Can someone tell me why this happens?
If you check my profile you can see my signature, but whenever [b]I[/b]check my posts I never see my signature...
Lately, I've been thinking that no one can see my signature... :(  :(  :(

Is there a way to fix this?! :(

yes, above the "submit" button, there is an option that says "attach signature."
alternatively, you can go to your preferences and set "always attach signature."


MATHCOUNTS
geometry
3D geometry
pyramid
octahedron
parallelogram
rectangle

I got finish with my own mock MATHCOUNTS test! :)  I will post them here when there are atleast 7 people interested. The level is a easy state/hard chapter.

I'm interested. please post it.

 :huh:  :ninja:

Me too.

(Come one, people.  Say you're interested before Saturday...)

Edit: Whoa, three people at the exact same time!!! :!:

Interested!please post!

If I triple post will you count it as another 3 people? :P

I'm interested. And even if not enough people were interested (which is unlikely), it wouldn't hurt to post it anyways.  :)

Thats already 7! I'll post it tomorrow to give other people a chance.

Ug
same, except a little more than 30 minutes...oh well, my fault..
ANyways, I'll start the discussion (I suppose it's here).
13. Consider a convex body having 6 vertices. For example, a cube is convex body having 8 vertices. How many distinct triangles can be constructed by connecting three different vertices of this body?

Wow...
wonder how I managed to miss that...that was idiotic.

17. At Doper’s University, 42 % of all athletes are honor students and 3 % of all honor students are athletes. Given that 58 athletes are not honor students, how many honor students are not athletes?

note that there are 100 athletes.

42 are honor students.

42=.03x
x=1400

Subtract 42
1358

I think thats because your answer is wrong. When slicing the cube, you get triangles in place of the vertices.
3*6=18 not 24

If there's a pyramid cut from each corner, there will be 8 edges on the base. $8*3=24$

o yeah theres 8 vertices on a cube :oops:

But if you pass planes through the [b]midpoints[/b], you'll end up with an octahedron, which has 12 edges, not 24.

[quote="Fanatic"]My answer was 12.

Also, how do you do #16?[/quote]

I assumed the parallelogram to be a rectangle :dry:. It worked...I remember there was this one formula...but i forgot.

3. Houses on the south side of Cherry Street are numbered with consecutive even integers from 1902 to 2192. How many houses are on the south side of Cherry Street?

 :blush:

[quote="goldendomer"]Here's one random fact. Nobody got number 22 right.[/quote]

I got that one right  :roll:

i got it wrong, mainly because of stupidity


algebra
polynomial
sum of roots
algebra unsolved

Determine all polynomials $ P_n(x)\equal{}x^n\plus{}a_1 x^{n\minus{}1}\plus{}...\plus{}a_{n\minus{}1} x\plus{}a_n$ with integer coefficients whose $ n$ zeros are precisely the numbers $ a_1,...,a_n$ (counted with their respective multiplicities).

[quote="moldovan"]Determine all polynomials $ P_n(x) \equal{} x^n \plus{} a_1 x^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 1} x \plus{} a_n$ with integer coefficients whose $ n$ zeros are precisely the numbers $ a_1,...,a_n$ (counted with their respective multiplicities).[/quote]

Very nice !

First consider that if $ P(x)$ fits the rule, $ x^kP(x)$ also fits. So let us look for polynomials with $ a_n\neq 0$

Then $ \prod a_i\equal{}(\minus{}1)^na_n$ and, since $ a_n\neq 0$, $ a_i\equal{}\pm 1$ $ \forall i\in[1,n)$

So $ P(x)\equal{}(x\minus{}1)^p(x\plus{}1)^q(x\minus{}a)$
Sum of roots gives $ p\minus{}q\plus{}a\equal{}\pm 1$
Sum of inverse of roots gives $ p\minus{}q\plus{}\frac 1a\equal{}\pm\frac 1a$ and so either $ p\equal{}q$ and so $ a\equal{}\pm 1$, either $ p\minus{}q\equal{}\minus{}\frac 2a$ and so $ a\in\{\minus{}2,\minus{}1,1,2\}$. So :

1) $ p\equal{}q$ and $ a\equal{}\pm 1$
$ P(x)\equal{}(x^2\minus{}1)^p(x\minus{}a)$ (with $ a\equal{}\pm 1$)
If $ p\equal{}0$, $ P(x)\equal{}x\minus{}a$ and the root need to be $ a_1\equal{}\minus{}a$, which is impossible if $ a_n\neq 0$
Then $ p>0$ and the coefficient of $ x^{2p\minus{}1}$ is $ \minus{}p$ and so $ p\equal{}1$ and $ P(x)\equal{}(x^2\minus{}1)(x\minus{}a)\equal{}x^3\minus{}ax^2\minus{}x\plus{}a$ and the roots are $ \minus{}a,a,\minus{}1$. But we know that the roots are $ \minus{}1,1,a$ and so $ a\equal{}\minus{}1$
So we got the solution $ P(x)\equal{}(x^2\minus{}1)(x\plus{}1)\equal{}x^3\plus{}x^2\minus{}x\minus{}1$

2) $ a\equal{}\minus{}2$ and $ p\minus{}q\equal{}1$
Then $ p\minus{}q\minus{}2\equal{}\pm 1$ and $ p\minus{}q\equal{}1$ and $ P(x)\equal{}(x^2\minus{}1)^q(x\minus{}1)(x\plus{}2)$ but then the coefficient of $ x^{2q}$ is $ q\plus{}2$ and the only solution would we $ q\equal{}0$
And we got the solution $ P(x)\equal{}(x\minus{}1)(x\plus{}2)\equal{}x^2\plus{}x\minus{}2$

3) $ a\equal{}\minus{}1$ and $ p\minus{}q\equal{}2$
Then $ P(x)\equal{}(x\minus{}1)^{q\plus{}2}(x\plus{}1)^q(x\plus{}1)$ $ \equal{}(x^2\minus{}1)^{q\plus{}1}(x\minus{}1)$ then the coefficients of $ x^{2q\plus{}1}$ is $ \minus{}(q\plus{}1)$ and $ q\equal{}0$
Then $ P(x)\equal{}(x^2\minus{}1)(x\minus{}1)\equal{}x^3\minus{}x^2\minus{}x\plus{}1$ and this does not respect the requirements

4) $ a\equal{}1$ and $ p\minus{}q\equal{}\minus{}2$
Then $ P(x)\equal{}(x\minus{}1)^p(x\plus{}1)^{p\plus{}2}(x\minus{}1)$ $ \equal{}(x^2\minus{}1)^{p\plus{}1}(x\plus{}1)$ then the coefficients of $ x^{2p\plus{}1}$ is $ \minus{}(pq\plus{}1)$ and $ p\equal{}0$
Then $ P(x)\equal{}(x^2\minus{}1)(x\plus{}1)\equal{}x^3\plus{}x^2\minus{}x\minus{}1$ and we already got this solution

5) $ a\equal{}2$ and $ p\minus{}q\equal{}\minus{}1$
Then $ P(x)\equal{}(x\minus{}1)^p(x\plus{}1)^{p\plus{}1}(x\minus{}2)$ $ \equal{}(x^2\minus{}1)^p(x\plus{}1)(x\minus{}2)$ but then the coefficient of $ x^{2p}$ is $ \minus{}(p\plus{}2)$ and so $ p\equal{}0$
Then $ P(x)\equal{}(x\plus{}1)(x\minus{}2)\equal{}x^2\minus{}x\minus{}2$ and this does not respect the requirements

Hence the solutions :
$ P(x)\equal{}x^n$
$ P(x)\equal{}x^{n\minus{}3}(x^2\minus{}1)(x\plus{}1)\equal{}x^n\plus{}x^{n\minus{}1}\minus{}x^{n\minus{}2}\minus{}x^{n\minus{}3}$
$ P(x)\equal{}x^{n\minus{}2}(x\minus{}1)(x\plus{}2)\equal{}x^n\plus{}x^{n\minus{}1}\minus{}2x^{n\minus{}2}$


inequalities
inequalities proposed
algebra

the same resault holds for $ 3$ variables:

let $ a,b,c$ be positive real numbers less than $ 1$ such that $ a \plus{} c\leq 1$,prove that:

$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} 1} \plus{} \sqrt {(1 \minus{} b)^2 \plus{} (a \plus{} c)^2} \geq \left(1 \plus{} \sqrt {5}\right)\left(a \plus{} bc\right)$

[quote="silouan"]Prove that for any  $ x,y\in \left[ 0,1\right]$  we have that

$ \sqrt {1 \plus{} x^2} \plus{} \sqrt {1 \plus{} y^2} \plus{} \sqrt {(1 \minus{} x)^2 \plus{} (1 \minus{} y)^2}\geq (1 \plus{} \sqrt 5)(1 \minus{} xy)$[/quote]
Already solved by me,see here:
http://www.mathlinks.ro/viewtopic.php?t=161686

[quote="BaBaK Ghalebi"]Erken:what about the second one I posted,could it be done by the same method? :maybe:[/quote]
I'll try it.

[quote=silouan]Prove that for any  $x,y\in \left[ 0,1\right]$  we have that

$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}\geq (1+\sqrt 5)(1-xy)$[/quote]
[url=https://artofproblemsolving.com/community/c6h161686p904762]here[/url]


inequalities
trigonometry
geometry proposed
geometry

Hello!

Let $\triangle ABC$ be a non-obtuse triangle. Prove that : \[\sum_{cyclic}a\cos^{3}A\leq{{abc}\over{4R^{2}}}.\]

Just use the following formulas: $S=\frac{abc}{4R}=\sum \frac{1}{2}R^{2}\sin{2\alpha}$ and $a=2R\sin{\alpha}$. Then the inequality transforms into:
\begin{eqnarray*}\sum R a \cos^{3}{\alpha}&\leq& \sum \frac{1}{2}R^{2}\sin{2 \alpha}\Longleftrightarrow \\ \sum 2 \sin{2\alpha}\cos^{2}{\alpha}&\leq& \sum \sin{2\alpha}\Longleftrightarrow \\ \sum \sin{4\alpha}&\leq& 0 \Longleftrightarrow \\-4\sin{2\alpha}\sin{2\beta}\sin{2\gamma}&\leq& 0 \end{eqnarray*}
And the last one is obviously true.


fie x,y,z 3 numere pozitive, cel mult unul dintre ele putand fi 0 si avand suma 1/2.
demonstrati ca:

suma ciclica din 1/(x^2 +2*y^2+3*z^2) e mai mic sau egal decat 3*suma ciclica din 1/(x+2y)

am analizat problema si se poate face mai usor decat credeam eu. n-am apucat ieri sa analizez toate caile de atac ca sa zic asa, pt ca facusem problema la scoala intr-o ora.voi pune alta problema, mult mai interesanta:

determinati toate multimile finite A incluse in Z cu cel putin 2 elemente, care au proprietatea ca oricare ar fi x,y din A, x diferit de y avem x+y+xy apartine lui A


Here is the IMO shortlist problems 2000, I got it from internet.


Let $ 0<x,y,z <1$Prove that:
$ x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x)\leq\frac {3}{4}$

I think this inequality is wrong because $ x \equal{} 1,y \equal{} 0,z \equal{} 0$  $ \implies$ $ x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x) \equal{} 1\leq\frac {3}{4}$ Contradiction!

