linear algebra
matrix
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linear algebra unsolved

Prove that the matrix $ A \equal{}\begin{pmatrix}1 &\frac{1}{2}&\cdots &\frac{1}{n}\\ \; &\; &\; &\;\\ \frac{1}{2}&\frac{1}{3}&\cdots &\frac{1}{n\plus{}1}\\ \cdots &\cdots &\; &\cdots\\ \frac{1}{n}&\frac{1}{n\plus{}1}&\cdots &\frac{1}{2n\minus{}1}\end{pmatrix}$ is $ \textbf{invertible}$ and all the entries of $ A^{\minus{}1}$ are integers.

Note:(actually matrix $ A_{n\times n}\equal{}\left\{a_{ij}\right\}$ where for $ 1\leq i,j\leq n$ we have: $ a_{ij}\equal{}\frac{1}{i\plus{}j\minus{}1}$)

One more [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146373]link[/url], with a full solution, along with several other related solutions. I showed the integer entries fact by writing the inverse of this matrix as a product $ A^{\minus{}1}\equal{}\begin{pmatrix}q_{0}& q_{1}&\cdots & q_{n\minus{}1}\end{pmatrix}\cdot\begin{pmatrix}1 & 0 &\cdots & 0\\ 0 & 3 &\cdots & 0\\ \vdots &\vdots &\ddots &\vdots\\ 0 & 0 &\cdots & 2n\minus{}1\end{pmatrix}\cdot\begin{pmatrix}q_{0}^{T}\\ q_{1}^{T}\\ \vdots\\ q_{n\minus{}1}^{T}\end{pmatrix}$, where $ q_{k}$ is the column vector of coefficients of the polynomial $ \frac{1}{k!}\frac{d^{k}}{dx^{k}}\left((x\minus{}x^{2})^{k}\right)$. Every entry of these matrices is an integer, so the product has integer entries.

With the relation to polynomials, I like this better indexed from 0 to $ n\minus{}1$ rather than from 1 to $ n$.

Hilbert matrix.

[Inserting key words to enable future searches.]

Wow nice solutions(also the links where really helpful :) ), and also thanks for the key word :)


AMC
AMC 12
AIME
inequalities
AIME I
\/closed

I hear that woot is a excellent program but could I get a more critical review for Woot?

Also I am wondering if it really is suitable to me...
For AMC12 I get about 108.5 or 109.5
For 2006 AIME I got about 8(I guessed on one)

I am a raising freshman for the coming school year which wil be my first highschool experiance.  So I am not sure whether I will really have enough time to do it because of all those APs that I have to prep for and etc.....

Also wouldn't taking classes through online classroom make a little harder?

My last question is does it really make a lot of difference?
(This sounds like I'm very negative, but I really wonder)

But I am certainly willing to spend money and take my time if it really is worth it!  :D

Based on what some of the administrators have said, it sounds like the scores you get should be enough for at least the lowest levels of WOOT.

Regarding amount of time, I don't think it would take up too much. Can you devote a few hours a week or less to math? Seeing as you might be doing math anyway, you might as well do it. And WOOT is a efficient program for sharpening up your math abilities  :wink:

So that brings me to another question: how good is WOOT really? Although I have never taken it personally (I plan to this year), it seems as if WOOT is a very good program that helps get you to the level where you can solve olympiad problems - problems that can be very difficult and that require great creativity, experience, and insight.

What I can say from experience, however, is that AoPS do help improve problem-solving ability. A class helps give something more than just simply practicing problem after problem; a class helps teach you key concepts, ideas, and methods that you might not have otherwise learned. This could even mean giving you a [i]specific[/i] set of problems that familiarize you with some concept. As well, as the AoPS site says left and right, the classes do not teach you to memorize facts; instead, they take a constructive approach to teaching by making you think about and gain insight into the topics discussed. So just consider whether you think that WOOT would be a useful way to spend a few hours of your time each week  :wink:

Although your AIME score is at the lowest level of WOOT, I don't think your AMC12 is.

Try doing some regional or national olympiad problems and see how you fare.

What do you mean when you said 'about'?  Did you actually write the tests officially?

[quote="climskywell"]thank you very much...

I did sign up for it after all

Are there any books that I should purchase along with WOOT?[/quote]

Well, no books are actually required or specifically recommended for WOOT. However, there are books that are particularly recommended for avid problem solvers in general. For one, AoPS Volume 2 or even Volume 1 for you would be particularly recommended.

[quote="PenguinIntegral"][quote="ShoeFactory"]Art and Craft of Problem Solving if you haven't gotten to it.[/quote]
Eww. Icky. Maybe from a public library, but definately not worth buying.[/quote]

I disagree. Although one might not be up to reading all of that volume, that book is a classic on problem solving, as well as a good guide to help one transition to olympiad-level problems.

[quote="kimby_102"]the Art and Craft of Problem Solving is totally worthbuying..
I have, along with Art and Craft, Problem Solving Strategies which also teaches many problem solving tactics with a lot of problems..[/quote]
Yeah, problem solving strategies is great, but I prefer [url=http://www.kalva.demon.co.uk/books/zeitz1.html]kalva's intereptation of ACoPS[/url], that the book skims through way too much material and is essiantially a collection of hard problems,which are easily found with the kalva olympiad problem database.

I have ACoPS, but I have never used it very much.  Engel, on the other hand, I now take with me wherever I go.  The first few problems are easy, but still challenging to those who have not worked with olympiad problems yet, although I believe that you would want additional resources for inequalities and geometry.

I thought they used to sell ACops here....

but apparently they do not..

If I want to purchase is it... do I just have to get it by myself on the internet..
My parents prefer not to do online shopping..

I think until the new edition comes, out, all you can do is seach used bookstores.

It comes out later this year, right?

The best bet would be to find a local college, and look through it's library. That is where I found my copies of both, without having to pay like 50 dollars each...

They really cost a LOT OF MONEY, so you should check local libraries.


Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=ab+bc+cd+da$. Prove that for every positive real constant $k$ then \[\sqrt[k]{a}+\sqrt[k]{b}+\sqrt[k]{c}+\sqrt[k]{d}\ge \min(4,2\sqrt[k]{2}).\]

[quote="Parkdoosung"]Thank you very much!  :lol:

I think $a,b,c,d$ must be non-zero because if $a=b=c=d=0$, $a^{1/k}+b^{1/k}+c^{1/k}+d^{1/k}=0$. :(

Let $f(x)=x^{1/k}$, then $f'(x) = \frac{1}{k}x^{1/k-1}$, and $f''(x) = \frac{1}{k}\cdot(\frac{1}{k}-1)x^{1/k-2}$.
(1) If $k>1$, $f(x)$ is concave on $(0,\infty)$, so $f(a)+f(b)\geq f(a+b-\epsilon)+f(\epsilon)$.($\epsilon$ is a very small positive real number),
but because $f(x)$ is continuous on $[0,\infty)$, $f(a)+f(b)\geq f(a+b)+f(0)$,
so $a^{1/k}+b^{1/k}+c^{1/k}+d^{1/k}\geq (a+c)^{1/k}+(b+d)^{1/k}$.
(2) If $k\leq 1$, $f(x)$ is convex on $(0,\infty)$, $f(a)+f(b)\geq 2f(\frac{a+b}{2})$, so $a^{1/k}+b^{1/k}+c^{1/k}+d^{1/k}\geq \frac{2}{2^{1/k}}(a+c)^{1/k}+\frac{2}{2^{1/k}}(b+d)^{1/k}$.

Therefore, we only need to prove that if $(a+c)(b+d)=(a+c)+(b+d)$, $(a+c)^{1/k}+(b+d)^{1/k}\geq 2\sqrt[k]{2}$.
Let $(a+c)=\frac{1}{p}$ and $(b+d)=\frac{1}{q}$, then $p+q=1$,
and let $g(x)=x^{-1/k}$, then $g'(x)=-\frac{1}{k}x^{-1/k-1}$, and $g''(x)=\frac{1}{k}\cdot(\frac{1}{k}+1)x^{-1/k-2}$,
so $g(x)$ is convex on $(0,\infty)$, so $g(p)+g(q)\geq 2g(\frac{p+q}{2})$.
It implies that $(a+c)^{1/k}+(b+d)^{1/k}\geq 2\sqrt[k]{2}$.  :)[/quote]

What for the case $a=b=c=d=1$? That makes one equality?   :wink:


Solved this system
$ x^2\plus{}y^2\plus{}z^2\equal{}xyz\plus{}4$
$ xy\plus{}yz\plus{}zx\equal{}2(x\plus{}y\plus{}z)$

hello, by solving the second equation to $ y$ you will get $ y\equal{}\frac{2x\plus{}2z\minus{}xz}{x\plus{}z\minus{}2}$ and inserting this in the first equation, you will get:
$ x^4\minus{}x^3(z^2\minus{}4)\plus{}x^2(z^3\minus{}3z^2\minus{}4z\plus{}4)\plus{}x(16\minus{}4z^2)\minus{}16\plus{}16z\plus{}4z^2\minus{}4z^3\plus{}z^4\equal{}0$ and you must solve this equation with respect to $ x$. Also you have to consider the case $ x\plus{}z\equal{}2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by solving the second equation to $ y$ you will get $ y \equal{} \frac {2x \plus{} 2z \minus{} xz}{x \plus{} z \minus{} 2}$ and inserting this in the first equation, you will get:
$ x^4 \minus{} x^3(z^2 \minus{} 4) \plus{} x^2(z^3 \minus{} 3z^2 \minus{} 4z \plus{} 4) \plus{} x(16 \minus{} 4z^2) \minus{} 16 \plus{} 16z \plus{} 4z^2 \minus{} 4z^3 \plus{} z^4 \equal{} 0$ and you must solve this equation with respect to $ x$. Also you have to consider the case $ x \plus{} z \equal{} 2$.
Sonnhard.[/quote]
Sorry.Dr Sonnhard Graubner,can you deatail it,it is quite strange,and i cant solve your equation,my solution is nice but hard


Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor {\sqrt{x}} \rfloor = 12$, find the probability that $\lfloor {\sqrt{100x}} \rfloor = 120$. ($\lfloor {v} \rfloor$ means the greatest integer less than or equal to $v$.)

$\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1$

Edit: Never Mind.  I thought x had to be an integer.

First of all, that would give $3$ values, since $x=146$ still works ($100(146)=14600<14641$). Second, it's all real values. ;)

Why is this topic posted twice?  The other one should probably be deleted.


Given a circle (C) with radius 1 and a square(S) with length side  2. Contruct a isosceles right triangle XYZ with XYis hypotenuse, when Z,X lies on  (C) and Y lies (S).

Is this necessarily possible? If we must have XZ = ZY, but XZ has a maximum value of 2, since they're points on the circle, the square could be far enough away from the circle so that ZY has a minimum value greater than 2.

You must discuss this problem, there are two case.
You can assume that we can contruct the triangle.Please post proof for this problem.


limit
trigonometry
geometry
modular arithmetic
probability
integration
logarithms

Evaluate the limit:

\[ \lim_{n \to \infty}{n \cdot \sin{1} \cdot \sin{2} \cdot \dots \cdot \sin{n}}.\]

Proposed to "Unirea" Intercounty contest, grade 11, Romania

The limit must be zero.

Here's an informal argument; it could be made rigorous with some more effort.

Fix some number smaller than $ 1$ - say we take $ \frac{\sqrt{3}}2.$ In the long run, we should have $ |\sin n|<\frac{\sqrt{3}}2$ half of the time; the rest of the time $ |\sin n|<1,$ Hence in the long run, $ \left|\prod_{k\equal{}1}^n\sin k\right|<\left(\frac{\sqrt{3}}2\right)^{\frac n2}.$

This product then goes to zero exponentially fast. Multiply that by $ n$ and it still tends to zero.

Dear  Kent

could you explain for me about your solution :blush:
When I chose n=281 then sin281=0.98
                                      sin 21800=0.999

it means that we can choose a subsequence of $ \{sinn\}_{n}$ which is arbitrary closed to 1. So how can you have $ |sinn| < \frac {\sqrt {3}}{2}$ "half of the time" ?

