Kind of messy unless you find a shortcut:
Two fractions are chosen from the following set and are multiplied: {11/14, 37/42, 5/6, 19/21, 6/7}. x/y, in lowest terms (meaning after being reduced), is the greatest possible product obtainable. Find x+y :lol:

Also find xy

The set, once ordered, is 11/14, 5/6,6/7,37/42,19/21. Since you want the division x/y to be large, take the largest fraction and divide it by the smallest. Thus, x=19/21, y=11/14. Therefore, x+y=71/42, and xy=209/294.

Yup, the basic rule for divison holds true no matter what unless on of them's negative.

:lol:, oh I don't know anymore, I mean I know that logically "negative numbers make mmm mmm good soup. oh oh booyah *pelvic thrust*" makes no sense nor is a legitimate joke, but it's [b]oh[/b] so funny! :lol:

Reminds me of the episode where Spongebob opens a bubble stand and constantly says about "techniques" I believe the "pelvic thrust" was one of them :lol:


Prove that:

$a^4+b^4+3 \geq a+b+3 \cdot \frac{3ab+1}{4} \cdot \sqrt[3]{\frac{3ab+1}{4}}$
for all real $a,b$


What continent do you live on?  Currently live on?  Used to live on?  Want to live on?  Feel free discuss continents here.

Note that some models use different names for the continents or combine some continents mentioned in the poll into one.  Pease use the model with the continents as named in the poll

What do middle eastern people put?  Central America?

Anyway, I'm from NC, so I guess I'll put North America.

BTW, someone lives in Antarctica  :roll:

no that would be 2 people. :rotfl:
maybe there aresome pengins on the forum :P

btw, middle eastern people are from asia, not central america :roll:

Let's see.. Australia, hmm.. where's Australia!?  :maybe:

Holy crap!  I forgot to put something for Australia... I thought I had... but maybe it didn't save that???  Oops...

I was wondering myself what Middle Eastern or Central American people would put...

[quote="tjhance"]I think he meant something along the lines of:

[quote]
What do middle eastern people put?
What do Central American people put?
[/quote][/quote]

oh, sorry, b-flat, i misunderstood you.  it still wouldve been pretty funny though.

North America.

@rd5493: nice signature. it reminds me of chapter team round when i was in 6th grade and i had to join a group of another school's alternates. there was pindrop silence and one person asked me, "what does 'factor' mean?".

i guess more mathlinkers (AoPS ers) are from Antarctica than Asia(where i'm from)! :rotfl:

Ya Antarctica has more users than South America, Africa, and Asia combined!!! :rotfl:  :rotfl:  :rotfl:

I find it interesting that Antarctica is more popular than Africa, Asia, and South America...

I live in North America

Life on Antarctica is very nice. No taxes, free airco for everyone, ...

want to say Europe but sadly, North America

Why don't you look at what my location is? It's on the left side profile bar.
I'd recommend Google Earth to find out, but there are other ways.

[quote="Arvind_sn"][quote="xscapezaer"]So, how is the climate like down there on Antarctica?[/quote]

Pretty good, actually. it's summer, so its sweltering; a scorching -30 degrees Celsius. We get to take off our 18th layer of clothing! Yeah!!!

JK but did you know there are more internet users in antarctica then there are in some african countries? that's weird.[/quote]

HaHaHaHaHaHaHaHaHaHaHa!! Good one Aravind_sn!!   HaHaHa!! :rotfl:

north america... like most everyone

8 ppl live in antarctica

[quote="hunter34"]North America,

I guess Antarctica is popular place now, with all the penguin movies.[/quote]

Haha penguins ftw! I think the population in Antarctica is like, 200 now. Yippee.

I live in North America.

I live in Hilbert Space!

Where's Australia? :(

[quote="Nerd_of_the_Ages"]
 :huh:
I live in california[/quote]
yes i know people live there but i was just answering the question in the way
[quote="mathrose"]Does anyone live in California???? :Smile:[/quote]
asked.
i think he meant does anyone here in ML?Aops lives in california

[quote="Nerd_of_the_Ages"]
 :huh:
I live in california[/quote]
but it seems u live in some other planet too  :P

[quote="Nerd_of_the_Ages"][quote="pardesi"][quote="mathrose"]Does anyone live in California???? :) [/quote]
no that place is haunted  :D[/quote]

 :huh:
I live in california[/quote]

How are those wildfires going? Or earthquakes/landslides? Pretty good? Bad? In-between? Must be pretty bad living in the land of movie stars and beaches...

[quote="7h3.D3m0n.117"]North America. (Strangely I live within a 25 feet radius from D3m0n Shad0w. Oh wait, she's my sister  :laugh:).[/quote]

I would say 12 mile radius within 7h3.D3m0n.117. He takes up a lot of space  :rotfl:

I live in North America, but I accidently clicked on Asia... Sorry

[quote="ProtestanT"]50% of the world's population is supposed to live in asia.
About 0% is supposed to live in antarctica...
however,from these statistics, there is equal amounts of people from asia and antarctica, and there are more north americans than both of those combined... (north america is supposed to be around 8%, asia + antarctica is around 50%)
That shows how uneven AoPS is :D[/quote]

Hmm...North America's population in the world: 30% (29% illegal immigrants :P)
Antarctica's population: 0.00001%
Europe (not including Russia): 20%
Australia: 10%
Asia: 39% of the world
Russia: 1% :P

Not entirely accurate. Making this stuff up...

Like I got in my fortune cookie the other day. 42.7% of statistics are made up on the spot.

I live in NA

if you guys didn't know, Central America is a part of North America and Middle East, some is Europe some is Asia

[quote="you-know-who"]I live in...nah, I'm not giving away my location...the place I live in is not available (*hint hint*) :lol:[/quote]

Found it. :P


geometry
parallelogram
geometry proposed

Three rectangular parallelograms ABDE, ACFH, BCKL, are erected on the sidelines AB, AC, BC respectively, of a given triangle ABC, outwardly of it.

     If M, is the intersection point of the lines EK, HL and N, the one of the lines through K, L and perpendicular to KF, LD respectively, prove that the points A, M, N, are collinear.

One year ago, a friend of mine Nikos Kyriazis, from Thessaloniki, submitted to me a clever solution, different than I have in mind, based on another old and well known problem which has already been discussed lately, in this forum.

See the last reply at: http://www.mathlinks.ro/Forum/viewtopic.php?t=118200

Kostas Vittas.

A solution of this problem, has already been posted in this forum.

http://www.mathlinks.ro/Forum/viewtopic.php?t=164548

Kostas Vittas.


Hello people,
I'm involved in the school's math club and it's very challenging for me because I don't have a good foundation of math.  It's really annoying because I just can't get a good score, it's almost discouraging.  What do you think I should do?

Well this is the place for you!  If you take everyone's advice here and become an active member you will become much better.  Just some questions:
What grade are you in?
What math class are you in?
How much math competition experience do you have and how long have you been competing?
Just about anything about your math experience would be helpful to tell you where to start.

I hope you have a fun time here!

Math is not a specatator sport!

You participate in it, and you get better.

At least, according to my math teacher  :mrgreen:

What grade are you in? 9

What math class are you in? Geometry and some Algebra II

How much math competition experience do you have and how long have you been competing? I had a few competitions in middle school.  Now I'm competing a few times per month.

Just about anything about your math experience would be helpful to tell you where to start.
I don't have people to talk to about math, they all know much more than me.  Also I'm a little bit careless, so that kills.
I didn't even make into AIME this year.
I don't know what else to say, but you can always ask.

[quote="nobrain"]I don't have people to talk to about math, they all know much more than me.[/quote]

Are those people who know more than you WILLING to talk to you? Or are you just shy about approaching them, because you think you would be bothering them?

The great thing about this site is that people who know more than I know about math (= just about everyone here) are very willing to help, and I feel I can ask informational questions whenever I need to. You may know people like that in person too, but just haven't tried asking them. Anyway, welcome aboard here, and feel free to learn from all the other people here. That's what we're all here for: encouraging learning math by mutual support of learners.

Well, there's one group that is not willing to help me.  Another group, they are willing to, but we rarely have time to talk about the problems.
Oh yes, that leads to my other problem (not math problem)--TIME.  I just can't squeeze time out for math, maybe I don't know how.  So how do you guys do it?

I just make time to do the math.  I do a little bit in the morning before school.  I get up at 5 and don't leave until 7 so I have about 1.5 hrs. to do whatever(2 minus shower, breakfast etc.).  Then I do some at school when my classes are doing nothing.  Since I have no track or cross country after school now I do some after my homework.  Of course I don't do math in every free space I have but I do do it when I am bored and such.

you should practive math whenever you have time for it! sometimes, you won't have any time at all. other days, you'll have all the free time you ever needed. for now, you might want to take it lightly because you'll probably have finals soon. but in the summer, there'll be plenty of time.
or you could be like my friend and do math (homework he didn't do at home) in english. i'm still trying to figure out how he got a 96 on our anne frank test...

[quote="JBL"]Social Studies class provides an excellent daily stretch of 40 minutes or so in which one can do math.
[/quote]

unless, you did A LOTof that earlier in the year and your SS teacher watches you extra close to make sure that you DON'T do maths in class...

I do math (olympiads) in my math (calculus) class.

I do math in my math(geometry) class too and at home

I do math whenever I have time.  Teachers usually don't bother on it since i'm a good student

I do maths in my maths class and whenever i am free...I like maths and physics ..though I hate Chemistry...its all so like a fairy tale goin on..


LaTeX
trigonometry
inequalities
geometry solved
geometry

Let a,b,c length of sides of a triangle and let m,n,p length of bisects of it. prove that :
(1:m)+(1:n)+(1:p)>(1:a)+(1:b)+(1:c)

Please forgive me for writing like that! Because I dont have latex yet! But I promise that I will get it as soon as I can! :blush:

You don't need to [b]have[/b] LaTeX in order to use it on the forum. Believe me, it's easy! Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 .

Here is your problem rewritten using LaTeX (and using my notations):

[color=blue][b]Problem.[/b] If a, b, c are the sidelengths of a triangle ABC, and $w_a$, $w_b$, $w_c$ are the lengths of its angle bisectors, then prove that $\frac{1}{w_a}+\frac{1}{w_b}+\frac{1}{w_c}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.[/color]

[i]Solution.[/i] Let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Then, by the half-angle formulas, $\cos^2\frac{A}{2}=\frac{s\left(s-a\right)}{bc}$. Since $\cos^2\frac{A}{2}<1$ (this is because $-1<\cos\frac{A}{2}<1$; note that this is a strict inequality, i. e. equality cannot occur, since $\frac{A}{2}<90^{\circ}$), we thus have $\frac{s\left(s-a\right)}{bc}<1$, so that s (s - a) < bc. Now, by the formula from http://www.mathlinks.ro/Forum/viewtopic.php?t=805 post #3, we have $w_a^2=\frac{4bc}{\left(b+c\right)^2}\cdot s\left(s-a\right)$, and thus $w_a^2<\frac{4bc}{\left(b+c\right)^2}\cdot bc=\left(\frac{2bc}{b+c}\right)^2=\left(\frac{2}{\frac{1}{b}+\frac{1}{c}}\right)^2$, so that $w_a<\frac{2}{\frac{1}{b}+\frac{1}{c}}$. Hence, $\frac{2}{w_a}>\frac{1}{b}+\frac{1}{c}$. Similarly, $\frac{2}{w_b}>\frac{1}{c}+\frac{1}{a}$ and $\frac{2}{w_c}>\frac{1}{a}+\frac{1}{b}$. Adding these three inequalities together, we get

$\frac{2}{w_a}+\frac{2}{w_b}+\frac{2}{w_c}>\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)+\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}$;

upon division by 2, this yields $\frac{1}{w_a}+\frac{1}{w_b}+\frac{1}{w_c}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, and the problem is solved.

  darij

Thank you for your sollution.you solved it too fast!
If it is possible please think about homogeneous hexagons and    solve it and send me your solution. :lol:

[quote="jensen"]Let $a,b,c$ length of sides of a triangle and let $w_{a},w_{b},w_{c}$ length of angle-bisectors of it, prove that
\[\frac{1}{w_{a}}+\frac{1}{w_{b}}+\frac{1}{w_{c}}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \]
[/quote]
This was conjecture, see :

Problem Gy 2188, Koz. Mat. Lapok, 1984, No. 3, p. 126.

In fact, we have the following inequality chain :
\[\left(\frac{1}{w_{a}}+\frac{1}{w_{b}}+\frac{1}{w_{c}}\right)^{2}\geq3\left(\frac{1}{w_{b}w_{c}}+\frac{1}{w_{c}w_{a}}+\frac{1}{w_{a}w_{b}}\right)\geq\frac{4}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}, \]
which was established in

Zhen Wang, Ji Chen, Sharpening two conjecture inequalities and some others (Chinese), [url=http://qks.zjnu.net.cn/Unit_View_3.asp?id=26]High School Teaching Research (Mathematics)[/url] (Jinhua), 1994, No. 7-8, pp. 51-53.

Another inequality chain :
\[\left(\frac{1}{w_{a}}+\frac{1}{w_{b}}+\frac{1}{w_{c}}\right)^{2}\geq\left(\frac{1}{b}+\frac{1}{c}\right)^{2}+\left(\frac{1}{c}+\frac{1}{a}\right)^{2}+\left(\frac{1}{a}+\frac{1}{b}\right)^{2}\geq\frac{4}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2},\]
was established in :

Ji Chen, Zun Shan, The extension of an angular bisector inequality (Chinese), [url=http://www.shuxuetongxun.com/]Bulletin of Mathematics (Wuhan)[/url], 1995, No. 11, pp. 17 - 18.


function
inequalities
inequalities proposed

Since I cannot recover from the shock of that idiot problem, I will present a very hard one. I really don't think this one is trivial. Even the case n=3 was very discussed when I first posted it. So:
    Find the minimum value of the expression $\sum_{i=1}^n {\sqrt{\frac{a_1\cdot...\cdot a_n}{1-(n-1)a_i}} }$ when $ a_i<\frac{1}{n-1} $ add up to 1.