I think $ x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x) \leq 1$

and proof of this inequality is easy :wink:

You can still take x, y, z arbitrarily close to 0,0,1 and obtain a contradiction.[/hide]

For showing the expression to be greater than zero, or

$ 0 < x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x)$,

[hide]

Let $ y \equal{} z \equal{} x$.

Then the expression can become $ 3x(1 \minus{} x) \equal{} 3x \minus{} 3x^2$.

Let $ x \equal{} \frac1n$ where $ n \ge 2$, for n a positive integer (even though

x, y, and z actually belong to the real numbers in the interval $ (0, 1)$.

Then $ 3x \minus{} 3x^2$ becomes $ \frac3n \minus{} \frac {3}{n^2}$.

The limit (as n -> oo) of $ (\frac3n \minus{} \frac {3}{n^2})$ is 0.
[/hide]


Hello!
I think I should put my question in this section.
"How to define a leap year?"

In the Gregorian calendar:  http://en.wikipedia.org/wiki/Leap_year#Gregorian_calendar

Thanks a lot for your info!  :)
******************************************************************************
In the past, some old people can calculate the day of a date with a palm ( in fact, doing come calculations with the aid of the fingers) very quickly.
I wonder if Western people also know this calculating method?
Note that I am not talking about the formulae given by some famous Mathematicians like C.F.Gauss.

I could. I just look at my knuckles and count.

A leap year usually occurs every year that is divisible by 4.
EXCEPT
The ones that end in 2 zeros are different.
Those are only leap years if they are divisible by 400 at least says my programming teacher

Was going to post about the "astronomical" defintion before but, I thought it would have been covered in the Wikipedia article.
Since the Earth takes $ 365.25 \ \text{days}$ to revolve around the Sun, or the "astronomical" year, this fourth of a day is made up every four years.
Sorry if that is redundant.  :wink:

Requirement:
1. Knowing how to read Chinese;
2. knowing the concept of modulus.

Link: http://163.27.14.11/dean/fac/46/junior/0304/030413.pdf

Enjoy the document!  :)


geometry
3D geometry
combinatorics unsolved
combinatorics

Show that every n cube has a hamiltonian circuit.

Please post problems like this in Combinatorics! Some combinatorics problems can be solved by superior algebra, but this is not a reason to post them all in teh Superior Algebra section.

For this particular problem, I posted a solution at http://www.mathlinks.ro/Forum/viewtopic.php?t=15416 .

  darij


search
number theory unsolved
number theory

Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n\plus{}b*3^n\plus{}c$ is a perfect square

Perhaps you meant $ 0$ instead of $ o$.
And, use \cdot instead of *, the difference:

$ \cdot,*$

[b]Conjecture:[/b]  Let $ s_n$ be a sequence satisfying a linear recurrence with integer coefficients such that $ s_n$ is square for all $ n$.  Then $ s_n \equal{} r_n^2$ where $ r_n$ also satisfies a linear recurrence with integer coefficients.  (This would generalize the well-known problem that's been posted before in the case that $ s_n$ is itself a polynomial.)

If the conjecture is true, then the answer is "no."

[quote="hxy09"]Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n \plus{} b*3^n \plus{} c$ is a perfect square[/quote]
See here: I think they are both interesting.http://www.mathlinks.ro/viewtopic.php?p=1382629#1382629

[quote="hxy09"]Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n + b*3^n + c$ is a perfect square[/quote]
It's very nice problem.
Hxy09, can you show me your solution?
And here is very hard problem (at least for me) similar to your prob
Does there exist $ a;b;c\in Z$ such that for every postive integer $ n$, $ a4^n+b6^n+c9^n$ is a perfect square.

[quote="tronghieu"][quote="hxy09"]Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n + b*3^n + c$ is a perfect square[/quote]
It's very nice problem.
Hxy09, can you show me your solution?
And here is very hard problem (at least for me) similar to your prob
Does there exist $ a;b;c\in Z$ such that for every postive integer $ n$, $ a4^n + b6^n + c9^n$ is a perfect square.[/quote]
$ a=b=0$, $ c$ is a perfect square, dear.

No,$ (a;b;c)\equal{}(k^2;0;0)$ is also solution. Allnames, can you solve it?

[quote="tronghieu"][quote="hxy09"]Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n + b*3^n + c$ is a perfect square[/quote][/quote]
It's very nice problem.
Hxy09, can you show me your solution?
And here is very hard problem (at least for me) similar to your prob
Does there exist $ a;b;c\in Z$ such that for every postive integer $ n$  $ a4^n + b6^n + c9^n$ is a perfect square

Very sorry that I have no solution for this problem :( (this is why I put it in "unsolved problem")
Indeed,it is just a conjecture for me,one thing should me mentioned is that for every $ n$ $ a*2^n + b*5^n + c$ is a perfect square,then $ a=b=0$.This is a problem from Berlarus.It can be easily proved by [b]limit[/b](similar with http://www.mathlinks.ro/viewtopic.php?p=1382629#1382629)
The main idea is $ 5>2^2$
But it failed when $ 3<2^2$
This method can't do with your problem: $ a4^n + b6^n + c9^n$ either   :(
By the way,can you post the solution of your problem?Thank you very much!

[quote="tronghieu"]No,$ (a;b;c) \equal{} (k^2;0;0)$ is also solution. Allnames, can you solve it?[/quote]
Sorry, trong hieu. But you asked "Does there exist" and I showed a triple satisfying that.

[quote="tronghieu"]I have an idea.
Consider the sequence $ b_n \equal{} 3x_n \minus{} x_{n \plus{} 1}$. Not hard to solve that for n is large enough then $ b_n > b_{n \plus{} 1}$. Morever, b_n>0 with every $ n\in N$. Thus, this sequence tend to a positive integer $ M$.
So $ 3x_n \equal{} x_{n \plus{} 1} \plus{} M$ for n is large enough.[/quote]

Sorry,but I think $ b_n < b_{n \plus{} 1}$ instead of $ b_n > b_{n \plus{} 1}$ :?:
Or maybe I mistook what you thought somewhere?

[quote="ywl"]dear hxy09,
using limit methods, I also came to a dead end on this problem.
There is, however, a nice theorem by vess that I believe is the thing you want.
http://www.mathlinks.ro/viewtopic.php?search_id=510938072&t=26797
[/quote]

Thank you so much for such a nice link :lol:
After the result of Pisot,all problems above are trivial  :D

Dear hxy09, that is my mistake. :blush: Thank yml  so much.
Allnames:  :oops:  ( you're right).

[quote="tronghieu"][quote="hxy09"]Does there exist $ a,b,c \in Z$ and $ ab \neq o$
such that for every positive integer $ n$
 $ a*2^n + b*3^n + c$ is a perfect square[/quote]
It's very nice problem.
Hxy09, can you show me your solution?
And here is very hard problem (at least for me) similar to your prob
Does there exist $ a;b;c\in Z$ such that for every postive integer $ n$, $ a4^n + b6^n + c9^n$ is a perfect square.[/quote]

Hi my friend,I posted a solution of this nice problem in http://www.mathlinks.ro/viewtopic.php?p=1722570#1722570
The same method goes for the initial problem,we only need a tricky start:let $ f(n) = a*2^{n} + b*3^{n} + c$ and consider the sequence $ f(2n) = a*4^{n} + b*9^{n} + c$,then apply the same method in the previous link :lol:

by the way,Merry Christmas!!!


f(0)=s

Note that s>-1 so s+1>0.

if x=0 then f(f(y)) = y+ys + s

so f(x) is one-to-one since s+1>0.

Now if you let y=0 then f(x+xf(x)+s) = f(x)

so xf(x) = s

hence f(x) = s/x and since f(x) is increasing we should have f(x)= - a/x

for some a>0. where f(0)= -a.

Once can easily see that this leads to s=0.

So f(x) = 0 for all x.

[quote="cefer"]Babak, I think your solution is false ,because it doesn't satisfy $(2)$[/quote]

but it doesn't say that f(x) is strictly increasing...  :!:

I'm having a hard time with this now:

If f(x) = -x/x+1

then we need f(x+xf(x)) = f(x)

or that x+xf(x) = x

or that xf(x) = 0 ????

[quote="kudlaty"]Let $S=(-1,\infty)$.Find all functions $f: S \mapsto S$ satisfaying the following conditions:
(1)  $f(x+f(y)+xf(x))=y+f(x)+yf(x)$ for all $x,y \in S$.
(2)$\frac{f(x)}{x}$ is icreasing in $(-1,0)$ and $(0,\infty)$.[/quote]

A such function does'nt exist ! :D

Warning, the correct IMO problem is  $f(x+f(y)+xf(y))=y+f(x)+yf(x),\ \forall x,y \in S$.
 :cool:

Sorry,I made an ugly mistake in first condition.I have corrected it.


3.14 is finally here! Any special plans?
I look forward to short talks about $\pi$ and eating pie.

edit- only 4 minutes late :(

Is 3.14 officially celebrated as $\pi$ day or is it just some common convention ??  :?

Happy 3.14159265358979323846264338327950288419716975 day!

It's probably just a common convention.

PS: Ironically, Albert Einstein's birthday is March 14.

Maybe it`s not just a convention because in Romania there is a contest called "Mister PI" which was held today. Anyways Happy $\pi$ day!

Does this mean on 6/28, we get double the pi, and double the pie?

Haha, today I was writing the date down on a test...then i saw 3/14 and I thought, "cool, pie. hey, this happens every year, maybe there should be a 'Pie Day.'" I remembered later that there was already something like this and that might explain why people were asking me if i ate any pie yesterday (i thought it was really weird at the time)

our school held a math pi day competition

i memorized pi by 30 digits XDXDXD

im such a loser  :whistling:

What someone got this topic up before me???? :what?: visit my blog:[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=200]www.artofproblemsolving.com/Forum/weblog.php?w=200[/url]

[quote]Go to:
http://oldweb.cecm.sfu.ca/projects/ISC/dataB/isc/C/1pi10000.txt[/quote]

wtf is this?? thats not pi... :?

[quote="Apprentice"]Go to:
http://oldweb.cecm.sfu.ca/projects/ISC/dataB/isc/C/1pi10000.txt[/quote]

What is that? Because it's definitely not $\pi$.

We should have an $e$ day, except it's over a month too late.

I hope everyone also has a good Ides of March tomorrow.

Pi day actually would have been 414 years ago (3/14/1592).

Hey did any of you guys eat pi? I didn't  :(

How about july 22?

After all that was our childhood pi :roll:

                      [size=200] [color=red] [b]22/7[/b][/color][/size]

[quote]How about july 22?

After all that was our childhood pi [/quote]

Hmm.....interesting idea.....I'm just wondering....here in India I came to know about this pi day convention only because of AoPS.......

At least where I'm located (Bangalore) math seems to have quite a bit to catch up with the world....  :(

How come this particular page looks different now ?? :?   The page seems to have kind of expanded towards the right ?? Or is it some problem with my browser ?  :(

I have a strange book containing the first $300,000$ digits of $\frac{1}{\pi}$.  :|  Who would want to print that?  :huh:

[quote="jhollenbeck"]I think I Posted too much pi ;) ill fix it now

Hooray!  I deleted 15,000 digits and the browser went back to normal :D[/quote]

But doesn't the forum have some automatic thing to limit the number of characters in a line ??  :?