Simple: that point you picked was from the [i]other[/i] half, and the only estimate I made for [i]that[/i] particular point was that $ |\sin n|<1.$ There area also plenty of values of $ n$ that you didn't name for which $ \sin n$ is [i]not[/i] close to 1.

Let $ x_n\equiv n\pmod{2\pi},$ chosen on the interval $ [0,2\pi).$ Then in the long run, $ x_n$ gives a uniform distribution on the interval $ [0,2\pi).$ For a point $ x$ chosen uniformly on $ [0,2\pi)$ the probability that $ |\sin x|<\frac{\sqrt{3}}2$ is $ \frac12.$

That's what I meant by "half the time." I meant this: if $ p(n)$ equals the number of values of $ k\in\{1,2,\dots,n\}$ such that $ |\sin k|<\frac{\sqrt{3}}2,$ then $ \lim_{n\to\infty}\frac{p(n)}n\equal{}\frac12.$

The particular values I used for this argument are by no means optimal and the sequence actually tends to zero quite a bit faster than I estimated it to. In fact: just as you can find some values of $ n$ for which $ \sin n$ is very close to $ 1,$ you can also find some values of $ n$ for which $ \sin n$ is very close to zero. The former values make little difference to the product; the latter send the product more rapidly towards zero.

very nice explanations, great "combination" of probability and analysis,  thanks ^_^

[quote="Kent Merryfield"]The particular values I used for this argument are by no means optimal and the sequence actually tends to zero quite a bit faster than I estimated it to.[/quote]  Just plugging it in to mathematica, it looks like $ \lim_{n \to \infty} 2^n \prod_{i \equal{} 1}^n \sin i$ could be bounded but not convergent.  I guess this is related to $ \int_0^1 \ln \sin x \, dx$?

[b]Kent[/b], how you get probability $ \frac {1}{2}$?

We have line $ y \equal{} a$, where $ 0\leq a\leq 1$. Then probability that $ |\sin x| < a$: $ p \equal{} \frac {4\cdot\arcsin a}{2\pi} \equal{} \frac {2\arcsin a}{\pi}$

In your choise $ a \equal{} \frac {\sqrt {3}}{2}$, so $ p \equal{} \frac {2}{3}$ :) Or maybe i don't understand something.

We can find this optimal value $ a$. Just need to minimize expression $ a^{\frac {2\arcsin a}{\pi}}$ and we get $ a_{\text{min}}\approx 0.39$.

If $ \lim_{n\to\infty}f(n)\cdot a_{\text{min}}^{n\cdot\frac {2\arcsin a_{\text{min}}}{\pi}}$ doesn't exist (in our case $ f(n)\equal{}n$), we can't solve problem this way.

You're right, Yuriy: 2/3 of the time, not half the time. The argument, with slight adjustments, still stands.

The official solution looks like Kents one :D. Quite ingenious, I think.

[hide="Here it goes"] We observe that the distance between two consecutive intervals of this type:

$ \left[ n \pi \plus{} \frac{\pi}{6} ; n \pi \plus{} \frac{5 \pi}{6}  \right]$ is $ \frac{\pi}{3}<2$ and the length of the interval is $ \frac{2 \pi}{3} >1$.

So, if we take any two consecutive natural numbers $ k, k\plus{}1, k\plus{}2$ at least one of them is in an interval like that.

Then

\[ |n \cdot \sin{1} \cdot \sin{2} \dots \sin{n}| \leq n \sin^{\left[ \frac{n}{3} \right]} {\frac{\pi}{6}} \leq \frac{n}{2^{\frac{n}{3}}} \to 0.\] [/hide]

It will be interesting to find approximation for the ${ \prod_{k = 1}^n \sin (kx)}$

This method might also work.

Let $ x_{n}\equal{}n\prod\limits_{k\equal{}1}^{n}\sin k$.

We have, based on AM-GM inequality that

$ 0<x_{n}^{2}\leq n^{2}\left(\frac{\sum\limits_{k\equal{}1}^{n}\sin^{2}k}{n}\right)^{n}$

$ \equal{}n^{2}\left(\frac{n\minus{}\sum\limits_{k\equal{}1}^{n}\cos(2k)}{2n}\right)^{n}\equal{}n^{2}\left(\frac{1}{2}\minus{}\frac{\sin(n)\cos(n\plus{}1)}{2n\sin(1)}\right)^{n}$.

Thus, for large $ n$ we get that

$ 0<x_{n}^{2}<\frac{n^{2}}{2^{n}}$

and the result follows.


Did anyone else get an AoPS certificate in the mail for making MC Nats?  Also, where will Mr. Rusczyk be?

I think I did last year... did they send them out last year?

I don't remember... hmmm... I think I do recall getting one last year, but I didn't know about this site until later this (school) year.

I didn't get one last year but I did this year.

Yeah i got one this year, but where will Mr.Rusczyk be?
Also, does anyone know how to redeem them?

There should be instructions with the gift certificates regarding redeeming them.  You can use them when you purchase books or enroll in classes.  If you have any questions about how to use them, please email me.

We didn't something a little different last year - we offered a free audited enrollment in a class, but only to state winners.  This year, we sent out gift certificates to all the students who made nationals.

I will be at Nationals.  I will be easy to find - I'll post more details soon!  I look forward to meeting you all.  Please don't be shy - come introduce yourselves.

hmm...i wonder if they want student speakers at Nats once in a while to tell ppl how great the thing is...I tend to think I do loads better on public speaking than on math :-( wat a shame...

i got 1 this year, but nothing last year. must've been because i didn't joing AoPS yet or something.

I got a thingie last year which is how I came to know about AoPS (and why I joined). hm, what are these gift certificate things? are they different from audited enrollment in a class? (j/w obviously, I couldn't get a gift certificate now, MC days are long gone :))

I only got this invitation to the website on red paper with black ink, but none of the stuff all of you are talking about... maybe i have to look harder.

Hmm that's odd... and you were state winner, right?

[quote="Chinaboy"]ya, but only through countdown... (hehehe...)

what did your mail look like?[/quote]

It was just a plain white paper with blue ink I think.  If your state's countdown is official, you should be getting it...

this is only recently? maybe that's why... thanks!

I just got it today (dated 4/21). (Along with a Regional certificate from 2/7!) I seems that my coach was hoarding all that stuff.


Let $ x$, $ y$ be two positive numbers for which $ x^2\plus{}y^2\equal{}1$.

Prove that $ \frac 1x\plus{}\frac {x}{1\plus{}y}\ge \sqrt 3\ .$ Try to extend it.

Multiplying both sides by $ x(1 \plus{} y)$  we get:

$ 1 \plus{} y \plus{} x^2 \geq \sqrt {3}x(1 \plus{} y)$ or
$ \sqrt {3}x(1 \plus{} y) \leq 1 \plus{} y \plus{} 1 \minus{} y^2 \equal{} (y \plus{} 1)(2 \minus{} y)$

So it is sufficient to prove that:

$ \sqrt {3}x \leq 2 \minus{} y$ and by squaring we get

$ 3x^2 \leq 4 \minus{} 4y \plus{} 4y^2$ or

$ 3(1 \minus{} y^2) \leq 4 \minus{} 4y \plus{} y^2$ or

$ 4y^2 \minus{} 4y \plus{} 1 \geq 0$ or

$ (2y \minus{} 1)^2 \geq 0$ which is true

[quote="bilot4"]$ \frac 1x \plus{} \frac {x}{1 \plus{} y}\ge \sqrt 3$ $ \Longleftrightarrow$ $ 1 \plus{} y \plus{} x^2 \equal{} 2 \plus{} y \minus{} y^2\ge x(1 \plus{} y)\sqrt 3$ $ \Longleftrightarrow$

$ x(1 \plus{} y)\sqrt 3\le (y \plus{} 1)(2 \minus{} y)$ $ \Longleftrightarrow$  $ x\sqrt {3}\le 2 \minus{} y$ $ \Longleftrightarrow$

$ 3x^2\le 4 \minus{} 4y \plus{} 4y^2$ $ \Longleftrightarrow$ $ 3(1 \minus{} y^2)\le 4 \minus{} 4y \plus{} y^2$ $ \Longleftrightarrow$

$ 4y^2 \minus{} 4y \plus{} 1\ge 0$ $ \Longleftrightarrow$ $ (2y \minus{} 1)^2\ge 0$ which is truly.

Is equality iff $ x \equal{} \frac {\sqrt 3}{2}$ and $ y \equal{} \frac 12\ .$[/quote]
[b]Very nice ![/b] [u]See an extension at[/u] http://www.mathlinks.ro/Forum/viewtopic.php?t=170549


conics
parabola
geometry
circumcircle
geometry proposed

constant lines $\ell_{1}$ , $\ell_{2}$ are tangent to a Parabola , an arbitrary line $\ell$ tangent to that Parabola creat a triangle with $\ell_{1}\ ,\ \ell_{2}$ .
Prove that all of the triangles are coaxal.

Let $E$ be, the focus of the given parabola and we denote as $M,$ $N$, its tangent points from the lines $\ell_{1},$ $\ell_{2},$ respectively.

Also, we denote the points $S\equiv \ell_{1}\cap \ell_{2},$ $A\equiv \ell_{1}\cap \ell$ and $B\equiv \ell_{2}\cap \ell.$

We draw the lines $AA',$ $BB',$ parallel to the axis of the given parabola and , by [b][size=109]Poncelet’s theorem,[/size][/b] we have that $\angle MAA' = \angle EAB$ $,(1)$ and  $\angle NBB' = \angle EBA$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $\angle EAB+\angle EBA = \angle MAA'+\angle NBB' = \angle ASB.$

So, we conclude that $\angle ASB+\angle AEB = 180^{o}$ and hence, the quadrilateral $SAEB,$ is cyclic.

That is the circumcircle of the triangle $\bigtriangleup SAB,$ passes through the focus $E,$ of the given parabola.

So, the segment  $SE,$ is the common cord of the circumcircles of all the triangles formed by the constant tangent lines $\ell_{1},$ $\ell_{2}$ of the given parabola, intersected from the variable one $\ell$ and the proof is completed.

Kostas Vittas.


AMC USA(J)MO USAMO geometry inequalities AIME induction

I personally thought that this was a strange USAMO. It had 1 number theory (#1), 1 number theory/algebra (#2) , 1 number theory/geometry (#4), 1 number theory/inequality (#6), 1 counting (#5), and 1 geometry (#3) problem. Too much number theory for me... And it didn't have your good old classic inequality, either.

I think I got problems 1, 4 and 5. In order of increasing difficulty I thought it went 4, 5, 1, 2, 6, 3.

Yeah, I'd have to agree with most of that.  I think I got 1,4,5, part of 3 but I just realized I messed up (somewhat) on #5...  maybe 3-5 pts if I'm lucky haha.  Anyways, solutions are up:

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-test.html

I hope 17 ish is enough for MOP...  and I think it was more like 1,4,5,2,3,6 in order of difficulty; this USAMO definitely wasn't geared towards my strengths.

Wow all their solutions are so elegant and short. I still don't know how I was able to write a 7 page solution to problem 1 :blush: ... hopefully they don't take off points for inelegancy...

Anyways, this year's test definitely seemed to be easier than last year's.

[quote="tekno10m"][quote="beta"]I solved #1 and 4, although in #4 at the very end I messed up(forgot when k_1+k_3 exceeds n we lose solutions :( ), I did have everything else proved(k_1+k_3 = k_2 + k_4), do you think I can receive a 5/7 on #4?[/quote]

I would guess they would score that as a 4, as they are probably quite strict. But it doesn't matter, because you are going to MOP anyways since you solved number 1.[/quote]

I heard they rarely give 3's or 4's.

They do rarely, but I think this case warrants a 4. Anyways, just my guess.

Ahh 8 more days until they post our results...

[quote="tekno10m"]Anyways, this year's test definitely seemed to be easier than last year's.[/quote]
I agree with you there. I also thought that it was interesting that there wasn't even one problem that dealt with inequalities, and that the focus of the test was with NT and Combinatorics...