[quote="harazi"]Since I cannot recover from the shock of that idiot problem, I will present a very hard one.[/quote]

Which problem are you talking about? :)

For Sasha: from the condition imposed we clearly have $ a_i>0 $ for all i. Yes, I know you will say that you could take $ a_1$ close to zero, but then allmost all other variables will be close to $\frac{1}{n-1} $ and thus you will have a nice and hard $\frac{0}{0} $ case. For Arne: the idiot problem was that with complex numbers. Boy, don't let me speak about it. What a shame! :blush:  :blush:  :blush:  :blush:

[quote="harazi"]For Sasha: from the condition imposed we clearly have $ a_i>0 $ for all i.[/quote] I do not see why... you better impose that condition.

EDIT: whoops, misread. forget about whatever i said =)

Hi!
I am possibly wrong, but take $a_i = e^{x_i}$.
Then $\frac{1}{\sqrt{(1-(n-1)e^x)}}$ is a convex function (I think so)...
We can take Jensen to forget about the numbers.
Then taking $x_1..x_n = t$, we have to find the minimum of that expression.
We could be able to do that, because we have some restrictions on $t$.

For Iugash: the conditions can be deduced from the given ones!
  For Dust: well, I would like to see a complete solution. I will verify later if you are right.

If my computations aren't wrong, then Dust you are not right, you won't deduce to minimum from there. But maybe you post a more detailed solution.

[quote="harazi"]
Find the minimum value of the expression $\sum_{i=1}^n {\sqrt{\frac{a_1\cdot...\cdot a_n}{1-(n-1)a_i}}}$ when $a_i<\frac{1}{n-1}$ add up to 1.[/quote]

Can anyone provide a good solution for this one now? :)

The problem is the problem 118 in old and new inequalities and it is  quite-very hard.
The minimal value is  $\frac{1}{\sqrt{n^{n-3}}}$ .

This could be reached using Suranyi's Inequality at first Holder at second and finally Jensen inequality

hi

what exactly is the Suranyi's inequality

never heard of it before :oops:

plz someone explain or send a link :lol:

thnx

[quote="toetoe"]is this book "Old and new inequalities" free to download ?[/quote]
You can by it through http://www.gil.ro.

[quote="mathmanman"]It was nicely showed by [b]harazi[/b]. :)
See :
http://www.mathlinks.ro/Forum/viewtopic.php?t=14008
http://www.mathlinks.ro/Forum/viewtopic.php?t=14906

@Silouan : Ok, thanks, now I understand why I didn't manage to find it... - If you have the time, could you please post the solution? (or, well, at least a sketch of it if it's too long ;))[/quote]

He need not do that:

http://www.mathlinks.ro/Forum/viewtopic.php?t=26131&postorder=asc&start=20


[quote="bos1234"]what is the identity used here?

$ (\cos\theta-1)(\cos\alpha-1)$=$ (-2sin^{2}\frac{\theta}{2})(-2sin^{2}\frac{\alpha}{2})$[/quote]

Typo fixed.

This identity should help:

[hide]$ \cos 2x=1-2\sin^{2}x$[/hide]

[hide]
$ (\cos\theta-1)(\cos\alpha-1) = \left( 1-2 \sin^{2}\frac{\theta}{2}-1 \right) \left( 1-2 \sin^{2}\frac{\alpha}{2}-1 \right) = \boxed{ \left(-2 \sin^{2}\frac{\theta}{2}\right) \left(-2 \sin^{2}\frac{\alpha}{2}\right) }$

Double Angle Formulae for Cosine
$ \cos ( \alpha \pm \beta ) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

Let $ \alpha = \beta$, then
$ \cos (2 \alpha ) = \boxed{ \cos^{2}\alpha-\sin^{2}\alpha}$

Use Pythagoreans
$ \boxed{ 2 \cos^{2}\alpha-1}$
$ \boxed{ \boxed{ 1-2 \sin^{2}\alpha }}$
[/hide]

[quote="bos1234"]what is the identity used here?

$ (\cos\theta-1)(\cos\alpha-1)$=$ (-2sin^{2}\frac{\theta}{2})(-sin^{2}\frac{\alpha}{2})$[/quote]

[hide="Proof"]\begin{eqnarray*}(\cos\theta-1)(\cos\alpha-1) &=& \left(1-2\sin^{2}\frac{\theta}{2}-1\right)\left(1-2\sin^{2}\frac{\alpha}{2}-1\right) \\ &=&\left(-2\sin^{2}\frac{\theta}{2}\right)\left(-2\sin^{2}\frac{\alpha}{2}\right) \end{eqnarray*}[/hide]
[hide="Extension (similar problem)"]Prove that:
$ (1+\cos\theta)(1+\cos\alpha) = 4\cos^{2}\frac{\theta}{2}\cos^{2}\frac{\alpha}{2}$[/hide]

[hide="HINT"]
Use the double angle formula for the cosine in the form of $ 2 \cos^{2}\theta-1$.
[/hide]

$ cos2\theta$ = $ 2cos^{2}\theta-1$

$ cos\theta$=$ 2cos^{2}\frac{\theta}{2}-1$

$ cos\theta+1$=$ 2cos^{2}\frac{\theta}{2}$

\\\ly $ cos\alpha+1$=$ 2cos^{2}\frac{\alpha}{2}$

multiplying, $ 4cos^{2}\frac{\theta}{2}cos^{2}\frac{\alpha}{2}$

kk i get it now

thanks for the helps


linear algebra
matrix
algebra
polynomial
linear algebra theorems

if S=span{ A^i | i= 1 2 3 ..., A is 5x5 Matrix }
what is dim(S) ?
is suppose that the answer should be no more then 5, and maybe it's equal to dim(A),
but I didn't success to prove it.
do you have any idea how to solve it?

The dimension is equal to the degree of the minimal polynomial unless $ A$ is singular; if $ A$ is singular, it's 1 less than that degree.

can you please write me a proof (or at last how to start)?


superior algebra
superior algebra unsolved

I had seen month ago an interesting algebraic, but simple, proof for:

$ \sqrt[n_1]{m_1}\plus{}\sqrt[n_2]{m_2} ~...\plus{}\sqrt[n_t]{m_t}$ for $ n_i, m_i \in \mathbb{N}$ is irrational, if at least one root is irrational.

It was by an Italian, I think..but I've forgotten his name, have searched for hours in the internet, but nothing found. Can anybody help me? Nucci, Nocchi, Nuchatelli or somethin like that was his name...

Thx, but that was not the proof-way, I searched for. It was not so standard, and, as I said, published by an italian in an american journal...sorry, I have no more information, it will be very difficult to find the original article


From a group of boys and girls, 15 girls leave.  There are then left two boys for each girl.  After this, 45 boys leave.  There are then 5 girls for each boy.  The number of girls in the beginning was:
\[ \text{(A)}\ 40 \qquad \text{(B)}\ 43 \qquad \text{(C)}\ 29 \qquad \text{(D)}\ 50 \qquad \text{(E) none of these} \]

[hide] $(A)$

I just created two equations, and instead of actually solving them, I just plugged in the numbers, and A happened to work!
These were the equations, as can be seen from the word problem

$x$ = boys  $y$ = girls
$2(y-15) = x$
$5(x-45) = y - 15$

Substituting 40 for y

$2(25) = x$
$x = 50$

Substituting 50 for x to see if y will come out to be 40

$5(50 - 45) = y - 15$
$25 + 15 = y$
$y = 40$

And that only takes a few seconds of mental math so it was easier than solving the equations imho.[/hide]

[hide]Let $g$ be the original number of girls, and $b$ the original number of boys.  We can write:
$2(g-15)=b$ and $g-15=5(b-45)$

Substituting for $b$ in the second equation and solving gives us $g=40$.[/hide]


calculus
integration
function
limit
symmetry
analytic geometry
graphing lines

Is this statement correct for a contour passing right through M poles:

$\oint f(z)dz=2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{i}}f(z)-\pi i\sum_{n=1}^{M}\mathop\text{BRes}_{z\;=\;z_{i}}f(z)$

that is, the integral around a closed contour which passes through poles can be interpreted as one which encloses the poles including the ones on the boundary but with the small half-circle contours around each boundary pole subtracted from the total thereby extending Cauchy's Residue Theorem?

Probably not.  Maybe depends on the pole or principal values need to be considered.   I need to look into it more . . .

The standard Cauchy residue theorem is just a simple consequence of the fact that for a small circle around the pole the integrals of the inverse powers of $z$ are all $0$ except the integral of $z^{-1}$, which is $2\pi i$. When you deal with simple poles only, you may, indeed, say that the integral over approximately half the circle of $z^{-1}$ is about $\pi i$ and the formula you proposed holds with the integral understood in the principal value sense. But when you have a multiple pole, you have to look at the integrals of $z^{-2}$, $z^{-3}$, etc. as well and you can no longer say that the integrals of these guys over your approximately half circle are close to $0$. In the case of $z^{-2}$, the corresponding term actually tends to infinity (which is not surprizing: try to integrate $1/z^{2}$ over the real line; the integral will diverge in any standard sense). So, no such simple formula holds for multiple poles on the boundary.

Thanks fedja.  What I posted above is incorrect.  Dang it!  I hate when that happens. :blush:

Alternatively then, the theorem concerning indented paths around simple poles on the real axis looks applicable for (simple) poles elsewhere.  Using this might I say instead:

$\oint f(z)dz=\text{P.V.}\oint f(z)dz+\pi i\sum_{n=1}^{M}\mathop\text{BRes}_{z\;=\;z_{n}}f(z)=2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{n}}f(z)$

where N represents all the poles and M the poles bysected by the contour.

This I brought up in regards to the integral:

$\int_{0}^{1}\frac{ln(x)}{x^{2}+1}dx$

which I evaluated by a half-circle closed contour passing through the pole at $i$ and concluded:

$\int_{0}^{1}\frac{ln(x)}{x^{2}+1}dx=1/2\left\{2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{n}}f(z)-\pi i\sum_{n=1}^{M}\mathop\text{BRes}_{z\;=\;z_{n}}f(z)-\text{P.V.}\int_{C}\frac{Log[z]}{z^{2}+1}dz-\int_{0}^{1}\frac{\pi i}{z^{2}+1}\right\}$

which reduces to:

$\int_{0}^{1}\frac{ln(x)}{x^{2}+1}dx=-1/2\;\text{P.V.}\int_{C}\frac{Log[z]}{z^{2}+1}dz$

where C is the curved portion of the contour.

I'll then state the following working hypothesis:

I[b]f an integral around a closed contour which intersects simple poles is interpreted in the principal sense then Cauchy's Residue Theorem still applies.[/b]

Why do I find that amazing?  How do I prove a contour through ANY simple pole will have a finite principal value?

[quote="shawtoos"]  How do I prove a contour through ANY simple pole will have a finite principal value?[/quote]
Just add a small "almost semicircle" to the cut contour (I assume that the contour is smooth) and note that the full integral is constant and the integral over the almost semicircle tends to some finite value. This implies that the integrals over the contours with cuts tend to a limit. Alternatively, investigate integration of the function $1/z$ over a smooth contour passing through $0$ (the remaining analytic part is, clearly, integrable).

Thanks again! I'll work on that.  I think integrating through a pole as well as a branch cut is interesting. For the record, here's an example. Let:

$f(z)=\frac{1}{(z-e^{\pi i/4})(z-e^{3\pi i/4})(z-i/2)}$

the function of course has 2 poles on the unit circle and one inside.  Therefore I suggest:

$\int_{C}f(z)dz=\text{P.V.}\int_{C}f(z)dz+\pi i(R_{1}+R_{2})=2\pi i(R_{1}+R_{2}+R_{3})$

where C is the unit circle and:

$R_{1}=\mathop\text{Res}_{\theta\;=\;\pi/4}f(z)$

$R_{2}=\mathop\text{Res}_{\theta\;=\;3\pi/4}f(z)$

$R_{3}=\mathop\text{Res}_{z\;=\;i/2}f(z)$

Numerical estimates are virtually identical to the value via residue integration.

I need to modify my equation above according to the length of the indented contours around the poles.  If the contour is a straight line then obviously this length it $\pi$.  But suppose the pole is at a corner of a box and the indentation is outward.  Then the length would be $3\pi/2$.  So to generalize the equation for ALL contours I need to move the "indentation angle" into the summation:

$\oint f(z)dz=\text{P.V.}\oint f(z)dz+i\sum_{n=1}^{M}\theta_{n}\mathop\text{Res}_{z\;=\;z_{n}}f(z)=2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{n}}f(z)$

And the equation would hold only if I can prove:

$\text{P.V.}\int_{C}\frac{dz}{z}\overset{?}{<}\infty$ for any contour passing through 0.

Also I'd like to determine if for any smooth contour, will the indentation angle approach $\pi$ so that the simpler form of the equation above could be used.  That is does:

$\mathop\lim_{\rho\to 0}\theta_{n}\overset{?}{=}\pi$ for smooth contours.

Not sure about any of these but I'm working on them.

Here goes . . .

[b]Theorem:[/b]

$\left|\text{P.V.}\int_{C}f(z)dz\right|<\infty$ for all simple arcs passing through simple poles of $f(z)$.

