calculus
integration
complex analysis
trigonometry
complex numbers
complex analysis theorems

Let $f(z)=z+\sum^{\infty}_{n=2}a_nz^n$ a serie with the convergence radius $R\geq 1$ and $a_n\in R$.
Show that if $f$ is injective on the disk of complex numbers with modulus < 1 then $|a_n|\leq n$.

Maybe you could give the first question of the problem, Killer, since like this the problem is a little bit too difficult. It is a theorem of Dieudonne, if I remember correctly and even if the proof is elementary, it is very difficult (IMHO) to find it. By the way, the conclusion remains true if we consider just complex numbers, but the proof is unbelievable difficult.

Do we have to use Cauchy's integral formula?

The generalized case($a_i \in \mathbb{C}$) has been solved only recently (around 1985) told me a friend.

This is a hard problem.  It is actually the real version of the Bieberbach conjecture.  The full version was only proved in recent memory as alekk mentioned.  A proof for the real version goes something like this:

Let $f=u+iv$ and $0<r<1$.  Cauchy's formula says:  $f^{(n)}(0)=\frac{n!}{2 \pi r^{n}}\int_{0}^{2 \pi}f(r e^{i \theta}) e^{-i n \theta}d\theta$.  So we have:

$c_{n}=\frac{1}{2 \pi r^{n}}\int_{0}^{2 \pi}(u(r e^{i \theta})+i v(r e^{i \theta}))e^{-i n \theta}d\theta$.

But we also have:

$0=\frac{1}{2 \pi r^{n}}\int_{0}^{2 \pi}(u(r e^{i \theta})+i v(r e^{i \theta}))e^{i n \theta}d\theta$ since $f(z) z^{n}$ is holomorphic in the unit disk.

Now just subtract the conjugate of the second equation from the first equation:

$c_{n}=\frac{i}{ \pi r^{n}}\int_{0}^{2 \pi}v(r e^{i \theta})e^{-i n \theta}d\theta$.

So $c_{n}=\Re c_{n}=\frac{1}{\pi r^{n}}\int_{0}^{2 \pi}v(r e^{i \theta}) \sin (n\theta) d\theta$.

Bounding this:  $|c_{n}| \leq \frac{1}{\pi r^{n}}\int_{0}^{2 \pi}|v(r e^{i \theta})| |\sin (n\theta)| d\theta \leq \frac{n}{\pi r^{n}}\int_{0}^{2 \pi}|v(r e^{i \theta})| |\sin (\theta)| d\theta$.

$=\frac{n}{\pi r^{n}}\int_{0}^{2 \pi}v(r e^{i \theta}) \sin (\theta) d\theta = \frac{n}{\pi r^{n}}(\pi r c_{1}) = \frac{n c_{1}}{r^{n-1}}=\frac{n}{r^{n-1}}$.

(We can remove the absolute values since $f$ is univalent meaning that it preserves the real line and $v(z)>0$ when $\Im z>0$ and $v(z)<0$ when $\Im z<0$).

Now just let $r\to  1^{-}$ and the result follows.

Finally, this proof is not original to me.

Oral exam, eh?  How did that go anyway?   :rotfl:


[quote="xxreddevilzxx"]I thnk that we should hold an official AoPS Contest, aiming at High school students, not yet at the level of taking the MathLinks contest. I tried to take MathLinks test, but they are WAYYY too hard.[/quote]
I think it will be very difficult to start a new nation-wide contest among high school students. Besides, you already have USAMO, AIME, and other contests...

[quote="bubala"]first, of all, I think reddevilz wants to make a contest for AoPS members.  The questions would be posted online and done in the honor system.  I think it's a great idea--for me (at least for now) the MathLinks contest is waaaaay beyond me.

second of all, I'd note that while difficult, it's not impossible to start a nationwide contest.  Mr. Rusczyk and his old pals Sandor Lehoczky and Sam Vandervelde started the Mandelbrot only a little more than a decade back--I think.  I could be wrong.[/quote]

There's a ton of things that have to be done to make a contest. Mostly comimg up with problems, but also orginazation, planning it, administering it, a ton of other stuff. You might not mean it, but it sort of sounds like you think starting a nation-wide contest isn't too hard.

It might be cool, but it would probably depend a lot on the level of interest in the contest.

It's impossible at this point to have a contest like your suggesting. Mainly because we already are dealing with a few online contests (USAMTS, MathLinks Contest) and because of other projects that are underway.

oh well.  I'll just keep improving my math skills and one day I'll be good enough for the MathLinks one.

[quote="bubala"]no no no, I never said it wouldn't be hard.  It would be a gigantic undertaking.  What I meant was it's not impossible.

At the beginning, I was saying that starting an AoPS contest wouldn't be terribly hard.  Then, I went on to say that although extremely difficult, starting a nation-wide contest is not impossible.

SOrry if that's unclear.[/quote]

Mmm, gotcha.  :)

Also, since you're a middle schooler, you haven't been exposed to a lot of math contests. I think you'll find that there are many more contests available once you're in high school, as long as you're part of your school's math team. Although this may vary from school to school.

There are already [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]many, many contests[/url] out there.


integration
algebra
function
domain
advanced fields
advanced fields unsolved

I'm trying to solve this ODE problem:

let $ x^{(1)} \equal{} \left( \begin{array}{ccc}1\\1\\1\end{array} \right)$,  $ x^{(2)} \equal{} \left( \begin{array}{ccc}1\\t\\t^2\end{array} \right)$,  $ x^{(3)} \equal{} \left( \begin{array}{ccc}1\\t\\t^3\end{array} \right)$

a) Find the Wronskian $ W(x^{(1)}, x^{(2)}, x^{(3)})$

b) You are told that $ x^{(1)}, x^{(2)}, x^{(3)}$ are solutions of a

linear homogeneous system $ x' \equal{} P(t)x$

i) what can you say about the discontinuities of the coefficient $ p_{i,j}(t)$?

ii) Find $ p_{1,1}(t) \plus{} p_{2,2}(t) \plus{} p_{3,3}(t)$.  (Hint: use your solution to part a).

==============================================

For part a), the Wronskian is W = determinant of $ (x^{(1)}, x^{(2)}, x^{(3)})$, which I found to be $ t^4 \minus{}2t^3 \plus{} t^2$.  I'm not sure how to do part b).  What discontinuities is the question refering to, and how do I find them?  How do I use my answer to part a) to solve b)ii)?  Any help is appreciated!


A pirate ship spots, 10 nautical miles to the east, an oblivious caravel sailing 60  degrees south of west at a
steady 12 nm/hour. What is the minimum speed that the pirate ship must maintain at to be able
to catch the caravel?

Note: I think the answer is 12, but could someone confirm this?

[hide="solution"]
The fastest rout between any two points is a straight line. So construct a triangle with one side length $ 10$ and the other length $ 12t$ and an angle between them of $ 120^o$
The distance that the pirate ship travels, $ c$ is $ c^2 \equal{} 10^2 \plus{} (12.t)^2 \minus{} 120.t.cos(120)$
So the speed it is traveling is $ \dfrac{c}{t}$
$ \displaystyle\lim_{t\to\infty}\dfrac{\sqrt {144t^2 \plus{} 60t \plus{} 100}}{t} \equal{} 12$

So $ 12$ m/h is correct[/hide]


m$(d^2 x$/$dt)$ - b$( dx/dt)$ = -a ( a and b and m are constants)

I got the CF as x=A+Be^ (-bt/m)
But i do not know how to  get the PI

Should this be in advanced physic ?I think it is under calculus .Anyway

the auxiliary equation is $mp^2-bp=0$ $\implies p=0,b/m$ So the CF is $x=A+Be^{bt/m}$

To choose PI , we notice that it is $-a$ which is a constant . Hence we should also pick a constant for PI. But notice that in CF, we already have a constant , hence we have to multiply it with $t$ , so our PI is $x=ct$ for constant $c$ . Plugging back into the original equation and solve it to get $c=\frac{b}a$ .

So our equation should looks like

$x=A+Be^{bt/m}+\frac{b}at$ where $A,B$ are arbitrary constant .


function
algebra
domain
Functional Analysis
calculus
calculus computations

I am not really sure if this question belongs here. Here it goes:

I was wondering what's the difference between the following two "mapping" symbols:
$f: A \to B$ vs $f: A \mapsto B$

Also, in how many and in what ways can you read the following notation:
$f \circ g$

$\to$ is used when the thing on the left is the domain and the thing on the right is (a set containing) the range.  Thus, we say $f: \bf R \to C$ is a function from the reals to the complexes.

$\mapsto$ is used for particular values of a function.  For instance, the function $f(x) = x^2 + i$ can be written $f: x \mapsto x^2 + i$.  If $x$ is only allowed to take real values, this function is a function $f: \bf R \to C$, as well.

So $\mapsto$ is used to specify the rule itself whereas $\to$ is used to specify the domain and co-domain?

The notational distinction JBL is making is news to me. Many people use $f: A\mapsto B$ just to show the domain and codomain.

As for swapnillium's second question, I would read $f\circ g$ only as the composition of functions - that is, $f\circ g(x)=f(g(x)).$

That's for writing functions on the left. If you for some reason write functions on the right, then you had better explain your notation at the beginning of what you're writing.

[quote="swapnillium"][quote="Kent Merryfield"]
As for swapnillium's second question, I would read $f\circ g$ only as the composition of functions - that is, $f\circ g(x)=f(g(x)).$
[/quote]
I know that $f\circ g$ means $f(g(x))$ but how do you read it? I mean, when I first saw it, I started calling it "f dot g," but is this the way most people read this notation?[/quote]

I say "f of g of x" as if it was just written as $f(g(x))$.

[quote="paladin8"]I say "f of g of x" as if it was just written as $f(g(x)).$[/quote]
Absolutely. Whatever you call it, you can't call it "dot" because it's not a dot. It's a circle (which is vaguely like an "o", which is the first letter of "of".

If you're feeling slightly more formal, say "f composed with g at x" or "f compose g/"

[quote="Kent Merryfield"]The notational distinction JBL is making is news to me. Many people use $f: A\mapsto B$ just to show the domain and codomain.[/quote]

Hmm, okay.  I feel like that's the way all of my professors have used the two notations, but I suppose that could be not true or just a coincidence.

For the second question, "f composed with g" or "f of g" is how I would say that out loud.  (I don't say "at x" unless there actually is an x there.)

[quote="amcavoy"]Yeah, that's basically what he said.  One thing though is that I don't think it should be called the co-domain as you just mentioned in your question.  If I remember the terms correctly, the co-domain of $f: A\rightarrow B$ is $\textrm{Im}\left(f\right)$ and $\textrm{Im}\left(f\right)$ is only equal to $B$ if $f$ is onto.[/quote]

You have the meaning of co-domain backwards. In $f: A \rightarrow B$, $A$ is the [i]domain[/i] of $f$, $B$ is the [i]co-domain[/i] of $f$, and $\textrm{Im}\left(f\right)$ is the [i]range[/i] of $f$. The co-domain is a [i]superset[/i] of the range.

If $\textrm{Im}\left(f\right) = B$, then $f$ is said to be [i]onto[/i] (or surjective)

[quote="amcavoy"]Yeah, that's basically what he said.  One thing though is that I don't think it should be called the co-domain as you just mentioned in your question.  If I remember the terms correctly, the co-domain of $f: A\rightarrow B$ is $\textrm{Im}\left(f\right)$ and $\textrm{Im}\left(f\right)$ is only equal to $B$ if $f$ is onto.[/quote]

Wait, are you sure? I mean, for example, let $f(x)=x^2$ and we can describe the input and output of this function as $f: \mathbb{R} \to \mathbb{R}$, right? Then, don't we then mean that $f: domain \to codomain$ since $\mathbb{R}$ is not really the range but the co-domain of $f$?

[quote]Kent Merryfield wrote:
[quote]The notational distinction JBL is making is news to me. Many people use  just to show the domain and codomain.[/quote]

Hmm, okay. I feel like that's the way all of my professors have used the two notations, but I suppose that could be not true or just a coincidence. [/quote]

I've seen this notational distinction explicitly stated in a book before.  So it's a real thing, but maybe it hasn't become standard yet.

I think Stein uses them as JBL said, but that's the only text where I remember any kind of distinction.

It doesn't really matter so long as the author makes clear the meaning of his notation when he first uses it (and if it isn't too bizarre).

Well, that makes me feel better :)

[quote="swapnillium"]anyone else have anything to say in this matter?[/quote]
Read the post above yours.

I am sorry for the confusion. Here's my main question:
In general when we write $f: domain \to  ?$, do we actually mean $f: domain \to codomain$ or $f: domain \to range$?

The co-domain is the set [i]into[/i] which a function maps its domain.