It seems not.

Anyway, after a while you can make it all up. Nobody will notice.
As for myself, i am happy with 3.141592!

[quote="Apprentice"]Go to:
http://oldweb.cecm.sfu.ca/projects/ISC/dataB/isc/C/1pi10000.txt[/quote]
Hey, I figured it out.
These are the first 10,000 digits of $\frac1{\pi}!$

I have found another pi day : April 31.
If you take 3-14, you may as well take 31-4 and no questions asked.

[quote="Klebian"]
Hey, I figured it out.
These are the first 10,000 digits of $\frac1{\pi}!$[/quote]

Err.......are you sure that stuff is accurate ??  :!:

It does not matter. Check the first five digits and that's all that counts.

Does anyone have an idea as to what value of $\pi$ is used for those ten thousand digits on that page ??


geometry
geometric transformation
rotation
linear algebra
matrix
conics
ellipse

How do I know that x^2-xy+y^2 = n/m is Elliptic equation?
And this equation has xy's term,how can I eliminate it? Can I do it by using rotation matrix? :?

[quote="blahblahblah"]a) Check the discriminant $\Delta=b^2-4ac$
b) Yes, a rotation matrix is the usual method.[/quote]

What is this rotation matrix?? I never across this term during my school or Engineering days :-((

Start with an ellipse with axes parallel to the coordinate axes:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

To rotate the ellipse, use different parameters: $\epsilon = \pi a b$ for the ellipse area and $\beta_0 = \frac a b$, $\gamma_0 =\frac b a$. Clearly, $\beta_0$ and $\gamma_0$ are not independent, because $\gamma_0 \beta_0 = 1$.

$\gamma_0 x^2 + \beta_0 y^2 = \frac{\epsilon}{\pi}$

This equation can be written in a matrix form:

$X^T \cdot E \cdot X = \frac{\epsilon}{\pi}$

where

$X = \left( \begin{array}{cc} x \\ y \\ \end{array}\right)$, $E = \left( \begin{array}{cc} \gamma_0 & 0 \\ 0 & \beta_0 \\ \end{array}\right)$

and the superscript T means transposed matrix (rows become columns and columns rows). Now it is easy to rotate the ellipse using a rotation matrix $R$, because the area $\epsilon$ does not change in the rotation.

$R(\phi) = \left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\sin{\phi} & \cos{\phi} \\ \end{array}\right)$

The equation of the rotated ellipse will be

$X'^T \cdot E' \cdot X' = \frac{\epsilon}{\pi}$, where $X' = R(\phi) \cdot X$

$X^T \cdot R(\phi)^T \cdot E' \cdot R(\phi) \cdot X = \frac{\epsilon}{\pi}$

from which you can see that

$E = R^T(\phi) \cdot E' \cdot R(\phi)$
$E' = R(\phi) \cdot E \cdot R^T(\phi)$

That's because for a rotation matrix, $R^{-1}(\phi) = R(-\phi) = R^T(\phi)$. If you now do the matrix multiplication, then

$E' = \left( \begin{array}{cc} \gamma & \alpha \\ \alpha & \beta \\ \end{array}\right) = \left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\sin{\phi} & \cos{\phi} \\ \end{array}\right) \cdot  \left( \begin{array}{cc} \gamma_0 & 0 \\ 0 & \beta_0 \\ \end{array}\right) \cdot \left( \begin{array}{cc} \cos{\phi} & -\sin{\phi} \\ \sin{\phi} & \cos{\phi} \\ \end{array}\right)$

where

$\gamma = \gamma_0\ \cos^2{\phi} +  \beta_0\ \sin^2{\phi}$
$\beta = \beta_0\ \cos^2{\phi} +  \gamma_0\ \sin^2{\phi}$
$\alpha = (\beta_0 - \gamma_0)\ \cos{\phi}\ \sin{\phi}$

$\gamma x'^2 + 2\alpha x'y' + \beta y'^2 = \frac{\epsilon}{\pi}$

Again, the parameters $\gamma, \beta, \alpha$ are not independent, because

$\gamma \beta - \alpha^2 = \text{det} E' = \text{det} R(\phi) \cdot \text{det} E \cdot \text{det} R^T(\phi) = 1 \cdot \gamma_0 \beta_0 \cdot 1 = 1$

For your particular equation, you have to normalize, in order to achive this. Multiply the equation by $\frac {2} {\sqrt{3}}$:

$\frac {2}{\sqrt{3}} x^2 - \frac{2}{\sqrt{3}} xy + \frac{2}{\sqrt{3}} y^2 = \frac{2}{\sqrt{3}} \frac n m$

Then $\gamma = \frac{2} {\sqrt{3}}$, $\beta = \frac {2} {\sqrt{3}}$, $\alpha = -\frac {1} {\sqrt{3}}$ and  $\gamma \beta - \alpha^2 = \frac 4 3 - \frac 1 3 = 1$. The ellipse area is therefore $\epsilon =  \pi \frac{2}{\sqrt{3}} \frac n m$.

[quote="blahblahblah"]Yetti, may I ask what your math background is?  It seems quite impressive.[/quote]

It would be impressive if I was a high school student, but I am an (experimental) physicist, who likes geometry. Rotating and squashing ellipses with constant area is important in the beam transport in particle accelerators. In this case, the variables x, y (usually denoted x, x') are the particle displacement x from the beamline axis and the particle divergence from the beamline axis, which is proportional to the particle radial velocity x'. The ellipse is then called the beam trace ellipse and its area the beam emittance. From a certain theorem in analytical mechanics (Liouville's theorem), it follows that the ellipse area has to remain constant, i.e., no focusing element in the beamline (such as a magnetic quadrupole lens) can decrease this area - these elements only rotate the ellipse or squash it without changing its area. Either the beam is focussed almost to a point, but the divergence increases, or the particles travel almost parallel to the beamline axis, but the beam cross-section area increases. To sucessfully transport beam in the beamline for huge distances, the beam emittance from a particle ion source must be small to begin with. Hence, the beam emittance is an important beam parameter to measure.

Oh, I wasn't particularly impressed by the fact that you know how to rotate an ellipse; I guess most people who know how to diagonalize a matrix can do that :)   I was more referring to the use of homothety, etc , in your geometry solutions, but happened to post in this thread.

I try to find the length of major axis from x^2 - xy + y^2 = n/m. In the article the length is  :sqrt:(2n/m),but I can't do it from my method.So I added an attachment with this reply. Please check it for me where they have something wrong and I'm not sure a picture of an ellipse correct too. Thank you.

You made two errors in signs. First, if $n = u^3 + v^3$, then

$n = (u + v)(u^2 - uv + v^2)$

while you used the plus sign in front of $uv$. But then you used the correct ellipse equation

$x^2 - xy + y^2 = \frac n m$

i.e., this sign error is just a typo. Next, you used a rotation matrix $R(\theta)$, such that $X = R(\theta) \cdot X'$, or

$\displaystyle \left( \begin{array}{cc} x \\ y \end{array}\right) = \left( \begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \\ \end{array}\right) \cdot  \left( \begin{array}{cc} x' \\ y' \end{array}\right)$

and tried to calculate the angle $\theta$. All the subsequent equations are correct, up to $\theta = 45^o$, which is suspicious. The equation $\cot{2\theta} = 0$ has another solution in the interval $[0, 360^o)$ (or $(-180^o, 180^o]$), namely $\theta = 315^o$ (or $\theta = -45^o$). You can make the cross-term disapper in two possible ways - by rotating the coordinate system to make the ellipse main axis coincide either with the x'-axis or with the y'-axis. Hence, you cannot pick one solution for the angle $\theta$ at random and ignore the other. You know that in the 1st quadrant of the x-y coordinate system, the ellipse should intersect the line $y = x$ at a point with coordinates $[x, y] = [p \frac{\sqrt 2}{2}, p \frac{\sqrt 2}{2}]$, where $p$ is the length of the ellipse main axis. In the x'-y' coordinate system, you want these coordinates to be $[x', y'] = [p, 0]$. Since a rotation by the angle $\theta$ followed by a rotatin by the angle $-\theta$ is an identity, you can write $R(-\theta) \cdot R(\theta) = I$ and $R^{-1}(\theta) = R(-\theta)$. Hence,

$X' = R^{-1}(\theta) \cdot X = R(-\theta) \cdot X$

$\displaystyle \left( \begin{array}{cc} x' \\ y' \end{array}\right) = \left( \begin{array}{cc} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \\ \end{array}\right) \cdot  \left( \begin{array}{cc} x \\ y \end{array}\right)$

and substituting $\theta = 45^o$, $\cos{\theta} = \sin{\theta} = \frac{\sqrt 2}{2}$ and $x = y = p \frac{\sqrt 2}{2}$

$\displaystyle \left( \begin{array}{cc} x' \\ y' \end{array}\right) = \left( \begin{array}{cc} \frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2} \\ \frac{\sqrt 2}{2} & \frac{\sqrt 2}{2} \\ \end{array}\right) \cdot  \left( \begin{array}{cc} p \frac{\sqrt 2}{2} \\ p \frac{\sqrt 2}{2} \end{array}\right) = \left( \begin{array}{cc} 0 \\ p \end{array}\right)$

Did I get the x'-coordinate greater than zero and the y' coordinate zero? No, I did not. So, I try the other solution $\theta = -45^o$, $\cos{\theta} = \frac{\sqrt 2}{2}$, $\sin{\theta} = -\frac{\sqrt 2}{2}$:

$\displaystyle \left( \begin{array}{cc} x' \\ y' \end{array}\right) = \left( \begin{array}{cc} \frac{\sqrt 2}{2} & \frac{\sqrt 2}{2} \\ -\frac{\sqrt 2}{2} & \frac{\sqrt 2}{2} \\ \end{array}\right) \cdot  \left( \begin{array}{cc} p \frac{\sqrt 2}{2} \\ p \frac{\sqrt 2}{2} \end{array}\right) = \left( \begin{array}{cc} p \\ 0 \end{array}\right)$

Ha! The solution $\theta = -45^o$ is the right one. BTW, you also have the line $x + y = m$ incorrectly drawn in fig. 1. This line has to cut segments of the length $m$ on both the x- and y-axes.


Let p odd prime divisor $a^2+9$, then p=1(mod 4). If n have odd divisor, then $2^n-1$ have prime divisor viev 3(mod 4), it give contradition. Therefore $n=2^k$.


trigonometry
calculus
calculus computations

Hi all,

Prove that whenever $ \minus{}\pi\leq \theta \leq \pi$, one has

\[ \left |1\minus{}e^{i\theta}\right | \geq \frac{2\left | \theta \right |}{\pi}.\]

Any hints?

Thanks!
Sev

$ |1\minus{}e^{i\theta}|\equal{}|e^{i\theta/2}(e^{\minus{}i\theta/2}\minus{}e^{i\theta/2}|$

$ \equal{}\left|(\minus{}2ie^{i\theta}2)\left(\frac{e^{i\theta/2}\minus{}e^{\minus{}i\theta/2}}{2i}\right)\right|\equal{}2\left|\sin\left(\frac{\theta}2\right)\right|.$

Can you finish it from there?

You could also think geometrically about two points on the unit circle, arriving in a similar place.