[b]Proof:[/b]

Let f(z) have a simple pole at $z_{0}$.  The Laurent series for $f(z)$ about $z_{0}$ is:

$f(z)=g(z)+\frac{B_{0}}{z-z_{0}}$ with $g(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$

then:

$\int_{C}f(z)dz=\int_{C}g(z)dz+B_{0}\int_{C}\frac{dz}{z-z_{0}}$

The first integral converges since $g(z)$ is analytic over the region of integration.  The second integral can be written:

$\int_{C}\frac{dz}{z-z_{0}}=\int_{C-z_{0}}\frac{dS}{S}$ for $S=z-z_{0}$

Convergence in the principal sense of this integral is completely determined by it's principal convergence in any small $\delta$-neighborhood of zero.  If this neighborhood is made sufficiently small, then any simple arc through the origin can be approximated to any degree of accuracy by two straight lines connected at the origin.  Defining $L_{1}(t)=t+im_{1}t$ as the first line and $L_{2}(t)=t\pm im_{2}t$ (as needed) as the second and substituting into the equation above, I obtain:

$\text{P.V.}\int_{L_{1}+L_{2}}\frac{dS}{S}=\text{P.V.}\left\{\int_{a}^{0}\frac{dt}{t}+\int_{0}^{b}\frac{dt}{ t}\right\}$

where $a$ and $b$ (positive or negative) are the starting and ending values of t.  Because of symmetry, this clearly implies:

$\left|\text{P.V.}\left\{\int_{t_{0}}^{0}\frac{dt}{t}+\int_{0}^{t_{1}}\frac{dt}{t}\right\}\right|<\infty$

Since the slopes $m_{1}$ and $m_{2}$ were arbitrary, the integral over any such linear contour has a bounded principal value which implies so too the integral over any smooth arc through the pole as well.

Still not too confident about this though. :(

Think I have the other one too.  I know, it took longer to draw the picture then to come up with the proof.  I like making plots though.  Assume contour is a differentiable function of x without loss of generality.

[b]Theorem:[/b]

$\mathop\lim_{\rho\to 0}\theta=\pi$

[b]Proof:[/b]

Refer to figure below.  Since f(x) differentiable then:

$\mathop\lim_{x\to x_{0}^{-}}f(x)=f(x_{0})\quad\text{and}\quad\mathop \lim_{x\to x_{0}^{+}}f(x)=f(x_{0})$

and:

$\mathop\lim_{x\to x_{0}^{-}}f'(x)=f'(x_{0})\quad\text{and}\quad \mathop\lim_{x\to x_{0}^{+}}f'(x)=f'(x_{0})$

The only way for a set of three points to satisfy these relations is for them to be co-linear.  Therefore:

$\mathop\lim_{\rho\to 0}\theta=\pi$

If f(x) is continuous but not differentiable at $x_{0}$ then the two derivatives are not equal at the limit and so $\theta$ reduces to the angle between the two limiting values for the right and left-hand derivatives.

I think this is it and I don't want to hog the forum.  So unless someone can point out faults in my proofs, I'll finish with stating what I believe to be the most general expression for extending Cauchy's Residue Theorem to a contour passing through a simple pole as:

$\oint f(z)dz=\text{P.V.}\oint f(z)dz+i\sum_{n=1}^{M}\theta_{n}\mathop\text{Res}_{z\;=\;z_{n}}f(z)=2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{n}}f(z)$

reducing, if the contour is smooth about the intersected poles, to:

$\oint f(z)dz=\text{P.V.}\oint f(z)dz+i\pi\sum_{n=1}^{M}\mathop\text{Res}_{z\;=\;z_{n}}f(z)=2\pi i\sum_{n=1}^{N}\mathop\text{Res}_{z\;=\;z_{n}}f(z)$

[quote="shawtoos"]I am wrong.  After checking the expression above further I've concluded it's not correct.  I apologize to everyone who waisted their time reading this thread.  Next time I'll do more work before saying anything. :blush:[/quote]

Dang it!  I've done more work so I'll say it:  I was evaluating the principal values incorrectly above.  When done properly, the equations as written now work with every example I've tested thus far.  Wish to inform the readers of this thread perhaps it wasn't a waste of time.

Jesus, I'm such a clown.  I'm sorry I ever started this thread!  Now can someone post something else to get my thread off the top please.


algebra
polynomial
modular arithmetic
induction
strong induction
algebra proposed

It is known of a polynomial over $Z$ that  $p(n)>n$ for every positive integer $n$ .Consider $x_{1}=1$ , $x_{2}=p(x_{1})$ ,
$x_{3}=p(x_{2})$ ,....We know that for any positive integer $N$ , there exist a term of the above sequence which is divisible by $N$ .Prove that $p(x)=x+1$

First, I will show that $p(1) = 2$.  Let $N = p(1)-1$.  Then $p(1) \equiv 1 \pmod{N}$.  Thus the terms $x_{1}$, $x_{2}$, $x_{3}$, etc., are all $1 \pmod{N}$.  But the hypothesis is that at least one of them is $0 \pmod{N}$.  The only way that can work is if $N = 1$.  In other words, $p(1) = 2$.

Next, I will show that $p(2) = 3$.  Let $N = p(2)-1$.  (By hypothesis, $N \ge 2$.)  Then $p(2) \equiv 1 \pmod{N}$.  Thus the sequence $x_{1}$, $x_{2}$, $x_{3}$, etc., is the alternating sequence $1$, $2$, $1$, $2$, etc., modulo $N$.  By hypothesis, one of these terms is $0 \pmod{N}$.  The only way that can work is if $N = 2$.  In other words, $p(2) = 3$.

By continuing the argument above using induction, we can prove that $p(n) = n+1$ for all positive integers $n$.  Because $p$ is a polynomial, that means $p(x) = x+1$ for all real numbers $x$.

Ravi B,

you posted "By continuing the argument above using induction..." **

But you DIDN'T complete the induction in your post, and referring to it without the proper step of "assume such-and-such, using the k's, (k+1)'s, show this such-and-such, and 'Now I have shown with ...'" renders your solution to the original
problem:  incomplete.

Also, all of your steps prior to, and following the ** line, I cannot vouch for, nor can I speak against at this time.

I agree that my solution is incomplete.  It was meant to be more of a sketch.  In an actual publication or contest, I would be more complete.  What follows is more detail.

Here is the induction hypothesis.  Let $S(n)$ be the statement $p(n) = n+1$.  We will prove that $S(n)$ is true for all positive integers $n$.  The proof will be by strong induction on $n$.

Let's assume that $S(k)$ is true for all $k < n$.  We will prove that $S(n)$ is true.  Set $N = p(n)-1$ (which is at least $n$ by hypothesis).  By the induction hypothesis, the sequence $x_{1}$, $x_{2}$, etc., modulo $N$ has a period of $n$: namely, it will be $1$, $2$, ... $n$, repeating over and over.  But we are told that there should be a $0 \pmod{N}$ in the sequence.  Thus $N$ must be $n$.  In other words $p(n) = n+1$.  We have settled the induction.

[quote="Ravi B"]  Thus the terms $x_{1}$, $x_{2}$, $x_{3}$, etc., are all $1 \pmod{N}$.

  Thus the sequence $x_{1}$, $x_{2}$, $x_{3}$, etc., is the alternating sequence $1$, $2$, $1$, $2$, etc., modulo $N$. [/quote]

Could you explain these two arguments ??

I am using the well-known fact that if $a \equiv b \pmod{N}$, then $p(a) \equiv p(b) \pmod{N}$.

In the first case, recall that we set $N = p(1)-1$.  Thus $p(1) \equiv 1 \pmod{N}$.  In other words, modulo $N$, the polynomial $p$ has a fixed point at $1$.  Applying the well-known fact above, we get $p(p(1)) \equiv p(1) \equiv 1 \pmod{N}$.  By repeating, we get $p^{(k)}(1) \equiv 1 \pmod{N}$ for every nonnegative integer $k$.  Thus $x_{k}\equiv 1 \pmod{N}$ for every positive integer $k$.

A similar argument works for the second case.  There we know that $p(1) \equiv 2 \pmod{N}$ and $p(2) \equiv 1 \pmod{N}$.  So, modulo $N$, the polynomial $p$ has an orbit of length $2$.

Thank you very much .Your solution is very nice.  :D
In the solution I had I supposed that $p(n)=n+k$ where k is an integer $k\geq 1$ and then I roved that $k=1$ only satisfay the condition.

The problem is from Iran TST 1998. Here is my solution :

We use induction on $n$ to prove that $x_{n}=n$. For $n=1$ it's the problem's hypothesis. If we have proved the statement for $n$ we will prove it for $n+1$. Now $p(x_{n})>x_{n}=n$. If $=x_{n+1}=p(x_{n})\neq n+1$ then $x_{n+1}\geq n+2$. Let $k=x_{n+1}-x_{n}$. Then $k|x_{n+2}-x_{n+1}|\dots|x_{N+1}-x_{N}$. Because there is a $t\geq n+1$ that $kx_{n+1}|x_{t}$ and because $x_{t}\equiv x_{n}\pmod{k}$ so ${x_{n+1}-x_{n}}|x_{n}$. Thus $x_{n+1}\leq 2x_{n}=2n$. Also because $n-1=x_{n}-x_{1}|x_{n+1}-x_{2}=x_{n+1}-2$ we conclude that $x_{n+1}\geq 2n$ and $x_{n+1}=2n$. Now $n-2=x_{n}-x_{2}|x_{n+1}-x_{3}=2n-3$. Contradiction prove that $x_{n+1}=n+1$, and we re done.


inequalities
function
inequalities solved

let 1>a, b,c>-1[tex] prove that ab+bc+ca+1>0[/tex](alt+x)

consider the function $f(x) = (b+c)x + bc + 1$

The problem asks us to prove $f(a)>0$

If $b+c=0$, we have $f(a) = 1 + bc = -b^2 + 1 >0$

If $b+c\neq 0$,

$f(1) = (1+b)(1+c) > 0$ and $f(-1) = (1-b)(1-c) > 0$. Hence $f(x)>0$ for $x\in (-1,1)$ because it is linear.

it seems that this problem appeared in my school exam ...  :maybe:

perhaps, do you have any inequality to post liyi?

[quote="tranminhhoang"]I think this inequality comes from these ones

(1-a)(1-b)(1-c) > 0
(1+a)(1+b)(1+c)>0[/quote]

Indeed!

Here is my proof:

We have to prove the inequality bc + ca + ab + 1 > 0 for any three real numbers a, b, c with -1 < a < 1, -1 < b < 1 and -1 < c < 1.

In fact, we begin with considering the case when one of the three numbers a, b, c is $\leq 0$ and the two other ones are $\geq 0$. We can WLOG assume that a is $\leq 0$, and that b and c are $\geq 0$. In this case, we have $b+c\geq 0$. We can even assume b + c > 0 (else, we had b = c = 0, and this case is actually trivial). This, combined with a > -1, gives a (b + c) > - (b + c). So we have

  bc + ca + ab + 1 = a (b + c) + bc + 1 > - (b + c) + bc + 1 = (1 - b) (1 - c) > 0,

since 1 > b and 1 > c yield 1 - b > 0 and 1 - c > 0. Thus, we have proven the inequality bc + ca + ab + 1 > 0 for the case when one of the three numbers a, b, c is $\leq 0$ and the two other ones are $\geq 0$.

In the case when one of the three numbers a, b, c is $\geq 0$ and the two other ones are $\leq 0$, we just replace a, b, c by -a, -b, -c and reduce this case to the precedent one.

Two cases remain: The case when the numbers a, b, c are all $\geq 0$, and the case when the numbers a, b, c are all $\leq 0$. But in both of these cases, the inequality bc + ca + ab + 1 > 0 is trivial, since $bc\geq 0$, $ca\geq 0$, $ab\geq 0$.

Hence the proof is complete.

  Darij

[quote="tranminhhoang"]I think this inequality comes from these ones

(1-a)(1-b)(1-c) > 0
(1+a)(1+b)(1+c)>0[/quote]

Why not just add them together?


How many colors are needed to paint the plane so that no two points a unit distance apart are painted the same color? If the same question is asked of the line, the answer is 2: Coloring [0, 1) red, [1, 2) blue, etc., ensures that no two unit-separated points have the same color. One can view the question as asking for the chromatic number X(E^2) (2 of the infinite unit-distance graph G, with every point in the plane a vertex, and an edge between two vertices if they are separated by a unit distance.
 ;)

This is a famous and unsolved problem.
Chromatic number is at least 4 (classic olympiad problem) and a hexagonal lattice coloring with 7 colors is possible.  The chromatic number is detected by finite configurations, so the next question is whether there is some finite subgraph that requires 5 colors.


If $a,b,c$ are positive numbers such that $a+b+c = 9$ then show that:
$\frac{a}{a^2-2a+4}+\frac{b}{b^2-2b+4}+\frac{c}{c^2-2c+4} < 2$

Note: this is a strict inequality, not a $\leq$ sign.

For positive numbers $a,b,c,$ using $A.M.-G.M.$ inequality we have $\frac{a}{a^2-2a+4}=\frac{1}{a+\frac{4}{a}-2}\leq \frac{1}{2\sqrt{a\cdot \frac{4}{a}}-2}=\frac{1}{2}.$
Simillarly we have

$\frac{b}{b^2-2b+4}\leq \frac{1}{2},\ \frac{c}{c^2-2c+4}\leq \frac{1}{2}.$ Adding these inequalities we have

$\frac{a}{a^2-2a+4}+\frac{b}{b^2-2b+4}+\frac{c}{c^2-2c+4}\leq \frac{3}{2}.$

The condition for which the equality holds is $a=b=c=2,$ however this condition gives $a+b+c=6,$ which contradicts to the given condition $a+b+c=9.$ thus the equality doesn't hold.

