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Okay here are the rules. I will have a sign up list of [u][b]10[/b][/u] people.
Each person will be given an egg, and one will be a bomb, which one it is I am not telling.
Every person will have a chance to throw the egg at someone, (including themselves?). When all the votes are up, I will post them up. The person that has the bomb thrown at them will die a sad and yolk filled death. We will continue on without the straggler and procede to the final showdown where each person will have 5 eggs. 1,2,3,4,5. 4 of those eggs will have bombs. The person with the most bombs dies and the last person will be the 'egg champion'!!!

Sign up list:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

Please Join!!!

/in

Isn't this like round one of Beat the mod?

That's what the title indicates. But it is adapted.
I think I'll leave myself out of this game though. Other people can have the fun of egg bombs. :D

/in Considering my history in this, it should be interesting.   :D

hrmm... revenge seems greener on this side of the forum... $ 1\equal{}2\in Game$

all the spots have been filled

REVENGE!!! [b]throw at cf249[/b]

[b]throw at 1=2[/b] just because he has already thrown his egg.

[b]Throw at lotsofmath[/b] for same reason as above.

Take eggs number: 2,3,5,7,11,13,17,19,4,9

Throw non-bombs at [b]Dojo[/b]
Throw bombs at [b]james4l[/b]

While james is offline, I will boil the eggs and make them on fire.
Also dodge eggs number 1,10,15,16, and 18

you can't dodge... hem hem...

anyway this is so unfair for ernie. fine pick 5 now, not 10

Cna't you make the finale a little more interesting? A story? Please?

EDIT: oh my bad, I thought that was the end or something. In my last post, change all 'james's to 'ernie'

[quote="Dojo"]Well sorry chen[b]h[/b]si.
Anyway,

dragon96 hurls at ernie and splatyango, the goddess of eggs pops up again...

ernie throws at the motionless james4l and kaboosh poached eggs...

james4l carelessly drops his egg and: whaoue;g;eawlugobuoelgboaug;ouoagbsugbesou;gb;sdubga. Dead in seconds.

Prepare for the final round!!!

I have changed the rules.
Each one of you will use your matrix skills. There will be 20 eggs and 9 of them have bombs. You choose 10 of the eggs and the person with more bombs will die.[/quote]

Huh I thought I fired at ernie?

@ Dojo:  See post #70.  james4l did throw at ernie, even if it wasn't explicitly clear.   :wink:

It was?
Ah...
Then ernie dies and james4l chooses.


Prove that the complex number $ z_{0}$  lies inside or on a side of the triangle with vertices $ z_{1} , z_{2} ,z_{3}$ in the complex plane if and only if

$ z_{0} \equal{}pz_{1} \plus{}qz_{2} \plus{}r{z_3}$ where p ,q , r are non-negative real numbers satisfying p+q+r=1.

Let the initial point $ O$ in a plane and $ z_4$ be the point such that $ z_2z_4 : z_4z_3 \equal{} t : 1 \minus{} t\ (0\leq t\leq 1)$ and let $ z_1z_0 : z_0z_4 \equal{} s : 1 \minus{} s\ (0\leq s\leq 1)$ we have

$ \overrightarrow{Oz_0} \equal{} \overrightarrow{Oz_1} \plus{} s\overrightarrow{z_1z_4}$

$ \equal{} \overrightarrow{Oz_1} \plus{} s\{(1 \minus{} t)\overrightarrow{z_1z_2} \plus{} t\overrightarrow{z_1z_3}\}$ with $ 0\leq s\leq 1,\ 0\leq t\leq 1\ \cdots\ [*].$

$ \equal{} \overrightarrow{Oz_1} \plus{} s\{(1 \minus{} t)(\overrightarrow{Oz_2} \minus{} \overrightarrow{Oz_1}) \plus{} t(\overrightarrow{Oz_3} \minus{} \overrightarrow{Oz_1})\}$

$ \equal{} (1 \minus{} s)\overrightarrow{Oz_1} \plus{} s(1 \minus{} t)\overrightarrow{Oz_2} \plus{} st\overrightarrow{Oz_3}.$

$ \therefore z_0 \equal{} (1 \minus{} s)z_1 \plus{} s(1 \minus{} t)z_2 \plus{} stz_3.$

Therefore setting $ p \equal{} 1 \minus{} s,\ q \equal{} s(1 \minus{} t),\ r \equal{} st$ and $ [*]$ gives $ p \plus{} q \plus{} r \equal{} 1$ with $ p\geq 0,\ q\geq 0,\ r\geq 0.\ Q.E.D.$

This is a basic result in convexity: given "vectors" $ z_1, z_2, \ldots, z_n$, the convex hull of this set is given precisely by the convex combinations $ \lambda_1z_1 \plus{} \lambda_2z_2 \plus{} \cdots \plus{} \lambda_nz_n$, where $ \lambda_i \geq 0$ for all $ i$, and $ \lambda_1 \plus{} \lambda_2 \plus{} \cdots \plus{} \lambda_n \equal{} 1$.


What kind(s) of chemical bonds are found in magnesium nitrate?  Why?

Thanks in advance!

the bond between magnesium and nitrate is ionic bond because magnesium is a metal and nitrate is a polyatomic molecule. magnesium loses two electrons; nitrate NO3-, so 2 nitrate gain two electrons from magnesium. there's losing and gaining electrons. thus, the bond between magnesium and nitrate is ionic.

nitrate is NO3 with the oxidation number of negative one. this is a covalent bond.

the oxidation number of magnesium is +2 and that of nitrate is NO3-. nitrate is the polyatomic molecule.

don't you mean charge?

my feeling is that oxidation number is something assigned to an individual atom, not the molecule...

[quote="I_thirst_for_knowledge"]nitrate is the polyatomic molecule.[/quote]

Nitrate is a polyatomic [b]ion[/b].

[quote="Elcric"]What kind(s) of chemical bonds are found in magnesium nitrate?[/quote]

Magnesium is an alkaline earth metal, while nitrogen and oxygen are non-metals: therefore, we will have ionic bonds between $ \ce{Mg^{2 + }}$ (ions) and $ \ce{NO3 - }$ (ions). However, in nitrate ion the three nitrogen-oxygen bonds are covalent (due to small electronegativity difference), and they are all equal, since nitrate ion is a ressonance hybrid.


for any graph $G$,we have $ \theta \leq (\frac {\omega}{2}+\frac {\Delta +1}{2})+1$

What does  :theta: means?
chromatic number?

Well, usually the chromatic number is denoted by $\chi$, whereas $\theta$ is used for the clique cover number...But I agree that according to the title it is unclear.

In another hand, $\omega$ is usually the clique number, so...

Pierre.

Yes i know, thats why it seem unclear.
If you know let me know because it seems interesting.


Find $ \sum_{r\equal{}1}^{n} \frac{1}{\sqrt{k\plus{}1}\plus{} \sqrt{k}}$.

[quote="AndrewTom"]Find $ \sum_{r \equal{} 1}^{n} \frac {1}{\sqrt {k \plus{} 1} \plus{} \sqrt {k}}$.[/quote]

$ \sum_{r \equal{} 1}^{n} \frac {1}{\sqrt {k \plus{} 1} \plus{} \sqrt {k}}\equal{}\frac {n}{\sqrt {k \plus{} 1} \plus{} \sqrt {k}}$

Thanks kunny. I meant to write the sum from k=1 to n (my CTRL key isn't working - again).

hello, hint:
$ \frac{1}{\sqrt{k\plus{}1}\plus{}\sqrt{k}}\equal{}\sqrt{k\plus{}1}\minus{}\sqrt{k}$.
Sonnhard.


limit
induction
trigonometry
calculus
calculus computations

show  :maybe:

$ \lim_{\textbf m\to\infty}\;\;\;2^{\textbf m+1}\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt 2}}}}}_{\textbf m}}\;\;\;\quad=\quad\;\;\;\boxed{\pi}$

By induction on $ n$ we have

${ \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt 2}}}}}_{\textbf m}}=2\cos\frac{\pi}{2^{m+1}}$.

Therefore that limit is equal to $ \lim_{m\to\infty}2^{m+1}\sqrt{2(1-\cos\frac{\pi}{2^{m+1}})}=\lim_{m\to\infty}2^{m+1}\cdot 2\sin{\frac{\pi}{2^{m+2}}=\lim_{m\to\infty}\left(\pi\cdot \frac{\sin{\frac{\pi}{2^{m+2}}}}{\frac{\pi}{2^{m+2}}}\right)=\pi.}$


Prove that $f_{n}$ and $f_{n+2}$ are relatively prime $\forall n$. where $f_{i}$ is the i-th Fibonacci number.

[quote="cuopbientcd"]I know $gcd(f_{m},f_{n})=f_{gcd(m,n)}$. and you can do it!OK :lol:[/quote]

where did you get this? it's a very beautiful fact. can you give me a proof? :D

[quote="cuopbientcd"]and you can do it!OK :lol:[/quote]

i made the problem so obviously i have a solution. i'm interested in second opinions. :D

I think it is hard posting it but it is posted before many times

can someone solve my problem [b]not[/b] using the fact stated above?

@Rust : But how can you say that $n > 2m$ ?
Can we have $F_{k}$ for $k$ having a negative value ?

It is well known that $(f_{k},f_{k-1}) = 1$...
PROOF for $k \geq 3$ : $(f_{k},f_{k-1}) = (f_{k-1}, f_{k}-f_{k-1}) = (f_{k-1},f_{k-2}) = \cdots = (f_{2},f_{1}) = 1$
Thus, $(f_{n+2},f_{n}) = (f_{n+2},f_{n+2}-f_{n}) = (f_{n+2},f_{n+1}) = 1$

Hm... Now let's look again the second problem :
$(f_{m},f_{n}) = f_{(m,n)}$


I know the remainder of 9/4 is 1.

But what is the remainder of 0/3, 1/2, 3/4.....

what is the formula for finding the remainder?

Any help is appreciated. thank you.

Remainder of 1/2 is 1
Remainder of 3/4 is 3

If a/b  and a < b Then remainder of a/b is a

the reminder of $p/q$ is $r$ if and only if $a*q+r=p$ for $a$ integer with $r<q$ if you are asking for the reminder of $p/q$ and $p<q$ obviously it will be $p$

Take $1\div 3$ for example.  When you try to divide, you get 0 (1 is less than 3, so 3 doesn't even go into 1 once!) with 1 left over.  That one left over is your remainder.


function
number theory proposed
number theory

Find all functions  $ f: N^{*}\to Z $ satisfy:
     i)if $ b|a $ then $ f(a) \geq f(b) $
     ii)$ f(ab)+f(a^2+b^2)=f(a)+f(b) $
let's go for it,good luck :)  :)

Come on , man , Your proof is false ! Do you beleave that $n>2$  is a divisor of $(n-1)^2 + 1$ ??  :P  :rotfl:
 :cool:

Diogene , the :rotfl: emoticon is quite offensive, really.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=53451[/url]

Please continue there, and don't post a problem again if it has been posted just a few days ago. (Obviously it's possible that you didn't know...)


Will there be an expansion of number of USAMO qualifiers in 2007?

I believe the USAMO expansion will continue as long as AoPS continues to fund it. So that would be a yes.


Can you tell me how to solve an equation like this?

cosec (x+10) = 3

Take the cosine of each side of the equation.  Then $ x\plus{}10\equal{}\cos{3}$ so $ x\equal{}\cos{3}\minus{}10$.

$ \csc \neq \cos^{\minus{}1}$.

Recalling that $ \csc x \equal{} \frac{1}{\sin x}$, rewrite as

$ \sin (x \plus{} 10) \equal{} \frac{1}{3} \implies$
$ x \plus{} 10 \equal{} \arcsin \frac{1}{3} \implies$
$ x \equal{} \arcsin \frac{1}{3} \minus{} 10$


number theory
greatest common divisor
number theory unsolved

$a$ and $b$ are non zere integers. Prove that :  iff $a^{2}-2b^{2}=1$ then $a$ and $b$ are coprime.


calculus
integration
calculus computations

hi everyone
this may be simple..but am not able to do it...

Integrate :
{(1+X^3)}^(-0.5)dx
thanx...

It's not simple; the antiderivative can be expressed in terms of an elliptic integral.

hi
thank u very much for such a prompt reply.... :lol:
Shyam


did  anybody solve Q.no 6th of rmo 2009 which is "ina book with page nos. 1 to 100, some pages are torn off . The sum of the nos. on the remaining pages is 4949. How many pages are torn off?
I came with the solution 6.

the problem is solved here: [url]http://www.mathlinks.ro/viewtopic.php?p=1695946#1695946[/url]

pfanni

actually 6 is for both sides of a sheet of paper. i was confusing with fact that 1 page for one side.
whatsoever thax for that link.

for this the main logic is if u take out the first page then second page comes out with it

now we find the maximum no of pages that can be torn
3+7+...+xn<=101    where n is the no of pages torn
=n/2*(3*2 +(n-1)*4)<=101
=n*(3+2n-2)<=101
=n(2n+1)<=101
for n=7 n(2n+1) is 105 and for n=6  it is 78 which is less than 101 thus n=1,2,3,4,5,6

for taking out one page the simple solution is 50 and 51 but they r not of the same page

let x1=1 x2=3  x3=5 and so on
then (2*x1+1) +(2*x2+1)+...=101
if there r even no. of pages it is not possibe as 101 should be divisible by 2

thus n=3,5
2x1+2x2+2x3+2x4+2x5+5=101
2(x1+x2+x3+x4+x50)=96
(x1+x2+x3+x4+x5=48
but x1,x2,x3,x4,x5 are all odd
thus n=5 is not possible

thus n=3
lets check
2x1+2x2+2x3+3=101
2(x1+x2+x3)=98
x1+x2+x3=49
this is possible (1,3,45 )and many more
(1+2)+(3+4) +(45+46)=101
thus ony 3 pages can be torn :D


limit
integration
linear algebra
matrix
algebra
polynomial
vector

Suppose $A, C$ are positive definite and $AX+XA=C$ has a unique solution $B$. Show that $B$ is positive definite too.