Visualize the graph of $ \sin\left(\frac{\theta}{2}\right)$ on the interval, and notice that it is concave downward on $ [0,\pi]$, so that it lies above the line $ y\equal{}\frac{\theta}{\pi}$.  Similarly, on $ [\minus{}\pi,0]$, it lies below the line $ y\equal{}\frac{\theta}{\pi}$, so $ \left|\sin\left(\frac{\theta}{2}\right)\right|\geq\frac{\left|\theta\right|}{\pi}$.


function
algebra
functional equation
algebra unsolved

Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ which satisfy \[ f(m + f(n)) = f(m) + n \] for all $m,n \in \mathbb{Z}$.

$m=0$ => $f(f(n))=n+f(0)$ (1)

Now, the hypothese implies that :

$f(f(m+f(n)))=f(f(m)+n)$

But $f(f(m+f(n)))=m+f(n)+f(0)$ with (1)
And $f(f(m)+n)=f(n)+m$ with the hypothese

So , $f(0)=0$ and so $f(f(n))=n$ (2)

Now, take $n=f(a)$ and $m=b$, you obtain:

$f(b+f(f(a)))=f(b+a)=f(b)+f(a)$ (3)

As (2) implies that $f$ is bijective,
(2)(3) implies that $f(n)=n$ because (3) is the Caushy's equation.

[quote="10000th User"]Sorry for bringing up an old post but I do see there is a missing function from Labiliau's solution. Please correct me if I am wrong.[/quote]

ie. $f(n) = -n$?

Causchy's equation implies : $f(n)=nf(1)$ :)
Get this in the original equation :
$f(m+f(n))=(m+f(n))f(1)=mf(1)+nf(1)^2=mf(1)+n$
So $f(1)^2=1$

Actually $f(n)-n$ is solution as well. We notice that by cauchy's functional equation we must have that $f(x)=cx$.
So we get that $f(m+cn)=cm+n$ which means that $c^2n=n$. Which means that $c=+/-1$ so we have that either f(x)=x or f(x)=-x.

let m=0 then f(f(n))=n+f(0) now let n=m-f(0) then it is proved that f is surjective
let x be a number such that f(x)=0 then let n=x    f(m)=f(m)+x then x=0  so f(f(x))=x now substitute n with f(n) so f(m+n)=f(m)+f(n) and now easy to say f(n)=kn and only f(n)=n or f(n)=-n is accepted

Let $P(m, n)$ denote the given assertion.
Now observe that $P(0, n)$ gives us $$f^2(n) = n + f(0),$$ and therefore $f$ is bijective.
This also gives us $f(0) = 0$ by $P(0, 0)$. so $f^2(n) = n$.
Now, $P(m, f(n))$ gives us $f(m+n) = f(m) + f(n)$, which is nothing but Cauchy's Functional Equation, and therefore $$f(x) = cx$$ for some constant $c$.
Subbing into the original equation, $c^2= 1 \implies c = 1, -1$, or $\boxed{f(x) = x, f(x) = -x}$. These clearly work, so we are done. $\square$


abstract algebra
superior algebra
superior algebra unsolved

Is there a number field other than $ \mathbb{Q}$ which has no unramified proper extensions?

Yes, $ Q(\sqrt{\minus{}1})$ for instance, but it's not immediate to prove it works: it's actually a good exercise.  :D

It has obviously no abelian unramified extension, but this doesn't help us much. Do I have to calculate a discriminant?


What are the values of $k$ such that the roots of the quadratic equation $2x^2+kx+12=0$, $p$ and $q$, satisfy $1<p<2$ and $3<q<4$?

[hide]There are none because the sum of the roots is $12/2=6$, but $4<p+q<6$!

Right??? :?[/hide]

[hide]

hmm so basically what i did was find the bounds on $k$ using the information I have for $p,q$

we know the graph is of a parabola, and that it is concave up

thus, at the first root,

$2(1) + k + 12 > 0 > 2(4) + 2k + 12$ (1)

and for the second root,

$2(9) + 3k + 12 < 0 < 2(16) + k(4) + 12$ (2)

in (1), we know that $1<p<2$, so we put $1$ and $2$ into our quadratic equation to find the bounds on the values of the equation, which then gives us bounds on $k$, giving us $-14<k<-10$

similarly, for (2) we know that $3<q<4$, so we put $3$ and $4$ into our quadratic equations, giving us $-11<k<-10$

but since we want both conditions to be satisfied, we take the interception of these two sets of bounds, so $-11<k<-10$ should satisfy both conditions...

i hope thats right

[/hide]

[quote="gotztahbeazn"][quote="chess64"]There are none because the sum of the roots is $12/2=6$, but $4<p+q<6$!

Right??? :?[/quote]

i think you mean the product of the roots is $12/2 = 6$, whereas the sum of the roots is $-k/2$[/quote]

oops, got them mixed up  :oops:

Yes.  However, I am much indebted to you for pointing out that $4<p+q<6$, as it led me to the following, much more concise (than I had earlier) [hide="solution."]First, we state that $k$ is real.  We know that $x^{2}-(p+q)x+pq=0$, so $k=-2(p+q)$.  Since $p$ and $q$ are real, then $k$ must be real.

Now we can use some quadratic identities.  We know that $\displaystyle p+q=-\frac{k}{2}$, and, as chess64 pointed out, $4<p+q<6$, so we have $-12<k<-8$.  However, since $p$ and $q$ must be real, we must have $k^{2}-96>0$, or (since $k$ is negative) $k<-4\sqrt{6}$.  Since $-12<-4\sqrt{6}<-8$, our final answer is $-12<k<-4\sqrt{6}$.
[/hide]EDIT :  No, it didn't either.  Here is my new solution, keeping the parts in the last one that $k$ must be real and $k<-4\sqrt{6}$.  Here is my new one :
[hide]We know that $\displaystyle 1<\frac{-k-\sqrt{k^{2}-96}}{4}<2$, or, by a little manipulation, $\displaystyle k+4<-\sqrt{k^{2}-96}<k+8$.  Each of these is negative, so we can square these and obtain

$k^{2}+16k+64<k^{2}-96<k^{2}+8k+16$, or $-14<k<-10$—WAIT A MINUTE—this looks suspiciously familiar.  I won't spell it out, but if we carry out this procedure for the other one, we get esssentially the same result as that given above, only in a more roundabout method.[/hide]

So, gotztahbeazn was right and I was wrong.  The method I just outlined (the right one), however, I find very easy to prove to provide necessary and sufficient boundaries.  Gotztahbeazn's however, I can prove in this particular case.  My question is now this :  gotztahbeazn, did you derive that this was necessary and sufficient specially for the problem, did you derive it earlier, or is it more widely known?

[quote="gotztahbeazn"]uh... how is ur solutiom more "concise"?

your bounds on $k$ are larger than mine

plus, if you try when $k=11$, you see that one of the roots is $4$

so yea i think some of ur reasoning might be wrong boy soprano[/quote]
You should check your claims as well. If $k=11$, $2x^2+11x+12=0$ does [b]not[/b] have $x=4$ as a root. Check what you are talking about before criticizing Boy Soprano's answer.

Yes, yes, yes.

Gotztahbeazn made a typo.  He meant $-11$, in which case 4 is indeed a root.  The fact is that I take a long time writing responses, and when I started to write my first response, the only other one was chess64's.  By "more concise," I meant "more concise than my previous solution."  Upon submitting my first solution, I saw gotztahbeazn's solution, checked with $-10$, and saw that I was wrong.  While I was editing my post, these other two popped up.  I have since completed my edit.  Gotztahbeazn was actually right.

However, I do think, gotztahbeazn, that you should have made it more clear that your method was guaranteed to produce necessary and sufficient conditions, unless, as I edited my post above to say, it is more widely known.  So, which is it?


geometry
geometric transformation
reflection
geometry unsolved

A quadrangle  $ABCD$   is inscribed in a circle $(O)$. Perpendiculars $AA_2 , BB_2$  are dropped on  $CD$  from points  $A , B$ . Perpendiculars $BB_1 , CC_1$ are  dropped on $DA$  from points  $B, C$  .Perpendiculars $CC_2 , DD_2$  are dropped on  $AB$ from points $C, D$ . Finally , from points  $D , A$  we drop perpendiculars  $DD_1 , AA_1$  on $BC$ . Prove that :

a)	The segments $A_1A_2 , B_1B_2 , C_1C_2 , D_1D_2$  are equal and lie on lines passing through the same point  $F$ , which is symmetric to point $O$ with respect to the centroid of the system of the points  $A , B , C , D$.
b)	The quadrangles  $A_1B_1C_1D_1$  and  $A_2B_2C_2D_2$ are similar to the quadrangle  $ABCD$ and are inscribed in circles with center $F$ .

[u]Babis[/u]

I'll just prove the concurrence of $A_1A_2,B_1B_2,C_1C_2,D_1D_2$ in $F$, since that's the part I found more interesting :). Sorry for the incomplete solution; I'm sure someone else will finish it.

Let $M,N,P$ be the midpoints of $AD,BC,CD$ respectively. $F$ is the reflection of $O$ in the midpoint of $MN$, so $MFNO$ is a parallelogram. In turn, this means that $MF\per BC,NF\perp AD$, so $F$ is actually the [url=http://mathworld.wolfram.com/Anticenter.html]anticenter[/url] of $ABCD$. It must then lie on the perpendicular from $P$ to $AB$ as well.

Now let $S,T$ be the points where $MF,PF$ cut $DD_2,DD_1$ respectively. The triangles $MDD_2,PDD_1$ are similar and isosceles, and the angles $\angle MSD,\angle PTD_1$ are both equal to the angle $\angle ABC$ (this follows immediately from the observation that $F$ lies on the perpendicular from $P$ to $AB$ made in the paragraph above), so, in particular, they are equal. This shows that the figures $MDD_2S,PD_1DT$ are similar, which implies that $\frac{D_2S}{SD}=\frac{DT}{TD_1}\ (*)$. Together with the fact that $SF,TF$ are parallel to the sides $DD_1,DD_2$ of $DD_1D_2,\ (*)$ implies that $SF,TF$ actually concur on $D_1D_2$, i.e. $F\in D_1D_2$.


You have a unit square. 1 point on each side is chosen. These points are connected. What is the minimum of the sum of the squares of the lengths of the sides of this new quadrilateral? The maximum?

[hide]By the Pythagorean Theorem, the sum of squares of lengths of sides of the quadrilateral is just sum of squares of lengths of fragments of sides of the square the points divided the sides into. The lengths of these fragments are just $a,1-a,b,1-b,c,1-c,d,1-d$ for arbitrary $0\leqslant a,b,c,d\leqslant 1$. Thus all we have to do is to minimize and maximize the quadratic $x^{2}+(1-x)^{2}$ for $0\leqslant x\leqslant 1$. It is easy to compute that the maximum is $1$ reached for $x\in\{0,1\}$ and the minimum is $\frac{1}{2}$ reached for $x=\frac{1}{2}$. Therefore the minimum of the sum of squares is $2$ and the maximum is $4$. [/hide]


Find the minimum and maximum value of $ \frac{\sum_{cyc}a^{2}\left( a\minus{}b\right) }{\left( \sum_{cyc}a^{2}b\right) \minus{}3abc}$

( for positive reals.)
taking a-->infinity you can see that there is no maximum
taking k for a lower bound
you arrive at k=3/((2)^(2/3))

You are right ,ywl

But

$ \max \frac{\sum_{cyc}a^2b\minus{}3abc}{\sum_{cyc}a^2(a\minus{}b)}\equal{}1$

and :

$ \text{inf}\;\frac{\sum_{cyc}a^2b\minus{}3abc}{\sum_{cyc}a^2(a\minus{}b)}\equal{}0$

Can you give your complete solution?
Thanks

Please, I need the solution :(
Can anyone post it?

