3}={2A^{2}+1\over 3}x$.
So ${2A^{2}+1\over 3}\in\mathcal{G}$ and therefore ${2A^{2}+1\over 3}\leq A$ and $0\geq 2A^{2}-3A+1=(A-1)(2A-1)$.
This and $A\leq{1\over 2}$ give $A={1\over 2}$.

My solution :D
By induction on $n$ we have $f(x)\geq u_{n}x\forall n\in\mathbb{N}$, where $u_{1}=\frac{1}{3},u_{n+1}=\frac{2u_{n}^{2}+1}{3}(n=1,2,...)$. Now, I see $\lim_{n\to\infty}u_{n}=\frac{1}{2}$.

My solution is [url=https://artofproblemsolving.com/community/q1h532939p3077799]here[/url]


Find the sum of all k digit palindromic numbers if k is odd.

[hide="hint:"]what is the average of all $k$ digit palindromes?[/hide]


[quote="Beat"]$a,b,c\ge 0$ and a+b+c=3. Prove that $4(a+b+c)^{3}\ge 27(ab^{2}+bc^{2}+ca^{2}+abc)$[/quote]
Why $a+b+c=3$ $?$ This superfluous.

Let $a=\min\{a,b,c\}.$ Then $4(a+b+c)^{3}\ge 27(ab^{2}+bc^{2}+ca^{2}+abc)\Leftrightarrow$
$\Leftrightarrow9a(a^{2}+b^{2}+c^{2}-ab-ac-bc)+(b+4c-5a)(a+c-2b)^{2}\geq0.$ :)
See also  the Sung-yoon Kim's idea:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

Prove me that if a+b+c=3 and a,b,c>0 then $ab^{2}+bc^{2}+ca^{2}+abc\le4$ in a different from arqady`s method

The problem is: If the sum of x,y,z non-negatives is 1 then prove x^2y+y^2z^+z^2x+xyz<=4/27
We should know that (see in the solved problems) x^2y+y^2z+z^2x<=4/27 and that was a quite hard question :([/hide]


[quote="Internet Explorer"] phpBB : Critical Error

Could not connect to the database

[/quote]

Anybody know what happened?...i got in now...but yea...was the server down for a bit?

Same here.
[quote="forum"]phpBB : Critical Error

Could not connect to the database [/quote]

Now I encounter even stronger error messages than the "Debug Mode" errors :)

Well the first time I saw the "Debug" error I thought "there's blue screen of death in AoPS too!? They're everywhere! Oh the horror, THE HORROR!" :)

Yeah, I know, I'm such a drama king.

Anyways, so after I saw the critical error I was like "oh (enter the person/object you worship here :) )! Not again!!!" :D

But, yeah, i got over it :P

lol...i just thought there was something wrong with the server for once (it doesnt happen often on math websites)


[quote="bubka"]A man who lives within half a kilometer from my house has proved the Beal's Conjecture and Fermat's Last Theorem using number theory only!
His lecture in the Hawaii International Conference on Statistics is available on the web at http://www.hicstatistics.org
Now where should I post this sensational news?[/quote]

[url=http://www.norvig.com/beal.html]Beal's Conjecture[/url]

What you say sounds like not true...


So I thought that I would make it but as of now I still have not gotten a letter from the AAPT. Is anyone in a similiar position to mine? Also, if you already got the letter, hom much of the semifinal exam did you solve?

There are so many topics regarding the USAPhO that perhaps no one noticed yours. I would suggest you choose one topic for your discussion instead of generating new topics for every stage of the USAPhO.

But not too easy. As you can see, there are so many USAPhO-related topics, they are going to have a tough time deciding which one is the best for their discussion. :rotfl:

they shud better discuss it in their Natinal forums :D  :rotfl:

Kamil might be angry Rituraj! He expected some information about the letter and we (not even remotely related to USAPhO) are raining spam in his topic.
 :rotfl:

No Kamil, i hav not got the letter (i m in a similar position...)

[hide]this shud help :P
[/hide]


Let $ n$ and $ k$ be given natural numbers, and let $ A$ be a set such that \[ |A| \leq \frac{n(n+1)}{k+1}.\] For $ i=1,2,...,n+1$, let $ A_i$ be sets of size $ n$ such that \[ |A_i \cap A_j| \leq k \;(i \not=j)\ ,\] \[ A=  \bigcup_{i=1}^{n+1} A_i.\] Determine the cardinality of $ A$.

[i]K. Corradi[/i]

I 'm a little bit confused about this problem. Is there any difference between the two $ A$?

Let's assume $ A\equal{}\{1,2,\dots,m\}$.
Let $ b_i$ ($ i\equal{}1,2,...,m$) denote the number of occurrences of $ i$ among all the sets $ A_j$ ($ j\equal{}1,2,...,n\plus{}1$)
we have
$ k\binom{n\plus{}1}{2}\geq \sum_{i>j}\left|A_i\cap A_j\right|\equal{}\sum_{i\equal{}1}^m \binom{b_i}{2}$

So, $ kn(n\plus{}1)\geq \sum_{i\equal{}1}^m (b_i\minus{}\frac12)^2\minus{}\frac{m}{4}\geq \frac{1}{m}(\sum_{i\equal{}1}^m(b_i\minus{}\frac12))^2\minus{}\frac{m}4$
As, $ \sum_{i\equal{}1}^m b_i\equal{}n(n\plus{}1)$
we get $ kn(n\plus{}1)\geq \frac{1}{4m}\left((2n^2\plus{}2n\minus{}m)^2\minus{}m^2\right)\equal{}\frac{n(n\plus{}1)(n^2\plus{}n\minus{}m)}{m}$
$ \Rightarrow mk\geq n(n\plus{}1)\minus{}m$
or $ m\geq \frac{n(n\plus{}1)}{k\plus{}1}$
so we conclude that $ |A|\equal{}\frac{n(n\plus{}1)}{k\plus{}1}$

Main part is proof
\[ |A| \geq \frac{n(n+1)}{k+1}.\]

$ |A|=\frac{n(n+1)}{k+1}$ This is equivalent to: every element of $A= \bigcup_{i=1}^{n+1} A_i$ belongs to the same number of subsets $A_1,A_2,...,A_{n+1}$. Can anyone give a construction for this?

@above I doubt what you meant to say/ask. The deduction in #3 already conclude that the answer is $\frac{n(n+1)}{k+1}$ and no construction is needed since the problem already given the existence of $A$.


THE BEST THING EVA: [youtube]iRcI-mFr6aQ[/youtube]

Haha yeah I remember that this was played on one of the "Funniest Home Videos" (or something close to it).


Let $g$ be a Fibonacci primitive root ($\mod p$). i.e. $g$ is a primitive root
($\mod p$) satisfying $g^2 \equiv g + 1 \mod p$. Prove that:

(a) Prove that $g - 1$ is also a primitive root ($\mod p$).

(b) If $p = 4k+3$, then $(g-1)^{2k+3} \equiv g-2 \mod p$ and deduce
that $g - 2$ is also a primitive root ($\mod p$).

P.S. a) is quite easy and i don't have a solution for b)

For (b) we have $X = (g-1)^{2k+3} = (g-1)^{(p+3)/2}$.   So $X^2 = (g-1)^{p+3} = (g-1)^4$ and $X = \pm (g-1)^2$.  Now $X$ is an odd power of a primitive root, and so not a square.  Since $+1$ is a square and $-1$ isn't, we have $X = -(g-1)^2 = -g^2 + 2g - 1 = g-2$.  Now $p-1$ and $(p+3)/2$ can't have any factor in common, since such a factor would divide $2.(p+3)/2 - (p-1) = 4$ and $(p+3)/2 = 4k+3$ is odd.  So $X$ is a primitive root, as it's a primitive root raised to a power coprime to $p-1$.


[quote="vishalarul"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]
[hide]Isn't it just $\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{10}\cdot\frac{7}{10}\cdot\frac{6}{10}\cdot\frac{5}{10}\cdot\frac{4}{10}\cdot\frac{3}{10}\cdot\frac{2}{10}\cdot\frac{1}{10}=\frac{567}{1562500}$.[/hide]
I know I did this wrong. I don't think it should be a big denominator.

wait... it should be
the first marble he picks doesn't matter
the second can only be 9/10 (not the one he already picked)
the third can only be 8/10 (not the two he already picked)
....
the tenth can only be 1/10 (not the nine he already picked)

so it is that big fraction thingy....
unless im being dumb

[quote="vishalarul"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]
[hide]The first marble can be any of the $10$. The second can be any of the other $9$. The third can be any of the other $8$, and so forth.

$\frac{10!}{10^{10}}=\frac{9\cdot7\cdot3\cdot3}{5^{6}\cdot10^{2}}=\boxed{\frac{567}{1562500}}$[/hide]

@ davidlizeng - With replacement.

Just wondering... Why do people like to answer the same solved problem again and again and again?

[quote="vishalarul"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]
[hide]
$\frac{10!}{10^{10}}$=$\frac{567}{1562500}$

[/hide]

[quote="i_like_pie"]Because people don't look at others' answers and/or solutions and post their own.[/quote]

And maybe because they want a higher post count.

i think u of all people want a higher post count.

[quote="jhredsox"][quote="i_like_pie"]Because people don't look at others' answers and/or solutions and post their own.[/quote]

no, it just helps to write out answers and people want to solve them on their own

u don't HAVE to look, that's why there's a hide button...

jorian[/quote]

The hide button is there so you can solve the problem yourself, not so you can just type up what they wrote, even if you don't know what they said. The purpose of the thread is discussion of the problem. Let's say that I were to have a conversation with some people. Person X starts with a comment that brings up a topic that we can discuss. Person Y makes a comment about the topic. But if I say exactly what person Y said, the conversation goes no where. If I disagree with person Y, however, then it contributes something to the conversation.

See my point?

What's with this higher post count theory? Is there a prize or something.

Seriously, is there any legit reasoning to that claim or is it just a rumor that could possibly be true.

+1 [jk]

[quote="vishalarul"]Bob has $10$ marbles. What is the probability that when he chooses $10$ marbles, with replacement, that he gets all $10$ different types of marbles in just $10$ tries? Assume that the marbles are all distinct and distinguishable. Express your answer as a common fraction.[/quote]

[hide]He will pick the first marble and there is a 100% chance of him not repeating. for the second it is 9/10, then 8/10, then 7/10... 1/10

multiplying those would mean

$\frac{9!}{10^{9}}$[/hide]


I'm new at inequalities, but here are my thoughts:

[hide]The denominators are positive because of the constraints given. The numerators are trivially positive. So the expression will always be greater than zero. At least we have a lower bound.

Perhaps if you could get $\frac{4x^{2}}{3y-1}>\frac{4x^{2}}{3y}$ and $\frac{9y^{2}}{2x-1}>\frac{9y^{2}}{2x}$
so that your have another bound $\frac{4x^{2}}{3y}+\frac{9y^{2}}{2x}=\frac{8x^{3}+27y^{3}}{6xy}$
$=\frac{(2x)^{3}+(3y)^{2}}{(2x)(3y)}$

This has been mathnerd314's random speculations.

Graagh, I'm really bad at inequality problems :wallbash:[/hide]

if you have something like:

$\frac{(a^{2}+1)^{2}}{b^{2}}+\frac{(b^{2}+1)^{2}}{a^{2}}$, you can do direct algebra probably...because then a,b are reals

[quote="Altheman"]let $a=2x-1$ and $b=3y-1$

then we have

$\frac{(a+1)^{2}}{b}+\frac{ (b+1)^{2}}{a}$

eventhough we are shunned for calculus...it is quite easy to solve with calc...i think the min is when a=b=1...

i think you can contrive some algebraic justification using that info...[/quote]
yes , minimum occurs when a=b=1  :)
now with the same denote , its equivalent to finding min of A=$\frac{(a+1)^{2}}{b}+\frac{ (b+1)^{2}}{a}$
on the other hand , $A \geq \frac{(x+2)^{2}}{x}$ where $x=a+b$
now $(x+2)^{2}\geq 8x$ then $A \geq 8$

[quote="vishalarul"]Is this too difficult for this forum? I know that it is definitely too hard for HSB...

