First one is a partial fraction decompostion:
[hide]
Fixify the denominators.
$ x^3\plus{}9x^2\plus{}11x\plus{}5\equal{}A(x\plus{}1)^3\plus{}Bx(x\plus{}1)^2\plus{}Cx(x\plus{}1)\plus{}Dx$
Rearrange. Warning: do not try this at home
$ x^3\plus{}9x^2\plus{}11x\plus{}5\equal{}(A\plus{}B)x^3\plus{}(3A\plus{}2B\plus{}C)x^2\plus{}(3A\plus{}B\plus{}C\plus{}D)x\plus{}A$
Equationify. | is just so you don't confuse the equations together.
$ A\plus{}B\equal{}1$ | $ 3A\plus{}2B\plus{}C\equal{}9$ | $ 3A\plus{}B\plus{}C\plus{}D\equal{}11$ | $ A\equal{}5$
Usually you would have to solvify for these, but I'm going to take a shortcut for this one.
Subtract 2 times the 4th equation from the 3rd equation.  $ A\plus{}B\plus{}C\plus{}D\equal{}1$
[/hide]

[quote="sami.d"]Hi
For 19:
X=2x²+12x+19
the equation becomes:
X²-4X+3=0
...
How we can use the latex languege here?thx[/quote]

yep its the same as $ x^4\plus{}bx^2 \plus{} c$ where you let $ \alpha\equal{}x^2$

--
"@sami.d" - to use latex scroll over the  $ x^4\plus{}bx^2 \plus{} c$  and you will see the code. But you need to use the dollar sign to activate it.. so dollarsign1234dollarsign and it will become like latex

for more info

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

I got the 2nd one, but for 17, I still am not sure how to approach the problem the correct way. If someone could explain that in further detail, that would be awesome =D

Like I said, it's a partial fraction decomposition.  Look it up or see my post above.

The format $ \frac {x^3 \plus{} 9x^2 \plus{} 11x \plus{} 5}{x(x \plus{} 1)^3} \equal{} \frac {A}{x} \plus{} \frac {B}{(x \plus{} 1)} \plus{} \frac {C}{(x \plus{} 1)^2} \plus{} \frac {D}{(x \plus{} 1)^3}$ is correct.
[color=red]It is essential that the format is correct, or else the methods used below will not work at all.[/color]

[hide="Solution"]First, multiply through $ x (x \plus{} 1)^3$:

$ x^3 \plus{} 9x^2 \plus{} 11x \plus{} 5 \equal{} A(x \plus{} 1)^3 \plus{} Bx(x \plus{} 1)^2 \plus{} Cx(x \plus{} 1) \plus{} Dx$

$ x : \equal{} 0 \implies \boxed{A \equal{} 5}$
$ x : \equal{} \minus{} 1 \implies 2 \equal{} \minus{} D \implies \boxed{D \equal{} \minus{} 2}$

We now subtract $ 5(x \plus{} 1)^3 \minus{} 2x$ from each side:

$ \minus{} 4x^3 \minus{} 6x^2 \minus{} 2x \equal{} Bx(x \plus{} 1)^2 \plus{} Cx(x \plus{} 1)$
$ \minus{} 2x (x \plus{} 1)(2x \plus{} 1) \equal{} Bx(x \plus{} 1)^2 \plus{} Cx(x \plus{} 1)$
$ \minus{} 2(2x \plus{} 1) \equal{} B(x \plus{} 1) \plus{} C$
$ \minus{} 4(x \plus{} 1) \plus{} 2 \equal{} B(x \plus{} 1) \plus{} C$

$ \boxed{B \equal{} \minus{} 4}$ and $ \boxed{C \equal{} 2}$[/hide]

[quote="archimedes1"][quote="erdogankerem123"]I thought partial fraction decomposition would make it ${ \frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \frac {A}x} + \frac {B}{(x + 1)} + \frac {C}{(x + 1)} + \frac {D}{(x + 1)}$
instead of
${ \frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \frac {A}x} + \frac {B}{(x + 1)} + \frac {C}{(x + 1)^2} + \frac {D}{(x + 1)^3}$
???[/quote]
 :?:
Why would you want to have three fractions with the same denominator?  I don't think that would help very much.[/quote]
I don't know; I'm pretty new to this whole partial fraction decomposition thingy. I just read the AopsWiki on it, and all the notation was somewhat confusing, but that's how I interpreted it. If I'm wrong could someone explain exactly what partial fraction decomposition is to me and how and when to use it?


geometry
inequalities
complex numbers
national olympiad

What do you mean by Greek TST 2005 ? :?  maybe a TST for the JBMO, but surely not for the IMO. Be more specific when you compose such posts.  ;)
I don't think that these problems are difficult.

[quote="silouan"]Problem 1
Examine if we can place 9 convex 6-angled polygons (the one next to the other) to construct a convex 39-angled.

Problem 2
Prove that: For each x,y,z in the R have power that
(x^2-y^2)/ (2x^2+1) + (y^2-z^2)/(2y^2+1)+(z^2-x^2)/ (2z^2+1)<=0

Problem 3
Let the midpoint M of the side AB of an 4sided, inscribed in circle,ABCD.Let P the  point of section of MC with BD. Let the parallel from the point C to the AP which intersects the BD at S. If CAD angle=PAB angle= BMC/2 angle, prove that BP=SD.

Problem 4
Find all the positive integers n , n>=3 such that n/ (n-2)!

Try to find the solutions of the problems. If you find it difficult i have the solutions.[/quote]

1.  Let us try to dissect a 39gon into 9 6-gons.  Each 6 gon can only occupy at most 5 sides of the 39 gon.  If we have 8  6gons occupying, then there are 8 sides in the middle.  If we have sone 6gons occupying less than 5 sides, we use all 6 gons covering the sides, and there are none left to cover the middle, which is concave.

2.  We prove for positive reals $x,y,z$, that $\sum \frac{x}{2x+1} \leq \sum \frac{y}{2x+1}$.  Obviously, $(x,y,z)$ and $(\frac{1}{2x+1},\frac{1}{2y+1}, \frac{1}{2z+1})$ are oppositely sorted.  Suppose $x \ge y \ge z$. By rearrangement, we have the result immediately.  Suppose $x \ge  z \ge y$.  We also have the result by rearrangement.

3.  no idea

4.  I dont understand the question

sorry it is for JBMO. For the IMO i also give the problems

hi, silouan!
now, i look over the problems.
did you have only 4 problems?

if i understood everything at the 4th, then n can be any positive integer that is not a prime except the number 4.
indeed if n is not a prime then
I.   n = a*b, with 2<=a,b<=n div 2< n-1  and a<>b so a|(n-2)! and b|(n-2)! so n=ab|(n-2)!
II. n=p*p, where p is a prime, p>2 then p|(n-2)! and p|2p< n-1 so 2p|(n-2)!
so n=p*p | p*2p | (n-2)!.
now if n= 4then (n-2)! = 2 and 4 isn't a divisor of 2.
and if n is a prime then (n,t)=1 any 1<=t<=n-2
i think this is a easy problem
something is un clear at the geometry problem(maybe i did a mistake).
i reformulte the condition and u tell me if it is ok"
ABCD - inscribed[u] CONVEX [/u]quadrilateral, thus A,B,C,D- concylclic
m is on (AB), such that BM=MA.
P is the intersection point of MC and BD. S is on BD, such that CS||AP.
(angle CAD) = (angle PAB) = 1/2*(angle BMC) then BP=SD.

the inequality is rather simple and indeed it's obvious to use re-arrangements inequality, although you may prove it in a different way. here is my solution.
set a=x^2 b=y^2 z=c^2.

weh have ab+bc+ca<=a^2+b^2+c^2 <==>
4(ab+bc+ca)<=2(a^2+b^2+c^2)+2(ab+bc+ca) <==>
12abc+(a+b+c)+4(ab+bc+ca)<=2(a^2+b^2+c^2)+2(ab+bc+ca)+12abc+(a+b+c)

now note that.
12abc+a+b+c+4(ab+bc+ca) = a(2b+1)(2c+1) +
                                              b(2a+1)(2c+1) +
                                              c(2a+1)(2b+1).

and 2(a^2+b^2+c^2)+2(ab+bc+ca)+12abc+a+b+c = b(2b+1)(2c+1)+
                                                                                c(2c+1)(2a+1)+
                                                                                a(2a+1)(2b+1).

dividing by (2a+1)(2b+1)(2c+1) the obtained inequality and subtracting the right side from the left side we get what is asked.

the first problem it is beautiful and suitablke for the for the first problem in a JBMO tst . i have an other solution.
well, the sum of the angles of a 39convex n gon is 37*180.
but the sum of the angles of a 6 gon is 4*180, so the sum of the angles of 9 6gons is 9*4*180 = 36*180<37*180.
that is a contradiction, because, if a convex Ngon is divided into convex Kgons, thatn the sum of the angles of all the Kgons must be greater or equal than the sum of the angles of the bin Ngon.
thus the answer is no.

Freemind you solve great the 1,2,4 problem. The geometry problem is quite difficult.I will post his solution the night. I hope that you will have your computer open. I also will post the 4 solutions of Moldova TST. Noow try to solve the TST Greek for the IMO. IT has very clever problems. See you!

thx silouan.
i don't think i'll be online this night.
tomorrow i think i'll enter on mathlinks just for a little.i'll think more at the geometry problem and i hope i'll enter on moldova's tst.
they were kinda easy, except some problems.
and i will also look this night (but not at the computer over the tst for imo problems).
hope i'll solve something.
do you really have only one TST?
bye
 keep in touch.

I will post my solutions for Greek TST JBMO problems.
The first is exactly  similar with yours freemind.
Problem 2
We post a=2*(x^2)+1, b=2(y^2)+1 , c= 2(z^2)+1 , so a,b,c>0
and (x^2)-(y^2)=(a-b)/2 , (y^2)-(z^2)=(b-c)/2 , (z^2)-(x^2)=(c-a)/2
sowe shpold prove that , (a-b)/(2a)+(b-c)/(2b)+(c-a)/ (2c)<=0      <=>
(a-b)/(a)+(b-c)/(b)+(c-a)/ (c)<=0              <=>     (b/a)+(c/b)+(a/c)>=3       <=>
(a^2)b +(b^2)c + (c^2)a>=3abc  <=> [(a^2)b +(b^2)c + (c^2)a] / 3 >= 3rd root of (abc)^3 which is the inequality of AGM.

Problem 3
It is a very good problem!
At first, at the triangle MAP the angle PMB is exterior, so  angle PMB=angle MAP+angle MPA. If angle MAP=x , the angle PMB=2x from the hypothesis. So 2x=x+angle MPA. So angle MPA=x. Triangle MPA isosceles with MA=MP.But MA=MB because M is midpoint. Now at the triangle PAB the MP which is medium =MA=MP. So from the theory the triangle PAB is a right-anlged triangle with angle BPA=90  . But from hypothesis we have that AP// CF(F is the point where the CS intersects the AD). So angle PSF=90.
Let MPB=m.So x+m=90.The triangle MBP is isosceles because MB=MP so angle MBP=m. Angle DBC=x because it is going to the same arc of the circle with angle DAC. So angle B=x+m=90. The ABCD is insribed angle B+angle D=180 so angle D=90. Because angle B=angle D=90 the diagonal AC pass through the O which is the center of the circle.So AC=2R, R is the ray of the circle.Now we make a trick. We extend the CS until CS intersects the circle in N. We also form the lines AN and ND. The angle SNA =90 because it is going to the middle of the circle. The APSN has all of his angles=90 so it is an right-angled parallelogram.So NS=AP. The triangles NSD and APB are equals because there are right-angled and they have: AP=NS and angle DNS=angle PAB=x(angle DNS=x decause it is going to hte same arc of the circle with angle DAC). Finally because   NSD and APB are  equals they have SD=BP. Good?