$\lim_{t\rightarrow \+\infty}e^{-tA}Ce^{-tA}=0$
$\int_{0}^{\infty}e^{-tA}Ce^{-tA}dt$ is convergent

We have (1)  $X=B=\int_{0}^{\infty}e^{-tA}Ce^{-tA}dt$ is solution of
$AX+XA=C$

Indeed $AX+XA=\int_{0}^{\infty}(Ae^{-tA}Ce^{-tA}+e^{-tA}Ce^{-tA}A)dt=\int_{0}^{+\infty}\frac{d}{dt}(-e^{-tA}Ce^{-tA})dt=[-e^{-tA}Ce^{-tA}]_{0}^{+\infty}=C$

With (1) can prove $B$ is positive definite

well, i didn't know how to solve out $B$...

My proof goes like this (hopefully it is correct)

First, take transpose of the given equation, we get $B^T = B$.

Suppose $\lambda$ is an eigenvalue of $B$ and $B\xi = \lambda\xi$, \$xi\neq 0$.
From $AB+BA=C$
we have
$\xi^T AB \xi + \xi^T BA \xi = \xi^T C \xi$
or
$\xi^T AB \xi + (B\xi)^T BA \xi = \xi^T C \xi$
which is,
$\lambda \xi^TA\xi + \lambda \xi^T A \xi = \xi^T C \xi$
$2\lambda \xi^TA\xi = \xi^T C \xi$
Since $A, C$ are both positive definite, $\xi^T A \xi$ and $\xi^T C \xi$ are positive, hence $\lambda$ is positive.
That means $B$'s eigenvalue is positive, therefore it is positive definite.

[quote="liyi"] $AX+XA=C$ [/quote]

I read sowhere this call  Lyapunov equation, russian no ?
Myth do you have russian material about this equation or similar.
thx

[quote="Moubinool"][quote="liyi"] $AX+XA=C$ [/quote]
I read sowhere this call  Lyapunov equation, russian no ?
[/quote]
Lyapunov is a famous russian mathematician of first half of 20th century.

[quote]Myth do you have russian material about this equation or similar.[/quote]
I haven't seen this equation before. I think that no one of my sources is concerning such kind of equations.

[quote="Myth"][quote="liyi"]First, take transpose of the given equation, we get $B^T = B$.
[/quote]
Are you sure? We don't know that $A^T=A$ or $C^T=C$.[/quote]

In most books here in China, a positive / negative / ... matrix is considered symmetric...

[quote="Moubinool"]liyi,

What is

the author
the title
the editor

of this advanced algebra texbook ?
thx[/quote]

The book is in Chinese written by two professors of my univ, Shanghai Jiaotong Univ (unfortunately, dept of math in this univ is not good at all)

Title : 大学代数 (University/College Algebra)
Author : 陆少华 沈灏 (Lu Shaohua & Shen Ying)
Published by 上海交通大学出版社 (Shanghai Jiaotong Univ Publishing house)

This book is not for those whose major is math but for those who are majoring engineering. It consists of polynomials, linear algebra and abstract algebra.

[quote="Myth"][quote="liyi"]First, take transpose of the given equation, we get $B^T = B$.
[/quote]
Are you sure? We don't know that $A^T=A$ or $C^T=C$.[/quote]

It is nothing difficult to prove that positively definite matrix must be hermit:
$x^TAx\geq0$ for all vector x, thus $x^TAx$ is real number for all x, so A must be hermit(as far as i know it is known fact from matrix analysis or something like)

[quote="Myth"]What will you say about following matrix:
1 0
2 2
What is "hermit" matrix?[/quote]

I mean that if $A^*=A$ then matrix A called hermit(in real case it means that A is symmetric)

to Moubinool:

Could you please explain us how you find out that's clear expression for B ???


function
algebra
polynomial
linear algebra
matrix
calculus
derivative

1. Let $f(x)=x^2+bx+c$, M = {x | |f(x)|<1}. Prove $|M|\leq 2\sqrt{2}$ (|...| = length of interval(s))

2. $f$ function, $f^{n}$ polynomial. Is $f$ polynomial?

3. what's the max dimension of linear subspace $V$ of $n \times n$ matrices so that forall $A,B$ in $V$,  $trace(AB)=0$ ?

4. $f$:$R\rightarrow R$ 3 times differentiable
 Prove that there exists $w \in [-1,1]$ such that
$ f^{'''}(w)/6 = f(1)/2-f(-1)/2-f^{'}(0)$

5. find all $r>0$ s.t. when $f$:$R^{2}\rightarrow R$ diff, $||grad \;f(0,0)||=1$, $||grad\; f(u)-grad \;f(v)||\leq ||u-v||$, then the max of $f$ on the disk $||u||\leq r$, is attained at exactly one point. (ugly question)

6. If $ p,q$ are rationals, $r=p+\sqrt{7}q$, then prove there exists a matrix
$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) \in M_{2}(Z)- ( \pm I_{2})$
 for which $\frac{ar+b}{cr+d}=r$

Hopefully I solved 1,2,3 :)

for problem 4:

Expand $f(1)$ and $f(-1)$ at $x=0$, respectively:
$f(1) = f(0) + f'(0) + \frac{f''(0)}{2}  + \frac{f'''(\theta_1)}{6}$
$f(-1) = f(0) - f'(0) + \frac{f''(0)}{2}  - \frac{f'''(\theta_2)}{6}$
Subtract them
$f(1)-f(-1) = 2f'(0) + \frac{f'''(\theta_1)+f'''(\theta_2)}{6}$
or
$\frac{f(1)}{2}-\frac{f(-1)}{2}-f'(0) = \frac{f'''(\theta_1)+f'''(\theta_2)}{2}\cdot \frac{1}{6}$

Notice that a derivative (it is $f'''$ here) has Darboux property, hence there exists $\xi \in [\theta_2,\theta_1]$ such that $f'''(\xi) = \frac{f'''(\theta_1)+f'''(\theta_2)}{2}$. The conclusion follows.

Isn't it too easy? Or do I miss something?

Maybe liyi. One iranian guy told me an alike thing.

I had your taylor expansions as well, but I never heard of Darboux, and thus didn't hand it in. It still seems counterintuitive to me that there is no continuity requirement of f''' to apply something that -to me- looks like Lagrange Mean Value Theorem... But if this is correct I will feel sorry for not handing it in :( since just the word taylor would bring you points yet...

Anyway, I found 3 a quite nice question. :)

I think that most book would mention (maybe as an exercise) that derivatives have two important properties:
(1) A derivative does not have a jump discontinuity
(2) A derivative has Darboux property (even it is not continuous)

Seems strange to me... still I think I am getting your point.

Well too bad I missed some points then. I'll really have to train my analysis. But far from what you said, we didn't even learn what darboux property is :?

Darboux property = intermediate value property.

Frankly, I don't understand absolutely anything from the statement of the problem 6. And indeed, problem 3 is very very nice.

3 is nice indeed :)

6: I made some typos there indeed. it should be this:

If p,q are rationals, r=p+sqrt(7)*q, then prove there exists a matrix[[a,b],[c,d]] with integer entries (not +- unit matrix) for which so that (ar+b)/(cr+d)=r.

I had to change number 1 a little but the rest is fine, thanks! :)

In no.2 $n{\geq}2$(  :D ).
In no.5 the maximum on the disk $|u|{\leq}r$.

no, u is 2x1 vector thus it's ||u||. On the number 2 you're right of course. A team leader proposed me the following variant on it: if f^n continuous --> f continuous.


limit
function
calculus
calculus computations

$ \lim_{x \to 0} \frac{arccos(1\minus{}x^2\plus{}3x^3\minus{}4x^4\plus{}5x^5)}{x}$

I don't think this limit can be computed without using Taylor series.

Assuming that you actually mean $ \lim_{x\searrow0}$, use $ \arccos x\equal{}\arcsin\sqrt{1\minus{}x^2}$. owk

[quote="FOURRIER"]$ \lim_{x \to 0} \frac {\arccos(1 \minus{} x^2 \plus{} 3x^3 \minus{} 4x^4 \plus{} 5x^5)}{x}$

[b][color=darkred]I don't think[/color] this limit can be computed without using Taylor series[/b].[/quote]
[color=darkblue]You can use the remarkable limit $ \lim_{t\nearrow 1}\frac {\arccos t}{\sqrt {1 \minus{} t^2}} \equal{} 1$

for $ t: \equal{} 1 \minus{} x^2 \plus{} 3x^3 \minus{} 4x^4 \plus{} 5x^5$ if exists $ \alpha > 0$ so that

for any $ |x| < \alpha$ , $ x\ne 0$ we have $ \left(1 \minus{} x^2 \plus{} 3x^3 \minus{} 4x^4 \plus{} 5x^5\right)\in ( \minus{} 1,1)\ !$[/color]

what do the diagonal arrows in the limit mean?

The up-pointing and down-pointing arrows arrows indicate limits from the left and right direction, respectively.  (Note that the limit Virgil Nicula gives is undefined from the right.)

So the same as the more familiar $ x \to 0^{\plus{}}$ and $ x\to x^{\minus{}}$ ?

[quote="Virgil Nicula"][quote="FOURRIER"]Ascertain $ \lim_{x \to 0} \frac {\arccos P(x)}{x}$ , where $ P(x)\equiv 1 - x^2 + 3x^3 - 4x^4 + 5x^5\ .$

[b][color=darkred]I don't think[/color] this limit can be computed without using Taylor series[/b].[/quote][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that $ |P(x)|<1$ $ \Longleftrightarrow$ $ \|\begin{array}{c}
f(x)=5x^5-4x^4+3x^3-x^2+2>0\\\
g(x)\equiv 5x^5-4x^4+3x^3-x^2=x^2\cdot h(x)<0\end{array}\ ,$

where $ h(x)\equiv 5x^3-4x^2+3x-1\ .$ The functions $ f$, $ g$, $ h$ are continue. Therefore,

$ \blacktriangleright$ Since $ f(0)=2>0$ results that exists $ \alpha >0$ so that for any $ |x|<\alpha$ , $ x\ne 0$ we have $ f(x)>0\ .$

$ \blacktriangleright$ Since $ h(0)=-1<0$ results that exists $ \beta >0$ so that for any $ |x|<\beta$ , $ x\ne 0$ we have $ h(x)<0$ , i.e. $ g(x)<0\ .$

Hence exists $ \epsilon =\min\{\alpha ,\beta\}>0$ so that for any $ |x|<\epsilon$ , $ x\ne 0$ we have $ \|\begin{array}{c}
f(x)>0\\\
g(x)<0\end{array}$ , i.e. $ |P(x)|<1\ .$

Show easily  that $ \frac {\arccos P(x)}{x}=$ $ \frac {\arccos P(x)}{\sqrt {1-P^2(x)}}\cdot \frac {\sqrt {1-P^2(x)}}{x}=$  $ \frac {\arccos P(x)}{\sqrt {1-P^2(x)}}\cdot \frac {\sqrt {-x^2h(x)f(x)}}{x}\Longrightarrow$

$ \boxed {\frac {\arccos P(x)}{x}=\frac {|x|}{x}\cdot \frac {\arccos P(x)}{\sqrt {1-P^2(x)}}\cdot \sqrt {-h(x)f(x)}}\ .$ Observe that $ \lim_{x\to 0}\sqrt {-h(x)f(x)}=\sqrt 2$ .

Since $ P(0)=1$ , exists $ \epsilon >0$ so that $ |P(x)|<1$ for any $ |x|<\epsilon$ , $ x\ne 0$ and $ \lim_{t\nearrow 1}\frac {\arccos t}{\sqrt {1 - t^2}} = 1$ $ \Longrightarrow$

$ \lim_{x\to 0}\frac {\arccos P(x)}{\sqrt {1-P^2(x)}}=1\ .$  Therefore, $ L_l\equiv\lim_{x\nearrow 0}\frac {\arccos P(x)}{x}=-\sqrt 2$ and $ L_r\equiv\lim_{x\searrow 0}\frac {\arccos P(x)}{x}=\sqrt 2\ .$

In conclusion, since $ L_l\ne L_r\ ,$ then the "direct" limit $ L= \lim_{x\to 0}\frac {\arccos P(x)}{x }$ has a real sense and it doesn't exist ! [/color]


[quote="arqady"][quote="rachid"]$ a,b,c \geq 0,$prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 4 \geq 2ab \plus{} 2ac \plus{} 2bc$[/quote]
Because $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2 \geq 2ab \plus{} 2ac \plus{} 2bc$ is true.[/quote]
Try $ a\equal{}b\equal{}c\equal{}2$  :wink:

[quote="rachid"][quote="arqady"][quote="rachid"]$ a,b,c \geq 0,$prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 4 \geq 2ab \plus{} 2ac \plus{} 2bc$[/quote]
Because $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2 \geq 2ab \plus{} 2ac \plus{} 2bc$ is true.[/quote]
Try $ a \equal{} b \equal{} c \equal{} 2$  :wink:[/quote]
I meant because $ 4 \plus{} abc\geq3\sqrt [3]{a^2b^2c^2}$ is true.
If $ a^2\equal{}x^3,$ $ b^2\equal{}y^3$ and $ c^2\equal{}z^3$ then, we need to prove that
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz\geq2\sum_{cyc}\sqrt{x^3y^3},$ which follows from Schur and AM-GM.