The range is the set [i]onto[/i] which a function maps its domain.

gauss202, you are not answering his question. If you see this $f: A\to B$, would you interpret it as

A) $f: \mathrm{domain}\to\mathrm{codomain}$;

B) $f: \mathrm{domain}\to\mathrm{range}$;

C) Both A and B

D) None of the above.

No explanation, just [size=150][b][u]CHOOSE[/u][/b][/size] one please. Thank you.

swapnillium: you can delete that kind of post with the 'x' box in the corner.

10000th user: I choose A. Often you want to discuss some kind of general function, such as an integrable function. Then you'll say, "Suppose $f: R \rightarrow R$ is integrable." I don't want to imply that the range of $f$ is all of $R$ -- I want $f$ to be as general as possible. Wouldn't it be a bit unwieldy if I always had to write "$f: R \rightarrow A$ with $A \subset R$" all the time?

[quote="Xevarion"]swapnillium: you can delete that kind of post with the 'x' box in the corner. [/quote]
I don't see an 'x.' Where exactly is it?

BTW, I also agree with A). I mean, that's what I see my professors doing all the time.  :lol:

[quote="10000th User"]If you see this $f: A\to B$, would you interpret it as

A) $f: \mathrm{domain}\to\mathrm{codomain}$;

B) $f: \mathrm{domain}\to\mathrm{range}$;

C) Both A and B

D) None of the above.
[/quote]

[b]A[/b]

[quote="swapnillium"]I don't see an 'x.' Where exactly is it?[/quote]

At the bottom right of anyone else's post, you should see "quote" and nothing else. At the bottom right of your own posts, you should see "quote" "edit" "[x]" -- that third box is the one I'm talking about.

See MareleG's recent post in Unsolved and Proposed problems for an example of this notation in action (using interpretation A, as seems to be popular).

[quote="Xevarion"]
At the bottom right of anyone else's post, you should see "quote" and nothing else. At the bottom right of your own posts, you should see "quote" "edit" "[x]" -- that third box is the one I'm talking about. [/quote]
Call me delusional, but I just see two icons ("quote" and "edit") on the bottom right corner of my post. However, when I move my mouse over the "edit" icon, it does display "Edit/Delete this post," but when I do click on the "edit" icon it just gives me the same old editing menu.  :huh:

Oh, I get it! You can't freely delete it anymore if someone has replied, I think. But if you click edit, there's still nowhere to delete it? Let me post this and then check on an older message of mine.

Edit: Sorry, you're right. I can't figure out how to delete the post if someone has already replied. I guess you're just supposed to replace the useless post with something witty.  ;)

But moderators can delete any post, whether replied to or not. Since you're all fussing about this, I just did that to swapnillium's post.

Fairly often, when I notice a post with no content, I'll delete it. But I don't get to everything.

[quote="Kent Merryfield"]The notational distinction JBL is making is news to me. Many people use $f: A\mapsto B$ just to show the domain and codomain.[/quote]
Surprising, it always seemed a common notation to me ! At last in Serge Lang's "Algebra" and "Real and functional analysis" the distinction (that of JBL) is clearly mentioned. Also French people standardly use this convention (For example, a french math teacher told me I've made a mistake when I wrote $f : A \mapsto B$). This distinction is introduced almost as early as functions are in the French Educational system (functions are introduced in [i]troisieme[/i] and this notation comes a year later in [i]seconde[/i]).


AMC
USA(J)MO
USAMO
number theory theorems
number theory

I want to know which theorems and fomulas are useful for math olympiad.Is there any body who can introduce me an electric book or pdf file or something like this to find them and their solutions.
Thank You.

Well, I'd be glad to see a recomendation as well.

Engles: "Problem solving stratagies" is really good, it'll show you all the tricks and has a mindboggaling amount of problems.

Zetiz: "The Art and Craft of problem solving" is very good aswell, it shows you whatg mindframe to be in and how to start looking at problems in the right light.

There the only 2 books i have, so i cant say much more!

I woudl say that mathematical olympiad chellenges is also an amazing book.  It has theorems and many problems from USAMO, IMO, Journals, etc.


linear algebra
matrix
function
linear algebra unsolved

Let $ X$ be any set, and let $ f$ be any function from $ X$ into $ \mathbb{C}$. Let $ x_1,x_2,\cdots,x_n$ be any points of $ X$. Prove that the matrix $ A$ defined by $ A_{ij}\equal{}f(x_i)\overline{f(x_j)}$ is nonnegative definite.

[hide="hint"]For $ \xi \in \mathbb{C}^n$, $ \xi^*A\xi \equal{} \sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^nf(x_i)\overline{f(x_j)}\xi _i \overline{\xi _j} \equal{} \eta^*J\eta$ where $ \eta_i \equal{} \xi _i f(x_i)$ and $ J$ is the matrix of all ones.  Hence to show $ A$ is positive semi-definite it suffices to show that $ J$ is, which is easy enough (show that $ J$ has $ n$ as an eigenvalue, and $ 0$ as an eigenvalue with multiplicity $ n \minus{} 1$).[/hide]

It is a special kind of Gram matrix. Thus... :)

JoeBlow: Your $ J$ is [i]not[/i] the all-ones matrix. It is simpler than that ;)

  darij


[quote="andi"]( a+b+c)2                    (a+b)2     (b+c)2      (c+a)2
-------------- +3   >     --------  + --------- + ---------
a2+b2+c2           ~       a2+b2     b2+c2       c2+a2

2 means squared.
Prove it for a,b,c real[/quote]
If are you mean we need to prove that
 $ \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} 3\geq\frac {(a \plus{} b)^2}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{c^2 \plus{} a^2}$ then, it's wrong.
Try $ a \equal{} b \equal{} c \equal{} 0.$ :wink:
For $ a,$ $ b$ and $ c$ such that $ |ab|\plus{}|ac|\plus{}|bc|\neq0$ we obtain:
$ \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} 3\geq\frac {(a \plus{} b)^2}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{c^2 \plus{} a^2}\Leftrightarrow$
$ \Leftrightarrow1\plus{}\sum_{cyc}\frac{2ab}{a^2\plus{}b^2\plus{}c^2}\geq\sum_{cyc}\frac{2ab}{a^2\plus{}b^2}\Leftrightarrow1\geq\sum_{cyc}\frac{2abc^2}{(a^2\plus{}b^2)(a^2\plus{}b^2\plus{}c^2)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(c^2\minus{}\frac{2abc^2}{a^2\plus{}b^2}\right)\geq0\Leftrightarrow\sum_{cyc}\frac{c^2(a\minus{}b)^2}{a^2\plus{}b^2}\geq0.$


analytic geometry
geometry proposed
geometry

[quote="Bugi"][hide="Easy, here is a hint"]
Put coordinate system
[hide="Hint 2"]
 such that A=(0,0)  B=(0,b)   C=(x,y)
[hide="Final"]
Now use the condition to get the equation of the circle and that's it
[/hide]
[/hide]
[/hide][/quote]
Ok :) I did it with complex numbers. Then $ 2|a\minus{}z| \equal{} |b\minus{}z|$ simplifies to $ \left | z \minus{} \frac{4a\minus{}b}{3} \right | \equal{} \frac{2}{3}|a\minus{}b|$ :)

And likewise $ |a\minus{}z| \equal{} k|b\minus{}z| \iff \left | z \minus{} \frac{k^2b\minus{}a}{k^2\minus{}1} \right | \equal{} \frac{k}{k^2\minus{}1} |a\minus{}b|$ .

How can this be proved synthetic?

[quote]Given to points $ A,B$, show that all points $ C$ with the ability $ 2AC \equal{} BC$ lies on a circle.

Generalisation: Given to points $ A,B$, show that all points $ C$ with the ability $ \frac {AC}{BC} \equal{} k$ for a given $ k \in \mathbb{R}_ \plus{}$, lies on a circle.[/quote]
[b]Maybe this prove is synethic enough[/b], dear [b]Mathias_DK[/b]
The internal bisector of $ \angle ACB$ wrt $ \triangle ABC$ intersects $ BA$ at $ D$, and the outner bisector of $ \angle ACB$ wrt $ \triangle ABC$ intersects $ BA$ at $ E$. It is obvious that $ \frac {CA}{CB} \equal{} \frac {DA}{DB} \equal{} \frac {EA}{EB} \equal{} k \equal{} const$, but $ E,D\in AB$ wich is fixed $ \Longrightarrow E,D$ are fixed. But $ CE\bot CD$ at $ C$
Therefore $ C\in$ the circle with diameter $ ED$- wich is the fixed circle.
The result is lead as follow.


Can someone pm or email me pre-1999 chapter tests? I would reaaaaaaly appreciate it. Pre-1996 state tests and answer keys to any nationals from 1991 and on would also be very appreciated.

Look at my last post here:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78829]www.artofproblemsolving.com/Forum/viewtopic.php?t=78829[/url]I need to figure out how to work the tex center or get a scanner or something.

PM WHATEVER YOU HAVE AND I WILL MAKE IT INTO PDF LATEX OR WHATEVER :wow: :wow: YOU ARE MY HERO AND SAVIOR

I'll just type it up on that pdf program ccy gave. I'll post it in the public site thread.

[quote="biffanddoc"]youre gonna type up that much!?
woah. If you need any help let me know[/quote]
nah i'll prob end up getting a scanner. I'm a slow typer.

[quote="xxm0rg07hxx"]Can you PM anything before 1991 that you have.  PLEASE!!![/quote]
I infer from this that you have stuff from after 1991. Can you email/pm me stuff you have from 1991-1996 in states? thanks


algebra
polynomial
function
Galois Theory
superior algebra
superior algebra unsolved

Let $ f$ be an irreducible polynomial over field $ K$ of degree p, p - prime. Prove that if the Galois group of this polynomial is solvable (i.e. Galois group of the extension of K to the splitting field of $ f$), then for every extension $ L$ of $ K$ $ f$ has exactly 0, 1 or p roots in $ L$.

Use this result due to Galois himself:
In order that an irreducible equation of prime degree be solvable by radicals, it is necessary and sufficient  that all its roots be rational functions of two roots.


Let $ a;b;c$ be positive real numbers satisfying $ ab\plus{}bc\plus{}ca\equal{}7$
Find min of $ T\equal{}\frac{8a^4\plus{}1}{a^2} \plus{}\frac{ 108b^5\plus{}1}{b^2} \plus{} \frac {16c^6\plus{}1}{c^2}$

[quote="N.T.TUAN"][quote="chien than"]Let $ a;b;c$ be positive real numbers satisfying $ ab \plus{} bc \plus{} ca \equal{} 7$
Find min of $ T \equal{} \frac {8a^4 \plus{} 1}{a^2} \plus{} \frac { 108b^5 \plus{} 1}{b^2} \plus{} \frac {16c^6 \plus{} 1}{c^2}$[/quote]
Why title of the topic is ab+bc+ca=7abc?[/quote]
if $ 7abc \equal{} ab \plus{} bc \plus{} ca$ => $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 7$
$ T \equal{} (8a^2 \plus{} \frac {1}{2a^2}) \plus{} (54b^3 \plus{} 54b^3 \plus{} \frac {2}{9b^2} \plus{} \frac {2}{9b^2} \plus{} \frac {2}{9b^2}) \plus{} (16c^4 \plus{} \frac {1}{4c^2} \plus{} \frac {1}{4c^2}) \plus{} (\frac {1}{2a^2} \plus{} \frac {1}{3b^2} \plus{} \frac {1}{2c^2})\ge 4 \plus{} 10 \plus{} 3 \plus{} \frac {1}{7}.(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2 \equal{} 24$
 :D


limit
trigonometry
calculus
calculus computations

[b]1[/b] $\mathop {\lim }\limits_{x \to 1} \frac{{x^{\frac{p}{q}} - 1}}{{x^{\frac{r}{s}} - 1}}$

[b]2[/b] $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x \cdot \cos 2x \cdot \cos 3x}}{{x^2 }}$

[b]3[/b] $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{\cos \alpha x}} - \sqrt[n]{{\cos \beta x}}}}{{x^2 }}$ $m,n \in N$

1. $\mathop {\lim }\limits_{x \to 1} \frac{{x^{\frac{p}{q}} - 1}}{{x^{\frac{r}{s}} - 1}}= \frac{ps}{qr}$

2. $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x \cdot \cos 2x \cdot \cos 3x}}{{x^2 }} = 7$

3. $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{\cos \alpha x}} - \sqrt[n]{{\cos \beta x}}}}{{x^2 }} = \frac{b^2m-a^2n}{2mn}$

[quote="kunny"]For positive integers $n,$ we can generalize
\[ \lim_{x\rightarrow 0} \frac{1-\cos x\cdot \cos 2x\cdots \cdots\cos nx}{x^2}=\frac{1}{12}n(n+1)(2n+1).  \][/quote]
This can be proved by induction.


I have 6 friends and during a certain vacation, i met them during several dinners. I found that I dined with all the 6 on 1 day, with every 5 of them on 2 days, with every 4 of them on 3 days, with every 3 of them on 4 days, and with every 2 of them on 5 days. Further every friend was absent at 7 dinners, and every friend was present at 7 dinners. How many dinners did I have alone?