$ P\equal{}\frac {\sum_{cyc}a^{2}\left( a \minus{} b\right) }{\left( \sum_{cyc}a^{2}b\right) \minus{} 3abc}$

But, bambang, do you want to complete your problem ? You can see above, $ \max P\equal{}\plus{}\infty,\text{inf}P\equal{}0$ :(


superior algebra
superior algebra unsolved

Let G be a group in which $ (ab)^{n} \equal{} a^{n}b^{n}$ for some fixed integer n>1 for all $ {a,b} \in G$, prove that:
	\[ (ab)^{n\minus{}1} \equal{} b^{n\minus{}1}a^{n\minus{}1}\]

Could someone show how to do this?
I can get it, but I'm not sure my steps are legitimate.

$ a^nb^n\equal{}(ab)^n\equal{}abab\ldots ab\equal{}a(ba)^{n\minus{}1}b$. Now we cancel the $ a$ and $ b$ to get $ a^{n\minus{}1}b^{n\minus{}1}\equal{}(ba)^{n\minus{}1}$.


analytic geometry
combinatorics unsolved
combinatorics

A kangaroo jumps from lattice poin to lattice point in the coordinate plane. It can make only two kinds of jumps: $ (A)$ $ 1$ to right and $ 3$ up, and $ (B)$ $ 2$ to the left and $ 4$ down.
$ (a)$ The start position of the kangaroo is $ (0,0)$. Show that it can jump to the point $ (19,95)$ and determine the number of jumps needed.
$ (b)$ Show that if the start position is $ (1,0)$, then it cannot reach $ (19,95)$.
$ (c)$ If the start position is $ (0,0)$, find all points $ (m,n)$ with $ m,n \ge 0$ which the kangaroo can reach.

Supose that the kangaroo do $ a$ jumps$ (A)$ and $ b$ jumps $ (B)$. Then we have $ a\minus{}2b\equal{}m$ and $ 3a\minus{}4b\equal{}n$. Solving this yields $ a\equal{}n\minus{}2m$ and $ b\equal{}\frac{n\minus{}3m}{2}$. In order for $ a$ and $ b$ to be nonnegative integers we clearly must have $ n\geq{3m}$ and $ m$ and $ n$ with the same parity. It is the case of $ (19,95)$, for wich $ a\equal{}57$ and $ b\equal{}19$, giving $ 76$ jumps.


Has anyone read the book? What do you think of the book?

Just borrowed a copy of the book from a friend. I really think it will be an interesting read for me.

It is a pretty awesome book. Although I would have been interested in more details about the TST, I have to say it is fairly accurate in describing the process of IMO and MOP. I can't speak for all the 2001 IMO team members, but based on my experience at MOP I have no reason to believe their depictions were inaccurate either.

The best part of the book is really the discussion of attitudes towards mathematics in general that puts the IMO into the big picture. The frequent shift in perspective cumulatively forms a broad view of what mathematics is to MOPpers. It even has a little mathematical content which I think successfully strikes a delicate balance in being basic enough for anybody to get the gist but still engaging the experienced problem-solver.

The books follows the 6 members of the 2001 US IMO team.

the book seems to explain in some detail the steps/insights taken by some team members for questions. For example, for Tiankai's problem 1, it tells of his troubles, strategies, beginning trials for the problem, etc.

I was just wanting to ask: How accurate are these depictions of the steps? Other opinions?

Just wondering cuz.... still trying to improve my own problem solving skill by observing the experts  :lol:

[quote="bpms"]What's the theme of the book?[/quote]

There are many.  :P  One of them is that people can be interested in math and still have fun, do sports, etc.  My favorite message from the book was related to psychology - the author stressed the "environmental" factor in intelligence and development; i.e., everyone basically starts out with the same tools.  It is the opposite of what we usually hear.

honestly, for me this whole thing, i mean anylising what factors made one a better mathmatician than the other is just a rubbish. imo, mathmatical talent got nothing to do with intelligence. one can be completely normal person just out of a sudden god decided to give him extraordinary mathmatical talent. no one can guess what is behind just becouse there's nothing behind it. sometimes it happens.

and personally, i would like to see other's approach to the problems. i mean -  you get a set of problems and what next? you just dont thinka about anything and start just solving it or you have some techniques? personally i dont think much, i just draw it/try on examples and try different techniques, tricks and so on. sometimes it works, sometimes not.

I read it and I enjoyed it very much. I especially liked the friendship between the team mates and the conflict.
The fact that there were Israeli and Korean members in the team made it all the more interesting to me because I'm a Korean but I lived in Israel for a long time.
The solutions of the competitors were really nice and gave me insight into the way they think.

It's an easy read and I recommend it to all


Euler
geometry
circumcircle
symmetry
geometric transformation
vector
invariant

Prove that on the circumcircle of a triangle there exist exactly [b]3[/b] points of whose [b][u]Simson lines[/u][/b] are tangent to the [b][u]Euler circle[/u][/b] of this triangle
and that points determine an equilateral triangle.

It's well-known that the envelope of the Simson lines of $ \triangle ABC$ is the Steiner deltoid $ \mathcal{S}$ tri-tangent to its Euler circle through $ X,Y,Z.$ Tangents of $ \mathcal{S}$ at  $ X,Y,Z$ (Simson lines) from an equilateral triangle due to the central symmetry of $\mathcal{S}.$ The poles of these Simson lines are the wanted points. Of course, such poles are not constructible (in general) since this equilateral triangle has the same orientation of Morley's triangle of $\triangle ABC.$

An alternate approach is using the transformation that I call Guzman's translation (sorry for such a random name) due to the spanish geometrician Miguel de Guzman.

Given a triangle $ \triangle ABC$ and its circumcircle $ (O).$ Let us define the transformation $ \mathcal{H}(A,v)$  that takes $\triangle ABC$ into another $ \triangle AB'C'$ with circumcircle $(O),$ such that $B'C' \parallel BC.$ The vector $ \vec{v}$ defines the translation that takes the  line $ BC$ into $ B'C'.$

It's easy to prove that $ \mathcal{H}$ has the following features:

1) Simson line $p$ with  fixed pole $ P$ on $ (O)$ WRT $ \triangle ABC$ is taken into another Simson line $p'$ with pole $P$ WRT $ \triangle AB'C'$ by the translation $ \vec{v}.$ As a result, orientation $p \parallel p'$ is invariant.

2) Euler circle $ \mathcal {E}$ of  $ \triangle ABC$ is taken into the Euler circle $\mathcal {E'}$ of $ \triangle AB'C'$ under the translation $ \vec{v},$ since the circumradius of $(O)$ is invariant and midpoints and feet of the altitudes from $ A$ to $ BC$ and $ B'C'$ lie constantly on guidelines of the translation.

3) $\triangle ABC$ can be transformed into an equilateral triangle $\triangle A'B'C'$ through the composition of (at least) two translations. This can be easily proved by angle chase. Hence, $ \mathcal {E}$ is taken into the incircle $ \mathcal {E'}$ of  $ \triangle A'B'C'$ and Simson lines  tangent to $ \mathcal {E}$  into the sidelines of  $ \triangle A'B'C'$ yielding the conclusion.

Historical note: Miguel de Guzman used this transformation to study  the Steiner's Deltoid of ABC. He showed synthetically that the orientation of the Steiner Deltoid is given by the sidelines of its Morley's triangle  and  its  shape only depends of the circumcircle.

See also [url]http://www.mathlinks.ro/viewtopic.php?t=244550[/url]. If Simson lines $ s_x, s_y, s_z$ with poles $ X, Y, Z \in (O)$ from that problem (such that $ s_x \parallel OX, s_y \parallel OY, s_z \parallel OZ$) concur at the 9-point circle center $ N,$ then Simson lines $ s_x' \perp s_x, s_y' \perp s_y, s_z' \perp s_z$ with diametrically opposite poles $ X', Y', Z' \in (O)$ are tangent to $ (N).$

Dear Maxim Bodgan and Vladimir, I'd like to attached the Guzman's investigation about the Steiner Deltoid. It  shows several  interesting  features about it and the proof for all the assertions that I mentioned above.

http://nonio.mat.uc.pt/PENSAS_EN02/experdescgeomet/10deltoide/deltoideacademia/00delten.htm

Perhaps a quick solution to this problem is using the result of  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266718]very nice and interesting[/url].

9-point center of $ \triangle ABC,$ which is incenter of the triangle $ \triangle XYZ$ bounded by these three Simson lines, is equidistant from the orthocenter and circumcenter of $\triangle ABC.$ This implies that $ \triangle XYZ$ is equilateral.

You already have to know that (at least) 3 Simson lines are tangent to the 9-point circle. Even then, this only implies that the triangle formed by these 3 Simson lines has the middle angle of $ 60^\circ.$

[quote="yetti"]You already have to know that (at least) 3 Simson lines are tangent to the 9-point circle. Even then, this only implies that the triangle formed by these 3 Simson lines has the middle angle of 60[/quote]
Indeed, disregard my latter comment.  :oops:

[b][u]Solution[/u].[/b] Let $ O$ (the circumcenter of $ \triangle{ABC}$) be the origin of the complex plane, and let $ a,b,c$ be the coordinates of $ A,B$ and $ C$, respectively.

Consider a point $ X(x)$ lieing on the circumcircle of $ \triangle{ABC}$ with the given property, $ R$ be the projection of $ X$ onto $ BC$,  and let $ M$ be the midpoint of $ [MH]$. Then $ MR$ is the [b][u]Simson line[/b][/u] of point $ X$ w.r.t $ \triangle{ABC}$. It's clear that $ MR$ is tangent to $ \omega  (\epsilon, \frac {R}{2} )$ ([b][u]Euler circle[/b][/u] of triangle $ ABC$)  if and only if $ \boxed{\epsilon M\perp MR}$

We have:
\[ \boxed{\begin{array}{lcl} h = a + b + c \ ; \ \epsilon = \dfrac{a + b + c}{2} \ ;\ m = \frac {x + a + b + c}{2} \\
 \\
r = \dfrac{1}{2}\cdot  (x + b + c - \dfrac{bc}{R^2}\cdot \overline{x} ) \\
 \\
r\neq m \ \Longrightarrow\ x\neq - \dfrac{bc}{a}\end{array}}\]
Now,

$ \epsilon M\perp MR \ \Longleftrightarrow \ \frac {m - \epsilon}{m - r}\in i\mathbb{R^*}$   i.e.  $ \frac {x}{a + \frac {bc}{R^2}\cdot \overline{x}}\in i\mathbb{R^*}$

Note that:

$ \bullet \ \ \frac {x}{a + \frac {bc}{R^2}\cdot\overline{x}} = \frac {x}{a + \frac {bc}{x}} = \frac {x^2}{ax + bc}$

$ \bullet \ \ \frac {x}{a + \frac {bc}{R^2}\cdot\overline{x}} = \frac {\frac {1}{\overline{x}}}{\frac {a}{R^2} + \frac {bc}{R^4}\cdot\overline{x}} = \frac {\overline{abc}}{\overline{x}\cdot (\overline{ax} + \overline{bc})}$

Hence:

$ \frac {x}{a + \frac {bc}{R^2}\cdot \overline{x}}\in i\mathbb{R^*} \ \Longleftrightarrow\ \frac {x}{a + \frac {bc}{R^2}\cdot \overline{x}} = - \overline{\frac {x}{a + \frac {bc}{R^2}\cdot \overline{x}}}$

$ \Longleftrightarrow\ \frac {x^2}{ax + bc} = - \frac {abc}{x(ax + bc)}\  | \cdot (ax + bc)\neq 0 \ \Longleftrightarrow\ \boxed{x^3 = - abc}$

So, there are exactly three points $ X_1 (x_1), \ X_2 (x_2), \ X_3 (x_3)$ with the property mentioned in the statement and they form an [b][u]equilateral triangle[/b][/u]  (because they are the [i]cubic roots[/i] of a complex number). Thus the problem is solved.