Nice proof arqady,here is mine:
[hide="Proof"]WLOG we can assume that $ (a \minus{} 2)(b \minus{} 2) \geq 0$,then the inequality is equivalent to $ (a \minus{} b)^2 \plus{} (c \minus{} 2)^2 \plus{} c(a \minus{} 2)(b \minus{} 2) \geq 0$ which is true.[/hide]


find x,y so that  sinx+cosy=3x
                        cosx+siny=3y :maybe:

$3(x-y)=(sinx-siny)-(cosx-cosy)=2sin\frac{x-y}2(2cos\alpha+2sin\alpha)=4\sqrt2sin\frac{x-y}2sin(\frac\pi4+\alpha)$ where $\alpha=\frac{x+y}2$.
$2\sqrt2<3,\, |sin(\frac\pi4+\alpha)|\leq1$.
But if $x\neq y$ then $|2sin(\frac{x-y}2)|<|x-y|$
So $x=y$ is the root of equation $sinx+cosx=3x$.


Well mister Albatross, I have your answers.

For number 1
[hide]
This sequence can be split into two sequences.
The first, $1/7 + 1/7^3 + ...$ has first term $1/7$ and ratio $1/7^2$, so its sum is $(1/7)/(1-1/7^2) = 7/48$
The second sequence is $2/7^2 + 2/7^4 + ...$ which has first term $2/7^2$ and ratio $1/7^2$ so its sum is $(2/7^2)/(1-1/7^2) = 2/48$
Adding them together, $7/48 + 2/48 = 9/48 = 3/16$
[/hide]

For number 2
[hide]
Call S the sum of the sequence. Then $S = 1/5 + 2/5^2 + 3/5^3 + ...$
$S/5$ , however, equals $1/5^2 + 2/5^3 + ...$
Thus, $S - S/5 = 1/5 + 1/5^2 + 1/5^3 + ... = (1/5)/(1-1/5) = 1/4$
$(1 - 1/5)S = 1/4$
$(4/5)S = 1/4$
$S = 5/16$
[/hide]

Yay! I really liked number 2. I'm not sure I got these completly right, though. Could someone check?

EDIT: I CAN'T DIVIDE!!!! *cries*

#2 seems a little large. I did the sum up to 9 on the calculator and got .3124 , so converging to 5 is kind of large.. hmm

EDIT: Ok, I see where you went wrong, on your last step you multiplied the fractions wrong, you should end up with $\frac{5}{16}$ I believe. But good job on the solution


geometry
geometric transformation
symmetry
analytic geometry
rotation
vector

Find the area of the polygon with vertices $ \left((\sqrt3\minus{}1)/2),(\sqrt3\plus{}1)/2\right), \left((3\sqrt3\minus{}1)/2,(\sqrt3\plus{}3)/2\right),\left(2\sqrt3\minus{}1,\sqrt3\plus{}2\right),$ $ \left((3\sqrt3\minus{}3)/2,(3\sqrt3\plus{}3)/2\right),\left((\sqrt3\minus{}3)/2,(3\sqrt3\plus{}1)/2\right),\left(\minus{}1,\sqrt3\right)$
Is there a better way to solve this than brute-forcing using Shoelace formula?

Either you want to draw a picture or make a rigid transformation to figure out what's going on.  It's not to hard to find the translation you need to move the center of symmetry of this polygon to the origin, at which point life starts getting easier.  (I found it by making the substitution $ \alpha \equal{} \frac{\sqrt{3} \minus{} 1}{2}$ to eliminate $ \sqrt{3}$, but there are lots of other ways you could go about it.)

How would you figure out the center of symmetry (so that you move it to the origin)? I did the substitution of $ a\equal{}\frac{\sqrt3\minus{}1}2$ to get $ (a,a\plus{}1),(3a\plus{}1,a\plus{}2),(4a\plus{}1,2a\plus{}3),(3a,3a\plus{}3),(a\minus{}1,3a\plus{}2),(\minus{}1,2a\plus{}1)$ but now what about the translation? Thanks.

Look for symmetries.  The coefficient of $ a$ in the $ x$-coordinates is 0, 1, 1, 3, 3, 4; how much do you need to shift it by to get it centered at 0?  Etc.  After the translation, (1) it's a lot easier to notice some of the nice properties of this polygon (like the fact that four of its vertices form a square), and (2) it's not hard to see what a nice rotation to apply might be (and the fact that there are lots of $ \sqrt{3}$s flying around means there aren't that many options, anyhow).

The x-coordinates are being symmetic, but the y-coordinates do not, as the coefficients of a in y-coordinates are 1,1,2,3,3,2 while the constant terms are 1,2,3,3,2,1. Should a rotation be done instead?

I assure you that this figure has a center of symmetry and that you can make it obvious by an appropriate translation.  I'm not sure how you think the numbers you just wrote show anything to the contrary.  Just try it!  You certainly could rotate it without translating, but it's much easier to see what you're doing if you've first centered the figure at the origin.

I tried to graph the 6 points relatively to one another and the 1st, 2nd, 4th, 5th point form a square. There are also 2 isosceles triangles. How would you realize this without sketching the points (after ordering their y-coordinates from smallest to biggest)? Also how would you figure out the rotation (I think we can pick a side of the square, treating it as a vector, and find its argument, but is there a better way than this if we do not have the diagram)?


summer program
MathPath
MATHCOUNTS
calculus
Duke
college
trigonometry

Definitely the only one I know, IN  with  41,38,35,34 at states, they were all in 7th

[quote="math92"]whoa!!!
i never made this topic!!?!?!!?!?!
where did it come from?????
seriously i never made a topic called "Who will win Nationals in 2006?"[/quote]

.............mabye it was ur sister.............

i asked her and she said she didn't do it.  She wasn't on today.
 :)

[quote="math92"]Who do you think will win Nationals newst year.
I think it will be me.  just kidding.

[color=red][size=75][Edited by your friendly forgetful moderator who isn't at Mathpath - solafidefarms][/size][/color][/quote]
oh, so that's where this came from. :)

[quote="Fanatic"]nat_mc:

who was the last guy who had 34? I remember Nathan, John Ryan, and sichen got 2nd, 6th, and 7th (so they're basically gauranteed to make it this year), but who was the last kid?[/quote]
good point, it was an eighth grader, who won't be back next year, but we'll probably have you next year

[quote="nat mc"][quote="GoBraves"][quote="236factorial"]It is very hard to predict something that has many surprises. Maybe Neal Wu will take it all next year.[/quote]

Neal Wu is definently the favorite for next year, but you can never predict these things, how many people actually predicted that Neal Wu would win last year? If Neal makes countdown(which is likely) , then im definently gonna say him.[/quote]
Nathans easily the top favorite and contender for written and masters and will probably be in the finals with Neal for countdown, and will probabl, hopefully win[/quote]

I doubt he will beat Neal in countdown, Neal is just amazing at countdown. As for written and masters, yes Nathan should do well.

[quote="236factorial"][quote="sirorange"]Certainly no one from Massachusetts. MA's gonna be AWFUL next year.[/quote]

Yeah, with out you and that other guy who did well at this years nationals.[/quote]

Yeah, and like ALL of the top 8 or 9.

Also, Neal isn't that good at countdown, anyone can dethrone him. Remember, Sergei almost beat him, and Sergei's terrible at countdown. And believe me, I beat Sergei twice in Countdown.

[quote="236factorial"][quote="math92"].  And also, i think that Neal has a good chance of winning it all becuase since he has won it already he has alot of confidence. :)[/quote]

He could be overconfident. I don't think there was ever a person who won mathcounts nationals twice in a row.[/quote]never. but MCrawford got 2nd twice (i think)

I believe he got 2nd in 7th and 1st in 8th. I'm pretty sure I saw something on this site that said he missed only one problem in two years of Nationals. :o

He would have to confirm that, though.

Po Ru Loh got second and first in seventh and eighth grade respectively. (this is after Countdown)

Howard Liu got first and third in seventh and eighth grade respectively.

Ryan Ko got first and fourth in seventh and eighth grade respectively.

And William Engel was first at Nationals as a 7th grader in 94 but didn't make it back to Nationals in 95!

As history has shown, it's never easy to repeat.

[quote="darktreb"]And William Engel was first at Nationals as a 7th grader in 94 but didn't make it back to Nationals in 95![/quote]

Huh??? :huh: He probably got [i]way[/i] too overconfident. :roll:

Yeah...or maybe in his state they used countdown to determine the top four, and he screwed up in the countdown round. That would really suck :(

Yeah, but he would still have to do bad enough on written to be $4^{th}$ or lower.

Edit: Fancy $\LaTeX$ fourth

[quote="236factorial"][quote="chess64"]I think 236! will make it into the top 12 at least.[/quote]

I like your humor.

I [i]hope[/i] that our team doesn't get picked through votes; I won't even make [i]chapters[/i] that way![/quote]

I wasn't joking. But what do you mean by getting picked through votes?? :?

As in, everyone on the team votes for certain people getting onto the team.

Weird--I've always grown up with the concept that if you win Nationals one year, you can't win Nationals the next year because you're disqualified. But maybe that was the spelling bee.

I'd be content with just getting to Nationals.

[quote="mathnerd314"]... But maybe that was the spelling bee....
[/quote]
Yes, you cannot come back after wining the spelling bee

[quote="JesusFreak197"]I believe he got 2nd in 7th and 1st in 8th. I'm pretty sure I saw something on this site that said he missed only one problem in two years of Nationals.[/quote]
It the yearbook thingy we got last year it said the names of the past champions. Mathew Crawford wasn't on the list. I do know that he made top 2 twice and only missed one problem in two years. So he must have gotten 2nd in 7th and 2nd in 8th.

[quote="JesusFreak197"]Either that, or he came from a state where in '95, the top four got perfect scores and he made a dumb mistake on one problem.[/quote]

Nah, he just didn't have the greatest day and got fifth. And yeah, he was a little overconfident from what I've heard, as I had the same coach he did one of my years at Nationals.

[quote="chess64"]Oh. That's stupid; we picked our team based on scores on the school competition.[/quote]

So do we, or so did we last year.


[quote]9.  During football season, 25 teams are ranked by three reporters (Alice, Bob, and Cecil).  Each reporter assigned all 25 integers (1-25) when ranking the 25 teams.  A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting 1 point for a 25th place ranking.  The Hedgehogs earned 27 total points from the three reporters.  How many different ways could the three reporters have assigned their rankings for the Hedgehogs?  One such way to be included is Alice - 14th place, Bob - 17th place, and Cecil - 20th place.[/quote]

For this problem, is it just $ \binom{26}{2}$?  Because say you have 27 objects.  To split the objects into three groups (for Alice, Bob, and Cecil), we must put 2 dividers between the 27 objects.  There are 26 places to put the dividers, for the answer of $ \binom{26}{2}$, right?

that works. another way to do it:

if, say, alice gave $ 25$ pts, there would only be $ 1$ way for the other two to add up to $ 2$.

if she gave $ 24$, there would be $ 2$ ways.

if she gave $ 23$, there would be $ 3$ ways.
....

$ 1\plus{}2\plus{}3\plus{}...25\equal{}\frac{25\cdot26}{2}\equal{}\boxed{325}$

[hide="short balls and urns"] Let the rankings be x, y, and z. All of x, y, and z have to be positive, which we fix by using balls and urns on (x+1), (y+1), and (z+1) since that makes all three positive. We then get $ x\plus{}y\plus{}z\equal{}24$, and balls and urns gives $ \binom{26}{2}\equal{}325$ ways.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

I just start learning algebra, and I have one question here,
Let G be the group generated by a, b, and c, and G=<a,b,c | ab=bc=ca>,
then is G finite or infinite?

Or, the subgroup generated by $ a$ is infinite.


modular arithmetic
number theory proposed
number theory

$p$ is a prime number greater than $3$. Let  $\displaystyle\frac{A}{B}=1+\displaystyle\frac{1}{2}+...+\displaystyle\frac{1}{p-1}$, where $A,B\in Z$ and $(A,B)=1$. Prove that $A\vdots p^2$.

I still remember the problem... It was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=41503 , where some other solutions were given. In an equivalent form, it was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=302 .

  darij

The solution is probably the same as i found, and i didn't think that anyone will solve it otherway.


geometry
perimeter
trigonometry
inequalities proposed
inequalities

[quote="nrs.sowrabh"]$ Prove$ $ that$ $ sinA \plus{} sinB \plus{} sinC > 2$ ,  in an acute $ \triangle$[/quote]
We have:
$ sinA\plus{}sinB\plus{}sinC\equal{}\frac{s}{R}$, s:semi-perimeter,R: circumradius.
But $ s>2R$!!