 :blush:

Let the friends be called $ A_i$, for $ i\equal{}1, \, 2,\,\ldots ,\,6$.
I had $ 1$ dinner with all of them $ A_1A_2A_3A_4A_5A_6$.
With every five of them, I should have exactly two dinners. All possible five of them are:
$ A_1A_2A_3A_4A_5$,
$ A_1A_2A_3A_4A_6$,
$ A_1A_2A_3A_5A_6$,
$ A_1A_2A_4A_5A_6$,
$ A_1A_3A_4A_5A_6$, and
$ A_2A_3A_4A_5A_6$
We see that each of the six groups of five friends were present when I had dinner with all six. So I ought to have one more dinner with each of the above groups to make the number of dinners to be two with every five of my friends. That makes the total number of dinner so far to be 7.
Further, we see that, by now, I already have three dinners with every four of my friends, four dinners with every three of them, and five dinners with every two of them. So the conditions given in the problem are all fulfilled, except the last condition which requires each friend to be present in exactly seven dinners and each of them to be absent in exactly seven of them.
If we again look at the above groupings, we see that $ A_1$ is so far present in exactly six of the dinners. In order for $ A_1$ to be present in exactly seven dinners, I must have one more dinner with him. In a similar way, I must have one dinner each with only $ A_2$, only $ A_3$, only $ A_4$, only $ A_5$, and only $ A_6$. So by now, I have a total of 13 dinners.
Finally, by now we see that $ A_1$ is not preent in exactly six dinners. The same goes true for each friend so far. Thus, to make the number of absents for each friend to be exactly seven, [b]I must  have one dinner alone.[/b]


the semifinalist list got posted today, if anyone here entered.
http://www.siemens-foundation.org/

I entered, and I am a semifinalists also. I pray that I make it to Regional Finalist, though I was very surprised to make it to semifinalist in the first place.

Out of curiosity, how on earth are regional finalists supposed to prepare for the poster session (I am referring to those who have to do it on Nov 4), considering that regional finalists aren't even announced until Nov 1.

good question, i'd like to know this as well.

by the way, anyone know many many semi's make it to regionals in each of the 6 regions???

EDIT: to answer that first question, I'd guess S-W contacts the regional finalists BEFORE the nov. 1 notification (by phone, email? i dunno). that way, they have some advance notice.  This is just a guess though, but it seems logical.

I'm sure that there are many people here on AoPS who have qualified for regional finalist or beyond. Does anyone know the answer? (are regional finalists notified before Nov 1?)

Out of curiousity, how do people get their research published? Do they just randomly find a scientific/mathematical journal and ask that their work be published? I mean, can high schoolers really do research that matches the work of people who spent the past thirty years of their life in some field? I would be rather depressed if I had a Phd in some field, and a high schooler like thirty years younger than me and without so much as a high school diploma published his/her research in the same journal that I was publishing mine.

[quote="**********"]
by the way, anyone know many many semi's make it to regionals in each of the 6 regions???
[/quote]

there is a maximum of 5 individuals and 5 teams for each region. If there are 2 -3 people per team, that means that a maximum of 15-20 people advance in each region.


ugh can you please explain? I looked at number one and didn't feel like trying to even attempt it.

[hide="1"]There are 128 painted faces and 512 cubes so the answer is 1/4 when you divide[/hide]

Wouldn't #3 be
[hide]
37/128

Because there are 170 cubes painted black if I'm not mistaken and 22 of those have 2 or 3 sides painted. 170 - 22 = 148 cubes painted 1 side

148 / 512 = 37 / 128[/hide]

If you think about it...

[hide]one face of an 8x8x8 cubes has 64 squares. For #3, you subtract two of the edges to get 7x7 squares = 49 squares then, you multiply that by 3 to get 147/512, which is in lowest terms, not 148/512 [/hide] :P

[quote="Quevvy"]Express probability in a common fraction please!

1. If a white 8x8x8 cube has 2 sides (non-adjacent) painted black and you cut the cube into 1x1x1 cubes and placed them into a bag, what is the probability that you will pick out one cube with one black face on it?

2. If 2 [i]adjacent[/i] sides are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?

3. If [i]3[/i] sides ([i]all touching the same corner[/i]) are painted black (on the same cube), what is the probability that you will pick out a cube with [i]exactly[/i] one face painted black?[/quote]
[hide="1"]$p=\frac{8\times8\times2}{8\times8\times8}=\boxed{\frac14}$[/hide]
[hide="2"]$p=\frac{8\times7\times2}{8\times8\times8}=\boxed{\frac7{32}}$[/hide]
[hide="3"]$p=\frac{7\times7\times3}{8\times8\times8}=\boxed{\frac{147}{512}}$[/hide]


Let $ n \geq 3$ be an odd integer. We denote by $ [\minus{}n,n]$ the set of all integers greater or equal than $ \minus{}n$ and less or equal than $ n$.
Player $ A$ chooses an arbitrary positive integer $ k$, then player $ B$ picks a subset of $ k$ (distinct) elements from $ [\minus{}n,n]$. Let this subset be $ S$.
If all numbers in $ [\minus{}n,n]$ can be written as the sum of exactly $ n$ distinct elements of $ S$, then player $ A$ wins the game. If not, $ B$ wins.
Find the least value of $ k$ such that player $ A$ can always win the game.

k>n+1, since if k<n+2 $ B$ can get only odd numbers

I can get $k\le\ n+1+\frac{n-1}{2}$,just by choosing the $k$ smallest number and their sum can
 not be more than $n$


Let ABC be right triangle with right angle at A. On the sides AB, BC, CA< choose D, E, F respectively such that $ DE \bot BC$ and DE = DF. M is the midpoint of EF. Prove that $ \angle BCM \equal{} \angle BFE$

Denote $ H$ the projection of $ D$ on $ BF$.
We get $ ADHF,ADEC$ are cyclic quadrilaterals.
$ \Rightarrow BE.BC\equal{}BD.BA\equal{}BH.BF \Rightarrow EHFC$ is cyclic.
$ \Rightarrow \angle FHC\equal{}\angle FEC\equal{}\angle FDM$
But $ DHMF$ is cyclic $ \Rightarrow \angle FDM\equal{}\angle FHM$
Therefore $ H,M,C$ are collinear.
$ \Rightarrow \angle BFE\equal{}\angle BCM$

$ \widehat{FEC} \plus{}\widehat{DEM} \equal{}90^0\equal{}\widehat{MDE} \plus{}\widehat{MED}$
$ \equal{}>\widehat{MDE}\equal{}\widehat{FEC}$
that: $ \widehat{FDM}\equal{}\widehat{MDE}$
$ \equal{}>\widehat{FDM}\equal{}\widehat{FEC}$


The convex quadrilateral ABCD is inscribed in the circle $S_1$. Let $O$ be the intersection of $\overline{AC}$ and $\overline{BD}$. Circle $S_2$ passes through $D$ and $O$, intersecting $\overline{AD}$ and $\overline{CD}$ at $M$ and $N$, respectively. $\overline{OM}$ and $\overline{AB}$ intersect at $R$, lines $ON$ and $BC$ intersect at $T$, and $R$ and $T$ lie on the same side of line $BD$ as $A$. Prove that $O, R, T,$ and $B$ are concyclic

That's soooo banal...

Using directed angles modulo 180, we have < OMD = < OND since the points O, M, N and D lie on one circle (namely the circle $S_2$), and we have < DAB = < DCB since the points A, B, C and D lie on one circle (namely the circle $S_1$). Hence,

  < ORB = < (OM; AB) = < (OM; AD) + < (AD; AB) = < OMD + < DAB
            = < OND + < DCB = < (ON; CD) + < (CD; BC) = < (ON; BC)
            = < OTB,

what immediately yields that the points O, R, T and B are concyclic. The condition that the points R and T lie on the same side of line BD as the point A is not necessary, of course.

  Darij


Show that the sequence $(a_{n})_{n\geq 1}$ does not contain a square of a integer:
$a_{n}=[n+\sqrt{n}+\frac{1}{2}]$

if $n$  is a square then $a_{n}=x(x+1)$ where $x^{2}=n$ so if n is  a perfect sq. we are done
let n not be a perfect square then let $x^{2}< n < (x+1)^{2}$
[b]case 1: [/b] $(\sqrt{n})=\sqrt{n}-[\sqrt{n}]<\frac{1}{2}$
then $a_{n}=n+x$
thus $x^{2}< x^{2}+x <n+x<(x+1)^{2}+x<(x+2)^{2}$
so if $a_{n}$ is a square then $n+x=(x+1)^{2}$ in other words $n=x^{2}+X+1$
but $\sqrt{n}=x+y$ where $0<y<\frac{1}{2}$
thus  $n=x^{2}+2xy+y^{2}<x^{2}+x+1$
so case 1. doesn't hold good
[b]case 2:[/b]let $(\sqrt{n}) \geq \frac{1}{2}$
then $a_{n}=n+x+1$
so $x^{2}<x^{2}+x+1<n+x+1<(x+2)(x+1)<(x+2)^{2}$
so again if $a_{n}$ is a square it has to be $(x+1)^{2}$
thus $n=(x+1)^{2}-(x+1)=x^{2}+x$
but since $(\sqrt{n}) \geq \frac{1}{2}$ we have $n > x^{2}+x$
so we are done


abstract algebra
group theory
function
superior algebra
superior algebra theorems

Any surjective endomorphism of $SL_2(Z)$ is bijective.

I forgot the algebra at all, even definitions. What is endomorphism?

Endomorphism is a homorphism from a group G to itself

This is really a magnificient problem with a splendid idea. It worths saying that it remains true in any $ SL_n(Z)$.

[b]Definition:[/b] A group $G$ is called Hopfian if every surjective endomorphism is an isomorphism.

The problem can be restated like this: $SL_n(\mathbb{Z})$ is Hopfian.
This however is not much of a help in actually solving the problem.

[b]Defintiion[/b] A group $G$ is called residually finite if for all $g\in G\setminus\{e\}$ there exists a finite group $F$ and $f_g:G\to F$ such that $f_g(g)\neq e$. I used $e$ for both the neutral element of $G$ and $F$.

[b]Lemma 1[/b] $SL_n(\mathbb{Z})$ is residually finite.
Proof: Let $A\in SL_n(\mathbb{Z})\setminus\{I_n\}$. Choose $p$ a prime greater than the module of any entry of $A$. There is a canonical morphism $\pi:SL_n(\mathbb{Z})\to SL_n(\mathbb{Z}_p)$. It is easy to see that because of the choice of $p$, $\pi(A)$ "looks" (the elements of $A$ are already reduced mod $p$)  just like $A$, so it can't be $I_n\in\SL_n(mathbb{Z}_p)$.

[b]Proposition 2[/b]: If $G$ is residually finite and $H$ is a subgroup of $G$, then $H$ is a residually finite group.
Proof: Let $g\in H\setminus\{e\}$. Because $G$ is residually finite, there exists a finite group $F$ and $f:G\to F$ such that $f(g)\neq e$. Let $u:H\to F$, $u=f|_H$ (the restriction of $f$ to $H$). $u(g)=f(g)\neq e$, so $H$ is residually finite.

[b]Proposition 3[/b]: $SL_n(\mathbb{Z})$ is a finitely generated group.
Proof: Leva1980 already proved, in a prblem posted by harazi([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36105[/url]), that $SL_n(\mathbb{Z})$ is generated by 2 elements.

[i]Notation:[/i] Let $G$ be a group. Denote by $H<_kG$ a subgoup of $G$ of index $k$.

[b]Theorem (Maltev)[/b]: A finitely generated residually finite group is Hopfian.
Proof: Let $G$ be a finitely generated residually finite group.
Assume $G$ is not Hopfian. Then there exists non-injective surjective endomorphism $f:G\to G$. Let $N=\ker f$. $N$ is a normal subgroup of $G$ and by the Fundamental Theorem of Isomorphism, $G/N\simeq G$.

Let $A_k=\{H| H<_kG\}$. Let $B_k=\{H| N<H, H<_kG\}$. Obviously $B_k\subseteq A_k$.
Because there is an isomorphism $G\simeq G/N$, there is a bijection between the set of subgroups of $G$ and the set of subgroups of $G$ containing $N$.
It can be proven but I have to check the proof again that $A_k$ is finite, meaning $A_k=B_k$ because they are finite, $B_k\subseteq A_k$ and there is an injective function $A_k\to B_k$.

Let $g\in N\setminus\{e\}$. There exists a finite group $F$ and a morphism $h:G\to F$ such that $h(g)\neq e$. $G/ \ker h\simeq F$. Let $m$ be the order of $F$. Then $\ker h<_mG$. Because $A_m=B_m$, $N<\ker h$. This means $h(N)=e$, contradicting $h(g)\neq e$ and $g\in N$.
Modulo the proof of $A_k=B_k$, this concludes the proof of Maltev's Theorem.