$\textrm{Given }x>\frac{1}{2}\textrm{ and }y>\frac{1}{3}\textrm{. Find the smallest value of}$
\[\frac{4x^{2}}{3y-1}+\frac{9y^{2}}{2x-1}\]
[hide="My start"]I tried two things, and they failed (at least for me)
1. AM-GM
2. Combine into 1 fraction, try to factor[/hide][/quote]
What an easy problem for so many posts!
Take $a,b$ as before. \[\frac{(a+1)^{2}}{b}+\frac{(b+1)^{2}}{a}=\frac{(a-1)^{2}+4a}{b}+\frac{(b-1)^{2}+4b}{a}\geq 4\left(\frac{a}{b}+\frac{b}{a}\right)\geq 8\]
Equality for the last is at $a=b=1$. Checking, this is the equality all along. Thus $x=1,y=1.5$ or something like that.

[quote="vishalarul"][quote="tunganh"]We have
$\frac{4x^{2}}{3y-1}+\frac{9y^{2}}{2x-1}\ge \frac{(2x+3y)^{2}}{2x+3y-2}$[/quote]

How did you get that?[/quote]
Cauchy Engel Form. Search mathlinks (or derive it yourself - its equiv. to cauchy (hint) )
[quote="tunganh"]
[hide="inequality occurs"]Inequality occurs if and only if $x=1$ and $y=\frac{2}{3}$ not when $x=y=1$ or $x=1$ and $y=1.5$[/hide][/quote]
Sorry I meant 2/3 but since I was doing it all in my head late at night I wrote 3/2. Both solutions are just as correct :D


1) Find all integers (m,n) satisfying :
 $ (m^2+n)(n^2+m) = (m+n)^3 $

  2) Find all integers (x,y) satisfying:
 $ (x+y^2)(y+x^2)= (x-y)^3 $

For the first problem.
[tex] (m^2+n)(n^2+m)=(m+n)^3 [/tex]
[tex] \Leftrightarrow mn(mn+1-3m-3n)=0 [/tex]
This is easy, [tex] 3|m|+3|n| \geq |mn|-1 [/tex].
For the second one:
The equation is equal to:
[tex] 2y^2+(x^2-3x)y+3x^2+x=0 [/tex] or [tex] y=0 [/tex]
And we must find $ x $ satisfying [tex] x^4-6x^3-15x^2-8x=z^2 [/tex]( if you don't want you can do by gen method). This is easy, right? :)


conics
parabola
analytic geometry
real analysis
real analysis unsolved

Write the equation of the parabola given the focus at (-3,-2) and the directrix is the line y=-6. Draw a sketch of what is given to help in choosing the correct formula. Be sure you have the correct formula to fill in. leave the equation in standard form.

Please list all steps and coordinates, so I can go back and teach myself.

Thank you.

Let $F(-3,-2), P(x,\ y)$, by the defnition of a parabola, $\overline{PF}$=the distance between the point $P$ and the directrix $y=-6\Longleftrightarrow |y+6|=\sqrt{(x+3)^{2}+(y+2)^{2}}\Longleftrightarrow (y+6)^{2}-(y+2)^{2}=(x+3)^{2}$, yielding the desired equation of the parabola, $(x+3)^{2}=4\cdot 2(y+4)$.


I'm in middle school so I contacted the nearby high school that offers the AMC and the lady there who organizes it said that every school has a seperate number and that they only administer the AMC to their students. I was almost sure that the people here said earlier at some point that the AMC should be taken at a high school for middle schoolers. Can someone clarify?

Last year, our high school had a girl from the middle school take the AMC 12 at our school, and then end up taking the AIME at our school...

I took the AMC 10 last year as an eighth grader and made AIME so I'm sure middle school people can take it at the nearby high school.

Is there any link on the AMC website that says something about how if you are in middle school you need to find a high school?
Also, when are we supposed to get our AMC 10 invitation for being a good scorer on the 8.

Just ask around, I'm sure you'll find something

well, my middle school has a AMC 10/12 club organized by volunteer parents that let you take AMC 10A and your choice of AMC 10B or AMC 12B.
 :lol:

[quote="13375P34K43V312"][quote="Go Around the Tree"]i think i'm taking 10a maybe[/quote]
That, my friend, would be spam.

Anyway, did you try any local universities? There are probably plenty of high schools in your area taking the AMCs, there's bound to be at least one letting you in. Also, you can talk to your middle school coach to see if he/she will accomidate it, if it's not too late.

Also, I never got an AMC 10 invitation, and I got a 25 on the 8.[/quote]

Please refrain from speaking,13375P34K43V312. You are being very annoying and disruptive. Try being a polite girl.

Anyways, bpms I'm pretty sure most schools offer it. You should just ask someone to join in. I did that since my high school doesn't offer it ( Mountview High School).

I was able to take the AMC at my school when I was in eighth grade, so I don't think there's any rule against that.

There's certainly no rule against middle schoolers taking it at a high school, as many others have said.  I'm another who took it at the local high school when I was in 8th grade.  It is possible, though, that your local high school is going to be a pain in the you-know-where and not let you take it; there's no rule that says they *have* to let you.  In that case, look for another high school, as there's also no rule that says you have to take it at the local school.  If you take it at another school, though, make sure you talk to them in advance about how to get your results.  And also keep in mind that you'll probably have to take AIME and maybe even USAMO at that school if you make it.

Sly Si is correct on all accounts.

No high school is obligated to take you.  Local rules differ by high school and local circumstances.  It also depends on the willingness of the Contest Adminstrator to do the extra work required (see below about the AIME, etc.)

You should ask around and find a willing high school, or local college or University (see the AMC website for a list of colleges and universities adminstering the AMC B contests) or a local home school group.

Be prepared in advance to get your score from the administering site and contest manager.  Also be prepared in advance to take the AIME and potentially the USAMO again at the administering site.  The AMC has no way of individually contacting and reporting back to your everyday school if  your everyday school does not adminster the contests.

Steve Dunbar
AMC Director

[quote="purple plume"][quote="13375P34K43V312"][quote="Go Around the Tree"]i think i'm taking 10a maybe[/quote]
That, my friend, would be spam.

Anyway, did you try any local universities? There are probably plenty of high schools in your area taking the AMCs, there's bound to be at least one letting you in. Also, you can talk to your middle school coach to see if he/she will accomidate it, if it's not too late.

Also, I never got an AMC 10 invitation, and I got a 25 on the 8.[/quote]

Please refrain from speaking,13375P34K43V312. You are being very annoying and disruptive. Try being a polite [b]girl[/b].

Anyways, bpms I'm pretty sure most schools offer it. You should just ask someone to join in. I did that since my high school doesn't offer it ( Mountview High School).[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="tokenadult"]If you are within reasonable driving distance of the Twin Cities metropolitan area of Minnesota, I am always very happy to accommodate you in my homeschoolers testing group. We have a full set of test kits for this year, and not all have been claimed yet.[/quote]
I'm pretty sure he lives in ...

[quote="lotrgreengrapes7926"][quote="tokenadult"]If you are within reasonable driving distance of the Twin Cities metropolitan area of Minnesota, I am always very happy to accommodate you in my homeschoolers testing group. We have a full set of test kits for this year, and not all have been claimed yet.[/quote]
I'm pretty sure he lives in[/quote]
You are correct. Thanks for the offer tokenadult. Can you edit your post, I do not want everyone to know where I live.


[url=http://www.savetoby.com/]http://www.savetoby.com/[/url]

[i]Moderator Note: This is NOT appropriate for viewers "of all ages". If you are under 13 years of age or if you are a parent with a child under 13 years old beside you, I [b]strongly[/b] discourage you from visiting the site.[/i]

That guy has a great way to make money, but if he doesn't get it, will he really eat the bunny?

Lol, awesome.

I wonder if he really has gotten that much money or if he just says so to get others to be more inclined to help Toby.  :D

I wonder if he only raises 49,999$ will he still eat him? :)

OK.... That is a little twisted. Smart! But twisted...

probing furthur onto the site, you find out that paypal has stopped accepting donations because of numerous PETA complaints

it almost makes you want to buy a shirt just for the utter... "WHAT?"-ness of it. :D even though i love my bunny to death and think that's awful. (his name is finley. we found him in our driveway not long after easter about two years ago. CURSE those **** people who think pets are disposable).

Well, unfortunately, I just found out that it isn't true.

[url=http://www.snopes.com/critters/crusader/savetoby.asp]Oh Snopes, always ruining my fun[/url]

yeah, but you could still possibly buy shirts....althought that could be fake too

I dont want to try.

who would buy a "save the rabbit" shirt if the rabbit didn't need to be saved?

Maybe I should try this...........
[img]http://www.goldfiles.com/images/screenshots/s1_frogblender.png[/img][img]http://www.goldfiles.com/images/screenshots/s1_hamstermicro.png[/img]
I bet I could earn money twice as fast with twice the number of animal lives at stake. (note to self-get better pictures)

Yay JoeCartoon!  That site is not nearly as well-known as it should be!

I definitely could earn some money off this!
[img]http://blog.trg.ru/img/shaved_cat.jpg[/img]


In the figure to the right, the largest circle has a radius of six meters. Five congruent smaller circles are placed as shown and are lined up in east-to-west and north-to-south orientations. What is the radius in meters of one of the five smaller circles?

2.

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

There's supposed to be a picture.  :wink:

It would be helpful to explain HOW it's 2.   :|

 :D  Please explain how.

Fine izzy... GOD!

The large circle has diameter 12.  Since there are 3 circles lined up, divide by 3.  Each circle has diameter 4, thus, they have radius 2.

There... you wanted it.

Umm...please explain how people are supposed to know how to solve it when all they have is an answer? They can see the answer themselves by reviewing the questions...what they want is an explanation.  :)

If they want an explaination, they should ask for it.  I believe mz94 already brought this up to you, izzy.  :D

Ok.

Please explain why a person would go to the Review Questions, go to a problem, and go to the TROUBLE of Asking On the Forum, when about 2 inches above the "Ask on Forum" button is the answer. People want to know HOW it's whatever it is. *sigh*

yea they wanna know how the problem is done

thats right izzy

they just dont wanna know certain random facts like factorials unless they specifically ask it in the problem or ask it on teh problem in teh forum

the picture is basically 2 concentric circles, the larger w/ radius 6, the smaller having 4 circles tangent to it that are congruent to it and such that a certain diameter of teh circle will pass thru 3 of the circles' centers

Wh00ps, my bad.  :blush:  :blush:  :blush:


i don't remember the problem too well, but i'll try.

there are three distinct prime numbers, e, f, and g. If $ e\equal{}f^2 \plus{} g^2$, what is the greatest possible value $ e$ such that e is less than 100?

[hide=" Answer"]
53[/hide]

Are there any quicker ways of doing this besides guess and check?

Thanks,
LHM

I would go through the possibilities of $ f$ and $ g$ and use those.  It's not too bad given 5-10 seconds :)

[hide]All primes except 2 are odd, so you know either $ f^2$ or $ g^2$ must be even. The only even prime is 2, so one of them has to be $ 2^2 \equal{} 4$.

So now you've got $ 100 \ge 4 \plus{} g^2$

so $ 96 \ge g^2$

That means $ g$ can be: 2, 3, 5, or 7. To maximize $ e$ you need to choose the greatest of these, which is 7.