[quote="nrs.sowrabh"][quote="NguyenDungTN"][quote="nrs.sowrabh"]$ Prove$ $ that$ $ sinA \plus{} sinB \plus{} sinC > 2$ ,  in an acute $ \triangle$[/quote]
We have:
$ sinA \plus{} sinB \plus{} sinC \equal{} \frac {s}{R}$, s:semi-perimeter,R: circumradius.
But $ s > 2R$!![/quote]
why does the last one hold?[/quote]
I can prove $ s \ge 2R\plus{}r$, but solution very long and complicated.
But simple solution:
We have $ sin{A},sin{B},sin{C} \in [0,1]$ ( because ABC be a acute triangle)
So: $ sin{A}\plus{}sin{B}\plus{}sin{C} \ge sin^2{A}\plus{}sin^2{B}\plus{}sin^2{C}$
and:
$ sin^2{A}\plus{}sin^2{B}\plus{}sin^2{C}\equal{}\frac{1\minus{}cos{2A}}{2}\plus{}\frac{1\minus{}cos{2B}}{2}\plus{}1\minus{}cos^2{C}$
$ \equal{}2\minus{}\frac{cos{2A}\plus{}cos{2B}}{2}\minus{}cos^2{C}$
$ \equal{}2\minus{}cos{(A\plus{}B)}cos{(A\minus{}B)}\plus{}cos^2{(A\plus{}B)}$
$ \equal{}2\minus{}cos{(A\plus{}B)}(cos(A\minus{}B)\plus{}cos(A\plus{}B))\equal{}2\plus{}2cosAcosBcosC >2$
Problem is solved!

[quote="nrs.sowrabh"]$ Prove$ $ that$ $ sinA \plus{} sinB \plus{} sinC > 2$ ,  in an acute $ \triangle$[/quote]
Let $ f(x)\equal{}\sin{x}$,so $ f''(x)\equal{}\minus{}\sin{x}\leq 0$.It is easy to undestand that $ \{\frac{\pi}{2},\frac{\pi}{2},0\}\succ\{A,B,C\}$.
So using Karamata's inequality(Hardy-Littlewood-Polya):we obtain the result:
$ \sin{A}\plus{}\sin{B}\plus{}\sin{C}\geq \sin{\frac{\pi}{2}}\plus{}\sin{\frac{\pi}{2}}\plus{}\sin{0}\equal{}2$.

My proof goes like this:

In an acute $ \triangle$

             $ Sin A > \frac{2A} {\pi}$
             $ Sin B > \frac{2B} {\pi}$
             $ Sin C > \frac{2C} {\pi}$

 Adding up, we are done!!

[quote="Erken"][quote="nrs.sowrabh"]

             $ Sin A > \frac {2A} {\pi}$

 [/quote]
But why does it hold?[/quote]
Consider the graph of sinx. Since x is acute,and y=sinx in $ (0,\frac{\pi}{2})$ lies above the graph of $ y\equal{}\frac{2x}{\pi}$,
 the result follows!! :)


algebra
polynomial
Rational Root Theorem

[quote="kimby_102"][quote="cincodemayo5590"]Find all $x \in \mathbb{R}$ such that $\sqrt{x^2-5x+14} + \sqrt{2x^2-16x-36} = x$.[/quote]
[hide]
Rearrange the terms:
$\sqrt{x^2-5x+14}=x-\sqrt{2x^2-16x-36}$.
Now square both sides, and u will get only one radical in the equation. squaring the equ. again would remove the radical and u can solve for $x$. I think this is how u do..unless there's a more elegant way of doing this pbm...
[/hide][/quote]

Yes, and eventually you will get the equation [hide]
 $0=4x^4-20x^3-65x^2-1100x-2500$
First, we see that $x \geq 0$ since $x$ is the sum of two radicals. Then, by Descartes' Rule of Signs, there must be only one positive real root.
So I dont know if theres another way to do this but i just fought at it with the rational root theorem until [i]finally[/i] i found one:[hide]$x=10$ works. Then the polynomial is $0=(x-10)(4x^3+20x^2+135x+250)$ so we are done, with only one solution at $\boxed{x=10}$[/hide][/hide]


group theory
abstract algebra
Ring Theory
superior algebra
superior algebra theorems

Let $R$ be a ring such that $x^n = x$ for all $x \in R$ and $n > 1$ fixed. Prove that $R$ is commutative.

See also http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Jacobson&t=24481 , there a more general statement is given (but without proof!).

I.N.Herstein, "Noncommutative Rings"

confer also http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/herstein

here, you can download an article http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.pjm/1102948670

If you want to prove this yourself, here are the steps from Herstein, [i]Noncommutative Rings[/i], pp. 70--74.

[b]Lemma 1.[/b] Let $D$ be a division ring of characteristic $p > 0$ and $Z$ be the center of $D$.  Suppose that $a \in D \setminus Z$ is such that $a^{p^n} = a$ for some $n \geq 1$; then there exists an $x \in D$ such that $xax^{-1} \neq a$ for some integer $i$.

[b]Theorem 2 (Wedderburn).[/b] A finite division ring is a commutative field.

[b]Lemma 3.[/b] If $D$ is a division ring of characteristic $p > 0$ and $G \subset D$ is a finite multiplicative subgroup of $D$ then $G$ is abelian (and so, cyclic).

[b]Lemma 4.[/b] Let $D$ be a division ring in which for any $a \in D$ there exists an integer $n(a) > 1$ such that $a^{n(a)} = a$. Then $D$ is a commutative field.

[b]Theorem 5 (Jacobson).[/b] Let $R$ be a ring in which for every $a \in R$ there exists an integer $n(a) > 1$, depending on $a$, such that $a^{n(a)} = a$ then $R$ is commutative.

For Thm. 5 use
[b]Theorem A.[/b] Let $R$ be a primitive ring*).  Then for some division ring $\Delta$ either $R$ is isomorphic to $\Delta_n$, the ring of all $n \times n$ matrices over $\Delta$ or, given any integer $m$ there exists a subring $S_m$ of $R$ which maps homomorphically onto $\Delta_m$.

[b]Theorem B.[/b] $R$ is semisimple if and only if it is isomorphic to a subdirect product of primitive rings.  (Corollary: A commutative semisimple ring is a subdirect sum of fields.)

*) A ring $R$ is said to be primitive if it has a faithful irreducible module.


Hi. Is there a way to see these problems in English, so we get the secret to how the Chinese team does so good?

Do you guys just do math since you are very young. That is what I have heard. That they like train people into different fields and concentrate on those only.

Just dropping in.

As far as I know, many competitors around me did this.

:(
That's ALL right as what you have seen, for the serious more about these.
Just since you're born in china,you must become a POLITICAL PROP by some SELFISH GUYS. :ninja:


What did everyone think of the Mock AIME 3 for 2007? (I'm not the writer, but I am just curious)

I'd probably put it just on the harder side...but very close to actual AIME level.

I never turned it in, but I either found it slightly easier, or else I am better at doing AIME's that I thought. I could do about 6 or 7 of the problems, which is slightly above average for me. I might, however, just have improved.

I said "impossible." But that applies to all the AIME's I've taken.  :|

at least my intent was to have the test slightly more difficult than the current aimes (so i am not sure what ranking that gets, the question should be more how does this compare to the other aime's)

the question seems very different for different people, it really depends on what problem solving level you are at, there are people who would think many of the problems were simple, and others will think they are very difficult, in my opinion, i thought that the later problems were pretty difficult, but that is because of my level in problem solving

i think one problem with difficulty is that there are certain techniques that i took for granted that people would know, the first few problems were just using standard problem solving techniques (that one should learn), with a little twist, i.e. the first one was eliminating the constant term, the second was factoring, the third was vieta's sums (a bit tricky though), the fourth was just complex conjugates,


prove that the absolute value of norm U equalling zero means that U = 0j + 0i

absolute value of norm U?

what does that mean? if you mean that the absolutely value of U is 0, than U has to be zero, which cancels your whole equation out. not sure i understand?

Uhh I believe norm=modulus? (Not sure...)

[hide="Weak Proof :( "]
If $ U\equal{}aj\plus{}bi$, the modulus of $ U$ is $ \sqrt{a^2\plus{}b^2}$.

We have $ \sqrt{a^2\plus{}b^2}\equal{}0$; however, $ a, b$ must be a real number, so $ a\equal{}0$ and $ b\equal{}0$.[/hide]

oh! okay. then yeah, refer to fanfan's proof.

By trivial inequality a^2,b^2>=0. Equality holds when a,b=0.


calculus
derivative
calculus computations

[quote="ProblemSets"]If x + y = xy, then dy/dx is:

A. 1/(x-1)
B. (y-1)/(x-1)
C. (1-y)/(x-1)
D. x+y-1
E. (2-xy)/y[/quote]

Solution:
[hide]Using implicit differentiation on $ x\plus{}y\equal{}xy$ with respect to $ x$:

$ 1 \plus{} \frac{dy}{dx} \equal{} y \plus{} \frac{dy}{dx} \cdot x$
$ \implies \frac{dy}{dx} \cdot \left( x\minus{}1 \right) \equal{} 1\minus{}y$
$ \implies \frac{dy}{dx} \equal{} \frac{1\minus{}y}{x\minus{}1}$

Therefore our answer is $ \boxed{C. ~~\frac{1\minus{}y}{x\minus{}1}}$.[/hide]


algebra
polynomial
conics
parabola
quadratics
special factorizations

Hi,

I'd appreciate if someone could help me understand how to solve these problems. I've seen proofs and all being done but I don't really understand the logic behind it. They are probably really simple questions tho.

1) Is there a number of the form a + b√2 (where a, b ∈ Q[rational #s]) whose square is equal to 7? (if yes, give an example and if no, prove it).

2) For the equation X + 2X + 16 = 0.
           a. Does it have a solution in Q[rational #s]?
           b. Does it have a solution in R[real #s]?
           c. Does it have a solution in C[complex #s]?
When it does have a solution, solve completely the equation. If it doesn’t, prove that it doesn’t.

Thanks for help in advance.

first, i want to tell you that all the equation with n degree will have n roots in complex field, but we have already proved that we just solved equation by square roots for the equation with 4,3,2,1 degree. in your equation, because it doesn't have roots in real field so will doesn't have in Q field (of course) :D have fun

First of all, $ \mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$, so #2, being a polynomial, has 2 complex roots (Fundamental Theorem of Algebra). I'm assuming you meant $ x^2\plus{}2x\plus{}16\equal{}0$, because otherwise it's linear and the solution is obvious. This parabola has a vertex above the $ x$ axis (completing the square can prove that it's positive for all $ x\in\mathbb{R}$), so all its roots are non-real.

You can solve it with the quadratic formula.

[quote="Ariana"]1) Is there a number of the form a + b√2 (where a, b ∈ Q[rational #s]) whose square is equal to 7? (if yes, give an example and if no, prove it).[/quote]
You know that $ \sqrt{2}$ is irrational, so argue by contradiction.  What would happen if $ (a \plus{} b \sqrt{2})^2 \equal{} 7$?


I'm totally lost on this one. Could someone please help me.

Thanks.

Two equally big squares with the sides 12 cm partly covers each other as the figure shows. One of the squares corner is in the other squares center. Decide the area of the shadowed part.

[url=http://g.imagehost.org/view/0352/Jobbigare_3][img]http://g.imagehost.org/0352/Jobbigare_3.jpg[/img][/url]

I make it $ 36 cm^{2}$:

Draw another $ 12 cm\times12 cm$ square with sides horizontal and vertical whose top left hand corner is the centre of the original square. Then rotationg the oblique square to fall onto this we lose a right -angled triangle above the third square and gain one to the left of the oblique square (if this makes sense). So the required area is a quarter of the area of any of te 3 squares.

Nice!
[hide] Any perpendicular cross passing through the center of a square cuts it into four equal pieces by symmetry. So rotate the square four times, we find that the shaded area is 1/4 of the total, which is 36. [/hide]

When steps are irrelevant, the solution is quite obvious.

[hide]You can make the two squares overlap in a way such that the shadowed part is a square too.
And the area of the square would be $ 6^2 \equal{} 36 cm^2$[/hide]

Ok thanks I understand how you think. I was about to try going like this

[url=http://g.imagehost.org/view/0617/fyrk][img]http://g.imagehost.org/0617/fyrk.gif[/img][/url]

It's not as fast, but here's the way I approached it (it also proves why the area is constant no matter how you rotate it).

[hide="My Solution"]
Naming points:
Center = O
Corner that is inside the other square = A
Point of intersection on bottom side = B
Point of intersection on right side = C
Midpoint on bottom side = D
Midpoint on right side = E

Draw the draw two line segments from the center of the square to the two sides intersected by the other square. This forms another square a quarter the size of the entire square. <DOB = <DOC - 90 = <COE. Both  triangles DOB and COE are right, so all their angles are congruent. Both of them have a side that is the length of the distance from the center to the midpoints of the sides, so they must be congruent. Thus, the shaded area is equal to the area of the square formed by the center, corner, and midpoints. 12^2/4 =

$ \boxed{36\text{ cm}^2}$[/hide]


quadratics
function
number theory
totient function
number theory proposed

Find all nonnegative integers $ a<2007$ such that the congruence $ x^2\equal{}\minus{}a(mod 2007)$ has exactly two different nonnegative integer solutions smaller than $ 2007$.

Let T(A): Totient function of A.
By Euler's theorem: if x is coprime to A, x^T(A)=1 mod p => x^T(2007) = 1 mod 2007 =>
x^1332 = 1 mod 2007
Given x^2 = -a mod (2007) => order of (-a) mod p = 666. No. of a values possible = T(666) = 216.