[b]Problem:[/b] $SL_n(\mathbb{Z})$ is Hopfian.
Proof: By Proposition 3, $SL_n(\mathbb{Z})$ is finitely generated.
By Lemma 1, $SL_n(\mathbb{Z})$ is residually finite.
By Maltev, $SL_n(\mathbb{Z})$ is Hopfian.

Super. By the way, I remember that at least $SL(2,\mathbb{Z})$ has a simple system of generators and relations. May the problem be solved in this terms, without using the representation as matrices?

harazi, what was your solution. The same ??

Oh, my god, no! In fact, it used the same ideas, but I didn't know all this. Everything which matters is the fact that $SL_n(Z)$ is finitely generated and that if you have a surjective endomorphism of a finitely generated group, then its kernel is included in the kernel of any morphism from this group into an arbitrary finite group. Then you just choose this finite field to be $ SL_n(Z/pZ)$ for a convenable prime p. Anyway, I see a real hurry to put difficult problems that deserve more discussion in the solved section area. So be it... :?

This is not solved section area.
I admit that 175 problems in the unsolved forum stress me out, but I thought that this problem looks better in this section.
There are some instructive proofs (Leva1980's proofs for example) that stay better in an easier to browse section like this one.

Ok, I understood, indeed it was a good idea. I apologise.

@amfulger: Ok, and what's about the proof that $A_k$ is finite? :-)

[quote="amfulger"]
Let $A_k=\{H| H<_kG\}$.
It can be proven but I have to check the proof again that $A_k$ is finite[/quote]

Let $H\in A_k$. Then $H$ is the kernell of the surjective map $G\to G/H$ onto a finite group with $k$ elements. [b][1][/b]

Let $\mathcal G(k)$ be the set of isomorphism classes of finite groups with $k$ elements. $\mathcal G(k)$ is finite.
Let $G(k)$ be the set of subgroups of $S_k$ with $k$ elements. $G(k)$ is obviously finite. By a proof for Cayley's Theorem, $G(k)$ contains a representative for any class in $\mathcal G(k)$.

Let $F_k=\{f|\ f\mbox{ is a group homomorphism }f: G\to T,\mbox{ where }T\in G(k)\}$.
Let $g_1,\ldots, g_n$ be a finite set of generators for $G$. Every group homomorphism $f: G\to G'$, where $G'$ is another group, is determined by $f(g_1),\ldots,f(g_n)$. The choices for $f(g_1),\ldots,f(g_n)$ may not be independent because $G$ may not be free.
However, given $T\in G(k)$, there are at most $k^n$ distinct group homomorphisms $G\to T$. This means that $|F_k|\leq |G(k)|\cdot k^n$.

The application $f\to\ker f$ induces a surjective mapping $\ker: F_k\to A_k$. The surjectivity of this $\ker$ map is given by [b][1][/b].
This means $|A_k|\leq |F_k|\leq |G(k)|\cdot k^n$, so $A_k$ is a finite set.


algebra polynomial function geometry incenter

The 15th PAMO is currently taking place in Algiers, Algeria.  Eight African countries are taking part.  The papers follow an IMO format but are easier.  They could be useful practice for guys who hope to make it to the IMO in the future but find IMO problems too hard right now.

For more info on the PAMO and other olympiad news from Africa and South Africa visit
[url=http://www.mathspage.co.za]www.mathspage.co.za[/url]

15th Pan African Mathematics Olympiad, Algiers

Day one, 1st August 2005
Time : four and a half hours

[b]Question 1[/b]

For any positive real numbers [tex]a,b[/tex] and [tex]c[/tex], prove:

[tex]\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{2}{a+b} + \frac{2}{b+c} + \frac{2}{c+a} \geq \frac{9}{a+b+c}[/tex]

[b]Question 2[/b]

Let [b]S[/b] be a set of integers with the property that any integer root of any non-zero polynomial with coefficients in [b]S[/b] also belongs to [b]S[/b].  If [tex]0[/tex] and [tex]1000[/tex] are elements of [b]S[/b], prove that [tex]-2[/tex] is also an element of [b]S[/b].

[b]Question 3[/b]

Let [tex]ABC[/tex] be a triangle and let [tex]P[/tex] be a point on one of the sides of [tex]ABC[/tex].  Construct a line passing through [tex]P[/tex] that divides triangle [tex]ABC[/tex] into two parts of equal area.

Day two, 2nd August 2005
Time : four and a half hours

[b]Question 4[/b]
Let [tex][x][/tex] be the greatest integer less than or equal to [tex]x[/tex], and let [tex]\{x\} = x - [x][/tex].
For example [tex][11] = 11[/tex], [tex][\pi] = 3[/tex] and [tex][-\pi] = -4[/tex].  Solve the equation: [tex][x]\{x\} = 2005x[/tex]

[b]Question 5[/b]
Noah has to fit 8 species of animals into 4 cages of the Arc.  He planes to put two species of animal in each cage.  It turns out that, for each species of animal, there are at most 3 other species with which it cannot share a cage.  Prove that there is a way to assign the animals to the cages so that each species shares a cage with a compatible species.

[b]Question 6[/b]
Let [tex]f: \mathbb{Z} \rightarrow \mathbb{Z}[/tex] be a function such that:  For all [tex]a[/tex] and [tex]b[/tex] in [tex]\mathbb{Z} - \{0\}[/tex], [tex]f(ab) \geq f(a) + f(b)[/tex].  Show that for all [tex]a \in \mathbb{Z} - \{0\}[/tex] we have [tex]f(a^n) = nf(a)[/tex] for all [tex]n \in \mathbb{N}[/tex] if and only if [tex]f(a^2) = 2f(a)[/tex]

Question 6 as it is stated above is how it appeared in the paper, but it sould have been
... if and only if [tex]f(a^2) = 2f(a)[/tex] for all [tex]a \in \mathbb{Z}[/tex].  It sounds like a number of contestants answered it that way and got marks for it.  It is possible to construct  counter example for the problem that was asked in the paper.

Problem three: Construct the incentre I and PI half the area.

where should we post the problems? Olympiad or Pre-Olympiad?

Problems are posted in the Pre-Olympiad forum. Please check them.


If $ k$ is an odd integer and $ n$ is a positive integer then prove that $ k^{2^n} \minus{} 1$ is divisible by $ 2^{n \plus{} 2}$.

[hide]$ k^{2^n} \minus{} 1\equal{}(k^{2^{n\minus{}1}}\plus{}1)(k^{2^{n\minus{}1}}\minus{}1)\equal{}(k^{2^{n\minus{}1}}\plus{}1)(k^{2^{n\minus{}2}}\plus{}1)(k^{2^{n\minus{}2}}\minus{}1)\equal{}\ldots\equal{}(k^{2^{n\minus{}1}}\plus{}1)(k^{2^{n\minus{}2}}\plus{}1)(k^{2^{n\minus{}3}}\plus{}1)\cdots(k^{2^{n\minus{}n}}\plus{}1)(k^{2^{n\minus{}n}}\minus{}1)$, which is a product of $ n\plus{}1$ even numbers. Also, we have $ k\plus{}1$ and $ k\minus{}1$ as factors, and one of these must be divisible by 4, adding another power of 2.[/hide]


It says in the pre-camp briefing that I should not bring a scientific calculator. Is this because teachers don't want you to play games on your calculator? I have a TI-83 that does not have any games on it. Would that be ok? If not, I also have a less-advanced, non-graphing texas instruments calculator (no games). but I think both of these are scientific calculators. would this be ok?

I think the concern is with games.  Some of the lectures and breakouts last summer actually required a calculator, and several campers had graphing calculators.


MIT college Harvard calculus AMC AIME USA(J)MO

I'm in 7th grade, and I really wanna go to Exeter for HS.  However, my parents said they won't let me go, unless I can (somehow) get a FULL scholarship.

Does anyone know about some merit-based scholarships that I would qualify for (I live in CA)?

WELL..

There is:

Caroline D Bradley Scholarship
Jack Kent Cooke Scholarship

and I can't think of any more...

But the jack kent one is partially need-based :(

Anyone else have some?

Why are you dying to go to Exeter?

I mean, good luck getting a full scholarship and all. But give it a good thought before committing. There are many, many advantages of going there over a public school near your house, but there are also many, many disadvantages.

[hide="THE GOOD PART OF GOING TO EXETER"]
EVERYTHING
[/hide]
[hide="THE BAD PART"]
:( my parents dont want me to go and it costs too much.
[/hide]

I have several friends here who went to Exeter and I'm not sure they necessarily had any better an experience than I did going to public school just because they went to Exeter.  Of course it will be different and of course there will be certain advantages, but pinning your future on going to Exeter is just as bad as pinning your future on going to MIT or Harvard - both viewpoints are extremely narrow and neglect to take into account the fact that [b]the top schools are not everything[/b] and that it is possible to prosper anywhere with the right resources.

Well, I got that just by being in an academic magnet program within a large urban school. There are a lot of good public schools in California, and a lot of resources beyond those schools as well. When I needed math classes beyond calculus, I could just go to the local CSU.

Yeah, but my local high school doesn't have that...our math team really stinks also.

only 8 qualified to aime

NOBODY to usamo for 2 years...

8 AIME qualifiers, and no USAMO qualifiers in the last two years? That's better than my school. Aside from me, the school has only had one student ever qualify for the USAMO. They participate in contests by the will of one teacher, but there's not much beyond that. It's still enough if you put some of your own effort in.

You know what the fall startup event is?  I got 2nd place as a 7th grader...with a 58.

Our math team meets 30 minutes per week.  At lunch.  So I don't even get to go.

Half the time they just watch NUMB3RS.

8 AIME qualifiers?  that's like amazing.  Last year, at my school, I was the only AIME qualifier, and I was in middle school, no one from the high school made to AIME.  I don't we've ever had a USAMO qualifier, at least not in the last 10 yrs.

Are your friends going to Exeter?  If they're not, won't you lose all your friends?  :(

I wanted to go to Exeter last year, but after my parents talked to my about it, I changed my mind. The main reason was the tuition, and also the distance from home.

Friends? I've attended three different public schools and I still made friends. Even now, homeschooled, I have several real ;) friends I talk with often.

[quote="AIME15"]What drew me in was the letter.

"Imagine being in a school where being smart was cool..."

You don't see that every day :P [/quote]

Even if you don't, I don't think it's a real big benefit. It's true that as a middle schooler, it can seem like a big deal (it definitely was for me as a 7th grader), but as you grow older, you'll see that being smart is actually something you can be proud of, even among your peers (well, okay---duh). Ignore anyone who looks at you with contempt for being smart---they're not worth your time. It's more important to earn the respect of other intelligent people like you. Being in an environment like Exeter where there are people smarter than you can be a bonus, but learning to socialize with average-intelligence (not trying to be offensive..) people and being able to push yourself are just as important.

Sorry.. it's past [my nonexistent] bedtime and my head isn't exactly clear.

Earlier today, I read somewhere -- [b]"College isn't a prize to be won, but a match to be made."[/b]

Very well said, undefined.  I don't go to an elite high school myself, but I'm sure you could get plenty of opinions here from those who do.

Another advantage to not going to an elite high school is the entire idea of comparison.  Odds are, if you can make it into Exeter, then you would be the very top in an average high school.  If you go to a good, non-elite high school, then you can make your own programs and schedule that can turn out just as good, but saving money, and seeming more impressive (or just as) than if you had gone to Exeter and been completely average, or possibly below average.  It's all perception.

For elite schools, if you just stick with the curriculum that they give you, you'll do fine, but if you do that with a lesser school, then you'll find yourself extremely outmatched in the real world, competing for spots in colleges, and jobs.  As undefined says, it's a matter of being able to drive yourself.

In some high schools, unfortunately mine included, aren't... "very academically oriented".  Then again, it VERY WELL could just be me, but I've found that the places I've felt I fit in most were at my classes in the city (NYC, what else?) and AMSP.  Not at school.  I hope my point got across.  Thanks for reading.

OK, thanks everyone.  I'll keep this in mind.

@iphailedusamo:  What friends?  :(

I go to a pretty elite high school in the Boston area--around 20-25% go Harvard-MIT-Princeton (note: I chose these as they are fairly math-oriented) each year. I think that your high school can affect your chances of success, but it's not worth it if you don't find a good match with it. Our college guidance's motto is "Match matter most," which I feel also applies nicely to the high school you choose to go to. The reason I chose to go here over other places was not the matriculation, or the amount of AIME and USAMO qualifiers, but that I felt more comfortable here than other places.

You really need to check out the environment of a school before seriously considering going--try and arrange a tour, or a "spend a day" with a current student. I think most private schools will offer this during the admissions process, but correct me if I'm wrong.