Problem 4
I will use the theorem of Wilson.This theorem says: The positive integer p>2 is a prime number if only if: (p-1)! ≡ -1(modp). So n is not a prime number.Now if n=ab the a,b will appear at the (n-2)! so if n=ab n/ (n-2)!
Finally if n=(q^2)   ( q>2) we have that q=sqrt(n). In addition 2q<n-1 <=> 2*sqrt(n)<n-1    <=> (n^2)-6n+1>0 which have power for n>=6 . So 2(q^2)/ (n-2)! and (q^2)/(n-2)!
After all these we conclude that n/(n-2)! when n complex number>=6

Freemind what is your opinion for the geometry?

Here freemind i put all the solutions. I wait for your comments. See you!

hi!
the solution at geometry is nice indeed, and it seems to be correct :))
the inequality is also nice.
but b/a +c/b + a/c >=3 results immediatelly from the AM-GM inequality because a/b*b/c*c/a=1.
and the 4th is correct.
just make a correction. don't use "complex numbers" but nonprime or smth like that.
because complex numbers is something totally different.

Freemind do you believe that the geometry problem was difficult? Also I want to tell you that in Greece we write only 4 problems. In two days i will sent you the problems of my national olympiad in your e-mail. See you.

the geometry was kind difficult. but it was beautiful as it ha an elementary non-algebraical solution :)
thx for the problems you sent me, tonight i'll try to solve them and i hope i will post tomorrow the solutions (if i can solve them  :D )
keep in touch

Ok freemind i wait your solutions in my e-mail. See you


I wish you a happy new year!

But let us remember the many victims of the desaster in Asia!
 :(  :(  :(

We will remember it for many many years...

[quote="Moubinool"]Over 100000 victims, may they'll receive a lot of help from rich countries quickly.[/quote]
From France, for example...

CNN reports the death toll at 135,000 and it should breach the 150,000 mark as remote regions report deaths.

A German organization which helps there even estimates more than 200,000 victims.  :(  :(  :(
I hate to see these videos from tourists in which you can see how the people try to save themselves, but you don't know if the did or not and you sit in front of the TV and can't do anything. The most horrible thing is that you know:
That wasn't an action film: THAT WAS TRUE. And maybe these people are dead now.

speaking to all german speaking people here, i would like to point your attention to this: http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=29686 , what is mainly a discussion of the article at http://derstandard.at/?id=1905744 . having no clue about physics and being not the best in geography either, i cannot comment, but what would you say about it?

  darij

WHO is also trying to stop the epidemics that are caused because of the tidal wave which could double the death toll.

[quote="Scrambled"]i read something about a "bidding war" on who can pay more between US and some other  countries in google news[/quote]

US raised it's 35 million figure to 350 million and I believe that's all that's going to happen now. UN's already got about 1.1 billion secured with more to come soon obviously as more countries get permission from federal gov't and etc. to donate.

i believe that one of those "other countries" is japan, not sure though

it's so depressing watching the ppl run for their lives and then just get swept away so helplessly........ :(

Yes, I was thinking of organizing the ASIA club at my school to actually do fundraisers.


[quote="Silverfalcon"][China 1996] In a soccer tournament, each team plays another team exactly once and receives 3 points for a win, 1 point for a draw, and 0 points for a loss. After the tournament, it is observed that there is a team which has earned both the most total points and won the [i]fewest[/i] games. Find the smallest number of teams in the tournament for which this is possible.

Enjoy!  :)[/quote]

[hide]
you never said it was unique most and unique fewest
so you could have 3 teams
each 1 beat one of the other teams and lost to one of the other teams.
each has 3 points, each has 1 win.

[/hide]

I typed up exactly what it said on the book.

And no, the answer is not 3.

[hide]I found a solution with 8 teams.

Name the 8 teams A,B,C,D,E,F,G,and H.

A defeats B and C, and draws against D,E,F,G,and H for 2 wins and 11 points.
B defeats C, E, and F for 3 wins and 9 points.
C defeats D, E, and G for 3 wins and 9 points.
D draws against A and defeats B, F, and H for 3 wins and 10 points.
E draws against A and defeats D, F, and H for 3 wins and 10 points.
F draws against A and defeats C, D, and G for 3 wins and 10 points.
G draws against A and defeats B, E, and H for 3 wins and 10 points.
H draws against A and defeats B, C, and F for 3 wins and 10 points.

Therefore, A has the fewest wins and the most points.

I believe 8 teams is the fewest for which this is possible.[/hide]

yes cf249 you're right,the answer is 8,I can wirte the official solution here if anybody is interested.

It'd be nice, since I don't quite know how to prove that it's the minimum.

ok here it goes:
[hide="the official solution"]
let $ W$ be the desired team,now let $ n$ denote the number of soccer teams which participated in this tournament,so there are $ \binom{n}{2} \equal{} \frac {n(n \minus{} 1)}2$ matches played among these teams,now note that each game gives at least $ 2$ points in total,so sum of the points of all teams is greater than $ n(n \minus{} 1)$ so the average is at least $ (n \minus{} 1)$.
now note that $ W$ has played $ (n \minus{} 1)$ games,and also we know that $ W$ has the most total points,so the total points of $ W$ is more than $ (n \minus{} 1)$ so $ W$ must have won at least one game...
now as the problem said $ W$ has won the fewest number of games,so every other team must have won at least $ 2$ games.thus every other team has at least $ 6$ points.hence $ W$ must draw at least $ 4$ games in order to have the most total points,but note that if team $ A$ had a draw with team $ W$ then the total points of $ A$ would be at least $ 7$,hence $ W$ must have drawn at least $ 5$ games.which means $ n\geq 7$.
now we show that $ n\not \equal{} 7$:
if $ n \equal{} 7$ then let $ S$ denote the set of all teams,now team $ W$ has won a single match and has drew $ 5$ matches,so the total points of $ W$ is equal to $ 8$,now note that every other team has won exactly two games,and drew at most one game.thus every other team has lost at least $ 3$ games,so there are at least $ 6\times 3 \equal{} 18$ looses in $ S\setminus W$,so there must be at least $ 18$ wins in $ S$,but we know that the number of wins in $ S$ is equal to $ 1 \plus{} 6\times 2 \equal{} 13$,which is a contradiction...
so $ n\geq 8$,now we will give an example for $ n \equal{} 8$:

let $ S \equal{} \{W,A_1,A_2\ldots , A_7\}$ be the teams which participated in the tournament,now by $ X\to Y$ we mean $ X$ won $ Y$ and by $ X\iff Y$ we mean $ X,Y$ drew in their match:

$ W\to A_1$
$ W\to A_2$
$ W\iff A_i$ for $ i\geq 3$

and for every $ 1\leq i\leq 7$:

$ A_i\to A_{i \plus{} 1}$
$ A_i\to A_{i \plus{} 2}$
$ A_i\to A_{i \plus{} 3}$
$ A_{i \plus{} 4}\to A_i$
$ A_{i \plus{} 5}\to A_i$
$ A_{i \plus{} 6}\to A_i \left(A_{i \plus{} 7} \equal{} A_i\right)$.

so $ W$ would have $ 11$ points in total,$ A_1,A_2$ would have $ 9$ points in total (with $ 3$ wins),and $ A_3,A_4,\ldots ,A_7$ have $ 10$ points in total (with 3 wins)...[/hide]

Yup, basically what he wrote (although he paraphrased it bit).  :)

[quote="BaBaK Ghalebi"][quote="Silverfalcon"]although he paraphrased it bit[/quote]

yeah because I have the translated version,so I had to translate it back to english again... :)[/quote]

Oh.

I have English version here so I could've typed it up and saved you time.  :D


Given a convex 2n polygon. (n is a positive integer )
Prove that exists at least one diagonal that is not parallel to any of the edges.

This has already been posted:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=polygon&t=85912[/url]


geometry
rhombus
number theory
least common multiple

On Saturday we had the inter-provincial in south africa and unfortunately Gauteng won :(  I say unfortunately cos im western province.  We usually win but we lost our two best members to the computer olympiad so we ended up coming 4th.  Well done Boland for coming 3rd!(they also lost their two best members to the CO).

Last 3 questions from individual paper:

13.  Find the area of the rhombus ABCD given that the radii of the circles circumscribed around triangles ABD and ACD are 12.5 and 25 respectively.

A) 50   B)100  C) 200  D) 400 E)800

14.  In a sequence consisting of 2006 digits, each number formed by two successive digits is divisible by 19 or 23.  If the first digit of the sequence is 1, the last possible digits are...

A) 2 and 5  B) 2 and 7  C) 3 and 5  D) 3 and 7  E) 9

15.  Which of the following is equal to 1/(sqrt(2006+sqrt((2006^2)-1))?

 (sqrt = square root   ^2 =squared.... think its pretty obvious but just in case...)

A)  (sqrt(2007)-sqrt(2005))/sqrt(2)

B)  (sqrt(2007)+sqrt(2005))/ sqrt(2)

C)  (sqrt(2006)-sqrt(2004))/sqrt(2)

D)  (sqrt(2006)+sqrt(2004))/ sqrt(2)

E)  1/2(sqrt(2006^2-1))

I know they're probably simple for most of you (especially since they're just multiple choice)... but normal school students who dont have any other training in olympiads take part so they dont have any written solution type questions.  The team paper has 10 questions and these require numerical or algebraic answers but no explanations.

The last 3 of this paper were:

8. Given triangle ABC with AB=5=AC and BC=6.  P is a point inside the triangle such that <CPA=90, <ABP=<BCP.  BP is produced to meet AC at D.  Calculate AD : DC.

9.  The sequence 5, 7, 10, 14, 15, 20, 21, 25, 28, 30, 35, 40.. consists of all the multiples of 5 and 7 in ascending order.  Determine the 100th number of this sequence.

10.  The number 18 is not the sum of any 2 consecutive inteers but is the sum of consecutive positive integers in at least 2 ways, since 5+6+7=18 and 3+4+5+6=18.  Determine a positive integer less than 400 that is not the sum of 11 consecutive positive integers but is the sum of consecutive integers in at least 11 different ways.

Do other African countries have similar olympiads??

[hide]For 9, think about the lcm of 5 adn 7.  This is 35.  Now think of the numbers less than 35, which have eitehr a factor of 5 or 7.  There are 11.

Next, we see that 99 is the closest number to 100, that is a multiple of 11.

So we take 35*99/11=315.  Then we take the first numebr larger than 315, which is a multiple of 5 or 7, which is 320.[/hide]


Let $ n$ be an odd positive integer. Take all positive integers less than or equal to $ n$ and relatively prime to $ n$. Show there are an equal number of even and odd integers.

[hide="Hint."]Consider what the mapping $ x\mapsto n \minus{} x$ does.[/hide]

So apparently this problem has become trivial? :maybe:

yes, 'cause $ (n,x) \equal{} (n,n \minus{} x)$ and $ x,n \minus{} x$ have different parity... not sure how to formulate a proof that would give me 7 pts though...  :ninja:

That's all you need to say.  This isn't a deep problem.

Let $ S$ be the set of positive integers that are less than and relatively prime to $ n$ that are even. Let $ T$ be the set of those that are odd.

Let us define a mapping $ f: S\rightarrow T$ that takes $ s\in S$ and maps it to $ n \minus{} s$.