$ 2007\equal{}3^2*223$, therefore $ x^2\equal{}a\mod 2007$ equavalent to
$ x_1^2\equal{}a_1mod 9,x_2^2\equal{}a_2\mod 223, a_1\equal{}a\mod 9,a_2\equal{}a\mod 223,x\equal{}x_1\mod 9,x\equal{}x_2\mod 223.$
If $ a_1\equal{}0$ we get at least 3 solution $ x^2\equal{}0\mod 9$ (x=0,3,6), therefore $ a_1\equal{}1$ or $ a_1\equal{}4$ or $ a_1\equal{}7$ and $ x_1^2\equal{}a_1\mod 9$ had 2 solutions $ x_1,\minus{}x_1$. If $ x^2\equal{}a\mod 2007$ equation$ x_2^2\equal{}a_2\mod 223$ must had unique solution. It give $ a_2\equal{}0$. Therefore $ x^2\equal{}a$ had 2 solution if and only if $ a\equal{}223$ or $ a\equal{}892$ or $ a\equal{}1561$.


Find all four-digit numbers $ \overline{ABCD}$ and $ \overline{DCBA}$ such that there exists a prime number $ p$ which satisfies that $ \dfrac{\overline{ABCD}}{p^2}$ and $ \dfrac{\overline{ABCD}\minus{}\overline{DCBA}}{37p}$ are non-zero integers.

[quote="campos"]Find all four-digit numbers $ \overline{ABCD}$ and $ \overline{DCBA}$ such that there exists a prime number $ p$ which satisfies that $ \dfrac{\overline{ABCD}}{p^2}$ and $ \dfrac{\overline{ABCD} \minus{} \overline{DCBA}}{37p}$ are non-zero integers.[/quote]

[hide="Outline Solution"]
$ \overline{ABCD}\equal{}10^3A\plus{}10^2B\plus{}10C\plus{}D$ (same for $ \overline{DCBA}$) we get,$ \overline{ABCD} \minus{} \overline{DCBA}\equal{}999A\plus{}90B\minus{}90C\plus{}999D$
Now, $ 37|999\implies 37|90(B\minus{}C) \implies 37|B\minus{}C\implies B\minus{}C\equal{}0 \iff B\equal{}C$
So,$ \overline{ABCD}\equal{}10^3A\plus{}D$
Again, $ p|999(A\minus{}D)$ From here we can get the values of the digits. (By simple checking, note that $ p\leq8$ )
[/hide]


LaTeX
quadratics
algebra
difference of squares
special factorizations
quadratic formula

z^ over z - y subtracted from y^ over z - y

^ meaning squared

that would be:

z(z) - Y(y)
over
z - y

correct?

5x over x - 1 PLUS x over x - 1

that would be:
x(5 + x)
OVER
x - 1

Correct?

Try using latex by putting $ \$$ signs around your mathematical text.

For example:

$ \frac{numerator}{denominator\minus{}5y} \plus{} \frac{obama\plus{}100x}{mccain}$

or

$ x^{z\minus{}y} \minus{} x^{5\minus{}x}$

Its hard to figure out what you're saying exactly.

grrr i can't get it to work

5x over x - 1... meaning 5x is the numerator and x - 1 is the demnominator

PLUS

x over x - 1... meaning x numerator and x -1 is denominator

in short the problem is combine 5x + x ( the numerator ) over x -1 ( denominator )

Are these rewritten correctly suave?

[b]Part 1[/b]: $ \frac {z^2}{z} \minus{} \frac {y^2 \minus{} y}{z} \minus{} y$

[b]Part 2[/b]: $ \frac {5x}{x \minus{} 1} \plus{} \frac {x}{x \minus{} 1}$

If this was what you meant for the second part then you're correct.

[b][Edit][/b] Ya, I was already noticing $ 6x$ while editing.  :oops:

Is this the first one $ \frac{z^{2} \minus{} y^{2}}{z \minus{} y}$, if so simply factor as a difference of squares, note that $ z \ne y$.

[quote="Patterns_34"]
$ \frac {5x \plus{} x}{x \minus{} 1} \equal{} \frac {x(x \plus{} 5)}{x \minus{} 1}$ [/quote]

$ 5x \plus{} x \equal{} 6x$  :P edit: you've probably noticed that already.

[quote="azjps"][quote="Patterns_34"]
$ \frac {5x + x}{x - 1} = \frac {x(x + 5)}{x - 1}$ [/quote]

$ 5x + x = 6x$  :P[/quote]

oh my god.... you can tell i've been at it for far too long.. 10 peices of paper an counting...

so   $ 6x
   ________$ x - 1

and the first problem.. there is no second y and no - y on the end just the y^2 over z-y

Is this what you mean?

$ \frac{z^{2}}{z \minus{} y} \minus{} \frac{y^{2}}{z \minus{} y} \equal{} \frac{z^{2} \minus{} y^{2}}{z \minus{} y} \equal{} \frac{(z \plus{} y)(z \minus{} y)}{z \minus{} y} \equal{} z \plus{} y$?

[quote="SnipedYou"]Is this what you mean?

$ \frac {z^{2}}{z \minus{} y} \minus{} \frac {y^{2}}{z \minus{} y} \equal{} \frac {z^{2} \minus{} y^{2}}{z \minus{} y} \equal{} \frac {(z \plus{} y)(z \minus{} y)}{z \minus{} y} \equal{} z \plus{} y$?[/quote]

yes thats the exact problem.. i'm guesing that's the correct answer then?

next one..

sorry i can't figure this syntax thing out, it's the same format as the others..

2b + 8 Over b multiplied times 2 Over 3b + 12

2b + 8
______

   b

and the other side

  2
_____

3b + 12

I got 4 over 2b + 3

4
__

2b + 3

however does the 4 and 2 cancel out meaning that it is = 2 over b + 3?

$ \frac {2b \plus{} 8}{b} \cdot \frac {2}{3b \plus{} 12} \equal{} \frac {2(b \plus{} 4)}{b} \cdot \frac {2}{3(b \plus{} 4)} \equal{} \frac {4(b \plus{} 4)}{3b(b \plus{} 4)} \equal{} \frac {4}{3b}$, as long as $ b \ne 0, \minus{} 4$

edit: To answer your second question, no, all terms must share a common multiple and since $ 3$ is not divisible by $ 2$ that answer is in lowest terms.

dang you saved my arse again...

ok just 2 more...

2y^2 - y - 1 over 4y^2 - 4y - 3 MULTIPLIED times 4y^2 + 12y + 9 over 2y^2 + y - 3

same format

I got it down to

  (2y+1) (y-1) X (2y+3) (2y+3)
___________________________
  (2y-3) (2y+1) X (2y+3) (y-1)

so i did the factoring for you.. now what does that come out to be?

-1?

Assuming you factored it correctly just eliminate that terms that appear in both the numerator and denominator to leave you with

$ \frac {2y \plus{} 3}{2y \minus{} 3}$, as long as $ y \ne \minus{} \frac {1}{2}, \pm \frac {3}{2}, 1$

Make sure you understand what I'm doing and go through the steps yourself too, so I'm not just giving you the right answer. I'm trying to go step by step, plug in some values to verify that the simplified form is true too. If your wondering why I have the values the variables can't be is because there would be division by zero which is not allowed, these will come up later when you study asymptotes.

ok stay with me i got that one right for once... stay with me

2a^2 - 3a - 5 X 3a^2 + 5a -2
__________________________

3a^2 + 2a -1 X 2a^2 + a - 15

i've goteen the first numerator and dominator to:

(2a - 5)  (a + 1)
 ___________________________
(3a - 1)  ( a + 1)

i'll race you  :rotfl:

As for typing in Latex (the fancy format of typing formulas and other expressions), here is a summary (not complete but some helpful statements to help begin out):
[hide="Some statements on Typing in Latex"]A first step and very important step would be to just use the $ \$$ sign (on the 4 key buttom) in between expressions you want to post. Example: if you type "w^5=x^2", just finish off by posting one $ \$$ sign on both ends of the expression like $ \$$w^5=x^2$ \$$, to get $ w^5 \equal{} x^2$.
Also, you can add other stuff like type in the expression \frac{n+1}{n-1}, and using $ \$$ signs to get the fraction $ \frac {n \plus{} 1}{n \minus{} 1}$.
You can learn more codes like \sum to get $ \sum$, \binom{a}{b} to get $ \binom{a}{b}$, etc.
Notice the use of {}. With that, instead of typing w^(2x+1), which gives this $ w^(2x \plus{} 1)$, use w^{2x+1} to get $ w^{2x \plus{} 1}$.
Characters like alpha, beta, sigma, and gamma can easily be typed by: \alpha for $ \alpha$, \beta for $ \beta$, \infty for $ \infty$,etc.
Two other commands that will be important in addition with those I've given is the use of _ and ^. For example, x_2 gives $ x_2$. Also, an example of typing these series of codes can include: \sum_{n=0}^{\infty}{n+1} to get $ \sum_{n \equal{} 0}^{\infty}{n \plus{} 1}$.
A way to learn more codes and ways to type an expression, one can always refer to clicking an expression from someone and see the codes he/she used (notice they don't show the $ \$$ sign). Overall, the $ \$$ sign is a must use.[/hide]

In addition, here are some links to [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]codes for symbols[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands]how to type certain codes[/url] if you need reference.

Irrational? 2a^2 + a -15 can't be factored?

Actually, that expression is factorable.

[hide="Factored"]$ 2a^2 \plus{} a \minus{} 15 \equal{} (2a \minus{} 5)(a \plus{} 3)$ (I've used the quadratic formula as a reference)[/hide]


#7 2001 State Sprint

Line WA divides rectangle WXYZ (Where A is somwhere on Line XY) into two parts whose areas are in the ratio of 7:1. What is the ratio of Line XA:LineXY? Express your answer as a common fraction.

[hide]1/3[/hide]

Let $ WX\equal{}x$,$ XA\equal{}a$,$ AY\equal{}b$.  Thus, $ \frac {7ax} {2}\equal{}\frac {(a\plus{}2b)x} {2}$  If we simplify this equation, we get that $ a\plus{}2b\equal{}7a$.  Thus, $ 6a\equal{}2b$, so $ \frac {XA} {XY}\equal{}\frac {1} {4}$.

Well, try to make the triangle a rectangle. So, place another triangle congruent to the first one such that its base is at the others tip. Then, the ratio of the smaller rectangle to the bigger is 1:3, and they have the same length so the ratio of their widths is still 1:3.

[quote="Math Geek"]Let $ WX \equal{} x$,$ XA \equal{} a$,$ AY \equal{} b$.  Thus, $ \frac {7ax} {2} \equal{} \frac {(a \plus{} 2b)x} {2}$  If we simplify this equation, we get that $ a \plus{} 2b \equal{} 7a$.  Thus, $ 6a \equal{} 2b$, so $ \frac {XA} {XY} \equal{} \frac {1} {4}$.[/quote]

actually, Thus, $ 6a \equal{} 2b$, so $ \frac {XA} {XY} \equal{} \frac {1} {3}$

Um, no, cuz it's asking for XA/XY, which is the ratio of a to a+b, not the ratio of a to b. :wink:

oh yeah your right
I hope this doesn't happen to me at nats :(

Ryuk actually posted it wrong. They asked for $ \dfrac{XA}{AY}$.


Να δωσετε ενα πολυωνυμο δυο μεταβλητων το οποιο ειναι μη αρνητικο
αλλα δεν μπορει να γραφει ως αθροισμα των τετραγωνων πολυωνυμων δυο μεταβλητων.

Προσωπικα το θεωρω πανεμορφο προβλημα. :)
Εκει που το βρηκα αναφερει επισης οτι ο Hilbert οταν ρωτηθηκε αν υπαρχει τετοιο πολυωνυμο
απεδειξε οτι υπαρχει χωρις ομως να βρει ενα παραδειγμα!

[quote="AlexandrosG"]Να δωσετε ενα πολυωνυμο δυο μεταβλητων το οποιο ειναι μη αρνητικο
αλλα δεν μπορει να γραφει ως αθροισμα των τετραγωνων πολυωνυμων δυο μεταβλητων.

Προσωπικα το θεωρω πανεμορφο προβλημα. :)[/quote]

Welcome to Hilbert's problems! Αυτό είναι το 17ο :)

Cheerio,

Durandal 1707

Πραγματι. Παντως το προβλημα που εθεσα λυνεται με ελαχιστες γνωσεις και ειναι πανεμορφο.
Μπορειτε να δωσετε ενα παραδειγμα? :)


Prove that
$sinA^{sinB}+sinB^{sinC}+sinC^{sinA}> 1+\frac{1}{e^{\frac{1}{e}}}$ for all triangle $ABC$

Hm. No one interested in it?
It's in MYG some years ago, and i think it's really hard. Please help.

The weaker result is $\frac{1}{2}+\frac{1}{e^\frac{1}{e}}$ in the $R.H.S$ and it's no so hard. Can you prove it?


linear algebra
matrix
function
modular arithmetic
number theory
least common multiple
algebra unsolved

Show that the Fibonacci sequence is periodic modulo $ 7$.

It is even possible to derive a formula for finding $ m$, the length of the period $ \mod n$ for any $ n \in \mathbb{N}$.