Let a, b, c > 0 sastify a+b+c  >=   a/b  + b/c + c/a. Prove that:

                              a^3c/b(a+c) + b^3a/c(a+b)  +  c^3b/a(b+c)  >= 3/2 :oops: [/hide][/code][/quote]

[quote="TV-TV-07"]Let $ a, b, c > 0$ sastify $ a\plus{}b\plus{}c  >\equal{}   a/b  \plus{} b/c \plus{} c/a$. Prove that:
$ a^3c/b(a\plus{}c) \plus{} b^3a/c(a\plus{}b)  \plus{}  c^3b/a(b\plus{}c)  >\equal{} 3/2$ :oops: [/quote]
By Cauchy-Schwarz inequality, we have:
$ \sum \frac{a^3c}{b(a\plus{}c)} \equal{}\sum \frac{a^2}{\frac{b}{c}(1\plus{}\frac{c}{a})}\ge \frac{(a\plus{}b\plus{}c)^2}{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}}$
$ \ge \frac{(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2}{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}}$
We will prove that:
$ \frac{(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2}{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}} \ge \frac{3}{2}$
$ \Leftrightarrow 2(\frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2})\plus{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c} \ge 3(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})$
Which is clearly true because by AM-GM:
$ \frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c} \ge 3$
$ 2(\frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2})\plus{}3 \ge 3(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})$

I'm sure it was a Romania 2005 [unused] problem.

Here's a solution.

Problem: For arbitrary 3 real numbers a,b,c satisfying $ a \plus{} b \plus{} c \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$, prove the following inequality.

$ \frac {a^3 c}{b(a \plus{} c)} \plus{} \frac {b^3 a}{c(b \plus{} a)} \plus{} \frac {c^3 b}{a(c \plus{} b)} \geq \frac {3}{2}$

Proof: Denote the left-hand side of the original inequality by LHS.
The Cauchy-Schwartz Inequality (degree 3) yields

$ ( \frac {a^3 c}{b(a \plus{} c)} \plus{} \frac {b^3 a}{c(a \plus{} b)} \plus{} \frac {c^3 b}{a(b \plus{} c)} ) ((a \plus{} c) \plus{} (a \plus{} b) \plus{} (b \plus{} c)) ( \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {a}{b} ) \geq (a \plus{} b \plus{} c)^3$

which implies

$ LHS ( \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {a}{b} ) \geq \frac {(a \plus{} b \plus{} c)^2}{2}$

however, $ a \plus{} b \plus{} c \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$, so this yields

$ LHS \geq \frac {(a \plus{} b \plus{} c)}{2}$.

Now, the original inequality is a direct consequence of $ a \plus{} b \plus{} c \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 3$ (by AM-GM inequality)

Equality holds in the original inequality iff $ a \equal{} b \equal{} c \equal{} 1$.


[b]JMC: Level Two: Round One: Questions[/b]
This first round has ten questions.  A correct answer to each question is worth two points.  A student may send in solutions to each question for an extra point.  The extra point will be awarded to solutions that show an extensive amount of creativity.  Calculators are permitted for this round, although they are not necessary.  The time given for this round is one week.

http://yashar.biz/jmc/jmc_files/JMC%20Level%20Two%20Round%20One%20Questions.pdf

[b]
Questions, comments, and solutions should be e-mailed to jiftedland at gmail dot com.[/b]

[b]When submitting solutions please have the subject in the form Level [X]: Round [X]: [Name]: [Grade][/b]

One point will be deducted for solultions not in that form.

Check out the JMC site for more information:
http://yashar.biz/jmc/

Released on 4/1/05 @ 12:00 AM Eastern Time. Solutions Due by 4/8/05 @ 12:00 AM Eastern Time.

Sending solutions to round one by April 3rd will get you a bonus point!

Neither of the links work for me....

Edit: Works now.

[quote="xxreddevilzxx"]do we have to TeX it?[/quote]

Oh god! I certainly hope not!

I'll try to get this in by Apr. 3 ... I've been missing a lot of these contests.

[quote="eryaman"][quote="xxreddevilzxx"]do we have to TeX it?[/quote]How would we tex an e-mail?  Attach a pdf or something?[/quote]

Since it's not mentioned anything about "texing," I suppose it's not necessary.

Although like in many contests, texing your answer wouldn't hurt.

By texing, you mean to attach a pdf document that contains $\LaTeX$ for your solutions.

I'm pretty sure that's what he means unless I didn't read it carefully.

Just note: I'll be thinking whether I'll sticky this or not because there are actually two contests (excluding mine) going on (hiitsme (sp?) and yours) now. Also, I read the thread concerning about contest called MSM (forgot who was in charge of that thread though) so I'm either going to build a thread with the links to the contests going on here by users (excluding the moderator :) ) and put the deadline beside it thus making it easier for people to see.

Another suggestion: Like what happened to hiitsme's (sorry I'm still spelling it wrong) contest, you might want to just in case check how people respond to your contest by putting in poll or thread with "Are you interested in competing my contest ~~~~" and so forth. This would give more time for people to know about your competition and not only this, it would give them time to decide whether they are interested or not.

At least that's why I put sign-up thread. :)

Hope this helps!

[quote="Silverfalcon"]Also, I read the thread concerning about contest called MSM (forgot who was in charge of that thread though) [/quote]

That would be me...

Also, to Farazy: On Level One, Problem One of your JMC, is question 7 supposed to be rounded, or am I just doing it wrong (probably).

I would just leave it as a fraction--but that's just my opinion :D

Fraction.  All answers are to be exact unless stated otherwise.  :D

Im suprised they havent stickied either thread...

TeXing is NOT required.  You can also write your answer as tex code that hasn't been processed.  Meaning you can do it how you would write your answer here on AoPS.

NUMBER 8 HAS BEEN CHANGED...I seemed to have typed it wrong... Amazingly I have worked all the problems 5 times and I never caught that.  :blush:

I think Tex code not processed is good, because Farazy can always check it out with the Preview button on forums :)

For Level 2, #10, it's not a trick is it? lol (don't answer if you can't give away info ;))

No it is not a trick.  There is a method to solving this problem.

 :D

One last question--he only flips the coins once right?

These aren't ordered by difficult are they like the AMCs? lol

Bob flip's two coins.  He flips each coin ONCE.  Nothing vague here.

There is no increasing order of difficulty.  Even with respect to the rounds!  You may find round two a lot easier than round one.  Everything is based on a point system anyway...So it doesn't really matter.  :P

Thanks :D I have a wary feeling that I've got that one wrong, but I'll have to stick with what I got :D

I'll send it tomorrow probably--around 12am right now.

I'm waiting for people to response on the thread for requesting certain contests to appear as stickied thread.

So far no response replied.

Wait little bit, I want to see what contests are active and put them on all on the stickied thread with hyperlinks. This would make things a lot simpler.

Solutions due by 4/8/05 @ 12:00 Eastern Time! BETTER GET MOVING IF YOU WANT TO GET IN!

aaaahhhh...I can't believe I missed number 9.  Should have checked over my work before e-mailing solutions

Arg ... lost 6 points on 2 silly mistakes. 1 question I didn't read and the other I didn't read.

[quote="Farazy"]JMC Round One finished yesterday

Here are the results:
http://yashar.biz/jmc/jmc_files/Level%20Two%20Scores.pdf

And here are the solutions:
http://yashar.biz/jmc/jmc_files/JMC%20Level%20Two%20Round%20One%20Solutions.pdf

 Good job to all that competed.  :D

Round two is coming soon!  :P[/quote]

Seeing that everyone is within 2 points of each other, hopefully it'll be harder next time...  :D

hey.... i got number 10 right but didnt get credit.... ???? i put 10.25... ok

and on number 3 lol i thought it was 11 symbols lol... i thought (,) was one symbol lol -_-;;; o well

and why did  i get 2 on #2. was my solution wrong??

oops lol messed u p9...

man i should have double checked my answers... i made a stupid division error on number 4 lol  :blush:


algebra
polynomial
function
trigonometry
real analysis
real analysis unsolved

Solve differential equation $ y'\equal{}\left(\frac{3x\plus{}y^3\minus{}1}{y}\right)^2.$

[hide="approach"] Multiply by $ y^2$ and notice that the right hand side is $ y'y^2 \equal{} \frac {1}{3}\frac {d}{dx}y^3$, so the ODE contains only references to $ y^3$.  After the appropriate substitution the ODE is still nonlinear, but can be manipulated into a linear one.  The general solution winds up being sums and products of polynomials and trigonometric functions with period $ \frac {\pi}{6}$.  Ugly.

If by chance you meant the ODE
\[ y' \equal{} \minus{} \left( \frac {3x \plus{} y^3 \minus{} 1}{y}\right)^2
\]
or perhaps
\[ y' \equal{} \left( \frac {3x \minus{} y^3 \minus{} 1}{y}\right)^2
\]
then there is a nice particular solution, $ y \equal{}  (3x)^{\frac {1}{3}}$.  Judging from your other 'easy' ODEs, I gather this is what you were going for :)[/hide]

Solution:
$ y^2 dy\equal{}(3x\plus{}y^3\minus{}1)^2 dx$
$ \frac{1}{3}d(y^3\minus{}1)\equal{}(3x\plus{}y^3\minus{}1)^2 dx$
Let $ z\equal{}y^3\minus{}1.$
$ \frac{1}{3}dz\equal{}(3x\plus{}z)^2 dx$
Now let $ 3x\plus{}z\equal{}t.$ Then $ \frac{1}{3}(dt\minus{}3dx)\equal{}t^2 dx$
$ (1\plus{}t^2)dx\equal{}\frac{1}{3}dt$
$ 3dx\equal{}\frac{dt}{1\plus{}t^2}$
We get $ \arctan t \equal{} 3x\plus{} C.$
Final answer:
$ y^3\plus{}3x\minus{}1\equal{}\tan (3x\plus{}C)$.  :wink:


AMC
AIME
USA(J)MO
USAMO
geometry
analytic geometry
graphing lines

Sorry, but it's to late right now to do the AMC and the AIME.  But there's always next year.