First, lets show that $ n \minus{} s$ is in $ T$. If $ s$ is even then $ n \minus{} s$ is odd. Furthermore, if $ (n,s) \equal{} 1$, then $ (n \minus{} s,s) \equal{} 1$ also. Also $ n \minus{} s$ is a positive integer that is less than $ n$. So $ f$ is well-defined as a function from $ S$ to $ T$.

Now we claim that $ f$ is a bijection:

check: $ f$ is onto: if $ t\in T$, then $ n \minus{} t$ is even and relatively prime to $ n$ (and between $ 0$ and $ n$) so $ n \minus{} t\in S$, so $ f(n \minus{} t) \equal{} t$.

check: $ f$ is one to one: if $ f(s_1) \equal{} f(s_2)$ then $ n \minus{} s_1 \equal{} n \minus{} s_2\implies s_1 \equal{} s_2$ so $ f$ is one to one

Therefore $ f$ is a bijection. Since $ S$ and $ T$ are finite sets, they have the same number of elements and we are done.


Use the vector representation of $ z_{1}$ and $ z_{2}$ on an Argand diagram to show that

(a) If $ |z_{1}| = |z_{2}|$, then $ \frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is imaginary

(b) If $ 0 < argz_{2}< argz_{1}<\frac{\pi}{2}$, and $ arg(z_{1}-z_{2})-arg(z_{1}+z_{2}) =\frac{\pi}{2}$, then $ |z_{1}|=|z_{2}|$

(b) Can be interpreted as, the perpendicular bisector of the hypotenuse of a right triangle has length equal to half the hypotenuse.


[quote="robinhe"][quote="Karth"]If $x$ and $y$ are acute angles, such that

$\tan x + \tan y = \sqrt{3}(1- \tan x \tan y)$, and $\sqrt{2} \sin y =1$, find $x$ in degrees.[/quote]

[hide]From the second equation, $y=45^\circ$

Substituting into the first equation gives $\tan x+1=\sqrt{3}-\sqrt{3}\tan x$
Rearranging gives $\tan x=2-\sqrt{3}\implies x=15^\circ$

How do you find out $x$ without a calculator :blush:  :? [/hide][/quote]

[hide="Without Calculator?"]Use the tangent formula for the addition of two angles.[/hide]

How to do it without a calculator :) :
[hide]
We have:

$\frac{\tan x + \tan y}{1 - \tan x \tan y} = \sqrt{3}$
$\tan(x+y) = \sqrt{3}$

by the first equation.

As a result of the second equation, we have $y = 45^\circ$.

So, $\tan(x+45) = \sqrt{3}$. Since $\tan 60 = \sqrt{3}$, $x + 45 = 60$, yielding $\boxed{x = 15}$.[/hide]

Or you can memorize the sin, cos, and tan of multiples of 15 degrees like my math teacher says.  It comes in handy.  (tan15=2-sqrt3, tan75=2+sqrt3.)


[quote="jhredsox"]I know the answer - See if you can figure it out!

You are standing in front of a room with one lightbulb inside of it. You cannot see if it is on or off. Outside the room there are 3 switches in the off positions. You may turn the switches any way you want to. You stop turning the switches, enter the room and know which switch controls the lightbulb. How?

-jorian[/quote]

umm do you only have one chance? if not just try each of them at a time  :D

[quote="Ignite168"]OMFG U HEARTLSES DESTROYER OF ICE CAPS.

WHY ARE YOU USING NOTASEFFICIENTASTHEYCOULDBE LIGHTBUKLBS THAT HEAT UP?? I CANT BELIEVE UW OULDNT USE ENERGY SAVER ONES

BOYCOTT THIS THREAD[/quote]

 :spam:  :spam:  :spam:

[quote="Lena"]you turn on 2 of the switches and wait 20 minutes

then turn off one of the switches and go into the room

if the light bulb is off and cold, the 3rd switch controlled it

if the light bulb is off and hot, the switch you turned off controlled it

if the light bulb is on,the switch you left on controlled it[/quote]

correct :D

-jorian

[quote="Ignite168"]OMFG U HEARTLSES DESTROYER OF ICE CAPS.

WHY ARE YOU USING NOTASEFFICIENTASTHEYCOULDBE LIGHTBUKLBS THAT HEAT UP?? I CANT BELIEVE UW OULDNT USE ENERGY SAVER ONES

BOYCOTT THIS THREAD[/quote]

well I have energy saver ones, and THEY DO HEAT UP. Why would you boycott a puzzle/teaser/joke?

[quote="Ignite168"]OMFG U HEARTLSES DESTROYER OF ICE CAPS.

WHY ARE YOU USING NOTASEFFICIENTASTHEYCOULDBE LIGHTBUKLBS THAT HEAT UP?? I CANT BELIEVE UW OULDNT USE ENERGY SAVER ONES

BOYCOTT THIS THREAD[/quote]

To protect your precious ice caps, i think you should hide in the jungle and be a barbarian. Since for every 20mins you stay in your house, you will destroy countless ice caps.

But you cant find computer in those rock caves, maybe you should carve your strange language on stones instead. :P

wow! that was neat! how did u think of the answer?

[quote="math92"][quote="asdf2345"]obviously he looked it up on google[/quote]
I've heard this riddle many many times.  He probably knew it from prior knowledge.[/quote]

I would like to refute this statement

1. I am not a HE last time i checked

and 2. As said this is a very known riddle, and many people know it , and its answer. So i did not look the answer up on google

there is a different version to this riddle

There are 3 switches and 3 lightbulbs, the switches and lightbubls are in different rooms, and you cannot see one room from inside the other

how do you know which switch controls which lightbulb

and for that the answer would be exactly the same

you turn on 2 switches, and wait a little, turn off 1 and go in the room
the lit bulb is the one you left on, the hot bulb is the one you turned off, and the cold bulb is the one you didnt touch

Shouldn't this be in puzzles/brainteasers?

[quote="Lena"][quote="math92"][quote="asdf2345"]obviously he looked it up on google[/quote]
I've heard this riddle many many times.  He probably knew it from prior knowledge.[/quote]
1. I am not a HE last time i checked
[/quote]
Sorry.  Its just that most people on this site are male.


Compute:

(1+3i)^8
______________
(-1+2i)^6*(-3-4i)

Didn't want to use parantheses cuz could get messy

If u don't understand, its a fraction

This problem is hokey, it only has an easy solution because the numbers work out nicely.

The answer is 16.

Huh? We don't want to convert to polar or anything like that since the angles would get messy, so we probably want to multiply it out the way it is. I don't see how the numbers work out well though.

They work out well because -3-4i is the square of (-1+2i). So both numerator and denominator are eighth powers. Upon further consideration, one realizes that (1+3i)/(-1+2i) = 1-i, which is even nicer. (1-i)^2 = -2i, (-2i)^2 = -4, (-4)^2 = 16.

Um, i tried using polar form the first time i did, and you end up that sine and cosine are greater than one, which makes the angles not only messy, it also makes them complex!
(complex--difficult, haha get it?)

(errr, i have no sense of humor)

haha. (is there such thing as a consolation laugh? :))


vector
function
calculus
integration
probability and stats

Random vector variable $ (X,Y)$ has following density function:
\[ g(x,y) \equal{} \frac {x}{(e \minus{} 2)(y \plus{} 1)^{2}}1_{D}(x,y)
\]
where $ D \equal{} \{(x,y) : 0 < x < e \minus{} 2 ; y > x\}$
1) Are variables $ X$ and $ Y$ independent?
2) Does $ X$ or $ Y$ belong to $ L^{1}$?
3)Compute $ E(X|Y)$
4)Are variables $ \frac {Y}{X}$ and $ X$ independent?

I don't quite see how I am supposed to compute marginal densities (densities of X and Y, to check whether $ g_{x}(x)g_{y}(y)\equal{}g(x,y)$. I mean how we compute density of $ X$, when D is $ D \equal{} \{(x,y) : 0 < x < e \minus{} 2 ; y > x\}$

$ D$ is just some region.  Note that $ e\minus{}2$ is a constant, so $ 0<x<e\minus{}2$ is just some infinite vertical strip.  Drawing a picture always helps when doing double integrals.

If $ X$ and $ Y$ are independent, then the region in the $ (x,y)$-plane on which the joint density is nonzero must be a "rectangle". (To be specific, it must be a simple Cartesian product $ A\times B$ of one-dimensional sets.)

The region $ D$ as described is not a rectangle. Therefore $ X$ and $ Y$ cannot be independent.

To illustrate this, ask what the joint density is at the point $ (1,.5).$ Since $ y<x,$ this point lies outside $ D$ and its joint density is zero. But the marginal density of $ X$ at $ x\equal{}1$ is the the integral in $ y$ of the joint density along the line $ x\equal{}1,$ which is clearly positive. And the marginal density of $ Y$ at $ y\equal{}.5$ is the integral in $ x$ of the joint density along the line $ y\equal{}.5.$ Since that line has a nontrivial intersection with $ D,$ that marginal density is also positive. So the product of the two marginal densities is positive at $ (1,.5)$ but the joint density is zero. Hence $ X$ and $ Y$ are not independent.

Note that this argument did not require the details of the joint density and did not require me to specifically compute the marginal densities - although such a computation looks feasible.


function
calculus
integration
derivative
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group. Suppose that $ f$ and $ g$ are homomorphisms of $ G$ such that $ g \circ f \equal{} 1_{G}$. Does it necessarily follow that $ f$ and $ g$ are automorphisms of $ G$?

The claim is certainly true when G is finite... What about its veracity for groups of infinite order? I haven't been able to find a proof yet.  :(

Take an infinite product (or sum) of the same nontrivial group.

Do you mean that the claim is false when G is a group of infinite order?  :(

OK... I'll further analyze your post.

Thanks for replying.

Let $ G \equal{} \mathbb{Z}^{\mathbb{N} }$ (for example) and let $ f$ take $ (a_1, a_2, ... )$ to $ (0, a_1, a_2, ...)$ and let $ g$ take $ (a_1, a_2, ...)$ to $ (a_2, a_3, ... )$.  Then $ g \circ f$ is the identity but neither $ f$ nor $ g$ are automorphisms.  ($ g$ has a left inverse and $ f$ has a right inverse, but not the other way around.)

I'd thought of the same counterexample, t0rajir0u! Thanks for taking the time to leave your comment.

A specific and very well-known manifestation of this phenomenon occurs in the additive group of $ C^{\infty}$ functions:  one has the fundamental theorem of calculus, $ \frac {d}{dx} \int_{a}^{x} f(t) \, dt \equal{} f(x)$ for fixed $ a$, but neither differentiation nor integration (with fixed lower boundary) are automorphisms.

This raises the following question:  does the result hold in a finitely generated group?

[quote="t0rajir0u"]A specific and very well-known manifestation of this phenomenon occurs in the additive group of $ C^{\infty}$ functions:  one has the fundamental theorem of calculus, $ \frac {d}{dx} \int_{a}^{x} f(t) \, dt \equal{} f(x)$ for fixed $ a$, but neither differentiation nor integration (with fixed lower boundary) are automorphisms.
[/quote]

That's very interesting, t0rajir0u!

[quote="t0rajir0u"]This raises the following question:  does the result hold in a finitely generated group?[/quote]

[b]Riquelme[/b] pointed out that it is certainly true for groups of finite order... I've just proved that it is also true when G = $ \mathbb{Z}$.