Therefore we obtain $\frac{a}{a^2-2a+4}+\frac{b}{b^2-2b+4}+\frac{c}{c^2-2c+4}< \frac{3}{2}.\ Q.E.D.$

according to kunny's solution I don't understand why is necessary to know the value of expression $a+b+c$..

because my solution requires that you know $a+b+c$ :P
The original problem is this:
If $a+b+c=s$ then prove that $\sum \frac{a}{a^2-2a+4} \leq \frac{s+3}{6}$
Where equality occurs iff $a=b=c=1$

Hendrata what about your solution ? can you write it please...

Davron

[quote="kunny"]$\sum_{cyclic}\frac{a}{a^2-2a+4}=\frac{1}{a+\frac{4}{a}-2}<\sum_{cyclic}\frac{1}{2\sqrt{a\cdot \frac{4}{a}}-2}=\frac{3}{2}.$[/quote]

Kunny you forgot here to put Summation symbol ... and also you used AM-GM after we have an expression which is smaller or equal but you wrote there smaller... anyway nice solution thanks ...

Davron

actually, I think the original problem with $\frac{s+3}{6}$ may be wrong....
nevertheless, this is my solution to my problem:
$a^2-2a+4 = (a-1)^2+1^2+1^2+1^2 \geq \frac{1}{4}(a+2)^2$ (by cauchy)

Then it's sufficient to show $\frac{4a}{(a+2)^2} \leq \frac{2}{3}$, which is equivalent to $a^2-2a+4 \geq 0$
But the equal sign cannot occur together.


For any quadrangle AB^2 + CD^2 = BC^2 + AD^2 - 2. AC.BD .cos(<BD∩AC)

I don't know if I figured it out all right, but if you denote the intersection of diagonals by S you obtain

AB^2 = AS^2+BS^2-2AS*BS*cos<ASB
BC^2 = BS^2+CS^2-2BS*CS*cos<BSC = BS^2+CS^2+2BS*CS*cos<ASB
CD^2 = CS^2+DS^2-2CS*DS*cos<CSD = CS^2+DS^2-2CS*DS*cos<ASB
DA^2 = DS^2+AS^2-2DS*AS*cos<DSA = DS^2+AS^2+2DS*AS*cos<ASB
(cosine law)

1+3-2-4 yields
AB^2+CD^2-BC^2-AD^2-2cos<ASB*(CS+AS)(BS+DS)
AB^2+CD^2 = BC^2+DA^2+2AC*BD*cos<ASB = BC^2 + AD^2 - 2*AC*BD*cos(<BD∩AC)

and while I don't know the name, just remember these lines and you can prove it when needed :P


videos geometry search probability MATHCOUNTS AMC AIME

What is the best method to study and/or review math at home?

Are math instructional videos or dvds sufficient to increase math knowledge?

I welcome different replies and methods?

Tell me, how do you (the reader) keep up with the enormous amount of math content needed to help people like me with questions regarding different areas of mathematics?

I try to do all problems by myself, it forces me to memorize methods.  If i still need help, then I ask someone else.
But there are a lot of topics like this around the site.

I'm not an AOPS tutor, but I have saved a few asses in high school mathematics. practice makes perfect, but so does understanding the deepest darkest roots of a subject. I personally like to ravage, plunder, and turn new subjects inside-out. It's effective.

Lucky for you, members of AOPS try to integrate those fundamentals into their replies. If people reply but don't explain their answer (or it is not obvious), then ask for an elaboration or even a PM. I know I wouldn't mind.

Hope that's some help.

I think you can better your understanding of a given topic by reading more than just one author's explanation.  You can find dozens of free eBooks on google which cover almost any topic in math you can think of.  Goto Google and type "math topic you want" filetype:pdf.
Heres an example of a search [url=http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=fse&q=probability+filetype%3Apdf&btnG=Search]probability filetype:pdf[/url]

Coming to this forum will also help a great deal.

[quote="sharkman"]What is the best method to study and/or review math at home?[/quote]

Buy some math books, especially from here, or search an author called Paul Zeitz.(He was invited to the IMO one or two times) Also, do a few problems from contests, like some Natonal level MathCounts contest(They consist of Sprint, Target, Team, Countdown, and Masters.), AMC 10/12, AIME, USAMO, MOSP, or if you are really brave, like some like 25 people on this website, do some IMO problems.

[quote="sharkman"]Are math instructional videos or dvds sufficient to increase math knowledge? [/quote]

Everyone on this website will say "No.". I don't recall seeing any one video that teaches you mathematics that you have never learned in school. You can use those tapes to brush up, or give to someone that urgently needs tutoring in that subject, but those are not enough. Refer to the answer above.

[quote="sharkman"]Tell me, how do you (the reader) keep up with the enormous amount of math content needed to help people like me with questions regarding different areas of mathematics?[/quote]

Read those books a lot. And refer to the first answer.

Also, come to this forum a lot. You never know when you may learn something.

[quote="max_tm"]...Interval?[/quote]

well... he is making himself quite obvious because hes referring to AoPS as a tutoring site again and also the similarity in questions... and notice how hes posting more and more and more. just like interval! :o .

anyhoo

if ur studying for a test like the SAT get an SAT prep book (duh) I like princeton review

if ur studying for a school final or something then just read ur text book and do the examples without looking at the answers first.  or u could just ask ur teacher.

OR if ur preparing for MATHCOUNTS then maybe you should start posting in the MATHCOUNTS forum

what test are you taking?

oh and btw, vids and dvds are helpful but you cant just rely on them

and to the last question, we dont answer every single one (except when im really bored like in the summer) and there are so many users that they all chip in to answer questions.  i try to answer the topics with 0 or 1 posts before i start with the others (just so i dont seem like i copy other peoples solutions)

See, he misspelled it twice so it can't possibly be sharkman.

Anyways, sometimes having actual human beings to answer your Q's and teach new topics is very valuable.  So go join a math club, if you already haven't.  It is different from just working through even a very good super book of problem solving.

this is kinda getting off topic so this will be the last post about sharkman being interval in this topic :P

look, sharkman i dont see why you are still trying to hide that you arent Interval.  too many things you do are similar to Interval like the difficulty of your topics, the bulk you post them in, the way you write etc...  there is no point in just changing usernames; your personality will always be the same.  maybe explain your situation like what you are trying to prepare for?...  almost all of your questions belong in the Mathcounts forum or below.

[quote="sharkman"]Who is intervol?

What is an intervol?[/quote]

We're not stupid. Everyone has told you countless times that we are not math tutors, and you are clogging up this forum with useless trivial exercises.

Someone mentioned that there were not any good videos...

However, there are two Simon's Favorite Factoring Trick videos  :D

[url=http://www.artofproblemsolving.com/Movies/simon1.php]First Video[/url]

[url=http://www.artofproblemsolving.com/Movies/simon2.php]Second Video[/url]

[quote="archimedes1"][quote="bpms"][quote="archimedes1"]See, he misspelled it twice so it can't possibly be sharkman.
[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=668873435&t=140972
intentional[/quote]
my attempt at sarcasm[/quote]
oops, I'm not good at reading sarcasm off the internet


How consistent are all of you in math competitions like the AMC12 and AIME? This question is directed to both intermediate and advanced students. For example, on the AIME, would you sometimes score a 4 one day on a test, but a 9 another day on a different test. And for the more experienced, olympiad-takers, could you qualify for USAMO 9/10 times you took a pair of AMC12 and AIME tests?

I'm not as consistent as I would like to be. I used to have nightmares before the AIME that I'd only get a 4 and wouldn't make USAMO. I knew I shouldn't fear getting a 4 because if I were to work slowly and make sure not to make arithmetical errors, I would get much higher than that. But still I feared the worst. Only in 2003 was I completely confident that I would qualify for USAMO. In 2001 and 2002 I felt that I should make it but something disastrous could happen, and I wouldn't. I think if I were to take an AMC12 and an AIME today I'd get at least 138/9, which would make USAMO, but on a good day I could get as much as 150/14 or so, which is a big difference.

I'm consistent all right, I only took AMC once :)

Im NEVER consistent!  :D

I can get 9's on AIME ( sadly, im not that advanced yet) and 3's on CML.

I think people are usually pretty consistent on tests, especially at the higher levels. There's apparently only been one person in the history of AMC who has made MOP one year and not made it back (except for seniors anyway, obviously there are many that have gone before that don't make it back), and that one case was a kid that missed USAMO by 1 point.

I think [i]at any given time[/i] people are generally fairly consistent, but it changes over time. In my experience, any fluctuation is due to stupid mistakes, rather than the ability to do the math. So far as I can tell, how many problems people are capable of attaining at a given time is very consistent, but the number of those they botch by misreading the problem or whatever varies substantially.

However, people do get better or worse over long periods of time, so obviously their score one year may be very different from their score the previous year (for example, my AIME score- 4 in 2002, 13 in 2003).

Oh... I will edit my original post. Do you know how to strikeout text on this board? I've tried some different things, and I haven't quite figured out how.

I think I heard that from one of the instructors at MOP this year, because I mentioned how afraid I was of not making it back next year. Maybe what she said was that she only [i]knows[/i] of one person that happened to.

That is really surprising though; that kind of epitomizes a bad test day.

[quote="ComplexZeta"]Where did you hear that story about MOP? I happen to know that it's not true. I know someone who made MOP 2000 (when he was a freshman), but then he got a 1 on AIME 2001 and didn't make USAMO. He then got honorable mention in 2002 (I think) and 2003.[/quote]

was he ill that day or somat. i really dont see how that is possible otherwise.

yeah...making MOP one yr and only making a 1 on the AIME the next yr is a really big difference. And even if he was sick the day he took the AIME and got a 1, I would think he could do better than honorable mention on the USAMO the next 2 yrs.

Personally, i tend to be consistent in scoring lower than what i am capable of because of all the careless errors i tend to make. On this past yrs AIME, i knew the correct method for solving 12 of the problems, but unfortunately made stupid mistakes on 4 of them. Hopefully someday that will change =).


Let $ X$ be a nonempty set having $ n$ elements and $ C$ be a colour set with $ p\geq 1$ elements. Find the greatest number $ p$ satisfying the following property: If we colour in an arbitrary way each subset of $ X$ with colours from $ C$ such that each subset receives only one colour , then there exist two distinct subsets $ A$ , $ B$ of $ X$ such that the sets $ A$,$ B$,$ A\cup B$,$ A\cap B$ have the same colour.

No, as I wrote $ A$,$ B$ should be distinct and $ A\cap B\equal{}\emptyset$ :wink:

That actually isn't what you wrote  :wink: (You left out the condition $ A\cap B \equal{}\emptyset$.)

sorry  :oops: , my meant was $ A\cap B\equal{}\emptyset$.

[quote="Peter Scholze"]Is it allowed that $ A\subset B$? The answer is quite obvious in that case...[/quote]

Believe it or not, but the AMM problem (E3188) was meant the way you read it. It [i]is[/i] obvious...

Adding the condition $ A\cap B\equal{}\emptyset$ may make it more interesting...

  darij

But then only $ A$, $ B$ and $ A\cup B$ should have the same color. Indeed, that may give an interesting problem.


trigonometry
algebra
polynomial
product of roots
sum of roots

bump.  this is very similar to something i just posted.

[quote="Ravi B"]Evaluate $\cot^{2}\frac{\pi}{7}+\cot^{2}\frac{2\pi}{7}+\cot^{2}\frac{3 \pi}{7}$.

Reminder: $\cot$ (short for cotangent) is the reciprocal of $\tan$ (short for tangent).[/quote]
Here is some progress I made on this one.

Let $A = \frac{\pi}{7}, B = \frac{2\pi}{7}, C = \frac{4\pi}{7}.$

$E=\cot^{2}\frac{\pi}{7}+\cot^{2}\frac{2\pi}{7}+\cot^{2}\frac{3 \pi}{7}$
$=\cot^{2}\frac{\pi}{7}+\cot^{2}\frac{2\pi}{7}+\cot^{2}\frac{4 \pi}{7}$
$= \cot^{2}A+\cot^{2}B+\cot^{2}C$

We know that $\tan(A+B+C) = \frac{\sum_{cyc}{\tan A}-\tan A\tan B\tan C}{1-\sum_{cyc}\tan A\tan B}$
$\because A+B+C = \pi,$ we get
$\sum_{cyc}{\tan A}= \tan A\tan B\tan C$
$\Rightarrow \sum_{cyc}\cot A \cot B = 1$

Now, I only need to evaluate $\sum_{cyc}\cot A = \cot \frac{\pi}{7}+\cot \frac{2\pi}{7}-\cot \frac{3\pi}{7}$ to compute my original expression, $E.$

Any ideas?

[quote="jhcreinhardt"]$\sum_{cyc}{\tan A}= \tan A\tan B\tan C \Rightarrow \sum_{cyc}\cot A \cot B = 1$

i dont understand this step.  could you elaborate?[/quote]
$\sum_{cyc}{\tan A}= \tan A\tan B\tan C$
$\Rightarrow \tan A+\tan B+\tan C = \tan A\tan B\tan C$
$\Rightarrow \frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}= 1$
$\Rightarrow \cot A \cot B+\cot B \cot C+\cot C \cot A = 1$

[quote="Ravi B"]FieryHydra, that's an interesting approach.  I'll have to think about a good way to finish it off.

The approach I had in mind was different.  The idea is to find an explicit polynomial whose roots are $\cot^{2}\frac{\pi}{7}$, $\cot^{2}\frac{2\pi}{7}$, and $\cot^{2}\frac{3 \pi}{7}$.[/quote]
I did briefly doodle over the approach you mentioned but somehow couldn't find a way to use the same. The advantage of your approach is it is more general than the one I tried employing and that it can be used to solve a wider range of problems. Well, I will go over it once again.