One may consider the Fibonacci sequence to consist not of single number, but of the ordered pairs $ (F_n,F_{n+1}).$ Then there is a single fixed mapping which sends $ (F_n,F_{n+1})$ to $ (F_{n+1},F_{n+2}).$ This is a linear mapping and we may write it explicitly as
\[ \begin{bmatrix}F_{n+1}\\F_{n+2}\end{bmatrix}= \begin{bmatrix}0&1\\1&1\end{bmatrix} \begin{bmatrix}F_{n}\\F_{n+1}\end{bmatrix}\]
This mapping is invertible (an important consideration), and its inverse is given by the inverse matrix, $ \begin{bmatrix}-1&1\\1&0\end{bmatrix}.$

If we're working mod $ m,$ we regard the pairs as belonging to the finite set $ (\mathbb{Z}_m)^2,$ where I'm using $ \mathbb{Z}_m$ to denote $ \mathbb{Z}/(m\mathbb{Z}),$ the set of equivalence classes mod $ m.$ This set has exactly $ m^2$ elements, but we also note that $ (0,0)$ cannot appear in the Fibonacci sequence, as that would lead to an infinite string of zeros in both directions. Hence, our function maps a set with $ m^2-1$ elements into itself. If we iterate the mapping, we must repeat an element (remember that each element is an ordered pair) after no more than $ m^2-1$ steps. Once that repetition occurs, the sequence must be periodic going forward - and also periodic going backwards, as the mapping is invertible.

The Fibonacci sequence mod $ m$ must be periodic with period no greater than $ m^2-1.$ In practice, this maximum is achieved only for $ m=2$ and $ m=3.$

What about $ m=7?$ Let's run the sequence, starting with $ F_0=0$ and $ F_1=1,$ until the next time a zero appears:
\[ 0,1,1,2,3,5,1,6,0\]
So we have $ F_8\equiv 0\pmod{7}.$ The next ordered pair is $ (0,6),$ which we can also write as $ (0,-1).$ That means that $ F_8$ through $ F_{15}$ will simply be the negatives of $ F_0$ through $ F_7,$ bringing us to $ F_{16}=0$ and $ F_{17}=-F_9=1,$ and we start over.

The period is $ 16.$

This is a quick-and-dirty way to find the periodicity for $ m$ not too large. Run the sequence until the next zero (after $ F_0$) appears; assume this takes $ k$ steps. Look at $ F_{k+1}.$ What is it's multiplicative order mod $ m.$ If $ F_{k+1}$ is of order $ r$ - that is, the smallest power of it that equals $ 1$ is the $ r$th power - then the period of the Fibonacci sequence is $ rk$ In the case of $ m=7,$ that was $ 2\cdot 8=16.$ For $ m$ composite, do this for each of the prime-power factors and use the Chinese Remainder Theorem to stitch it back together.

I've outlined a fairly unsophisticated method; there are more tools than that which can be applied.

This is known as the [url=http://mathworld.wolfram.com/PisanoPeriod.html]Pisano Period[/url].

Another, somewhat similar solution:
Consider the criteria for repetition. Since each Fibonacci number is based off the previous two, and there are only seven values that they can take on in $ \mod{7}$ while there are an infinite number of values in the entire sequence, there must be some point that a pair of consecutive terms is repeated; i.e. there has to be some repeating part. Since we can find the previous pair to any given pair, if only a subset of the sequence repeated, then every part of the Fibonacci sequence would have to be included within that subset. Therefore the enitre Fibonacci sequence is periodic. This argument certainly generalizes to any $ \mod{n}$.

If want to see the period, here it is:

$ \boxed{1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0}, 1, 1,2,3,5,\dots$


Let $ x$ be a positive integer. Find a formula for the number of digits in $ x^{x}$ in terms of $ x$.

[hide]$ x^{x}\equal{}10^{x\log x}$. This implies that the number of digits in $ x^{x}$ is $ \lfloor x\log x\rfloor\plus{}1$.[/hide]


probability
limit
function
integration
real analysis
geometry
3D geometry

We choose one natural number. What is the probability that the number is odd. How to prove ??
I know that the probability in the set {1,2,...,2n} is 0.5. But how we can prove to the set {1,2,3,...}

Interesting question.

From one point of view, it seems that the probability is obviously .5.  But somebody could argue that there are twice as many even counting numbers as odd counting numbers, for example by pairing up odds and evens as follows:

$ 1 \leftrightarrow 2,4$
$ 3 \leftrightarrow 6,8$
$ 5 \leftrightarrow 10,12$
$ 7 \leftrightarrow 14, 16$
$ \ldots$

In other words, any odd positive integer $ n$ gets paired up with $ 2n, 2n \plus{} 2$.

From that point of view, it seems like the probability of picking an odd number is $ \frac13$.

This seems to be a paradox, and no doubt mathematicians puzzled over it and had violent debates and duels for centuries over it.  I believe the resolution to the paradox is that the original question in fact is not well posed.  What do you mean, "choose a natural number at random"?  By what method are you picking a natural number?  Perhaps if that were more well specified, the question would be answerable.

The question is not ill posed.  Considering infinite sets, we may pose the following question: when choosing at random a real number, what is the probability of the number being rational?  Since the rationals are numerable, but the reals are not, the probability is actually 0.  This type of questions are more related to measure theory than to probability.

For the particular case at hand, we may simplify the way to find a solution as follows: call $ p_N$ the probability of a number chosen randomly (uniform distribution) out of $ \{1,2,...,N\}$ being odd.  The probability is $ \frac {N \plus{} 1}{2N}$ if $ N$ is odd, and $ \frac {1}{2}$ if $ N$ is even.  Take $ N$ to infinity, and we find $ \frac {1}{2}$ regardless of whether $ N$ is odd or even.

[quote="Singularitarian"]
From one point of view, it seems that the probability is obviously .5.  But somebody could argue that there are twice as many even counting numbers as odd counting numbers, for example by pairing up odds and evens as follows:

$ 1 \leftrightarrow 2,4$
$ 3 \leftrightarrow 6,8$
$ 5 \leftrightarrow 10,12$
$ 7 \leftrightarrow 14, 16$
$ \ldots$

In other words, any odd positive integer $ n$ gets paired up with $ 2n, 2n \plus{} 2$.

From that point of view, it seems like the probability of picking an odd number is $ \frac13$.
[/quote]

Please note that somebody else, using a similar argument, could argue that there are 5 times as many odd numbers as even numbers, for example with these pairings:
$ 2 \leftrightarrow 1, 3, 5, 7, 9$
$ 4 \leftrightarrow 11, 13, 15, 17, 19$
$ 6 \leftrightarrow 21, 23, 25, 27, 29$
$ 8 \leftrightarrow 31, 33, 35, 37, 39$
$ \ldots$

From this point of view, it would seem that the probability of picking an odd number is $ \frac{5}{6}$.

However, we know this doesn't make much sense - we know there are equally many odd and even numbers since we can create a bijection from one set to another.

[quote="Singularitarian"]What do you mean, "choosing at random a real number"?  Didn't atomicwedgie show that this statement is meaningless?

Even if there were such a thing as picking a natural number "at random", and such a thing as the probability that the natural number you pick "at random" is odd, what makes you think that this probability is equal to $ \lim_{N \to \infty}p_N$?[/quote]

It is not meaningless, but you cannot use finite probability distributions such as the probability of getting head or tails on a coin, or getting each one of the 6 possible results in a standard die.  Consider for example the case where you take "at random" (meaning that you use a uniform probability DENSITY function), a real number from 0 to 1.  Intuition tells you that the probability that it is $ \geq\frac{1}{2}$ is half, and that it is $ \leq\frac{1}{2}$ is also half.  Strict calculations will in fact yield that the probability that the chosen number belongs to interval $ (a,b)$, where $ 0\leq a\leq b\leq 1$ is actually $ b\minus{}a$.  But the probability that it is exactly $ \frac{1}{2}$, or exactly $ a$, or exactly $ b$, is 0 (1 favourable outcome divided by an infinity of possible outcomes, if you want to look at it that way).  The definition of probability for set with infinite cardinal (whether numerable or not numerable) is not obvious, and unless the set of possible outcomes is an interval of real numbers (such as the situation where you pick any real number from 0 to 1), it may require some basic understanding of measure theory...

An axiomatic definition of probability may be found for example in http://en.wikipedia.org/wiki/Probability_axioms, where the $ \sigma$-additivity is precisely introduced in order to cope with situations such as the one where you consider an infinite (but countable) set of possible outcomes.  Check also (if your level of math allows for it) measure theory, it deals with infinite sets where probabilities may be introduced.

PS: just out of curiosity, what grade are you currently studying?

Hmm... I changed my mind, now I'm in the $ \frac56$ camp.   :lol:

Just joking.  Mornik, we know there are equally many prime numbers and even numbers, since we can create a bijection from one set to another.  Does this mean the probability of picking an even number is equal to the probability of picking a prime number?

daniel73, I have taken some courses on measure theory, so feel free to use it.

My grade?  I'm an engineer with a master's in applied math from UCLA.  But I don't pretend this gives me any kind of authority on AoPS.  I've been owned by 7th graders on For the Win too many times for a delusion like that.   :)

[quote="Singularitarian"]daniel73, I have taken some courses on measure theory, so feel free to use it.

My grade?  I'm an engineer with a master's in applied math from UCLA.  But I don't pretend this gives me any kind of authority on AoPS.  I've been owned by 7th graders on For the Win too many times for a delusion like that.   :)[/quote]

I actually ment jdai_ming's grade and knowledge of measure theory, since he is the one that originally asked the question and the one that may be interested in the answer...  Anyway, I am also an engineer with a masters in math, so that makes us even  :wink: I have also been owned by kids on math competitions, although that becomes rarer and rarer (experience seems to count for something after all), so we are even again there  :wink:

The importance of having a measure is that if you have a universe with a finite measure, you can always define a probability, just divide the measure of a subset into the measure of the entire set (universe), and you have the probability of choosing an element of the given subset when randomly choosing an element from the set.  Note that the measure does not need to be the "usual" measure, ie, when choosing a reals interval $ (a,b)$ as your universe, you do not need to define the Stieltjes measure, you may use a weight function $ f(x)$ defined in $ (a,b)$ such that
\[ \mu((c,d)) \equal{} \int_c^df(x)dx
\]
for any $ (c,d)\subset(a,b)$.  This is equivalent to picking a non-uniform probability distribution...  If the measure of a given subset is finite in a set with infinite measure, you may still wave your hands and say that the probability of the subset is 0, even though you may not define a probability for all sets, or in fact you may not be able to define a probability as such (see further below).

Cardinality sets an even easire criterium, meaning that, with "usual" measures, sets of higher cardinalities have measures "infinitely larger".  This allows us to say that the probability of picking a rational out of the reals is 0 (watch my hands waving as I say this!)

What happens however when you pick a set with infinite measure and want the probability of a subset with infinite measure, and both set and subset have the same cardinality?  Well, the definition of a strict probability for all subsets of the set may not be possible, but "especial" sets may be assigned a probability, based on a concept of "density".  For example, we know that the "density" of primes with respect to naturals tends to zero as we pick higher and higher primes.  To a certain extent, this is equivalent to saying that, if you pick the entire set of naturals, the primes are so far appart from one another in any infinite interval of naturals, that their effective density is 0, so the probability of picking a prime when randomly choosing a natural is 0 (unless, of course, you do something like picking a measure that assigns a finite weight to primes and 0 weight to non-primes!).  A similar argument allows us to wave our hands and say that the probability of choosing an even number is 1/2, or in general the probability of choosing a number congruent $ k$ modulus $ d$ is $ 1/d$.  This is possible since the "density" of such numbers in the set is constant when you pick a sufficiently large interval.  Irregularly distributed infinite sets, however, may not be assigned in general a probability, so the whole axiomatic definition may fail, while you may still define a "probability" for particular subsets (waving your hands like crazy while you do, since as far as I know, no actual estrict definition or calculation may be actually performed in many cases where a "probability" is still assigned).

In conclusion:

1) You may always define a probability in any set, as long as there is a finite measure defined in it (using SOME measure, not necessarily the "usual" or "natural" measure of the set).  The measure basically determines the probability properties of the set and its subsets, and is equivalent to introducing some non-uniform probability density if the measure is not the "natural" one.
2) When the cardinality of a subset is lower than the cardinality of the set, the probability of choosing an element of the subset may always be considered as 0.  If the cardinality is the same, you need other arguments (measure, density or the like) in order to determine a probability.
3) Even when cardinality or measure are not available as criteria for establishing probabilities, you may still use density arguments in order to assign probabilities to certain particular subsets, although an actual probability satisfying all three axioms may not always be defined.  Note that in this case I claim that you can only ASSIGN a probability to SOME subsets, not DEFINE it for EVERY subset...

If someone reads this and finds something wrong, please feel free to point it out, I don't consider measure theory to be my forte, so I will not be offended  :wink: In fact, you may help me correct some deep-seeded wrong concept  :oops:

[quote="atomicwedgie"]The question is ill-posed, because "at random" implies a uniform distribution.  But for such a distribution to be a valid probability distribution, it must have finite support.[/quote]

you must not be a very big fan of lebesgue measure on $ [0,1]$ then

When is there a canonical "uniform distribution" on a set? We get a notion of uniformity from a nice enough group action; $ aS$ should have the same measure as $ S$. For example, consider the sphere acted on by the group of rotations $ SO_3$. The action is transitive, and it feels right to have a rotated version of a region have the same measure (area) as the original region.
What is "nice enough" here? The group should be a locally compact topological group, and the action should be continuous and transitive. This lets us construct a unique (up to scaling) translation-invariant measure. The measure of the whole set is finite if and only if the set is compact.
If our set is a nice subset of some larger set with such an action, we can build the measure on the larger set and restrict to the subset.

For the natural numbers, start with the integers acting on themselves by addition. The translation-invariant measure simply counts cardinality. This can be restricted to the natural numbers, but $ \mathbb{N}$ has infinite measure and this cannot be scaled to a probability.
For this reason, there are no canonical probability distributions on the reals or the integers, and questions like "what is the probability that a natural number is even" or "what is the probability that a positive real number is $ >1$" have no standard answer. The rationals in the reals are a little different- since the rationals have measure zero, they have probability zero in every (absolutely) continuous distribution, and we most likely intend to use one.