You just need to get at least the qualifying score for the AMC to take the AIME, which is t the highest, 120 for AMC 10, and 100 for AMC 12, but it can go lower.  Since you're going to be a sophomore, you just need the floor score on the AIME, which usually goes from 6-9 I think. If you get that as a sophomore, then you can do the USAMO.

And most people here would say that the AoPS books are the best.

[quote="Calculus"]I am a freshman this year, and i really want to make it to the USAMO. What type of scores would i need on the amc and aime to qualify?

Also, what is the best prep for amc/aime. Like books, summer camps, programs, etc.

Thanks for your help.[/quote]

If you took AIME this year, a 9 would put you in USAMO contention range.

Start with the AoPS series. There is a list of helpful resources on this site. Click the "Resources" link at the left side of this screen.

There are plenty of discussions about summer camps going on in Other Programs forums.

thanks, sorry for the lack of clarity, i meant when i take it as a 10th grader. Does the AOPS series cover all the math covered on the amc/aime?

[quote="MithsApprentice"]AMC, yes. AIME, not quite.[/quote]

Hmm... Really?

All the AIME I've seen can relate to a principle taught by Aops, but is there one that is out there? Probably didn't do that problem then - what is an example.

This year's #14 explores some principles not found in AoPS (although there could be a solution which relies on strategies that are found within the pages of said books). For example, there is a nice complex numbers transformation solution...uh...ask zabelman for that.

AoPS could probably take you to the point where you'd qualify for USAMO. But beyond that, you'd want to look into something a bit deeper.

All the tools you need for AIME you can get from AoPS, imo.  That certainly does not mean that by reading the AoPS series that you will be able to ace the AIME.  Knowing the tools is only part of the battle.  You must also be creative, playful, calm, and patient.  Also, you need lots of experience with solving AIME type problems.  This experience comes from practice!

As for #14 this year, looking at the solution in the math jam I don't see why you would say that AoPS doesn't cover those principles.  The analytic geometry section in Volume 2 is quite good.  The only tools that you really need for the problem are slope and distance from a point to a line (both of which are certainly in AoPS V2).  Of course, just knowing those two things won't solve the problem.  So in a sense, you are right that just reading AoPS wounldn't be enough.  But if you are looking for a book that would have prepared you for #14 I don't know of any better than AoPS.

A 115 can get you in the AIME (from the AMC 10).  I did it last year.

Sean, that number changes from year to year, depending on how well the everyone did overall. The only sure way to get into AIME by way of AMC10 is by getting a 120.

It's no gaurantee, but a 115 should do it, as the test gets 1.15 points harder every year.

This year i believe you had to get a 120 or higher for amc 10.

After the AOPS series is it recommended that i do lots of practice amc/aime problems and also work through the art and craft of problem solving? What other things should i try doing?

And also does anyone know about how long each aops book takes to complete?

[quote="joml88"] You must also be creative ,playful, calm, and patient[/quote]

I strongly agree. This year on AIME I couldn't be calm and patient and I ended up with a pathetic(compare to most people) 9. If I had been calm and patient, I would've gotten like a 12 or 13(assuming 1 or 0 careless mistake) this year.

So the most important thing for AIME other than knowledge, be calm and patient! You have to play around and be creative to solve some of the harder problems! And you won't be able to do that if you are not calm and patient!

A comment to the above poster: a 9 is never a pathetic score on the AIME. Based on the polls, you beat a majority of the people on the forum. For your information, the sum of my freshman and sophomore years' scores combined was a 9 :) .

As for the topic of this thread, the best way to do well in AMC/AIME, as far as I know, is to do the AoPS books. I did not get the books until last summer, but working on Volume 2 a few hours a day for a few weeks seemed to help a lot.

Another good source of practice is the Kalva website. This has both AIME and olympiad problems. If you use it to practice for AIME, make sure you solve the problems before looking at Kalva's solutions (even if you need help from someone else on some of them). Understanding how to get to the solution is more important than understanding the solution.

do not always rely on Kalva's solution when you correct.

ask us. and make sur eyou  keep up witht he book work, because that is where you learn most of your mathematical knowledge. Sure, practice makes perfect, but you gotta know your math to do that, and AoPS2 is perfect just for getting down the fundamentals.

[quote="Calculus"]After the AOPS series is it recommended that i do lots of practice amc/aime problems and also work through the art and craft of problem solving? What other things should i try doing?

And also does anyone know about how long each aops book takes to complete?[/quote]

The Art and Craft of Problem Solving is definitely a reasonable follow-up, but it's also quite a step higher than the difficulty of the AoPS books, in my opinion at least. Personally, I'd recommend you not worry about Art and Craft until you have your eyesights set on scoring well in USAMO, but I'm guessing there will be some that disagree.

Practice is of the utmost importance (and I'm being a huge hypocrite in making this statement) so make sure to do loads and loads of problems.

As for AoPS book 1, it took me about 2 years on and off to finish it. As for book 2, well...I'm still not quite done. You should be able to get through both books in a year, working everyday at a reasonable pace. I'd say a chapter or two a week is reasonable...but it's not as easy at it sounds.

[quote="JGeneson"]A comment to the above poster: a 9 is never a pathetic score on the AIME. Based on the polls, you beat a majority of the people on the forum. For your information, the sum of my freshman and sophomore years' scores combined was a 9 :) .
[/quote]

But a 9 get me to USAMO, I'll live(but it probably won't). But for now, I'll say it's a horrible score because most people I know from this site beat me.

[quote="origamimasterjared"]Even a zero's a pretty good score.[/quote]

eh not if your goal's USAMO...

[quote="beta"]But a 9 get me to USAMO, I'll live(but it probably won't). But for now, I'll say it's a horrible score because most people I know from this site beat me.[/quote]

Many times, it's not fair to compare yourself to members of a certain community. At MOP last summer, I began to feel like I was mentally challenged. (And playing cards every night with USAMO Winners and HM didn't help.) Then I came home and went off to a language studies and writing camp in China and seeing my classmates struggle with some pre-calc or algebra they were doing over the summer made me feel oodles better. If you look around your normal real life environment, I'm sure you'll find your position on the mathematical hierarchy quite satisfying. However, if you intend to come here and compare scores with people, a good portion of whom have experience in the USAMO already, be ready to feel slightly uncomfortable at your position. It has nothing to do with how good you are, but with who you're playing against.

A 9, by most predictions, will get you into the USAMO. That means you've made it into the ranks of the top 250, and you should be proud. But even if it doesn't, 9 is a perfectly respectable score. I'm sure there are plenty of people besides yourself that are proud of it (your parents for one).

Yeah - I wasn't patient either and scored a nasty 1 on the AIME. Instead, I was looking around at my classmates and making faces at them as my classmates gave up too. It was a hilarious scene with my classmates even starting to eat and getout their CD players but I really should have tried to focus. *sigh* Had I been calm, I may have made 2 or 3 this year (it sounds so pathetic compared to AoPS but statistically, it gets me to the top few of sophomores).

I'm in the same school as Fred, Dawson, Sunil, Richard... smart ppl. Makes me want to work harder though.

thanks a lot for the advice, more is still greatly appreciated

[quote="beta"]
But a 9 get me to USAMO, I'll live(but it probably won't). But for now, I'll say it's a horrible score because most people I know from this site beat me.
[/quote]

dont worry. I got a 9 too, but i am already eyeing towards USAMO. i am sure you and i will qualify. 9 is a high score, considering the high percentile of AoPSers among the testakers.


Sorry about the confusion yagron!a, b, c are real numbers such that a+b+c=abc .
Prove that 1/a^2+1/b^2+1/c^2<=1-(12/abc)^2 :mad:


inequalities
function
inequalities proposed

With $ a,b,c$ are sides of traingle. Prove that
 1.$ 3(abc)^2 \geq (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a^3\plus{}b^3\plus{}c^3)$
 2.$ \frac{a^3}{c(a\plus{}b\minus{}c)}\plus{} \frac{b^3}{a(b\plus{}c\minus{}a)}\plus{} \frac{c^3}{b(c\plus{}a\minus{}b)} \geq \sqrt{ \frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}}$

1) Use SOS, we have: $ Sa \equal{} \frac {3(b \plus{} c \minus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \minus{} \frac {a \plus{} b \plus{} c}{abc} \ge \frac {2(b \plus{} c \minus{} a)}{abc}$. Easily to see that $ Sa, Sb, Sc > 0$. We have done  :)

[quote="leedt26"][quote="2424"]With $ a,b,c$ are sides of traingle. Prove that
 1.$ 3(abc)^2 \geq (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3)$
 2.$ \frac {a^3}{c(a \plus{} b \minus{} c)} \plus{} \frac {b^3}{a(b \plus{} c \minus{} a)} \plus{} \frac {c^3}{b(c \plus{} a \minus{} b)} \geq \sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}$[/quote]
I think 2. must be
$ \frac {a^3}{c(a \plus{} b \minus{} c)} \plus{} \frac {b^3}{a(b \plus{} c \minus{} a)} \plus{} \frac {c^3}{b(c \plus{} a \minus{} b)} \geq 3\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}$[/quote]

$ < \equal{} > \sum \frac {a^3}{c} \plus{} \sum \frac {2a^3}{a \plus{} b \minus{} c} \ge  \sum \frac {a^3}{c}\plus{}2(a^2\plus{}b^2\plus{}c^2) \ge  3\sqrt {(a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)}$
It's easy SOS :P

[quote="can_hang2007"]

In addition, I found that the best constant $ k$ (the smallest value of $ k$) such that the inequality
$ (k \plus{} 3)a^2b^2c^2 \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3 \plus{} kabc),$
holds if $ k \equal{} 1.$ And there is a nice proof with AM-GM Inequality for this value of $ k$, I will wait for someone to figure it out.[/quote]

For $ k\equal{}1$ , it is equivalent to Gerretsen ;)

How about your proof can_hang2007 ??? Thanks you very much :)

[quote="can_hang2007"][quote="2424"]With $ a,b,c$ are sides of traingle. Prove that
 1.$ 3(abc)^2 \geq (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3)$
 [/quote]
This inequality is not true, dear Vuong. Let $ a \equal{} b \equal{} 1,$ the inequality becomes
$ 3c^2 \ge c^2(2 \minus{} c)(2 \plus{} c^3),$
or equivalently,
$ c^2(c \minus{} 1)^3(c \plus{} 1) \ge 0.$
Thi is not true for $ c \equal{} 0.9,$ (and many other values of $ c$ so that $ a,b,c$ are the side of a triangle).