Wonder whether it also holds when G is a direct sum of 2 or more finitely generated abelian groups.  :D

Well, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237902]this[/url] would imply it in the finitely-generated abelian case. It comes down to the same question: if a homomorphism $ \phi$ from a finitely generated group $ G$ to itself is surjective, $ \phi$ is also injective.


geometry
circumcircle
incenter
perpendicular bisector
projective geometry
geometry proposed

In an acute triangle let circle O be its circumcircle and A' is on ray AO with $ { \angle BA'A\equal{} \angle CA'A}$  and $ {A'A[1]}$ ⊥$ {AC}$ ; $ {A'A[2]}$⊥ $ {AB}$;$ {A'H}$ ⊥ $ {BC}$ , the foot are points $ {A1,A2,H}$ . $ {R[A]}$ is radius of the circumcircle of triangle $ {HA1A2}$ , $ {R[B],R[C]}$ are defined similarly . Prove that :$ {\frac {1}{R[A]} \plus{}\frac {1}{R[B]}\plus{}\frac {1}{R[C]}\equal{} \frac {2}{R}}$ ;($ {R}$ is radius of the circumcircle of triangle $ {ABC}$).

Let $ N_a \equiv AA' \cap BC$ and define the points $ N_b,N_c$ cyclically. Since $ A'A$ is the bisector of $ \angle BA'C$ and $ O$ lies on the perpendicular bisector of $ BC,$ it follows that $ O$ is the midpoint of the small arc $ BC$ of the circumcircle $\odot(BA'C).$ Therefore, $ (O)$ cuts $ AA'$ at  the A'-excenter $A$ of $ \triangle BA'C$ and its incenter $I'$ $\Longrightarrow$ $ BA$ and $ CA$ become external bisectors of $ \angle A'BC$ and $ \angle A'CB$ $ \Longrightarrow$ $ \angle A'CA_1 \equal{} \angle ACB,$ $\angle A'BA_2 \equal{} \angle ABC.$

Quadrilaterals $ A'HCA_1$ and $ A'HBA_2$ are cyclic on account of the right angles at $ H$ and $A_1,A_2,$ respectively. Thus

$ \angle A_1HA' \equal{} \angle A'CA_1 \equal{} \angle ACB \ , \ \angle A_2HA' \equal{}\angle A'BA_2 \equal{} \angle ABC$

$\Longrightarrow \ \angle A_1HA_2 \equal{} \angle ABC \plus{} \angle ABC \equal{}  \pi \minus{} \angle BAC \equal{} \angle A_2A'A_1$

This implies that circumcircles of $ \triangle HA_1A_2$ and $ \triangle A'A_1A_2$ are congruent. Hence, the diameter $ AA'$ of $ \odot(A'A_1A_2)$ is twice the length of the radius $ R_a.$ Now, from the harmonic division $ (A,I',N_a,A'),$ we obtain

$ OA^2 \equal{} R^2 \equal{} ON_a \cdot OA' \Longrightarrow \  AA' \equal{} OA' \plus{} R \equal{} \frac {R^2}{ON_a} \plus{} R$

$ \Longrightarrow \ \frac {R}{AA'} \equal{} \frac {R}{2R_a} \equal{} \frac {ON_a}{AN_a}$

The cyclic sum yields:

$ \frac {R}{R_a} \plus{} \frac {R}{R_b} \plus{} \frac {R}{R_c} \equal{} 2 \left (\frac {ON_a}{AN_a} \plus{} \frac {ON_b}{BN_b} \plus{} \frac {ON_c}{CN_c} \right)=\frac{2}{R}$


[hide="Not a trick question?"]$574$[/hide]

What kind of solution???????????? I added.

[quote="funcia"]Is there a trick to this? I'd say this s below MC but who knows....[/quote]
I agree. No trick. Although there was a problem on a MC worksheet once that was just addition...

No, it is for sure not below MC considering it's above chapter level.

And this is state/nationals countdown level.

Adding them up is SLOW as well as easy to make a mistake.

The way I would do it uses Hockey Stick

[quote="i_like_pie"]If you get to nationals, I'm pretty sure you don't want to make any easy mistakes in addition.[/quote]

Then no one will ever get to Nationals.

Also, the fact is, adding it is still much slower.

[hide]the answer is 574. Please do not post boring questions. :P[/hide]

[size=75][color=darkred]Added hide tags.
-nebula42[/color][/size]

nutz, please hide your answers and provide solutions. Don't spam just to raise your post count.

Is this really above chapter? When did they have a summing problem? I would like to know.

[quote="Ignite168"][quote="i_like_pie"]If you get to nationals, I'm pretty sure you don't want to make any easy mistakes in addition.[/quote]

Then no one will ever get to Nationals.[/quote]
If no one goes to nationals then there wouldn't be a nationals...
That would be very scary...

If no one went to nationals no one would go to chapter. Then MC wouldn't be there.

Everyone, please keep your posts [b]relevant[/b] to the topic. If not, I may have to lock this thread.
Thanks for your cooperation.

[quote="bpms"]back on topic, recognizing hockey stick and then using it would take way longer than adding. The only reason you see hockey stick is because you wrote it... Adding is the best way to go...[/quote]

Well obviously deriving it would take awhile but if you already knew Hockey Stick to sum triangular numbers and the multiplying two digit numbers by 11 rule, it would be much faster not adding.

[quote="bpms"]but you would have to realize that first... Which might take a while...[/quote]

I'm saying if you realize it then adding it is quicker. But if you already know, then it's quicker.

[hide="Alternate Method"]
Do 12*12*2 to get 288 and then you have 11 triangular numbers. By Hockey Stick it is 13C3 which is 26*11. Use rule for multiplying by 11 which is 2 2+6 6 giving you 286+288.[/hide]

[quote="moogra"][quote="Ignite168"]Mr. Gleason has 24 jelly beans. Mr. Peters has 25 jelly beans. Mr. Slywotzky has 27 jelly beans. Mr. Tippy has 30 jelly beans. Ms. Venter has 34 jelly beans. Mr. Merideth has 39 jelly beans. Mr. Harpin has 45 jelly beans. Mr. Czepiel has 52 jelly beans. Mr. Weisner has 60 jelly beans. Mr. Faitsch has 69 jelly beans. Ms. Cox has 79 jelly beans. Mr. Calderone has 90 jelly beans. How many jelly beans do they have together?[/quote]

add them up.[/quote]
Really??? :|

why would you want to do that?

[hide]
574[/hide]

it could have been placed in classroom math :wink:

to make it faster, and to make this thread of some use...

round up, add them up, and subtract by the sum of the difference between the rounded numbers and the actual numbers...


A coin is tossed 10 times. Find the probability of not having two consecutive tails.

Please, use the search function!
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1707883597&t=68728[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1707883597&t=76521[/url]

I particularly like Altheman's solution in the first one.

sorry, I didn't even know this was an AIME problem


when you have a line segment with 4 distinct pts. (2 endpts. and 2 interior), how do you assign weights to the pts. so that you can accurately calculate ratios of sides?

Act as if two of the smaller segments are one segment.


calculus
integration
topology
inequalities
function
limit
Functional Analysis

If f is in L^2[0.1], and

Tf(x) = Integral(0,x){ Tf(y) dy }

show T is compact.

A hint is to use the fact that the Fourier coeffs. obey the bound

|T^f(n)| <=  C||f||/n   ||f|| is the L^2 norm on [0,1]

where 0 < C < infinity

I can prove the bound but I don't know how to show the operator is compact, i know i must show that image of the closed unit ball is compact but how to do this?

I suppose you meant \[Tf(x) = \int_{0}^{x}f(y)\, dy\] Think of $L^{2}[0,1]$ as a space of square summable sequences ($\ell_{2}$) - the Plancherel theorem says that you can. Since the image of the unit ball under $T$ is contained in the set $Q=\{x\in\ell_{2}\colon |x_{n}|\le C/n\}$, it suffices to prove that $Q$ is compact. Hint: think of $Q$ as an infinite-dimensional cube.

Do you use Tychnoffs thm to show Q is compact.

Uh-huh.

Of course, you'll have to check that the product topology agrees with the norm.

you say that i need to show that the product topology agrees with the norm, what do you mean and how do i prove this?  I am studying for a qualifying exam in analysis and this is of great importance to me, thank you for all your help.

Tychonoff's theorem says that the product of the intervals $[0,1/n]$ compact when equipped with the product topology. This is the topology generated by sets of the form $U_{1}\times U_{2}\times\dots$ where each $U_{i}$ is an open subset of $[0,1/n]$ and only [b]finitely many[/b] of $U_{i}$s are proper subsets. (I have read somewhere that this topology was deemed ridiculously weak by Tychonoff's colleages.) The norm topology on $Q=\{x\in\ell_{2}\colon |x_{n}|\le 1/n\}$ is generated by intersections of open balls with $Q$. To prove that both topologies coincide, you must verify two things:
1) if $x\in \prod U_{i}$ with $U_{i}$ as above, then there is $r>0$ such that $\prod U_{i}$ contains the open ball $B(x,r)$.
2) if $x\in Q$ and $r>0$, then there is a product $\prod U_{i}$ of the above form such that $a\in \prod U_{i}\subset B(a,r)$.

Schwarz inequality shows that:
$|Tf(y)-Tf(x)|=|\int_{[x,y]}f(t) dt| \leq \parallel f\parallel _{2}\sqrt{|y-x|}$
Hence one immediately sees that the image of the closed unit ball
of $L^{2}$ is a uniformy continuous subset of continuous functions on the compact $[0;1]$. Ascoli theorem
says that this set is relatively compact for the uniform norm $\parallel .\parallel _{\infty;[0;1]}$.
Hence it is relatively compact in $L^{2}([0;1])$.

Try to prove it is also true for:
$Tf(x)=\int_{[0;1]}K(x,y) f(y) dy$
where $K(.,.)$ is any function in $L^{2}([0;1] \times [0;1])$

Just wondering, is this way right:

I will use next: Let $A:\mathcal{H}\to \mathcal{H}$ is linear operator. Then $A$ is compact iff for every $x_n \overset{w}{\to} 0$ implies $Ax_n \overset{s}{\to} 0.$ ($\mathcal{H}$ is notation for Hilbert space)

We want to show that $T:L^2[0,1] \to L^2[0,1]$ is compact operator, where $\ Tf(x) =\int_{0}^{x}f(t)\, dt $. Let $f_n$ be sequence which weakly converges to $0$. That means that for every function $g \in L^2[0,1]$ is true that $\lim_{n \to +\infty} \langle f_n, g \rangle=0$, or $\lim_{n \to +\infty} \int_0^1 f_n(t) \overline{g(t)}\, dt= 0. \qquad (1)$

 Let $g_x(t)=\chi_{[0,x]}(t)$. Obviously, $g_x \in L^2[0,1]$. Then $(1)$ is giving us $\lim_{n \to +\infty} \int_0^x f_n(t)dt=0$, that is $(Tf_n)(x) \to 0$ when $n \to +\infty$.

We want to prove that $\| Tf_n - 0 \| \to 0$ as $n \to +\infty$. This is same as proving that $\| Tf_n \|^2=\int_0^1 |(Tf_n)(x)|^2\, dx \to 0$ as $n \to +\infty$.

So actually we want to swap $\lim$ and $\int$ and then, because $(Tf_n)(x) \to 0$ as $n \to +\infty$, it is obvious that $\| Tf_n \|^2 \to 0$, or $Tf_n \overset{s}{\to} 0$, and then $T$ is compact operator.