Exeter has a very generous financial aid package -- full ride for students with family income less than 75K.
http://www.exeter.edu/admissions/147_9580.aspx

Exeter provides unusual opportunities for serious math students such as learning exclusively from problem sets and having a critical mass of strong math peers.
http://www.exeter.edu/news_and_events/news_events_10538.aspx

Exeter has a good record of both getting math students from the wannabe stage to USAMO qualifer stage and producing (or at least acquiring and training) MOSP types.
See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31725&search_id=1682055875

What kind of things should you do before considering Exeter?
AMC scores, AIME scores etc etc...
I'm interested in going in about 11th or 12th grade.
Actually to tell you the truth, I have never done a MATHCOUNTS (my school isn't interested) and I am taking the AMC8 for the first time this year. I'm in 7th grade. Do I even stand a chance?

[quote="mathemonster"]perfect628 where do u go?[/quote]
Roxbury Latin


I was wondering how everybody knows what their their scores are. Where did you get the answer key? And could you please post the answers:)

The answer keys are generally compiled by students comparing answers. Official answers are released a few days after the test.


limit
logarithms
function
inequalities
calculus
calculus computations

Let $ M(x,\mathbf{a}) \equal{} \left( \frac {\sum_{j \equal{} 1}^{n} a_j ^x}{n} \right) ^{\frac {1}{x}}$ for $ x \in \mathbb{R}^*$ and $ a_j \in \mathbb{R}^ \plus{} , \; 1 \leq j \leq n$.

Show that $ \lim_{x \to 0} M(x, \mathbf{a}) \equal{} GM(\mathbf{a}) \equal{} \left( \prod _{j \equal{} 1}^n a_j \right) ^{\frac {1}{n}}$

You've neglected the (important) condition $ a_i > 0$ for all $ i$.  Also, of course, the top expression isn't defined when $ x \equal{} 0$.

We can use L'Hôpital's rule here.

$ \ln M(x,\mathbf{a}) \equal{} \frac {\ln\left(\frac1n\sum_{j \equal{} 1}^n a_j^x\right)}{x}$

By L'Hôpital,

$ \lim_{x\to 0}\ln M(x,\mathbf{a}) \equal{} \lim_{x\to0}\frac {\frac1n\sum_{j \equal{} 1}^na_j^x\ln a_j}{\frac1n\sum_{j \equal{} 1}^na_j^x}$

Since $ \lim_{x\to0}\frac1n\sum_{j \equal{} 1}^na_j^x \equal{} 1,$ this becomes

$ \lim_{x\to 0}\ln M(x,\mathbf{a}) \equal{} \frac1n\sum_{j \equal{} 1}^na_j^0\ln a_j \equal{} \frac1n\sum_{j \equal{} 1}^n\ln a_j.$

Exponentiating this gives the desired result.

Other things to prove include:

$ M(x,\mathbf{a})$ is an increasing function of $ x,$ and $ M(x,\mathbf{a}) < GM(\mathbf{a})$ for $ x < 0$ and $ M(x,\mathbf{a}) > GM(\mathbf{a})$ for $ x > 0.$ (This is assuming that $ a_j\ne a_k$ for some $ j,k.$ Else replace these inequalities with equalities.)

$ \lim_{x\to\infty}M(x,\mathbf{a}) \equal{} \max_{1\le j\le n}a_j.$

$ \lim_{x\to \minus{} \infty}M(x,\mathbf{a}) \equal{} \min_{1\le j\le n}a_j.$

Thanks Professor Merryfield!

Missed the "+"  :oops: , thanks JBL.

at first use asymptotical expression of $ a_{j}^{x}$ when $ x \minus{} > 0$
it's $ a_{j}^{x}$~$ 1 \plus{} x\ln a_{j}$
and in seccond use the next fact :
$ (1 \plus{} \frac {x \ln \prod_{1\leq j \leq n } a_{j}}{n})^{1/x} \equal{} \exp (\frac {\ln \prod_{1\leq j \leq n } a_{j}}{n})$ when $ x \minus{} > 0$
or $ (1 \plus{} xA)^{1/x} \equal{} \exp ( A)$ when $ x \minus{} > 0$


linear algebra
matrix
calculus
function
number theory
greatest common divisor
least common multiple

Let $X_1,X_2,\ldots,X_m$ a numbering of the $m=2^n-1$ non-empty subsets of the set $\{1,2,\ldots,n\}$, $n\geq 2$. We consider the matrix $(a_{ij})_{1\leq i,j\leq m}$, where $a_{ij}=0$, if $X_i \cap X_j = \emptyset$, and $a_{ij}=1$ otherwise. Prove that the determinant $d$ of this matrix does not depend on the way the numbering was done and compute $d$.

ok, let's try...
first of all, by the (axiomatic) definition of determinant of a matrix, we have that if $d'$ is the determinant of a matrix with two exchanged rows or columns, with respect to a matrix with determinant $d$, we have $d' = -d$. (*)
now, instead of showing that any for any permutations of the $m$ sets $d$ remains constant, we can just show that for a transposition (i mean a permutation $\sigma$ such that $\sigma(X_i) = X_j$, $\sigma(X_j) = X_i$, $\sigma(X_k) = X_k \forall k \neq i,j$), we have $d_\sigma = d$, where $d_\sigma$ is the determinant of the matrix associated to the set after permutation $\sigma$.
but this easily follows from the (*) above.
now, we're trying to show that $d_n = -1 \forall n > 1$, and $d_1 = 1$.
first of all, cases $n=1,2$ are trivial, and determinants are $d_1 = 1$, $d_2 = -1$ so the given formula seems to be working. ;)
let's call $M_{n-1}$ the matrix associated to a fixed numbering of the $2^{n-1} - 1$ subsets of $\{1,\cdots n-1\}$.
we can get $M_n$ by adding $X_{2^{n-1}} = {n}$, and $X_{2^{n-1}+i} = {n} \cup X_i$.
now, the matrix associated to this permutation gets obviously:
\[ \left( \begin{tabular}{ccc|c|ccc} & & & 0 & & & \\ & M_{n-1} & & \vdots & & M_{n-1} & \\& & & 0 & & & \\ \hline 0 & \cdots & 0 & 1 & 1 & \cdots & 1 \\ \hline & & & 1 & 1 & \cdots & 1 \\ & M_{n-1} & & \vdots & \vdots & \ddots & \vdots \\ & & & 1 & 1& \cdots & 1 \end{tabular}\right) \].
by subtracting the $2^{n-1}$-th row, and applying Laplace's rule for the calculus of $d_n$ , we get that $d_n = (-1)^{2^{n-1}-1}(d_{n-1})^2$, because we can get to a matrix divided into three square boxes, the bottom-left of which is $0$. but, for inductive hypothesis, $ d_{n-1} = -1$ so we have our nice result.

It occurs to me to ask what happens if you replace the lattice of subsets of $[n]$ with a more general lattice with a bottom element $\mathbf 0$?  The definition of $a_{ij}$ would then become $a_{ij} = 0$ if $X_i \wedge X_k = {\mathbf 0}$, otherwise 1.  (I don't have an answer BTW.)

I don't want to think about the determinant of an infinite matrix, so let's restrict to finite lattices.

If there is some pair $x,y$ such that for every $z$, either both or neither of $x\wedge z$ and $y\wedge z$ are $0$, then the determinant is zero. In the nontrivial case, each element is completely determined by the set of elements it has nonzero intersection with. We can extend this another step, by considering the set of minimal nonzero elements; $x$ and $y$ have nonzero intersection if and only if there is some minimal element $z$ with $z\le x$ and $z\le y$.

The nontrivial case then reduces to considering subsets of the power set of the minimal nonzero elements.

This still leaves room for variation; the lattice $0,a,b,c,abc$ gives a determinant of $-2$.

In Lovasz's book [i]Combinatorial Problems and Exercises[/i], one of the problems expresses the determinant of the lattice matrix in terms of the Mobius function of the lattice.

I believe this is called Smith's determinant, after the case(s) where $a_{ij}$ is the GCD or LCM of $i$ and $j$ (which Smith computed in the 1800's).


Prove that
$ \frac{a}{(b\plus{}1)(c\plus{}1)}\plus{}\frac{b}{(c\plus{}1)(a\plus{}1)}\plus{}\frac{c}{(a\plus{}1)(b\plus{}1)} \ge \frac{3}{4}$
where $ a,b$ and $ c$ are positive real numbers satisfying $ abc\equal{}1$

[quote="Litlle 1000t"]Prove that
$ \frac {a}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {b}{(c \plus{} 1)(a \plus{} 1)} \plus{} \frac {c}{(a \plus{} 1)(b \plus{} 1)} \ge \frac {3}{4}$
where $ a,b$ and $ c$ are positive real numbers satisfying $ abc \equal{} 1$[/quote]

By Cauchy Schwarz ineuqlaity we have

\[ \left( {\sum {\frac{a}{{\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right)}}} } \right)\left( {\sum {a\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right)} } \right) \ge \left( {\sum a } \right)^2
\]

We will prove

\[ 4\left( {\sum a } \right)^2  \ge 3\left( {\sum {a\left( {b \plus{} 1} \right)\left( {c \plus{} 1} \right)} } \right)
\]

\[ \Leftrightarrow 4\left( {\sum a } \right)^2  \ge 3\left( {3abc \plus{} 2\sum {ab}  \plus{} \sum a } \right)
\]

true since

\[ 3\sum {a^2 }  \ge 3\sum a
\]

\[ 2\sum {ab}  \plus{} \sum {a^2 }  \ge 9
\]

[quote="Litlle 1000t"]Prove that
$ \frac {a}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {b}{(c \plus{} 1)(a \plus{} 1)} \plus{} \frac {c}{(a \plus{} 1)(b \plus{} 1)} \ge \frac {3}{4}$
where $ a,b$ and $ c$ are positive real numbers satisfying $ abc \equal{} 1$[/quote]

The above inequality equivalents to:
$ 4\sum a(a\plus{}1) \ge 3(a\plus{}1)(b\plus{}1)(c\plus{}1)$

$ \Leftrightarrow 4(a^2\plus{}b^2\plus{}c^2) \plus{}4(a\plus{}b\plus{}c) \ge 3abc\plus{}3(ab\plus{}bc\plus{}ca)\plus{}3(a\plus{}b\plus{}c)\plus{}3$

$ \Leftrightarrow 4(a^2\plus{}b^2\plus{}c^2) \plus{}(a\plus{}b\plus{}c) \ge 6\plus{}3(ab\plus{}bc\plus{}ca)$

Which is clearly true by AM-GM inequality:
$ a^2\plus{}b^2\plus{}c^2\plus{}a\plus{}b\plus{}c \ge 6$ and $ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ca$

I think your inequality should be :

$ \sum \frac{a}{(a\plus{}1)(b\plus{}1)}$ ??

Because $ abc\equal{}1$

We may set
$ a\equal{}\frac{x}{y}$
$ b\equal{}\frac{y}{z}$
$ c\equal{}\frac{z}{x}$

So we need to prove :

$ \sum \frac{yz}{(y\plus{}z)(z\plus{}x)} \ge \frac{3}{4}$
$ \sum \frac{yz(y\plus{}z)}{(y\plus{}z)(z\plus{}x)(x\plus{}y)} \ge \frac{3}{4}$

$ 4 \sum_{sym} x^2y \ge 3 (\sum_{sym} x^2y \plus{}2xyz)$
$ \sum_{sym} x^2y \ge \sum_{sym} xyz$

Which is obvious..


With this I start a series of motto's. They are my own creation and I post them as discution topics, essay topics...

So here it comes:
"Surprises only come to those not expecting them"

that's absolutely true..
if you expect it, it would become a suprise.... isn't it?

A surprise also comes when an expected surprise does not come! :)

you do not expect that what you expected won't come...
so when it really does not come, it is a surprise to you.

The mottos are a very good idea [img]http://www.ectona.org/Mysmilies/Smilies/eclipsee_steering.gif[/img]

As liyi said, they wouldn't be considered surprises otherwise, right? This similar in concept with one of Murphy's laws: "The place where yoyu find something is the last one you're searching." (or something like that :D). That's actually not surprising at all, since you obviously stop looking after you've found what you were looking for.

I think these are truisms: of course it's like that, we don't really need to make mottos out of them... :D:D

In this motto I see a warning: If you don't want an unpleasant surprise spoil your fun, then you should better be cautious.
That's if you don't have a blind trust in life and think you will only be pleasantly surprised. Or if you are pessimist and think you can only get nice surprises if you think for the worst possible future.

I am realistic and cautious.


What is the value of
(n)  +  (n) + ...  + (n)
(1)     (2)                      (n)

Please give me an easy way to[b]solve[/b] the problem?
I dont just need an answer.

It's #11 on the Mock Sprint.  Don't worry, we helped him over in the Mock Results thread.