In addition, I found that the best constant $ k$ (the smallest value of $ k$) such that the inequality
$ (k \plus{} 3)a^2b^2c^2 \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3 \plus{} kabc),$
holds if $ k \equal{} 1.$ And there is a nice proof with AM-GM Inequality for this value of $ k$, I will wait for someone to figure it out.[/quote]
I am very sorry,you true! :blush:
Thank you very much,Can_hang2007!
@anh Can:[hide] anh Can oi,e xin loi vi da khien anh mat thoi gian cho bai toan sai nay  :blush: .E xin tang anh bai nay coi nhu ta loi nha:CMR voi moi $ a,b,c > 0$ thi bdt sau luon dung.
 $ \frac {a}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {8(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \frac {12}{a \plus{} b \plus{} c}$. :) .
Bai nay k kho nhung dep ca ve hinh thuc va loi giai,e danh tang rieng anh.Dung chem' e nha! :lol:[/hide]

[quote="can_hang2007"]If $ a ,b,c \ge 0,$ then
$ 4a^2b^2c^2 \ge (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)(a^3\plus{}b^3\plus{}c^3\plus{}abc).$[/quote]
This inequality is [b]Problem 2.61[/b], page 115 of new book (where it is presented with a simple solution by AM-GM):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="can_hang2007"]

 I found that the best constant $ k$ (the smallest value of $ k$) such that the inequality
$ (k \plus{} 3)a^2b^2c^2 \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3 \plus{} kabc),$
holds if $ k \equal{} 1.$ [/quote]
Yes! We can get easily this result by $ uvw$ - method:
Let $ a \equal{} y \plus{} z,$ $ b \equal{} x \plus{} z,$ $ c \equal{} x \plus{} y,$ where $ x,$ $ y$ and $ z$ are non-negatives,
$ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2$ and $ xyz \equal{} w^3.$
Hence, $ (k \plus{} 3)a^2b^2c^2 \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3 \plus{} kabc)\Leftrightarrow f(v^2)\geq0,$
where $ f(v^2) \equal{} 9(k \plus{} 3)u^2v^4 \plus{} (18 \minus{} 10k)uw^3v^2 \plus{} (k \plus{} 3)w^6 \minus{} 48u^3w^3.$
But $ f$ is increasing function (we can assume $ k\leq1.8$ :wink: ).
Hence, $ f$ gets a minimal value, when $ v^2$ gets a minimal value, which happens when two numbers from $ \{x,y,z\}$ are equal
or  when maybe $ v^2 \equal{} 0$ (for the last our inequality is obviously true).
Id est, it remains to check what happens when $ y \equal{} z \equal{} 1.$
In this case $ (k \plus{} 3)a^2b^2c^2 \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a^3 \plus{} b^3 \plus{} c^3 \plus{} kabc)\Leftrightarrow$
$ \Leftrightarrow(x \minus{} 1)^2((k \minus{} 1)x^2 \plus{} 2(k \minus{} 1)x \plus{} k \plus{} 3)\geq0,$ which gives $ k_{best} \equal{} 1.$


I am new here.I want to find some very famous geometry
problems,what should I do?
I want national geometry problems ,Could you tell me how to download from Internet
我是中国大陆学生，想从这里批量下载一些平面几何练习题，请各位高手告诉我该怎么办

Some problems are available in the Resources Section.

屏幕右上角Resources里面有许多题。


inequalities
floor function
ceiling function

For how many integers $x$ does a triangle with side lengths $10$, $24$ and $x$ have all its angles acute?

For a triangle to be acute, if we arrange the side lengths into $a \lte b \lte c$, $a^2 + b^2 > c^2$.

We notice that the only possible integers for $x$ are between $15$ and $33$.  So for the inequality to hold when $14 < x < 24$,
$x^2 > c^2 - b^2$
$x^2 > 525$
$x > 21$
So only $x=22,23$ works when $x < 24$.  We also see it works when $x=24$.  When $x > 24$,
$x^2 < 676$
$x < 26$
So only $x=25$ works.
So the only values of $x$ that make it acute are between $22$ and $25$, so the total number is $4$.

Since $x$ can't be the smallest side (or else $x+10<24$), let $10\le x\le24$.
Then $x\ge\left\lfloor\sqrt{24^2-10^2}\right\rfloor+1$ or $x\ge22$.
The upper bound occurs when $10\le24\le x$.
And $x\le\left\lceil\sqrt{24^2+10^2}\right\rceil-1$ or $x\le25$.

So the only possible values are $22\le x\le25$, which make $4$ total.


[quote="1/(ln x)"]I there a printer friendly feature on AoPS? It seems wasteful to print the borders every time.[/quote]
There was, but with the new server, no.

You could copy and paste what you want into Word and then print it out.

I doubt you can copy the $\LaTeX$ to Word ;)
At least you need an interpreter and some work.

LaTeX is just an image.

I even tried it; the LaTeX copied fine.

Next problem: not everyone uses Word. Same for any other programm, and forcing people to install something is not good.

[quote="Xevarion"][quote="7h3.D3m0n.117"]Doesn't everyone have WordPad installed though?[/quote]
How about people who don't have Windows?[/quote]

haha touche  :D ...but i thought Linux had something?

Untill you guys manage to get your printer friendly version

you can use print screen buttom, Click it then paste it in the paint software and  edit it as a usual picture, cut the sides and then print it.

I am not sure why Valentin didnt answered Darij, here ?:maybe:
http://www.mathlinks.ro/Forum/viewtopic.php?t=125784


I know that both shoot electron beams which distinguishes them from other microscopes that shoot light through a specimens, but what is the main difference between these two?

SEM can view outer surfaces of cells whereas TEM cannot.

SEM gives a nice 3-D image while TEM is a thin cross section.


Is it true that for any a_1,...,a_2003 different positive integers we can find x_1,...,x_2003 integers such that the sum of digits of a_1*x_1+...+a_2003*x_2003 is 9?

No, take all a_i divisible by 99. Then sum (a_i x_i) is divisible by 99 and therefore has sum of digits >=18 (the sum of digits is divisible by 9, but may not be equal to 9, since the number is divisible by 11).

Similarly, one can prove that for any finite set M we can find x_1,...,x_2003 such that whenever a_1,...,a_2003 are integers the sum of digits of a_1*x_1+...+a_2003*x_2003 does not belong to M.

However prove that S(S( \sum {1...2003} a i x i )) does belong to M, where S(E) is the sum of the digits of E.  :)


If you are given a certain amount of fence, the greatest area you can make with it is the area you can make with a square. (For example, if given 36 ft of fence, the most area one can make is 9 by 9)

Same principle allows with volume and stuff, but it should be a cube.

And the same with the volume, with a sheet of paper with a restricted size the largest volume is created with a cube.

Thanks, I was really only wondering about the 2d - and from the 3d one, one could guess, but I was just wanting to confirm.

Actually you guys, the most efficent use of fence is a CIRCLE.

pn12345 is right. But what do you mean by 'find the best area possible'? Is the greatest? smallest?

pn is right. Sphere is greatest for volume...

[quote="i_like_pie"]How can u make a sphere out of fence?[/quote
It said similar to above, what is the best for volume...

[quote="lotrgreengrapes7926"][quote="stevenbao"]I. Best Area Formula
Let's say we have...
x = amount of gate we have
What is the formula to find the largest area possible of a fencing with perimeter $x$?

II. Best Volume Formula
Similar to above. What is the largest volume possible with a paper of area $x$?[/quote]

No one's answered the questions yet.  :wink:[/quote]
I didn't see the word a paper... Still the answers are circle and sphere, but if a sphere is impossible then cube...

Well, in general, if we have a given perimeter, $P$, and we construct a regular $n-gon$ having this perimeter, as $n$ increases, so does the area. As $n$ approaches $\infty$, the area approaches the maximum possible area, which is $\pi \cdot (\frac{P}{2\pi})^{2}$. Likewise, we can generalize this for $n$ dimensions. The 3D case is just a special case for the $n$ dimensional case. In the 3D case, we would maximize the volume. Note that a circle has "infinitely" many sides. Think about it this way... think of a square. Then a pentagon. Then a decagon. Then a 1000-gon. Then a 1000000-gon. (I think you get the idea) This ends up looking more and more like a circle. As the number of sides increase, the polygon looks more and more like a circle. Eventually, as the number becomes very large, the area of the polygon is close enough to its circle. :wink:

[quote="stevenbao"]I. Best Area Formula
Let's say we have...
x = amount of gate we have
What is the formula to find the best area possible of a fencing with perimeter x?

II. Best Volume Formula
Similar to above. What is the best volume possible?[/quote]

hm... how many sides... its kinda hard if you don't say


Sequence
algebra
recurrence relation
Inequality
IMO Shortlist

The set $ \{a_0, a_1, \ldots, a_n\}$ of real numbers satisfies the following conditions:

[b](i)[/b] $ a_0 \equal{} a_n \equal{} 0,$
[b](ii)[/b] for $ 1 \leq k \leq n \minus{} 1,$ \[ a_k \equal{} c \plus{} \sum^{n\minus{}1}_{i\equal{}k} a_{i\minus{}k} \cdot \left(a_i \plus{} a_{i\plus{}1} \right)\]

Prove that $ c \leq \frac{1}{4n}.$


Find the best constant $k \in R^{+}$ such that for all $x,y,z >0$ satisfy $xyz=1$ we have
              $\sum \frac{1}{(x+k)^2} \geq \frac{3}{(k+1)^2}$

$\frac{1}{{\sqrt 3 - 1}}$
see   http://www.mathlinks.ro/Forum/viewtopic.php?highlight=scpajmb&t=67330


I paint, but I have to clean it up afterwards.

one of the parents in my 5th grade was a professional artist, so he thought us how to use oil and acrylics.

I use to paint when I was young, but I gave up few years ago.

I painted murals for awhile. Of course I used wall paints from Home Depot. But the techniques generally carry over despite what type of paint you use (i.e. oils, watercolor,etc.). By technique I guess I don't mean the best brush strokes or how to paint in the style that you want, but simply the ideas behind color, blending, shadows, perspective,etc. However, I m also a realist, so I can't really speak for someone who's a surrealist or an impressionist or whatever.

I love painting!

And I love a lot paintings by other artists as well.
I like the Starry Night by Van Gogh, especially.

I love to paint, I have been going to an art school once a week for half of my life...
I am almost as good as painting as i am at math XD

which is sad since i just started math competitions last year

I do oil painting mostly... with some water color

i LUV LUV LUV to paint! although i'm pretty bad at it...especially my people that i paint. o my gosh, the horrors. i like painting flowers, though. 'cause those are easy. and pretty.


inequalities
logarithms
induction
inequalities unsolved

Suppose $a>2$, $\{x_{n}\}$is a sequence: ${x_{1}}=a, x_{n+1}=\frac{x_{n}^{2}}{2(x_{n}-1)}$.
1. If $a \le 3$, prove that $x_{n}\le 3$. (I have worked this out)
2. If $a > 3$, prove that if $n \ge \frac{\ln a-\ln 3}{\ln 4-\ln 3}$, thus $x_{n+1}< 3$. (How?)

Proof:
1.
$x_{n+1}=\frac{x_{n}^{2}}{2(x_{n}-1)}=\frac{1}{2}({x_{n}-1+\frac{1}{x_{n}-1})+1 \ge{2}}$.
$x_{1}=a>2$.
$\therefore \forall n \in \mathbb{N}^{+}, x_{n}\ge 2$
Use mathematic induction to prove $\forall n \in \mathbb{N}^{+}, x_{n}\le 3$:
i) If $n=1$, thus $x_{n}=x_{1}=a \le 3$, satisfies this statement.
ii) If $n=k, k \in \mathbb{N}^{+}$, the statement is true, i.e. $x_{k}\ge 2$.
Thus, while $n=k+1, x_{k+1}=\frac{1}{2}{x_{k}-1+\frac{1}{x_{k}-1}+1 \le \frac{1}{2}({3+\frac{1}{2}})+1 < 3}$.
$\therefore \forall n \in \mathbb{N}^{+}, x_{n}\le 3.$

So sorry that I forgot the statement:
It should be prove $x_{n}\le 3$, not $x_{n}\ge 2$