[quote="beta"]I heard they rarely give 3's or 4's.[/quote]
They probably do give out 3's and 4's very rarely, but as tekno10m said, even if you got 1-2 points, your going to MOP. I imagine a lot of people who don't have that much experience in problem solving would get 0's...

Anyway, even though I wasn't able to take the test, I definetly liked this year's USAMO due to the fact that I like both NT and Combinatorics and it seemed to be easier than last year's test. The problems that I thought were the most interesting were problems #4 and #6...

I got #1 and #4 without many problems, and I somehow miscopied #2 when I went to work with scratch paper.  Now that I look at #3 I really should have made some progress on it.  I'm hoping for a point from writing "Show that B1, C1, and A are collinear."  #5 also wasn't that bad, but #6 was too scary to even look at :P.  I definitely think this USAMO was weird, and it didn't have any inequalities-focused problems!

I didn't specifically show why exactly $k_1+k_3=k_2+k_4$ but I explained why it had to be so using words (at least somewhat). What do you think my score will be? (I got everything else right)

[quote="tekno10m"][quote="MithsApprentice"]Hmmm...on #5, do you guys think we would've gotten full credit for citing the solution in Discrete and Computational Geometry?[/quote]

I have no idea what is meant, but w/e.[/quote]

[i]Discrete and Computational Geometry[/i] (which I should've put in italics) is a book that has a solution to this problem. Indeed that solution, as the AMC website says, is stronger than the requirements.

#6 was surprisingly easy. (#3 too but my natural aversion to Geometry made it questionable from the start. I thought something tricky with Pascal's or something would be necessary)

For #6 I had the RHS and the first Lemma for the LHS with about 2 hrs to go ... I don't understand why I wasn't able to finish it off. Still an easy USAMO (too easy :()

I posted this elsewhere, but I think it might be more apppreciated here:  A proposed rewording of question 1 - it's a limerick!

I once had an n-height square table
That I wanted to stand on a gable.
So I hired a plumber
Who cut a whole number
From each leg, so the table stood stable.

Assuming each leg is distinct before cutting, how many different ways could the plumber have cut my table legs?

(note that a gable is a protruded upper-floor window with a roof that may or may not be slanted.)

That's most certainly true. Plus, we consider the high-scoring Canadians and non-Winner seniors who won't get to go. I'd say cutoff is around the same place. Lower for high-scoring WOPpers. Higher for WOPpers who neglected their duty to their math team. (Guilty as charged.) If you completely solved 3 problems, you have an incredible chance at making MOP. If you completely solved 2, you're borderline and partials might push you in. Below that, you gotta be frosh, have a killer WOP score, a high APMO score, or perhaps MOP experience might help too.

In any case, good luck to all aspiring MOPpers this year!

I have a question on the scoring of #5:

I tried doing it by induction and did about three of the four main parts, and then wrote a conjcture (which I was unable to prove) which helped me prove the fourth, and most important part.

Given this, even though I solved this problem in a different manner than the solution given, do you think will I get a point or two for my partial induction solution?

JB

[quote="beta"][quote="white_horse_king88"]Hm, I doubt people are that lazy after making USAMO. I mean, I think they would take the initiative to read the rulebook, you know? And I do think most freshmen got a positive score, based on this year's questions. But, that's just my opinion. Perhaps I am judging too much based on USAMO freshmen on this forum, but then again, probably more than 1/3 of freshmen participants post here. I think.[/quote]

So far I know at least 6 freshmen including myself who at least essentially solved #1 and #4. But looking on the positive side(something that many people said that I couldn't do :lol: ), last year's #1 and #4 aren't that much harder than this years and yet there are still 22 0's(and quite a few 9th grade 0's).[/quote]

whoa...feiqi looking on the positive side...I'm going to faint

I think within the next week.

Feiqi--you're being too optimistic.  Stop it, you're out of place.  Go back to your pessimism (just kidding :D).

Don't worry Alex, I'll spend the next few days worrying ;)

You won't have to. You'll get a call tomorrow inviting you to MOP and then the rest of us will have peace at least.

Are you guys sure they always call on the weekend for MOP? My 9th grade year I didn't get a call until Tuesday I think. I thought it was because people were declining spots, but Zuming told me I was on the original 30 person list and thye just didn't get around to calling me.

darktreb, that could easily be the case. Or maybe they tried calling and you weren't home or whatnot. The point, though, is that if you made it you made it. If you didn't, you didn't. No use stressing out.

Did anyone notice that the USAMO [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/questions/Day2/05USAMOday2.pdf]Day II Exam[/url] was extremely blurry?

When my teacher gave it to me, I thought that my school's printer had gone bad.

Yeah, I suspect that they scanned the thing or something.  The blurriness was slightly annoying.

But at least now people have a good reason for why they didn't do well on the test.  ;)

well...I think we used the HTML version...

Yeah, I've noticed the AMC often makes little errors like typos, blurriness, not labelling the questions as 4, 5, and 6, etc.

However, the quality of the exams and the hard work that goes into them is far more important than these minor details.


Six shirts are to be chosen from 3 red, 2 green, 4 white, and 2 blue shirts. Find the probability that more red than blue shirts are chosen (with no blue a possibility) or that the same number of green and white shirts (with at least one green and one white) are chosen.

[hide="I hate this; tell me theres a better way"]
$ \bullet$ $ R > B$:

1R 0B-$ 3 \left[{4 \choose 4} {2 \choose 1} \plus{} {4 \choose 3} {2 \choose 2}\right] \equal{} 18$,

2R OB-$ 3\left[{4 \choose 4} {2 \choose 0} \plus{} {4 \choose 3} {2 \choose 1} \plus{} {4 \choose 2}{2 \choose 2} \right] \equal{} 45$,

3R OB-$ 1 \times \left[ {4 \choose 1} {2 \choose 2} \plus{} {4 \choose 2} {2 \choose 1} \plus{} {4 \choose 3} {2 \choose 0}\right] \equal{} 20$,

2R 1B - $ 3 \times 2 \left[ {4 \choose 1} {2 \choose 2} \plus{} {4 \choose 2} {2 \choose 1} \plus{} {4 \choose 3} {2 \choose 0}\right] \equal{} 120$,

3R 1B - $ 1 \times 2 \left[{4 \choose 0} {2 \choose 2} \plus{} {4 \choose 1} {2 \choose 1} \plus{} {4 \choose 2} {2 \choose 0}\right] \equal{} 30$,

3R 2B - $ 1 \times 1 \times (4 \plus{} 2) \equal{} 6$. Sum = $ 18 \plus{} 45 \plus{} 20 \plus{} 120 \plus{} 30 \plus{} 6 \equal{} 239$.

$ \bullet$ $ G \equal{} W, R \le B$: If there are $ 2$ of each, then $ 1 \times {4 \choose 2} [3 \cdot 2 \plus{} 1] \equal{} 42$. If there is $ 1$ of each, then $ 2 \times 4 \times \left[{3 \choose 2}{2 \choose 2}\right] \equal{} 24$.

Thus the total number of ways is $ 239 \plus{} 42 \plus{} 24 \equal{} 305$, and total possibilities is $ {3 \plus{} 2 \plus{} 4 \plus{} 2 \choose 6} \equal{} 462$. so the answer is $ \frac {305}{462}$?

[/hide]

edit: hmm, this question was given at my school


calculus
integration
limit
function
inequalities
real analysis
real analysis unsolved

This problem is similar to one posted by Virgil Nicula a couple of days ago, although it's easier.  I considered putting it in the Calculus computations and tutorials section, actually.

Let $f$ be Riemann integrable on $[0,A]$ for every $A<\infty$ such that $\lim_{x\to\infty} f(x)=1$.

Prove that
\[ \lim_{t\to 0} t\int^{\infty}_0 \exp(-tx)f(x)dx=1  \]
where the limit is taken over $t>0$.

Extremely closely related problem:

Suppose $\lim_{n\to\infty}a_n=L.$ Prove that

$\lim_{r\to1^-}(1-r)\sum_{n=0}^{\infty}a_nr^n=L.$

You ought to be able to find an argument that, apart from the obvious notational shifts, works in both cases.

One side issue: the hypotheses of blahblahblah's problem do imply that $f$ is a bounded function; you can use that, if necessary.

Solution to Kent's problem:
Since $a_n \to L$ , for any $\epsilon>0$ there exists $N$ such that $L-\epsilon < a_n<L+\epsilon,\forall n>N$.
Now, for any $\epsilon >0$we can find $N$ such that for any $n>N$ we have:
\[{{ (r-1)[(a_1+a_2r^2+...a_Nr^N)+(L-\epsilon)(r^{N+1}+r^{N+2}+....})]<(r-1)\sum_{n=0}^{\infty}a_nr^n <(r-1)[(a_1+a_2r^2+...a_Nr^N)+(L-\epsilon)(r^{N+1}+r^{N+2}+....})] \]
Now, tending r to 1 we get that $L-\epsilon<\lim_{r\to1^-}(1-r)\sum_{n=0}^{\infty}a_nr^n<L+\epsilon$. Since $\epsilon$ was chosen arbitrarly we have that $\lim_{r\to1^-}(1-r)\sum_{n=0}^{\infty}a_nr^n=L$

Based on the same ideea:

Since $\lim_{x \to \infty}f(x)=1$ , for any $\epsilon>0$ we can find $N$ such that $1-\epsilon<f(x)<1+\epsilon,\forall x>N$. $t\int^{\infty}_0 \exp(-tx)f(x)dx= t\int_{0}^N \exp(-tx)f(x)dx+t\int^{\infty}_N \exp(-tx)f(x)dx<$
$t\int_{0}^N \exp(-tx)f(x)dx+(1+\epsilon)t\int^{\infty}_N \exp(-tx)dx=t\int_{0}^N \exp(-tx)(f(x)-1)dx+(1+\epsilon)t\int^{\infty}_0 \exp(-tx)dx$.
Similarly $t\int^{\infty}_0 \exp(-tx)f(x)dx>t\int_{0}^N \exp(-tx)(f(x)-1)dx+(1+\epsilon)t\int^{\infty}_0 \exp(-tx)dx$. Tending t to 0 in both inequalities, since $\lim_{t \to 0} t\int_0^\infty e^{-tx}dx=\lim_{t \to 0} \int_0^\infty e^{-x}dx=\int_0^\infty e^{-x}dx=1$, we find that $1-\epsilon< \lim_{t\to 0} t\int^{\infty}_0 \exp(-tx)f(x)dx<1+\epsilon$ and since $\epsilon$ was chosen arbitrarly we have that  $\lim_{t\to 0} t\int^{\infty}_0 \exp(-tx)f(x)dx=1$.

Just for completeness, my proof.

Fix $\epsilon>0$ and take $N$ large enough that $|f(x)-1|<\epsilon$ for $x\geq N$

For every $t>0$, a simple computation yields
\[ \int^{\infty}_0 t\exp(-tx)dx=1  \]
Thus it is enough to show that
\[ \lim_{t\to 0+}t \int^{\infty}_0 \exp(-tx)|f(x)-1|dx=0  \]
We have
\begin{eqnarray*} \left| t \int^{\infty}_0 \exp(-tx)|f(x)-1|dx\right|&=&\left| t\int^N_0 \exp(-tx)|f(x)-1|dx\right|+\left| t\int^{\infty}_N \exp(-tx)|f(x)-1|dx\right| \\ &\leq& tM+t\epsilon \\ \end{eqnarray*}
where $M=\int^{N}_0 |f(x)|dx+N$.  Clearly $M<\infty$, since $f(x)$ is Riemann integrable on $[0,N]$.  As $t\to 0+$, it is clear that
\[ t \int^{\infty}_0 \exp(-tx)|f(x)-1|dx\to 0  \]

Here's blahblahblah's argument (only expanded out) applied to my (easier!) variation on the problem:

Straigtforwardly, we have that $(1-r)\sum_{n=0}^{\infty}Lr^n=L.$

Hence it suffices to show that $\lim_{r\to1^-}(1-r)\sum_{n=0}^{\infty}|a_n-L|r^n=0.$

Given $\epsilon>0,$ find $N$ such that for $n>N,\,|a_n-L|<\frac{\epsilon}2.$

Then$(1-r)\sum_{n=0}^{\infty}|a_n-L|r^n=(1-r)\sum_{n=0}^N|a_n-L|r^n + (1-r)\sum_{n=N+1}^{\infty}|a_n-L|r^n$

<$(1-r)\sum_{n=0}^N|a_n-L|r^n + (1-r)\sum_{n=N+1}^{\infty}\frac{\epsilon}2r^n$

<$(1-r)\sum_{n=0}^N|a_n-L|r^n + (1-r)\sum_{n=0}^{\infty}\frac{\epsilon}2r^n$

=$(1-r)\sum_{n=0}^N|a_n-L|r^n+\frac{\epsilon}2.$

Now, let $M=\sum_{n=0}^N|a_n-L|.$ This is clearly finite.