This is all a slightly generalized version of Haar measure.

Another question (which has been asked and answered before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=187407]here[/url]):

Does there exist a probability distribution on $ \mathbb{Z}$ such that the probability of a number being divisible by $ p$ is $ \frac1p$ for all primes $ p$? All integers $ p$?


there are 1000 locks.there are 1000 persons numbered1,2,....1000.each one closes & opens(like a person opens if it is closed& locks it if it is open)  the locks on the basis of the multiples of his number.find the number of locks opened?

[hide="Hint"]
perfect squares have odd number of divisors
[/hide]

A lock will be opened by every person whose number is a factor of the lock, so each lock will be changed the number of times that is equal to its number of factors. The only way a lock will be open at the end is if it has an odd number of factors.
The only numbers with an odd number of factors are perfect squares. The largest perfect square up to 1000 is $ 31^2 \equal{} 961$, so $ \boxed{31}$ locks are open at the end of this process.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14739[/img]

 :(

The concur point is the Fermat point of triangle ABC.
See at [url]http://www.mathlinks.ro/viewtopic.php?t=208618[/url]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14744[/img]

 :thumbup:

And what about an equilateral triangle as the problem states, inscribed ( instead of circumscibed ) to the given triangle $ \bigtriangleup ABC$ ?

Kostas Vittas.

[quote="vittasko"]And what about an equilateral triangle as the problem states, inscribed ( instead of circumscibed ) to the given triangle $ \bigtriangleup ABC$ ?[/quote]
That equilateral triangle and ABC will be orthologic and so also the perpendicolar from the vertex of the equilateral triangle to the side of ABC are concurrent and so we must find a point that have the pedal triangle equilateral, that is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208291]this[/url]


quadratics
geometry
rectangle
function
symmetry
analytic geometry
algebra

Well the point of finding the maximum and minimum is simpler than how to find it - to explain anyways.

In "real life" situations, it is often necessary to find the greatest or least amount of something when it changes based on the change of some other factor that is under your control. For example, let's say you have 100 meters of string. You want to use that string to make a rectangle with the greatest possible area. What should the dimensions of that rectangle be?

Using simple algebriac concepts you can [i]isolate[/i] the variable for area on one side of the equation, and make it a function of the width.

You have a type of quadratic equation.

A=W(100-2W/2)=W(50-W)=50W-W^2

Now you are trying to find the largest possible area you can enclose with your 100 meters of string. Basically, a general way of doing this is just through logic. 50W-W^2 would be largest when W^2 is the smallest and 50W is the largest. W=25.

[quote="banblink182"]Now you are trying to find the largest possible area you can enclose with your 100 meters of string. Basically, a general way of doing this is just through logic. 50W-W^2 would be largest when W^2 is the smallest and 50W is the largest. W=25.[/quote]

I think what he is wondering is how you got W=25.

In every quadratic equation, the quadratic plotted on a graph is always symetric. The line of symmetry is called the axis of symmetry. That axis always passes through the vertex of the graph, which is the minimum or maximum. In any quadratic equation in the form ax 2 +bx+c, the axis of symmetry will be a line with he equation x=-b/(2a). Now that you know the x coordinate of the vertex, you can plug that into the original equation to get the y coordinate.

[quote="rrusczyk"]Have you learned completing the square?[/quote]
Elaborating... in case you know completing the square

\[(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\]

\[ax^2+bx+c=0\]

\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]

\[x^2+\frac{b}{a}+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\]

\[(x+\frac{b}{a})^2+\frac{c}{a}-\frac{b^2}{4a^2}\]

It can be seen that the minimum value will be when \[x=-\frac{b}{a}\], because a square is never negative, so it's minimum is 0. If the coeffiecient of \[x^2\] was negative, then we would be looking for the maximum, and would factor out -1 at the beginning, then multiply it in at the last step:

\[-(x+\frac{b}{a})^2-\frac{c}{a}+\frac{b^2}{4a^2}\]

I know how to complete the square. So thats basically all it is? But what I'm concluding from this is that the maximum and minimum are the same. Also it looks like you can only have either the min or max at one time because when the minimum is zero than you have to factor out the -1 to find the maximum. But since your just factoring out the -1 than wouldn't the maximum be the same as the minimum?

A quadratic equation will only have a maximum or a minimum.

It will be it's vertex.

Inside polynomials of a higher degree there can be multiple maximum's and minimum's (more correctly local maximums and local minimums) which is where slope changes.

It is useful for finding out how to find things like the maximum area achieveable of say a rectangle and circle if you have 10m of wire.

[quote="Archimedes27"]I don't understand how to find the minnimum and maximumum of quadratic equations. I'd appreciate it if someone could provide me with some intuitive examples. I also don't understand what their used for or the point of finding them.Thank you.[/quote]

Okay. You have a varying number, f(x), whose value depends solely on the value of x.

You aren't finding the maximum/minimum of the EQUATION. This is the wrong terminology. I could get the function:

f(x) = x^2 + 4x + 4

And make it "larger" by writing:

f(x) + 1000 = x^2 + 4x + 1004

This is silly. x VARIES, between negative and positive infinity, and f(x) varies depending on this. So you can say f(x) is the dependant variable, and x is the independant variable (sorry, this is coming from a wannabe physicist!).

So what do you want to maximise? f(x).
x can only be OPTIMISED. The optimum value of x is the one which gives the MAXIMUM, or MINIMUM value of f(x).

eg.

f(x) = 2 - x^2

If we want to maximise f(x), is 4 an optimum value of x? No, as f(2) = -2, and f(0) = 2. Since 2>-2, f(0) > f(2), and 0 is a more optimum value of x.

In quadratics, the maximum or minimum point of the curve is defined at the point where the f(x) value is at the highest/lowest (respectively) value it can possibly be. This is when the curve has a GRADIENT of ZERO. Think about this... at the top of a parabola, a tangent to the curve is horizontal.

But that's the calculus way of finding max/min points (this way can be used with lots of curves, not just quadratics like completing the square... unless you can complete the cube, which is horiffic).

In terms of doing it algebreically, think again of the general quadratic:

f(x) = (a)x^2 + (b)x + (c)

Where a, b and c define the nature of the equation and are "constants", as they do not vary like x and f(x) do in one equation. (they do vary between equations, eg. a larger value of c will shift the curve upwards and change the equation in hand).

We want to know which value of x (given the infinite values it can take in a quadratic) will give the largest.smallest value of f(x).

We complete the square. You know the procedure:

(a)x^2 + (b)x + (c) = 0
x^2 + (b/a)x + (c/a) = 0
(x + (b/2a))^2 +(4ac/4aa) - (bb/4aa) = 0
(x + (b/2a))^2 + (4ac - bb)/(4aa) = 0

Now I was simple taught as a fact that the largest value f(x) can take is (4ac-b^2)/(4a^2), which gave a value of (-b/2a).

I think this is right. And sorry for the lack of explanation on the last part, blame my GCSE teacher.

Here's an example for anyone who doesn't know how to complete the square, I bolded the portion of this exercise that would probably be relevant to the poster at the bottom.

Example...

The maximum OR minimum can be found by "completing the square", which occurs as follows:

given some quadratic, y = 4x^2-57x+27 is in the form y=ax^2+bx+c (a=2, b=-57, c=27), factor out the 4 from the first two terms only (you can factor it out from the third term but this is slightly quicker)

thus

y=4(x^2-14.25x)+27

We want to make x^2-14.25x a perfect square, so we add the square of half of 14.25, then we must subtract to keep the equation the same.

(14.25/2)^2 = 3249/64

Thus,

y=4(x^2-14.25x+3249/64-3249/64)+27

x^2-14.25x+3249/64 is a perfect square and factors to (x-57/8)^2, however, we must take the 3249/64 out of the brackets, so we multiply it by the term outside the brackets (4)

y=4(x-57/8)^2+3249/16+27
y=4(x-57/8)^2+3681/16

[b] This is now in "vertex form", or y=a(x-h)^2+k, where (h,k) represents the vertex. [/b]

In this case, the vertex is at (57/8, 3681/16) and thus, the optimum (or min/max) value is 3681/16.

[b] In this case, since a is positive, the value is a minumum. If a were to be negative, the value would be a maximum. [/b]

Hope that helps.

Just for kicks, another way of thinking about -b/(2a).

The parabola is symmetric across the aptly named line of symmetry, which intersects the parabola at the vertex.  Which means that the two zeros of the parabola will be equidistant from the vertex.  In other words, the x-coordinate of the vertex will be the midpoint, or average, of the two x-coordinates of the zeros.  Recall the quadratic formula (-b:pm: :sqrt: (b^2-4ac))/(2a).
Averaging the x-coordinates given by the formula gives us -b/(2a).

Unless that parabola is not vertical ;-), and for horizontal parabola's the vertex will have a y coordinate equidistant between the y-intercepts but for those parabolas that are neither, then some rotation is needed.


abstract algebra
group theory
linear algebra
linear algebra unsolved

If $M$ is a finite abelian group then $M$ is narurally a $\mathbb{Z}$-module. Can this action be extended to make $M$ into a $\mathbb{Q}$-module ?

your statement is of course correct about the abelian groups, but you don't need it to be finite

usually you can't make it a $\mathbb{Q}$ module

suppose $z= \frac{1}{2} a$ with a in M

now $z + z= \frac{1}{2}a +\frac{1}{2} a =(\frac{1}{2}+\frac{1}{2} )a =1 a =a$

so every element has a half then

it is clear than in many (finite abelian) groups this is not the case:

$\mathbb{Z}/\mathbb{Z}_{4}$ and the multiplicative group of the rationals are examples

An abelian group $M$ can be "extended" to a $\mathbb{Q}$-module $M$ if and only if $M$ is a divisible and torsion-free abelian group. If $M$ is finite, this works only for $M=0$.


They just sent me to attend after my junior year at High School. From the letter, I am assuming that Clarkson School is a subsidiary of Clarkson University. Has anybody heard of them?

I visited Clarkson when I was looking at colleges myself (granted, this was nearly 20 years ago, and I was interested in engineering then).  They are (or were) primarily known as an engineering school.  They were one of the first schools, back in the 1980s, to require that all undergraduates own a computer.

The big downside, for me, is that it's a fairly small school in a fairly small town in the middle of nowhere in far northern New York.  If that's your cup of tea, it might be a good option, especially if you're interested in engineering.  Or if you like to ski.


limit
integration
inequalities
real analysis
real analysis unsolved

Let $f\in L^{2}((0,+\infty))$. Evaluate
$\lim_{x\to+\infty}\frac{\int_{0}^{x}|f(t)| dt}{\sqrt x}$

Given $\varepsilon>0,$ find $M$ sufficiently large that

$\int_{M}^{\infty}|f(t)|^{2}\,dt<\left(\frac{\varepsilon}2\right)^{2}.$

We can do this because by the Monotone Convergence Theorem,

$\lim_{M\to\infty}\int_{0}^{M}|f(t)|^{2}\,dt=\int_{0}^{\infty}|f(t)|^{2}\,dt.$

Let $A=\int_{0}^{M}|f(t)|\,dt.$ This is finite because by the Cauchy-Schwarz inequality,

$A<\left(\int_{0}^{M}1^{2}\,dt\right)^{\frac12}\left(\int_{0}^{M}|f(t)|^{2}\,dt\right)^{\frac12}< \sqrt{M}\|f\|_{2}^{2}.$

Now we have that for $x>M,$

$\frac1{\sqrt{x}}\int_{0}^{x}|f(t)|\,dt= \frac1{\sqrt{x}}\int_{0}^{M}|f(t)|\,dt+\frac1{\sqrt{x}}\int_{M}^{x}|f(t)|\,dt$

The first term is equal to $\frac A{\sqrt{x}}.$ We can find $x$ sufficiently large that it is less than $\frac{\varepsilon}2.$

We then apply Cauchy-Schwarz to the second term.

$\frac1{\sqrt{x}}\int_{M}^{x}|f(t)|\,dt \le\frac1{\sqrt{x}}\left( \int_{M}^{x}1^{2}\,dt\right)^\frac12 \left(\int_{M}^{x}|f(t)|^{2}\,dt \right)^{\frac12}$

$\le\frac{\sqrt{x-M}}{\sqrt{x}}\cdot\frac{\varepsilon}2 < \frac{\varepsilon}2$

Hence by making $x$ sufficiently large, we can make

$\frac1{\sqrt{x}}\int_{0}^{x}|f(t)|\,dt<\varepsilon.$

The limit is zero.


Is there any easy way to find the number of distinct ways to color each edge of a n vertice complete graph one of two colors? What about with more than 2 colors?

It depends what you mean by "distinct."  If vertices are labeled, then the answer is just $c^{\binom{n}{2}}$ where $c$ is the number of colors.  If vertices are not distinct, then the two-color version is equivalent to asking for the number of graphs on $n$ vertices.  That sequence is available [url=http://www.research.att.com/~njas/sequences/A000088]here[/url], in the OEIS.  I haven't thought about multiple-color versions of this problem.

I meant with unlabled verticies. Is there any general formula? How do you find the answer?