We will use dominated convergence theorem: It's not hard to show that $|Tf(x)| \leqslant \| f \|_2 \sqrt{x}$ for every $f \in L^2[0,1]$ (use Cauchy-Schwarz inequality). From that we see that $|Tf_n(x)|^2 \leqslant \| f_n \|^2_2 x.$ We are almost here, we just don't want that $n$ in index. But, we can use fact that every weakly convergent sequence is bounded, so there is some constant $M$ such that $\| f_n \|_2 \leqslant M$ (for all $n$). So $|Tf_n(x)|^2 \leqslant M^2 x$ and our dominant is $M^2 x$, which of course belong to $L^2[0,1]$.

Is this correct?


calculus
integration
probability
expected value
calculus computations

Evaluate $\int_{0}^{a}(\frac{x^{2}}{2n}-2 \pi floor(\frac{x^{2}}{4n \pi}) ) * (\frac{1}{\sqrt{2 \pi}\sigma}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^{2}}) dx$.

$n, \mu, \sigma \in \mathbb{R}^{+}$.

Either as it is, or limit of that as $a \to \infty$, whichever is do-able (if not both).

No one?

Anyone at least have any ideas?

Well, i do have some ideas:

- notice that one of the factors of the integrand is equal to $N(\mu, \sigma^{2})$, so I suggest it has something to do with propability.
- Moreover, the integral is the expected value of $\frac{x^{2}}{2n}-2\pi floor \frac{x^{2}}{4n\pi}$ when $x$ is a stochastic variable which is normally distributed with mean $\mu$ and variance $\sigma^{2}$ for $x$ between $0$ and $a$, and 0 elsewhere.

Well, yeah.

That's where the integral CAME FROM.

I'm trying to find the expected value of $\theta$ (but just the part between $0$ and $2 \pi$).


On a circle, there are $ n$ $ (n>3)$ numbers with a total sum of 94, such that each number is equal to the absolute value of the difference between the two numbers which follow it in clockwise order. What is the possible value of $ n$?  :)

The numbers are non-negative, as they take absolute value.
Let them be $ a_1, a_2, ..., a_k$ arranged in clockwise direction.
Then $ |a_{k\minus{}2}|\equal{}|a_k\minus{}a_{k\minus{}1}| \le max(a_k, a_{k\minus{}1})$ and equality holds only when one of the $ a_k, a_{k\minus{}1}\equal{}1$.
Now choose $ a_i\equal{}k$ such that $ a_i$ is the largest of the numbers.Then it's easy to see that $ \begin{cases} a_{i\plus{}1}\equal{}0 \\ a_{i\plus{}2}\equal{}k \end{cases}$ or $ \begin{cases} a_{i\plus{}1}\equal{}k \\ a_{i\plus{}2}\equal{}0 \end{cases}$ are the only possible values of $ a_{i\plus{}1}$ and $ a_{i\plus{}2}$. Both of them gives patterns like $ k,k,0,k,k,0,k,k,0,...,k,k,0$

$ 94\equal{}47 \times 2\equal{} 94 \times 1$, so $ 1,1,0,1,1,0,...,1,1,0$ is the only possible answer , which gives $ n\equal{}141$

[quote="stephencheng"]The numbers are non-negative, as they take absolute value.
Let them be $ a_1, a_2, ..., a_k$ arranged in clockwise direction.
Then $ |a_{k \minus{} 2}| \equal{} |a_k \minus{} a_{k \minus{} 1}| \le max(a_k, a_{k \minus{} 1})$ and equality holds only when one of the $ a_k, a_{k \minus{} 1} \equal{} 1$.
Now choose $ a_i \equal{} k$ such that $ a_i$ is the largest of the numbers.Then it's easy to see that $ \begin{cases} a_{i \plus{} 1} \equal{} 0 \\
a_{i \plus{} 2} \equal{} k \end{cases}$ or $ \begin{cases} a_{i \plus{} 1} \equal{} k \\
a_{i \plus{} 2} \equal{} 0 \end{cases}$ are the only possible values of $ a_{i \plus{} 1}$ and $ a_{i \plus{} 2}$. Both of them gives patterns like $ k,k,0,k,k,0,k,k,0,...,k,k,0$

$ 94 \equal{} 47 \times 2 \equal{} 94 \times 1$, so $ 1,1,0,1,1,0,...,1,1,0$ is the only possible answer , which gives $ n \equal{} 141$[/quote]

Thanks Very Much  :)


[quote="ashegh"]$ABC$ and $A'B'C$are regular triangles with the same orientation.let  $P,Q,R$be the midpoints of the segments
$AB',BC,A'C$
respectively.show that: triangle$PQR$is regular[/quote]
In space is not correct. :?


If you know them than answer these questions:
Cubes are just like exponents.
1=
2=
3=
4=
5=
6=
7=
8=
9=
10=
11=
12=
13=
14=
15=
16=
17=
18=
19=
20=

i know that $ 13^3\equal{}2197$ and $ 15^3\equal{}3375$.

forgot 14... think it might be 2744 or something but im not sure

meh too lazy to get my calc

ya i posted it cuz my math counts teacher made me memorize squares and cubes.

I've found that for MathCounts, knowing them up to 12 is sufficient. You usually don't need to know 19^3 or something.

meh i memorized squares to 41...

i need to memorize more...

I learned a neat trick from "How to Calculate Quickly" by Henry Sticker, and it involves learning how to square numbers ending in 5... with hardly any work at all.

Suppose you have the number A5 where A is some string of digits.

The trick to squaring A5 has just two steps:
1:  multiply A(A+1)
2:  Concatenate (attach) 25 to the right end of the resultant number
Viola :)

Below, I use [] to mean concatenation... can't remember if there's standard notation for it!

$ 15^2\equal{}1\cdot(1\plus{}1)[]25\equal{}225$
$ 25^2\equal{}2\cdot(2\plus{}1)[]25\equal{}625$
$ 35^2\equal{}3\cdot(3\plus{}1)[]25\equal{}1225$
$ 125^2\equal{}12\cdot(12\plus{}1)[]25\equal{}15625$ etc

Good exercise to try and prove this little factoid!  Good luck!

Factorization is the key to why it works. Let A5 be 10n+5. So,

$ (10n\plus{}5)^2\equal{}100n^2\plus{}100n\plus{}25\equal{}100(n^2\plus{}n)\plus{}25\equal{}100n(n\plus{}1)\plus{}25$

Now, it's pretty clear.

To me squares are harder after 30 but i still have to memorize them cubes is much easier to me

RightAnnihilator Said: I learned a neat trick from "How to Calculate Quickly" by Henry Sticker, and it involves learning how to square numbers ending in 5... with hardly any work at all.

Suppose you have the number A5 where A is some string of digits.

The trick to squaring A5 has just two steps:
1: multiply A(A+1)
2: Concatenate (attach) 25 to the right end of the resultant number
Viola

Below, I use [] to mean concatenation... can't remember if there's standard notation for it!

 etc

Good exercise to try and prove this little factoid! Good luck!

myruggy89 (myself) says: That is a Factoid Indeed. Good Job on Figuring this Out

Another one, which I think is in Henry Sticker's book also, which generalizes the previous one I posted:

If AC and AD represent two numbers where C and D are single digits, and C+D=10, then we have:

$ (AC)\cdot (AD) \equal{} A\cdot (A \plus{} 1)[]C\cdot D$

Where [] is again the concatenation (attaching) symbol.

Again, examples:
24(26)=[u]6[/u]24
92(98)=[u]90[/u]16
123(127)=[u]156[/u]21

If you proved the last trick, this one will be nearly as easy!  Of course, 5+5 is 10, so this proof would cover the previous one as well...

How to square AB quicker:

$ AB\equal{}10A\plus{}B$

$ AB^2\equal{}(10A\plus{}B)^2\equal{}100A^2\plus{}20AB\plus{}B^2$

And that should simplify squares after 41.

Well, that's just the expansion of a binomial squared. So technically, you could do this with any number, like

$ 53^2\equal{}(40\plus{}13)^2\equal{}40^2\plus{}2(40)(13)\plus{}13^2$.

Although, separating it the first way would be much easier.

Awesome Work Everyone, I will eventually post similar stuf like this to elaborate on this maybe perfect squares but only numbers between 40-100 than harder to see how you do

for squaring and multiplying in general, putting some algebra into it makes it easier, for eample

$ 41^2\equal{}(40\plus{}1)^2\equal{}40^2\plus{}2(40)(1)\plus{}1^2\equal{}1600\plus{}80\plus{}1\equal{}1681.$

its' not as hard as it looks. another example, using algebra slightly differently is

$ 102*106\equal{}(100\plus{}2)(100\plus{}6)\equal{}100^2\plus{}(2\plus{}6)(100)\plus{}12\equal{}10000\plus{}800\plus{}12\equal{}10812$
$ 92*108\equal{}(100\minus{}8)(100\plus{}8)\equal{}100^2\minus{}8^2\equal{}10000\minus{}64\equal{}99936$
$ 52*89\equal{}(50\plus{}2)(90\minus{}1)\equal{}4500\plus{}180\minus{}50\minus{}2\equal{}4680\minus{}52\equal{}4628$

theres also a concept called the vinculum(i think?), where the digits 6 thru 9 are not really used. for example, 48 would be written as 52, with a line over the 2 to represent 50 - 2. 9801 would be 10000-199, so would be written as 10199 with a line over the 199. it takes some practice to get used to the new arithmetic properties, but it is an interesting alternative to conventional arithmetic.

I get what the concept of what you're saying is, but I looked up "vinculum" on Wikipedia, and it said that a vinculum is just a bar placed over a mathematical expression (as in repeating decimals), generally meaning that it is a group.

http://en.wikipedia.org/wiki/Vinculum_(symbol)

A couple of examples of using algebra have been given in some of the previous posts.

And this is not just for squaring, as your examples show. The difference of 2 squares is useful in particular.

I got on my first test a 31 out of 60 in 5 minutes. Not bad Right? :blush:

On a test of cubes?

Well, in general, I wouldn't be happy with a 50% in any case (or whatever your percent is).
But I guess cubes are kinda hard to do in 5 minutes.


function
number theory proposed
number theory

For the $t(n)$ function defined for non-negative whole numbers we have:$t(0)=t(1)=0$, $t(2)=1$.
If $n \gneqq 2$, then $t(n)$ is the smallest positive whole number that doesnt divide $n$.
Lets have :
$T(n)=t(t(t(n)))$.
Find the value of $S$ if :
$S=T(1)+T(2)+T(3)+...+T(2005)+T(2006)$


ARML
conics
geometry
probability
Science Olympiad
number theory

What have you acccomplished so far this summer? What do you have left to accomplish? (It doesn't only have to be in math.) My summer vacation is more than half over and I want to do more before school starts.

For me,

My accomplishments:
    1. Reviewed old math competitions
    2. Organized old school papers
    3. Went on vacation and relaxed   :)

ugh not much else

Left to do:
     1. Start up a sport
     2. Finish summer homework
     3. More math
     3. Practice piano more efficiently

What about you?

1) Math
2) Computer class
3) Slept all day

...actually, not really. I hogged the computer all day because there is nothing better at our house.

did arml, slept, did math not really, surfed the internet, read some web comics (life on forbez, girl genius, questionable content), practiced piano and cello kinda, studied earth science, made the usa international earth science olympiad team (stop laughing), got a superior rating slash trophy at M.A.M.A in A division piano solo, went to the Omaha Zoo after doing that, and that was this weekend. have not started on summer reading.

ARMLed, pianoed, violined, mathed, and summer homeworked

USA International Earth Science Olympiad TEam?  :huh: Congratulations!

1. Call of Duty 4 and Halo 3. Lots of it.

2. Summer reading

3. [Currently on a] Vacation!  :pilot:

4. [Currently doing] CTY online chem course.