I don't understand....
O
wow...
I was wondering what the parentheses were about.
Just use binomial theorem.


inequalities
induction
inequalities unsolved

Problem is very easy to find the [b]minimum[/b] ,I can post my solution if you want.

And for the maximum ,I'm not sure if it is bounded from above.

Experimenting with numbers $ a_1 ,a_2,a_3,a_4$ ,we have

$ a_1 \leq a_2 \implies \frac {a_1}{a_2}\leq 1$

$ a_1 \plus{} a_2 \leq a_3 \implies \frac {a_1}{a_3} \plus{} \frac {a_2}{a_3} \leq 1$

$ a_1 \plus{} a_2 \plus{} a_3 \leq a_4 \implies \frac {a_1}{a_4} \plus{} \frac {a_2}{a_4} \plus{} \frac {a_3}{a_4} \leq 1$

$ \left(\frac {a_1}{a_3} \plus{} \frac {a_1}{a_4} \plus{} \frac {a_2}{a_4} \right ) \plus{} \left(\frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} \frac {a_3}{a_4}\right) \leq 4$

To maximise $ \left(\frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} \frac {a_3}{a_4}\right)$ we need to minimise $ \left(\frac {a_1}{a_3} \plus{} \frac {a_1}{a_4} \plus{} \frac {a_2}{a_4} \right )$

By Am-Gm

$ \left(\frac {a_1}{a_3} \plus{} \frac {a_1}{a_4} \plus{} \frac {a_2}{a_4} \right ) \geq 3 \sqrt [3]{\frac {a_1}{a_3} \cdot \frac {a_1}{a_4} \cdot \frac {a_2}{a_4}}$

and the equality occur if and only if $ \frac {a_1}{a_3} \equal{} \frac {a_1}{a_4} \equal{} \frac {a_2}{a_4} \iff a_1 \equal{} a_2$ and $ a_3 \equal{} a_4$ ,contradiction .

So I think we can not minimise $ \left(\frac {a_1}{a_3} \plus{} \frac {a_1}{a_4} \plus{} \frac {a_2}{a_4} \right )$

therefore we can not maximise $ \left(\frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} \frac {a_3}{a_4}\right)$

----------------------------------------------------

Another thing that makes me thinking that  it is not bounded from above is this :

By induction we can easily show that

$ a_k \geq 2^{k\minus{}2}a_1$

Let $ S\equal{}\frac{a_1}{a_2} \plus{}\frac{a_2}{a_3} \plus{}...\plus{}\frac{a_{2008}}{a_{2009}}$

Then

$ S\leq\frac{a_1}{a_1} \plus{}\frac{a_2}{2a_1} \plus{}\frac{a_3}{4a_1} \plus{} ...\plus{}\frac{a_{2009}}{2^{2007}a_1} \leq \frac{a_{2009}}{a_1} \left(1\plus{}\frac{1}{2} \plus{}\frac{1}{4} \plus{} ...\plus{}\frac{1}{2^{2007}}\right)$

To minimise We need to have $ \frac{a_{2009}}{a_1} \leq x$ ,not $ \frac{a_{2009}}{a_1} \geq 2^{2007}$


geometry
3D geometry
sphere
inequalities
circumcircle
analytic geometry
graphing lines

can anyone post the answer for #22 and #25?

Tomorrow would probably be better to post answers, I got everything but #23,24 & 25, though I found 24 later. This should tie me in my school from who I've talked to, although it's not as good as I would have liked.

25 is D and 22 is C.

For 25, consider the aerial view of the whole structure and find the distance between the centres of the cone and the sphere.  Then consider a side 2D view of one half-cone (right triangle) and the sphere (circle).  Use similar triangles to knock off the problem.

For 22, it's just some sort of a greedy distribution into:  6x_1 + 5x_2 and 2y_1 + 3y_2.  I got 9:43, I'm not sure if there's an earlier time though.  It seems right  ;) .

22. Quincy and Celine have to move 16 small boxes and 10 large boxes. The chart indicates the time that each person takes to move each type of box. They start moving the boxes at 9:00 a.m. The earliest time at which they can be finished moving all of the boxes is
(A) 9:4a a.m.   (B) 9:42 a.m.   (C) 9:43 a.m.    (D) 9:44 a.m.    (E) 9:45 a.m.

(A chart is given with the following information:
- Celine moves a small box in 2 minutes and a large box in 6 minutes
- Quincy moves a small box in 3 minutes and a large box in 5 minutes)

[hide="Comments"]I have a problem with the wording of the question. It leaves questions like "Can more than one person move a box at the same time?" and "If only one person can move a box at a time, and Quincy finishes moving all of his boxes, and Celine has been moving a large box for three minutes, can Quincy take over to save 30 seconds?" I assumed that each box had to be moved completely by one person, and the question should have contained this clause.[/hide]

[hide="Solution"]C is the answer. We will consider pairs (C,Q), where C is the time Celine takes, Q is the time Quincy takes. We are trying to minimize max(C,Q). We have the inequality max(C,Q) >= avg(C,Q). C+Q is minimized when Celine moves all the samll boxes, and Quincy moves all the large boxes; avg(32,50) = 41. Now, starting from this situation, consider "switches" of boxes, a switch being Quincy moving an additional small box, or Celine moving an additional large box. Each switch increases the average by 0.5. We can make 3 switches (1 small, 2 large), to get (42,43), with max=43. Can we do better? Not with more than 2 switches, since then avg(C,Q) >= 42.5, so max(C,Q) >= 43 (since the max is an integer). Thus we are limited to 1 switch or 2 switches. There are only 5 possibilities. Computing them we get (38,45) and (30,52) for 1 switch; (44,40), (36,48), and (28,56) for 2 swtiches. These yield max's 45, 52, 44, 48, 56. Thus 43 is the best we can do.[/hide]

(The attachment is part of my solution, not part of the question.)

25. Three identical cones each have a radius of 50 and a height of 120. The cones are placed so that their circular bases are touching each other. A sphere is placed so that it rests in the space created by the three cones. If the top of the sphere is level with the tops of the cones, then the radius of the sphere is closest to
(A) 38.9   (B) 38.7   (C) 38.1   (D) 38.5   (E) 38.3

[hide="Solution"]Consider the three circular bases. The centres of the three circles form an equilateral with side length $100$. Let the circumcircle of this triangle have centre D. Its radius is $\frac{100}{\sqrt{3}}$.

Now consider the attached diagram, where O is the centre of the sphere radius $r$, and Q is the point of contact of the sphere with one of the cones. D is the point of contact of the perpendicular from O to the plane containing the bases of the cones. BD = $\frac{100}{\sqrt{3}}$.
AC has slope $-\frac{12}{5}$, so QO has slope $\frac{5}{12} \Rightarrow OT = \frac{5r}{13}, \ QT = \frac{12r}{13}.$

$AP = SO + OT = \frac{18r}{13}.$ By similar triangles, $\frac{PQ}{AP} = \frac{50}{120} \Rightarrow PQ = \frac{15r}{26}.$

$PT = PQ + PT = \frac{15r}{26} + \frac{12r}{13} = \frac{3r}{2}.$ But $PT = BD$, so

$\frac{3r}{2} = \frac{100}{\sqrt{3}} \Rightarrow r = \frac{200}{3\sqrt{3}} \approx 38.4900 \Rightarrow \boxed{D} \ .$[/hide]

almost everyone in my class (even my teacher) got # 10 wrong...

Steven Karp is my hero.
I left the last one blank .. :(. Pretty sure I got everything else though.

Yeah, #10 was a trickster, like 20, and 22.

[quote="kool_dudy"]What was the answer for #10?  I put 8 intersections but then realized after it is probably 4.  Similarly, my teacher thought the same thing.[/quote]

It's 8.  See AIME 1999 #4.  I got pwned on this question.  The funny thing is, I was working on this question the night before on the particular AIME.  But nah...my brain decided to imagine that the answer was 4. :(

EDIT:  Pwnage repealed.  Thank goodness for stupidity.   :rotfl:

[quote="Effervescence"]why would it be 4?? Almost everyone at my school put 8, unless they're all wrong[/quote]
That's exactly what I thought too! But, I think the answer is 4. Since it says "[b]any two rectangles[/b]" in the question, you have to take the minimum of all the possibilities. -> No matter what the sizes of the two rectangles are, you can always have four points of intersection at corner of the bigger rectangle (you place the smaller rectangle at of other corners of the big rectangle so that they have a common vertex and two sides touching each other [see attachment], and then you turn the small rectangle until you get fours points of intersection), and that's the maximum. You can have 8 points under ideal conditions, but it doesn't apply to [i]any two rectangles[/i].

I am not sure if that's the right answer, so correct me if I am wrong.  :P

Well it asks for the maximum possible points of intersection, and I think the "any" just means that the rectangles are not restricted to any conditions, meaning like, you take any rectangles of your choice and try to find a max number of points of intersection (I don't know if that makes sense, I'm explaining really poorly). It shouldn't make you think this much if it's only part A  :huh:

If you argue it your way, wouldn't it be correct to say that the minimum number of intersections is 0, if they're not touching?, or 2, like they showed in the example.

[quote]If you argue it your way, wouldn't it be correct to say that the minimum number of intersections is 0, if they're not touching?, or 2, like they showed in the example.[/quote]
just added a little bit more to my reply I poseted..  you can have a look again..
(for the diagram on the contest, you can still make it to have 4 points of intersection...)

[quote]It shouldn't make you think this much if it's only part A  :huh:[/quote]
I said that to myself too...  :D

Now that I think about it, chilly_snow seems to have put up a valid argument.  Even if we have say...a microscropic rectangle vs. a rectangle the size of Colorado, the smaller rectangle can wedge out 4 intersections from the corner of the larger one.  Drawing a picture confirms this.  However, 8 intersections can only occur if the rectangles are closer in size to each other...in ideal conditions as chill_snow puts it.  Kinda hard to get 8 intersections with given a really tiny and a really huge rectangle.

...Does this mean I got #10 right?  Haha...man! :roll:

thanks a lot guys...it seems like i'm doing worse every year...for this year, i already messed up amc 12..and now fermat is just adding to the list...sigh..

can anyone post how to find # 23? the tom and jerry one?
i totally guessed on it  :?

[quote="tzhu"]I just checked the CEMC website and the solutions to this year's Fermat is out. Question 10 turns out to be D, so it wasn't a trick question after all.  :lol:[/quote]
yaya... I just checked it.. the answer is 8..!!!
that means I did it right~!
but I still think the answer should be 4...
o well...

I owned(150). :D
The thing was such a joke: I had over 25 min for question 25. :rotfl:
I think they are making contests easier and easier by the year.

[quote="rem"]I think they are making contests easier and easier by the year.[/quote]This is definitely true - I have heard it directly from someone who formulates questions and decides which should be included on the contests.

I can't believe thay still haven't posted the results. It shouldn't take too long to mark everything and to do the statistics. :mad:

[quote="Effervescence"][quote]I owned(150).
The thing was such a joke: I had over 25 min for question 25.
I think they are making contests easier and easier by the year.[/quote]

Not everyone is superman like Alex though. (This is Tai by the way  :P ). Thinking back I probably should've been able to do 25. Had a brain fart with about 30 minutes left on 22-23, then in a flash Burns was like, "5 minutes guys".
I think people are getting smarter too, or they need to make contest easier so more people can take it.[/quote]
it's just because human r getting smarter...Darwin was right hehe

may i ask you when the results will be posted??

The results should have been posted about a week ago. I have no idea why it is taking them so long. maybe it is because many more people wrote the contest compared to last year.

[quote="nirvanatear"][quote="carpo"]So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_results.pdf

Haha my school came top 5 in Fermat. svg tech. My teacher will be happy.[/quote]
its the last year's result lol[/quote]They looked the same :?

I fixed the link.

[quote="carpo"]So finally the results book is out. http://cemc.uwaterloo.ca/english/contests/past_result/2006/pcf_results.pdf

Haha my school came top 5 in Fermat. svg tech. My teacher will be happy.[/quote]
HAHA my school got 1st in both Fermat and Pascal :)

[quote="rem"]In group 4 fermat, students names are written first name first last name last, while in the other groups it's the other way. :rotfl:  :rotfl:[/quote]

"first name first last name last"

Do you actually expect a human to be able to parse that?  It took me quite a few seconds before I actually realized what you were trying to say.


I don't understand Modular Arithmetic.  Can someone explain it to me?  And possibly give an example of how it can be used?

Thanks :lol:


let [tex]f: R^n \rightarrow R^n [/tex] be differentiable, with [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0 [/tex].  show that [tex]\omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n  [/tex]
is exact.