$ e \equal{} 2^2 \plus{} 7^2 \equal{} 4 \plus{} 49 \equal{} \boxed{53}$[/hide]

yeah I guess to think _all_ of this out during a cd would take too much time.. I think it gets faster once you grasp it well.


inequalities
ratio
invariant
function
Cauchy Inequality
inequalities proposed

Let $a,b,c \ge 0$ then prove

                $\sum_{cyc}^{a,b,c} \frac{a^2+b}{b+c} \ge\frac{\displaystyle \sum_{cyc}^{a,b,c} a +3}{2}$

[hide="My solution"]
Substitued $\begin{cases} a+b=x\\[.1in] b+c=y\\[.1in]c+a=z \end{cases} \longrightarrow \begin{cases}a=\frac{x+z-y}{2}  \\[.1in]b= \frac{y+x-z}{2} \\[.1in]c= \frac{z+y-x}{2} \end{cases}$

$\longrightarrow$

$\sum_{cyc}^{x,y,z}\frac{(\displaystyle\frac{x+z-y}{2})^2+\frac{y+x-z}{2}}{x}\ge \frac{\displaystyle\frac{x+y+z}{2}+3}{2}=\frac{x+y+z+6}{4}$

$\longrightarrow
\sum_{cyc}^{x,y,z}\displaystyle\frac{(x+z-y)^2}{x} +\sum_{cyc}^{x,y,z} 2(1+\frac{y}{x}-\frac{z}{x}) \ge x+y+z+6$

$\longrightarrow
 \sum_{cyc}\displaystyle\frac{(x+z-y)^2}{x}+\sum_{cyc}^{x,y,z}\frac{y}{x} \ge (x+y+z)+\sum_{cyc}^{x,y,z}\frac{z}{x}$

Which we devide to :
$1.\sum_{cyc}^{x,y,z}\displaystyle\frac{(x+z-y)^2}{x} \ge x+y+z$[hide="Solution"]
From cauchy inequality
$(\sum_{cyc}^{x,y,z} \displaystyle \frac{(x+z-y)^2}{x})( \sum_{cyc}^{x,y,z} x) \ge (\sum_{cyc}^{x,y,z} (x+z-y)=\sum_{cyc}^{x,y,z} x)^2$

$\longrightarrow  \sum_{cyc}^{x,y,z}\displaystyle\frac{(x+z-y)^2}{x} \ge x+y+z$[/hide]
$2.\sum_{cyc}^{x,y,z}\frac{y}{x} \ge \sum_{cyc}^{x,y,z}\frac{z}{x}$[hide="rearrangement"]
Assume $x\ge y \ge z \longrightarrow \frac1z \ge \frac1y \ge \frac1x $
By rearrangement :
$\sum_{cyc}^{x,y,z}\frac{y}{x} \ge \sum_{cyc}^{x,y,z}\frac{z}{x}$[/hide]
[/hide]

[quote="lomos_lupin"]Let $a,b,c \ge 0$ then prove

                $\sum_{cyc}^{a,b,c} \frac{a^2+b}{b+c} \ge\frac{\displaystyle \sum_{cyc}^{a,b,c} a +3}{2}$[/quote]

The inequality is wrong (take a = 0.0001, b = 0.0002, c = 0.0003). So is the solution (you can't assume $x\geq y\geq z$ in the proof of the inequality 2 since it is not symmetric). But some additional conditions can make the inequality correct. Here is a proof under the condition $a+b+c\geq\frac15$:

[color=blue][b]Theorem 1.[/b] Let a, b, c be three nonnegative reals such that $a+b+c\geq\frac15$. Then,

$\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq\frac{a+b+c+3}{2}$.[/color]

[i]Proof of Theorem 1.[/i] First we show a helpful inequality:

[color=blue][b]Theorem 2.[/b] For any three nonnegative reals a, b, c, we have

$(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\geq 3(c^2b+a^2c+b^2a)$.[/color]

[i]Proof of Theorem 2.[/i] Here is the ugliest possible proof of a 3-variable inequality.

Since the inequality in question is cyclic, we can WLOG assume that a is the smallest among the numbers a, b, c. Then, $a\leq b$ and $a\leq c$, so that $b-a\geq 0$ and $c-a\geq 0$. Denote b - a = X and c - a = Y; then, $X\geq 0$ and $Y\geq 0$. Now, some boring computation shows that

$(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a)$
$=a((b-c)^2+(c-a)^2+(a-b)^2)+(X^3+2X^2Y-3XY^2+Y^3)$.

Now, we have to show that $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\geq 3(c^2b+a^2c+b^2a)$, i. e. that $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a)\geq 0$. Since $a((b-c)^2+(c-a)^2+(a-b)^2)$ is trivially $\geq 0$, this will be proven once we show that $X^3+2X^2Y-3XY^2+Y^3\geq 0$. But this is true, since AM-GM yields $X^3+X^2Y+X^2Y+\frac{Y^3}{4}+\frac{Y^3}{4}+\frac{Y^3}{4}+\frac{Y^3}{4}\geq 7\sqrt[7]{X^3\cdot X^2Y\cdot X^2Y\cdot\frac{Y^3}{4}\cdot\frac{Y^3}{4}\cdot\frac{Y^3}{4}\cdot\frac{Y^3}{4}}$, what simplifies to $X^3+2X^2Y+Y^3\geq 7\sqrt[7]{\frac{1}{4^4}}XY^2$. But $7\sqrt[7]{\frac{1}{4^4}}>3$, and thus $X^3+2X^2Y+Y^3\geq 3XY^2$, so $X^3+2X^2Y-3XY^2+Y^3\geq 0$, and Theorem 2 is proven.

Now, we equivalently transform the inequality in question:

$\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq\frac{a+b+c+3}{2}$
$\Longleftrightarrow\ \ \ \ \ (\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})+(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})\geq\frac{a+b+c}{2}+\frac32$
$\Longleftrightarrow\ \ \ \ \ (\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}\geq\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})$.

First note that, if the right hand side of this inequality, $\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})$, is < 0, then the inequality is trivial, since the left hand side, $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}$, is always $\geq 0$ (in fact,

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}=\frac{(a+b+c)((b+c)(b-c)^2+(c+a)(c-a)^2+(a+b)(a-b)^2)}{2(b+c)(c+a)(a+b))}$
$\geq 0$).

Hence, in the following, we will assume that the right hand side, $\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})$, is $\geq 0$. Now,

$(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}-5(a+b+c)(\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}))$
$=\frac{(a+b+c)((a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)-3(c^2b+a^2c+b^2a))}{(b+c)(c+a)(a+b)}\geq 0$,

since $(a^3+b^3+c^3)+2(b^2c+c^2a+a^2b)\geq 3(c^2b+a^2c+b^2a)$ by Theorem 2. Thus,

$(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}\geq 5(a+b+c)(\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}))$.

On the other hand, $a+b+c\geq\frac15$ implies $5(a+b+c)\geq 1$, what, together with $\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})\geq 0$, yields

$5(a+b+c)(\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}))\geq\frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})$.

Thus, $(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b})-\frac{a+b+c}{2}\geq \frac32-(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b})$, so that Theorem 1 is proven.

  Darij

Perfect ,I couldnt expect a better answer(you are good) ,I am going to show darij`s solution to my teacher and say [b]At least check your problems before giving them to us[/b] :mad:

I managed to reach untill here :

$\sum_{cyc}^{,x,y,z} \frac{a^2+b}{b+c} - \sum_{cyc}^{x,y,z} \ge 3 - \sum_{cyc}^{x,y,z} \frac{b}{b+c}$
And the case that [b]LHS[/b] be negative ,But i have a question :
How can you do these computations ,I couldnt even think about doing these computations ,cause i will get lost in them . :(  :(,look at your solution man ,You perfectly just played with these numbers,Welldone.

Well, let me explain some things about the proof.

At first, look at your inequality:

$\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq\frac{a+b+c+3}{2}$.

It contains some terms of degree 1 and some terms of degree 0. So it is natural to bring all the degree 1 terms on the one side and the degree 0 terms on the other side:

$\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)-\frac{a+b+c}{2}\geq\frac32-\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)$.

Now, if you decrease the numbers a, b, c, keeping their ratios b / c, c / a, a / b fixed, then the degree 0 terms remain invariant, but the degree 1 terms decrease and get arbitrarily close to 0. Thus, if such an inequality is generally true, then we must have

$0\geq\frac32-\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)$.

But this is nonsense, since a simple calculation shows that

$\frac32-\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)=\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{2\left(b+c\right)\left(c+a\right)\left(a+b\right)}$,

and if this is $\leq 0$ for some number triple (a, b, c), then it is $\geq 0$ for the permutation (a, c, b) of this number triple.

This shows that your inequality, in the form given, cannot be true. So it needs an additional condition. I have looked at conditions of the form $a+b+c\geq k$ for a constant k, and found that $k=\frac15$ is more or less the optimal constant (there exists a slightly smaller constant, but if I am not mistaken, it is a root of an ugly irreducible polynomial). Now, of course, having the condition $a+b+c\geq\frac15$, we can homogenize the inequality into the form

$\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)-\frac{a+b+c}{2}\geq 5\left(a+b+c\right)\left(\frac32-\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)\right)$,

in which it contains only terms of degree 1 and thus is homogeneous. Now, after a simple computation (I have done it using computer algebra, but it is well manageable at hand: only b + c, c + a and a + b ocur as denominators, and lots of things are divisible by a + b + c, so the terms will not get very long) we arrive at Theorem 2.

Remains to prove Theorem 2. Well, define the function

$f\left(a;\;b;\;c\right)=\left(a^3+b^3+c^3\right)+2\left(b^2c+c^2a+a^2b\right)-3\left(c^2b+a^2c+b^2a\right)$.

Then, we have to show that $f\left(a;\;b;\;c\right)\geq 0$ for any three nonnegative a, b, c. Now, if we decrease the numbers a, b, c by a given positive number d, then the value of the function f decreases:

$f\left(a;\;b;\;c\right)-f\left(a-d;\;b-d;\;c-d\right)=d\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)\geq 0$.

Hence, if we succeed to prove that $f\left(a-d;\;b-d;\;c-d\right)\geq 0$ for some d, then we automatically get $f\left(a;\;b;\;c\right)\geq 0$. But in fact, if we WLOG assume that a is the smallest of the numbers a, b, c, and take d = a, setting b - a = X and c - a = Y, this comes down to showing that $f\left(0;\;X;\;Y\right)\geq 0$ for nonnegative reals X and Y; since $f\left(0;\;X;\;Y\right)=X^3+2X^2Y-3XY^2+Y^3$, it remains to show that $X^3+2X^2Y-3XY^2+Y^3\geq 0$. Now this is straightforward.

  Darij


combinatorics solved
combinatorics
Sequences

To every natural number k \geq 2 corresponds a sequence a_n(k) according to the following rules:

a_0 = k, a_1 = \gamma (a_0), a_2 = \gamma (a_1), ..., a_n = \gamma  (a_(n-1)) in which \gamma (a) is the number of different divisors of a.

Find all k for which the sequence a_n(k) does not contain the square of integer numbers !

For k= 2:  an=2 is not perfect square.

It is clear if n>2 then \gamma (n) < n ==> for some m am=2. Let m be the minimal such number ==> am-1=p>2 where p is prime number ==> p is odd number ==> if m>2 then am-2 is a perfect square.

Thus k must be prime number.


I also posted this on the mathcounts.org forum, but then I decided I'd like to be able to see replies in less than say, a week (any of you who frequent that forum undoubtedly know about the screening).

Is there anyone who wouldn't mind kind of giving me a general run-down of ARML? The website is very unhelpful. I believe Washington State does send an ARML team sometimes, but I have not yet had a chance to get involved. I'm just curious because just about everyone at MOP had gone to ARML, and it seems like something I should get involved in as well.

In particular, what are the different rounds and how do they work? I know what the round names are (from the problems of arml.com), but I don't know much about them. Also, how does the individual competition work? Obviously since the indiv. test is out of 8, there are lots of ties, and I know that there is some sort of "ARML Countdown" do determine the champion, but do they rank everyone top to bottom somehow as well?

[color=cyan]Okay, this is fun.  A team is 15 people.  The competition is four rounds:
Team, Power, Individual, Relay.
Team: 20 minutes.  Calculators are okay.  The entire team may work together.  10 short-answer questions ranging from very easy to quite difficult. 4pts / question
Power: the highlight of ARML.  1 hour.  The entire team may work together.  Calculators are okay, but rarely useful.  A series of proofs which build upon each other.  Very much fun.  Earlier problems are often just computation.  Later problems can be very difficult proofs.  40 points total.
Individual: 8 questions given in pairs, with 10 minutes per pair.  No calculators.  No team work.  Each question is worth 1 point, making the entire deal worth 120, or half of the overall total for the competition.
Relay: The relay portion of the contest is really bizarre.  Every team is broken up into 5 groups of 3 people.  Within each group, each person gets a separate question.  The answer to the first question is necesary for the second question, the asnwer for the second question is necessary for the third, and only the third answer counts for points.  For each group that gets it right within 3 minutes, 4 points are awarded.  There is also an opportunity to hand in an answer after 6 minutes, for which 2 points are awarded.  The whole procedure is carried out twice.  Thus the entire round is worth at most 5*4*2 = 40 points.  I enjoy the relays.  NYC A is not any good at them.