On [url=http://www.q-e-d.be]a Belgian forum[/url] someone [url=http://www.q-e-d.be/index.php?p=forum&topic=3756]posted[/url] a general formula, i.e. \[\sum_{i = 1}^{n}\cot^{2}\left(\frac{i\pi}{2n+1}\right) = \frac{n(2n-1)}{3}.\] Anyway, no-one found a solution yet (maybe he did, dunno), but his hint was [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=489026#p489026]this question[/url].

[quote="Kurt Gödel"]On [url=http://www.q-e-d.be]a Belgian forum[/url] someone [url=http://www.q-e-d.be/index.php?p=forum&topic=3756]posted[/url] a general formula, i.e. \[\sum_{i = 1}^{n}\cot^{2}\left(\frac{i\pi}{2n+1}\right) = \frac{n(2n-1)}{3}.\] .[/quote]

This is a  cute way to show that $\sum (1/n^{2}) = (\pi^{2}/6)$.

Hint: [hide]Use $( cos x+i sin x)^{n}$  for odd $n$  to find a  formula for $sin(nx)$   and  see  when  $(sin (nx)/(sin(x))^{n}$   ( which  is  a  polynomial in $\cot^{2}x$ )   becomes zero.
[hide]Here $\frac{\sin(7x)}{\sin^{7}x}= 7 \cot^{6}x-35 \cot^{4}x+21 \cot^{2}x-1 = 7y^{3}-35 y^{2}+21 y-1$ (If $y=\cot^{2}x$ )
What are the sum of the roots of $y$ ?
(Another similar popular problem asked is related  to  the product of roots of y) [/hide]

 [/hide]

This is a  cute way to show that $\sum (1/n^{2}) = (\pi^{2}/6)$.

(Once you know sum of cot^2, you also know some of cosec^2 (which is 1+ cot^2)  and 1/x^2 lies between these two - so you can get upper and lower bound for  1/x^2 (and the sum)  etc. for a finite n etc..)
[hide="details"]
$\sum_{i = 1}^{n}\cot^{2}\left(\frac{i\pi}{2n+1}\right) = \frac{n(2n-1)}{3}$
So
$\sum_{i = 1}^{n}\csc^{2}\left(\frac{i\pi}{2n+1}\right) = \frac{n(2n-1)}{3}+n = \frac{n(2n+2)}{3}$
So
$\frac{n(2n+2)}{3}> \sum_{i = 1}^{n}\left(\frac{2n+1}{i\pi}\right)^{2}> \frac{n(2n-1)}{3}$
or $(\frac{\pi^{2}}6) (\frac{(2n)(2n+2)}{(2n+1)^{2}}) > \sum_{i = 1}^{n}\left(\frac 1{i^{2}}\right) > (\frac{\pi^{2}}6) \frac{(2n)(2n-1)}{(2n+1)^{2}}$
When n goes to infinity, both side limits tend to $\pi^{2}/6$ [/hide]

we can consider the equation

$(z+i)^{7}=(i-z)^{7}$ where $z$ is a complex number

then we have

$\left(\frac{z+i}{z-i}\right)^{7}=1$
$\implies \frac{z+i}{z-i}=e^{\frac{2\pi k}{7}}$ where $k=0,\pm 1,\pm 2,\pm 3$

or $z=\frac{i(e^{\frac{2\pi k}{7}}+1)}{e^{\frac{2\pi k}{7}}-1}$

some algebra show us that $z=\cot\frac{k\pi}{7}$ where $k=0,\pm 1,\pm 2,\pm 3$

hence $z^{2}=\cot^{2}\frac{k\pi}{7}$ ,$k=1,2,3$ are roots for equation

$(z+i)^{7}=(z-i)^{7}$ $\Longleftrightarrow 7z^{6}-35z^{4}+21z^{2}-1=0$

and sum of roots is $35/7=5$ .

Here is what amounts to the completion of FieryHydra’s approach to this problem.  (I have adopted his/her notation.)

Switching the $\frac{3 \pi}{7}$ argument to  $\frac{4 \pi}{7}$ achieved more than $A+B+C=\pi$.  We also now have

$B=2 A$
$C=2 B$
$A=2 C (mod\ \pi)$

Using $B=2A$, the cotangent formula

$\cot{2A}=\frac{ \cot^{2}A-1}{2 \cot A}$

Becomes

$2 \cot A \cot B = \cot^{2}A-1$

Because the factor-of-two relationship is cyclic in our problem, we also have

$2 \cot B \cot C = \cot^{2}B-1$
$2 \cot C \cot A = \cot^{2}C-1$

Adding these three equations and adding 3 to each side of the result yields

$3+2 \sum_{cyc}\cot A \cot B = \sum_{cyc}\cot^{2}A = E$

FieryHydra already showed that

$\sum_{cyc}\cot A \cot B = 1$

Therefore $E = 5$.

Note: kudos to syhong for his/her slick roots-of-unity solution.[/quote]

Yes, kudos to [b]syhong[/b] for his nice solution and thanks to [b]likely[/b] for completing my partial solution. :)


superior algebra
superior algebra theorems

Let $k \geqslant 2$, find out the biggest integer $n$ s.t. for all prime $p$ and all $i\in \{1, p^{k}-1 \}$ $p^{n}$ divides $\binom{p^{k}}{i}$.

P.S: I'm note sure this is the right place for it, and it must be well known but I couldn't find it on this forum.

${0\equiv \left( \begin{array}{c}p^{k}\\ i \end{array}\right) \bmod p^{n}}$
${0\equiv \frac{p^{k}!}{i! \left(p^{k}-i\right)!}\bmod p^{n}}$

${\frac{p^{k}!}{i! \left(p^{k}-i\right)! p^{n}}=K}$

${\frac{p^{k}\left(p^{k}-1\right)!}{i! \left(p^{k}-i\right)! p^{n}}=K}$

${\frac{p^{k}}{p^{n}}=S}$

${k=n}$

i din't understand your question

binomial are every integers..
so the biggest $n$ is $k$

Hi,

The answer is not $k$.
To see it, for instance, $5$ divides $\binom{5^{2}}{5}$, but $5^{2}$ does not divide $\binom{5^{2}}{5}$.


What is the value of $ 5(3\cdot 2) \minus{} 3(2\cdot 5)$?

hello, you can write the difference in the form
$ 5\cdot(3\cdot2)\minus{}3\cdot(2\cdot5)\equal{}2\cdot3\cdot5\minus{}2\cdot3\cdot5\equal{}0$.
Sonnhard.

Or you can just use the Asssociative property.


We have a circle which is inscribed inside a square. Another circle is then inscribed about the square, and another square is inscribed about that circle, and so on until we have 3 circles and 3 squares. What is the area of the region between the largest square and the largest circle?

We don't have any measures....  :(  do you want a ratio or something?

oops - you can tell I made it up as I typed. The smallest circle has radius 1. :)

[hide]
smallest circle has radius 1
smallest square has half length of side=1
medium circle has radius $\sqrt{2}$
largest circle has radius $\sqrt{\sqrt{2}+\sqrt{2}} $
Largest square has side $2\cdot\sqrt{2\cdot\sqrt{2}} $
Area of largest circle $2\sqrt{2}\pi$
Area of largest square $8\sqrt{2}$

$8\sqrt{2}-2\sqrt{2}\pi$?[/hide]

[quote="JesusFreak197"]How did you go from step 3 to step 4? Try checking your work. ;)[/quote]

oops, :oops:
[hide]
smallest circle has radius 1
smallest square has half length of side=1
medium circle has radius $\sqrt{2}$
largest circle has radius $\sqrt{4} $
Largest square has side $2\cdot\sqrt{4} $
Area of largest circle $4\pi$
Area of largest square $4\cdot4=16$

$16-4\pi$?[/hide]

[quote="biffanddoc"]Hmmm... You seem to not be able to spell largest :P[/quote]

Woops!  :oops: I'll edit that.


algebra
polynomial
Vieta
number theory unsolved
number theory

Different lattice points of the plane have different distances from $( \sqrt{2}, \frac{1}{3})$

You want to show that integers $(x,y)$ are uniquely defined by $\left(x-\sqrt{2}\right)^2+\left(y-\frac{1}{3}\right)^2 = x^2-2\sqrt{2}x+2+y^2-\frac{2}{3}y+\frac{1}{9}$.
So let a distance of type $t+2u\sqrt{2}$ with $t,u$ rational be given.
Since $a+b\sqrt{2}=c+d\sqrt{2}$ for rational $a,b,c,d$ only happens for $a=c, b=d$, we get that $x=u$ is uniquely defined by the distance (there is just one term with $\sqrt{2}$: $2x\sqrt{2}$).
Also we can find out $y$ from it by $\left(y-\frac{1}{3}\right)^2 = t-u^2-2$. But the two solutions of this equation differ by $\frac{2}{3}$ by Vieta, so only one can be an integer, thus there is only one $y$.
So $x,y$ are uniquely defined by the distance.

There is just one such point/pair/number.

PS: "distance" is above the square of the distance.


is there a quick way to caculate stuff like cos (7x)-or any large multiple of an angle. without having to go through cos(3x+4x)..and expand

[hide]
$e^{i\theta}= \cos{\theta}+i\sin{\theta}$

$(e^{i\theta})^{7}= (\cos{\theta}+i\sin{\theta})^{7}$
$e^{7i\theta}= \cos{7\theta}+i\sin{7\theta}$

$(\cos{\theta}+i\sin{\theta})^{7}= \cos{7\theta}+i\sin{7\theta}$

expand and equate the real and imaginary portions to get $cos{7\theta}$ and $sin{7\theta}$
[/hide]


algorithm
modular arithmetic
number theory

Does anyone know a quick way to find base numbers such as:

145 in base 3

How can I do that quickly?

A proof or some sort of explanation would be appreciated.

Also, I do not understand the concept of "mod". I think it is called modulant in full. How to find this easily and what are the applications of this? An simple example problem and/or proof would be nice.  :D

This is the algorithm I was taught:
When dividing $145$ by $3$, the quotient is $48$ and the remainder is $1$. Proceed this way:
$145=3\times 48+\boxed{1}$
$48=3\times 16+\boxed{0}$
$16=3\times 5+\boxed{1}$
$5=3\times 1+\boxed{2}$
$1=3\times 0+\boxed{1}$
Then
$145=3(3(3(3(1)+2)+1)+0)+1=1\times 3^{4}+2\times 3^{3}+1\times 3^{2}+0\times 3^{1}+1\times 3^{0}$
and $\boxed{145_{(10)}=12101_{(3)}}$
(notice they are the boxed numbers in inverse order).
After a little practice, one only writes the important numbers
$145$
$48$, $1$
$16$, $0$
$5$, $1$
$1$, $2$
$0$, $1$

As for the other question, two numbers $a$ and $b$ are equivalent $\mod c$ if $a-b$ is a multiple of $c$,
or (it's exactly the same) if dividing $a$ and $b$ by $c$ you obtain the same remainder $r$ (often called the representative),
so sometimes it's written $a\mod c\equiv b\mod c\equiv r$.
For instance $14\mod 3\equiv 8\mod 3\equiv 2$.

"Mod" usually refers to "modulo" or "modulus." An explanation was provided in the previous post. Mods are extremely useful in number theory, for example:

Determine the remainder of $2^{100}$ when divided by $101$. A well-known theorem (Fermat's Little) relating to mods would tell us that $2^{100}\equiv 1 \pmod{101}$.

Another example:

Find all integers $x, y, z$ such that

$x^{2}+y^{2}+z^{2}= 8007$.

If we consider all possible values of squares $\pmod{8}$, for example, we can make a nice conclusion about this problem.

Mods show up all the time on the forum, too, so if you want to learn more just hang around.

For paladin's problem:

$8007\equiv 7 \pmod{8}$

Since, all squares end in $0, 1,$ or $4\pmod{8}$, there exist no such combination of three where $0,1,4$ can add up to 7.

Hence, there are no integer solutions to the above equation.


Ακολουθώντας την ιδέα του Μπάμπη, εδώ βάζω τις μεταφράσεις όλων των θεμάτων τις άλγεβρας του 2000 από το IMO Compendium. (Ελπίζω να μην έχω κάνει οποιοδήποτε λάθος στην μετάφραση.)

Περιμένω τις λύσεις σας! Αν έχετε διαφορετική λύση απο αυτές που ήδη έχουν γραφτεί βάλτε την και αυτή.

[b]2000 Α1 (USA):[/b] Δίνονται θετικοί πραγματικοί αριθμοί $ a,b,c$ με $ abc \equal{} 1$. Να δειχθεί ότι $ \left(a \minus{} 1 \plus{} \frac{1}{b} \right)\left(b \minus{} 1 \plus{} \frac{1}{c} \right)\left(c \minus{} 1 \plus{} \frac{1}{a} \right) \leqslant 1$.

[b]2000 Α2 (GBR):[/b] Δίνονται θετικοί ακέραιοι $ a,b,c$ με $ b > 2a$ και $ c > 2b$. Να δειχθεί ότι υπάρχει πραγματικός αριθμός $ t$ με την ιδιότητα ότι και οι τρεις αριθμοί $ ta,tb,tc$ έχουν το κλασματικό μέρος τους στο διάστημα $ (1/3,2/3]$.

[b]2000 Α3 (BLR):[/b] Να βρεθούν όλα τα ζεύγη συναρτήσεων $ f: \mathbb{R} \to \mathbb{R}$, $ g: \mathbb{R} \to \mathbb{R}$ έτσι ώστε \[ f(x \plus{} g(y)) \equal{} xf(y) \minus{} yf(x) \plus{} g(x)\]   για κάθε $ x,y \in \mathbb{R}$.