-------------------------------------
here's what i got so far:

[tex]\omega[/tex] is a n-1 form, since the "^" on the [tex]dx_i[/tex] indicates that this term is omitted.  f is defined on [tex]R^n[/tex] which is an open, star-shaped set.  because of this, we can use poincare's lemma, which states that on an open starshaped set, every closed form is exact.  therefore we only need to show that [tex]\omega [/tex] is closed.  that is, dw = 0.

when i write out dw, i get

[tex]dw = \sum_{i=1}^n \sum_{\alpha = 1}^n \frac{ \partial{f_i}}{\partial{x_\alpha}} dx_\alpha \Lambda dx_1 \Lambda... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n [/tex]

how do i simplify this? how do i make this equal zero? if [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0 [/tex], does this mean that the expression i wrote for dw equals zero?  i don't really see the connection here, since [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} [/tex] has an extra factor of [tex](-1)^i[/tex]...i'm not sure about the subscripts either.  i hope someone can help me with this.

Hi, man,

I don't know how to write formulas, so sorry if you can't understans smth.

Look at your formula for the dw. Do you understand that all if alpha is equal to any index except the index of the omitted member, the item is zero? And if it's equal to i than it will require exactly i transpositions to put x_alpha to the place of the omitted member, that's how you'll get (-1)^i. The rest is obvious.

P.S. It would have been easier to understand if you started with 1-form on R^2 and have written it all without double sum formulas but as is.

here's what i got so far:

[quote]when i write out dw, i get

[tex]dw = \sum_{i=1}^n \sum_{\alpha = 1}^n \frac{ \partial{f_i}}{\partial{x_\alpha}} dx_\alpha \Lambda dx_1 \Lambda... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n [/tex][/quote]

Since $dx_i\wedge dx_i=0$, then
\begin{eqnarray*}
dw &=& \sum_{i=1}^n \frac{\partial{f_i}}{\partial{x_i}}dx_i\wedge dx_1\wedge\cdots \hat{dx_i}\wedge\cdots\wedge dx_n\\
& = & \left(\sum (-1)^i \frac{\partial{f_i}}{\partial{x_i}}\right)dx_1\wedge\cdots \wedge dx_i\cdots\wedge dx_n=0
\end{eqnarray*}


Let triangle ABC with each angle less than 90 degree.  let p be the FERMAT point of triangle ABC .
THEN  we  will  get   three smaller triangles.  prove that the  RULER LINE of these three triangles meet within one  point.
And when the triangle has one angle larger than 90 degree .Does the problem be all   right.

This problem has been post  before
see http://www.mathlinks.ro/viewtopic.php?search_id=916513781&t=6188


AMC
USA(J)MO
USAMO
geometry
geometric transformation
reflection

Tiankai Liu
Matt Ince
Alison Miller
Aaron Pixton
Tony Zhang
Oleg Golberg

Alternate: Anders Kaseorg
2nd Alternate: Brian Lawrence (he actually started out in blue zuming picked him as alternate because he totally pwned everyone on the TST)

Congratulations to everyone on the team. And I'm looking forward to meeting you guys in Greece.

Interesting how half of the U.S. team is from the same school (Exeter) and only one contestant is from a public high school (Aaron Pixton).  Congrats and good luck at IMO.

[quote="MysticTerminator"]Oleg Goldberg does not attend Exeter.[/quote]

It says so on the AMC website.

[quote="JS1527"]Is one of those people you MysticTerminator?[/quote]

No.

[quote] Are any of those people from AoPS?[/quote]

Occasionally Alison and Aaron.

At least four of the people on the team have accounts on AoPS.

Congratulations Alison!! Have a great time in Greece! We will all be cheering for you!

Good luck to you also Bill!! We will be cheering for you too.

One good reason there are so many Exeter students on the Team this year is that Zuming Feng, the US Team Coach, is a teacher there. He has attracted a lot of very good students to Exeter recently; and I'm sure they get GREAT practice for Olympiads throughout the year.

Oleg Goldberg, who is a student there now, is a recent immigrant from Russia. He was on the Russian IMO Teams in 2002 and 2003, and received Gold Medals (with nearly perfect scores) in both years.

Notice the team this year is also all seniors. So 6 completely new students will be competing next year.

No. They didn't have any on last year's team. Tiankai decided to go to RSI instead.

oops, must've been something else i saw then. i don't follow this much and i just saw something about the IMO or something on a magazine some time ago.

[quote="JS1527"]Is one of those people you MysticTerminator?  Are any of those people from AoPS?[/quote]

hehe not this year, but I know at least 2 of the people are active on AoPS.

[quote="Fierytycoon"][quote="yif man12"]didn't exeter also have like 3 people on last year's team?[/quote]

Actually, I think there were none last year, heh. Although if I remember correctly, Tony was an alternate.[/quote]

Actually, Alison was the alternate last year. Congrats to her and all the others who made the team this year.

[quote="314159265358979323846"]I believe Alison wasn't a USAMO winner last year.[/quote]

I wouldn't bet my life on it, but I'm pretty still pretty sure she was the alternate.

Being an alternate, or even team member, does not require being a USAMO winner. USAMO winners = top 12 USAMO scores, including Canadians. But the candidates for the IMO team = top 12 people eligible to be on the US team only (citizens or green card holders). So however many Canadians there are, there will be that many extra US people eligible for the US IMO team. Anders wasn't a winner this year either, but he's the alternate.

Again, I am not sure, but I believe that there were in fact two alternate last year because of a scoring tie. I'm pretty sure Tony was one of them, but Alison may have been the other.

I thought Boris Alexeev was an alternate last year.

USAMO perfect score but did not attend MOP (opted for RSI):
Tiankai Liu [Exeter]

US IMO team 2003:
Dani Kane
Anders Kaseorg
Mark Lipson
Po-Ru Loh [also perfect USAMO]
Aaron Pixton
Yan Zhang

Alternates (tied by USAMO + TST, tie broken by MOP score):
Tony Zhang (1st alternate) [Exeter]
Boris Alexeev (2nd alternate)

Rounding out the top 12:
Jae Bae
Po-Ling Loh
Kwok Fung Tang [Exeter]

"Blue" group at MOP (27 on the USAMO -- top 12 cutoff was 28):
Jongmin Baek
Ricky Biggs
Matt Ince
Nate Ince
Alison Miller

All of this should be correct, except at worst one mistake in "Blue".  Note that indeed nobody from Exeter was on the team last year.

First of all, for those casually browsing this discussion and seeing that the team in my prior comment resembles last year's team ... that's because it is.  The first comment, to the best of my knowledge, actually reflects this year's team.

[quote="MCrawford"]Last year Canadians were ineligible, though Tiankai Liu (and possibly others) declined the invitation to MOsP, so at least one USAMO Honorable mention took the selection exams last year.[/quote]

Last year only the USAMO Top 12 were eligible for the US IMO team.  Unfortunately, you can't extrapolate an exceptional situation from this year onto previous years (or even later years, although the AMC might now be compelled to use the rule from this year in the future).

For example, I'd be interested to learn whether any honorable mentions were eligible for the team the year Gabriel Carroll opted for RSI.

well, I'm not totally hallucinating. Alison was an alternate. But it was 2002, not 2003.

[quote="Celeborn"]The second alternate this year is Brian Lawrence, a freshman who got a 13 on the USAMO but got the highest score (31?) on the TST.

Crazy...

-Oleg R[/quote]

No one scored higher than a 31 on the TST? I would have thought that there would have been higher scores, since students score even higher on the IMO.

[quote="ComplexZeta"]How can one be an alternate without being a USAMO winner?[/quote]

(After edit:) Because the Canadians who were USAMO winners can't be on the USAMO team, they went deeper into the depth chart after USAMO, I think, to establish eligibility for the IMO team. [Checking this year's [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamowinners.html]USAMO winner's page on the Web[/url]:] Yes, even Anders wasn't a winner this year, although he was an honorable mention. But the name mentioned above, Brian Lawrence, as the second alternate doesn't seem to appear on the USAMO winner's page on the Web, although it does appear on the [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html]USAMO qualifier's page[/url]. Maybe the team selection test is counting for more than it used to these days.

[quote="tokenadult"]
Because the Canadians who were USAMO winners can't be on the USAMO team, they went deeper into the depth chart after USAMO, I think, to establish eligibility for USAMO. [/quote]

No, Brian's score on TST was exceptional, so Zuming made him alternate anyway; it had nothing to do with the Canadians.

However, it is not going to matter (unless two people on the team get murderplied or something).

Yes, I was the alternate two years ago and did not place last year.

Brian Lawrence's score was a 33.  This year's TST was weird in that it was not really harder than the IMO, but one of the problems was computationally intensive, and a number of people misread another.

[quote="tokenadult"]Congratulations to Alison, who has been working toward being on the team for YEARS. (Her mom is the very first person who ever told me about competition culture, back when my oldest son was first-grade age.) Matt Ince is a homeschooler too, so that's a pretty good representation of homeschoolers on this year's team, what with Anders Kaseorg being an alternate.[/quote]

Perhaps the team will include 3 homeschoolers this year. Two team members are having difficulty obtaining visas to go to Greece at the moment and Anders is next in line.

btw, at the awards, aaron pixton got 2nd for black mop tests scoring like 20 pts higher than everyone and oleg golberg just pwned everyone (something like 106 points on 8 combined MOP tests) awesomeness. On a lighter note, I was "most likely to be arrested" and runner-up for something else I cannae remember. Dunbar was voted most late to class :) (after having been about an hour late to one and missing another entirely)

Probably not on the internet. The graders probably still have the list though.

moppers.kaseorg.com

check there after a bit


The incircle $ (I)$ of triangle $ ABC$ touches side $ BC,CA,AB$ at $ A_1,B_1,C_1$ respectively.The circumcircle of $ ABC$ meets the lines $ AI,BI,CI$ at $ A_2,B_2,C_2$ respectively. Denote $ S,S_1,S_2$ be the areas of triangles $ ABC,A_1B_1C_1,A_2B_2C_2$ respectively. Prove that $ S^2\equal{}4S_1S_2$

$ (O, R)$ is the circumcircle, $ (I, r)$ is the incircle. $ \triangle A_1B_1C_1$ is pedal triangle of $ I$ WRT the $ \triangle ABC,$

$ S_1 \equal{} \frac {p(I, (O))}{4R^2}\ S \equal{} \frac {R^2 \minus{} IO^2}{4R^2}\ S \equal{} \frac {R^2 \minus{} (R^2 \minus{} 2rR)}{4R^2}\ S \equal{} \frac {r}{2R}\ S.$

(see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=38164[/url]). $ \triangle A_1B_1C_1 \sim \triangle A_2B_2C_2$ with circumcircles $ (I), (O)$ are (centrally) similar, having parallel their vertex circumradii,

$ \frac {S_2}{S_1} \equal{} \frac {R^2}{r^2}\ \Longrightarrow\ 4S_1S_2 \equal{} \frac {4R^2}{r^2}\ S_1^2 \equal{} S^2.$


Hi Guys
Let f be strictly increasing on the subset S of R
IF f(S) is open  then [b]prove  f is continuous on S[/b]
Thank u in advance

Decompose $f(S)$ into a bunch of disconnected open intervals.  Now consider $x\in\text{bdd}(S)$.  For small enough $\epsilon$, $f(x+\epsilon)\not\in f(S)$.  Then if $f(x)\in f(S)$, $f(x)\in\text{bdd}(f(S))$.  But this is absurd, because $f(S)$ is open and hence does not contain its boundary.  So $S$ does not contain any of its boundary.  In particular, that means that it is open.


1 Suppose that g(x)=x^2 +4x and that f(z) =z square root . Let h(x)=f(g(x)) Calculate h(2). Use two decimals, for example 6.21

2 Which of the following functions will have an inverse?
Välj minst ett svar
	 a. f(x) = x^2 with domain R
	 b. f(x) = x^2 with domain x > 0
	 c. f(x) = x^2 with domain x > 0 AND x = 0
	 d. f(x) = ln(x) (x > 0)
	 e.  for domain R
	 f. f(x) = x - ln(x) for x > 0

3 Suppose that f(1) = 1, f(-1) = 2 and f(3) = 2. Let g(x) = (f(x) + f(-x))*f(3x). Calculate g(1).

What are you having trouble with specifically?

Need help to find solutions to all of them please

1.
$ g(2)\equal{}12$, I think that $ f(x)\equal{}\sqrt{x}$ right if so, $ \sqrt{12}\equal{}\boxed{3.46}$


integration
calculus
Euler
function
limit
calculus computations

$\int_{a}^{b}\frac{xdx}{\sqrt{(x-a)(b-x)}}$
 caculus intergal

Do the Euler substitution $\sqrt{(x-a)(b-x)}= t(x-a)$, i.e. $t = \sqrt{\frac{b-x}{x-a}}, \, x = \frac{b+a t^{2}}{1+t^{2}}, \, dx = \frac{2(a-b)t}{\left( t^{2}+1 \right)^{2}}$!