At the very end of the competition, the number of 8's and 7's determines how many people will compete in a tiebreaker round, which is carried out simultaneously at the three ARML sites.  The fastest correct answer among the 8s earns first place; if there are fewer than 5 or 10 8's, the sevens also take part in the tiebreaker, and the highest scorer among the 7's earns a place after the lowest-ranking 8.  There is no ranking system for individual scorers not competing in the tiebreaker.

Okay, I think that's it.  I have no idea if Washington sends a team, but there is a list of teams which competed for the past few years on the AMRL website.[/color]


Given that $ a \otimes b \equal{} (a^2 \plus{}b) \div 2$. What is the value of $ 5 \otimes 3$?

plugging in $ 5$ and $ 3$ for $ a$ and $ b$, we get

$ (5^2 \plus{} 3)/2 \equal{} 28/2 \equal{} \boxed{14}$


Let $ x,y,z>0$ prove that
\[ \frac{x}{4x^2\plus{}y^2\plus{}z^2}\plus{}\frac{y}{x^2\plus{}4y^2\plus{}z^2}\plus{}\frac{z}{x^2\plus{}y^2\plus{}4z^2}\leq\frac{3(x\plus{}y\plus{}z)}{10(xy\plus{}yz\plus{}zx)}\]

[quote="encyclopedia"]Let $ x,y,z > 0$ prove that
\[ \frac {x}{4x^2 \plus{} y^2 \plus{} z^2} \plus{} \frac {y}{x^2 \plus{} 4y^2 \plus{} z^2} \plus{} \frac {z}{x^2 \plus{} y^2 \plus{} 4z^2}\leq\frac {3(x \plus{} y \plus{} z)}{10(xy \plus{} yz \plus{} zx)}
\]
[/quote]
Sorry but it is not true, try $ a \equal{} b \equal{} 0,4$ and $ c \equal{} 1$.

Let $ x,y,z \ge 0$ then

$ \frac {x}{{4x^2 \plus{} y^2 \plus{} z^2 }} \plus{} \frac {y}{{x^2 \plus{} 4y^2 \plus{} z^2 }} \plus{} \frac {z}{{x^2 \plus{} y^2 \plus{} 4z^2 }} \le \frac {{x \plus{} y \plus{} z}}{{2\left( {xy \plus{} yz \plus{} zx} \right)}}$

By AM-GM we have

$ \frac {x}{{4x^2 \plus{} y^2 \plus{} z^2 }} \le \frac {x}{{2x^2 \plus{} 2xy \plus{} 2zx}}$

$ \Rightarrow \sum {\frac {x}{{4x^2 \plus{} y^2 \plus{} z^2 }}} \le \sum {\frac {x}{{2x^2 \plus{} 2xy \plus{} 2zx}}} \equal{} \frac {3}{{2\left( {x \plus{} y \plus{} z} \right)}}$

It suffices us to show that

$ \frac {3}{{2\left( {x \plus{} y \plus{} z} \right)}} \le \frac {{x \plus{} y \plus{} z}}{{2\left( {xy \plus{} yz \plus{} zx} \right)}}$

which is obviously :).

no i think that true for $ a\equal{}b\equal{}0.4$ and $ c\equal{}1$
by caushy shwartz :
$ (4x^2\plus{}y^2\plus{}z^2)(\frac{1}{4}\plus{}1\plus{}1)\leq(x\plus{}y\plus{}z)^2$
$ \iff$          $ \frac{1}{4x^2\plus{}y^2\plus{}z^2}\leq\frac{4}{9(xy\plus{}yz\plus{}zx)}$
we have to prove now :
$ \frac{4(x\plus{}y\plus{}z)}{9(x\plus{}y\plus{}z)^2}\leq\frac{3(x\plus{}y\plus{}z)}{10(xy\plus{}yz\plus{}zx)}$

$ \iff$   $ 27(x\plus{}y\plus{}z)^2\geq40(xy\plus{}yz\plus{}zx)$
$ \iff$  $ 27(x^2\plus{}y^2\plus{}z^2)\plus{}14(xy\plus{}yz\plus{}zx)\geq0$

we are done  :wink:

[quote][b]mehdi cherif wrote:[/b]
$ (4x^2 \plus{} y^2 \plus{} z^2)(\frac {1}{4} \plus{} 1 \plus{} 1)\leq(x \plus{} y \plus{} z)^2$
$ \iff$      [/quote]it is >= and not =<  :wink: .

My inequality isn't flase but not nice, I will post solution after three days!

[quote="Sudoku"][quote="Honey_S"]
$ \sum {\frac {x}{{2x^2 \plus{} 2xy \plus{} 2zx}}} \equal{} \frac {3}{{2\left( {x \plus{} y \plus{} z} \right)}}$[/quote]
You are wrong.[/quote]
I think it's true. :wink:


Find the maximum value of  $ b \minus{} \frac{1}{a^2 \plus{} b^2}$  when $ a,b > 0$ verify$ a^3 \plus{} b^5 \leq a^2 \plus{} b^2.$

I must be misreading this but

$ b \minus{} \frac {1}{a^2\plus{}b^2}$ will not have a maximum, because as b gets large, $ \frac{1}{a^2\plus{}b^2}$ will get smaller.

The inequality is false too. The only way that that would make sense is if $ a,b < 0$, then the question would make sense

[quote="SimonM"]I must be misreading this but

$ b \minus{} \frac {1}{a^2 \plus{} b^2}$ will not have a maximum, because as b gets large, $ \frac {1}{a^2 \plus{} b^2}$ will get smaller.

The inequality is false too. The only way that that would make sense is if $ a,b < 0$, then the question would make sense[/quote]
Maybe i'm wrong too,but i think that means:
$ \frac{b\minus{}1}{a^2\plus{}b^2}$,not $ b\minus{}\frac{1}{a^2\plus{}b^2}$

Still doesn't make any difference to the problem in terms of an answer


Two angles of a triangle measure $ 55^{\circ}$ and $ 47^{\circ}$. What is the number of degrees in the measure of the supplement of the remaining angle of the triangle?

The supplement of the remaining angle is equal to the sum of the 2 other angles, so it's $ 55\plus{}47 \equal{} \boxed{102}$.

We first note that the supplement of the remaining angle is equal to the sum of the other two angles (the supplement is 180-$ n$, where n is the degree measure of the other angle, and the sum of the angles of any triangle is $ 180^{\circ}$). So, $ 55^{\circ}\plus{}47^{\circ}\equal{}\boxed{102^{\circ}}$.


calculus
integration
inequalities
inequalities proposed

$ \bigstar$ Let $ a,b,c\ge 0$. Prove that:
\[ \frac {4}{3a} \plus{} \frac {4}{3b} \plus{} \frac {4}{3c} \plus{} \frac {3}{2a \plus{} b} \plus{} \frac {3}{2b \plus{} c} \plus{} \frac {3}{2c \plus{} a}\ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {6}{c\plus{}2a}\plus{}\frac {6}{a \plus{} 2b} \plus{} \frac {6}{b \plus{} 2c}
\]
For Argady  :wink:

[quote="zaizai-hoang"]$ \bigstar$ Let $ a,b,c\ge 0$. Prove that:
\[ \frac {4}{3a} \plus{} \frac {4}{3b} \plus{} \frac {4}{3c} \plus{} \frac {3}{2a \plus{} b} \plus{} \frac {3}{2b \plus{} c} \plus{} \frac {3}{2c \plus{} a}\ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {6}{c \plus{} 2a} \plus{} \frac {6}{a \plus{} 2b} \plus{} \frac {6}{b \plus{} 2c}
\]
For Argady  :wink:[/quote]
Because $ \sum_{cyc}(4x^3\plus{}3x^2y\minus{}6x^2z\minus{}xyz)\geq0$ is true.

[quote="arqady"][quote="zaizai-hoang"]$ \bigstar$ Let $ a,b,c\ge 0$. Prove that:
\[ \frac {4}{3a} \plus{} \frac {4}{3b} \plus{} \frac {4}{3c} \plus{} \frac {3}{2a \plus{} b} \plus{} \frac {3}{2b \plus{} c} \plus{} \frac {3}{2c \plus{} a}\ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {6}{c \plus{} 2a} \plus{} \frac {6}{a \plus{} 2b} \plus{} \frac {6}{b \plus{} 2c}
\]
For Argady  :wink:[/quote]
Because $ \sum_{cyc}(4x^3 \plus{} 3x^2y \minus{} 6x^2z \minus{} xyz)\geq0$ is true.[/quote]
Can you explain me? mr.arqady?

[quote="Sunjee"]
Can you explain me? mr.arqady?[/quote]
Of cause!
The harasi's integral method:
Let $ t > 0$ and $ x \equal{} t^a,$ $ y \equal{} t^b$ and $ z \equal{} x^c.$
Hence, $ \sum_{cyc}(4t^{3a} \plus{} 3t^{2a \plus{} b} \minus{} 6t^{2a \plus{} c} \minus{} t^{a \plus{} b \plus{} c})\geq0,$ which gives the following inequality:
$ \int_0^1\left(\sum_{cyc}(4t^{3a \minus{} 1} \plus{} 3t^{2a \plus{} b \minus{} 1} \minus{} 6t^{2a \plus{} c \minus{} 1} \minus{} t^{a \plus{} b \plus{} c \minus{} 1})\right)dt\geq0,$
which gives the original inequality.


linear algebra
matrix
complex numbers
linear algebra unsolved

$A,B$ -diagonalizable matrices from $M_n(C)$, such that $e^A=e^B$ prove that $A=B$

The result is false. Let

\[A_n=\left[\begin{array}{cc}0&-2n\pi \\ 2n\pi&0\end{array}\right], n\in\mathbb{Z}\]
Then each $A_n$ is diagonalizable over the complex numbers, but $e^{A_n}=I$ for each $n$.

[quote="Moubinool"]Is there a result like if $A,B$ diagonalisable in $M_n(C)$ and $f$ injective
analytic $f(A)=f(B)$ then $A=B$ ?[/quote]

Dear Kent do you heard some result like this ?
thx

2Moubinool: 1). It is some kinda training/select contest. 2). I don't know anything about you generalization-maybe you know something about truth or non-trus of this fact-in the contest was proposed problem,posted by me
2grobber: Maybe my question stupid:as far as i understand you proved that each eigenvalue of A is equal to some eigenvalue of B-how it will do if for example every
$a_i=b_1$ -in that case it's not neccesary to be equal A and B --- it seems to mke that it's stupid, i think that i don't understand you solution

I had based my.. "solution" on the following false fact: $diag(e^{a_i})=S\cdot diag(e^{b_i})\cdot S^{-1}\Rightarrow a_i=b_i,\ \forall i$, but we can simply take $a_i$ as we please ($a_1\ne a_2$), and take $S$ to be the matrix having $1$ on $(1,2),(2,1), (i,i),\ \forall i\ge 3$ and $0$ everywhere else.

To eugene: that was no solution: I was wrong, and Kent has given a counterexample.

No, Moubinool, I don't know any result like that. The best I can say is that I haven't yet spotted any obvious reason for it to be false (now that you've included the word "injective.")

Does what I wrote up there count for a counterexample to Moubinool's question?

ok, sorry everyone for posting wrong problem
thanks to Kent for counterexample

If $A,B \in M_n(R)$  diagonalisable such that $e^A=e^B$
then $A=B$

This one was an Oral Examination Ecole Polytechnique

RMS 4 (Revue de Mathmatiques Spciales)
Anne 2002-2003
page 307

[quote]If $ A,B \in M_n(R) $ diagonalisable such that $ e^A=e^B $
then $ A=B $ [/quote]

For this to have any chance of being true, we must assume that by "diagonalisable" you mean that $\exists P$ such that $P^{-1}AP=D$ where $P$ and $D$ have only real entries and $D$ is diagonal. That rules out my previous counterexample which has imaginary eigenvalues.