[b]2000 Α4 (GBR):[/b] Η συνάρτηση $ F$ ορίζεται στο σύνολο των μη αρνητικών ακεραίων και παίρνει μη αρνητικές ακέραιες τιμές ικανοποιώντας τις ακόλουθες συνθήκες:
Για κάθε $ n \geqslant 0$:
(i) $ F(4n) \equal{} F(2n) \plus{} F(n)$;
(ii) $ F(4n\plus{}2) \equal{} F(4n) \plus{} 1$;
(iii) $ F(2n\plus{}1) \equal{} F(2n) \plus{} 1$;

Να δειχθεί ότι για κάθε θετικό ακέραιο $ m$ , ο αριθμός των ακεραίων $ n$ που ικανοποιούν $ 0 \leqslant 2^m$ και $ F(4n) \equal{} F(3n)$ είναι $ F(2^{m\plus{}1})$.

[b]2000 Α5 (BLR):[/b] Δίνεται θετικός ακέραιος $ n \geqslant 2$ και θετικός πραγματικός $ \lambda$. Αρχικά, $ n$ ψύλλοι βρίσκονται πάνω σε μια οριζόντια ευθεία αλλά όχι όλοι στο ίδιο σημείο. Ορίζουμε μια κίνηση διαλέγοντας δυο ψύλλους που βρίσκονται στα σημεία A και Β, με το Α στα αριστερά του Β, αφήνωντας τον ψύλλο στο Α να πηδήξει πάνω από τον Β στο σημείο C έτσι ώστε $ BC/AB \equal{} \lambda$.

Να βρείτε όλες τις τιμές του $ \lambda$ έτσι ώστε για κάθε σημειο Μ της ευθείας, και για κάθε αρχική διάταξη των $ n$ ψύλλων, να υπάρχει μια ακολουθία κινήσεων που να μεταφέρει όλους τους ψύλλους στα δεξιά του Μ.

[b]2000 Α6 (IRE):[/b] Ένα μή κενό σύνολο Α ονομάζεται $ B_3$-σύνολο αν οι συνθήκες $ a_1,\ldots,a_6 \in A$ και $ a_1 \plus{} a_2 \plus{} a_3 \equal{} a_4 \plus{} a_5 \plus{} a_6$ συνεπάγονται ότι οι τριάδες $ (a_1,a_2,a_3)$ και $ (a_4,a_5,a_6)$ είναι ίσες (ως προς μία μετάθεση).

Δίνονται άπειρα σύνολα $ A \equal{} \{a_0,a_1,a_2,\ldots\}$ και $ B \equal{} \{b_0,b_1,\ldots \}$ με $ 0 \equal{} a_0 < a_1 < \ldots$, $ 0 \equal{} b_0 < b_1 < \ldots$ και $ D(A) \equal{} D(B)$ όπου $ D(X) \equal{} \{|x\minus{}y|: x,y \in X \}$. Αν το Α είναι ένα $ B_3$-σύνολο, να δειχθεί ότι $ A \equal{} B$.

[b]2000 Α7 (RUS):[/b] Για ένα πολυώνυμο $ P$ βαθμού 2000 με διαφορετικούς συντελεστές, το $ M(P)$ είναι το σύνολο όλων των πολυωνύμων που μπορούν να παραχθούν από το $ P$ με μετάθεση των συντελεστών του.  Ένα πολυώνυμο $ P$ θα ονομάζεται $ n$-ανεξάρτητο αν $ P(n) \equal{} 0$ και μπορούμε για κάθε $ Q \in M(P)$ να πάρουμε ένα πολυώνυμο $ Q_1$ με $ Q_1(n) \equal{} 0$ εναλλάσσοντας το πολύ ένα ζεύγος συντελεστών του $ Q$. Να βρεθούν όλοι οι ακέραιοι $ n$ για τους οποίους υπάρχουν  $ n$-ανεξάρτητα πολυώνυμα.

Μπραβο για τη δουλεια  Δημητρη  :thumbup:
Για την ανισοτητα βρηκα μια πολυ ωραια και απλη λυση απο το βιβλιο των Στεργιου-Σκομπρης "Κλασικες και νεες ανισοτητες", και ετσι την βαζω
Το σημειο κλειδη της ασκησης ειναι να κανεις την κλασικη αντικατασταση οταν εχεις γινoμενο $ 1$....
Εχουμε λοιπον
$ a\equal{} \frac{x}{y},b\equal{} \frac{y}{z}, c\equal{} \frac{z}{x}, x,y,z>0$
Αρα η ανισοτητα μας γραφεται
$ (\frac{x}{y}\minus{}1\plus{} \frac{z}{y}) (\frac{y}{z}\minus{}1\plus{} \frac{x}{z})(\frac{z}{x}\minus{}1\plus{}\frac{y}{x}) \leq 1$

$ \leftrightarrow (x\plus{}y\minus{}z)(x\minus{}y\plus{}z)(\minus{}x\plus{}y\plus{}z)\leq xyz$
Εαν δεν ειναι και οι $ 3$ παραγοντες $ (x\plus{}y\minus{}z),(x\minus{}y\plus{}z),(\minus{}x\plus{}y\plus{}z)$ θετικοι τοτε ευκολα συμπεραινουμε την ανισοτητα....
Εστω οτι ειναι και οι $ 3$ θετικοι...
Τοτε εχουμε να λυσουμε ,μια πολυ κλασικη ανισοτητα,τη
$ (x\plus{}y\minus{}z)(x\minus{}y\plus{}z)(\minus{}x\plus{}y\plus{}z)\leq xyz$
Εστω $ x\plus{}y\minus{}z\equal{}k,x\minus{}y\plus{}z\equal{}n,\minus{}x\plus{}y\plus{}z\equal{}m$.Tοτε
$ x\equal{} \frac{k\plus{}n}{2}, y\equal{} \frac{k\plus{}m}{2}, z\equal{} \frac{n\plus{}m}{2}$
Αρα αρκει $ (k\plus{}n)(m\plus{}n)(m\plus{}k) \geq 8knm$
Που φυσικα ισχυει απο την θεμελειωδη $ a\plus{}b \geq 2 \sqrt{ab}$
H ισοτητα ισχυει οταν $ k\equal{}n\equal{}m$ αρα $ x\equal{}y\equal{}z$ ,αρα $ a\equal{}b\equal{}c$ Και επειδη το $ abc\equal{}1$
$ a\equal{}b\equal{}c\equal{}1$

Θα περιμένω ακόμη μια βδομάδα πριν να βάλω τις επίσημες λύσεις. Προς το παρόν, βάζω σε νέο άρθρο μερικά καινούργια προβλήματα.

Έμειναν αρκετό καιρό γι'αυτό βάζω τις λύσεις:

[u][size=150]2000 A2 GBR:[/size][/u]

Παρατηρούμε ότι το κλασματικό μέρος του $ ta$ ανήκει στο $ (1/3,2/3]$ αν και μόνο αν $ t \in I_k \equal{} \left(\frac {k \plus{} 1/3}{a},\frac {k \plus{} 2/3}{a}\right]$ για κάποιο ακέραιο κ. Ομοίως πρέπει $ t \in J_m \equal{} \left(\frac {m \plus{} 1/3}{b},\frac {m \plus{} 2/3}{b}\right]$ και $ t \in K_n \equal{} \left(\frac {n \plus{} 1/3}{c},\frac {n \plus{} 2/3}{c}\right]$ για κάποιους ακεραίους m,n.

Πρέπει να δείξουμε λοιπόν ότι $ I_k \cap J_m \cap K_n \neq \emptyset$ για κάποιος ακεραίους k,m,n. Επειδή $ \frac {2}{3c} < \frac {1}{3b}$, κάθε $ J_m$ τέμνει κάποιο $ K_n$. (Τα $ J_m$ έχουν μήκος $ \frac {1}{3b}$ και κάθε δυο διαδοχικά $ K_n$ έχουν απόσταση $ \frac {2}{3c}$.)

Αρκεί λοιπόν να δείξουμε ότι $ J_m \subseteq I_k$ για κάποιους ακεραίους k,m. Θέτω $ u_m \equal{} \frac {m \plus{} 1/3}{b}, v_m \equal{} \frac {m \plus{} 2/3}{b}$. Έχουμε $ av_m \equal{} au_m \plus{} a/3b < au_m \plus{} 1/6$.

Αρκεί λοιπόν να βρούμε ακέραιο m ώστε το κλασματικό μέρος του $ au_m$ ανήκει στο $ [1/3,1/2]$.

Έστω $ a \equal{} d \alpha, b \equal{} d\beta$ με $ (\alpha,\beta) \equal{} 1$. Θέτω $ m \equal{} d \mu$. Τότε $ au_m \equal{} \frac {\alpha \mu}{\beta} \plus{} \frac {\alpha}{3 \beta}$. Αρκεί να δείξω ότι μπορώ να διαλέξω μ ώστε το κλασματικό μέρος του $ \frac {\alpha \mu}{\beta} \plus{} \frac {\alpha}{3 \beta}$ ανήκει στο $ [1/3,1/2]$.

Επειδή $ (\alpha,\beta) \equal{} 1$, αρκεί να διαλέξω $ j$ ώστε το κλασματικό μέρος του $ \frac {j}{\beta} \plus{} \frac {\alpha}{3 \beta}$ ανήκει στο $ [1/3,1/2]$.

Αν $ \beta \geqslant 6$ τότε αυτό σίγουρα γίνεται επειδή $ 1/2 \minus{} 1/3 \equal{} 1/6 \geqslant 1/\beta$.  Αν $ \beta \leqslant 5$, τότε η συνθήκη $ 2a < b$ δίνει ότι οι μόνες πιθανές τιμές του α/β είναι 1/3,1/4,1/5,2/5. Παρατηρούμε ότι τα μ = 1,1,2,3 αντίστοιχα δουλεύουν.

Επειδή η μετάφραση παίρνει τελικά αρκετή ώρα:

Υπόδειξη για 2000Α3:
1) Δείξτε ότι υπάρχει y ώστε g(y)=0
2) Δείξτε ότι υπάρχει $ c \neq \minus{}1$ ώστε $ f(x) \equal{} \frac{cx\minus{}c^2}{1\plus{}c}, g(x) \equal{} cx \minus{} c^2$.

Υπόδειξη για 2000Α4: Δείξτε ότι αν $ n \equal{} \sum a_k 2^k$, με $ a_k \in \{0,1\}$, τότε $ F(n) \equal{} \sum a_k F_{k\plus{}1}$, όπου $ F_1,F_2,F_3,\ldots$ είναι η ακολουθία Fibonacci.

Υπόδειξη για 2000Α5: $ \lambda \equal{} 1/(n\minus{}1)$.

Υπόδειξη για 2000Α7: Υπάρχουν 0-ανεξάρτητα και 1-ανεξάρτητα πολυώνυμα.
Το ακόλουθο λήμμα βοηθάει για να δείξετε πως δεν υπάρχουν ν-ανεξάρτητα πολυώνυμα για $ |n| \geqslant 2$. (Για ν=-1 πρέπει να δουλέψετε ξεχωριστά.)

Λήμμα: Για κάθε $ n \geqslant 2$ και για κάθε πολυώνυμο P βαθμού m με διαφορετικούς συντελεστές το σύνολο $ \{ Q(n): Q \in M(P)\}$ έχει τουλάχιστον $ 2^m$ στοιχεία.


At a picnic of at least three people, a plate of 26 cookies is passed around. Each person takes one cookie and passes the plate to the next person. Jana gets the first cookie and the next to last cookie. What is the median of all possible numbers of people at the picnic?

As she gets 1st and 25th cookie, the number of people is a divisor of 25-1=24, so it's 3,4,6,8,12 or 24. Median: (6+8)/2=7


When simplified, $ (x^{ \minus{} 1} \plus{} y^{ \minus{} 1})^{ \minus{} 1}$ is equal to:

$ \textbf{(A)}\ x \plus{} y \qquad\textbf{(B)}\ \frac {xy}{x \plus{} y} \qquad\textbf{(C)}\ xy \qquad\textbf{(D)}\ \frac {1}{xy} \qquad\textbf{(E)}\ \frac {x \plus{} y}{xy}$

[hide="Click for solution"]
$ (x^{\minus{}1}\plus{}y^{\minus{}1})^{\minus{}1}\equal{}\frac{1}{\frac{1}{x}\plus{}\frac{1}{y}}\equal{}\frac{xy}{x\plus{}y}$, or $ \boxed{\textbf{(B)}}$.
[/hide]


An integer is chosen at random from $ \{x: 100<x<300\}$. Find the probability that this integer is divisible by 7 or 11.

[quote="sunehra"]An integer is chosen at random from $ \{x: 100 < x < 300\}$. Find the probability that this integer is divisible by 7 or 11.[/quote]
$ 7 \times 15 \equal{} 105$
Upto $ 7 \times 42 \equal{} 294$ gives $ 42 \minus{} 15 \plus{} 1 \equal{} 28$ favourable events.
Similarly for 11.
Hence,
total=$ 28 \plus{} 18 \equal{} 46$
Answer:
$ \frac {44}{199}$

[hide] There are $ \lfloor{\frac{300}{7}}\rfloor\equal{}42$ multiples of 7 under 300 while there are 14 under 100. Hence, there are 28 numbers divisible by 7 in our range. Similarly, there are $ \lfloor{\frac{300}{11}}\rfloor\equal{}27$ multiples of 11 under 300, and 9 under 100, so there are 18 in our range. We are overcounting the numbers divisible by both 7 and 11 in the range, which are 154 and 231. There are 299-101+1=199 numbers in the range, so our answer is $ \frac{44}{199}$.[/hide]

I probably messed up.

exactly, about to post about overcounting, then i realized that tinytim and xpmath already did.