1.  Survived the boyfriend's family on multiple occassions (including the scary aunt)
2.  Found, and will move into a nice apartment
3.  Getting a kitten on Saturday!
4.  Joined a choir as a piano player
5.  Started and memorized two small piano pieces to perform for my family when I finally go home again.
6.  Surviving work...can't wait till school starts in the fall.
7.  Finally went up the CN tower =P
8.  Consistently played on the dodgeball team, 2nd/19 teams!  Yay!

[quote="7h3.D3m0n.117"][quote="Sunny"]1.  Survived the boyfriend's family on multiple occassions (including the scary aunt)

7.  Finally went up the CN tower =P
[/quote]

1.) Well that must be fun  :P . It's worse though when you're the guy and you have to visit your girlfriend's father  :| .

7.) I went up to the CN tower and sat on the floor glass - it was creepy. Although I did go to Tokyo Tower two days ago and did pretty much the same thing.[/quote]

Hahahaha, yeah...Dad's are scary.  =)  Not mine though, my boyfriend has it ridiculously easy in that department hahaha.  With my family, he suffers the risk of being stuffed to death with food if anything!

Yeah, the glass floor is really disconcerting, especially with everyone jumping on it!  I actually thought the elevator ride up would be scary...but it wasn't at all.
Oh Tokyo Tower eh?  That's pretty cool!

umm...

Stuff for fun:
1. went crazy in New York
2. watched Rent on Broadway (that leaves on September 7)
3. ate ginormo Philly cheesesteaks at Reading terminal
4. CTY LANCASTER 08.1!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
5. made a facebook and got addicted (ohnoes!)

Stuff for academics/preparation for school:
1. CTY LANCASTER 08.1!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
2. practiced violin nearly 2 hrs a day (thats a lot for me cuz i haven't practiced in a long time)
3. read about 10 uber boring books (Great Expectations, Jane Eyre, etc.)
4. Learned AP Statistics
5. Learned Spanish II (my sis was taking a course, and i was just using her books)

woww.. i thought my list of fun would be much longer, but never mind! :D

i'm proud of myself.

went through aops intro to geometry and intro to counting and probability almost done with number theory and i feel really smart  although it probably woud have ehlped to study  alittle more and wut i have left is to do intro to algebra to enhance my algebra skills and get a start on my java book and if anyone here makes it to thomas jefferson let me know ill probably be there 2 :alien:  :starwars:  :spam: is actually a type of  food

went to ARML, GIV Math Camp, volunteered at church summer camp, taking Inter. Alg, doing other math, piano, guzheng, facebook, computer, aim, tv, the usual summer time wasting...

1. Picked up golf, can't hit it past 150 yards with the longest club.  :(
2. Joined AoPS
3. Practiced flute
4. Finally wrote a 4-page story
5. practiced writing C# a little.. can't get past console applications without my dad's help :(
6. taught my sister how to speak english  :lol:
7. nothing in particular

i haven't done much...

1. slept (yay! who hasn't?)
2. went to summer skool (world history is boring)
at summer skool i...
~got eaten, stabbed, and shot at by a crazy guy that likes hitting me on the head with history books
3. ran after a bunch of hyperactive kids at a summer camp (arg i'm tired)
4. taken an english class at sabio (very fun)
5. partied with my friends!!!!! (yay!!)

hmm i think i did more than that...ahh well can't think :D


inequalities
function
inequalities proposed

$A \ge B$

how can  I proof that?

I can proof that finding $C$ such as $A \ge C$ so i proof that $C >B$?

$A-B\geq 0$. If $A\geq B>0$, then $\frac{A}{B}\geq 1$.

err...

you dont understand

A,B are functions

i just want to know the ways I can proof an inequality.

Even $A,\ B$ are functions,  my opinion holds. :wink:


geometry
IMO Shortlist
reflection
circumcircle

Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$.

Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.

If  K = AE^GB one can see that B is the orthocenter of triangle AKF. Then KF == d is the polar of P = EG^AB wrt $\Gamma$ and EG is the polar of D wrt $\Gamma$. Therefore DG is tangent to $\Gamma$  at G. It is easy to see that GCDF is cyclic, then <DCF = <DGF. So if G’ = CF^$\Gamma$, it holds that <G’CA = <DCF = <DGF = <GCA, which implies that <G'BA = <GBA or the thesis.

Here is [i]my solution[/i]:

In the following, we will use directed angles modulo 180° and directed segments (hereby, we will denote directed segments by XY and non-directed segments by |XY|).

In order to prove that the reflection H of the point G in the line AB lies on the line CF, we have to prove that < ACH = < ACF.

Now, the circle $\Gamma$ is symmetric with respect to the line AB (since AB is a diameter of $\Gamma$). Since the point G lies on $\Gamma$, the reflection H of G in the line AB must therefore also lie on $\Gamma$. Hence, by the chordal angle theorem, < ACH = < ABH. But since the point H is the reflection of G in the line AB, we have < ABH = - < ABG. Thus, < ACH = - < ABG = - < (AB; BG) = < (BG; AB). Since the point G lies on the circle $\Gamma$ with diameter AB, we have < AGB = 90°, or, equivalently, < (AG; BG) = 90°. Hence,

  < ACH = < (BG; AB) = < (AG; AB) - < (AG; BG)
        = < (AG; AB) - 90°.

On the other hand, < (d; AB) = 90°, since the line d is perpendicular to AB. Thus,

  < ACH = < (AG; AB) - 90° = < (AG; AB) - < (d; AB)
        = < (AG; d) = < AFD = - < DFA.

Hence, instead of showing < ACH = < ACF, it will be enough to establish - < DFA = < ACF.

After the tangent-chordal angle theorem, we have < DEB = < EAB, so that < DEF = < EAB. On the other hand,

  < EFD = < (BE; d)
        = < (BE; AE) + < (AE; AB) + < (AB; d).

Since < (BE; AE) = 90° (this is because < BEA = 90°, as the point E lies on the circle $\Gamma$ with diameter AB), < (AE; AB) = < EAB and < (AB; d) = 90° (since the line d is perpendicular to AB), this becomes

  < EFD = 90° + < EAB + 90° = 180° + < EAB
        = < EAB     (directed angles modulo 180° !);

together with < DEF = < EAB, this yields < EFD = < DEF. Hence, the triangle DEF is isosceles with |DE| = |DF|. Consequently, $DE^2=DF^2$. But after the intersecting tangent and chord theorem, we have $DE^2=DC\cdot DA$ (with directed segments!). Hence, $DF^2=DC\cdot DA$, and thus DF : DC = DA : DF. Since also, trivially < FDC = - < ADF, we can conclude that the triangles FDC and ADF are oppositely similar. Hence, < DCF = - < DFA, so that - < DFA = < DCF = < ACF, and this completes the proof.

  Darij

Anyway I found a cute solution using inversion:

First of all it’s not difficult to see that:
1. $\angle{EAO}=\angle{OEA}=\angle{EFD}=\angle{DEF}$ hence $ED=DF$.
2. $\triangle{EOD}$ and $\triangle{EAF}$ are similar hence $\angle{EOD}=\angle{DOG}$, and
      $ED=DG$.
From 1. and 2. $DG=DF $.
Let $CF$ intersect $\Gamma$ at $G_1$, and $AG_1$ intersect $d$ at $F_1$.

Consider the inversion $I$ of pole A and power ${AG}\cdot{AF}$. Since $I(F)=G$ and $I(C)=I(D)$, line $ CF$ transforms to circumcircle of $\triangle{AGD}$ since $F_1$ is on this circumcircle: $\angle{AF_1D}=\angle{DGF}=\angle{DFG}$ hence $AF_1=AF$ and obviously we obtain that $AG_1=AG$ and we are done.

just angle chasing the obvious angles (= and some similar triangles... and done. i suspect its the same as darij's proof above although i haven't read it.

At a QED seminar a nice solution was presented using Pascal's Theorem. Ty it !  :D

do you need some preliminary result? because I couldn't find the hexagon (degenerate or otherwise)... unless I haven't been looking hard enough.

I'm afraid, but I can't find my notes on it right now :( . As far as I remember you have to apply Pascal at least twice. You're right, that the degenerate version plays an important role; furthermore you also have to consider a point at infinity once. Sorry for the lack of details, but I will try to reconstruct it this afternoon.

EDIT:
I think I've managed to reconstruct the proof that was given at the seminar (although I can't give a guarantee, that I didn't do a mistake somewhere and that the proof is entirely identical to the original one   :) )

Some Definitions:
-$\Phi$ is the intersection of two distinct parallels of $d$
-$Y: =AE\cap BG$
-$S': =AB\cap G'G$
-$F': =EB\cap G'C$
-$G'$ is the intersection of the circle with the parallel of $d$ through $G$.
With these definitions it is enough to prove that $G'C$ and $d$ intersect in $F$

First, take the (degenerate) hexagon $AAEBBG$. By applying Pascal we get that $\Phi,Y,F$ are collinear. Thus $Y\in d$
Secondly, take the (degenerate) hexagon $ABGG'EA$. By applying Pascal we get that $\Phi,Y,S'$ are collinear. By the previous we get that $S'$ is on $d$ as well. Thus $S'=S$
Finally, take the (degenerate) hexagon $G'EEBAC$. By applying Pascal we get that $S,D,F'$ are collinear. By the previous we get that $F'$ is on $d$ as well. Thus $F'=F$, which proves the theorem.

We had this at an unofficial tst type test in december.  I used 2 inversions to complete my proof then...
Anyway, I remember we also had G8 from the isl in our test, but I don't seem to remember it right now and neither find the test sheet, and moreover I don't know if I am allowed to post it.

Let $\angle EAB=\alpha$, $AB$ intersects $d$ at $T$,

$G'$ be the refleciton of $G$ by $AB$, $GG'$ intersects $AC$ at $X$.

then, cuz $DE$ is tangent to circle $\Gamma$, $\angle DEF= \alpha$

And cuz $\angle AEB= \angle ATF = 90 \DEG$, $A, E, D, F$ are cocircle.

so $\angle EFD=\angle EAB=\alpha$.

therefore, from $\angle DEF=\angle DFE = \alpha$, we get $DE=DF$

so $DF^2 = DE^2 = DC \times DA$, or ${{DC} \over {DF}}={{DF} \over {DA}} = {{XG} \over {XA}} = {{CX} \over {G'X}}$

cuz ${{DC} \over {DF}} = {{CX} \over {G'X}}$ problem is proved. $Q.E.D.$

This problem was used as problem 2 of the final exam of the 3rd TST of Taiwan 2005.

Let  O be the center of the circle. Draw OE. $\angle BOE = 2 \angle DEB$. Let AB intersect $d$ at $H$, EHDO is cyclic hence
$\angle HDE = \angle HOE =2 \angle DEB$. Now by exterior angle of the triangle DEF we see that
$\angle DFE = \angle HDE -\angle DEF = 2\angle DEB- \angle DEB = \angle DEB$
Hence $\triangle DEF$ isosceles so $DE=DF$.

By Power of a Point $DE^2 = DF^2 = DC*DA$, or $\frac{DF}{DC}=\frac{DA}{DF}$

Hence SAS similarity $\triangle DFC \sim \triangle DAF$
so $\angle DFC =\angle DAF$.
Now since $FC$ intersect $\Gamma$ at another point $X$.
$\angle FXG = \angle DAF = \angle DFC$ so $d$ is parallel to $GX$ , which implies $X$ is the reflection of $G$ over $AB$, and the reflection is unique the desired result follows

[quote="MindFlyer"]There's a cute projective solution.
Send $d$ to infinity, in such a way that $\Gamma$ becomes an ellipse, and $AB$ becomes its axis.
Then apply the affine transformation that sends $\Gamma$ again to a circle (just scale in the direction of $AB$).
Now, we have a circle with two sets of parallel lines, and the result follows trivially (with many little additional corollaries, indeed!).[/quote]

Does anyone know of a decent reference on general projective transformations? I've read some notes on perspectivities (preserving cross ratios) and projectivities, so I've covered the basics of mappings from lines to lines, but I certainly wouldn't be sure that the first transformation given in this post is valid (how do you know there exists a transformation sending d to infinity that keeps a circle as an ellipse?).