ARML I-8 ?1997

let cbrt((cbrt(2))-1)=cbrt(a)+cbrt(b)+cbrt(c), where a, b, and c are rational integers. Evaluate a+b+c

[quote="maokid7"][hide]
cube both sides and get
$\sqrt[3]{2}-1=a+b+c+3(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac})+6\sqrt[3]{abc}$
since $-1$ is the only term w/o a $\sqrt[3]{x}$ term i think it is safe to say that
$a+b+c=-1$
[/hide]
unless i made some incorrect assumption. :blush:[/quote]

The cube rooted terms are not guaranteed to not be integers.

[quote="maokid7"][hide]
cube both sides and get
$\sqrt[3]{2}-1=a+b+c+3(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac})+6\sqrt[3]{abc}$
since $-1$ is the only term w/o a $\sqrt[3]{x}$ term i think it is safe to say that
$a+b+c=-1$
[/hide]
unless i made some incorrect assumption. :blush:[/quote]
in any case, thats not the right answer (and yes, i did mean what hydro said)

[quote="maokid7"][hide]
cube both sides and get
$\sqrt[3]{2}-1=a+b+c+3(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac})+6\sqrt[3]{abc}$
since $-1$ is the only term w/o a $\sqrt[3]{x}$ term i think it is safe to say that
$a+b+c=-1$
[/hide]
unless i made some incorrect assumption. :blush:[/quote]

this expansion is not correct also

[hide]
It's rather difficult to cube something and get a $\sqrt[3]{2}$ without a $\sqrt[3]{4}$, so $\frac{1}{\sqrt[3]{2}-1}=1+\sqrt[3]{2}+\sqrt[3]{4}$ is nicer to look at. Also 3's are common (from $(x+1)^3=x^3+3x^2+3x+1$) so it makes sense to look at multiples of 3.

It turns out that $3+3\sqrt[3]{2}+3\sqrt[3]{4}=1+3\sqrt[3]{2}+3\sqrt[3]{4}+2=\left( 1+\sqrt[3]{2} \right)^3$. So this means

\[ \sqrt[3]{\frac{3}{\sqrt[3]{2}-1}}=1+\sqrt[3]{2}\Rightarrow \sqrt[3]{\sqrt[3]{2}-1}=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}  \]

which simplifies to $\frac{\sqrt[3]{12}-\sqrt[3]{6}+\sqrt[3]{3}}{3}$, so a possible ordered triple is $\left( \frac{4}{9}, -\frac{2}{9}, \frac{1}{9} \right)$ (wait, is it supposed to integers or rationals?) $\Rightarrow \boxed{\frac{1}{3}}$.
[/hide]


Count to 100000 by adding the post number of the poster above you to the total, starting with mine.

[quote="r15s11z55y89w21"]How about let's not?[/quote]

Because then we'd only have about 450 posts? Sure. Mine is number 6.

[quote="r15s11z55y89w21"]@sdkudrgn88: If we don't count at all, we could have 0 pointless posts, how nice, right?[/quote]

I agree, but the purpose of the Fun Factory is to be pointless. 21

I'll just stop posting here...

sorry to be unclear... I meant the number of posts that user had...at least I put it in the FF

[quote="sdkudrgn88"]I agree, but the purpose of the Fun Factory is to be pointless. 21[/quote]

There was a time when it's purpose was to create fun (ya, rly). Seems today's youth has no clue about "fun"  :|  :wink:

[quote="ZetaX"]There was a time when it's purpose was to create fun (ya, rly).[/quote]
Yes, but that time is over. Now the "Games" forum is to fulfill that purpose.
Damn, I broke my promise. But I'm just replying to a quote... 9738

restart? plz don't delete my post

0+1355=1355

no you add how many posts you have

I had 1355, you have 557, so a total of 1912

I add 1355 again --> [b]3267[/b]

edit: and lets say double posting is allowed

3267+1356=4623

(how did I gain a post lol)

4623 + 1357 = 5980

if we get fransliszt to post 10 times we can get to 100000 in 10 posts B))

jom said i should post here :/
bluecarneal pr0 instructor

5980+ 1721 = 7701

where am I gaining these posts???

ohhh we add the user above us's post count... ahha! I understand now

[quote name="pog" url="/community/p19001858"]
12693
this is a lot easier when everyone on aops has 1k+ posts nowadays

(easier than in 2009 lol)
[/quote]

*me who has 581*
pEnSiVe

12693+1721=14414 according to google

18387+1274=19661
[s]me being amazed by this is what my aops instructor did 11 years ago when i was barely alive
in fact this thread was posted on my bday


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let $ ABC$ be a triangle and the points $ D\in (BC)$, $ E\in (CA)$, $ F\in (AB)$ such that \[ \frac {BD}{DC} \equal{} \frac {CE}{EA} \equal{} \frac {AF}{FB}.\] Prove that if the circumcenters of the triangles $ DEF$ and $ ABC$ coincide then $ ABC$ is equilateral.

Let us a take a complex plane with the circumcenter of the triangles $ DEF$ and $ ABC$ as origin, and let us denote with $ x$ the afix of the point $ X$.
By denoting \[ \frac {BD}{DC} \equal{} \frac {CE}{EA} \equal{} \frac {AF}{FB}\equal{}k>0\]
we have that : \[ d\equal{}\frac{b\plus{}kc}{1\plus{}k} , e\equal{}\frac{c\plus{}ka}{1\plus{}k} , f\equal{}\frac{a\plus{}kb}{1\plus{}k}\]
Because we have chosen the circumcenter of the two triangles as origin, it results that :
\[ |a|\equal{}|b|\equal{}|c| \Leftrightarrow a\bar{a}\equal{}b\bar{b}\equal{}c\bar{c}\]
and
\[ |d|\equal{}|e|\equal{}|f| \Leftrightarrow d\bar{d}\equal{}e\bar{e}\equal{}f\bar{f}\]
The last one is equivalent to \[ (b\plus{}kc)(\bar{b}\plus{}k\bar{c})\equal{}(c\plus{}ka)(\bar{c}\plus{}k\bar{a})\equal{}(a\plus{}kb)(\bar{a}\plus{}k\bar{b})\]
and using the fact that $ a\bar{a}\equal{}b\bar{b}\equal{}c\bar{c}$ it follows :
\[ a\bar{b}\plus{}b\bar{a}\equal{}b\bar{c}\plus{}c\bar{b}\equal{}c\bar{a}\plus{}a\bar{c}\]
This is equivalent to \[ |a\minus{}b|^{2}\equal{}|b\minus{}c|^{2}\equal{}|c\minus{}a|^{2}\]
so the triangle $ ABC$ is equilateral.

Let a line through D parallel to AC that meet AB on D', a line through D' parallel to BC that meet AC on E, a line through E parallel to AB that meet BC on E', a line through E' parallel to AC that meet AB on F, a line through F parallel to BC that meet AC on F', a line through F' parallel to AB that meet BC on D, all for Thales.
Call O the circumcircle of ABC, then we have: $ \overline{OF} \equal{} \overline{OE} \equal{} \overline{OF'}$ so the circumcircle stay on the axis of FF' and so the axis of FF' and of BC are the same and $ AB\equal{}AC$.
Equaly we obtain $ BC\equal{}BA$ and $ CA\equal{}CB$ so the triangle is equilateral.

Given points [b]A[/b], [b]B[/b], [b]C[/b] in the complex plane, what is the circumcentre of triangle abc in terms of [b]A[/b], [b]B[/b], and [b]C[/b]?

Here's a different way to approach this very interesting problem.

Let $R$ be the circumradius of $\triangle ABC$ and $r$ be the circumradius of $\triangle DEF$.  Furthermore, let $O$ be the common circumcenter of both triangles.  By Stewart's Theorem on isosceles triangle $OAB$ with cevian $OF$, \begin{align*}R^2(AF)+R^2(BF)=r^2(AF+BF)+r(AF)(BF)\implies (R^2-r^2)(AF+BF)=r(AF)(BF).\end{align*}  Now let $\dfrac{AF}{FB}=k$, so $AF=k\cdot FB$.  Then our equation transforms into \[(R^2-r^2)(BF+k\cdot BF)=R(k\cdot BF)(BF)\implies (R^2-r^2)(1+k)=Rk\cdot BF.\]  Applying this cyclically, we see that $(R^2-r^2)(1+k)$ is also equal to $Rk\cdot AE$ and $Rk\cdot CD$, so $BF=AE=CD$.  It is now obvious that $AB=BC=CA$ as desired.  $\blacksquare$

Heh. Let the isotomic of $D$ wrt $BC$ be $D'$. Then note that $D'E \parallel AB$. But clearly if $O$ is the circumcentre then $OD = OD' = OE = OF$ (reflection through perpendicular bisector) so done.

Notice a couple of things:

If $ABC$ is equilateral, then $DEF$ is also equilateral. Furthermore, if the centroids of them coincide, then $ABC$ must be equilateral as well (This problem can be seen at Goutham's blog). In general, this type of construction has very interesting properties, makes u wonder for what other points $P$ might it be such that if they coincide the triangle is equilateral.


inequalities
trigonometry
function
real analysis
real analysis unsolved

For any $x\in \left( 0,\frac{\pi}{2}\right)$ the folowing inequalities are truly:

$1.\ \ 2\tan x+3x\cos x>5\sin x.$

$2.\ \ 3\tan x+2x\cos x>5x.$

$3.\ \ \frac{2}{\pi}\cdot \tan \frac x2<\frac 1x -\frac{1}{\tan x}<\frac 23\cdot \tan \frac x2.$

Expanding by taylor series the functions obtain:

$\cos x>1-\frac{x^2}{2}\Rightarrow 2x \cos x> 2x- 2\frac{x^2}{2}$
$\tan x>x+\frac{x^3}{3}\Rightarrow 3 \tan x>3x+3\frac{x^3}{3}$
Adding this equations the second inequality follows.

For the first inequlity we must write more terms from taylor expansion of that trig functions:

$2\tan x+3 x\cos x>2(x+\frac{x^3}{3}+\frac{2x^5}{15})+3x(1-\frac{x^2}{2!}+\frac{x^4}{4!})>5(x-\frac{x^3}{3!}+\frac{x^5}{5!})>\sin x$.

Can you solve these inequalities without the Taylor's expansion ?


trigonometry
geometry
3D geometry
pyramid
Pythagorean Theorem
trig identities
Law of Cosines

Consider a pyramid $P-ABCD$ whose base $ABCD$ is a square and whose vertex $P$ is equidistant from $A$, $B$, $C$, and $D$. If $AB=1$ and $\angle APB=2\theta$, then the volume of the pyramid is

$\text{(A)}\ \frac{\sin \theta}{6} \qquad \text{(B)}\ \frac{\cot \theta}{6} \qquad \text{(C)}\ \frac{1}{6\sin \theta} \qquad \text{(D)}\ \frac{1-\sin 2\theta}{6} \qquad \text{(E)}\ \frac{\sqrt{\cos 2 \theta}}{6\sin \theta}$

Here's my approach if you get stuck.

[hide="An Approach"]Use the law of cosines to find the length of the edges, then apply the Pythagorean theorem to get the height.[/hide]

As far as I can tell, the fifth option should be corrected yet again, since the answer is
[hide]${{\sqrt{\cos 2\theta}}\over{6\sin\theta}}$[/hide]

I got the same answer as farenhajt.  It was easy to get that expression, but I spent a lot of time trying to see if it simplified to one of those expressions...

[quote="scorpius119"][hide]Draw in a slant height from $P$ to side $AB$. Since triangle $PAB$ is isosceles, this is also a median and angle bisector. If the slant height intersects $AB$ at $M$, then $\angle APM=\theta \Rightarrow PM=\frac 12 \cot \theta$.

Now draw in the height of the pyramid. If $H$ is the projection of $P$ onto square $ABCD$, then $MH=\frac 12$. Using Pythagorean Theorem, $PH=\frac 12 \sqrt{\cot^2 \theta -1}$.

Now $\sqrt{\cot^2 \theta -1}=\frac{1}{\sin \theta} \sqrt{cos^2 \theta-\sin^2 \theta}=\frac{\sqrt{\cos 2\theta}}{\sin \theta}$, so $PH=\frac{\sqrt{\cos 2\theta}}{2\sin \theta}$.

The volume is 1/3 times the area of the base, 1, times the height, which is $\boxed{E}$.[/hide][/quote]That was rrusczyk's solution at the SDMC. :)


Let $A = (0,0)$ and $B = (200,0)$. For each point $C$ in the plane with the area of $\triangle{ABC}= 2006$, consider the point $D$ on line $BC$ with segment $AD$ an altitude to triangle $\triangle{ABC}$. The set of such points $D$ together with point $A$ enclose a region in the plane. Find the area of this region.

yeah, um, great, everyone's focused on solving some stupid code instead of this problem :mad:

The answer's $10000\pi$ btw ;)

He's stated the problem incorrectly. The problem should say:

Let A = (0,0) and B = (200,0). For each point C in the plane with the area of ABC equal to 2006, consider the point D on line BC with segment AD an altitude to triangle ABC. The set of such points D together with point A enclose a region in the plane. Find the area of this region.

[quote="mysmartmouth"]Can someone please post a solution to this problem?  I don't get it.  Thanks![/quote]

I'll give you a hint instead and YOU have to try and figure it out :)

An inscribed right triangle has what relationship with a circle? Or in other words, what does the diameter have to do with an inscribed right triangle.

[quote="mysmartmouth"]The diameter of a circle is just the hypotenuse of an inscribed right triangle, right?[/quote]

yep and so......................what else do you know?

well, doesn't the altitude creat a RIGHT angle ;)

Furthermore, doesn't the altitude partition the triangle into two right triangles?  ;)