[quote="tinytim"]Xpmath is correct. Rajiv should have subtracted 2 from 46 because the numbers 154 and 231 were counted twice (because they are mutliples of both 7 and 11).[/quote]
Yup right! I didn't think about that! I'll correct it...
Thanks.


The 1x1 squares in opposite corners of a chessboard are removed. Show that the chessboard cannot be tiled completely with 1x2 rectangles without overlap.

is this from one of texas state u.'s math explorer magazines?  I was just reading that this afternnon.  :lol:

I read it in a book called Fermat's Enigma. I was just reading while my dad was browsing Barnes & Noble.

This is really similar to a problem I did at a math olympiad involving a checkerboard.  :lol:

The problem is easy if you think about it:

In an 8*8 chessboard opposite squares have the same color. If you remove two opposite squares, instead of having 32 white squares and 32 black sqrares the chessboard has 32 of one color and 30 of the other.

A 1*2 rectangle, by definition, has to 1/2 of it on a white square and 1/2 of it on a black square. However, there are more of one color that of the other. Therefore, it is impossible to completely cover the board with 1*2 rectangles. QED.

the rectangle covers 1 square of one of the colors, and 1 square of the other color because the rectangle covers 2 vertically or horizontally adjacent squares, which are of different colors.

[quote="perfect_radio"][quote="NoSoupForYou"]
A 1*2 rectangle, by definition, has to 1/2 of it on a white square and 1/2 of it on a black square.[/quote]

If you consider that a 1*2 rectangle is just a rectangle(no colors) does the result still hold?
(my guess is yes, but can't prove it)[/quote]

Yeah it doesn't have to be a checkerboard, but even if isn't, the coloring technique makes the problem easy! :) For example, "The 1x1 squares in opposite corners of a 8x8 board are removed. Show that the board cannot be tiled completely with 1x2 rectangles without overlap" could be solved simply by coloring the board like a chessboard, and then using NSFY's method.


Can you reveal the source of problems from previous rounds?


Bobby Shen, as the National MATHCOUNTS Champion, appeared on the NBC Today Show on May 11, 2009! As a guest there, he got to play a Math-Off mini countdown style competition with the hosts, and won!

Here's a link to the video:

[url]http://vids.myspace.com/index.cfm?fuseaction=vids.individual&VideoID=57219129[/url]

EDIT: date typo fixed. thanks, mathlete5. idk why i keep mixing may with march :blush:

Okay wow...that video is so....fail. Bobby just owned them.  :rotfl:

[quote="echen"]Is this what they think math is? Adding large numbers.[/quote]

Quite. I was really disappointed/angry at them for displaying those problems to perhaps millions of people watching. Not only are those problems "trivial" and demean the field of mathematics in general, but they also lessened Bobby's accomplishment. By showing such easy problems that anyone can solve, they just ruined the whole thing.

Idk, thats just me, but all those problems were computational, without a hint of problem-solving whatsoever.

I kinda agree with echen. A lot of people who aren't so aware in math usually think math is all about large numbers, which I get pissed off on a lot. Many people in my school think that I am good at math just because they think I can add large numbers in my head quickly, which I clearly can't do EXTREMELY well, and neither can many other mathy people I know.

[quote="echen"]Is this what they think math is? Adding large numbers.[/quote]

That's better than a lot of people I know, who think that math is about solving word problems that are essentially 3+5 or something in very complicated ways. So they think I suck at math since I solve those by saying "the answer is 3+5" and they use unnecessary overkill...

uh, actually I thought those questions were extremely typical of mathcounts. if its creativity in the problems you want, perhaps you should look elsewhere.

[quote="PowerOfPi"]I think math is a tool to use to benefit you, not to show off and be the best adder. They seriously made math look like a game and made it look useless. Also, MathCounts is not the point of math, it is a motivation for you not to lose interest in it, as many other people do. It is to show to you that it can be fun.[/quote]

I wouldn't describe MathCounts problems as fun. They are definitely less boring than school math, but they aren't exactly that fun.

[quote="alex960531"][quote="echen"]Another reason I do not like this video is because Bobby Shen is a great mathematician. This video did not show that and if anything, made him look worse then he really is. By the problems that were given, you cannot see just how good Bobby really is. If they at least showed some clips of Nats CD or something, it would be better.[/quote]

Or they should've put Daivd and Bobby put together and make it really competitive.  :wink:[/quote]

I like this suggestion. But add in USAMO #1 level problems.

Was there only this on the actual show? Or was there more? Since.. 4 minutes seems a bit short for a show.

No, I think there was also an interview. They just cut out that part and only video recorded the math-off part.

[quote="levans"]Wow, this video made me sick. It shows awful math problems and the hosts were loud and obnoxious the whole time. The worst part, however, is when the hosts all but said that math is useless. They did this when they talked about how bad they are at math. If you can make 2 million dollars a year hosting a morning television program while being bad at math, what is to convince others that they should learn math?

I wish instead of holding the silly little mock competition that they instead showed some of the actual countdown round and then did more of an interview with Bobby. They hardly talked to Bobby at all.[/quote]

That's true. In the last year's Darryl Wu interview, (Forget the show.) the people humiliated Darryl and said, "Take that Wu!". Calling someone by their last name isn't very respectful.


topology geometry invariant function calculus integration Gauss

I have a question that I think falls into the realm of topology, the fundamental group, and maybe Morse theory, but I am a novice in these areas and I need some help!

Consider the following image:

http://www.geocities.com/yapquack/Slide1.jpg

(I've also included the image as an attachment.)

Consider a loop (map from the circle) in a four dimensional manifold, like in the picture above. (The picture shows a 2-manfold, but I can't exactly draw a 4-manifold ...) Assume also that the 4-manifold is multiply connected. In the picture, the loop phi3 can be continuously deformed to a point. However, phi2 cannot, because it has a "handle' or a "hole' blocking the way. Phi1 also cannot, because it has *two* holes blocking the way.

Actually I have two questions. First: remembering that we are dealing with loops in 4-space, what is the name of the topological invariant that we give to the "number of holes" that are in the way?

Second: Is it possible to define a function f(x1, x2, x3, x4) over the  4-manifold such that the integral of f around any given loop equals the "number of holes" as discussed above??

straycat

By the way, I've looked into trying to apply the Gauss-Bonnet formula, or Green's theorem, or Stokes' theorem, or Cauchy-Goursat, or the residue theorem of complex analysis, but I can't quite figure out how to apply them to my problem. By the way #2: I think that I wish to assume that my function f is everywhere analytic and nowhere singular.

-s

Your question is addressed by a field of mathematics called Cohomology. You can find a good introduction to the field in a book called [u]From Calculus to Cohomology[/u] by Madsen and Tornehave. It presupposes only a knowledge of Algebra and Multivariable Calculus.


My solution.
change the nesbit inequality's RHS-LHS to SOS form
and then change 2-(6(ab+bc+ca)/(a+b+c)^2)to SOS form
subtract and multiply 2(a+b)(b+c)(c+a)(a+b+c)^2
to arrive at
sigma (a-b)^2((a^2+b^2-c^2)(a+b))>0
and Wlog a<b<c
S_b+S_c>0 can be shown easily.


AMC
USA(J)MO
USAMO
summer program
PROMYS
ARML
geometry

Calls start tomorrow for MOP... anyone that gets a call, put what they found out in the poll, and post if you feel like it, so we can all contratulate what shall be the AoPS delegation in Lincoln.

How do you think I knew that the calls were coming out? :-)

I also know my score is in the 28-34 interval, which is pretty much as expected.

[quote="tetrahedr0n"]how did you do Alison?[/quote]

If you are a winner, then don't you get to go to MOP automatically as well?

[quote="polymorphic"]wow, at least 1/3 of the winners from here?  hmm...well if it's true, that's very impressive[/quote]

Doubt it. Most people are just abusing the poll. If they are indeed winners, then they should and would have left some comments.

You should probably add another option: "MOP? In my dreams!"  (this is what would fit me :-)).  This is, of course, only if you can (I have never started a poll before, so I don't know).

[quote="intrepia"][quote="Alison"]The calls for the winners are going out now.  It looks like we have another one here -- who would that be?[/quote]

Mysterious persons, of course. ;-)[/quote]

At least, two of the three anonymous winners are mysterious persons... :-)

Have any ninth graders gotten calls yet?

[quote="apix"]At least, two of the three anonymous winners are mysterious persons... :-)[/quote]

Maybe two of the two anonymous winners? ;-)

Have no honorable mentions or regular MOPpers from this board recieved calls yet? I can't imagine the calls haven't been made, but it seems like there'd at least be some of the 9th graders.

No call to my house today... no big deal I guess, but it seems like making it Sophomore year should have meant I'd make it this year, but apparently not. I think I'm going to go to sleep and try to forget about it. Though I don't think it's going to work.

I know of 2 "regular MOPers" who got their calls today, so at least some of the calls have been made. I'm guessing the USAMO cut-off was around 14-16.

I think I got 14-16, maybe 17. Maybe my proofs were worse than I thought. Argh, I wanted this more than anything; I don't know what to do now. I keep trying to think of reasons why maybe I will be called but they haven't got to me yet, but I think it's wishful thinking...

[quote="Kamior"]Actually, at least one ninth grader (me) "made it the hard way," assuming you were just referring to exluding the freshman bonus. I didn't make HM (scored 17), but I qualified once the HM seniors and Canadians were eliminated.[/quote]

That's interesting to know. By the way, have you met forum user Picounts? He was looking for you earlier, as you and he are both Mathcounts veterans from last year.

Congratulations.

[quote="Kamior"]Actually, at least one ninth grader (me) "made it the hard way," assuming you were just referring to exluding the freshman bonus. I didn't make HM (scored 17), but I qualified once the HM seniors and Canadians were eliminated.

-Adam Hesterberg[/quote]

Congrats. At least we'll have some freshman representation at the "normal" MOP. Hmm...maybe Sherry Gong got close to your score and was also invited to the "normal" MOP?

I got a 17 as well.  I'm still kicking myself for forgetting to put equality conditions for #1  :shock:

With them I would have 18 points and the Honourable Mention  :P

   -Oleg Rudenko, FL

No, I haven't met Picounts - I didn't know he was looking for me.

I would be really surprised if Sheery Gong didn't have around a 17, given that she had an 18 at the IMO. Maybe she had a bad day...

[quote="Kamior"]
I would be really surprised if Sheery Gong didn't have around a 17, given that she had an 18 at the IMO. Maybe she had a bad day...[/quote]

Hmm...I didn't know she went to IMO...

But in any case, it'd be good to have her in the MOP too... :-D but then that makes only 23 freshmen in the freshmen MOP...

I wonder if they invited 7 other people...

Are the two MOPs going to be separate? Because if so, I don't know which one I will end up in... since I'm a sophomore who made freshman MOP.  :?

Anyway, did any of you receive any notification by mail or e-mail from them yet?

[quote="MithsApprentice"][quote="zfzf21"]... since I'm a sophomore who made freshman MOP.  :?
[/quote]

A sophmore??...that made freshman MOP?? *confused*[/quote]

They didn't have enough freshmen, so they took 5 extra sophhomores.

[quote="zfzf21"]
They didn't have enough freshmen, so they took 5 extra sophhomores.[/quote]

oh!...so that's what they did to fill the ranks!...see?...class of 06 got pretty lucky anyways...

was it like the top 5 soph USAMO scorers not already going or wat?

[quote="zfzf21"]Are the two MOPs going to be separate?...[/quote]
I want to know that too. Does anyone have an answer?

My guess is that it will be like MOP 2002, where the two MOPs are roomed together, have the same instructors and graders, but have classes and tests separately.  I hope Steve Dunbar comes around and gives an official answer on this shortly.

[quote="Alison"]My guess is that it will be like MOP 2003, where the two MOPs are roomed together, have the same instructors and graders, but have classes and tests separately.  I hope Steve Dunbar comes around and gives an official answer on this shortly.[/quote]

Do you mean 2002?

In any case, that would make sense. Putting freshmen USAMO qualifiers and people like Tiankai in the same class and expecting them to be able to solve problems of the same difficulty level would be quite ridiculous.

But it would be nice to room together, hang out, chill etc. without segregation.

I wouldn't know, because I didn't go last year, but I heard (in 2003):
Black: IMO Team
Blue: Honorable mentions and other winners that did not go on to IMO team.
Red: others invited to MOP

(Yellow: Freshmen?)

Can anyone confirm this?

yay!...MOP invitations came in today...(along with the forms)...

Yes, I meant MOP 2002.  Edited now.

Nukular, the color distinctions you had for 2003 are roughly right.  To be precise it was
Black: USAMO winners (or, after TSE, IMO people)
Blue: other people who had been to a small MOP before, or had been in Black or Blue group (winners / honorable mentions) in 2002.
Red: everyone else.

My guess is pretty much the same as Nukular's, that they will add a Yellow group this year that will include the freshmen.

Last year, the dividing lines were somewhat flexible, so it's possible that there will be some interchange between the groups as people find their level.


Or, write $\left(\begin{matrix}2 & 1\\ 1 & 0 \\ \end{matrix}\right)=E+B$ and we see $B^{2}=2E$.


abstract algebra
group theory
advanced fields
advanced fields unsolved

Let $G$ be an abelian group and $T = tG$ the torsion elements of $G$. Suppose that for some $n \in \mathbb Z$, $ng = 0$ for all $g \in T$. Then show that $\text{Ext}^{1}(G/T, T) = 0$, and use this to prove that $T$ is a direct summand of $G$.