$(1-r)\sum_{n=0}^N|a_n-L|r^n<(1-r)\sum_{n=0}^N|a_n-L|=(1-r)M.$

By choosing $r$ sufficiently close to $1$, we can make this as small as we please, such as less that $\frac{\epsilon}2.$ Thus by choosing $r$ sufficiently close to $1,$ we can force the original sum to be within $\epsilon$ of $L.$

I think that we all worked based on the same ideea. :P


calculus
integration
function
limit
analytic geometry
graphing lines
slope

$f$ function $R+\rightarrow R$ wich $C^{2}$ such that
$\int_{R+}|f|$ and $\int_{R+}f''^{2}$ are convergent

prove that $\lim_{x\rightarrow+\infty}f'(x) =0$

Suppose that there exists $\epsilon>0$ such that $f'(x)\ge\epsilon$ for an infinite (and spread out) sequence of $x$s. Then we can find infinitely many intervals $[a,b]$ such that $f'(a)=f'(b)=\epsilon/2$ and $\max_{[a,b]}f'\ge\epsilon$. Since $\int_{a}^{b}|f''(x)|dx\ge\epsilon$, the Cauchy-Schwarz implies $\int_{a}^{b}f''(x)^{2}dx\ge \frac{\epsilon^{2}}{b-a}$. Let $c=\frac{a+b}{2}$. If $f(c)\ge 0$, then $\int_{c}^{b}f(x)dx\ge (b-c)^{2}\frac{\epsilon}{4}=(b-a)^{2}\frac{\epsilon}{16}$, because the graph of $f$ has slope $\ge \epsilon/2$ on $[c,b]$. If $f(c)\le 0$, then $\int_{a}^{c}f(x)dx\le-(b-a)^{2}\frac{\epsilon}{16}$ for the same reason.

Finally,  $\max\left(\frac{\epsilon^{2}}{b-a},(b-a)^{2}\frac{\epsilon}{16}\right)\ge \frac{\epsilon^{5/3}}{16^{1/3}}$ using a generalized geometric mean. Thus, either $\int |f|$ or $\int |f''|^{2}$ diverges.

i am very confused with one thing. Sims for me that if $\int_{R}^{+}|f|$ is zero, hence f'(x) for x->infty is also zero, since that if we transform this integral in a infinit series, to satisfy convergence, we might have that  the area of a integral from a to b, where a,b>n where n is arbitrally big have to tend to 0, and using this way of thinking we just have to have the first integral converging to have the limit of its derivative to be 0.
Could someone  do an example such that $\int_{R}^{+}|f|$ converges and $lim f'(x)=a$a for x->infty and a<>0?

There are no such examples. But there are examples for which $\int_{R}^{+}|f|$ converges and $\lim_{x\to\infty}f'(x)$ does not exists.

With cauchy schwarz $|f'(x)-f'(y)|\leq \sqrt{|x-y|}\sqrt{|\int_{x}^{y}f''^{2}|}$

I wonder if someone can proof that $f'(x)$ verify cauchy criterion at +oo
for any $b>0$ there exist $A>0$ such that for any x,y >A then
$|f'(x)-f'(y)|\leq b$ ?


MATHCOUNTS
AMC
AIME
AMC 10
AMC 8
Pascal\u0027s Triangle

Find the number of positive integers, who when expressed in binary, have 10 digits with no one's next to each other.

This is a simplification of a problem from the mathcounts school handbook (I think) and [hide="This might be intimidating"]AIME. (I did the first step in the solution already.)[/hide]

I know what binary numbers are, but I don't understand what the problem is asking. Can you make it more clear? :(  :(

Oops, I made a mistake. It should say a sequence of 0s and 1s instead of binary number. (I forgot that binary numbers can't start with zeroes.)

The conditions are that the sequence must be 10 numbers long, and no 1s can be next to each other.
For example 0001010100 is a sequence that satisfies the conditions, but 1100101011 doesn't. Count the number of sequences that satisfy the condition.

The possible places where a 1 could go when there are 5 zeros and 5 ones would be the - marks on:
-0-0-0-0-0-
We have to put 5 ones in, and there are 6 places to but them, so C(6,5)=6.
If there are 4 ones and 6 zeros, it would look like:
-0-0-0-0-0-0-
where - is a possible place to put a one. There are
C(7,4)=35 combinations of places to put ones for that. WE evaluate for:

C(6,5)=6
C(7,4)=35
C(8,3)=56
C(9,2)=36
C(10,1)=10
And the string 0000000000=1

Adding these together we get 144.
 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

[quote="RC-7th"]That's correct.

Congraduations! You just solved one of the harder problems of a contest that many high schoolers have trouble on! :first:[/quote]
high schoolers?
isnt mathcounts a middle school competitioN?

AIME is a national competition. Anyone who took that year's AMC test can enter.  However, you have to score above a 120 on the AMC 10 or above a 100 on the AMC 12.

This means that a sixth-grader who took the AMC 6 would then have to take the AMC 8 and the AMC 10 (in that order, assuming he/she did well enough on both to go to the next level).

Once a student scores in the range for acceptance into the AIME, they do not take any further AMC tests (e.g., if you scored a 132 on the AMC 10 you wouldn't take the AMC 12).

High scorers on the AIME are invited to take the USA Mathematical Olympiad.

[hide="click text"]That's probably why the AIME is mostly high-schoolers.[/hide]

[quote="ln(dx/dy)"]The possible places where a 1 could go
. . .
C(6,5)=6
C(7,4)=35
C(8,3)=56
C(9,2)=36
C(10,1)=10
And the string 0000000000=1

Adding these together we get 144.[/quote]
Very nice job, ln(dx/dy).  This is an excellent, straight-forward, solution to a problem that can be pretty hard.

This seems like a good opportunity to point out that if you replace the final combination with C(11,0)=1, the combinations you have added are in an oblique, but straight line through Pascal's triangle.  The sum of the numbers along the lines of that particular slant through Pascal's triangle are members of the Fibonacci Sequence.  The final answer is a member of the Fibonacci Sequence.  (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...).

I will also mention, that a standard solution to this AIME problem involves recursions, resulting in the Fibonacci Sequence.  And that's probably why this problem is considered to be "hard".  (Such a discussion is too advanced for the Middle School forums.  If you want to discuss such a solution, please take the discussion to another forum, but leave a pointer, here.)

I really like ln(dx/dy)'s simple solution!

How'd you get the quantity 6C5 when you only had ten spots? Also, how'd you get the amount of spots that you could fit, for example if there were 5 ones you said there were six spots that you could put them in.... Please explain :? ...

-0-0-0-0-0- There are five zeroes. In addition to the spaces in between the zeroes, you have to count the spaces outside of them. There are only ten numbers to place, but if you look at the diagram, there are eleven spots to fill (5 are taken by zeroes).

If you'll notice, in all the combinations C(m,n) that were used in this problem, m+n=11. While  the number of zeroes plus the number of ones always adds up to ten (okay, a bit obvious there), there is always one more space than there are zeroes (each series of alternating zeroes and spaces starts and ends with a space).

It's probably not the best explanation, but this is my first post here so.... :?

[quote="Vitek"]
This means that a sixth-grader who took the AMC 6 would then have to take the AMC 8 and the AMC 10 (in that order, assuming he/she did well enough on both to go to the next level).

Once a student scores in the range for acceptance into the AIME, they do not take any further AMC tests (e.g., if you scored a 132 on the AMC 10 you wouldn't take the AMC 12).
[/quote]

There is no such thing as the AMC 6.  Also, there is no rule that you have to take the AMC 8 before going on to the AMC 10.  I know many people in 8th grade that took the AMC 10 or 12 without taking the AMC 8.  Finally, you could take the AMC 12 in the same year as the AMC 10 in order to try to get a higher score (or a score that qualifies for the AIME).


Hi everybody,

p*q = x^2 + a* (y^2)

where p and q are prime
x, y, a integers > 0

Is there any way to find a general solution?
Can we write all the semi-prime as x^2 + a* (y^2) ?

Thanks for any clue.

If $ a$ is fixed, not every prime is of type $ x^2 \plus{} ay^2$; in fact for $ a > 1$ you can't even get $ 2$. If $ a$ can be chosen, too, then take $ x\equal{}y\equal{}1$ and choose $ a$ accordingly.

[quote="ZetaX"]If $ a$ is fixed, [b]not every prime [/b]is of type $ x^2 \plus{} ay^2$; in fact for $ a > 1$ you can't even get $ 2$. If $ a$ can be chosen, too, then take $ x \equal{} y \equal{} 1$ and choose $ a$ accordingly.[/quote]

Hi,

I'm talking about semi-prime not prime.

If semi-prime doesn't include prime for you, then you will still easily find another counterexample.

[quote="aadil"]what is a semi-prime?[/quote]

A semi-prime is a product of two prime.

[i]In mathematics, a semiprime (also called biprime or 2-almost prime, or pq number) is a natural number that is the product of two (not necessarily distinct) prime numbers. The first few semiprimes are 4, 6, 9, 10, 14, 15, 21, 22, 25, 26, ... (sequence A001358 in OEIS).[/i]

http://en.wikipedia.org/wiki/Semiprime


search
number theory proposed
number theory

Find all integers $x,y$ such that $7^x - 3^y = 4$.

[quote="Dong Gyu Lim"]:P Help me solve this problem. $7^x - 3^y = 4$ ($x,y$ are integers) Hurry~![/quote]
See here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=66637[/url]

I had that link too, but there is a simple reason why I didn't post it: the simple try of searching for $7^x - 3^y = 4$ works very well (and gives also other threads like http://www.mathlinks.ro/Forum/viewtopic.php?t=72452 where one can see how often it was posted).
Thus the author didn't even try and should learn to do so.


Ean x kai y einai thetikoi pragmatikoi arithmoi,deikste oti:

                                                 $ x^y\plus{}y^x > 1$

Για κοίτα [url=http://www.mathlinks.ro/viewtopic.php?t=200413]εδώ[/url] και [url=http://www.mathlinks.ro/viewtopic.php?t=204371]εδώ[/url] :)

oops  :blush: ...eprepe na to fantasto oti tha exei ksanampei...ayto omos den tin kanei mi periergi..lol


Let $a,b,c,d\ge0$ and $2(ab+ac+ad+bc+bd+cd)+abc+bcd+cda+dab=16$
Prove that $a+b+c+d\ge\frac{2}{3}(ab+ac+ad+bc+bd+cd)$

Have a look here for a more general result (the given statement is the specianl case $n=4$) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=3463

Pierre.