I'd like to claim that this is true. I am missing one step: I need that $e^A=e^B$ implies that $e^Ae^{-B}=e^{A-B}.$ (Which would follow from $AB=BA.$) If I had that, I'd be done.

I think the bad solution I had given for the first one works for this one. Assume WLOG that $A=diag (a_i)$ and $B=S\cdot diag(b_i)\cdot S^{-1}$ where $S=(s_{ij})$ is invertible. The condition $e^A=e^B$ translates to $diag(e^{a_i})=S\cdot diag(e^{b_i})\cdot S^{-1}\iff diag(e^{a_i})\cdot S=S\cdot diag(e^{b_i})$, and this is equivalent to $(e^{a_i}-e^{b_j})s_{ij}=0,\ \forall i,j$, and from here and the injectivity of $e^x$ on $\mathbb R$ we easily get $(a_i-b_j)s_{ij}=0,\ \forall i,j$, which is equivalent to $A=diag(a_i)=S\cdot diag(b_i)\cdot S^{-1}=B$.


Putnam geometry probability conics parabola symmetry integration

A dart, thrown at random, hits a square target.  Assuming that any two parts of the target of equal area are equall likely to be hit, find the probability that hte point hit is nearer to the center than any edge.

I've never liked this question very much -- it's just kind of ugly.

The basic approach is to temporarily ignore the fact that you have a square and just ask about the point being closer to the center than to one side.  Then you realize that you have the classical description of a parabola, and coordinatize and integrate and add or subtract a few triangles as necessary.

[quote="justdudxit"]Ohhh... the center is a focus, and the sides are directrixes (directrici?).  Makes sense.[/quote]

Right, exactly.  Up until that point, the idea is very nice (and who'd have thought you'd ever again use the fact that a parabola is the locus of points equidistant from a fixed line and point?).  But the actual calculation after that point is not so pretty.

In fact, using your brains you can really avoid the calculation. ;) Take a square with sidelength $2$ for easiest calculations. Draw the diagonals and consider the lower part of the four. Obviously the parabola here is $y=\frac12(x^{2}-1)$ (filling in 2 well-chosen points gives you the unique one -- you don't even need a pencil for that).

To find the requested probability, it's sufficient (by symmetry) to calculate the area above the parabola in the right half of the lower part: \[\text{probability}= 2 \int^{\sqrt{2}-1}_{0}-x-\frac12(x^{2}-1) dx =\frac{4\sqrt{2}-5}{3}.\]


induction
inequalities
algebra
polynomial
number theory unsolved
number theory

Find the greatest real number $ \alpha$ for which there exists a sequence of infinitive integers $ (a_n)$, ($ n \equal{} 1, 2, 3, \ldots$) satisfying the following conditions:
1) $ a_n > 1997n$ for every $ n \in\mathbb{N}^{*}$;
2) For every $ n\ge 2$, $ U_n\ge a^{\alpha}_n$, where $ U_n \equal{} \gcd\{a_i \plus{} a_k | i \plus{} k \equal{} n\}$.

Is $ \alpha\equal{}\frac{1}{2}$ correct answer?I've proved that $ \alpha\leq\frac{1}{2}$ but I don't know whether this value can be achieved... :blush:
I tried to construct an example,but I failed...

I think $ \alpha\equal{}1$ is a solution.  Take $ a_n\equal{}1998n$.  Then if $ i\plus{}j\equal{}n$ then $ a_i\plus{}a_j\equal{}1998(i\plus{}j)\equal{}1998n$.  So $ U_n\equal{}1998n\equal{}a_n$.

[quote="cosinator"]I think $ \alpha \equal{} 1$ is a solution.  Take $ a_n \equal{} 1998n$.  Then if $ i \plus{} j \equal{} n$ then $ a_i \plus{} a_j \equal{} 1998(i \plus{} j) \equal{} 1998n$.  So $ U_n \equal{} 1998n \equal{} a_n$.[/quote]
There is a serious mistake in the statement of the problem above,there should be $ a_n> 1997^{n}$ instead of $ a_n>1997\cdot n$.

Note that $\alpha=0$ works for any sequence with $a_n>1997^n$ for all $n$, so we can assume $\alpha>0$.

If $\alpha\ge1$ then $a_n\le a_1+a_{n-1}$ shows that $a_n=O(n)$, contradicting the fact that $a_n>1997^n$ for all $n$, so $\alpha<1$. Now $a_{2n}\le a_n+a_n$ shows that $a_{2^k}\le 2^{\alpha^{-1}}a_{2^{k-1}}^{\alpha^{-1}}$, so by induction
\[a_{2^k}\le2^{\alpha^{-1}+\cdots+\alpha^{-k}}a_1^{\alpha^{-k}}\implies 1997^{(2\alpha)^k}\le 2^{1+\alpha+\cdots+\alpha^{k-1}}a_1<2^{1/(1-\alpha)}a_1.\]Thus $2\alpha\le1$.

To find a construction for $\alpha=1/2$, we use "sum-to-product" identities (motivated by Chebyshev polynomials, Fibonacci numbers, etc.): set $\alpha=N+\sqrt{N^2-1}$ (so $\alpha^{-1}=N-\sqrt{N^2-1}$) for some large integer $N$ and define $b_n=\alpha^n+\alpha^{-n}$ for all $n\ge0$ (clearly these are all positive integers); letting $a_n=\alpha^{2n}+\alpha^{-2n}$ for each $n\ge1$, we have
\[a_i+a_j=\alpha^{2i}+\alpha^{2j}+\alpha^{-2i}+\alpha^{-2j}=(\alpha^{i+j}+\alpha^{-i-j})(\alpha^{i-j}+\alpha^{j-i})=b_{i+j}b_{|i-j|}.\]For $N$ sufficiently large, clearly $a_n>\alpha^{2n}>1997^n$ for each $n$, and
\begin{align*}
\gcd\{a_i+a_j\mid i+j=n\} \ge b_n &= \alpha^n+\alpha^{-n} \\
&= \sqrt{\alpha^{2n}+\alpha^{-2n}+2} > \sqrt{\alpha^{2n}+\alpha^{-2n}} = a_n^{1/2},
\end{align*}as desired.

[b]Edit:[/b] Cleared up some things in the first part.

There is you only took the case n>=1 and find a contradiction. It does not mean that n should be smaller than 1/2. Can you explain it?


how are team scores calculated? i am ignorant yes.. is it like top 3 of each school

yeah top 3 of your school on the same test (i.e. can't mix A and B)

[quote="nat mc"]Just wondering whast everyone got

We got 129+130.5+144=403.5[/quote]

who got the 144???

o.O Phil, Carmel almost beat you guys. If I hadn't screwed up #1, we would have.

135+135+132=402. (I think since I haven't really checked, if we didn't get that, then someone probably made a careless mistake).


geometry
power of a point
radical axis
cyclic quadrilateral
geometry open

I found very interesting Miquel theorems. Most interesting for me was this one: http://www.gogeometry.com/miquel_pentagram1.htm
Inspired of that theorem using geogebra I found the configuration on the picture.

To be easier for you - take a look at the file attached.
It is given a pentagram (five rays star). The vertices of the pentagram lies on a circle. There are constructed five smaller circles.
Prove that the five lines, each formed by inersecting point of two small circles and opposite star vertex, intersects at a common point.

My questions are:
1. Is it a famous theorem or a problem from math competition?
2. How to prove that statement?
3. What is its level of difficulty?
4. Is it a beatiful problem?

I'm sorry for the quality of the picture if the problem is not clear I can restate it in a better way.

It is a nice problem. I don't know if this will work, but

let $ A,B,....E$ be the points of the large circle, and $ A_1,B_1..E_1$ be the opposite points on the smaller circle. By radical axis, it would suffice to show that $ ABA_1B_1$ is cyclic. (If the concurrency is true, then this is clearly true by power of a point).

[b]Solution: [/b]
First we will show that $ EB_1A_2B$ is cyclic. $ (1)$
$ \Leftrightarrow \angle EBA_2 \plus{} \angle EB_1A_2 \equal{} 180^o$
$ \Leftrightarrow \angle EBA_2 \plus{} \angle B_2C_2B \plus{} \angle B_2DA_2 \equal{} 180^o$ (right)
Therefore $ (1)$ is true. Similarly we get $ EB_1A_2E_2B, AD_1E_2A_1D$,... are cyclic.
We will show that $ \angle A_1B_1B \equal{} \angle A_1AB$
$ \Leftrightarrow \angle DB_1B \minus{} \angle DB_1A_1 \equal{} \angle DAB \minus{} \angle DAA_1$
$ \Leftrightarrow \angle DC_2B \minus{} \angle DA_2A_1 \equal{} \angle DAB \minus{} \angle DE_2A_1$
$ \Leftrightarrow \angle DAB \plus{} \angle EBA \minus{} \angle DE_2A_1 \minus{} \angle A_2A_1E_2 \equal{} \angle DAB \minus{} \angle DE_2A_1$
$ \Leftrightarrow \angle EBA \equal{} \angle A_2A_1E_2 \equal{} \angle A_2CE_2.$
The last is right because $ ABCE$ is a cyclic quadrilateral.
So $ ABA_1B_1$ is a cyclic quadrilateral.
Similarly we can show that $ AEA_1E_1$ is a cyclic quadrilateral. On the other side, $ EB_1E_1B$ is also a cyclic quadrilateral then by the theorem about radical center, we obtain $ AA_1, BB_1, CC_1$ are concurrent.
Similarly our problem is solved.

[b]Remark:[/b]
Denote $ X,Y,Z$ be the second intersections of $ (B_2A_2D), (EB_2C_2), (CE_2A_2)$ and $ (O)$. Show that $ DX, YZ, AB$ are concurrent.

HAPPY LUNAR NEW YEAR!  :)

Very very good idea and solution. Just to remark there is small technical issue. From the quadrilaterals cyclic mentioned it follows that AA1, BB1 and EE1 meet at common point - but this does not change the result, we can take suitable other cyclic quadrilaterals and show by analogy that AA1, BB1, CC1; AA1, BB1,DD1 meet also at common point and we are done.

I would like to thank to livetolove212 for the excellent solution. I think he is a great mathematician! There must me more people like him.
If the problem is new you can call
"Pentagram open problem 1" and
"Pentagram open problem 2" with the names:
Mirchev-Kazakov and Mirchev-Lynh theorems by your choice.
I would like to mention - both problem can be solved using complex numbers.
georgi111 is very busy person but some day he can post solutions using complex numbers.
If you want I can publish more open problems. Some of them are related to the pentagrams and pentacles.
For all mathlinkers - you can feel free to give an opinion for my last unsolved open problem:
http://www.mathlinks.ro/viewtopic.php?t=313384

HAPPY LUNAR NEW YEAR!  :wink:


probability videos AoPSwiki articles Alcumus

PIE counting is when you overcount something then you subtract those .

Example:

How many numbers less than $ 50$ are multiples of $ 2$ and $ 3$ but not $ 4$?

Solution: We count $ 6, 12, 18,..., 48$ or $ \frac{50}{6} \est 8$. But we also counted some multiples of $ 4$. So we subtract the multiples of $ 12$, which are $ 4$. So the answer is $ 8 - 4 = \boxed{4}$.

I know it as the Inclusion-Exclusion Principle so I guess PIE stands for the Principle of Inclusion and Exclusion.

[quote="KingJamesMvP"]I have seen PIE in many solutions and i cant understand excactly what it is. Could someone help me. Thx[/quote]

PIE is Principle of Inclusion/Exclusion. It is in Intermediate Counting/Probability but don't just memorize the formula because a lot of upper level contest problems require:

1) A [b]thorough[/b] understanding of PIE, and
2) A good [b]insight[/b] on how to use it!

Look [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]here[/url], [url=http://www.artofproblemsolving.com/Alcumus/Videos/Video.php?video_id=37&m=af6aa14999718745da45a47a8075b91b]here[/url] and [url=http://www.artofproblemsolving.com/Alcumus/Videos/Video.php?video_id=71&m=86ef46c49a5de3839873fbebe49af597]here[/url].  The first one is an AoPSWiki article, and the other two are Alcumus Videos.


[quote="2424"]With $ a,b,c > 0$  Prove that $ \sum_{cyc}\frac {\sqrt {a^2 \plus{} bc}}{(a \plus{} b)\sqrt {a \plus{} b}} \geq \frac {3\sqrt {3}}{2\sqrt {a \plus{} b \plus{} c}}$[/quote]
IT's very interesting! :P

Do you meant:
$ (a\plus{}b\plus{}c)^3(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab) \geq\ \frac{27}{64}(a\plus{}b)^3(b\plus{}c)^3(c\plus{}a)^3$
 :maybe:

@nguoivn:[hide]E nghe noi anh va anh Can co newbook sap ra.E dang to mo ve nhung ki thuat moi su dung AM-GM cua anh.
Khoang bao jo thi 2 anh cho xuat ban quyen do ak?[/hide]

[quote="nguoivn"][quote="2424"][quote="nguoivn"]Do you meant:
$ (a \plus{} b \plus{} c)^3(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \geq\ \frac {27}{64}(a \plus{} b)^3(b \plus{} c)^3(c \plus{} a)^3$
 :maybe:[/quote]
yes,It's true and interesting! :P
I have nice sulotion for it!
 :P[/quote]
Can you share it for every body? Thank you  :)
Indeed, I don't see any proof without pqr  :oops:
@2424: [hide]Khoang? dau` thang' 9 cuon "Bat dang thuc va n~ loi giai dep" cua tui anh se~ ra mat' ban. doc. em a. Tui. anh vua` moi ki' hop dong` chinh' thuc' hom qua ^^![/hide][/quote]
@2424:[hide]Bai nay e dung mot bai cua e da post len dien dan nhung chua ai jai,bdt nay chi la truong hop dac biet cua bdt do voi k=2 thui a ak.E muon xem a co cach jai bat ngo nao bang AM-GM cho no k thui a ak.:).E se post loi jai cua minh sau vai hum nua :).E ngoi wan nen ngai post loi giai lem.anh ah,khi nao quyen do ra thi anh bao vs e 1 tieng dc k ak?E rat to mo ve AM-GM cua anh tn do :roll: Thanks anh Quoc Anh nhiu nha! :) [/hide]


limit
logarithms
real analysis
real analysis unsolved

evaluate the limit ${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\left(\frac{1}{n i}+\frac{1}{n}+\frac{1}{i}\right)^{i+n}\right)}$

[hide="Hint"]
Remember Euler' Limit
[/hide]

Let $a_{i}= \left ( \frac{a}{ni}+\frac{1}{n}+\frac{1}{i}\right )^{i+n}$

$\boxed{\text{Case}\ i=1}$
$\lim_{n \to \infty}a_{1}= \lim_{n \to \infty}\left ( \frac{a}{n}+\frac{1}{n}+1 \right )^{n+1}= \lim_{n \to \infty}\left ( \frac{a+1}{n}+1 \right )^{n+1}= e^{a+1}$

$\boxed{\text{Case}\ i\ge2}$
$\lim_{n \to \infty}a_{i}= \lim_{n \to \infty}\left ( \frac{a}{ni}+\frac{1}{n}+\frac{1}{i}\right )^{n+i}= \lim_{n \to \infty}\left ( \frac{a+i+n}{ni}\right )^{n+i}= \lim_{n \to \infty}\left ( \frac{\frac{a}{n}+\frac{i}{n}+\frac{n}{n}}{\frac{ni}{n}}\right ) = \lim_{n \to \infty}\left ( \frac{\frac{a}{n}+\frac{i}{n}+1}{i}\right )^{n+i}= 0$

$\therefore \ \lim_{n \to \infty}\left ( \sum_{i=1}^{n}\left ( \frac{a}{ni}+\frac{1}{n}+\frac{1}{i}\right )^{i+n}\right ) = \lim_{n \to \infty}\left ( a_{1}+\sum_{i=2}^{n}a_{i}\right ) =e^{a+1}$

This Problem is case of $a=1$.
Thus, $\boxed{\boxed{\lim_{n \to \infty}\left ( \sum_{i=1}^{n}\left ( \frac{1}{ni}+\frac{1}{n}+\frac{1}{i}\right )^{i+n}\right ) = e^{2}}}$

correctly :)

just a small remark. you forgot $n+1$ in this line.

$\lim_{n \to \infty}\left ( \frac{\frac{a}{n}+\frac{i}{n}+\frac{n}{n}}{\frac{ni}{n}}\right ) = \lim_{n \to \infty}\left ( \frac{\frac{a}{n}+\frac{i}{n}+1}{i}\right )^{n+i}= 0$

but the solution is correct, so this does not matter

You're right, I made a typing error:(
Thanks for the input.

I don't think that you can pass to the limit inside the sum
you have that $ a_{i}\to 0,i\ge 2$ but you have infinitely many $ a_{i}$s when you take the limit as $ n\to\infty$ and you get smth like $ \infty\cdot 0$.

Kouichi Nakagawa is rigth, we can do this yes.

see

$ {i\geq 2}$
$ {\lim_{n\to \infty }\, n \left(\frac{\frac{i}{n}+\frac{1}{n}+1}{i}\right)^{i+n}=0}$

your indertemination does not exist, because the limit is zero.

You can not take the limit into the sum and after that sum up infinitely (because the sum is from 2 to $ n\to\infty$ ) many zeros and say that it is zero.
What if, for example $ a_{i,n}=\frac1{i\log n}$ for $ i=1,2,...\ldots,n$
then $ \lim_{n\to\infty}a_{i,n}=0,\forall i\ge 1$
but $ \lim_{n\to\infty}\sum_{i=1}^{n}a_{i,n}=1$

see my previos post.

that limit is greater than the sum.

i didnt want to post it. because my last limit explain everything..

$ {i\leq n\land i\geq 2}$
$ {\left(\frac{1}{2}+\frac{3}{2 n}\right)^{n+2}\geq \left(\frac{1}{n i}+\frac{1}{n}+\frac{1}{i}\right)^{i+n}\geq \left(\frac{1}{n^{2}}+\frac{2}{n}\right)^{2 n}}$

$ {n\left(\frac{1}{2}+\frac{3}{2 n}\right)^{2+n}\geq \sum_{i=2}^{n}\left(\frac{1}{n}+\frac{1}{n i}+\frac{1}{i}\right)^{n+i}\geq n\left(\frac{1}{n^{2}}+\frac{2}{n}\right)^{2 n}}$

$ {\lim_{n\to \infty }\, n \left(\frac{1}{2}+\frac{3}{2 n}\right)^{n+2}=0}$
$ {\lim_{n\to \infty }\, n \left(\frac{1}{n^{2}}+\frac{2}{n}\right)^{2 n}=0}$

So, just the first term matters.

Q.E.D


In the sequence $ (a_n)$ with general term $ a_n \equal{} n^3 \minus{} (2n \plus{} 1)^2$, does there exist a term that is divisible by 2006?

$ n\equal{}361$ Is a possible solution.
In fact there are many more, because of the chinese reminder theorem.
If you want to get all the solutions work $ mod 2, 17, 59$ and you will get all solutions. :)

Daniel


This is a problem from the All-Time Greatest MathCounts Problems:
In a 4x4 chessboard, what is the maximum number of squares that a knight can visit, starting from the top left corner without going to any square more than once?

This problem, in my opinion leaves a lot to luck.  I still can't understand the book's solution, so could anyone explain clearly how to get the answer?  Is there some [b]MathCounts-level[/b] algorithm to do this?  I know that this is a really common question in mathematics, so I really need help.  Thanx! :D

Hmmmm...I remember our mathCounts teacher told us to figure out a pattern that would get the most squares.

I think this is luck. I mean, after a few tries i would just put down the largest number i got.

[hide]
hm, is it 13?

i dont think its pure luck, it has a bit of logic in it.[/hide]

the answer's 15. i remember this problem very well.

Yup I just got 15.

Yeah i dont think theres a straightforward method to this problem... does anyone have a logical approach?

Ha, the first two topics are "Knights tour" and "Knights not getting much of a tour" :lol:

anyways, if the grid is
1  2   3  4
5  6   7  8
9 10  11 12
13 14 15 16
it has to start 1,10,16,7, then it's random luck time.
it turns out to be 1,10,16,7,9,15,8,2,11,13,6,12,3,5,14.
I remember putting 14 for this :oops:

[quote="emerson woerner"][quote="mihail911"]i think my reasoning is very off, but isn't it kinda obvious the answer is 15. i mean, if the knight starts at square 1 in the top left corner, he can just keep going all the way across the first row, then down and all the way across the second row, till he gets to the last square at the bottom left corner, for a total of 15 squares. i thought it would be (total number of squares)-1.
im probably wrong.[/quote]
First a knight doesnt move that way...(it doesnt move horizontally or vertically by 1 step)_

Second, if it did move the way you are tlaking about, then it would be 16 squares, because the first square he is on he visits too

So yes you are totally wrong..the only reason why you think is 15 because indianamath already posted that answer, and you are trying to find a solution thjat complements the answer..[/quote]

I am such a moron!
when the question asked for knight, i didn't know it meant the actual chess figure..then yes, ur right!. srry, for the stupid mistake!  :wallbash_red:

I got the right answer on the the first try.

Playing chess has its rewards...


modular arithmetic
real analysis
real analysis unsolved

Prove that the sequence $ |sin n|+|sin n^3|$ has no limit.

Upps, sorry I do have a solution for this one, so it would be better to move it to proposed section.

It follows rather easily from the fact that $\{n\},\{n^3\}$ are uniformly distributed in $(0,\pi)$, where $\{\}$ is the "fractional part" $\pmod \pi$. Did you use weaker results to prove it?


I was working through some inequality sheets and came across the term $ inf [tx^{2}\plus{}(y^{2})/t]$. What does $ inf$ mean please? $ max$ which i know stands for maximum was used in a similar situation if that helps clarify things further.

$ inf\equal{}infimum$, it means that if $ a$ is $ inf$ of $ S$ and $ a'>a$ then there is an element of $ S$, may be $ x_{0}$, such that $ x_{0}<a'$


group theory
abstract algebra
graph theory
topology
superior algebra
superior algebra theorems

Can somebody tell me why the subgroup of a free group is free as well?

A very elegant proof of this theorem is using algebraic topology. You can find a proof of it in most textbooks, (you can get Hatcher here for free http://www.math.cornell.edu/~hatcher/AT/ATpage.html  its on page 94). The basic idea is to use graph theory and show that its fundamental group is free. Because there is a Galois connection between the fundamental group and covering spaces (and the fact that covering spaces of a graph are graphs), we can deduce that subgroups are also free.

Tanks !
It must be dumb question but it seems strange to me thazt proves are so uneasy:
(dumb question: on) why couldn't we just say that being free means that there are no relations between elements and so if there is no reltions in a group there can't be any in a subgroup either ?(dumb question: off)

It's not that easy as there are dependences (one can directly "check" them, but still, there are some).
You still need to show that you can find independent generators.
Try to think of free _abelian_ groups, there it isn't completely trivial to find independent generators, too.


Can someone provide a solution for this question because when i looked into the solution it was in my opinion not sufficient for hlping understand how to solve the actual problem.

:Three friends, Ralph, Emerson, and Waldo, each select a number from the set {1,2,3,11,12,13,21,22,23} and remove it. Then they add their three numbers together. If they put their numbers back a repeat this process for all possible combinations, how many different sums will they get?

ok the answer that they gave would be helpfull

[hide]
Answer: 30

Solution:

The way i would aproach this problem is to find the number of possible combinations that they could draw 3 of the numbers from the set and then subtract from that the number of ways that the sums would be the same.

The total number of combinations, not permutations because order does not matter is just $ \frac{9\times8\times7}{3\times2\times1}$ therefore there are 84 possible sums of the numbers.

Now to find the ways that they could be the same.

you could either write them all out or just use logic which i dont have cause i just wrote them all out!!!
This really might take a while so i wont write all of the overlaps.  just trust me that there are 54 overlaps.  I see now that it would have been easier to just list the possible sums which i now find to be 6, 14,15,23,24,25,26,16,17,27,33,34,35,36,44,45,46,18,28,37,3847,39,48,54,55,56,57,58,66

so there are 30 different sums that he could form with that set.[/hide]

Please tell me there is an easy way to do this, but i dont know what it is!!!if someone could tell me i would be so gratefull!!!

maybe it would be possible to find cases for the ten's column such as if the tens column was 2, then the ones digit could be.....
and im pretty sure you would only have to do cases from 0 to 3 since 4-6 would be symmetrical to 2-0 respectively.

Ok so if the tens digit were the #'s before the ----, then the ones digit could be the #'s after the ----
0---6
1---4,5,6,7,8
2---4,5,6,7,8
3---3,4,5,6,7,8,9
4---4,5,6,7,8
5---4,5,6,7,8
6---6

that totals to 29 possibilities

Edit: all references to zolojetto were deleted.

Find the impossible sums between $ 6$ and $ 66$.  We obviously know that 6 is the least possible sum and that the next possible sum will be 11+1+2=14.

Continue like this.

[quote="stevenmeow"]please check you work zolojetto thanks![/quote]

and what makes you think that your so great.  I made a careless addition mistake, sorry for any confusion.

Because. He's bobby shen... And if someone misses something, he'll assume they did something stupid.  :wink:

ok for clarification issues the answer is [b]29[/b]

[quote]maybe it would be possible to find cases for the ten's column such as if the tens column was 2, then the ones digit could be.....
and im pretty sure you would only have to do cases from 0 to 3 since 4-6 would be symmetrical to 2-0 respectively. maybe it would be possible to find cases for the ten's column such as if the tens column was 2, then the ones digit could be.....

[/quote]'
I kind of understand your solution but how would u know "and im pretty sure you would only have to do cases from 0 to 3 since 4-6 would be symmetrical to 2-0 respectively" that they are symmetrical?

The set $ S\equal{}{1,2,3,11,12,13,21,22,23}$ has a symmetric look to it =).  I mean seriously, all the units digits are $ 1,2,3$, Simply increase by $ 10$ to find the next subset of $ 3$ elements.


Three sox are drawn at random(without replacement) from a drawer cotaining equal numbers of   white red and black sox.If the number of sox is large and p is the probability that at least two of the sox drawn are the same colour prove that p is approximately $\frac{7}{9}$.

[hide]
$p=1-q$, where $q$ is the probability that all socks drawn are the same colour.

The number of different possibilities of drawing one of each colour is $3!=6$ (WRB, WBR, RBW, RWB, BRW, BWR).

The probability of one if these occurring is approximately $(1/3)(1/3)(1/3)=1/27$ (as the  number of socks is large, we can ignore the fact that they are not replaced, and still get a fairly accurate answer).
So, $q=(1/27)(6)=2/9$.
$p=1-2/9=7/9$.
[/hide]


So, the AIME is over.  I was just curious to know what kinds of scores the top students at these schools received (there is a lengthy discussion about magnet schools and regular high schools in the College Forum).  This might be useful in speculating about the AIME cutoff (i.e. lots of 15s might mean that good math students across the nation  felt the exam was easier, lots of low scores might mean the test was harder, etc.)  Anyway, if kids from these schools who read this forum can post what the higher scores from their schools are going to be, that would be pretty interesting (I'm sure we're all curious about that).

[quote="Idiot_without_a_village"]Rumor's that biggie smalls missed one.[/quote] Who is that?

Ricky Biggs.  At TJ.

http://activities.tjhsst.edu/vmt/ranking.phtml

haha
biggie smalls is also Notorious B.I.G., a dead rapper

I can tell you about stuy:

THe highest grade was I believe a 9, and only one person got that.
Next came one 8, or was it 2...
There were a couple of 7's.
Some 6's (like me)
and the rest is 5's, 4's, 3's, etc

I CAN tell you that the senior math team captains at Stuy didn't do too well. NOt well at all.

Do you know how Sho did?  How about Hooyung?

ouch.  stuy died this year.  hopefully tj didn't do too well, and yeah.  not sure about the scores in my area.  there's at least two 11/12 and quite a few 9/10...  yeah

AAST has 1 13 (Jae) , 1 9 (Hyun-Soo), and 2 8's (Myself, Fan).. then a handful of sevens and lower.

Hmm, no one from Exeter and TJ are on this board?


The results have been posted!

http://www.math.umd.edu/highschool/mathcomp/2007.shtml

Congratulations to all of the Winners!

Including you, of course.  :wink:

Anyway, I'm pretty disappointed.  I dropped from 3rd to 11th on Part II.  :(  At least I got Baltimore County winner.

So, congratulations to everyone!

Good job to everyone!
Everyone says congratulations to the winners. How about
*sympathy for the non-winners :| *

As for me,
96+110=206=#4
(very silly mistake on part 2 question 1 cost ~10 critical pts)


LaTeX
geometry
geometric transformation
reflection
similar triangles

Hi, i went to this math competition known as Fall Classics today and i got stuck in some problems.  :(  So can someone please tell me how to do these problems so i dont mess up next time.  :D

1. A game is played in which n players sit around a table and each possesses one penny. Play proceeds in a clockwise fashion. The first player passes one penny to the left, then the next player passes two pennies to the left, the next player passes one penny to the left, the next player passes two pennies to the left and so forth. Once a player possesses no more pennies he is removed from the game. If play continues indefinitely, what is the smallest number n for which one player will NOT end up with all the pennies?

a. 5              b. 6             c.7             d.8                  e.NOTA

2. Give that angle MON = 20degrees and A is a point on segment OM such that OA = 4 square root 3. If D is a point on segment ON, with OD = 8 square root 3, and B and C are arbitrary points on segment OD and AM respectively, find the minium distance |AB|+|BC|+|CD|.

a. 10            b.12            c. 14            d.16               e.NOTA

3. How many paths are there from the point (1,2,3) to the point (6,5,4) if each path must consist of a sequence of unit length movements in the positive direction parallel to any of the three axes, and no path can travel through the point (4,4,4)?

a.318             b.320             c.322             d.324               e.NOTA

4. Let x, y, z be positive real numbers satisfying the system:

x^2+xy+[(y^2)/3]=25
[(y^2)/3]+z^2=9
z^2+zx+x^2=16

Find xy+2yz+3zx .

a. 24 square root 3         b. 18 square root 3         c. 12 square root 3
d. 6 square root 3           e. NOTA

NOTA stand for non of the above.

PS: how do u type up things like square root of 9? is there a special program or something like that?

You can type mathematical symbols and equations using a program called Latex, I believe. I unfortunately do not have it, but you can look around this site.

1.[hide]We can note that the only way this game will end is if there are 2 people left after the first "go-around," because that way one person will always be passing the same amount, so the person passing 2 pennies at a time will eventually run out of pennies. Going back to the beginning of the game, we can test each choice to see how many people are left after the first go-around. If any of the choices will leave 3 people remaining, then that's the answer we want.[/hide]

2.[hide]Are you sure that angle is 20 degrees? It would make much more sense if it were 30 degrees. Anyways, we know that the shortest distance from a point to the line is when it's perpendicular. So try to draw a diagram with all 3 of your lines AB, BC, and CD perpindicular to either OM or ON. Then look for similar triangles[/hide]

3.[hide]We know that we have a total of 6-1 + 5-2 + 4-3 = 9 moves. Instead of drawing out each path and counting all of them, we realize that [i]the path we take is determined by when(out of the 9 steps) we move in the direction of the x/y/z axis[/i]. This is extremely useful, because these kinds of problems come up a lot. So basically we can use combinations to find out the possible times we move in the z, the possible times when we move in the x, and the possible times we move in the y. We would have 9 choose 5 for the possible combinations of Z axis movements. Out of the 4 remaining moves, we have 4 choose 3 for the y, and finally 1 c 1 for the x. BTW, it does not matter whether i do this in the order of z,y,x, or x,z,y, or whatever. So 9C5 * 4C3 * 1C1 will give us the total paths. But since we cannot travel through (4,4,4) we can subtract the number of paths from (1,2,3) to (4,4,4).[/hide]

I'm really sorry it's so wordy. Perhaps if you still don't understand someone else can drop in and help clarify for me.

[quote="iaintsosmart"]
PS: how do u type up things like square root of 9? is there a special program or something like that?[/quote]
$\LaTeX$
http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 :)

To add to mathmanan's its really very simple, if you are typing on this site, ... really, other than putting a  $\$$  around what you  want to type, rest is almost natural... try it for example ... below all I did was to put the $\$$ signs  around what you wrote: and
[quote]x^2+xy+[(y^2)/3]=25
[(y^2)/3]+z^2=9
z^2+zx+x^2=16
[/quote]
looks like:
[quote]$x^2+xy+[(y^2)/3]=25$
$[(y^2)/3]+z^2=9$
$z^2+zx+x^2=16$
[/quote]

More ,of course, is on the  link given above, but it takes very little time to start ...

[quote]Are you sure that angle is 20 degrees? It would make much more sense if it were 30 degrees. Anyways, we know that the shortest distance from a point to the line is when it's perpendicular. So try to draw a diagram with all 3 of your lines AB, BC, and CD perpindicular to either OM or ON. Then look for similar triangles

[/quote]

Actually making it $20^o$ makes the answer very simple!  Hint:
[hide]Think of a  $30^o- 60^o - 90^o$   right angle triangle  with sides: $4\sqrt 3, 8\sqrt 3, 12$ ... the answer would be $12$ ...

[hide]$OXY$ is the  triangle, $OX=4\sqrt 3$ , $OY=8\sqrt 3$  now trisect  $\angle XOY$ so that  $\angle XON = \angle NOM = \angle MOY = 20^o$
Think of reflections (AB=XB, CD=CY etc)  and remember for shotest distance  between point $X,Y$ (the sum)  would have to be equal to line $XY=12$ (When line XB+BC+CY is shortest - It has to be the straight line XY )  [/hide][/hide]

BTW there is a beautiful theorm/result  (though a little less known) ... some times given as a problem:

Given a triangle ABC find point  X(on BC), Y(on CA) and Z (on AB) such that sum of  XY+YZ+ZX is minimum.


What is your personal favorite math book?

Mine is: ACoPS!

Are we talking textbooks, biographies, problem books,... or everything?  I'd have to say my favorite textbook is Spivak's "Calculus."  It's just so well-written, complete, and has great problems.  For non-textbooks, I thought both "Fermat's Enigma" and "The Man Who Knew Infinity" were quite good.

[quote="abcak"]um...AoPS II[/quote]

yea, mine's the same

Even though I don't have either of them(may get vol. 1 soon) the classics are the best of all the textbooks because they thoroughly cover a variety of topics and have more problem solving strategies than the subject books.

[quote="xpmath"]Heh, Volume 1  :wink: .[/quote]
Me too
The message contains only quote and code tags. Please add some text of your

[i]Surreal Numbers[/i] by Donald E. Knuth

The only math book I have read is AoPS Volume 1, but I am only half way through. I have no other choice...

AoPS Volume 1

I like word problem skills k-12 by Anita Hardanek

i like mcdougal littell algebra i textbook :P

[i]Another Fine Math You've Got Me Into[/i] by Ian Stewart.


Euler
modular arithmetic
number theory unsolved
number theory

The great mathematician Euler was said to be very religious, but one night he came in a situation where he had to gamble with the devil. The devil opened a bag of beans and placed 23 cups in a circle, 22 white cups and a black one. In the latter he put one bean.

The game was as follows: the first player would put one bean in the cup next to the black one. The second player would tale two beans and put one in each of the next two cups. Then the first player would take four beans, and put one in each of the next four cups, and so forth. In general, every player would take as many beans as there are in all cups together; starting with the cup adjacent to the one last "served", he would put one bean in each of the following cups in the circle.
If it happens that a player puts his "last" bean in the black cup, he loses the game. The devil started.