Thus,
\begin{eqnarray*}I &=& \int_{a}^{b}\frac{x}{\sqrt{(x-a)(b-x)}}\, dx \\ \ &=& \int_\infty^{0}\frac{\frac{b+a t^{2}}{1+t^{2}}}{\frac{t(b-a)}{1+t^{2}}}\cdot \frac{2(a-b)t}{\left( t^{2}+1 \right)^{2}}\, dt \\ \ &=& 2 \int_{0}^\infty \frac{b+a t^{2}}{\left( t^{2}+1 \right)^{2}}\, dt \\ \ &=& \int_{-\infty}^\infty \frac{b+a t^{2}}{\left( t^{2}+1 \right)^{2}}\, dt .\end{eqnarray*}

Consider the function $f(z) = \frac{b+a z^{2}}{\left( z^{2}+1 \right)^{2}}$. We integrate this from $-R$ to $R$ on the real line and then over a semicircle of radius $R$, centered at the origin, traced counter-clockwise. Call this contour $\gamma(R)$.

The first part clearly goes to $\int_{-\infty}^\infty \frac{b+a t^{2}}{\left( t^{2}+1 \right)^{2}}\, dt$ (which is finite) as $R \to \infty$. The second part can be written as $\int_{0}^\pi \frac{b+a R^{2}\exp \left( 2 i \theta \right)}{\left( 1+R^{2}\exp \left( 2 i \theta \right) \right)^{2}}\cdot i R \exp \left( i \theta \right) \, d \theta$. It should be pretty clear that this goes to $0$ as $R \to \infty$.

Hence, $I = \lim_{R \to \infty}\int_{\gamma(R)}f(z) \, dz$.

We can compute the last integral via the residue theorem. The only pole inside $\gamma(R)$ is $i$. The residue there is
\begin{eqnarray*}\textrm{Res}(f,i) &=& \frac1{1!}\cdot \lim_{z \to i}\frac{ d \left( \frac{b+a z^{2}}{(z+i)^{2}}\right)}{dz}\\ \ &=& \lim_{z \to i}\frac{2az(z+i)-2 \left( b+a z^{2}\right)}{(z+i)^{3}}\\ \ &=& \frac{a+b}{4i}.\end{eqnarray*}
Hence, $\int_{\gamma(R)}f(z) \, dz = 2 \pi i \cdot \textrm{Res}(f,i) = \frac{\pi \left( a+b \right)}2$.
Therefore, $I = \frac{\pi \left( a+b \right)}2$.

PS: There are many other ways to compute this... and I think I chose the most tedious one.

i think the following may be an easier way

put $x=acos^{2}t+bsin^{2}t$ with appropriate limits. sorry i dont have more time to work out the full solution

The easiest (???) solution is, I think, $y = x-\frac{a+b}2$:
\[I = \int_{\frac{a-b}2}^{\frac{b-a}2}\frac{y+\frac{a+b}2}{\sqrt{\left( \frac{b-a}2 \right)^{2}-y^{2}}}\, dy = \left. \frac{a+b}2 \cdot \arcsin \left( \frac{2y}{b-a}\right) \right|_\frac{a-b}2^\frac{b-a}2 = \frac{\pi \left( a+b \right)}2 . \]
(The other part of the integral is $0$, since we're dealing with an  odd function and the improper integral obtained is convergent.)


A man arrives at an airport from 11:00 to 12:00, waits for his wife to pick him up for 20 minutes, and goes home if she doesn't. His wife arrives at 11:00 to 11:40, waits for him for 10 minutes, and drives home if he is not there after 10 minutes. What is the probability that the wife picks him up?

let the times that they arrive be x, and y for the man and woman respectively
y-x <20 or x-y<10, then draw a box on a cooridinate grid that has 0 to 60 for x, and 0-40 for y and find the ratio of the area that is shaded to the ratio of the whole box
as this for being geometry, not really, it is pretty easy to find areas of these triangles, it is more algebra than anything, this is a lot like linear programming


In how many ways can we select 'r' objects from 'n' distinct objects placed in a circle such that no two of these selected objects are placed together.

cut the "necklace" of objects somewhere, say between consecutive objects $ A$ and $ B$, and make a line out of them: $ A, ..., B$. how many ways are there to choose $ r$ objects from this line so that no two are next to each other? well, $ \binom{n \minus{} r \plus{} 1}{r}$. but now back to the necklace: it may be that we chose both $ A$ and $ B$ from the line, which is not allowed for the necklace. so the question is, how many ways are there to pick $ r$ objects from the line, two of which are $ A$ and $ B$, so that no two are next to each other? well, $ \binom{n \minus{} 4 \minus{} (r \minus{} 2 \minus{} 1)}{r \minus{} 2} \equal{} \binom{n \minus{} r \minus{} 1}{r \minus{} 2}$. (take $ A$ and $ B$ and their neighbors out of the picture (4 objects in total), then use the same formula as above, with $ n$ replaced by $ n \minus{} 4$ and $ r$ replaced by $ r \minus{} 2$. so the answer is $ \binom{n \minus{} r \plus{} 1}{r} \minus{} \binom{n \minus{} r \minus{} 1}{r \minus{} 2}$.

Arithmetypo: $ \minus{} 4 \minus{} ( \minus{} 2 \minus{} 1) \equal{} \minus{} 1$ ;).

Alternatively, choose the first element in $ n$ ways; then we must choose $ r \minus{} 1$ elements from a line of $ n \minus{} 3$.  We can do this in $ \binom{n \minus{} 3 \minus{} (r \minus{} 1) \plus{} 1}{r \minus{} 1} \equal{} \binom{n \minus{} r \minus{} 1}{r \minus{} 1}$ ways.  However, this counts every choice $ r$ times, so we have in total $ \frac {n}{r} \binom{n \minus{} r \minus{} 1}{r \minus{} 1}$ choices.

[quote="JBL"]Arithmetypo: $ \minus{} 4 \minus{} ( \minus{} 2 \minus{} 1) \equal{} \minus{} 1$ [/quote]

now that's a word worthy of any linguist's lexicon ;)


modular arithmetic
number theory proposed
number theory

For positive integers $ n,k$, define $ F(n,k)\equal{}\displaystyle\sum_{r\equal{}1}^{n}r^{2k\minus{}1}$. Prove that: $ F(n,1)$ divides $ F(n,k)$.


This is pretty much where you make up your own awesome math competition, and then you make a cool acronym and stuff!

:rotfl: , looks like this thread is failing... I would call mine: the Mathematics Elements Listing Examination(MELE) :D !

It would be like a qualifying exam, sorta like AIME!

The Critical Objects of Mathematical Information Contest (COMIC)

The International Math Competition for Elementary Schoolers, or IMCES for short.  It would be sort of like Mathcounts, except I'd have 5 different rounds(and for once an elementary schooler would get a chance to participate in a real competition) The "Speed" round is where the competitors try to answer 250 word  problems ranging from "3rd grade ISTEP+ level" to hard MC states" level  in 40 minutes(That way there shouldn't be so many ties).  Then a "Skills" round where 20 problems are handed out in sets of fours with 4 minutes per set of problems(calculators allowed).  Following that would be a "Stretch" round where the third through fifth graders group in teams of four and answer 20 pretty difficult questions in 25 minutes.  Then the 50 highest scorers would participate in a "Semi-Final" round where all the students are(while on a stage), asked a problem and given 45 seconds to answer.  Each student must write out his/her answer and if they get 2 questions wrong, they're out.  Finally, the "Oral Round" is a bracket style tournament for the top 16 students very similar to an MC countdown round.  And the best part is, the competition would take place in Athens, Greece!

I can't think of anything, so I'll say NIMO. (pronounced NYE-moh)

All of you who said NIMO run NIMO. It is an actual competition that is run by you guys... Not bad... I'm thinking of doing something similar :wink:

Oh, come on, you didn't even say what NIMO meant (the National Internet Mathematical Olympiad).

No wonder this thread is failing.


The sides of unit square $ ABCD$ have trisection points $ X$, $ Y$, $ Z$ and $ W$, as shown. If $ AX: XB \equal{} BY: YC \equal{} CZ: ZD \equal{} DW: WA \equal{} 2: 1$, what is the area of the shaded region? Express your answer as a common fraction.

[asy]fill((.9,.3)--(2.7,.9)--(2.1,2.7)--(.3,2.1)--cycle,gray(.7));
draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1));
draw((0,0)--(3,1),linewidth(1));
draw((3,0)--(2,3),linewidth(1));
draw((3,3)--(0,2),linewidth(1));
draw((0,3)--(1,0),linewidth(1));
label("A",(0,3),N);
label("X",(2,3),N);
label("B",(3,3),N);

label("W",(0,2),W);
label("Y",(3,1),E);
label("D",(0,0),W);
label("Z",(1,0),S);
label("C",(3,0),E);[/asy]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252209]Click because I'm lazy[/url] :P


[color=cyan]If  1 + sin(2x) = 2cos(2x), compute all possible values of tan(x)[/color]

would u use the LAw of tangents for this?,

or sum of sine-cosine laws?

Transform this into sin (90-2x)/(1+cos (90-2x)) = 1/2, hence tan (45-x) = 1/sqrt(2). From this it is very, very easy to determine tan x, but I won't, because it's too hard!


pigeonhole principle
combinatorics proposed
combinatorics

[quote="felixmann"]In a school there are $ 10$ rooms. Each student from a room knows exactly one student from each one of the other $ 9$ rooms. Prove that the rooms have the same number of students (we suppose that if $ A$ knows $ B$ then $ B$ knows $ A$).[/quote]

[hide]Suppose there are two rooms $ M$ and $ N$ such that $ M$ has more students than $ N$.  Each student in $ M$ has exactly one friend in $ N$.  But since $ \# (M) > \#(N)$, then by pigeonhole, there is at least two students in $ M$ who have the same friend in $ N$.  However, this cannot be, because $ N$ would have more than one friend.  Hence, all rooms must have an equal number of students.[/hide]


Since I recently found out what the China Syndrome is, I asked a bunch of people if they knew what it was. No one did. So, I decided to move the poll here for fun. So, do you know what the China syndrome is?

[quote="t0rajir0u"]Well, [url=http://en.wikipedia.org/wiki/China_Syndrome]now I do.[/url][/quote]

Lol if you put "Don't Google it" people will Google.

I immediately thought of the China effect. Aww... well.

I thought of lead poisoning or something  :wink: Sorry to you people in China.

Believe it or not, I immediately thought of chow mein ( :D ). This is because I got confused with something I read somewhere about some kind of Chinese food effect thingy, which is about how Chinese food is not a good idea for dieting.

Well yes, if I put "don't Google it," people will probably Google it, but who cares. This is just for fun.

It was on the news, "The China Syndrome" about how they got into the car building business?


Can you draw the picture about   the function : f-0(x)=e^x

I'm sorry . But I don't know how to input the expressions into the messeage ,could you tell me??

Oh,I'm sorry .this is the first time that I write in the forum.Thanks for your help!  $f_0(x)=e^x$


geometry
pigeonhole principle
induction
symmetry
inequalities
number theory

For the first time since I can remember, there is no discussion of National Camp on this forum.  Is anyone here coming this year? July 6-12 at John Abbott College in Montreal - yay!  :D

what (olympiad?) topics are we going to learn there?

We have two sessions on Number Theory (mods and diophantine equations), two on Geometry (circle geometry and triangle geometry), one on proofs and problem solving techniques (Pigeonhole, induction, wolog, etc), a session on Parity and Symmetry, a session on Combinatorics, and possibly a session on either Inequalities or something else.  We focus on developing good-quality solution-writing, and help transition into being able to solve (more) olympiad-level problems.  We also have "fun talks" about cool things in math that don't necessarily relate to competitions (past fun talk topics have included Ramsey Theory, Surreal Numbers, Cryptography, etc).  And the whole experience is as much about meeting other math students across Canada as it is about learning math.

If you are Heinrich from Windsor (which I assume you are, because I don't know too many other Heinrichs), you should be able to ask numerous National Camp alums about what it's like.  And while you're at it, send my regards to Mr. White!

ahhhh I want to go again. Pretty awesome camp there=)

=)

Safe to say it was a fun, successful year!!!  Let's see how many national campers we bring to AoPS this year eh?

[quote="Sunny"]Let's see how many national campers we bring to AoPS this year eh?[/quote]
Well, you did convince me.....
Took a look on here for the first time today
It's a forum.... but dedicated for nerdy, mathy people.... *gasp*
This is almost as exciting as that 755-page IMO book we pirated off the internet XD

[quote="richrichardw"]
It's a forum.... but dedicated for nerdy, mathy people.... *gasp*
This is almost as exciting as that 755-page IMO book we pirated off the internet XD[/quote]

Hahaha, Hi Richie!  You mean, the book you pirated when you were supposed to be sleeping?  =)

Probably the IMO Compendium from 1959 to 2004... correct me if I'm wrong.  :D

:D It was a really fun camp!!:)

I just want to remind everyone that there's a facebook group for the national camp, too!

Come on up and join!!!!

We need more people :wink: