Same bracket group from last year; ID #84309; Password 'mathlinks' at [url]http://tournament.fantasysports.yahoo.com/t1[/url].

Now [i]this[/i] sounds like fun.  Count me in.

Come on people, join! There were like 15 AoPSers in this last year (which is almost double this year's count of 8 brackets so far).

Also, the logo seems to be biased towards a specific college...

I cant seem to join...it says that the group is not private...and it goes to the Second Chance thingy.

I have two brackets on espn.com

Right now I have 7/8 in one bracket and 6/8 in the other

Sorry, it's already too late, since the first game was on Thursday at about noon.

Good luck on everyone's predictions, though, whether you're in the group or not!

(darn you Temple)

Dang this would've been nice to find before...I don't usually go in this forum though.

My bracket is failing in an epic fasion.

[url=http://games.espn.go.com/tcmen/entry?entryID=3266060]here[/url]

Yeah I just made that account just for this. Also 496 was obviously a completely random number I just happened to choose, along with my password.

rofl...i'll REALLY lol if north carolina gets knocked out before the final four
probably would screw up every1's bracket :P

here is my bracket

http://games.espn.go.com/tcmen/entry?entryID=1492928

So I had Wake Forest in the final 4  :(

Interesting Statistic. Day 1 of Round 1: 15/15. Day 2: 8/17.  :huh:

So did I. :(

Hmmm... I remember I was dead last early on, when my predicted Mississippi St. upset messed up, but I somehow managed to predict all of the 5-12 "upsets" correctly.

So w00t first place but not for long because of Wake Forest. Hopefully, I'll get most of the third round picks right. Also, it would be pretty funny if all #1 seeds made it to final four again.

Edit: It would be even funnier if Arizona beat Louisville.

How many points do you all have and how is yours scored?

I think yahoo's is scored like ESPN's, except divided by ten (1 point for first round, 2 points for 2nd, 4 points for 3rd etc.).
So the top 7 in our group is (copied directly from the website, sorry for bad formatting)

1 	anderson 	25 	28 	- 	- 	- 	North Carolina 	53 	169
2 	Mewto's bracket 	24 	28 	- 	- 	- 	North Carolina 	52 	180
2 	hopefully... (popsticker muncher?)	24 	28 	- 	- 	- 	North Carolina 	52 	180
4 	random  (zx.gary)	23 	28 	- 	- 	- 	North Carolina 	51 	179
4 	cf249's Bracket 	25 	26 	- 	- 	- 	Oklahoma 	51 	179
4 	Math (lawrencewu)	25 	26 	- 	- 	- 	Louisville 	51 	179
7 	rawr go aime15 	22 	28 	- 	- 	- 	Louisville 	50 	178

Place, then bracket name, then first round score, second round score, predicted winner, points, total possible points (mine is lower than everyone else's due to the unfortunate Wake Forest pick).

popsticker muncher is me

sorry i made it so nebulous (SAT WORD!)

What's interesting is that my bracket is better than 85% of all brackets, yet it's only tied for 4th in the AoPS group.

[quote="CatalystOfNostalgia"]That's because math wins when it comes to these things. One of my friends wrote  a program using Markov chains and his bracket is like 3rd out of 30 (in a different group).[/quote]

um...math helps a lot but ultimately u have to know the teams really well (and get lucky)
i'm PRETTY SURE the program didn't put wake forest out in the first round... :wink:

[quote="spiderboy95"]I think UCONN to win...[/quote]

BOO
lol
i hope arizona upsets louisville (probably not going to happen...but oh well)

woot! go arizona!!!! lets win!!! (well, we prbly wil lose, but still.......)


function
quadratics
trigonometry
analytic geometry
graphing lines
slope
rotation

(1) Let $ x$ be real numbers such that $ x^3 \plus{} \frac {1}{x^3} \equal{} 52$, Find the value of $ x^4 \plus{} \frac {1}{x^4}$.

(2) Given a point $ P$ in the inside of the  regular triangle $ ABC$ such that $ PA \equal{} 1,\ PB \equal{} 2,\ \angle{APB} \equal{} 120^\circ$. Find the length of $ PC$.

(3)Find the least a positive number which has 28 divisor.

(4) Find the real numbers $ x$ for which the function $ f(x) \equal{} \sum_{k \equal{} 1}^{100} |kx \minus{} 1|$ is minimized.

[hide="Number 1"]

Let $ n \equal{} x \plus{} \frac{1}{x}$

$ n^3 \equal{} x^3 \plus{} 3(x \plus{} \frac{1}{x}) \plus{} \frac{1}{x^3}$

$ n^3 \equal{} 52 \plus{} 3n$

From inspection, $ n\equal{}4$

$ n^4 \equal{} x^4 \plus{} \frac{1}{x^4} \plus{} 4(x^2 \plus{} \frac{1}{x^2}) \plus{} 6 \equal{} 4^4$

$ x^4 \plus{} \frac{1}{x^4} \equal{} 250 \minus{} 4(x^2 \plus{} \frac{1}{x^2})$

To fill in the term on the right,
$ n^2 \equal{} x^2 \plus{} \frac{1}{x^2} \plus{} 2 \equal{} 4^2$
$ x^2 \plus{} \frac{1}{x^2} \equal{} 14$

$ x^4 \plus{} \frac{1}{x^4} \equal{} 250 \minus{} 4(14) \equal{} \boxed{194}$
[/hide]

Good! :)

[hide="My solution"]

$ x^3\plus{}\frac{1}{x^3}\equal{}52\Longleftrightarrow \left(x\plus{}\frac{1}{x}\right)^3\minus{}3\left(x\plus{}\frac{1}{x}\right)\minus{}52\equal{}0$

$ \Longleftrightarrow t^3\minus{}3t\minus{}52\equal{}(t\minus{}4)(t^2\plus{}4t\plus{}13)\equal{}0$ where $ t\equal{}x\plus{}\frac{1}{x}$.

Since $ t^2\plus{}4t\plus{}13\equal{}(t\plus{}2)^2\plus{}9>0$, yielding $ t\equal{}4$.

$ \therefore x^4\plus{}\frac{1}{x^4}\equal{}\left(x^2\plus{}\frac{1}{x^2}\right)^2\minus{}2\equal{}(t^2\minus{}2)^2\minus{}2\equal{}\boxed{194}$.[/hide]

[quote="kunny"](1) Let $ x$ be real numbers such that $ x^3 \plus{} \frac {1}{x^3} \equal{} 52$, Find the value 0f $ x^4 \plus{} \frac {1}{x^4}$.

(2) Given a point $ P$ in the inside of the  regular triangle $ ABC$ such that $ PA \equal{} 1,\ PB \equal{} 2,\ \angle{APB} \equal{} 120^\circ$. Find the length of $ PC$.

(3)Find the least a positive number which has 28 divisor.

(4) Find the real numbers $ x$ for which the function $ f(x) \equal{} \sum_{k \equal{} 1}^{100} |kx \minus{} 1|$ is minimized.[/quote]
[hide="1 Er this was messy..."]
Move 52 over to get x^3-52+1/x^3. Multiply by x^3, solve for x^3 with quadratic equation to get x^3=26+/-15root3. x^4+1/x^4=(26+15root3)^4/3 or (26-15root3)^4/3
[/hide]EDIT: nevermind i'm wrong.
[hide="2 Really messy"]If ABC is regular, it is equilateral. Using the law of cosines, and the given lengths/angle, we get AB=BC=AC=$ \sqrt {7}$ We can use law of cosines again to find that cosPAB=3/rt7, so sinPAB=2/rt7 (a little pythagorean theorem work gets us there). The altitude from P to BC is 2sinPBC. PBC=sin(ABC-ABP). We know sinABC=sin60=rt3/2. cos60=1/2, cosABP=3/rt7, sinABP=2rt7. Using the angle subtraction formula, we get sinPBC=sin60cosAPB-cos60sinAPB=(rt3/2)(3/rt7)-(1/2)(2/rt7)=((3rt3-2)rt7)/14. The altitude is then (3rt3-2)/rt7. The segment with endpoints at the foot of the altitude and B can be found by pythagorean theorem. It is $ \frac {\sqrt {12\sqrt {3} \minus{} 4}}{\sqrt {7}}$. To find PC we subtract that from root 7, square it, add the altitude we found, and take the square root. The answer is $ \frac {76}{7} \minus{} 2\sqrt {12\sqrt {3} \minus{} 4}$
Hopefully I didn't make an error :([/hide]

[hide="3"]
This number either equals
$ p_1^{27}$, $ p_1^{13}p_2^{1}$, $ p_1^6p_2^3$, or $ p_1^6p_2^1p_3^1$, where the $ p_i$ are distinct primes. Of these, the last is the least, with $ 2^6\cdot3\cdot5\equal{}960$.
[/hide]

[hide="4"]
The graph $ |kx\minus{}1|$ has a vertex at $ x \equal{} \frac{1}{k}$; to the left of here it has slope -k, to the right it has slope k.

We want to make the slope of this sum close to 0, since we know it increases gradually.

We start with a slope of $ \minus{}100\minus{}99 \minus{} \ldots \minus{} 1$ at 0. When we cross x=1/100, the slope becomes $ 100 \minus{} 99 \minus{} \ldots$, and when we cross -1/99, the slope becomes $ 100\plus{}99\minus{} \ldots$. Generally, immediately to the right of $ x \equal{} \frac{1}{n}$, the slope becomes $ 100 \plus{} \ldots n \minus{} (n\minus{}1) \minus{} \ldots \minus{} 1 \equal{} \frac{100(101)}{2} \minus{} 2 \frac{(n\minus{}1)(n)}{2} \equal{} 5050\plus{}n\minus{}n^2$. We want to find when this switches from negative to positive.

Solving the equation for 0, we get $ n \equal{} \frac{1}{2} (1 \plus{} \sqrt{20201}) \approx 71.5651 \ldots$

This means, after x = 1/72, the slope is still negative, and after x = 1/71, the slope becomes positive. This means that the minimum value occurs at $ \boxed{x \equal{} \frac{1}{71}}$

A quick number crank gives the minimum value to be $ \frac{2920}{71} \approx 41.127$
[/hide]

[quote="kunny"]Good! :)

What do you think how the problem 4 is difficult in U.S.A.?[/quote]
I think when you ask them to find with x is an integer or give the sum of several integer variables x_1,x_2,...,x_n and then find max! The restriction of problem 4 is not enough strict to make it hard!

As you said, we usually see a problem such that :

[color=blue]Find the positive number $ n$ for which $ |n\minus{}1|\plus{}|n\minus{}2|\plus{}\cdots \plus{}|n\minus{}100|$ is minimized.[/color]

1959 Kyoto University entrance exam/Humanities

[quote="kunny"]As you said, we usually see a problem such that :

[color=blue]Find the positive number $ n$ for which $ |n \minus{} 1| \plus{} |n \minus{} 2| \plus{} \cdots \plus{} |n \minus{} 100|$ is minimized.[/color]

1959 Kyoto University entrance exam/Humanities[/quote]

Right. These kinds of problems are especially approachable if one has seen something similar before, or if one is familiar with increasing-decreasing arguments. As such, someone who has taken calculus may be quicker to intuitively solve the problem than someone who has not, even though there is no calculus required -- of course, there's nothing stopping someone who hasn't taken calculus from figuring it out  :D

The difference between problem 4 and the problem above is that for the problem above, one can take the terms pairwise, namely |n-k| + |n-101+k|, and it works out easier, whereas for the problem 4 one likely has to consider all 100 terms.

I find it curious that those going into the humanities need to be able to solve problems like this... in America, lots of people going into the humanities shy away from math (or run away from math)... unless economics counts as humanities?

[quote="delta"][hide="2 not messy at all"] Rotate around A by 60 degree clockwise to get PC=sqrt3[/hide][/quote]
What? I don't quite get how you're getting that...


algebra
polynomial
geometry
3D geometry
Vieta

1. Compute the sum of the squares of the roots of the polynomial $f(x)=x^{3}+14x^{2}-30x+15$. What is the sum of the cubes of the roots?

2. Determine the sum of the reciprocals of the roots of the polynomial $f(x)=x^{3}+6x+15$.

k for the cubes of the roots I get a different answer..

[hide]

Using Newtons sum we have

$s_{3}+14s_{2}-30s_{1}+3(15)=0$
$\Rightarrow s_{3}=30s_{1}-14s_{2}-45$
$\Rightarrow s_{3}=30(-14)-14(256)-45=\boxed{-4049}$ [/hide]

Yeah, sorry about that, I think I used the wrong numbers when I calculated the sum  :blush: I fixed it and got $-4049$ as well.

ya i saw that too but i think he still gets a different answer after correcting it (probably a sign mistake or something)

[hide="2"]
Reverse coefficients of polynomial:

$15x^{3}+6x^{2}+1$

This is the polynomial with inverse roots.  The sum is thus $-2/5$ by Vieta.
[/hide]


The circles C1 and C2 lie inside the circle C, and are tangent to it at M and N, respectively. C1 passes through the center of C2. The common chord of C1 and C2, when extended, meets C at A and B. The lines MA and MB meet C1 again at E and F. Prove that the line EF is tangent to C2

$ NA $ meets $ c_2 $ at point $Q$ .  the tangent from $Q$ on $C_2$ passes $F $ and it is tangent on $ C_1$ on $f$.(?)
$ \rightarrow O_1F \perp FQ $ and $ o_2Q \perp FQ $.(1)
$ M$ is the homogeneous center of $C_1$ and $ C$ $ \rightarrow EF $ is parallel to $AB$. (?)
$ O_1O_2 \perp AB \rightarrow O_1O_2 \perp EF$ (2)
$ EF$ and $O_1O_2 $ intersects each other at point $W$.
$ \rightarrow $ if we prove that $\O_2W=r_2 $ the problem is solved!
$ O_2WFQ $ is an inscribed quadrilateral $\rightarrow$ if we prove $O_2Q=O_2W $ the problem is solved.(?)
so we should prove that \[ \angle WFO_2=\angle O_2FQ \]
\[ O_1O_2 \perp EF \rightarrow EW=WF \rightarrow \angle WO_1F=\angle WO_1E \rightarrow \angle O_2FW=\angle O_2FQ \] (?).
so we proved that $O_2W=O_2Q=r_2 \rightarrow EF $ is tangent on $C_2$
 :D


Does anyone know where I can get past Indian Physics Olympiad or any other countries' olympiad papers??? I already have all the international olympiads online!!

There's a web page with Slovak physics olympiad papers - all the rounds, past four years. Unfortunately, it's in Slovak. But if you're interested in something specific, I could translate a problem or two for you....

Please give me the link!!! I'll see if I find something interesting ( by seeing the diagram etc.) and inform you accordingly!!!!  :lol:

Any other suggestions??? What about Indian past papers????

No problem, it's

http://fpv.utc.sk/fo/archiv.htm

You'll be interested in category A - that's the highest category (for people planning to go to IPhO) - and round 3 is the national round, although that doesn't necessarily mean that round 3 questions are the thoughest. There's quite a few tasks with no pictures / formulas though so I'm not sure how much you can figure out...

Are you aware of some online Slovak to English translators (free!!)... I won't like to disturb you!!! - you were right - not many diagrams!!!

I use http://slovnik.cyberspace.sk/ but I think it's going to be really hard to translate the text just like that. I could translate the names of the problems for you though and then if you liked anything....

I have the website for Indonesian Physics Olympiad.
The Qs are in indonesian.
Because Malay is quite similar to Indonesian,i manage to understand the Qs :P
Inform me if any1 need help on translating the Qs ;-)

Here it is :
http://www.tofi.or.id/q22003/id/services/download.php

Asian Physics Olympiad

http://www.apho.org

How do you enter this??? Is India there?

Hm, I looked through this and found out that India doesn't seem to participate any more:

http://www.apho.org/en/olympiad/part.php?origin=aboutus&charencode=en

But you can at least download all the problem papers (except from the 3rd APhO)...

Ya, it looks that way to me too!! Maybe kennylcc can give more info!!! Too little opprtunities in India!!! Fiirst IYPT and now APhO  :mad:  :mad:

Does anyone know a web for Physics Olympiad  in[size=150] FRENCH[/size] :?:

http://apho.org/en/aboutus/enroll.php?origin=aboutus&charencode=en

"The APhO is not an open event. The APhOs are organized at the governmental level and always an invitation from the Organizers is necessary.
Each year on the occasion of APhOs, the organizing commitee of the next Olympiad will come along with invitations / Olympic booklet as well as the previous Organizer will bring the proceedings of the previous Olympiad.
Here every Official teams, as well as Observers and Guests may ask the next Organizing committee of the next Olympiad for an invitation."

I don't really know how..
But i think u can contact ur IPhO trainers/leaders to ask them about APhO.
I heard that the next APhO will be held in Taiwan.

BTW,
do u have good materials to prepare for IPhO ?
Fundamentals of Physics seem to be too "fundamental" for IPhO ^^"
Any good book to recommend?
Is Feynman's Lectures series suitable for IPhO ?


For $a,b,c$ sides of a triangle prove that

$(a+b+c)^3 \geq (5a-b-c)(5b-c-a)(5c-a-b)$

Here is, at least, an ugly proof: Since a, b, c are the sides of a triangle, we can substitute a = y + z, b = z + x and c = x + y with positive x, y, z. And then we get

$\left( a+b+c\right) ^{3}-\left( 5a-b-c\right) \left( 5b-c-a\right) \left( 5c-a-b\right) $
$=8\cdot \left( 2\left( x^{3}+y^{3}+z^{3}-3xyz\right) +3\left( x\left( x-y\right) \left( x-z\right) +y\left( y-z\right) \left( y-x\right) +z\left( z-x\right) \left( z-y\right) \right) \right) $.

And this is $\geq 0$, since $x^{3}+y^{3}+z^{3}-3xyz \geq 0$ by AM-GM and $x\left( x-y\right) \left( x-z\right) +y\left( y-z\right) \left( y-x\right) +z\left(
z-x\right) \left( z-y\right) \geq 0$ by Schur. Hence the inequality is proven.

  Darij

I simply do not see why AM-GM does not work!

You are right, as usual. The "GMA" in the source irritated me...  :blush:

And note that your solution, Harazi, shows that a, b, c can be arbitrary positive numbers, not necessarily the sides of a triangle.

  Darij

Indeed, these do not seem from GMA. Perhaps you meant GMB, manlio? Cuz I've seen some things from GMA while ago, and they were piece by piece much much harder than this... :)


Hi,

can anyone confirm results of the German team ? Thanks.   :)

Here are the complete results of the German team. The k-th number in the squared brackets means the number of points achieved by the student on the k-th problem.

Germany 1 = [b]Darij GRINBERG[/b]. [6; 7; 0; 7; 7; 0] - altogether 27 points, silver.
Germany 2 = [b]Annika HECKEL[/b]. [6; 2; 1; 7; 0; 1] - altogether 17 points, bronze.
Germany 3 = [b]Matthias OHST[/b]. [7; 1; 2; 0; 0; 1] - altogether 11 points, honorable mention.
Germany 4 = [b]Christian SATTLER[/b]. [6; 5; 7; 7; 1; 3] - altogether 29 points, silver.
Germany 5 = [b]Peter SCHOLZE[/b]. [7; 5; 1; 7; 7; 4] - altogether 31 points, silver, one point more would be gold.
Germany 6 = [b]Michail SHKOLNIKOV[/b]. [6; 1; 0; 7; 1; 0] - altogether 15 points, honorable mention, one point more would be bronze.

  Darij


vector
analytic geometry
trigonometry
linear algebra
matrix
calculus
calculus computations

Hello!

In cartesian coordinates, the vector gradient is :
[img]http://img151.imageshack.us/img151/1507/cartsiennescp4.jpg[/img] [b]{A}[/b]

In polar coordinates, we write :
[img]http://img151.imageshack.us/img151/7106/polairesvu0.jpg[/img] [b]{B}[/b]

I have forgotten how we demonstrate B from A... Can you help me?

Thank you!

There should be another term in the second of these. Two vecotrs are insufficient to form a basis of three dimensional space.

Either that, or reduce the first one to the two dimensional version.

Well the direct prescription is very straight forward:
We note that a point $ \(x,y,z\)$ when carried over to spherical polar coordinates is given by $ \(r \sin \theta \cos \phi , r \sin \theta \sin \phi, r \cos \theta \)$. Using the three relations above one can for example transform coordinates of $ U$ from rectangular to spherical system. Moreover the unit vectors $ \hat{i} , \hat{j}, \hat{k}$ can be written in terms of $ \hat{r} , \hat{\theta}, \hat{\phi}$ using the orthogonal matrix:

$ \begin{bmatrix}A_{r} \\ A_{\theta} \\ A_{\phi}\end{bmatrix}$ = $ \begin{bmatrix}\sin \theta \cos \phi  & \sin \theta \sin \phi &  \cos \theta \\ \cos \theta \cos \phi  & \cos \theta \sin \phi &  - \sin \theta \\ \sin \phi & \cos \phi & 0 \end{bmatrix} \cdot \begin{bmatrix}A_{x} \\ A_{y} \\ A_{z}\end{bmatrix}$

Where $ A_{x}$ means the scalar component of $ A$ along $ x$ direction and so on and so forth.

I hope that to some extent answers the question posed.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Prove that:
\[ (3\minus{}a)(3\minus{}b)(3\minus{}c)\geq8\]

[hide]Consider the function $ f(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c) \equal{} x^3 \minus{} (a \plus{} b \plus{} c)x^2 \plus{} (ab \plus{} bc \plus{} ca)x \minus{} abc$.
$ f(x) \equal{} x^3 \minus{} px^2 \plus{} qx \minus{} r$ where $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, and $ r \equal{} abc$.
$ p^2 \equal{} (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ac$ and
$ 2q \equal{} 2(ab \plus{} bc \plus{} ca) \equal{} 2ab \plus{} 2bc \plus{} 2ca$ so
$ p^2 \minus{} 2q \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.
Solving for $ q$, we get
$ q \equal{} \dfrac{p^2 \minus{} 3}{2}$.
$ r \equal{} abc \le \left(\displaystyle\sqrt {\dfrac{a^2 \plus{} b^2 \plus{} c^2}{3}}\right)^3 \equal{} \left(\displaystyle\sqrt {\dfrac{3}{3}}\right)^3 \equal{} 1$ by QM-GM.
$ f(3) \equal{} (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \equal{} 27 \minus{} 9p \plus{} 3q \minus{} r$
$ \equal{} 27 \minus{} 9p \plus{} 3\left(\dfrac{p^2 \minus{} 3}{2}\right) \minus{} r$
$ \equal{} 27 \minus{} 9p \plus{} \dfrac{3p^2}{2} \minus{} \dfrac{9}{2} \minus{} r$
$ \equal{} \dfrac{45}{2} \minus{} 9p \plus{} \dfrac{3p^2}{2} \minus{} r$
$ \equal{} 9 \plus{} \left(\dfrac{27}{2} \minus{} 9p \plus{} \dfrac{3p^2}{2}\right) \minus{} r$
$ \equal{} 9 \plus{} \dfrac{3}{2}(9 \minus{} 6p \plus{} p^2) \minus{} r$
$ \equal{} 9 \plus{} \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r$
We have $ (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \equal{} 9 \plus{} \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r$.
Since $ \dfrac{3}{2}(p \minus{} 3)^2 \ge 0$ and $ \minus{} r \ge \minus{} 1$,
and adding these inequalities together we have
$ \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r \ge \minus{} 1$,
$ 9 \plus{} \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r \ge 8$.
$ (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \equal{} 9 \plus{} \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r \ge 8$, with equality when $ a\equal{}b\equal{}c\equal{}1$,
so we are done.[/hide]

Nice proof, Mij! :lol:
I think, your method does not help for the following inequality.

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ (5 \minus{} 2a)(5 \minus{} 2b)(5 \minus{} 2c)\geq27
\]

Universal method of solving: $ f(a,b,c)\equal{}(3\minus{}a)(3\minus{}b)(3\minus{}c)\plus{}\lambda (a^2\plus{}b^2\plus{}c^2\minus{}3)$

$ f'_a \equal{} \minus{}(3\minus{}b)(3\minus{}c)\plus{}2\lambda a\equal{}0$
$ f'_b \equal{} \minus{}(3\minus{}a)(3\minus{}c)\plus{}2\lambda b\equal{}0$
$ f'_c \equal{} \minus{}(3\minus{}a)(3\minus{}b)\plus{}2\lambda c\equal{}0$

Solving this system gives permutation of the $ (a,b,c)\equal{}(3\minus{}2\lambda,3\minus{}2\lambda,2\lambda)$ and $ a\equal{}b\equal{}c$. After putting this into condition $ a^2\plus{}b^2\plus{}c^2\equal{}3$ first gives complex values of the $ a,b,c$, which does not satisfy our problem, second gives $ a\equal{}b\equal{}c\equal{}1$. Easy to see that it will be minimal value of the function $ f$. Anyway it is a not method for the students of school.

wrong solution edit--

for arqady's next one

taking ln of both sides it becomes
\[ ln(5 \minus{} 2a) \plus{} ln(5 \minus{} 2b) \plus{} ln(5 \minus{} 2c)\geq 3ln3
\]
from jensens inequality;
\[ LHS\geq 3.ln(5 \minus{} 2(a \plus{} b \plus{} c)/3)
\]
then from cauchy's inequality we find a+b+c<3 so we are done.

[quote="mehmetcantu"]for arqady's next one

taking ln of both sides it becomes
\[ ln(5 \minus{} 2a) \plus{} ln(5 \minus{} 2b) \plus{} ln(5 \minus{} 2c)\geq 3ln3
\]
from jensens inequality;
\[ LHS\geq 3.ln(5 \minus{} 2(a \plus{} b \plus{} c)/3)
\]
then from cauchy's inequality we find a+b+c<3 so we are done.[/quote]
I have obtained $ \left(\ln(5 \minus{} 2a)\right)'' \equal{} \left(\frac {2}{2a \minus{} 5}\right)' \equal{} \frac { \minus{} 4}{(2a \minus{} 5)^2}.$  :wink:

[quote="mehmetcantu"]so, function is convex inequality is true , is not it?[/quote]
No, we have a concave function and your inequality must be reversed.

wow, nice proof.
i got as far as $ (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \equal{} 9 \plus{} \dfrac{3}{2}(p \minus{} 3)^2 \minus{} r$ but then got stuck there.  :surrender:

[quote="mehmetcantu"]arqady i know that lnx is concav so shouldnt be ln(-x) convex?[/quote]
No! If $ f(x) \equal{} \ln( \minus{} x)$ then, $ f$ is a concave function too. :wink:

[quote="mehmetcantu"]arqady i know that lnx is concav so shouldnt be ln(-x) convex?[/quote]

If $ f(x)$ is concave, then $ \minus{}f(x)$ is convex, not $ f(\minus{}x)$. In fact $ f(\minus{}x)$ would be convex iff $ f$ was odd.

[hide="Interesting"]

For your second inequality, arquady, it turns out that if we substitute $ x\equal{}2(a\plus{}1)$ and so on for $ y,b$ and $ z,c$, then multiply everything out, then we get [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=255483]this[/url] inequality which I posted in my blog - pretty weird. The inequality I posted, of course, can be fairly solved with a suitable application of AMGM (credit to math154 for that solution; I was blinded enough by my weird derivation of the inequality to not realize that).[/hide]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ (3 \minus{} a)(3 \minus{} b)(3 \minus{} c)\geq8
\]
[/quote]
I haven't solved the 2nd one yet,but here's for the 1st one
Let a+b+c=S ,ab+bc+ac=P, abc=Q.
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3 \leftrightarrow\ S^2\minus{}2P\equal{}3$
$ (3 \minus{} a)(3 \minus{} b)(3 \minus{} c)\geq8$
$ \Leftrightarrow 3S^2  \minus{} 18S \plus{} 29 \minus{} 2Q \ge 0$
$ \Leftrightarrow 3(S \minus{} 3)^2  \plus{} 2(1 \minus{} Q) \ge 0$(true) :)


limit
analytic geometry
graphing lines
slope
calculus
calculus computations

Hi. I have this question, which I cannot figure out. I don't really understand the notation with the V (it's not just an interval like some others I've done).
Let f be continuous on V(c) = {x : |x − c| < a} and differentiable on
V[size=75]d[/size](c) = {x : 0 < |x−c| < a}. Suppose that lim f'(x) exists and equals L. Prove that f is differentiable at c and f'(c) = L. Thanks in advance for your help. It's greatly appreciated.

$ V$ is just an open interval containing $ c$, also called a neighborhood of $ c$. I suppose "d" in $ V_d$ stands for "deleted": it's a neighborhood of $ c$ from which $ c$ itself is removed.
[url=http://en.wikipedia.org/wiki/Mean_value_theorem]Mean Value Theorem[/url] is all you need here. Indeed, we are to prove that $ \lim_{x\to c}\frac{f(x)\minus{}f(c)}{x\minus{}c}\equal{}L$. By the Mean Value theorem we have $ \frac{f(x)\minus{}f(c)}{x\minus{}c}\equal{}f'(t)$ where $ t$ is a point between $ c$ and $ x$. As $ x$ approaches $ c$, so does $ t$, and this implies $ f'(t)\to L$. (You may want to use the epsilon-delta language for a rigorous proof.)

[quote="BrainMan1"]Would you mind showing me the epsilon-delta notation?  I don't really see how it works here.  Thanks for responding. Plus, how do I show f is differentiable?[/quote]

You show that $ f$ is differentiable at $ a$ by showing that the limit
\[ \lim_{h\rightarrow 0} \frac{f(a\plus{}h)\minus{}f(a)}{h}\]
or (the equivalent)
\[ \lim_{x\rightarrow a} \frac{f(x)\minus{}f(a)}{x\minus{}a}\]
exists, and that $ f^\prime (a)\equal{}s$ if one of the above limits is $ s$. (Observe that $ \frac{f(x\plus{}h)\minus{}f(x)}{h}$ is the slope of the line joining the points $ (x,f(x))$ and $ (x\plus{}h,f(x\plus{}h))$, so if it tends to something, it would tend to the slope of the tangent at $ x$.)

Also, the $ \delta ,\epsilon$ notation is used for the formal definition of limit, which you might not have encountered.

According to the defintion, we say that $ \lim_{x\rightarrow c} g(x)\equal{}L$ if for every positive $ \epsilon$, there exists a positive $ \delta$ with the property: if $ x\neq c$ and $ |x\minus{}c|<\delta$, then we must have $ |g(x)\minus{}L|<\epsilon$.

This seems like a lot of notation; what it means in plainer English is that we can get $ g(x)$ as close as we want to $ L$ by making $ x$ close enough to $ c$ (except that we don't care what actually happens when $ x\equal{}c$ or even whether $ g(c)$ is defined).


linear algebra
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algebra
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complex numbers
linear algebra unsolved

Prove that any algebra of finite dimension over the set of complex numbers which has no non-zero nilpotent elements is isomorphic with a certain $ C^k$.

I PM-ed Killer a long time ago asking whether the algebra was commutative, and he said it was, so I'll use this :).

To each element $a$ of the algebra $A$ associate the $n\times n$ (where $n$ is the dimension of $A$) matrix of the linear transformation $x\mapsto ax$ with respect to some basis. Since the matrices we have commute, we can make them upper-triangular through a similarity, so from now on assume they are upper-triangular.

Every matrix is uniquely determined by its diagonal, for if two unequal matrices had the same diagonal, their difference would be nilpotent, which would make $x\mapsto a^kx$ be the null homomorphism for some $a$ in the inverse image of the difference matrix and some positive integer $k$, and this means that $a^{k+1}=0\Rightarrow a=0$. This shows that we can restrict our attention to the diagonals of our matrices (i.e. we can assume the matrices were diagonal to begin with), and the correspondence between the elements of the algebra and these diagonal matrices is an algebra isomorphism.

The conclusion is now easy to draw: the matrix algebra consisting of diagonal $n\times n$ matrices is an $n$-dimensional subalgebra of $\mathbb C^n$, so it must actualy be $\mathbb C^n$.

That was acutally the greatest problem I had, since I prove it in the same way for commutative algebra ( actually, I've taken a basis $ e_1,...,e_n$ and considered the Dunford decomposition of the endomorphisms $ f_i(x)=e_i x$ and proved they are diagonalisable and then followed the reasoning from the topic with boolean rings), but without this (very strong) assumption, I have no idea if it remains true or not and if it remains true, how to prove it.

I think I have the solution for general case ( when algebra does not nessecerily considered to be commutative ) , and I hope I didn't make stupid mistakes, because I have no intuition in non-commutative algebra.

 So it is enough to prove that every finite dimensional algebra over $ \mathbb{C} $
without nilpotents must be commutative. First of all, if we have
an algebra $ A $ without nilpotents, we can add to it a formal
unit, i.e look on $ \widetilde{A} := \mathbb{C} \oplus A $ with
law $ (a,x) + (b,y) = (a+b,x+y) $ and $ (a,x) (b,y) =
(ab,ay+bx+xy) $. Then the element $ (1,0) $ is a unit of $
\widetilde{A} $. We will write $ a+x $ instead of $ (a,x) $. Now
let us show that $ \widetilde{A} $ is also without nilpotents.
Assume that element $ a+x $ is nilpotent. Then  if $ a = 0 $, then
$ x \in A $ is nilpotent, so $ x = 0 $. Now, if $ a \neq 0 $, then
denote $ y=-a^{-1}x $. Then we have that $ 1-y $ is nilpotent.
Therefore $ y(1-y) = y-y^{2} \in A $ is nilpotent, so $ y = y^{2}
$. Therefore $ (1-y)^{2} = 1-y $, so $ (1-y)^{n} = 1-y $ for every
$ n $. Because $ 1-y $ is nilpotent, we get that $ 1-y = 0 $,
contradiction.
 So now it is enough to prove that if we have a finite dimensional
algebra $ A $ with unit $ e $, then it is isomorphic to $
\mathbb{C}^{k} $. We will prove it by induction on the dimension
of $ A $. If every element $ a \in A $ is of the form $ \lambda e
$, then $ A $ is isomorphic to $ \mathbb{C} $. Now, if we have
some element $ a \in A $ not of the form $ \lambda e $, then there
exists some $ x = 1 + \lambda a $, $ \lambda \neq 0 $, such that
the operator $ L_{x} : A \rightarrow A $, $ L_{x}(y) = xy $ is
invertible. Then it is not a scalar. On the other hand it is
diagonalizable, because if $ P(t) $ is a minimal polynomial of $ x
$, then all it's roots are without multiplicities, because $ A $
does not contain nilpotents, so $ P(L_{x}) = 0 $, so $ L_{x} $ is
diagonalizable. Suppose it has eigenvalues $
\lambda_{1},\lambda_{2}, \ldots , \lambda_{m} $ (different), $ m
\geqslant 2 $. Then we can write $ L_{x} = \lambda_{1} L_{1} +
\ldots + \lambda_{m} L_{m} $, where $ L_{i}^{2} = L_{i} $ and $
L_{i}L_{j} = 0 $, $ i \neq j $, where $ L_{i} $ is  a projector to
eigenspace of eigenvalue $ \lambda_{i} $. Because $ L_{x} $ is
invertible, we have that $ L_{1} + L_{2} + \ldots + L_{m} =Id $.
It is well known that we can write $ L_{i} = P_{i}(L_{x}) $ for
some polynomials $ P_{1},P_{2}, \ldots , P_{m} $. Therefore if we
denote $ x_{i} = P_{i}(x) $, we will have $ L_{i} = P_{i}(L_{x}) =
L_{P_{i}(x)} = L_{x_{i}} $. Therefore $ L_{x_{i}^{2}} =
L_{x_{i}}^{2} = L_{x_{i}} $, $ L_{x_{i} x_{j}} =
L_{x_{i}}L_{x_{j}} = 0 $, $ L_{x} = \lambda_{1} L_{x_{1}} + \ldots
+ \lambda_{m} L_{x_{m}} = L_{ \lambda_{1} x_{1} + \ldots +
\lambda_{m} x_{m} } $ and $ Id = L_{e} = L_{x_{1}} + \ldots +
L_{x_{m}} = L_{ x_{1} + \ldots + x_{m} } $, so we get that $
x_{i}^{2} =x_{i} $, $ x_{i}x_{j} = 0 $, $ \lambda_{1} x_{1} +
\ldots + \lambda_{m} x_{m} = x $, $ x_{1} + \ldots + x_{m} = e $,
and $ m \geqslant 2 $. Take now some $ y \in A $. Then we can
write $ y = eye = ( x_{1} + x_{2} + \ldots + x_{m}) y ( x_{1} +
x_{2} + \ldots + x_{m}) = \sum_{i,j = 1}^{m} y_{ij} $, where $
y_{ij} = x_{i}yx_{j} $. Note now that $ y_{ij}^{2} =
x_{i}yx_{j}x_{i}yx_{j} = x_{i}y 0 yx_{j} = 0 $ for $ i \neq j $,
so because there are no nilpotents, we get that $ y_{ij} = 0 $.
Therefore we have that $ y = x_{1}yx_{1} + x_{2}yx_{2} + \ldots +
x_{m}yx_{m} $. Denote now by $ A_{i} $ the set of elements of the
form $ x_{i}yx_{i} $, where $ y \in A $. Then it is easy to see
that $ A_{i} $ is subalgebra of $ A $, with unit equal to $ x_{i}
= x_{i} e x_{i} $. As we saw, every element $ y \in A $ can be
represented as $ y = y_{1} + y_{2} + \ldots + y_{m} $, where $
y_{1} \in A_{1}, \ldots , y_{m} \in A_{m} $. Also there is no any
dependence between $ A_{1}, \ldots , A_{m} $, because if $ y_{1} +
y_{2} + \ldots + y_{m} = 0 $, where $ y_{i} = x_{i} z_{i} x_{i} $,
then $ 0 = x_{i} ( y_{1} + y_{2} + \ldots + y_{m} ) = y_{i} $.
Finally, for $ i \neq j $ , for $ y_{i} = x_{i} z_{i} x_{i} \in
A_{i} $ , $ y_{j} = x_{j} z_{j} x_{j} \in A_{j} $ , we have $
y_{i}y_{j} = x_{i} z_{i} x_{i} x_{j} z_{j} x_{j} = 0 $. Therefore
$ A \cong A_{1} \oplus A_{2} \oplus \ldots \oplus A_{m} $ as
algebras. Then, by induction on the dimension of $ A $, all $
A_{i} \cong \mathbb{C}^{k_{i}} $, so $ A \cong A_{1} \oplus A_{2}
\oplus \ldots \oplus A_{m} \cong \mathbb{C}^{k} $, where $ k =
k_{1}+k_{2} + \ldots + k_{m} $.

I hope it is correct.

I've checked in my book and my fears became true. Actually, this is how the problem was put in admission exam, in the most general form. Thank you Leva for the beautiful solution. It is so hard to think in non-commutative algebras.  :(

It follows from Wedderburn's characterization of semisimple finite-dimensional algebras over fields that the result stays true if we replace $\mathbb C$ with any algebraically closed field $K$:

The largest nilpotent ideal in our algebra $\mathcal A$ must be trivial, so we can apply Wedderburn's theorem, stating that it must have the form $\mathcal A=\mathcal M_{n_1}(D_1)\times\ldots\times\mathcal M_{n_t}(D_t)$, where $D_i$ are division algebras over $K$. Since any matrix ring $\mathcal M_{n_i}(D_i)$ of order $n_i\ge 2$ has a non-zero nilpotent element, we must have $n_i=1,\ \forall i$, i.e. $\mathcal A=D_1\times\ldots\times D_t$. Suppose now that for some $i$ we have $D_i\ne K$, and take $x\in D_i\setminus K$. Then, if $n-1\ge 0$ is the largest $u$ such that $1,x,\ldots,x^u$ are linearly independent over $K$, $x$ satisfies a polynomial equation of degree $n$ over $K$, so $K(x)$ is an extension field of $K$ of degree $>1$ over $K$. This is impossible because $K$ is algebraically closed, so $D_i=K,\ \forall i$.


trigonometry
LaTeX
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real analysis

I want to see, that $ \sum_{n \equal{} 1}^{t}{sin(n)}$ dont increase more than 2 and dont fall under -0.2

With a little Programm I've seen, that the maximum must be in the near of 1.95815868 (thats for t=51996) and the minimum in the near of -0.12767 (thats for t=52351). Experimental Ive seen that these are approximations of lim sup and lim inf of this series.

First: How to see the boundedness of this series ?

T.

You may use $ s_n (x) \equal{} \sum_{k\equal{}1}^{n} \sin k x \equal{} \frac{\cos \left(\frac{x}{2}\right)\minus{}\cos \left(\frac{x}{2} \plus{} nx \right)}{2 \sin \left(\frac{x}{2}\right)}$, together with the fact that $ \{ \sin (n\plus{}\alpha) \, | \, n \in \mathbb{N} \}$ is dense in $ [\minus{}1, 1]$ for any $ \alpha$, to deduce that

$ \limsup_{n} s_n (1) \equal{} \frac{\cos \left(\frac{1}{2}\right) \plus{} 1}{2 \sin \left(\frac{1}{2}\right)} \equal{} 1.958158682 \cdots$ and
$ \liminf_{n} s_n (1) \equal{} \frac{\cos \left(\frac{1}{2}\right) \minus{} 1}{2 \sin \left(\frac{1}{2}\right)} \equal{} \minus{}0.1276709606 \cdots$

Thank you, I had to see that my own, Ive asked about this formula last time ...

The situation completely changes, if we examine $ \sum_{n\equal{}1}^{t}{sin(\sqrt{n})}$, experimantly theres no min or max, why not? Are there an closed formula for this series?

Note: use \sin inside $ \text{\LaTeX}$- it looks better.

There's no closed form for that one, but we can actually estimate it reasonably well.
$ \int_0^N f(x)\,dx=\frac12f(0)+f(1)+f(2)+\cdots+f(N-1)+\frac12f(N)- \frac1{12}\sum_{k=1}^N f''(c_k)$ for some $ c_k\in (k,k+1)$; this is the trapezoid rule, with error estimate.
Now, let $ f(x)=\sin(a\sqrt{x})$. We have $ f'(x)=\frac{a}{2\sqrt{x}}\cos(a\sqrt{x})$ and $ f''(x)=-\frac14a^2x^{-1}\sin(a\sqrt{x})-
\frac14ax^{-\frac32}\cos(a\sqrt{x})$, so the sum of these isn't too big. The sum of the $ c_k^{-\frac32}$ terms is bounded, and the sum of the $ c_k^{-1}$ terms is at worst $ O(\ln N)$.

That leaves $ \sum_{k=1}^N f(k)=\int_0^N f(x)\,dx-\frac12f(0)+\frac12f(N)+ \frac1{12}\sum_{k=1}^N f''(c_k)$
$ =\int_0^N f(x)\,dx+O(\ln N)$.
$ \int_0^N \sin(a\sqrt{x})\,dx=\int_0^{\sqrt{N}}2t\sin(at)\,dt= \left[-\frac{2}{a}t\cos(at)+\frac{2}{a^2}\sin(at)\right]_0^{\sqrt{N}}$
$ =-\frac2a \sqrt{N}\cos(a\sqrt{N})+\frac2{a^2}\sin(a\sqrt{N})$
Combining with the previous parts, $ \sum_{k=1}^N \sin(a\sqrt{k})= -\frac2a \sqrt{N}\cos(a\sqrt{N})+O(\ln N)$. Since the logarithm is smaller than the square root in the limit, this is unbounded both above and below.

A more careful estimate might get within $ O(1)$, but this is good enough for the question you asked.


When going to the Forum Index I got this website:
http://www.artofproblemsolving.com/Forum/portal.php

Is there a bug? Or is it something new?

For me, this only happens when
After making a post, on the "Your message has been sent" page, I click on the small Home button (Hey, it used to be Forum Index!) in the upper left corner.  Then it takes be to MathLinks instead of AoPS.  My profile has been set to AoPS mode.

Yeah, I have seen it in the MOEMS, and Classroom Math section too.

[quote="b-flat"]Related to the portal page:

The recent posts section does not seem to be working properly, because it does not show either of the last two posts of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=911809#911809]this topic[/url].

Is it restricted to some particular forums, etc?[/quote]Yes, it is restricted to the Olympiad Section and College Playground forums right now, because that's what most people use it for anyway.


Suppose the integers $ 1,2,3$.....$ 10$ are split into two disjoint collections $ a_1,a_2,a_3,a_4,a_5$ and $ b_1,b_2,b_3,b_4,b_5$ such that
$ a_1 < a_2 < a_3 < a_4 < a_5$
and $ b_1 > b_ 2 > b_3 > b_4 > b_5$
(i) Show that the larger number in any pair $ \{a_j,b_j \}$, $ 1 \le j \le 5$, is atleast 6.
(ii) Show that $ \sum |a_i \minus{} b_i| \equal{} 25$ for every such partition.

(i)if both of $ a_j,b_j\leq 5$ then
$ a_j,a_{j \minus{} 1},...,a_1,b_j,b_{j \plus{} 1},...b_5$ are $ \leq 5$, But there are $ 6$ numbers($ a_j,a_{j \minus{} 1},...,a_1,b_j,b_{j \plus{} 1},...b_5$) in total. Contradiction.

The solution there looks too messy.
I proved that one of $ a_j,b_j$ is $ \geq 6$ similerly it can be proved that one of $ a_j,b_j$ is $ \leq 5$.
From each pair $ a_j,b_j$ take the greater number in to a set $ A$ and the smaller number into a set $ B$.
$ A \equal{} \{6,7,8,9,10\}$ and $ B \equal{} \{1,2,3,4,5\}$
$ |a_i \minus{} b_i| \equal{} \max(a_i,b_i) \minus{} \min(a_i,b_i)$. But $ \max(a_i,b_i)\in A$ and $ \min(a_i,b_i)\in B$. Thus the sum is $ ($sum of elements of $ A) \minus{} ($sum of elements of $ B) \equal{} (6 \plus{} 7 \plus{} 8 \plus{} 9 \plus{} 10) \minus{} (1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5) \equal{} 25$

[quote=Rahulmca]for the sequence 'a' there are j-1 elements smaller.  for b; there are 5-j smaller numbers ; also in (a,b) the smaller number is smaller than the larger number. so we hv atleast 5 numbers smaller than the number larger among (a,b)

similarly u can show that the smaller number is at most 5.

now ur sum becomes  SUM(larger numbers) -SUM(smaller numbers)[/quote]

Similar to official solution


[quote="mr.danh"]Show that the equation $ x^5 \minus{} x^3 \plus{} x \minus{} 2 \equal{} 0$ has a real root $ \alpha$ satisfying $ 3 < \alpha^6 < 4$[/quote]

$ P(x)\equal{}x^5 \minus{} x^3 \plus{} x \minus{} 2$
Notice that for $ x\equal{}(\frac{5}{4})^6$ , $ x<4$ and $ P(x)>0$
and for $ x\equal{}(\frac{301}{250})^6$, $ x>3$ and $ P(x)<0$

This is an INdian National Maths Olympiad problem, if I remember well...

[quote="mszew"]
$ P(x) \equal{} x^5 \minus{} x^3 \plus{} x \minus{} 2$
Notice that for $ x \equal{} (\frac {5}{4})^6$ , $ x < 4$ and $ P(x) > 0$
and for $ x \equal{} (\frac {301}{250})^6$, $ x > 3$ and $ P(x) < 0$[/quote]
I think what you meant to say is:

For $ x\equal{}\frac{5}{4}$, $ P(x)>0$ and $ 3 < x^6 < 4$, and for $ x\equal{}\frac{301}{250}$, $ P(x) < 0$ and $ 3 < x^6 < 4$. You don't want to evaluate $ P$ at $ (\frac{5}{4})^6$ since this is outside the range. Also, it's not enough to show that $ P$ is negative [i]somewhere [/i]to the left of $ 4^{1/6}$; to prove that there's a root between $ 3^{1/6}$ and $ 4^{1/6}$ you must show that $ P$ is both negative and positive [i]in that range[/i].

This solution is correct, but it's not very clear how one is supposed to find those values, and worse, even showing that $ (\frac{301}{250})^6>3$ requires some effort. I thing the intention here was to use estimates to show that $ P(4^{1/6}) > 0$ and $ P(3^{1/6}) < 0$ (the former is easy, the latter requires more work), or to use the fact that it's easy to solve $ P'\equal{}0$ and thus more easily investigate the roots.


AMC USA(J)MO USAMO Princeton college Harvard geometry

Longfellow Middle School had 2 in '98 according to last year's chart.

Gregory Price has been corrected from TJ to Longfellow. The other 8th grader, Joshua Dezube, definitely took the tests at TJ and apparently even claimed 9th grade. He stays for now.

Oops- I forgot to update the total when I moved Gregory Price. It's now a tie.

[quote="Ubemaya"][quote="alex960531"]I can't believe that Longfellow is now 2nd.

Longfellow > Miller, because VA>CA.

Longfellow Rocks!!![/quote]

? Did you notice that 3rd was another school from CA?

Also, I wonder what would happen if there was a magnet school in the Bay Area that only drew from like a 20 mile radius... it would have easily 30+ qualifiers I think (and if you double that radius, probably 50-60).[/quote]
A 15 mile radius circle around the spot a little east of cupertino would get around 50. If you increase it to 30, you get I think 60 something.

I believe TJ's attendance area is for the most part limited to Fairfax County.  So let's see what we get if we imagine a magnet school drawing only from Santa Clara County:

6 from Cupertino
1 from Los Altos (homeschool)
8 from Gunn
6 from Paly
2 from Leland
6 from Lynbrook
2 from Miller
1 from Piedmont Hills
5 from Harker
9 from Saratoga
2 testing with EPGY

That's 48 potential qualifiers.  If even half of those actually attended the hypothetical magnet school, it would still beat TJ's 18 by a significant margin.

In fact, if Santa Clara County were its own state, it would rank second in the number of USAMO qualifiers (the rest of California would be in first place, by a margin of 61-48).  We even beat out New York and Illinois.

Why is WV so sad?

The only school in there is Morgantown,and they have only had 8 in the last 10...
Time to take a stand! :P  :P  :P

Additional homeschoolers:  Matthew Babbitt and Jay White

Also, AAMC had 1 qualifying student in 2000, 2001, 2002, and 2004, and had 2 qualifying students in 2003.

Since the editing horizon is past, here's the relevant table updated:
[code]                8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0 0 0 0 0
Group     State 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Tot
Homeschool   -- 0 0 0 0 0 0 0 0 0 1 3 4 4 8 6 8 7 A 6 9 9 A D  98
ICAE         MI 0 0 0 0 0 0 0 0 3 1 3 4 4 4 4 5 2 2 2 C 5 ? ?  51
AAMC         NY 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 1 0 1 2 4 5  18[/code]

There are probably other lines that could be added, but I don't have the information. Even ICAE and AAMC have only 9 students each listing them as their site; in the AAMC case, all 9 are homeschoolers.

[quote="timwu"][code]Parkview     GA 0 0 0 0 0 1 1 2 2 2   8
[/code]

Yay, we are also listed! (Although we seem to be the only school in Georgia...)[/quote]
Heh, a couple years from now and Walton should go up a bit on that list.

West Virginia has had a total of 18 qualifiers (with multiplicity) in the 23 years I have recorded (The rule guaranteeing one qualifier per state was instituted in 1996). The last time a non-Morgantown student made it was 2002, and 2000 is the only year in which the state has had multiple qualifiers.
The weakest state by far is Wyoming, with only nine USAMO qualifiers total. Even the one per state rule isn't enough if there are no AIME qualifiers in the state.

In response to some stuff brought up in last year's thread:

Schools that did exceptionally well compared to previous performance:
- University (Irvine) HS in CA, with 7 this year and 8 total in all previous years.
- Torrey Pines in CA, with 6 this year and 8 total in all previous years.
- Walton in GA, with 5 this year and 2 total in all previous years.
- Fairmont Prep in CA (listed as ECC Academy), with 3 this year and one previously.
- First Colony MS in TX, Marianopolis College in Quebec, Brookwood in GA, Central Academy (Des Moines) in IA, and St. Mark's (Southborough) in MA with 2 each this year and none previously. Honorable mention to Langley HS in VA, with two this year and one in 1992.
Other schools that posted their highest single-year totals ever: TJ (18, record for all schools), Stuyvesant (13), Saratoga (9), Andover (8, tie), Gunn (8), Detroit Country Day (7, tie), Harker (5, tie), Mason (3), South Eugene (3, tie), Acton-Boxborough (3, tie), Belmont (3, tie) and many with 2 or 1.
If this list looks California-heavy, it shouldn't be a surprise. The state has nearly doubled its number of qualifiers since 2006.

On the flip side, Naperville North was shut out this year for the first time since 1997. They have the highest numbers in the last 10 years or total of any school with fewer than 2 qualifiers this year.

Is that maybe the result of the AIMEs this year? I thought that at least the II required having a larger arsenal of tricks than in past years, which favors the schools with real training programs.

Ladue Horton Watkins High School (MO) has had 10 qualifiers (or 9, depending how you count!) in the past 10 years (1 in 04, 1 in 05, 3 in 06, 1 in 07, 1 in 08, 2/3 in 09).

What do you mean "depending how you count?"

A qualifier this year attends half his classes at the high school, half at the middle school, took the test at the high school, but wrote on his AIME he attends the middle school, but on the AMC 10/12, is listed as from the high school.

I have the high school listed at 9 (visible in the table as LH Watkins) and the middle school at 1. I'll leave it that way, but it may be possible to convince me otherwise.

I am the qualifier listed as from Ladue Middle School (Victor Wang). Both Max Schindler and I are in 8th grade and took it at the high school anyway, so I don't see a reason why we should be in separate categories. I'm also pretty sure most middle schoolers will take the test at the high school in the future, so it might be more organized to keep everything in the High School category.

All right. I've changed it for future versions. "More organized" isn't such a big deal; I've got over 1300 schools in the system anyway.

Darn. My school seems to enjoy having 1 USAMO qualifier 9/10 years >.>

We had 2 people miss by 1 AIME problem :( oh well.

The 2004 entry (Sway Chen) is listed as Lexington. That's not the only middle-school aged student listed as Lexington- they also have one in 1990 and 1991.

My school and one other one in my state are there. (It's a combined middle and high school, so nobody in the ms has made it ever). Nice list there! Thanks for the info!


Is anyone using AOL Explorer?  I just recently downloaded it and it seems nice.  It has all the nice features of Firefox that I like to use (like tabbing and better popup blocking).

I personally don't like it. It's just a rip off of Firefox.

And I personally HATE aol so I'm not using it  :mad:

You guys are pretty hostile :P

It is pretty much a rip off of Firefox.  And since it has the AoL search bar, instead of google, I would have to say Firefox is better :)

My Opinion and probably everyone else's too:

Best Internet Browser: Mozilla Firefox

Best Search Engine: Google

Best Math Forum: Art of Problem Solving

Best Free E-Mail Provider: Gmail (if anyone wants an account just PM me and I'll invite you!  :D )

i am using aol explorer right now. i like it.

I repeat: What does AOL Explorer have that Firefox doesn't have? I haven't seen it.

I think it looks cool. Dunno about the features though, it doesnt provide anything new really, thats not to say that its not good. So what if it doesnt have anything MORE than Firefox - 1. maybe it does (security maybe - firefox isnt completely idiotproof, or maybe it has a feature we dont know about yet) 2. So? If its the same as firefox, and both are free, and AOL explorer LOOKS cooler.. what are you going to take?

Well, I went to the AOL explorer site and I didn't see a link for "extensions." I like this feature of firefox, you can customize it by downloading the extensions you like (for example, I use the Adblock, Web developer, Google toolbar, and NoScript extensions, among others). Firefox is also open source, which allows it to be perfected by a community of avid programmers, while AOL explorer may end up rotting away like IE... :(

i like AOL explorer. i know its basically a rip, but it's more aethetically pleasing than firefox, so i use it. :gleam:

Well, if you want an aesthetically pleasing user interface, go here: [url=https://addons.mozilla.org/themes/?application=firefox]https://addons.mozilla.org/themes/?application=firefox[/url]
You can choose between many themes instead of being constrained to just one.

now that's a lot better. thanks jmadsen.


Let  ABC be a triangle that AB=c,  AC=b,  BC=a  and  a<=2<=b<=3.
What is the maximum area of  ABC  triangle?

Um, the area definitely can't be that large -- it would give an altitude of length at least 12, but the triangle has sides of length at most 5.  You should really show your work.

The area can be expressed as $\frac{1}{2}ab\sin{\theta}$.  Clearly $\theta$ does not depend on the lengths of a or b.
Thus the maximum is given when $a=2;\ b=3;\ \theta=\frac{\pi}{2}$ so $A_{max}=3$


yes. they are.

I am excited because I have recruited ten-ish people from my school to come this year. as opposed to two last year.

Good luck to all.


Define for x,y,z real positive numbers

A=x+y+z
B=xy+yz+zx
C=xyz

Prove the inequalities

1) (AB)^2 +2ABC +9C^2 \geq 4A^3C

2) (AB)^2 +4ABC \geq 4A^3C + 9C^2

1) is equivalent to :
 \sum x 4 y 2  +  \sum x 3 y 3  +  \sum x 2 y 2 z 2   \geq   \sum x 4 yz + 2 \sum x 3y  2 z, where the sums are symmetrics.

But, since (4,2,0) majorizes (4,1,1), we deduce from Muirhead's inequality that :  \sum x 4 y 2  \geq  \sum x 4 yz.
And, in another hand, as noticed in one of other submissions, \sum x 3 y 3  +  \sum x 2 y 2 z 2   \geq  2 \sum x 3y  2 z is just Schur's inequality for a = xy, b = yz and c = zx.

Summing, you get the desired inequality. Note that equality occurs iff x = y = z.

Pierre.

For 2), we use the inequality AB 2  + 3BC  \geq  4A 2 C from one of your other submissions (post : 'degree 4,5,6').
Thus (AB) 2  + 3ABC  \geq  4A 3 C.

It suffices to prove that AB  \geq 9C.
But this last inequality comes from Mac-Laurin's inequalities :
A 2   \geq  3B and B 3   \geq  27C 2 .
Note that equality occurs iff x = y = z.

For a proof of Mac-Laurin's inequalities (in french), go to :
http://www.animath.fr
then 'les cours de l'olympiade franaise de mathmatiques'.
Then choose the paper on inequalities.

Pierre.


Micah places coins in the order of penny, nickel, dime, penny, nickel, dime, and so on, so that each row contains one more coin than the previous row, as shown. What is the number of cents in the value of all coins in the 13th row?

[asy]unitsize(0.4cm);
for (int i=0; i<4; ++i) {
draw(Circle((3i,0),1),linewidth(0.7));
}
for (int i=0; i<3; ++i) {
draw(Circle((1.5+3i, 3),1),linewidth(0.7));
}
for (int i=0; i<2; ++i) {
draw(Circle((3+3i,6),1),linewidth(0.7));
}
draw(Circle((4.5,9),1),linewidth(0.7));
label("1",(0,-1),N);
label("5",(3,-1),N);
label("10",(6,-1),N);
label("1",(9,-1),N);
label("1",(1.5,2),N);
label("5",(4.5,2),N);
label("10",(7.5,2),N);
label("5",(3,5),N);
label("10",(6,5),N);
label("1",(4.5,8),N);[/asy]

The third row contains 1 of each and nothing more.  The sixth will have 2 of each and nothing more.  The ninth three of each, and the twelfth 4 of each.  Hence the thirteenth row will have 4 of each, and one more penny.  So that's $ (16\times 4)\plus{}1\equal{}65\mbox{ cents}$.


Find the distance from P1 to 2.

P1 = (2, -3, -3) and P2 = ( 4, 1, -1)

[hide="Solution"]Use the distance formula:
$ \sqrt{(2\minus{}4)^{2}\plus{}(\minus{}3\minus{}1)^{2}\plus{}(\minus{}3\minus{}(\minus{}1))^{2}} \\
\equal{} \sqrt{4\plus{}16\plus{}4} \\
\equal{} \sqrt{24} \\
\equal{} \sqrt{4} \cdot \sqrt{6} \\
\equal{} \boxed{2 \sqrt{6}}$[/hide]


geometry
AMC
AIME
inequalities
LaTeX
3D geometry
trigonometry

you guys can discuss the problems in about 15 mins
about 7:15

and i should start grading :)

here are the answeres:
[hide="answers"]
1)C
2)E
3)B
4)A
5)D
6)B
7)B
8)B
9)C
10)A
11)D
12)A
13)C
14)E
15)C
16)A
17)A
18)D
19)A
20)A
21)B
22)C
23)B
24)A
25)D
[/hide]

unfortuantely not many people compteted but here are the results:

alan: 132.5
no_one_2000:124
eryaman:123
Andreas:119.5
behemoth571:117
Wizard:115

congrats to Alan for first place. :)
i have not made a solution packet, but i can provide solutions for any of the problems...

Well, I did this too late, but if anyone wants to know my score would have been 119.5.  It might have been higher, but I stopped half an hour early, and didn't really feel like finishing it.

What level do you guys think this is, 10, or 12?

i just finished taking it on my own. i got a 110 with two stupid mistakes (left out an $i$ and divided wrong). i felt the test was somewhat easy for a mock test (or maybe i'm just getting better at these things).

Wow... my score would have been 129. :o I wasn't planning on doing it, but figured I would look over the problems I knew how to do, and got 19 of them in about 50 minutes. O_O This must bode well for the AMC's this winter...

Somewhat easy for a mock test, but I thought it was very appropriate.  A decent AMC 10 I'd think.

For question 21, I don't really get it... unless those brackets were different.  I interpreted the answers as on the closed interval of [5,10],[1.5], etc.  In that case, no number would work... I'd think.

I'd like to know how to do 12, 18, 21, 22, 23, and 25.

I PMed my answers like an hour too late... ah well.  I got a 124 I think.  Not geometry heavy (a very good thing)

I think there should be a Mock AIME for this one.

dam, i should have gone with my instincts for 3, 18, and 24, but chose to omit them because i wasn't sure >_< oh well.

[hide="3"]
$a_0=\frac {1}{a_1+0}$
$a_1=\frac {1}{a_2+1}$
$a_2=\frac {1}{a_3+2}$

if you substitute each $a_n$ with the subsequent $a_{n+1}$, you find that this process never ends, so every $a_n$ is irrational.  i think
[/hide]

[hide="22"]
apply AM-GM twice
$\frac{a^2+b^2+c^2+d^2}{4}\ge \sqrt[4]{a^2b^2c^2d^2} =1^{\frac 24}=1$ so $a^2+b^2+c^2+d^2 \ge 4$
$\frac{ab+ac+ad+bc+bd+cd}{6}\ge \sqrt[6]{a^3b^3c^3d^3}=1^{\frac 36}=1$ so $ab+ac+ad+bc+bd+cd \ge 6$
$4+6=10$

edit: actually, now that i think about it...one application of AM-GM would have done just fine
[/hide]

Why is it that I keep forgeting these things!???!!??  :mad:  :(

[quote="Pakman2012"]A decent AMC 10 I'd think.[/quote]

Not really. This test seems to span about AMC-10 to AMC-12 from beginning to end. The final problems were definitely harder than those on the AMC-10, because, IIRC, I got all 5 of the final problems on the AMC-10 right this last year, and, while the first of the final five on this test was easier, the other four were harder.

I don't see what's so hard about #21... for all integers $n$ ending in a $6$, $n^3$ will end in a $6$, and for all integers $n$ ending in a $1$, $n^3$ will end in a $1$. Therefore, for $n=6,11,16$, $\frac{n^3-1}{5}$ will be an integer, and for $n=6$ you will have a prime, so the answer was $B$.

Is there any way to do #12 without brute-forcing it?

[quote="JesusFreak197"]
Not really. This test seems to span about AMC-10 to AMC-12 from beginning to end. The final problems were definitely harder than those on the AMC-10, because, IIRC, I got all 5 of the final problems on the AMC-10 right this last year, and, while the first of the final five on this test was easier, the other four were harder.

I don't see what's so hard about #21... for all integers $n$ ending in a $6$, $n^3$ will end in a $6$, and for all integers $n$ ending in a $1$, $n^3$ will end in a $1$. Therefore, for $n=6,11,16$, $\frac{n^3-1}{5}$ will be an integer, and for $n=6$ you will have a prime, so the answer was $B$.[/quote]
That's a nice way of solving it.  There is another way, however.  Notice that $n^3 -1$ factors as $(n - 1)(n^2 + n + 1)$.  5 is prime, therefore $5 | n -1$ or $5 | n^2 + n + 1$.  Running through $n = 0,1,2,3,4$, we can see that the latter is never true, so we must have $5 | n - 1$.  We also need $n - 1 = 5$, as any other multiple of 5 would leave another factor behind, giving a non-prime result.  So we must have $n = 6$, and checking we find that $6^2 + 6 + 1 = 43$ is indeed prime, so we have our solution.

22 is nice if you know a bit of inequality footwork.  Apply the inequality on AoPS's LaTeX help page and you're on your way:

$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \geq \sqrt[4]{abcd} = 1 \Rightarrow a^2 + b^2 + c^2 + d^2 \geq 4$.

Similarly,

$\frac{ab + ac + ad + bc + bc + cd}{6} \geq \sqrt[6]{(abcd)^3} = 1 \Rightarrow ab + ac + ad + bc + bc + cd \geq 6$.

Combining these we find our solution:

$a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bc + cd \geq 10$

The funny thing is, that solution can be reached quite easily by just setting each integer equal to $1$, which would, I assume, be the only possible way to attain that minimum. The only problem is that you're doubting whether or not you got it right, but say it's the only question you aren't sure about... why not go for the 150? :lol:

As for #12, it's probably easier to brute force, as it solves pretty quickly, but there's another way...

$I_{25}=I_{22}+I_{23}=I_{19}+2I_{20}+I_{21}=I_{16}+3I_{17}+3I_{18}+I_{19}...$

Notice the pattern? It ends up as $I_{25}=I_1+8I_{2}+28I_{3}+56I_{4}+70I_{5}+56I_{6}+28I_{7}+8I_{8}+I_{9}$, I think. So, you just need to find up to $I_9$.

Here's my solution to #25; it's kind of sketchy, but I think it makes sense:

Note that if $n$ is prime, then $d(n) = 2 \Rightarrow d(d(n)) = d(d(d(n))) = 2$; therefore all primes are nice.  If $(d)$ were true, it would follow that the cubes of all primes are nice.  However, this is not true, for $d(p^3) = 4 \Rightarrow d(d(p^3)) = 3 \Rightarrow d(d(d(p^3))) = 2$.  Therefore $(d)$ must be false.

Just out of curiosity, I set to proving the other statements given.  We already know that all primes are nice, giving us an infinite number of odd nice numbers and proving $(c)$.  We can also see that if $n = p_1^{p_2 - 1}$, then $d(n) = p_2 \Rightarrow d(d(n)) = d(d(d(n))) = 2$, so that we have an infinite number of perfect powers of 2 and 3, satisfying $(a)$ and $(b)$.  Finally, $(e)$ is easy, for we can simply take the divisor of both sides of $d(d(d(n))) = d(d(n))$ to find that $d(n)$ also satisfies the original equality.

Interestingly, only numbers of the form $n = p_1^{p_2 - 1}$ are "nice", as for a number to be "nice" we must have $d(n) = p \Rightarrow d(d(n)) = 2$ (because $n = d(n)$ holds true only for $n = 2$, excluding 1, which cannot occur as $d(n)$ for $n > 1$).

Hmm, for 22, I think just one application of AM GM is needed:

\[\frac{a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd}{10}\ge\sqrt[10]{(abcd)^5}\\ \\ \Rightarrow a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd\ge 10.\]

I woulda done it but the timing was awful.

Maybe I'll do it another time and find out my score.

[quote="alan"]
[hide="3"]
$a_0=\frac {1}{a_1+0}$
$a_1=\frac {1}{a_2+1}$
$a_2=\frac {1}{a_3+2}$

if you substitute each $a_n$ with the subsequent $a_{n+1}$, you find that this process never ends, so every $a_n$ is irrational.  i think
[/hide]
[/quote]

So does this mean that each term of the sequence is an infinite continued fraction of some irrational number? i.e.

$a_0 = 1 + \frac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4 + ...}}}$

$a_0 = \frac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3 + ...}}}$

and so on?

how to solve the geom ones 19 and .. i forget which number but basically any geom question that appears in the last 8 questions

[quote="zanttrang"]
That's a nice way of solving it.  There is another way, however.  Notice that $n^3 -1$ factors as $(n - 1)(n^2 + n + 1)$.  5 is prime, therefore $5 | n -1$ or $5 | n^2 + n + 1$.  Running through $n = 0,1,2,3,4$, we can see that the latter is never true, so we must have $5 | n - 1$.  We also need $n - 1 = 5$, as any other multiple of 5 would leave another factor behind, giving a non-prime result.  So we must have $n = 6$, and checking we find that $6^2 + 6 + 1 = 43$ is indeed prime, so we have our solution.[/quote]

exactly the solution i had in mind :)

Oh I just realized my mistake for 21...
The question said on which interval did it lie, not find the solution set or something..
Dang.
That was a stupid mistake

I took it in like 50 minutes too.
I would've at least attempted 25, it looked interesting.

A very nice AMC.

Wow, I guess I didn't do too badly this time. :)

I missed #3 and #11.  I don't really know how I missed #11, it must have been a computational mistake.  I just tried same the method I used before and got it right, so that's probably it.

For #3, I was thinking:

[hide="My idea"]
$(a_{n+1}+n)a_n=1$
$a_{n+1}+n=1/a_n$
$a_{n+1}=1/a_n-n$

Let $a_1=2$, so $a_2=-3/2$ and the sequence ends.  So $a_1$ is the only positive rational value in the sequence (of your choosing), thus the answer is 1, C.  But you could also let $a_1=-2$ or $a_1=-1337$ if you wanted, which would automatically end the sequence and give you zero.  So then the answer could also be B... For this problem, I almost felt like there weren't enough guidelines; like it was too open to your interpretation.  Hopefully I had the right idea, though.

...But I could be reading into it incorrectly. :)
[/hide]

Anyway, I omitted #13, #23, #24, and #25.  I would have had #13, but I was careless and forgot that $0!=1$.  Otherwise, I did that problem correctly.  The others, I omitted because they looked ugly.  I don't think I could have solved those in the given time limit.  Maybe #25, but not #23 or #24.

Anyway, I liked this AMC a lot.  It wasn't extremely difficult, but it wasn't super easy either.  It was fun.

I made one stupid mistake on putting wrong letter (which won't happen in real AMC-12 of course) so assuming that I got that one right, my score is 126.5.

This one was indeed easier than real AMC-12 but I wouldn't say it's decent AMC-10.

I bruted forced number 15. Is there anyway you can simplify it without brute forcing? (I believe there is except I was too lazy to actually do on the test).

Just for sidenote, #20 was extremely easy in my opinion.  :| I thought it was easier than #3 (since I got it wrong  :P ). I've seen many type of #20 but the question would've been better if you asked the "sum" of the divisors of $1374 \cdot 1375 \cdot 1376$.  :D (Of course, you can again apply the formula though)

#25 worked out nicely with 2 like zanttrang said. I didn't bother to prove other choices so I just put D.

I think this one was slightly easier than mine but I really liked it. I rather have bit easy Mock AMC than having terribly hard one where I can only answer couple questions.  ;)

And I just wished there were more trig questions.  :D

Questions that I need on how to do is:

17
18
19
23
24

And I agree, there should be a Mock AIME for this one.  :)

[quote="amirhtlusa"]no the way to show there are no rational solutions is a proof by contradiction.
if you assume that there exist sume rational term like $\frac{p}{q}$, then you will see that after a few terms you will get a negative number, but the sequence was defined a sequence of positive numbers, so there is no rational element in this sequence

i can show the proof if you want...[/quote]

Ah, no, that makes perfect sense.  Wouldn't that normally be a bit futher into an AMC 10/12 than problem 3, though? ;-)

[quote="joml88"][hide="Number 24"]
Just notice that in the product you have terms like $(x_1-x_1)$ and such...therefore the whole thing equals 0.  All the coefficients are therefore 0.[/hide][/quote]

Yep, it was basically a trick question  :D I spent a few minutes playing with it before I realized that you could have the same two terms in one part.

[quote="Silverfalcon"]
Questions that I need on how to do is:

17
18
19
23
24

And I agree, there should be a Mock AIME for this one.  :)[/quote]

The easiest way to do 17 is to note that $35!$ must be divisible by three, so the sum of the digits must be divisible by three.  If you sum all of the known digits, you find $S = 138$.  We want some $0 \leq n \leq 9$ such that $3 | 138 + n$.  $n=6$ is the only such integer that provides this property, so we have our solution.

18 is basic combinatorics.  The number of ways to find one event happening or the other event happening is to add the number of ways for both things to happen together.  So, we have

$N = \binom{17}{4} + \binom{17}{3} =  \binom{18}{4}$, which follows from the property

$\binom{a}{b} + \binom{a}{b+1} = \binom{a+1}{b+1}$

One nit with that: if it just has to be divisible by $3$, then you could have $n=3$, $n=6$, or $n=9$. However, If the digits sum to $9$, it is divisible by $9$, and we know that $35!$ is divisible by $9$, so we can narrow it down to $n=6$. ;)

#15: Notice that you have $\frac{n^2}{n!}-\frac{(n-1)^2}{(n-1)!}$. The next number will be $n+1$, so you have $\frac{(n+1)^2}{(n+1)!}-\frac{n^2}{n!}$. Summing the two gives $\frac{(n+1)^2}{(n+1)!}-\frac{(n-1)^2}{(n-1)!}$. Therefore, we can simplify the sum to $\frac{5^2}{5!}-\frac{1^2}{1!}=-\frac{19}{24}$. :)

[quote="alan"]you should have took out the second digit (the zero) instead, to make us look at it mod 11 as well :P[/quote]Hmmm...I did the $11$ approach.  I thought of the three but decided not to try it because of the divisible by 9 deal thing.  :P Ah well, it was a nice finger excersie on my calculator.  :P

[quote="amirhtlusa"]actually zanttrang, what you are saying is incomplete.
notice that 3|138+3 (i don't think i had 3 on there as achoice), but rather than looking at 35! modulo 3 you have to look at it Mod9, then THE ONLY possible value is 6[/quote]

Oops, you're right; guess I should be more careful about my assertions  :blush:

140.5 here. Hehe.... i just realized i could have gotten 24 right... that would have given me 146.5, the best ive ever had (whether it was mock or real)


Determine all positive integers n which are not perfect squares and for which n2 is a multiple of [ :sqrt: n ] :^3: , where [x] denotes the integer part of x.

Let m^2 be the closest perfect square less than n. Then [Sqrt(n)]^3 = m^3. Now n = m^2+k for some 0<k<2m+1, so n^2 = m^4 + 2km^2 + k^2.

So now we need to find when m^3|m^4 + 2km^2 + k^2. It is obvious that m^3|m^4, so we need to find when m^3|2km^2 + k^2=k(2m^2 + k).

If m is even, then k must be even.

..
Ok, this is turning out harder than I thought it would, and my bag of tricks just ran out, so I think I'll just leave what I have here and maybe finish it later (..or someone else could always help out).

That's a great start.  How about this to continue it:

we have for some positive integer c that
c*m3 = 2km2 + k2, so that c*m3 - 2km2 = k2, so that m2(cm - 2k) = k2

Well we could do that, but I don't know what that really does for us. We could try comparing the factors on either side to see what terms are divisible by what other terms, but there are too many unknowns. We don't even know whether m or k is larger (or whether k always has to be either smaller than, equal to, or larger than m at all).

No, it tells us a lot -- (cm - 2k) is an integer, so we know m2 | k2, and then you're nearly done!

Oh, right. So we can let k = am. Since k<2m+1, a must be either 1 or 2, so we have two cases:

1) k = m. Then (m^2)(cm-2k)=k^2 becomes k(c-2)=1. But since k and c-2 are both positive integers, they are both equal to 1: k=m=1 and c=3. Therefore, n=m^2+k=2.

2) k = 2m. Then (m^2)(cm-2k)=k^2 becomes m(c-4)=4. Since k and c-2 are both positive integers, we have m=1,c=8,k=2,n=3 or m=2,c=6,k=4,n=8 or m=4,c=5,k=8,n=24.

Mmm, that was nice practice for the AIME. I think I could see it as a tough AIME problem. Although everything used in the solution is very basic, the solution itself is sorta long, not to mention the number of variables involved (which was why I was sorta hesitant to introduce c into the mix of things as Joel suggested, since I was confused enough already with all the new variables I had defined and needed to keep track of).

..

Or maybe I'm just weird.

[quote]Mmm, that was nice practice for the AIME[/quote]

I got it from the 98 New Zealand IMO team selection test  :mrgreen:

[quote]I got it from the 98 New Zealand IMO team selection test [/quote]

lol!

Well, maybe it was problem 1... :D


Counting with Combinations Part 1

We introduce combinations and using them to solve basic counting problems.  We also invite our viewers to Pizza Monday, or at least to buy us the pizza!

No it is a real number
[url]http://www.beachcalifornia.com/reca/restaurants/alpine/1-vinnys-brooklyn-pizza-619-445-5995.html[/url]


calculus derivative inequalities geometry articles CEMC FTW

I'm absolutely certain that you learned math for contests, especially for the ones that went (or are going) to CMO/IMO.

I can tell you that I did learn math, thinking it would be of use in math contests somehow. But I'm absolutely certain that some of you learned a lot more math than I did for contests.

COMC, 2008: Partial derivatives (I even asked the supervisor if I could use partial derivatives)
AMC 12B, 2009: Lagrange multipliers
Purple Comet, 2009: Jacobians in general (although I learned the most commonly used Jacobians before the Purple Comet came, $ r$ and $ \rho^2sin\phi$)

Just a few clarifying questions:

You used partial derivatives on the recent COMC, where no problems involved maximization?

And you used Lagrange multipliers in AMC 12B?

And I cannot even begin to conceive how one would use Jacobians on the Purple Comet.

Just a few examples to clarify would be fantastic.

 :)  :ninja:

Not that these are math I actually used in contests, just saying that I had to do a bit of independent learning before a contest.

hmm.. I didn't really study anythin in particular..
Honestly, all I did were past contest problems of this and that. :blush:

But, I just started studying some inequalities (one of my weakest area) by reading Mildorf's, and some number theory (my other weakest).

I'm also planning to buy the book, "Problem Solving Strategies" for COMC and CMO (hopefully :D) prep.

I think this is an article targeted at people like you: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php[/url]

Maybe I fell headlong into the calculus trap! It may explain the entire story that was first intended to be about an IMO participation but whose characters ended up going to the final stage of the $ \sqrt{ANS}\,M.C.$ which is not much more than a multi-variable calculus version of Mathcounts.

For me personally, I do reasonably well on math contests (not the very top, but quite well)

I have never
-had a tutor
-listened in math classes
-taken math in elementary school (the teachers put me somewhere else to do w/e I want because the math was too easy)
-studied anything on my own
-learned from other people like parents
-studied calculus
-read math books

I have
-Done many of the past contests on the cemc
-Done a few past AMCs from here
-played FTW once in a while

If you're wondering what reasonably well is, I got 64 on COMC this year, and 118.5 on the AMCs (12)

eh? I learn math from mooseness. ex squared plus why squared equals zed squared.

ON A MORE SERIOUS NOTE

go do porblems instead of wasting time on AoPS


You can change it as much as you want!!

Yukon, AK

Antarctica. Where? somewhere all you can see is white.

Thiruvananthapuram, India.

How's that for random?

[b]Destruction Bay[/b], Yukon Territory

or maybe

[b]Moose Factory[/b], Ontario

or perhaps

[b]Shanghai[/b], West Virginia

[quote="jjfun1"]Moscow, Russia[/quote]

???pfffft

Marjkovic, Germany

FORKS WASHINGTON :P
stole jjfun's location xD

Somewhere in this world....

lol straight from k00lpeoples land.

lol musicnmath.
dunno, inoit's location :o

lulz
Houston texas.
No it's not random, but that was random to me...anyway

my best friend's second best friend's third best friend's best friend's bedroom
 :spider:  :spider:  :spider:

the bedroom's bed's spring's base's dust.


Find all positive integers $ a$, $ n$ such that:
\[ a^{n\plus{}1}\minus{}\left(a\plus{}1\right)^n\equal{}2001\].

for n=1 gives
$ a^2 \minus{} a \equal{} 2002 \\
a(a \minus{} 1) \equal{} 2002 \\
a(a \minus{} 1) \equal{} 2.7.11.13$ clearly no solution in positive integer
for n=2
$ a^3 \minus{} a^2 \minus{} 2a \equal{} 2002 \\
a(a^2 \minus{} a \minus{} 2) \equal{} 2002 \\
a(a \plus{} 1)(a \minus{} 2) \equal{} 2002$
a=13 is a solution
now for n=3
$ a^4 \minus{} a^3 \minus{} 3a^2 \minus{} 3a \equal{} 2002 \\
a(a^3 \minus{} a^2 \minus{} 3a \minus{} 3) \equal{} 2002$
no solution
for n=4
$ a^5 \minus{} a^4 \minus{} 4a^3 \minus{} 6a^2 \minus{} 4a \equal{} 2002 \\
a(a^4 \minus{} a^3 \minus{} 4a^2 \minus{} 6a \minus{} 4) \equal{} 2002 \\
a(a \plus{} 1)(a^3 \minus{} 2a^2 \minus{} 2a \minus{} 4) \equal{} 2002 \equal{} 2.7.11.13$
no solution

we can easily see that afterwards $ a^{n\plus{}1}\minus{}(a\plus{}1)^n$ will be much much greater than 2002
because only a's possible are 2,7,11,13 which are all primes

The above isn't true. $ a^{n+1}-(a+1)^n$ doesn't have to be greater than 2002, much less positive.

[hide]
Taking the equation $ \pmod{3}$, we see \begin{eqnarray*}a\equiv0\pmod{3}\qquad&\Rightarrow&\qquad a^{n+1}-(a+1)^n\equiv2\pmod{3}\\a\equiv1\pmod{3}\qquad&\Rightarrow&\qquad a^{n+1}-(a+1)^n\equiv1-2^n\pmod{3}\\a\equiv2\pmod{3}\qquad&\Rightarrow&\qquad a^{n+1}-(a+1)^n\equiv2^{n+1}\pmod{3}\end{eqnarray*} Since $ 2001\equiv0\pmod{3}$, then it follows that $ a\equiv1\pmod{3}$ and $ n$ is even. Consider the equation $ \pmod{a}$. Then \[ a^{n+1}-(a+1)^n\equiv -1\pmod{a}\equiv2001\pmod{a}.\] Therefore, $ 2002\equiv0\pmod{a}$ and $ a$ divides $ 2002$. Consider the equation $ \pmod{a+1}$. Then \[ a^{n+1}-(a+1)^n\equiv-1\pmod{a+1}\equiv2001\pmod{a+1},\] so $ a+1$ divides $ 2002$ as well. Since $ 2002=1\cdot2\cdot7\cdot11\cdot13$, the other pairs of consecutive integers dividing $ 2002$ are $ 1,2$ and $ 13, 14$.

$ a=1$ clearly does not have any solutions, but $ a=13$ has \[ 13^3-14^2=2197-196=2001.\] The only solution is $ \boxed{(a,n)=(13,2)}$.[/hide]


advanced fields
advanced fields unsolved

Prove that the stable category of an exact category is a suspended category.

Could you post the definition of a suspended category?


Hello
If we know that $ F\left( {F\left( x \right)} \right) \equal{} {e^{{e^x} \minus{} 1}} \minus{} 1,\forall x \in R,F\left( 0 \right) \equal{} 0$
and also that F is strickly increasing, ca we prove that unique type of F is $ F\left( x \right) \equal{} {e^x} \minus{} 1,x \in R$?
Tanx

It's not. There are infinitely many solutions; we can start with an arbitrary strictly increasing function mapping $ [1,e\minus{}1]$ onto $ [e\minus{}1,e^{e\minus{}1}\minus{}1]$ and tile everything else to match the functional equation.

It is the only analytic (given by a power series) solution.

thx for your reply
So is the only analytic solution and the way to prove it is by using series? Cant we avoid series and prove it with a more elementary way?

No, because the solution I'm thinking of is completely elementary from the definition; just follow the recursion, in which we simply solve for the coefficients one at a time.

[Edit: there is also an analytic decreasing solution]

Mhmhm...
tanx again
Im looking for a condition to add so we can have an elementary solution


algebra
polynomial
search
number theory unsolved
number theory

If $ x, y, z$ and $ n$ are positive integers satisfying the above equation, are necessarily $ x^{\frac{1}{n}}$, $ y^{\frac{1}{n}}$ and $ z^{\frac{1}{n}}$ integers?

[quote="feliz"]If $ x, y, z$ and $ n$ are positive integers satisfying the above equation, are necessarily $ x^{\frac {1}{n}}$, $ y^{\frac {1}{n}}$ and $ z^{\frac {1}{n}}$ integers?[/quote]
$ x \equal{} 18$, $ y \equal{} 32$, $ z \equal{} 98$, $ n\equal{}2$

[quote="feliz"]Smart.

And supposing $ x$, $ y$ and $ z$ have no common divisors...?[/quote]
Well.. Then it's a bit harder..

Ideas:
Assume that they are integers.
For $ n=2$: $ x^{1/2} + y^{1/2} = z^{1/2}$ $ \iff$ $ x - y = (zx)^{1/2} - (zy)^{1/2}$.
If $ x \neq y$ then $ zx$ and $ zy$ is squares. Which means $ x=y=1$.

Which may lead to the final solution. Note that:
$ x^{1/n} = z^{1/n} - y^{1/n}$ $ \iff$ $ x^{1/n}(z^{(n-1)/n} + z^{(n-2)/n}y + ... + zy^{(n-2)/n} + y^{(n-1)/n}) = z-y$
Or just:
$ \sqrt[n]{xz^{n-1}} + \sqrt[n]{xyz^{n-2}} + ... + \sqrt[n]{xy^{n-2}z} + \sqrt[n]{xy^{n-1}} = z-y$
Since $ RHS$ is an integer, we know $ xz^{n-1}, xyz^{n-2}, ... , xy^{n-1}$ is must be $ n^{th}$ powers. So $ x,y,z$ must be $ n^{th]}$ powers. And then $ x^{1/n}, y^{1/n}, z^{1/n]}$ is integers. So if $ (x,y) = (y,z) = (z,x) = 1$ then yes.

You're assuming a more general result than what you want to prove.

The basic idea is that the two numbers $ z^{ \frac{1}{n} }$ and $ x^{ \frac{1}{n} } \plus{} y^{ \frac{1}{n} }$ cannot have the same [url=http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)]minimal polynomial[/url] unless they are integers.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2055504702&t=228729]this post[/url]; the details are somewhat messy.


Let $ a$ denote a real root of equation $ z^{2007}\equal{}1$. What's more, $ z_{1}, z_{2}, ... , z_{2006}$ are imaginary roots of the equation. find $ a\minus{}b$ when $ b\equal{}(1\plus{}z_{1})(1\plus{}z_{2})(1\plus{}z_{3})...(1\plus{}z_{2006})$

ps. this problem is "excavated" from piles like Mt. Everest in my room. so, it'll be time-taking work to find answer key.

$ b\equal{}\frac{1}{2}(1\plus{}z_{1})(1\plus{}z_{2})(1\plus{}z_{3})...(1\plus{}z_{2006})(1\plus{}z_{2007})$, so after expanding, and pwning with vieta, we get $ b\equal{}\frac{1}{2}(1\plus{}1)\equal{}1$, hence $ a\minus{}b\equal{}0$

[quote="rofler"]$ b \equal{} \frac {1}{2}(1 \plus{} z_{1})(1 \plus{} z_{2})(1 \plus{} z_{3})...(1 \plus{} z_{2006})(1 \plus{} z_{2007})$, so after expanding, and pwning with vieta, we get $ b \equal{} \frac {1}{2}(1 \plus{} 1) \equal{} 1$, hence $ a \minus{} b \equal{} 0$[/quote]
It would be nice to explain to me why $ b \equal{} \frac {1}{2}(1 \plus{} 1) \equal{} 1$.I mean how you get $ b\equal{}\frac{1}{2}(1\plus{}1)$

$ a \equal{} 1$, so after dividing out the rest of the polynomial is $ z^{2006} \plus{} z^{2005} \plus{} ... \plus{} 1 \equal{} 0 \equal{} Q(x)$.  Now I claim that $ b \equal{} Q(\minus{}1)$.  Why?

I think you meant $ Q(z)$ instead of $ Q(x)$.


calculus
derivative
quadratics
algebra
system of equations
arithmetic series

The formula which expresses the relationship between $x$ and $y$ as shown in the accompanying table is:
\[ \begin{tabular}[t]{|c|c|c|c|c|c|}\hline
x&0&1&2&3&4\\\hline
y&100&90&70&40&0\\\hline
\end{tabular}\]
$\textbf{(A)}\ y=100-10x \qquad
\textbf{(B)}\ y=100-5x^2 \qquad
\textbf{(C)}\ y=100-5x-5x^2 \qquad\\
\textbf{(D)}\ y=20-x-x^2 \qquad
\textbf{(E)}\ \text{None of these}$

[hide="Solution"]notice that the difference between the values of y is increasing by 10 as x increases by1.As for x=0 the value of y is 100 and y is decreasing A is the right answer.[/hide]

@Cronus
$ 70 \ne 100 \minus{} 10(2) \equal{} 80$

[hide="Solution"]
Plugging into A, we notice it doesn't hold as soon as we reach 2.
B Does work with 2 either.
C holds for all values so C is the correct answer.
[/hide]
[hide="non-brute force solution"]
Notice that the points create a parabolic curve. So the equation of the points must be in the form $ y \equal{} ax^{2} \plus{} bx \plus{} c$

Since the y-intercept is $ (0, 100)$, that means $ c \equal{} 100$

Now to set up a system of equations
$ 90 \equal{} a \plus{} b \plus{} 100$
$ 0 \equal{} 16a \plus{} 4b \plus{} 100$

Solving this yields $ a \equal{} \minus{}5, b \equal{} \minus{}5$
[/hide]

[hide="Ok.What I meant"]Let the value of y when x is equal to k be a_k.
Then a_(k+1)=a_(k)-10(k+1) where k=0,1,2,3,...(1)
A(x):=Sum. a_k*x^k k E N U 0.
multply 1 by x^k and sum over all values ...
A(x)-100/x=A(x)-(10x)/(1-x)^2-10/1-x.
Find A(x) in a fractionlike value then expanding it yields the same answer.[/hide][/hide]

Brute force may always work faster with simple multiple choice questions like this one.
Otherwise you could see the arithmetic series in that sequence.
y(x) = 100 - S(x)  with S(x) being the arithmetic series:
S(x) = 0 + 10 + 20 + ... + 10x which we know is
S(x) = 5x(x+1)

The most brute force solution eva:
Take a derivative of each of A, B, C, D. Compare differences quick. done. C is the answer.

[quote="SnipedYou"]
Notice that the points create a parabolic curve. So the equation of the points must be in the form $ y \equal{} ax^{2} \plus{} bx \plus{} c$[/quote] Well tbh, there are an infinite number of equations it could be. However, because all of the answers are either linear or quadratic, we only have to try that.

The second differences of the y values are constant being -10. This means the equation is quadratic. Also note that because the second difference is -10 the coefficient of the x-squared term will be half of it or -5. This narrows it down to B, C, or E. The y-intercept is 100, so we are missing one constant. If we plug 1 into both equations, only C comes out with 90. Therefore the answer is C.

[hide= My Solution]I saw that y always equals 100 - 10 * the sum of all the previous values of x, or $ 100 - 10[x(x+ 1) / 2] $, which simplifies to
$100 - 5x^2 - 5x$ [/hide]

Ultimately, because option E exists, you really do have to check that all five points work...unless you can find a good reason that you don't have to. And there is a good rason you don't have to.

The first differences are linear:  $-10, -20, -30, -40$.  That means the second differences are constant:  $-10, -10, -10$.  When the $n$th differences are constant, you can fit an $n$th degree polynomial to your data exactly.

Thus, the data should all fit a quadratic.  Since one and only one (rectilinear) quadratic passes through $3$ noncollinear points, we can just test $3$ points in each equation to see which one works.  So, all you need to do is eliminate $A$ for not being a quadratic, $B$ for not having the right x-intercept, and $D$ for not having the right y-intercept.  That leaves $C$ as the only possibility.  Since $C$ fits $(0, 100), (1, 90), (4, 0)$ perfectly, it must fit the other two points.

Note that even if you had random numbers for $y$, the fourth difference would be constant...since there's only one fourth difference. Therefore, you could definitely fit a fourth degree polynomial to the function.  This makes sense - given $n+1$ points that are not all collinear, you can fit an $n$th degree polynomial to fit it.


geometry
parallelogram
trigonometry
geometric transformation
reflection
circumcircle
trapezoid

Given a point $P$ in parallelogram $ABCD$ such that $\angle{PAB}=\angle{PCB}.$ Prove that $\angle{PBC}=\angle{PDC}.$

Put $PBC=x$ and $PDC=y$ then easily by sine rule we find: $\frac{\sin x}{\sin (\alpha-x)}=\frac{\sin y}{\sin (\alpha-y)}$ where $\alpha = ABC$ so $x=y$

Let A' be a reflection of A in PB. Then $\angle PA'B = \angle PAB = \angle PCB$ and the quadrilateral A'PBC is cyclic with a circumcircle (O). Let a parallel to A'C through B meet (O) again at D'. A'D'BC is a cyclic trapezoid, isosceles, with equal diagonals CD' = A'B. But A'B = AB = CD, so that CD' = CD. Since $AB \parallel CD$ and $D'B \parallel CA',$ $\angle DCA' = \angle ABD'.$ From the circle (O), $\angle A'CP = \angle A'BP,$ and form the isosceles triangle $\angle ABA'$ with angle bisector BP, $\angle A'BP = \angle PBA.$ Consequently, $\angle DCP = \angle DCA'+\angle A'CP = \angle PBA+\angle ABD' = \angle PBD'.$ From the circle (O), $\angle PBD' = \angle PCD'.$ It follows that $\angle DCP = \angle PCD'.$ Combining with CD = CD', D' is a reflection of D in PC, which means that $\angle PDC = \angle PD'C.$ But $D' \in (O),$ so that $\angle PD'C = \angle PBC,$ which finally yields $\angle PDC = \angle PBC.$


angle < 45 degrees
Tuned to new pattern
Just in case its wrong
[hide]
>45 degrees  :P   :rotfl:   :blush:
[/hide]

Choices

a) Less than 45
b) Greater than 45
c) Equal to 45
d) Cannot be predicted

And for christ's sake give a proper reason da fool. !!!!

DONT MAKE ME CURSE YOU DA  :D
my answer $ \rightarrow$ a) correct i get +3, wrong i get -1 so no feelings da  :D   :rotfl:

Why do you choose A). Please think and give me a reason, its only a kinematics question which I am sure all the 11th stders will also answer and inside IIT you have to do physics.......;

[quote="madness"]bcos of momentum conservation  :maybe:
let $ m_{b}$ be the mass of the bullet and $ m_{c}$ be mass of the cannon. initial momentum = final momentum.
$ \implies$ $ m_{b} \cdot v_{b} \equal{} m_{c} \cdot v_{c}\cdot cos\alpha$
Hence the angle has to be smaller fer cosine to be larger bcos momentum along that of cannon has to be conserved  :maybe:[/quote]

Nice Bull s.h.i.t. I dont really understand why that implies anything, you want the answer??

plz da im not really in the mood fer phy so lemme try later  :D  i will try ok?  :)

the question is not clear....wat do u mean arm of the cannon???
isn't the cannon a one-piece machine, withe the cannon ball moving straight out wrt the cannon. how can the arm make an angle with the body?

I tink that the answer is >45 degrees .... because of the equal and opposite force exerted in the horizontal direction  by the canon ball when it just leaves the arm...

Answer is greater than 45. Here's the simple explanation;

W.r.t to Cannon The ball is at 45 so we have to find it wrt to ground.
V(wrt cannon) = Vx + Vy (But Vx = Vy)
Velocity of the cannon = -v
V(wrt ground) = (Vx-v)i + Vy.
tan(theta) = Vy/Vx-v

which is greater than one.
QED

Hell_ever: see http://en.wikipedia.org/wiki/Cannon the picture. The part which protrudes out is the arm of the cannon.


Vi har en følge $a_n$, der $a_0=m$ og $a_{n+1}=a_n^5+487$. Finn m slik at følga inneheld flest mogeleg kvadrattal.

Det første som er lurt å observere, er at $487=2\cdot3^5+1$. Ved å sjå på uttrykket $a_{n+1}=a_n^5+2\cdot3^5+1$ kan ein gjette seg fram til ei løysing med $m=3^2$. Då har ein $a_1=(3^5+1)^2$.

Så går det an å vise at $a_2,a_3....a_n$ aldri er kvadrat. Observer at $487\equiv3(mod4)$. Det går også an å finne ut at alle 5-potensar er $\equiv0,1,3(mod4)$. Når vi så vel en m, vil $a_1$ få verdien 0,2 eller 3 modulo 4. 2 og 3 er aldri kvadrat modulo 4. Dersom $a_1$ har verdien 0, vil $a_2$ få verdien 3 modulo 4, og det er heller ikkje kvadrat. Når ein fortsetter med verdiane 2 og 3 vil ein ende opp med 2 og 3 i det uendelege. Det følgjer at kun $a_1$ og $a_0$ kan vere kvadrattal.

Då må vi vise at det ikkje fins andre $a_0$ og $a_1$ som begge er kvadrat. Då har vi: $a_0=x^2$ og $a_1=(x^5)^2+487$. Hvis $a_1$ og er eit kvadrat, kan vi bruke konjugatsetninga:

$a_1^2-(x^5)^2=487$ som har ei løysing $x=3$

Her er en litt anna løsning (Geir sin): Du viser på samme måte at $a_n$ ikke er et kvadrat for $n\geq2$. Hvis nå både $a_0$ og $a_1$ er kvadrater, har vi $a_0 = x^2$, og $a_1 = y^2 = x^{10} + 487.$ Siden $x^{10} < y^2$ har vi at $x^5 < y$. Siden begge tallene er heltall, har vi at $x^5 + 1 \leq y$. Kvadrerer vi, får vi at
\[ x^{10} + 2x^5 + 1 \leq y^2 = x^{10} + 487 \Rightarrow 2x^5 \leq 486 \Rightarrow x \leq 3. \]
Nå holder det å sjekke at $a_1$ ikke er et kvadrat for $x=1$ og $x=2$, og dermed har vi vist at $x=3$, det vil si $a_0 = 9$, er eneste løsning.

Fordelen med det her beviset er at du slipper å sjekke at 487 er et primtall...


Let m,k be fixed positive integers. Prove that for sufficiently high $ n$, each $ k$-colored graph $ K_{n,n}$ has one-colored sub $ K_{m,m}$. Here we denote by $ K_{n,n}$ the full graph with $ 2n$ vertices and 2-pole.

You are right : it is a direct consequence of Ramsey theorem... :D

Pierre.

Is it direct? I don't think so, I'll write my solution some later.


Assume that the radiation emitted from the Sun moves radially outward from the Sun and that no radiation is absorbed between the Sun and Earth. How much energy in the form of radiation will fall per second on an area of $1\text{m}^2$ on Earth, if that area is perpendicular to the straight-line path of the radiation?

The distance from the Sun to Earth is $1.5\times 10^{11}\text{m}$.

I don't know how much total radiation the sun emits, but the answer would be

$\mbox{total radiation} \cdot \frac{1}{9\pi\cdot 10^{22}}$

You can use the Stephan-Boltzmann emission law for a black body (or whatever its name is in English):

P/S = \sigma .T^4

P/S is power through area S, T is the temperature of the surface and sigma is the Stephan-Boltzmann constant = 5.67 E-8 W.m^-2.K^-4.

Total energy dissipated by the sun is therefore

P = \sigma . T^4 . 4 . \pi . R^2

R is the sun radius, approx. 695500 km
T is the temperature of Sun's surface, approx. 5770 K

Then the power through one square meter in distance D from sun will be

P/(1 m^2) = \sigma . T^4 . (R / D)^2
P/(1 m^2) = approximately 1350 W


function
limit
real analysis
real analysis unsolved

Is there an increasing function $f: \mathbb{R}\rightarrow [0,1]$ that has uncountably many discontinuities?

It is well known (although still at too high a level for the "Computations and Tutorials" forum), and the answer is no. Let's wait for someone who doesn't already know this supply the argument.

If $f$ is increasing, one can show that $f$ is discontinuous at $a$ if and only if $\left(f(a-),f(a+)\right)\neq \emptyset$, where $f(a-)=\lim_{x\rightarrow a-}f(x)$ and $f(a+)=\lim_{x\rightarrow a+}f(x)$. One also can show that $\left(f(a-),f(a+)\right)\cap \left(f(b-),f(b+)\right)=\emptyset$ for $a\neq b$ where $a,b$ are poitns of discontinuity of $f$. If $(f(a-),f(a+))\neq\emptyset$ then this interval contains a rational number. It follows that the number of discontinuity points of $f$ cannot exceed the cardinality of $\mathbb{Q}$.

[quote="Peter VDD"]There is a [i]much[/i] simpler way... ;)[/quote]
What is it? I am quite curious...

Well... I hope it is right: (we denote discontinuity here as 'jump', since it's increasing each discontinuity is a jump)

There can be at most 1 jump of size in $]1,\frac12]$
There can be at most 3 jumps of size in $]\frac14,\frac12]$
...
There can be at most $2^n-1$ jumps of size in $]\frac1{2^n},\frac1{2^{n-1}}]$

Every jump/discontinuity is in one of those intervals, there can only be a finite amount of jumps per interval, and the amount of intervals is countable. Thus, the number of jumps is countable.

uhm... function is increasing, so if you have say two points of discontinuity $x<y$ such that both $f_+(x) - f_-(x), f_+(y) - f_-(y) \in \left(\frac12; 1\right)$, you'd have that $f_+(y) - f_-(x) > 1$, which is obviosly absurd, since $f(\mathbb{R}) \subset \left[0;1\right]$.

of course this proof can't work for increasing $\mathbb{R}\rightarrow\mathbb{R}$ functions, but it's easy to "correct" it even for that case..

[quote="liyi"][quote="Peter VDD"]Well... I hope it is right: (we denote discontinuity here as 'jump', since it's increasing each discontinuity is a jump)

There can be at most 1 jump of size in $]1,\frac12]$
[/quote]
Why?[/quote]If there were 2, then you'd have jumped a distance 1, so you'd have crossed your entire codomain.
Which means the function is constant everywhere else, thus not increasing.

Come to think of it, can't we just generalise this to $\mathbb{R}\rightarrow\mathbb{R}$ saying that for each such interval the number of jumps of length in it is countable?

You're not saying what you want to: "jumps of size " doesn't mean anything without a number there.

jumps, with the size of the jump in ]1,1/2]. So e.g. left limit is is 0.3 and right limit is 0.9, so a jump of 0.6. ok? (the wording sucks I know)


Find all functions $ f: \mathbb R^\plus{} \rightarrow \mathbb R^\plus{}$ such that
\[f(xf(y))\equal{}f(xy)\plus{}x\]
holds for all $x,y\in \mathbb R^\plus{}$.

[quote="DCTPKTCSPHN"]Find all $ f: R^ \plus{} \rightarrow R^ \plus{}$ such that
$ f(xf(y)) \equal{} f(xy) \plus{} x$ $ \forall x,y\in R^ \plus{}$[/quote]

$ x\equal{}1$ $ \implies$ $ f(f(y))\equal{}f(y)\plus{}1$ and so $ f(x)\equal{}x\plus{}1$ $ \forall x\in f(\mathbb R^\plus{})$
Let $ u\equal{}f(1)$ and $ x>u$ : $ f((x\minus{}u)f(\frac{1}{x\minus{}u}))\equal{}f(1)\plus{}x\minus{}u\equal{}x$ and so $ x\in f(\mathbb R^\plus{})$ and so  $ f(x)\equal{}x\plus{}1$

So : $ \exists u>0$ such that $ \forall x>u$ : $ f(x)\equal{}x\plus{}1$

Let then $ v\equal{}\frac{u^2}{u\plus{}1}$. If $ x>v$, we get $ x>\frac{u^2}{u\plus{}1}$ and so $ x\frac{u\plus{}1}{u}>u$ and so $ \frac{x}{u}>u\minus{}x$.
Let then $ a>0$ such that $ \frac{x}{u}>a>u\minus{}x$ and $ b\equal{}\frac{x}{a}$. We have $ b>u$ and $ ab\plus{}a>u$

We have $ f(af(b))\equal{}f(ab)\plus{}a$. Since $ b>u$, $ f(b)\equal{}b\plus{}1$ and $ af(b)\equal{}ab\plus{}a$ and, since $ ab\plus{}a>u$, $ f(af(b))\equal{}f(ab\plus{}a)\equal{}ab\plus{}a\plus{}1$, and so $ ab\plus{}a\plus{}1\equal{}f(ab)\plus{}a$ and so $ f(x)\equal{}x\plus{}1$

So : $ \forall x>\frac{u^2}{u\plus{}1}$ : $ f(x)\equal{}x\plus{}1$

So, let the sequence $ a_0\equal{}u$, $ a_{n\plus{}1}\equal{}\frac{a_n^2}{a_n\plus{}1}$. We have $ f(x)\equal{}x\plus{}1$ $ \forall x>a_n$ $ \forall n$

So, since $ a_n$ is a decreasing sequence whose limit is 0, we get :

$ f(x)\equal{}x\plus{}1$ $ \forall x>0$

here is another solution. [url]http://www.mathlinks.ro/viewtopic.php?t=266578[/url]


I tried downloading Go2PDF (That thing in the LaTeX downloads) and it won't work.  I tried to do $\sqrt{5}$ first, so first I typed it enclosed in $\$$ signs, and it only showed \sqrt{5} enclosed in dollar signs all over again.  Then I tried \sqrt{5}, and it still didn't show up as $\sqrt{5}$.  Then I tried the hello world thingy, and it STILL didn't work.

Any thoughts?

Thanks. :huh:

You don't want Go2PDF if you want to typeset with $\LaTeX$.  You want to download mixtex and TeXniccenter to do that (links are on the $\LaTeX$ pages)

You need miktex and texniccenter to do that.  Go2pdf just converts a document to pdf format.


algebra unsolved
algebra
inequalities
highschoolmath

let a, b, c are positive numbers. Prove that:
[code]\frac{a^{2}}{b^{2}+c^{2}}+frac{b^{2}}{c^{2}+a^{2}}+frac{c^{2}}{a^{2}+b^{2}}\gre\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}[/code]

$\{a,b,c\}\subset (0,\infty)\Longrightarrow \sum \frac{a^2}{b^2+c^2}\ge \sum \frac{a}{b+c}.$

In http://www.mathlinks.ro/Forum/viewtopic.php?t=50773 , I wrote:

[quote="darij grinberg"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=17207 and http://www.mathlinks.ro/Forum/viewtopic.php?t=34563 and the generalizations at http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 .[/quote]

  darij


[size=150]1984 AIME #10[/size]

A function $f$ is defined for all real numbers and satisfies
$f(2 + x) = f(2 - x)$ and $f(7 + x) = f(7 - x)$
for all real $x$. If $x = 0$ is a root of $f(x) = 0$, what is the least number of roots $f(x) = 0$ must have in the interval $-1000 \le x \le 1000$

-Contest Problem Book V


inequalities
trigonometry
geometry
circumcircle
function
angle bisector

Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.

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$ OA=OB=OC=R$

$ OB^{2}+BC^{2}-2\bullet OB\bullet BC\cos\angle OBC=OC^{2}$
$ R^{2}+BC^{2}-2\bullet R\bullet BC\cos\angle OBC=R^{2}$
$ BC^{2}=2\bullet R\bullet BC\cos\angle OBC$
$ \cos\angle OBC=\frac{BC}{2R}$
(We could also find this by drawing diameter BB' of the circumcircle, since $ \angle BCB'$ would be a right angle).
Similarily, $ \cos\angle OBA=\frac{AB}{2R}$
Because BC>AB, $ \cos\angle OBC>\cos\angle OBA\implies \angle OBA>\angle OBC$ since the cosine function is decreasing over $ [0,\frac{\pi}{2}]$
Thus the angle bisector of $ \angle ABC$ is in the interior $ \angle OBA$ instead of the interior of $ \angle OBC$

Likewise, the angle bisector of $ \angle BAC$ is in the interior $ \angle OAB$ instead of the interior $ \angle OAC$

Thus the two angle bisectors intersect in the interior  of $ \triangle AOB$ at I, so OI is a cevian of $ \triangle AOB$ and intersects segment AB.

Extend CO to meet AB at $ C_{1}$ and AO to meet BC at $ A_{1}$

We get, using the method above, that the angle bisector of $ \angle ACB$ is in the interior of $ \angle OCA$ instead of the interior of $ \angle OCB$. Thus I is in the interior of $ \triangle ACC_{1}$. We already know it is in the interior of of $ \triangle AOB$ so I is in the interior of $ \triangle AOC_{1}$, so OI is in the interior of $ \angle AOC_{1}$

$ \angle AOC_{1}$ and $ \angle COA_{1}$ are vertical angles, OI is in the interior of $ \angle COA_{1}$ so it intersects segment $ A_{1}C$ which is part of $ BC$

QED

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