[b]a)[/b] did the game terminate. if so, who won?

[b]b)[/b] the game took more than half an hour. The devil requested a second game, in which Euler should start, and - in order to shorten the game - the number of cups should be decreased. How many cups did Euler remove so that he would not only win the game, but - to irk the devil - the game would take about twice as long as the first one (ie: when the game ended there were twice as many beans used in total including the first bean put in the black cup as a preparation for the game) ?

[b]c)[/b] the devil was not very amused, and decided to changed the rules, so that a player would win, if he is in a position to put his first bean in the black cup. what happened ?

Thank you.

My thoughts (may be completely wrong)...
[hide="Introduction"]
The number of beans doubles with every play, so it's not hard to see that, in the nth play, it is equal to $ 2^{n}$. Therefore the ending condition can be rephrased as $ 2^{n} \equiv 1 \pmod{23}$
[/hide]

[hide="Part A"]
It's easy to see that it does terminate. If you make a table for $ 2^n \pmod{23}$, you'll see that $ 2^{11} \equiv 1 \pmod{23}$. Therefore the game ends on the 12th play, made by the first player. So our hero wins this one.
[/hide]

[hide="Part C"]
The ending condition is now $ 2^{n} \equiv 0 \pmod{23}$. However $ 2^n$ is never a multiple of 23. So this game doesn't terminate.
[/hide]


The grading on the 2008 Team Selection Test is finished and based on the combined scores from the USAMO and the TST the 2008 USA IMO team will be:

Paul Christiano
Shaunak Kishore
Evan O'Dorney
Colin Sandon
Krishanu Sankar
Alex Zhai

Congratulation to the team!

Steve Dunbar
MAA Director, American Mathematics Competitions

My friend knows one of those kids.

Good luck USA!

EDIT: I wonder how Evan O'Dorney will do

[quote="#H34N1"]Wow Evan O'dorney *beast*[/quote]
Indeed.

Go USA and congratulations to the team.

my predictions were pretty close

dude

maybe I should work instead of posting

yeah ok good idea

good job to you guys

well of course I suppose I could just tell you? oh well

Good luck everybody. Nice job Evan on making the team as a freshman.

Hahaha! WIN! Half the team is from SFBA!
PEW PEW PEW

Congrats to all! Evan O'Dorney... wow.  :)

lol, wow. It's unbelievable that Evan is beast at BOTH math and spelling.

We've known this for a while, ya know...

Ah. My bad. It's 1/3 SFBA. Still, that's pretty legit.

Good job team! What, nealth didn't make it again? how much does that stink? I feel bad for SamE, considering what an amazing year this has been for him.

I only know one Indian who's really not asian, and it's not you, #H34N1, sorry :P

Congrats to Evan.. I'm almost jealous :) maybe he'll cure cancer or something someday.

You guys are going *way* off-topic - I'll delete some posts.

Auh#..... I missed by ONE PROBLEM. :ewpu:
The race is CLOSE and...... Delong and I still have next year.

who is evan o dorney? and why does everyone have a spaz when they see his name. ive seen at least 1000 comments about him on this website

It's Eric Larson who didn't make it this year but made it last year.

Good luck to all members of USA IMO Team.

I'm glad I had a chance to meet one of the members, Shaunak Kishore. I do not know him personally and I doubt if he knows me but I was in same Lehigh Valley team for ARML and recalled him at Mathcounts States Round (obviously, he made to nats and I didn't though). So, quite impressive but somewhat expected.

I live really close to Alex.  He probably has no idea who I am though.  :P

HOW DOES EVAN O'DORNEY DO THAT?!!??

Sorry.
Wish I was insane like him.

[quote="#H34N1"]He made IMO as a 9th grader and won the NSB.[/quote]

yeah but people still had spazes before this thread was made

he also was the only eighth grader in blue mop last year.

interesting:

out of the 5 seniors, 4 are going to MIT.  yay yay yay.


[quote="puuhikki"]Let $ABCD$ be a convex quadrilateral drawn around a given circle. Rays $AB$ and $DC$ intersect at the point $E$ and rays $DA$ and $CB$ intersects at the point $F$. Let $X$, $Y$, and $Z$ be the centers of circumcircles of the triangles $AFB$, $BEC$, and $ABC$. The line $XZ$ intersects the lines $EA$ and $ED$ at the points $K$ and $L$ in this order and the line $YZ$ intersects the lines $FC$ and $FD$ at the points $M$ and $N$. Show that $EK=EL$ if and only if $FM=FN$.[/quote]

You probably have a typo in your problem because in this form you can easily see that $\angle{EKL}=90^{0}$, and,therefore,the condition $EK=EL$ is impossible :!:.


The odds in favour of a Pythag-Air-US airlines flight being on time are 3:1.
a) What is the probability that this airline's next eight flights will be on time?
b) What is the expected waiting time before a flight delay?

I'm horrible at probability. [hide="a."] I think the probability of a flight being on time is $\frac{3}{4}$. So, the answer is $(\frac{3}{4})^8=\frac{6561}{65536}=\boxed{0.1}$ :D [/hide]

That's correct for the first part.  I still don't get expected values though...

This is posted in the other 2 high school forums (which it shouldn't be). I think there may be solutions there.


abstract algebra
induction
linear algebra
linear algebra unsolved

Here's a well-known fact: if $ f,f_1,\ldots,f_n$ are linear functionals in normed space s.t. $ \bigcap\limits_{i \equal{} 1}^{n} ker(f_i)\subset ker(f)$, then $ f$ is a linear combination of $ f_i$. I remember there was a very short and nice argument which proves it, but i forgot it.
Any ideas?(please no standart solutions like induction etc).

if $ \phi: E \to \mathbb{R}^{n+1}$ is defined by
\[ \phi(x) = (f(x), f_1(x), \ldots, f_n(x))\]
then $ P=(1,0,\ldots, 0)$ is not in the closure of $ \phi(E)$. Hence one can find an hyperplane that separates $ P$ and $ \phi(E)$: there exist $ b, a_0, \ldots, a_n$ such that:
\[ a_0 f(x) + a_1 f_1(x) + \ldots + a_n f_n(x) \geq b\]
for all $ x \in E$. This gives the conclusion.


algebra
polynomial
trigonometry
AMC
AIME
AIME II

Depends on how easily evaluatable you want the formula to be.

$x^{2a} - s_a x^{a} + 1 = 0$
$x^{a} = \frac{s_a \pm \sqrt{s_a^2 - 4}}{2}$
$x^{b} = \left( x^{a} \right)^{ \frac{b}{a} }$
$x^{b} + \frac{1}{x^{b}} = \left( \frac{s_a \pm \sqrt{s_a^2 - 4}}{2} \right)^{ \frac{b}{a} } + \left( \frac{2}{s_a \pm \sqrt{s_a^2 - 4}} \right)^{ \frac{b}{a} }$

 :P

In practice, this is easier to do problem-by-problem.  A general formula wouldn't help much, in my opinion.

[quote="calculuslover800"][quote="nat mc"]If we know $x^a+\frac1{x^a}=s_a$ can someone find or derive a formula for $x^b+\frac1{x^b}$ in terms of $a$, $b$, and $s_a$?

What is it? Show your work.

I am just asking this due to some rising interest of mine from seeing some related problems recently.[/quote]

This is sort of a wild guess, but if the type of problems you're looking at involve polynomials of the form $ax^4 + bx^3 +cx^2 + bx + a = 0$, and then solving by dividing the equation by $x^2$ and looking at regrouping the polynomial like this:

$a(x^2 + \frac1{x^2}) + b(x + \frac1x) + c = 0$

and then using a substitution like $y = x + \frac1x$, $y^2 - 2 = x^2 + \frac1{x^2}$, then, I remember at one point I found that it isn't too hard to come up with formulas for expressing $x^n + \frac1{x^n}$ in terms of lower powers, and I was able to show that all polynomials of this form can be reduced in the above manner to a polynomial of half the degree.  I think amir's formula pretty much does that for you.

But idk if that will help you at all.[/quote]

You can show that $x^n + \frac{1}{x^n}$ can be written as a polynomial with integer coefficients in terms of $x + \frac{1}{x}$.

The type of problem he was probably looking at was exactly the problem that he gave, with specific values of $a, b, s_a$.

So $x$ is a positive real now?

[hide="proof"]
We have $x^2-2x\cos\alpha+1=0$ so \[ x = \cos\alpha\pm\sqrt{\cos^2\alpha-1}=\cos\alpha\pm i\sin\alpha. \] Thus $x^n=\cos n\alpha+i\sin\alpha$ and $x^{-n}=\cos n\alpha\mp i\sin n\alpha$ so \[ x^n+\frac{1}{x^n}=2\cos n\alpha. \]
[/hide]

See [url=http://www.kalva.demon.co.uk/aime/aime00b.html]AIME II 2000 #9[/url] for a problem that is trivialized by this result.

yes I was wondering about the positive reals

[quote="nat mc"][quote="chess64"]\[ \boxed {\text {If } x+\frac{1}{x}=2\cos \alpha \text{ then } x^n+\frac{1}{x^n}=2\cos (n\alpha)}.  \] I think that should help.[/quote]
Umm... I dont think this works. Because $2\cos\alpha\leq2$ and $x+\frac1x\geq2$[/quote]We can choose $\alpha$ to be a complex number to make this approach work.  Alternatively, we can use hyperbolic cosine:
\[ x + \frac{1}{x} = 2 \cosh \alpha \implies x^n + \frac{1}{x^n} = 2 \cosh(n\alpha) \, .  \]


15.  Define f(1) = 1, f(n) = n/10 if n is a multiple of 10 and f(n) = n+1 otherwise. For each positive integer m define the sequence a1, a2, a3, ... by a1 = m, an+1 = f(an). Let g(m) be the smallest n such that an = 1. For example, g(100) = 3, g(87) = 7. Let N be the number of positive integers m such that g(m) = 20. How many distinct prime factors does N have?

i think there is a trick to this, i'm not quite so sure what it is...

[quote="nukular"]
So yeah the basic idea is to work backwards. Start at 1 then multiply by 10 (denoted by B) or subtract 1 (denoted by A) at each step. Then you get a 19 letter word of A's and B's, with the following conditions.

1. The word starts with a B (you must go to 10 first)
2. The word does not start with BAAAAAAAAA (B, 9 A's) (otherwise you hit 1. But 20 is the first spot where you hit 1, so no good).

3. The word does not have AAAAAAAAAA (10 A's), Since you cant do 100 -> 99 -> ... -> 91 -> 90, as 90 can only go to 9, not 91.

This is now a simpler counting problem.

Consider only the remainder of the word after the B (18 letters). Then there are 2^18 possibilities. But some do not work. They fall into two nonoverlapping categories.

1. Starts with 9 A's but has no 10 consecutive A's.
2. Has 10 consecutive A's.

1) The word is of form AAAAAAAAABXXXXXXXX, where each X can be anything without violating any conditions. This makes 2^8 words of this form.

2) Either the word starts with 10 A's (2^8 words), or has a BAAAAAAAAAA (B, 10A's) in the word. There are 8 ways to choose where this goes in the word (pick the number of letters before it from 0 to 7), and 2^7 ways to pick the other letters (7 spots remain), making 8 * 2^7 = 2 ^ 10 words.

Therefore, there are 2^18 - 2 * 2^8 - 2^10 = 2^18 - 3 * 2^9 = 509 * 2^9 words. The sum of its prime divisors is 509 + 2 = 511 as desired.
[/quote]


[quote=chien than]Let be given positive numbers a;b;c satisfying the condition $abc>1$.Prove that:
$\frac{a}{\sqrt{b+\sqrt{ac}}}+\frac{b}{\sqrt{c+\sqrt{ab}}}+\frac{c}{\sqrt{a+\sqrt{bc}}}\geq \frac{3}{\sqrt{2}}$[/quote]
[url]https://artofproblemsolving.com/community/c4h2565013p21994721[/url]