Describe the effect of multiplication by n on this Ext group. owk

My guess to to some how use the following sequence, but I am not sure how.
\[0 \to tG \to G \to nG \to 0 \]
Then use the fact that $nG$ is torsion free some how, but again I am not sure what to do.

Torsion-freeness does not tell much about Ext groups, as we have learned recently ;-) owk


Now that G&FF post don't count, many people have had their post counts devastated.  Which AoPS members have had the biggest decrease in post count?  I'll start off with....Bubka!  He's down to 37 posts now.  :rotfl: Who else?


[quote="Stewie"]Let $ P(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ e $ Q(x) \equal{} x^2 \plus{} px \plus{} q$. Prove that there's a $ t \in \mathbb R$ s.t. $ P(t) < Q(t)$.[/quote]

That's obviously wrong.

Take $ P(x)\equal{}x^4\plus{}1$ and $ Q(x)\equal{}x^2$ : $ P(t)>Q(t)$ $ \forall t\in\mathbb{R}$

Sorry, under the hp. that there exists at least an $ s \in \mathbb R$ s.t. $ P(s)<0$?

[quote="Stewie"]Sorry, under the hp. that there exists at least an $ s \in \mathbb R$ s.t. $ P(s) < 0$?[/quote]
$ P(x) \equal{} 4x^4 \minus{} 4x^3$
$ Q(x) \equal{} \minus{} 4x^2 \minus{} 1$
Then $ P(x) > Q(x) \forall x\in R$
But $ P(\frac {1}{2}) < 0$
This problem I have solve but it is
Let $ P(x),Q(x)$ is negative on $ (x_1,x_2)$ such that$ x_2\minus{}x_1>2$and positive in other$ [\minus{}\infty,x_1),(x_2,\plus{}\infty)$
Prove that exist t such that $ Q(t) < P(t)$


inequalities
trigonometry
inequalities unsolved

If A, B, C are the angles of a triangle, prove that

$\Bigl|\sum\tan\frac{C}{2}\tan\frac{A-B}{2}\Bigr|<\frac{1}{16}$

I think it's related with the "Tangents Theorem" so that inequality becomes

$\Bigl|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\Bigr|<\frac{1}{16}$

Then I saw that is equivalent to

$\Bigl|\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\Bigr|<\frac{1}{16}$

WLOG I may assume $a>b>c$ so that I must show

$\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}<\frac{1}{16}$

What can I do now?

look here: http://www.kalva.demon.co.uk/ibero/isoln/isol893.html, my solution is another but my time is really short. :wink:

sorry:     http://www.kalva.demon.co.uk/ibero/isoln/isol893.html
 :blush:


geometry symmetry circumcircle conics parabola ratio trigonometry

Let $ D,E,F$ be the symmetry points of points $ P$ to the three edges $ BC,CA,AB$ of $ \triangle ABC$ respectively.Let $ O$ be the circumcenter of $ \triangle ABC$ and $ H$ be the orthocenter.Prove that:
     $ \frac {S(\triangle HDE)}{S(\triangle HDF)} \equal{} \frac {S(\triangle OAB)}{S(\triangle OAC)}$

Let $ H_a \in BC, H_b \in CA, H_c \in AB$ be the altitude feet. Denote $ a \equiv BC, b \equiv CA, c \equiv AB$ and $ h_a \equiv AH_a, h_b \equiv BH_b, h_c \equiv CH_c.$ Let $ P_a \in BC, P_b \in CA, P_c \in AB$ be the pedals of $ P.$

First, assume $ P \in BC, P_a \equiv P.$ Circles $ \odot(ABH_b), \odot(ACH_c), \odot(HBH_c), \odot(HCH_b)$ concur at $ H_a$ $ \Longrightarrow$ $ b, c, h_b, h_c$ are all tangents of a parabola $ \mathcal P_1$ with focus $ H_a.$ Circle with diameter $ AH_a$ cuts $ CA, AB$ at $ K_b, K_c.$ Since it goes through the parabola focus $ H_a$ and intersection 2 tangents $ A \equiv b \cap c,$ $ K_bK_c$ is also parabola tangent. As $ H_aK_b \perp b, H_aK_c \perp c,$ it is the parabola vertex tangent. Let $ AH_a$ cut $ H_bH_c$ at $ J$ and $ K_bK_c$ at $ L.$ From the harmonic cross ratio $ (H_a, J, H, A) = - 1,$ from centrally similar $ \triangle H_aK_bK_c \sim \triangle HH_bH_C$ with similarity center $ A,$ and from the lemma at [url]http://www.artofproblemsolving.com/viewtopic.php?t=280996[/url], $ L$ is midpoint of $ H_aJ$ $ \Longrightarrow$ $ H_bH_c$ is directrix of $ \mathcal P_1.$ Let $ l$ be parallel of $ BC$ through $ L$; since $ l \perp H_aL$ at $ L,$ $ l$ is also tangent of $ \mathcal P_1.$ Let $ (Q)$ be circumcircle of cyclic $ AP_bPP_c,$ $ A, H_a \in (Q)$ $ \Longrightarrow$ $ P_bP_c$ is also a tangent of $ \mathcal P_1.$  Let $ PH$ cut $ l$ at $ X,$ let $ AX, H_aX$ cut $ BC, PA$ at $ Z, Q^*.$ Using Van Aubel for $ \triangle APH_a$ and the above mentioned  lemma again,

$ \frac {\overline{AQ^*}}{\overline{Q^*P}} = \frac {\overline{AX}}{\overline{XZ}} - \frac {\overline{AH}}{\overline{HH_a}} = \frac {\overline{AL}}{\overline{LH_a}} - \frac {\overline{AH}}{\overline{HH_a}} = 1$ $ \Longrightarrow$ $ Q^* \equiv Q$ is midpoint of $ PA$ and center of $ (Q).$

$ H_aU$ be a diameter of $ (Q),$ $ AU \parallel BC.$ Let angent $ t_A$ of $ (O)$ at $ A$ cut $ (Q)$ again at $ V.$ From $ \angle P_cAU = \angle BAU = \angle B$ and $ \angle H_aAV = \angle A + \angle C$ $ \Longrightarrow$ $ P_cU = P_bV$ and $ UV \parallel P_bP_c.$ Let $ (R)$ be a circle with diameter $ H_aX,$ $ L \in (R).$ Let $ (R)$ cut the parabola vertex tangent $ K_bK_c$ again at $ Y,$ then $ XY$ is also a tangent of $ \mathcal P_1.$ Since $ (R), (Q)$ are internally tangent at $ H_a$ and $ K_bK_c \parallel H_bH_c \parallel t_A,$ $ K_bK_c$ is tangent of $ (R)$ at $ L$ $ \Longrightarrow$ $ XY \parallel UV \parallel P_bP_c.$ The 2 parallel tangents $ XY \parallel P_bP_c$ of $ \mathcal P_1$ are therefore identical, $ X \in P_bP_c.$ Lines $ b, c, l$ are 3 fixed tangents of the parabola $ \mathcal P_1,$ cutting its variable tangent, as $ P$ moves on $ BC,$ at $ P_b, P_c, X.$ It follows that the ratio $ \frac {\overline{XP_b}}{\overline{XP_c}}$ is constant for any $ P \in BC.$ [hide]This is consequence of the following theorem: 4 fixed tangents of a circle (or any conic) cut its fifth variable tangent in a constant cross ratio. Projecting one of the fixed tangents to infinity takes the circle (or any conic) to a parabola and yields the desired result: 3 fixed tangents of a parabola cut its fourth variable tangent in a constant ratio.[/hide]

Let $ S \in BC$ be another arbitrary point (in addition to $ P$) and let $ SH$ cut the parallel $ AU \parallel BC$ at $ T.$ In the $ \triangle PST,$ $ H \in ST$ is foot of cevian $ PX$ and let $ M \in PT, N \in BC$ be feet of cevians $ SX, TX.$ Exactly as above, by Van Aubel theorem,

$ \frac {\overline{TM}}{\overline{MP}} = \frac {\overline{TX}}{\overline{XN}} - \frac {\overline{TH}}{\overline{HS}} = \frac {\overline{AL}}{\overline{LH_a}} - \frac {\overline{AH}}{\overline{HH_a}} = 1$ $ \Longrightarrow$ $ M$ is midpoint of $ PT.$

Next, assume $ P_a \in BC$ is fixed and $ P$ moves on a perpendicular $ p \perp BC$ from $ P_a.$ Let $ p$ cut $ CA, AB$ at $ C', B'$ and consider $ \triangle AB'C'$ with its ortocenter $ H'$ and altitude feet $ H_a', H_b', H_c'.$ Circle with diameter $ AH_a'$ cuts $ C'A, AB'$ at $ K_b', K_c'.$ $ H_b'H_c', K_b'K_c'$ cut $ AH_a'$ at $ J', L'.$ Let $ l'$ be parallel to $ p \equiv B'C'$ at $ L'.$ In exactly the same way as above, $ b \equiv C'A \equiv CA, c \equiv AB \equiv AB', l'$ and $ P_bP_c$ are all tangents of a parabola $ \mathcal P_2$ with focus $ H_a',$ vertex tangent $ K_b'K_c'$ and directrix $ H_b'H_c'.$ As before, the 3 fixed tangents $ b, c, l'$ of the parabola $ \mathcal P_2$ cut its variable tangent, as $ P$ moves on $ p \equiv B'C',$ at  at $ P_b, P_c, X$ and the ratio $ \frac {\overline{XP_b}}{\overline{XP_c}}$ is constant for any $ P \in B'C'.$ In the compete quadrilateral $ (a, b, c, p),$ $ P_aHH'$ is the orthocentric line. But in $ \triangle AB'C'$ alone, $ P_a$ is just an arbitrary point on $ B'C'$; in other words, $ h_a, P_a, H$ in $ \triangle AB'C'$ play the role of $ AU, S, T$ in $ \triangle ABC.$ It follows that $ P_aX$ cuts $ PH$ at its midpoint $ M.$ When $ P, P_a$ coincide, lines $ PH, P_aM$ also coincide $ \Longrightarrow$ ratio $ \frac {\overline{XP_b}}{\overline{XP_c}}$ is the same constant for $ P$ moving on $ BC$ or on $ p \perp BC.$

Any point in the plane $ (ABC)$ can be reached by first moving $ P_a$ on $ BC$ and then $ P$ on the perpendicular $ p \perp BC$ at $ P_a.$ Consequently, if $ M$ is midpoint of $ PH$ and if $ P_aM$ cuts $ P_bP_c$ at $ X,$ then the ratio $ \frac {\overline{XP_b}}{\overline{XP_c}}$ is constant for any $ P.$ When $ P, O$ coincide, $ P_a, P_b, P_c$ become midpoints of $ BC, CA, AB,$ midpoint $ M$ of $ PH$ becomes the circumcenter of the medial $ \triangle P_aP_bP_c.$ Let $ AO$ cut $ BC$ at $ A^*.$ Line $ P_aM$ then cuts $ P_bP_c$ in the ratio $ \frac {\overline{XP_b}}{\overline{XP_c}} = \frac {\overline{A^*B}}{\overline{A^*C}} = - \frac {[OAB]}{[OAC]}.$  Since this ratio is constant for any $ P,$ $ \frac {[HDE]}{[HDF]} = \frac {[MP_aP_b]}{[MP_aP_c]} = - \frac {\overline{XP_b}}{\overline{XP_c}} = \frac {[OAB]}{[OAC]}$ for any $ P.$  QED.
________________
It is a nice problem.

My solution:

Let $ \text{R} $ be the radius of $ \odot (ABC) $ .

From the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=432213]How to prove this one?[/url] we get

$ \text{S} [HDE]=\frac{|{\text{R}}^2-{OP}^2|}{2} \cdot \sin 2C , \text{S} [HDF]=\frac{|{\text{R}}^2-{OP}^2|}{2} \cdot \sin 2B $

$ \Longrightarrow \frac{\text{S} [HDE]} {\text{S} [HDF]}=\frac{\sin 2C}{\sin 2B}=\frac{\text{S}[OAB]}{\text{S}[OAC]} $ .

Q.E.D


if $ c\equal{}b\equal{}(1/2)a$ and $ P$ is a point on $ BC$ such that $ BP: PC\equal{}1: 3$ prove that $ <ACP\equal{}2<APC$

hello
[hide="Doubt"]
What r a,b and c?As they can't be sides of a triangle[/hide]
Thank u.

Hello
In a triangle sum of two sides should be greater than third side.
c+b=2c=a
that's why they can't be sides of a triangle.
thank u.

[quote="ishfaq420haque"]if $ c \equal{} b \equal{} (1/2)a$ and $ P$ is a point on $ BC$ such that $ BP: PC \equal{} 1: 3$ prove that $ < ACP \equal{} 2 < APC$[/quote]
It's obviously a degenerated triangle so $ \angle ACP \equal{} 2 \angle APC \equal{} 0$. (Btw. instead of $ <$ you can use $ \angle$ \angle)


[quote="Silverfalcon"]If $x = \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots }}}$, find $x$.[/quote]

$x=\sqrt{2+x} \Rightarrow x^2=x+2 \Rightarrow x=2$?

I'll elaborate a bit on that...

Since $x = \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots }}}$, $x^2 = 2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots }}}$. By substitution, $x^2 = 2 + x$. Now you can factor it to $(x + 1)(x - 2) = 0$, so $x = -1$ or $x = 2$

How do you know? I can't prove it, but maybe imaginary numbers work that way. :lol:

[quote="nat mc"]Try this one
$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2...}}}}}$[/quote][hide="1"]1[/hide]

Here's another i posted a while ago:
[hide][quote="Xantos C. Guin"]
What a coincidence.
I rot my brain with that game.
about
$\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+....}}}}$ hours a day
Yay!
My $LaTeX$ worked[/quote][/hide]