The largest number by which the expression $ n^3 \minus{} n$ is divisible for all possible integral values of $ n$, is:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

[hide]Factoring gives $ n(n\plus{}1)(n\minus{}1)$. By a parity argument, this is divisible by 2. Also, since the product is of three consecutive integers, it is divisible by 3. Hence, $ n^3\minus{}n$ is divisible by $ 2\cdot 3\equal{}6\Longrightarrow\textbf{(E)}$[/hide]

[hide="Solution"]
$ n^3 \minus{} n \equal{} n(n^2 \minus{} 1) \equal{} n(n \minus{} 1)(n \plus{} 1)$

According to the factorization, one of those factors must be divisble by $ 2$ and one of those factors must be divisible by $ 3$. Thus, the largest number which can divide the above expression is $ \boxed{(E)6}$.
[/hide]


[quote="Art of Owna"]How many 5 digit numbers with all five digits different (and not starting with 0) are divisible by 25?[/quote]
[hide]So we automatically have the last digit in place. we just need the other 4. In order to be divisible by 25 the last two digits have to be divisible by 25. So the only two that can work are 2 and 7. Now we have two cases.

First. case, the fourth digit is 2. That leaves the other 3 open for a choice. We start from the first one to the third. Since they have to de different digits there are only 7 choices for the first digit. The second digit has also 7 choices because we can use 0. The third digit has 6. So 7*7*6=294.

Second case, the fourth digit is a 7. This should exactly result like the first case. So it has 294 ways.

We add both up to get 588 as our final answer.[/hide]

[quote="chickendude"][hide]
to be divisible by $25$, the last two digits must be $00$, $25$, $50$, or $75$

The last two digits can't be $00$ because the digits must be unique

for the last two digits being $25$ or $75$
First digit has $7$ possibilities (10 digits, two already used, can't be zero)
Second digit has $7$ leftover possibilities (now that 0 can be used)
Third digit has $6$ possibilities
4th/5th digit combo has 2 possibilities: 25, 75
$7 * 7 * 6 * 2 = 588$

If the last two digits are $50$
there are
$8 * 7 * 6 = 336$ possibilities for the first three digits

$336+588 = \boxed{924}$
[/hide][/quote]
Oops, I missed the other two options.  :blush:
I think your answer is right.


in triangle ABC let  A1 , B1 ,C1 be the foot of the bisectors of A ,B ,C . let D  be the intersection
of AA1 and C1B1  and E the intersection of CC1 and A1B1 . prove that BB1 is the bisector of
angle <DBE

Let $ W,K,L$ be the feet of the angle bisectors issuing from $A,B,C$ and $ D,E,F$ the points where $CL,AW,BK$ meet $WK,KL,LW,$ respectively. By cevian nest theorem, we deduce that $ \triangle EFD$ and $ \triangle ABC$ are perspective and the perpective axis coincides with the perspective axis of $ \triangle ABC$ and $ \triangle WKL.$ Hence, the lines $ LK, DF$  and $ BC$ concur at the foot $P$ of the A-external bisector.  If $ H \equiv AW \cap DF,$ $ X \equiv AF \cap BC,$ $ Y \equiv AD \cap BC,$ then $ (F,H,D,P)=-1$ implies that $ A(X,W,Y,P)=-1.$ Since $ AP \perp AW,$ then $AW$ bisects $ \angle XAY.$

[quote="v235711"] [color=darkred]In $ \triangle ABC$ let  $ D$ , $ E$ , $ F$  be the feet of the bisectors from $ A$ , $ B$ , $ C$ . Denote

$ X\in BE\cap DF$ , $ Y\in CF\cap DE$ . Prove that $ AD$ is the bisector of $ \widehat {XAY}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ \left\|\begin{array}{ccc}
m(\angle XAB) & = & x\\\
m(\angle YAC) & = & y\end{array}\right\|$ . Prove easily  that $ \left\|\begin{array}{ccc}
\frac {BD}{BF} & = & \frac {b+a}{b+c}\\\\
\frac {CD}{CE} & = & \frac {c+a}{c+b}\\\\
\frac {AE}{AF} & = & \frac {a+b}{a+c}\end{array}\right\|$ . Therefore,

$ \left\|\begin{array}{c}
\frac {b+a}{b+c}=\frac {BD}{BF}=\frac {XD}{XF}=\frac {AD}{AF}\cdot\frac {\sin\left(\frac A2-x\right)}{\sin x}\\\\
\frac {c+a}{c+b}=\frac {CD}{CE}=\frac {YD}{YE}=\frac {AD}{AE}\cdot\frac {\sin\left(\frac A2-y\right)}{\sin y}\end{array}\right\|\ (: )\ \implies$ $ \frac {a+b}{a+c}=\frac {AE}{AF}\cdot\frac {\sin \left(\frac A2-x\right)}{\sin x}\cdot\frac {\sin y}{\sin\left(\frac A2-y\right)}$ $ \implies$

${ \sin \left(\frac A2-x\right)\cdot\sin y}=\sin\left(\frac A2-y\right)\cdot\sin x$ $ \implies$ $ \cos\left(\frac A2+y-x\right)=\cos\left(\frac A2+x-y\right)$ $ \implies$ $ x=y$ .[/color]

my solution uses MD and NE , parallel to BB1, with M and N on sides AB and CB , respectively.
the goal is to prove that triangles MDB and NEB are similar.
then it´s just simple computations :lol:

[color=darkred][b][u]An easy extension[/u].[/b]

Let $ ABC$ be a triangle and denote the foot $ D\in (BC)$ of the $ A$ - bisector.

For a point $ P\in (AD)$ denote $ E\in BP\cap AC$ , $ F\in CP\cap AB$ and

$ X\in BE\cap DF$ , $ Y\in CF\cap DE$ . Prove that $ \widehat {DAX}\equiv\widehat {DAY}$ .

[b][u]A strong extension[/u].[/b]

Giving are $ \triangle ABC$ and an interior point  $ P$ . Denote $ \left\|\begin{array}{c} D\in AP\cap BC \\
 \\
E\in BP\cap AC \\
 \\
F\in CP\cap AB\end{array}\right\|$ , $ \left\|\begin{array}{c} X\in BE\cap DF \\
 \\
Y\in CF\cap DE\end{array}\right\|$

and $ \left\|\begin{array}{ccc} m(\angle BAX) \equal{} x & ; & m(\angle CAY) \equal{} y \\
 \\
m(\angle PAX) \equal{} u & ; & m(\angle PAY) \equal{} v\end{array}\right\|$ . Prove that $ \frac {\sin (x \plus{} u)}{\sin (y \plus{} v)} \equal{} \frac {\sin x\sin v}{\sin y\sin u}$ .

For $ x\plus{}u\equal{}y\plus{}v\equal{}\frac A2$ obtain the easy extension.[/color]

[quote="Virgil Nicula"][color=darkred]

[b][u]A strong extension[/u].[/b]

Giving are $ \triangle ABC$ and an interior point  $ P$ . Denote $ \left\|\begin{array}{c} D\in AP\cap BC \\
 \\
E\in BP\cap AC \\
 \\
F\in CP\cap AB\end{array}\right\|$ , $ \left\|\begin{array}{c} X\in BE\cap DF \\
 \\
Y\in CF\cap DE\end{array}\right\|$

and $ \left\|\begin{array}{ccc} m(\angle BAX) = x & ; & m(\angle CAY) = y \\
 \\
m(\angle PAX) = u & ; & m(\angle PAY) = v\end{array}\right\|$ . Prove that $ \frac {\sin (x + u)}{\sin (y + v)} = \frac {\sin x\sin v}{\sin y\sin u}$ .[/color][/quote]

Apply [u][b]Ceva[/b]'s theorem (trigonometrical form)[/u] to :

$ \left\| \begin{array}{cccccc} X/\triangle ABD & : & \frac {\sin x}{\sin u}\cdot\frac {\sin(\angle ADF)}{\sin(\angle BDF)}\cdot\frac {\sin(\angle DBP)}{\sin(\angle PBA)} = 1 & \Longleftrightarrow & \frac {\sin x}{\sin u} = \frac {\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac {\sin(\angle ABE)}{\sin(\angle CBE)} & (1) & \\
Y/\triangle ADC & : & \frac {\sin v}{\sin y}\cdot\frac {\sin(\angle ACP)}{\sin(\angle DCP)}\cdot\frac {\sin(\angle CDE)}{\sin(\angle EDA)} = 1 & \Longleftrightarrow & \frac {\sin v}{\sin y} = \frac {\sin(\angle BCF)}{\sin(\angle ACF)}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)} & (2) & \\
P/\triangle ABC & : & \frac {\sin(x + u)}{\sin(y + v)}\cdot \frac {\sin(\angle ABE)}{\sin(\angle CBE)}\cdot\frac {\sin(\angle BCF)}{\sin(\angle FCA)} = 1 & \Longleftrightarrow & \frac {\sin(x + u)}{\sin(y + v)} = \frac {\sin(\angle ABE)}{\sin(\angle CBE)}\cdot\frac {\sin(\angle BCF)}{\sin(\angle FCA)} & (3) & \end{array}\right\|$

$ \stackrel{(1)\ \odot\ (2)}{\implies}\ \frac {\sin x}{\sin u}\cdot\frac {\sin v}{\sin y} = \frac {\sin(\angle ABE)}{\sin(\angle CBE)}\cdot\frac {\sin(\angle BCF)}{\sin(\angle FCA)}\cdot\frac {\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)}$

$ \color{white}{.}$   $ \stackrel{(3)}{\Longleftrightarrow}\ \ \ \ \frac {\sin x}{\sin u}\cdot\frac {\sin v}{\sin y} = \frac {\sin(x + u)}{\sin(y + v)}\cdot\frac {\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)}\ \ (4)$

Also we have :

$ \left\|\begin{array}{ccc} \frac {BF}{FA} = \frac {BD}{AD}\cdot\frac {\sin(\angle BDF)}{\sin(\angle ADF)} \\
 \\
\frac {AE}{EC} = \frac {AD}{DC}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)} \end{array} \right\|\bigodot\implies$ $ \frac {BF}{FA}\cdot\frac {AE}{EC} = \frac {BD}{DC}\cdot\frac {\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)}$

$ \stackrel{Ceva}{\Longleftrightarrow}\frac {\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac {\sin(\angle ADE)}{\sin(\angle CDE)} = 1\ \ (5)$

$ \stackrel{(4)\ \wedge\ (5)}{\implies} \boxed{\frac {\sin x}{\sin u} \ \cdot \ \frac {\sin v}{\sin y} \ = \ \frac {\sin(x + u)}{\sin(y + v)}}$ .


Let $S$ be the set of points in the Cartesian plane that satisfy
\[|\parallel x|-2|-1|+|\parallel y|-2|-1|=1. \]
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b},$ where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number.  Find $a+b$.

hint: break x,y into intervals to get rid of the absolute value signs

[quote="usaha"]you have 2 problems in 1..[/quote]

huh?

my hint -> do it -> make a graph (actually try...if x=<-3 then...if -3<x<-2)


Ένα φώς τροχαίας λειτουργεί με ν διακόπτες. Κάθε διακόπτης έχει τρεις στάσεις (κόκκινο,κίτρινο και πράσινο). Οι διακόπτες είναι με τέτοιον τρόπο συνδεδεμένοι ώστε

- Αν όλοι οι διακόπτες αλλάξουν στάση, τότε και το φως της τροχαίας αλλάζει χρώμα
- Αν όλοι οι διακόπτες βρίσκονται στην ίδια στάση, τότε και το φως έχει το ίδιο χρώμα

Να δειχθεί ότι υπάρχει ένας διακόπτης "δικτάτορας", ο οποίος (αυτός και μόνο αυτός) ρυθμίζει το φως της τροχαίας.

Εξηγεις λιγο καλυτερα. Για παραδειγμα αν ν=4 και ο πρωτος διακοπτης δειχνει κοκκινο ενω οι αλλοι 3 δειχνουν πρασινο το φως δεν θα μπορουσε να ειναι κιιτρινο? Σε αυτη τη περιπτωση δεν υπαρχει διακοπτης δικτατορας.

Εκτος αν δεν εχω καταλαβει πως εννοεις τον διακοπτη δικτατορα. Εγω τον καταλαβαινω σαν ενα διακοπτη του οποιου το χρωμα το εχει παντα και το φαναρι...

[quote="filosofimenos"]Εξηγεις λιγο καλυτερα. Για παραδειγμα αν ν=4 και ο πρωτος διακοπτης δειχνει κοκκινο ενω οι αλλοι 3 δειχνουν πρασινο το φως δεν θα μπορουσε να ειναι κιιτρινο?
[/quote]

Όχι. Αν όλοι είναι κίτρινοι τότε το φως είναι κίτρινο. Αν τώρα όλοι αλλάξουν χρώμα και γίνει ο πρώτος κόκκινος και οι άλλοι τρεις πράσινοι, τότε η πρώτη συνθήκη λέει πως το φως πρέπει να αλλάξει χρώμα και άρα δεν θα είναι κίτρινο.

[quote="filosofimenos"]Εκτος αν δεν εχω καταλαβει πως εννοεις τον διακοπτη δικτατορα. Εγω τον καταλαβαινω σαν ενα διακοπτη του οποιου το χρωμα το εχει παντα και το φαναρι...[/quote]

Σωστά το κατάλαβες.

Ομολογω οτι δεν καταλαβαινω το προβλημα, κατι θα χανω :( . Για παραδειγμα αν [b]αρχικως[/b] οι μισοι διακοπτες ειναι πρασινοι και οι μισοι κοκκινοι ποιος μας λεει οτι το φως δεν ειναι κιτρινο?? Υπο αυτη την προυποθεση θα μπορουσε να μην υπαρχει διακοπτης δικτατορας αφου αν καναμε τους πρασινους κοκκινους και τους κοκκινους κιτρινους το φως ισως γινοταν πρασινο κτλ. Με αυτον τον τροπο κανενας διακοτης δεν εχει το ιδιο χρωμα με το φως...

Τι χανω? :fool:

...και καλη ανασταση ευχομαι σε ολους. :)

ΟΚ καταλαβα τι χανω απλα το προβλημα ειχε μια επιπλεον δυσκολια στο να το κατανοησεις πριν το λυσεις. :)

Νομιζω οτι πρεπει να παρουμε περιπτωσεις και να ανακαλυψουμε παραδοξα στην περιπτωση που δεν υπαρχει δικτατορας. Επισης αν υπαρχει δικτατορας αυτος θα ειναι μονο ενας γιατι αν ηταν πρισσοτεροι θα μπορουσαμε(ευκολα) να επιλεξουμε τις στασεις τους ετσι ωστε το φαναρι να εχει 2 χρωματα ταυτοχρονα, ατοπο.

Να βάλω κάποια (βοηθητικά ελπίζω) υποερωτήματα

α) Να δειχθεί ότι σε κάθε στάση των διακοπτών υπάρχει τουλάχιστον ένας διακόπτης που βρίσκεται στην ίδια στάση με το φως.

β) Να δειχθεί ότι υπάρχει μια στάση διακοπτών ώστε ακριβώς ένας διακόπτης να βρίσκεται στην ίδια στάση με το φως.

γ) Να δειχθεί ότι ο διακόπτης του (β) είναι δικτάτορας.

Ayto fainetai na exei ksexastei. Tha balei kapoios mia lysi...

Δίνω τις απαντήσεις στα υποβοηθητικά ερωτήματα

α) Έστω το φως είναι κόκκινο αλλά όλοι οι διακόπτες είναι κίτρινοι και πράσινοι. Κάνω όλους τους διακόπτες κόκκινους. Τότε το φως πρέπει να αλλάξει στάση (αφού όλοι οι διακόπτες άλλαξαν στάση), άτοπο. Ομοίως και για τις άλλες περιπτώσεις.

β) Έστω $ k$ ο μικρότερος ακέραιος ώστε σε κάθε στάση των διακοπτών υπάρχουν τουλάχιστον $ k$ διακόπτες που έχουν το ίδιο χρώμα με το φως. Υποθέτω ότι $ k \geqslant 2$. Έστω μια στάση όπου ακριβώς $ k$ διακόπτες έχουν το ίδιο χρώμα με το φως, ας πούμε (χωρίς βλάβη της γενικότητας) κόκκινο. Αν αλλάξουμε όλους τους κίτρινους/πράσινους διακόπτες σε κόκκινους ένα κόκκινο διακόπτη σε κίτρινο και τους υπόλοιπους κόκκινους διακόπτες σε πράσινους τότε το φως αλλάζει χρώμα και παίρνουμε μια στάση όπου είτε 1 είτε κ-1 διακόπτες έχουν το ίδιο χρώμα με το φως. Άτοπο. Άρα κ=1.

γ) Χωρίς βλάβη της γενικότητας υποθέτω ότι υπάρχει μια στάση όπου το φως είναι κόκκινο και μόνο ο πρώτος διακόπτης είναι κόκκινος. Κάνω τον πρώτο διακόπτη πράσινο και τους υπόλοιπους κόκκινους. Τότε δεν μπορεί το φως να είναι κόκκινο. Δεν μπορεί να είναι ούτε κίτρινο διότι όταν όλοι οι διακόπτες είναι κίτρινοι τότε το φως είναι κίτρινο. Άρα όταν ο πρώτος διακόπτης είναι πράσινος και όλοι οι υπόλοιποι κόκκινο τότε το φως είναι πράσινο. Ομοίως όταν ο πρώτος είναι κίτρινος και όλοι οι υπόλοιποι είναι κόκκινοι τότε το φως είναι κίτρινο. Άρα όταν ο πρώτος είναι κόκκινος και οι υπόλοιποι πράσινοι/κίτρινοι (σε οποιοδήποτε συνδυασμό) τότε το φως είναι κόκκινο. Με τον ίδιο τρόπο δείχνουμε ότι όταν ο πρώτος είναι πράσινος και όλοι οι υπόλοιποι κόκκινοι/κίτρινοι τότε το φως είναι πράσινο και όταν ο πρώτος είναι κίτρινος και όλοι οι υπόλοιποι κόκκινοι/πράσινοι τότε το φως είναι κίτρινο.

Έστω τώρα μια στάση όπου ο πρώτος διακόπτης είναι κόκκινος. Κάνοντας τον πρώτο διακόπτη κίτρινο, όλους τους άλλους κόκκινους πράσινους, και όλους τους άλλους πράσινους το χρώμα αλλάζει. Αλλά ξέρουμε ότι το χρώμα γίνεται κίτρινο, άρα το αρχικό χρώμα δεν ήταν κίτρινο. Ομοίως το αρχικό χρώμα δεν ήταν πράσινο, άρα ήταν κόκκινο. Ομοίως όταν ο πρώτης διακόπτης είναι κίτρινος/πράσινος.


Starting on January 1, what month will it be when exactly one-fifth of the year has passed?

Assume a non-leap year with $ 365$ days. Then, exactly $ \frac{365}{5}\equal{}73$ days will have elapsed when a fifth of the year has passed. Thus, the month will be March.  Assuming a leap year does not change the answer.


Do not discuss the School or Chapter rounds until they have been posted on the MATHCOUNTS website...this is usually the first week of March.

The only exception is discussion of places, but please do not associate AoPS screen names with real names.

Again, make sure [b]nothing about difficulty, scores, or specific problems[/b] is discussed on the forum.

[quote="AIME15"]...usually the first week of March.[/quote]

Reminder:  DO NOT discuss the competitions until Moderators or AoPS Administrators have given their clearance.  Normally, this will happen as soon as MATHCOUNTS posts the competitions to their web site.  Normally this is sometime during the week following the last scheduled Chapter competition.

However, there have been heavy snowfalls in some parts of the country.  There is at least one report that a Chapter competition has been delayed into March.  So...

[color=red][b]DO NOT discuss the competitions until Moderators or AoPS Administrators have given their clearance![/b][/color]

I received this update, today, Thursday, March 4, from the MATHCOUNTS Foundation National Staff:

[quote="David Dumas, National MATHCOUNTS Program Coordinator"]We do have several competitions that have been delayed until this weekend.  Assuming there are no further delays we should be releasing the tests on our website Monday or Tuesday.  [ [i]Ed:  March 8 or March 9[/i] ][/quote]
So, that gives us the *earliest* possible dates.  When MATHCOUNTS actually releases the competitions determines when we will permit discussions.  Meanwhile...

[b][color=red]DO NOT discuss the competitions until Moderators or AoPS Administrators have given their clearance![/color][/b]


algebra
polynomial
integration
college contests

Consider the polynomial $P_{n}(x)=\frac{1}{2^{n}.n!}.\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}$
Prove that if $f(x)$ is a polynomial with degree $m(m<n)$ then:$\int\limits_{-1}^{1}f(x)P_{n}(x)dx=0$


:(  i can not figure out how to solve this type of problem using this formula Pn=Pn-1+d

Problem #5 consider a population tht grows according to a linear growth model. the initial population is Po=8, and the population in the 10th generation is P10=38.

a) find the common difference d.

b)find P50

c) give an explict description of the population

I don't know what a "linear growth model" is but I'll assume it means that the population is a linear function of the number of generations (correct me if I'm wrong).

Make sure you understand what the formula means.  It's only saying that each year the population increases by a common difference $d$.  It's not really necessary to use the formula.

If $p_0=8$ and $p_{10}=38$, then the population increases by 30 in 10 years, so $10d=30\Rightarrow d=3$.

From there you can write a function for the population:
$y=3x+8$  (b/c the common difference is 3 and the y-int. is 8) and find that when x=50, y=158.

[hide="Solution"]
We have $P_0 = 8, P_{10} = 38$.  A linear growth model is of the form $P_n = P_{n-1} + d$, as you have already stated.  This gives the following:

$P_1 = P_0 + d$
$P_2 = P_1 + d = P_0 + 2d$
$P_3 = P_2 + d = P_0 + 3d$

And generally, $P_n = P_0 + nd$.  We have $P_{10} = P_0 + 10d$, or, with the numbers in, $38 = 8 + 10d$, and solving gives us $d = 3$, so our equation is $P_n = P_{n-1} + 3$.

$P_{50} = P_0 + 50d = 8 + 50 \times 3 = 158$.

The population is growing by three people per generation.  (This is a very unlikely scenario  :)  )

Also, this is a little easy to be in Intermediate topics.
[/hide]


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
Columbia

Hah, I am first.

not sure about the cutoff, lower than 224 though, higher than 219.5.

I'm pretty sure I got a 228, and my name is not on the list. What makes you think it was 224?

The qualifying score is at most a 9 too.

hmmm...really, now?

a friend of mine (Fountainhead) got a 124 on the 12A and 10 on AIME, his name is on the list.  my 139.5 + 8 didnt make it.

that's how i got the upper/lower bounds.

"qualifying score is XXX" <--- according to website...i guess they dont have stats compiled yet.

It's less than 224, since I got a 224 and am in.

the cut-off is above 220.5 (based on a student of mine who didn't make it)

Actually I know a 229 that didn't make it...

Do you think (*hoping aimlessly*) there may have been a mistake? I made USAMO in tenth grade, missed it last year, and was reasonably confident with 138+9=228 that I had a good chance this year... Wow, I hope there was a mistake.

The cutoff is greater than 230 for Juniors and Seniors.

The ppl who made it above had indices lower than the cutoff, but since they're 9th and 10th graders, all they needed was a 9 on the AIME score.

My sister (8th grader) got a 8 on the AIME, and she didn't make it, so the floor value for the AIME (for underclassmen) is a 9.

Not sure what the qualifying index for upperclassmen is, but someone from our school with a 230 didn't make it.

My index was a 241, and someone with a 240 also made it, so the cutoff is between 231-239.

Cutoff is 233 for AIME I and 220.5 for AIME II. It's on the site now.

yuo sure the floors a nine? I know someone(and hes not junior or seinor) who got an nine and isnt on the list... mayby he made a bubbling error  :(

joml88 made it! THis inspires me as he only made 1 on the AIME last year!

How much did you study, joml? DId you attend any special summer progs?

in that case, i suspect AMC is plain on CRACK or something...or their stats are messed up...

the person with the 224 who qualified is an 11th grader.

i just found out somebody who made it with 136AMC12 and 8AIME, giving him only 216... what the heck???? :?

[quote="xxreddevilzxx"]If you look at the title bar of your explorer on the Qualies site, it says:

2005 - Qualifiers for the [b]2004 [/b]United States of America Mathematical Olympiad

wow this yr AMC must be on CRACK because this is just weird.... :?[/quote]

I see someone about to be kicked out of USAMO.  :D

No, they aren't. It's just from personal connections that we get indices.

[quote="paladin8"][quote="tetrahedr0n"]Maybe that was the state high index?[/quote]

No, I think I know who xxreddevilzxx is referring to... he's in WA and that's definitely not the state-high index if that's what he really got. So maybe he guessed right or something silly like that...[/quote]

yea i think you know who i am talking about, jeffrey. It is just not fair that I thought i was the lone one at my school, and now i find out someone else is... he told me he didnt do well, but o well.

btw, he is not very concerned w/ USAMO, so i guess i can finally beat him :D

[quote="MithsApprentice"]Generally, derogatory comments concerning the mental health of the CAMC is not tolerated on these forums. They work very hard all year long, and making small errors, especially typos, is not a big deal.

As for selection, we do not know for sure if such a person did qualify. The information you give us is an indirect source that has probably gone through several other indirect sources. However, even if such a situation did occur, either the CAMC will resolve it or it was meant to be.[/quote]

I am sorry for such comments. Please erase it for me, moderators. It will not happen again. I am very sorry.

you can delete it yourself btw... click the EDIT button.

[quote="jackb1115"]I'm confused ... are official scores (AMC, AIME, index) of USAMO qualifyers available?  If so, where do I find them

Thx
JB[/quote]

There'll be a book published with the high AMC (but not AIME) scores. From there, you can look up what every USAMO qualifier got or just an individual, if you're that patient. Otherwise, I'm not sure how you could obtain everyone's scores unless you've got connections inside the CAMC.

AIME scores are published.

no, we just had like 25 last year, this year they've got, what, 38?

this line is unnecessary

[quote="MysticTerminator"]AIME scores are published.[/quote]

O.O Oh realli...hmm...must've overlooked.

[quote]
no, we just had like 25 last year, this year they've got, what, 38?[/quote]

Er...I think the 25 is normal. This 38 is being disproportional to previous results.

Yeah, all the fault of the stupidly easy AIME :P.


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204817]here[/url].  You're talking about the same user (probably a spambot.)

I have seen their posts in another thread also.


$P$ is a point inside of a rectangle $ABCD$, such that $PA = 3$, $PC = 5$, $PD = \sqrt{18}$.
Find $PB$.

Apply the theorem of the median in the triangles $APC$ and $BPD$ and obtain the relation $PA^{2}+PC^{2}=PB^{2}+PD^{2}\ .$

[Moderator edit: See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=143458#143458]http://www.mathlinks.ro/Forum/viewtopic.php?t=22321 post #2[/url] Lemma 1 for a more detailed proof. - Darij]

[hide="click to reveal hidden content"]since 3 squared + 5 squared = $\sqrt{18}$ squared + PB squared

PB therefore = $\sqrt{16}$ = 4

i heard it's called the british flag theorem.

I hope I didn't do anything wrong. [/hide]


Find the greatest power "a" of number 3 so that $3^a$  is divisor of $100!$

[hide="hint"]
all you have to do is to look for how many numbers 1-100, 3 is a divisor of.[/hide]

[quote="amirhtlusa"]
all you have to do is to look for how many numbers 1-100, 3 is a divisor of.[/quote]
Yes but i don't understand why? The number 3 divides 33 numbers 1-100. This is $a$?

Many of these numbers contain more than 1 factor of 3, though... (i.e. 11 multiples of 9 contribute 11 more 3's).
[hide="Answer..."]48[/hide]

Account for all threes by determining the largest power of 3 which divides 100 and then dividing by each power of three so as to account for all threes which occur as so:

$\left\lfloor {\frac{100}{3}}\right\rfloor+\left\lfloor {\frac{100}{(3^2)}}\right\rfloor+\left\lfloor {\frac{100}{(3^3)}}\right\rfloor+\left\lfloor {\frac{100}{(3^4)}}\right\rfloor=33+11+3+1=48$

[quote="Pakman2012"]
$\left\lfloor {\frac{100}{3}}\right\rfloor+\left\lfloor {\frac{100}{(3^2)}}\right\rfloor+\left\lfloor {\frac{100}{(3^3)}}\right\rfloor+\left\lfloor {\frac{100}{(3^4)}}\right\rfloor=33+11+3+1=48$[/quote]

this is exactly what i was talking about ;)


superior algebra
superior algebra unsolved

Here is the problem:
Prove that every element in A_n (the alternating group of degree n) is a product of n-cycles.

Check it by hand for small $n$ and then justify that you can write every $3$-cycle as product of two $n$-cycles. Now use that $A_n$ is generated by $3$-cyles when $n\geq 5$.

Yes that's what I tried earlier, but I haven't succeeded in writing (a b c) as a product of two n-cycles  :(
Could you help?

Yeah thx :)
I see now that it works. I have just started with these permutation groups so haven't got a good feel for it yet.


Hi

I was wondering what your guys' opinions on graduating from college early were...like really early...like at 11 years old...

http://www.nbclosangeles.com/news/local/11-Year-Old-Graduates-From-LA-College.html?yhp=1

Personally, I think it is the worst idea ever invented since the Watergate plot with Richard Nixon.  When you graduate at that young of an age, what are going to do?  I'm pretty sure companies would much rather take a 22 year old who has gone through a rigorous high-school and college curriculum than an 11 year old from a no-name college.  It might seem cool and awesome at first but after you graduate, now what?  Not to mention the social aspect of it too.

But that's just my opinion.

-Johnny

Its not about 'getting a job'. People like him go to college that young dont do it because they are looking 'to get a good job'. They do it because they are just working at the rate of which they learn and/or want to learn. Its simply not about money, its about further ones knowledge, about learning new things.

The boy commented on him liking "astrophysics", and he could certainly apply for numerous positions in top research faculties.

I would stay in school longer and learn more things.

I'm going to agree with jli. Going to college early and/or skipping a large number of grades is a horrible idea. Regardless of what people say, life is about happiness. This kid, going through school at such a rapid rate, will lose the childhood that he could've had, if he had just gone to a more advanced private school or just taken classes outside of school.

And that's another thing.  If the kid decided at age 8 that instead of playing with Legos, he would rather go to college because the elementary school curriculum was boring, then fine.  That is awesome.  There are bound to be kids like that, who are eager to excel all by themselves and require no parental pressure, like David Yang :) .

But I mean c'mon, how many 11 year olds do you hear saying this "I feel it's a waste of time playing video games because it's not helping humanity in any way" about video games?  Even David Yang talks about Starcraft nonstop.

There is clearly a fair amount of pressure from the parents.  What I'm trying to say is this: if the kid genuinely wanted to go to college, I applaud that.  But it makes me sick to see parents pushing their kids like this.

Also, my friend went to an early entrance program and started his first year of college after 7th grade.  And I quote my friend on this "Some of the kids in that program were a little bit strange, if you know what I mean".  By no means is he trying to be rude.  But when these kids get into the real world, things will be especially tough for them since they will not have had the time to mingle with others that are their own age.  Nowadays, colleges are looking for people who can talk, not just academic geniuses which shows the importance of being able to communicate with others. I just don't think going through college at that young of an age could teach a kid that.

For what it's worth, it seemed to work out OK for Charles Fefferman.

Touche Dr. Merryfield, touche.

However, there will always be success stories.  A lot of people make money from gambling, but it doesn't mean you should do it.  Bill Gates dropped out of college and became a billionaire, doesn't mean you should drop out of college.

Okay. I think that both sides can agree that there are both 'cases', people that work for both sides of the argument. (And kids, Bill Gates was special. Don't try dropping out of college. Seriously. )

I agree that if a person wanted to pursue higher education at a younger age, that is entirely his or her option.  If they wish to go through school at the 'normal'/'preferred' pacing, that is also their choice. The reprecussions have shown to be different, but this, again, depends largely on the type of person.  If they lack the proper social skills to go through life due to an accelerated program, this is one con.  If they stretch themselves to their full potential by being exposed to more of their peers (even with an age difference), this could be a positive aspect to this argument.

I wouldn't call 'accelerated learning' the worst idea ever since...Nixon's blunder. If it is truly beneficial to a child's learning and the cons outweigh the pros, if a child is truly motivated to do that, they go them! IF they can pursue that and are completely willing to, I think they should have a strong support program around them, which is generally helpful; especially for those that argue about the possibilities of a type of social gauche.

I can give two examples:
A looong time ago, my parents were presented with the idea of accelerating through school.  I would skip most of elementary, enter middle school and high school at an age where reaching the top shelf of my locker would be a stretch (still kinda is, but...).  However, my parents decided that in order for my 'intellectual development' to fully mature, I should be with kids my own age. They have since taken that idea back, but feel that I shouldn't rush high school, and from where I stand, I'm starting to feel the same.  However, at times, I do feel limited, but I supplement my learning by challenging myself outside of school. Like with AoPS.

Another example is a kid from my school.  He is 'accelerated' in some, but not all of his classes by choice.  He has already taken Honors Biology and Physics, is taking Calculus 3 at a local college, starting this fall, Chemistry H., some AP classes, etc.  He has scored well on stuff like the ACT and SAT, and has shown himself to be an excellent musician and very creative with some of the more abstract concepts of mathematics.  This is where he was not entirely accelerated, he will share some of the 'required' classes from he school.  He is actually considered to be in my grade, and although he has friends from all grades, he can get along with those around his age level the best, save for the minor gaffe. I would call his social levels about the same as mine, close enough to cooperate, but not completely attached.  Well enough to get around in the "real" world, though.

Are both 'standard' cases? No. I believe every case should be taken individually, as with this new 11 year old graduating from college.  If that's his decision, those around him should respect it, and not have the media make a spectacle of his graduation.

Yeah, see it says he graduated from a community college...
Seems to me that he kinda screwed up his childhood, like going to college instead of having fun as a kid. Honestly, i have never met an 11 year old who hates video games and says they are useless.
Also, i agree that instead of rushing through school, i would rather really learn more.

[quote="eli140"]
Also, i agree that instead of rushing through school, i would rather really learn more.[/quote]

But you have to consider that sometimes this is the only choice for people at schools where the teachers don't even know the most well known mathematics Olympiad in their country! For example, my teachers didnt even know that there was an Australian Mathematical Olympiad, and as such i was not aware of it, so the only way for me to further myself mathematically was through acceleration, but luckily there is still time for me to be able to do these things now that i have found out  :lol:

Yeah.
The people are my school aren't really mathematically aware nor inclined.  Some just do it because they can and/or are decent. It's kind of...disappointing?

Anyways, back on the topic, I am interested to see if he will be classified as a prodigy or a genius.
To me, they're kind of different as a prodigy is one who absorbs quickly what other people know and/or have discovered; a genius is where you come with things that people have never really thought about before or look at something in a particularly inventive way, a unique thing.

...I would say it's hard to tell at his stage...

Yeah, i would kinda like to meet him.
One more thing is that sometimes, younger kids just dont think the way older kids do, like i remember when i was little, i was so excited to know what DNA stood for and i just didn't care what it meant, i was just excited to know the word. Also, it seemed to me that when i was little, i didnt retain much information, i just forgot.
I don't know if this is the same situation for the rest of you, probably not for the 11 year old who graduated college but it was for me.

[quote="eli140"]Honestly, i have never met an 11 year old who hates video games and says they are useless.
[/quote]

Well, I'm 12 and I've never really enjoyed video games (they [i]are[/i] useless), but it doesn't mean I don't enjoy doing anything fun. I still like watching sci-fi movies and reading for pleasure and having sleep-overs with friends, etc. The point I'm trying to make is that just because he doesn't like video games doesn't mean he never has any fun.

That said, I do agree that his childhood is pretty messed up. If you started college when you were eight years old, you can't have too many friends that you can relate to. I also have a feeling that at least one reason he went to a community college is that as an eight - to - eleven year old, you can't really be expected to take care of yourself for 3-4 years, and thus he went to a college that would allow him to live at home/go home frequently. If I were him, I would have waited at least until I hit my mid-teens to start college, so that I could go to the best school I would be able to get into.

If a kid is able to and genuinely wants to start taking community college courses at 8 years old, then forcing him/her to stay back and be "normal" is going to make the kid really unhappy. It's not like this kid's experience with kids his age would be the same as a "normal" kid's experiences with kids their age would be, so you can't make policy recommendations based on what a "normal" kid would be missing out on, or the "childhood he could've had".

If I haven't made my point obvious enough, there's a difference between kids who are normal and kids who are not normal, and this difference needs to be taken into consideration before making policy recommendations.

It seems he'll take a year off to do martial arts, play piano, and possibly mingle with friends, before going on to study at a top tier institution. Sounds good to me. Good luck to him!

About the parents -- it's their job to guide their kid in the right direction without compromising his goals, dreams, and enjoyment. If they discourage him from playing video games, and he doesn't feel robbed, then what could be wrong with that? I'd discourage my kids from playing video games, that doesn't make me a Nazi or something.

I think it is a good idea, but only if you are ready. I don't like video games, because they are useless (I don't mean to offend anybody). They don't benefit you in any way. I do play them, but only on airplanes, or on long trips. In those cases, if I don't play video games, I am not doing anything better than playing video games. If I am at home, then I never play video games.

About the early college thing, I think it is good if you don't rush it. For example, if you start studying it from when you are in middle school, and you review it every year, you are likely not to forget it. If you study it once for one month, you are more likely to forget it. The point is, you have to put the same amount of time, but you have to start early.

I will say that on a fairly regular basis I hear about extremely young college students - usually tales relayed to me by people who want to make an "OMG, what an amazing genius" point - and often the child is supposed to be a math or physics or astronomy major. At the same time, I not only know about but also work with those students in my area who form the elite on high school mathematics contests (which means, by definition, that they aren't in college yet). The latter group of students often do take college math courses - but not a whole college schedule.

I'm generally more impressed with the latter group than the former group. Admittedly that may be a bias engendered by personal relationships - but at the same time I think it's a lot easier to get an A in a college course than it is to get a 10+ on the AIME.


Given $ \bigtriangleup ABC$ ,and  $ \angle B \equal{}30^{ \circ}$ ,$ \angle A \equal{}30^{ \circ}$,$ \angle C \equal{} 30^{ \circ}$,$ D \in (BC)$,$ E\in (AC)$ such that $ AC\equal{}DC$,$ EC\equal{}BD$.Find $ \angle EBC$

_______________________________________
Azerbaijan Land of the Fire    :P

[quote="Pirkuliyev Rovsen"]Given $ \bigtriangleup ABC$ ,and  $ \angle B \equal{} 30^{\circ}$ ,$ \angle A \equal{} 30^{\circ}$,$ \angle C \equal{} 30^{\circ}$,[/quote]  :huh:

Could you fix this? thanks

Sincerely T.H

Given $ \bigtriangleup ABC$ ,and  $ \angle B \equal{}30^{ \circ}$ ,$ \angle C \equal{} 20^{ \circ}$,$ D \in (BC)$,$ E\in (AC)$ such that $ AC\equal{}DC$,$ EC\equal{}BD$.Find $ \angle EBC$
 :?:

Take $ F$ on (CA, produced so that $ CF \equal{} BC$, see that $ AF \equal{} BF$ and $ AF \equal{} BD$, so $ CE \equal{} BF$. Construct the equilateral triangle $ \triangle BFG$, $ G$ inside the $ \triangle BFC$, see that $ \triangle BCE \sim \triangle CBG$, hence $ \measuredangle CBE\equal{}10^\circ$.

Best regards,
sunken rock


You are given a game of Tic Tac Toe

You have to write a function in which you pass the whole game and name of a player. The function will return whether the player has won the game or not. First you to decide which data structure you will use for the game. You need to tell the algorithm first and then need to write the code.

 Note: Some position may be blank in the game। So your data structure should consider this condition also.

Well the obvious choice is an array

Another possibility is to represent the board as a binary number, allowing you to use bitmasks to determine victory.

I don't think you can use binary. I think you'd have to use base 3, because there are three states: 'X', 'O', and empty square.

The easiest to think up is a 2-D array, if you don't know how to use bitmasks.

It's almost certainly faster to use 2 bits per square then to attempt bit packing.

Besides, even with 2 bits per square, it will still fit comfortably inside a 32 bit word. If you're on a 16 bit system I guess you'd have to be more creative.


Let $ P$ be the point inside triangle $ ABC$. The line through $ P$ perpendicular to $ AP$ intersects $ BC$ by the point $ A_1$. Define the same for $ B_1,C_1$. Prove that: $ A_1,B_1,C_1$ are collinear. :lol:
[b]Reference[/b]:http://forumgeom.fau.edu/FG2003volume3/FG200301.pdf

Circles $ (X), (Y), (Z)$ with diameters $ AA_1, BB_1, CC_1,$ centered on midlines $ B'C', C'A', A'B'$ of $ \triangle ABC,$ intersect at $ P.$ Powers of $ P$ to $ (X), (Y), (Z)$ are all zero and powers of orthocenter $ H$ to $ (X), (Y), (Z)$ are all equal to half the power of $ H$ to the circumcircle $ (O)$ $ \Longrightarrow$ $ (X), (Y), (Z)$ are coaxal, $ X, Y, Z$ collinear. By Menelaus theorem for $ \triangle ABC$ and its medial $ \triangle A'B'C',$ $ \frac {\overline{A_1B}}{\overline{A_1C}} \cdot \frac {\overline{B_1C}}{\overline{B_1A}} \cdot \frac {\overline{C_1A}}{\overline{C_1B}} \equal{} \frac {\overline{XC'}}{\overline{XB'}} \cdot \frac {\overline{YA'}}{\overline{YC'}} \cdot \frac {\overline{ZB'}}{\overline{ZA'}} \equal{} 1$ $ \Longrightarrow$ $ A_1, B_1, C_1$ are collinear. $ A_1B_1C_1$ is orthotransversal of $ P.$ (Posted before.)

Another solution (Using inversion:D)
Consider the inverseion pole $ P$, power $ k^2$, where $ k$ is some real number, we get, $ I(P,k): \overline {ABC_1} \mapsto (A'B'C'_1P)$, $ \overline {ACB_1}\mapsto (A'C'B'_1P)$, $ \overline {BCA_1}\mapsto (B'C'A'_1P)$. Then, it is easy to notice that $ PC'\bot PC'_1$, $ PA'\bot PA'_1$, $ PB'\bot PB'_1$. Therefore, in order to prove the fact $ A_1,B_1,C_1$ are collinear, we just need to prove that $ P,A_1,B_1,C_1$ are concyclic. Hence, we will prove the bisectors of $ PB'_1, PA'_1, PC'_1$ concurrent.
Denote $ O_1,O_2,O_3$ be the circumcenters of $ (A'C'B'_1P)$, $ (B'C'A'_1P)$, $ (A'B'C'_1P)$, respectively. Because $ O_2O_3\bot PB'$, $ PB'_1\bot PB'$, hence $ O_2O_3 //PB'_1$. Therefore, the bisector of $ PB'_1$ is the perpendicular line from $ O_1$ onto $ O_2O_3$. With the same argument, we have the bisectors of  $ PC'_1,PA'_1$  is the rest altitudes of triangle $ O_1O_2O_3$, therefore, they concurrent at the orthocenter of triangle $ O_1O_2O_3$, which implies the result of the problem :lol:


Sorry Virgil,but I think this problem is not true :( .
Becous if we take $a\to 0,b=c\to \sqrt{2}$=>$(a+b+c)\to 2\sqrt{2}<3$ :(

I am sorry ! I forgot a condition. I shall come back immediately.

Tiks, here is the transforming (in a algebraic inequality) of the geometrical inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=63801 (I think I didn't err !):

$\blacksquare\ 1^{\circ}.$ [b][u]Acute[/u][/b] $\triangle ABC\Longrightarrow\sum \sqrt{\frac{b+c-a}{a}}\ge 3$
$(2x=b+c-a,\ 2y=c+a-b,\ 2z=a+b-c)$
$\blacksquare\ 2^{\circ}.\ x,y,z>0,\ \frac{xyz}{x+y+z}\le \min \{ x^2,y^2,z^2\}\Longrightarrow\sum \sqrt{\frac{x}{y+z}}\ge \frac{2\sqrt 2}{2}.$
$(x,y,z>0,\ x+y+z=1)$
$\blacksquare\ 3^{\circ}.\ x,y,z>0,\ x+y+z=1,\ xyz\le \min \{x^2,y^2,z^2\}\Longrightarrow$$\sum \sqrt{\frac{x}{1-x}}\ge \frac{3\sqrt 2}{2}.$
$(x=\frac{m^2}{m^2+2},\ y=\frac{n^2}{n^2+2},\ z=\frac{p^2}{p^2+2})$
$\blacksquare\ 4^{\circ}.\ m,n,p>0,\ \frac{mnp}{\sqrt{(m^2+2)(n^2+2)(p^2+2)}}\le \min \left \{ \frac{m^2}{m^2+2},\frac{n^2}{n^2+2},\frac{p^2}{p^2+2}\right \},$
$\sum \frac{1}{m^2+2}=1\Longrightarrow m+n+p\ge 3.$

[quote="Virgil Nicula"]Tiks, here is the transforming (in a algebraic inequality) of the geometrical inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=63801 (I think I didn't err !):[/quote]
Yes,Virgil.I see it :) .


inequalities
induction
algebra solved
algebra

consider the sequence $a_n$ defined by

\[
a_n= \prod_{k=1}^n\sqrt[2^{k-1}]{k}
\]

(that is
\[
a_n=\sqrt{2\sqrt{3\sqrt \hdots \sqrt{n}}}
\]
)

show that $ a_n $ is increasing and bounded (easy) and find its limit for $n \rightarrow +\infty$ (dunno, but i think it's quite hard)

I think I'm missing something... 1st half

[hide]

It's always increasing since if it wasn't then the xth root of a number greater than one would be less than one which is inpossible.  Proof its bounded by contradiction.

Suppose it was unbounded.  Then for some n, there would be a term greater than 10.

sqrt(2)*sqrt(sqrt(3))*sqrt(sqrt(sqrt(4)))*sqrt(sqrt(sqrt(...sqrt(n)...))) > 10

In the sequence g_x = ((g_(x-1))^2)/x and g_2 = 10, it is clear that g_n will always be greater than n.  But if you take the inequality

sqrt(2)*sqrt(sqrt(3))*sqrt(sqrt(sqrt(4)))*sqrt(sqrt(sqrt(...sqrt(n)...))) > 10

and square and divide repeatedly eventually you will reach n > g_n, which isn't true, so all a_n musrt be less than ten, so it's bounded.[/hide]

seems ok

notice that there is another (maybe simplier) way to prove that the sequence is bounded..

in fact, one can prove by induction that

$\sqrt{m\sqrt{(m+1)\sqrt{\hdots\sqrt{n}}}}<m+1\quad \forall m<n$, and hence

$\sqrt{2\sqrt{3\sqrt{\hdots\sqrt{n}}}}<3$

wich is also a stronger bound ;)

ok, now the second half

Are you sure that the limit can be computed? I mean, are  you positive that the problem shouldn't go in the "Open questions" section? Does your friend have a solution?

errrr...

actually i don't know if the limit can be computed...

i think that you're right, the problem should go to the "open questions" section :blush:

i think the moderators can move the whole discussion...right?


function
limit
advanced fields
advanced fields unsolved

let f:R to R be a lebesque measurable function. Let E denote the set of all points x in R at which f is differentiable
ie.
E={x in R :lim (f(x+h)-f(x))/h exists} as h goes to 0.

prove that E is lebesque measurable.

more generally, if $f_n$ is a sequence of functions, the set of points where $\lim_{n \to  \infty} f_n(x)$ exists is measurable. (try to write it as an intersection of set for example with the Cauchy property of convergence ..)