Let $ H$ be the intersection of $ FC$ and $ \Gamma$, and let $ I$ be the intersection of $ d$ and $ AB$. By Pascal's theorem on cyclic hexagon $ EEBACH$, we have that $ F$, $ D$, and $ HE \cap AB$ are collinear, that is, we have that $ I$, $ E$, and $ H$ are collinear.

Construct point $ F'$ so that $ AIF'$ is directly similar to $ AHB$. Since $ F'I \perp IA$ and $ DI \perp IA$, we have that $ F'$, $ D$, and $ I$ are collinear. A spiral similarity centered at $ A$ maps $ F'I$ to $ BH$, so if we let $ X \equal{} F'B \cap IH$, we have that $ ABXH$ and $ AIF'X$ are cyclic. But this means that $ X$ lies on both $ \Gamma$ and $ IH$, implying that $ X \equal{} E$. Hence, since $ F'$ lies on both $ BX$ and $ ID$, i.e., both $ BE$ and $ d$, $ F' \equal{} F$. Thus, $ \triangle ABH \sim \triangle AFI$.

It is clear that $ \triangle AFI \sim \triangle ABG$, so $ \triangle AGB \sim \triangle AHB$. But since the two triangles share side $ AB$, they are congruent; it follows trivially that $ H$, which lies on $ CF$, is the reflection of $ G$ across $ AB$, which completes our proof.

Let $K = AE \cap d$ Observe that, since $BE \perp AK$, and $AB \perp d$, $B$ must be the orthocenter of $KFA$, hence $KB \perp AF$, but since $BG \perp AF$, we must have $K, B, G$ collinear, in other words $GB \cap AE \in d$. By Pascal's on $GGBEEA$, we see that the tangent to $\Gamma$ from $G$ intersects the tangent to $\Gamma$ from $E$ on the line $d$, which means the $DG$ is tangent to $\Gamma$. Let $H$ be the reflection of $G$ over $AB$. Since $HAGB$ is trivially harmonic (since it is a kite), it is sufficient to prove, by taking perspective at $F$, that $ECGA$ is harmonic. But observe that the tangents from $E$ and $G$, intersect at $D$, and since $A,C,D$ are collinear, $ECGA$ is harmonic. Hence, $H, C, F$ are collinear, as desired.

Let $G'$ be the reflection of $G$ and Pascal on $ACBEEG$ and $G'CBBEG$ implies that $\overline{G'C}\cap\overline{BE}$ lies on $d.$ $\square$

Let FC meet $\Gamma$ at S. we will prove S is reflection of G in AB. note that for that we need to prove SG ⊥ AB or in fact SG || DF.
Let O be center of $\Gamma$ and let AB meet DF at K. ∠OED = ∠OKD = 90 so OEKD is cyclic. ∠KDE = ∠KOE = 2∠BAE so ∠DFE = ∠BAE = ∠FED so
DE = DF. By power of point D we have DC.DA = DF^2 so DF is tangent to circle ACF. ∠GSC = ∠GAC = ∠CFD so DF || GS.
we're Done.

I randomly proved that $DG$ is tangent???

[img width=50]https://media.discordapp.net/attachments/925784397469331477/953121070657982494/Screen_Shot_2022-03-14_at_7.42.45_PM.png[/img]

Let $G'$ be the intersection of $FC$ and $\Gamma.$ We claim that $DG$ is tagent to $\Gamma.$ Let $G$ be intersection of $AF$ and $\Gamma.$ Note that $\angle EFD=\angle ABF-90^\circ=90^\circ-\angle ABE=\angle BAE=\angle FED$ so $FD=DE.$ Thus, $FD^2=CD\cdot AD$ so $(ACD)$ tangent to $DF.$ Also, $\angle AGC=90^\circ+\angle CAB=\angle ADF$ so $FGCD$ is cyclic. Then, $\angle CGD=\angle CFD=\angle CFD=\angle CAF$ which implies $DG$ tangent to $\Gamma$ as desired.

Apply Pascal's theorem on $AGEEBC$ to get $GE\cap BC,D,F$ are collinear, so the intersection of $GE,BC$ lies on $d.$ Then note that Pascal on $BBCG'GE$ gives that the intersection of $GE,BC,$ $F$ and the intersection of $GG'$ and the tangent at $B$ is on $d.$ However, the tangent at $B$ is parallel to $d.$ Thus, $GG'||d,$ as desired.

Let the other tangent from $D$ to $\Gamma$ meet $\Gamma$ at $G'$. Let $AG' \cap BE=F'$ and $AE\cap BG'=X$. Note that by pascal on $AG'G'BEE$ we have that $F', D, X$ are collinear. Furthermore $\angle F'EX= \angle XG'F'=90^{\circ}$ Hence $B$ is the orthocenter of $\triangle AF'X \implies AB \perp F'D\implies F'=F \implies G'=G$. Hence  $DG$ is tangent to $\Gamma$. Let $FC \cap \Gamma=H$. Then by Pascal on $CHGGAA$, we have that $F,HG\cap AA, D$ are collinear, but since $FD \parallel AA$, this is possible iff $GH \parallel d \perp AB$ id est $H$ is the reflection of $G $ across $AB$. $\blacksquare$

Let $X$ be the intersection of $AB$ with $d$. Let $P$ be the intersection of $AE$ and $d$. Let $G'$ be the reflection of $G$ around $AB$.  Note $\angle AEB=\angle AGB=\angle AG'B=\angle ACB=90$. In addition, because $B$ is the orthocenter of $FPA$, we know $P$, $B$, and $G$ are collinear.

We have $\angle APX=90-\angle XAP=\angle EBA$.

In addition, because $DE$ is tangent to $\Gamma$, $\angle PED=\angle EBA=\angle APX=\angle EPD$. Thus, $\triangle DPE$ is isosceles. Because $\triangle FPE$ is a right triangle, $D$ is the midpoint of $FP$. Because $\angle BGA$ is right and $P$, $B$, and $G$ are collinear, $\angle FGP$ is right. Since $D$ is the midpoint of $FP$ and $\triangle FPG$ is a right triangle, $\triangle FDG$ is isosceles with $\angle DFG=\angle DGF$.

We have $\angle DFG=\angle XFA=90-\angle XAF=\angle GBA=\angle GCA$. So, $\angle GCD=180-\angle GCA=180-\angle DFG$. Therefore, $GCDF$ is cyclic. Therefore,
$$\angle DCF=\angle DGF=\angle DFG=\angle GCA=\angle G'CA.$$
Therefore, $F$, $C$, and $G'$ are collinear.

Pascal's on $CAGEEB$ to get $\overline{GE} \cap \overline{BC} \in d$. Then, Pascal's on $GG'CBBE$ finishes.

Alternatively, let $O$ be the center of $\Gamma$ and $P=\overline{AB} \cap \overline{GE}$. Brokard's on $AGBE$ implies that $F$ and $\overline{AE} \cap \overline{BG}$ form a line perpendicular to $\overline{OP}$, which means $\overline{AE} \cap \overline{BG} \in d$. Then, Pascal's on $ACBGGE$ yields $\overline{AC} \cap \overline{GG} \in d$, i.e. $\overline{DG}$ is tangent to $\Gamma$, hence $AGCE$ harmonic. To finish, let $C \neq Q=\overline{FC} \cap \Gamma$, so we have
$$-1=(A,G;C,E)\stackrel{F}{=}(A,G;Q,B),$$
which implies $\tfrac{AG}{AQ}=\tfrac{BG}{BQ}$, hence $Q$ is the reflection of $G$ over $\overline{AB}$.

[hide = Proof] First we apply Pascal's theorem on hexagon $EEBCAG$. Thus we have $BC \cap GE \in d$. Now applying Pascal's theorem $CGG'EBB$ gives us
\begin{align*}
&CG' \cap EB = F'\\
&GG' \cap BB = P_{\infty} \\
&GE \cap BC = P,
\end{align*}
are collinear. Hence $CG' \cap EB$ lies on $d$ and so does $BE \cap AG$. Hence these lines concur at $F$ and we are done. $\blacksquare$
[/hide]

This is fun, to get some motivation its helpful to assume what he problem asks for.
Redefine $G$ as the second tangent from $D$ to $\Gamma$, $F$ as the intersection of $AG$ and $BE$, let $BG \cap EA=K$ and $GE \cap BA=J$ and $d \cap AB=X$, now we take polars w.r.t. $\Gamma$
[b][color=#f00]Claim 1:[/color][/b] $FK$ is just the line $d$
[b][color=#f00]Proof:[/color][/b] By brocard $\mathcal P_J=FK$ and since $J$ lies in $AB$ we have that $\mathcal P_J \perp AB$ and also note that $J \in \mathcal P_D$ so by La'Hire $D \in \mathcal P_J$ and all of those tell us that $FK=\mathcal P_J=d$
[b][color=#f00]Finishing:[/color][/b] Now its easy to see that $B$ is ortocenter of $\triangle AKF$ and by ortocenter config we get that $FGEK$ is cyclic and has center $D$ so $DF=DG$, now by the perpendiculars we get $XDCB$ cyclic so by miquel theorem in $\triangle XFA$ we get $FDCG$ cyclic so by arcs we get $\angle DFG=\angle FCD$ so $\angle ACG'=\angle GCA$ which means arcs $AG,AG'$ are equal in $\Gamma$ so $AG=AG'$ and since $AB$ is diameter we get that $G,G'$ are symetric w.r.t. $AB$ thus we are done :D

Pascal's theorem on $AGEEBC$ shows that $\overline{BC} \cap \overline{GE}$ lies on $d$.
[asy]
size(10cm); draw(unitcircle); pair O = origin;
pair A = dir(180); pair B = dir(0); pair D = (1.6,0.7); pair C = 2*foot(O,A,D)-A; pair M = 1/conj(D); pair E = IP(M--(M+dir(M)*dir(-90)),unitcircle); pair F = extension(B,E,D,conj(D)); pair G = 2*foot(O,A,F)-A; pair G1= conj(G); pair X = extension(C,B,G,E); draw(F--X); draw(A--D--E--F--A--B); draw(G1--F, dotted); draw(G--X--C, lightred+dashed); dot(X);
dot("$A$", A, dir(180)); dot("$B$", B, dir(0)); dot("$D$", D, dir(0)); dot("$C$", C, 1.9*dir(85)); dot("$E$", E, dir(-90)*dir(M)); dot("$F$", F, dir(0)); dot("$G$", G, dir(100)); dot("$G'$", G1, dir(-90));
[/asy]
Let $G'$ be the reflection of $G$ over $\overline{AB}$. Then applying Pascal's theorem to $CG'GEBB$ forces $\overline{CG'} \cap \overline{BE}$ to lie on $d$, so the intersection must be the point $F$.

Let $GB \cap AE$ at $P$. Now $AB$ is diameter $FB \perp AP$ and $PB \perp AF$ therefore $B$ is Orthocenter of $\triangle AFP$.As $AB \perp FP$ but $AB \perp \ell$, therefore P lie of line $\ell$.

$ED$ is tangent to $(AEBG)$ by [b]Lemma 1.44 (Three Tangents)[/b] in EGMO, We get $D$ as midpoint of $FP$ and $DG$ is tangent to circle $(AEBG)$.

Its easy to see $FDCG$ is cyclic, $GG' \perp AB$ so we get $\angle DFG = \angle G'GA = \angle GG'A = \angle GCA$ therefore $(FDCG)$ is cyclic.
$DF$ tangent to circle we get $\angle CFD = \angle DGC = \angle GAC = \angle GG'C$.

$GG'$ is parallel to $FD$ and  $\angle DFC =\angle GG'C$ we get $F-C-G$

Denote $G'$ as the second intersection of $FC$ with $\Gamma$. Pascal's on hexagon $EEBACG'$ gives $D$, $F$, and $BA \cap G'E$ collinear, so $BA \cap G'E \in \ell$.

Using Pascal's again on hexagon $ABBEG'G$, we have $BA \cap G'E$, $BB \cap GG'$, and $F$ collinear, so $BB \cap G'G \in \ell$.

However, since the tangent at $B$ is parallel to $\ell$, it follows that $GG'$ is too, from which we get the desired. $\blacksquare$

Let $R$ be the intersection of $BC$ and $GE$; note that from Pascal on $EEBCAG$, then we find that $R$ lies on $d$.
Let $S$ be the intersection of $AE$ and $BG$; note that from Pascal on $AEBBGA$ then $S$ also lies on $d$.
Finally, let $T$ be the intersection of $AB$ and $d$.

Note that $B$ is the orthocenter of $AFS$, and $T$, $E$, and $G$ are the feet; in particular, $(SF;RT) = -1$.
Also note that $D$ is the midpoint of $SF$ from Three Tangents Lemma, a fact I found that we will not use (though it does implies $ACGE$ harmonic).

Let $FC$ intersect $(AGCBE)$ at $U$; it suffices to show that $AGBU$ is harmonic. Projecting,
\[ (AB;GU) \stackrel{E}{=} (SF;R, UE \cap d) \]
So it suffices to show that $U$, $E$ and $T$ are collinear. But this is true from Pascal on $ABCUEG$. $\blacksquare$

Normally I don't post my solutions on AoPS, but I couldn't help but notice the SEVERE lack of complex bashes here. Please enjoy my mathematical equivalent of Star Wars Episode IX.

[hide=nobody's using hide tags but it still feels wrong to post solutions without hide tags]
We will first set some results with complex numbers we will use.
[list]
[*] If $A,B$ are on the unit circle, then $C$ is on $AB$ if and only if $c + ab\overline{c} = a + b$. To prove this, note that $\frac{a - c}{a - b}$ must be real, and set it equal to its conjugate.
[*] For any point $x$, the second intersection of the line through $x$ and $-1$ through the unit circle is $\frac{x + 1}{\overline{x} + 1}$. To prove this, let the second point be $y$ and plug in $a = -1$ and $b = y$ into the above equation.
[*] The intersection of $AB$ and $CD$ for $a,b,c,d$ on the unit circle is $\frac{ab(c + d) - cd(a + b)}{ab - cd}$. This is well-known.
[*] For $A,B$ on the unit circle, the intersection of the tangents to $A$ and $B$ from the unit circle is $\frac{2ab}{a + b}$. Get this from $a,a,b,b$ into the above equation.
[/list]

We will set $\Gamma$ as the unit circle, and let $b = 1$ and $a = -1$. Then line $d$ is vertical, so we can say that $x$ lies on the line if and only if $x + \overline{x} = k$ for a positive real number $k$ greater than $1$. Henceforth, we will refer to the line as $\delta$ to avoid confusing it with point $D$. (Edited to add: upon finishing, I never referred to the line again lol)

[b]Part 1:[/b] $G$ is the second tangency point of $D$ to $\Gamma$.

We will find everything in terns of $E$. Since $F$ is on $BE$, we have $f + e\overline{f} = 1 + e$. Since $\overline{f} = (k - f)$, we have $f(1 - e) = 1 + e - ke$ so
\[f = \frac{1 + e - ke}{1 - e}.\]
Thus,
\[g = \frac{f + 1}{\overline{f} + 1} = \frac{\frac{2 - ke}{1 - e}}{\frac{2e - k}{e - 1}} = \frac{ke - 2}{2e - k}.\]

Now, since $D$ lies on the tangent from $E$, we have $2e = d + e^{2}\overline{d}$. Since $\overline{d} = (k - d)$, we have $d(1 - e^{2}) = 2e - ke^{2}$ so
\[d = \frac{2e - e^{2}k}{1 - e^{2}}.\]

Now note that the intersection of the tangents from $G$ and $E$ to the unit circle is
\[\frac{2ge}{g + e} = \frac{2e \cdot \frac{ke - 2}{2e - k}}{\frac{2e^{2} - 2}{2e - k}} = \frac{ke^{2} - 2e}{e^{2} - 1}\] which is equal to the value of $D$ we found before, as desired.

[b]Part 2:[/b] Finishing the problem.

Now we will write everything in terms of $g$ and $e$, since it makes $c$ less annoying. We have $d = \frac{2ge}{g + e}$, so
\[c = \frac{d + 1}{\overline{d} + 1} = \frac{2ge + g + e}{2 + g + e}.\]

We can also find $F$ using the intersection of chords formula. Since $F = AG \cap BE$, we find
\[f = \frac{-g(e + 1) - (e)(g - 1)}{-g - e} = \frac{2ge + g - e}{g + e}.\]

It suffices to prove that $c,f,\overline{g}$ are collinear. To do this, we want $\frac{c - f}{f - \overline{g}}$ to be real. We first find that
\begin{align*}
    c - f &= \frac{(2ge + g + e)(g + e) - (2ge + g - e)(2 + g + e)}{(g + e)(2 + g + e)} \\
    &= \frac{(g + e)(2e) - 2(2ge + g - e)}{(g + e)(g + e + 2)} \\
    &= \frac{2e^{2} - 2ge - 2g + 2e}{(g + e)(g + e + 2)} \\
    &= \frac{2(e - g)(e + 1)}{(e + g)(e + g + 2)}.
\end{align*}
We can also find
\[f - \overline{g} = \frac{2g^{2}e + g^{2} - eg - g - e}{g(g + e)}.\] Thus,
\[\frac{c - f}{f - \overline{g}} = \frac{2g(e - g)(e + 1)}{(e + g + 2)(2g^{2}e + g^{2} - eg - g - e)}.\]
We want to prove this is real, that is, that it's equal to its conjugate.

We expand the denominator to get
\[2g^{2}e^{2} + g^{2}e - e^{2}g - eg - e^{2} + 2g^{3}e + g^{3} - eg^{2} - g^{2} - eg + 4g^{2}e + 2g^{2} - 2eg - 2g - 2e.\] Combining terms gives
\[2g^{2}e^{2} + 2g^{3}e + 4g^{2}e - e^{2}g + g^{3} - 4eg - e^{2} + g^{2} - 2e - 2g.\]
Its conjugate is equal to
\[\frac{2g + 2e + 4ge - g^{2} + e^{2} - 4g^{2}e - g^{3} + ge^{2} - 2g^{3}e - 2g^{2}e^{2}}{g^{3}e^{2}}.\] Conveniently, the numerator is the negative of the original! (Though we should expect this).

Thus,
\begin{align*}
    \frac{c - f}{f - \overline{g}} &= \frac{2g(e - g)(e + 1)}{(e + g + 2)(2g^{2}e + g^{2} - eg - g - e)} \\
    &= \frac{\frac{2g(e - g)(e + 1)}{g^{3}e^{2}}}{\frac{(e + g + 2)(2g^{2}e + g^{2} - eg - g - e)}{g^{3}e^{2}}} \\
    &= \frac{2\overline{g}\overline{(g - e)}(1 + \overline{e})}{-\overline{(e + g + 2)(2g^{2}e + g^{2} - eg - g - e)}} \\
    &= \overline{\left(\frac{c - f}{f - \overline{g}}\right)}
\end{align*}
as desired.
[/hide]

Some remarks: I saw one earlier complex bash post (#25) by [b]suryadeep[/b]. It does mostly the same thing as me (albeit in a different order) except when checking collinearity. I think the key here is getting the most symmetric-looking expression for $f$ possible (particularly, just looking at the expression for $f$ in #25 means that the collinearity check will probably suck, since $f$ is the sum of two fractions). Particularly, noticing that the line doesn't matter anymore is helpful, since the line is mostly just annoying.

Finally, the standard way of checking collinearity (determinant) is MUCH more annoying, unless you're OP at row reducing. I think the fact that the denominators of $e + g$ and $e + g + 2$ are similar should inspire just finding the differences of points instead of doing determinant things.


A four-digit number is of the form $ abab$ when $ a$ and $ b$ are its digits. What is the greatest prime factor of the number?

Hm, never mind.  I was asking for a differenet question, which I'm sure my answer was correct, but apparently it was a different question...


let MDOM' be a parallelogram .draw line OC is wich ,C is the mid point of MM'.the interior bisector and the exterior

bisector of the angle COD,intersect side MD in points A and B,respectivly.

prove that:

                MD/MA=MB/MD

Extend $DC$ to meet $DM$ at $M''$. Since $M'C=CM$, so $M''M=M'D=MD$. Let $M''M=M'D=a$, $MA=x$, $MB=y$.
Use the angle bisector theorem.
\[ \dfrac{a+x}{a-x}=\dfrac{M''O}{OD}=\dfrac{y+a}{y-a} \]
Simpify we get $a^2=xy$.

an easy and nice solution u wrote  :) .thanks shobber :D


[size=200]Note: ANYONE may hurt and heal, not just the participants. This game is spicy like that.[/size]

1. You cannot heal yourself.
2. You cannot hurt yourself.
3. You cannot double-post in this thread.
4. Once you hurt and heal, you must recharge (wait) for 1 hour (at least) before hurting & healing again.
NOTA BENE: This recharge period be independent of the other hurt&heal recharge period!

The next batch of people would like to start getting stabbed and bandaged.

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 20
#H34N1: 20
13375P34K43V312: 20
mustafa: 20
ckck: 20
now a ranger: 20

Hurt #H34N1, heal now a ranger

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 20
#H34N1: 19
13375P34K43V312: 20
mustafa: 20
ckck: 20
now a ranger: 21

thanks ckck!

Hurt #H34N1, heal ckck

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 20
#H34N1: 18
13375P34K43V312: 20
mustafa: 20
ckck: 21
now a ranger: 21

Hurt #H34N1, heal now a ranger

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 20
#H34N1: 17
13375P34K43V312: 20
mustafa: 20
ckck: 21
now a ranger: 22

[quote="Ignite168"]CAN I BE IN WITH 10 POINTS[/quote]

Okay.

Hurt ckck, heal #H34N1

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 20
#H34N1: 18
13375P34K43V312: 20
mustafa: 20
ckck: 20
now a ranger: 22
Ignite168: 10

hurt jli, heal ckck.

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 18
13375P34K43V312: 20
mustafa: 20
ckck: 21
now a ranger: 22
Ignite168: 10

hurt #H34N1, heal ckck

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 17
13375P34K43V312: 20
mustafa: 20
ckck: 22
now a ranger: 22
Ignite168: 10

Hurt #H34N1, heal now a ranger

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 16
13375P34K43V312: 20
mustafa: 20
ckck: 22
now a ranger: 23
Ignite168: 10

hurt ckck, heal Ignite

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 17
13375P34K43V312: 20
mustafa: 20
ckck: 21
now a ranger: 22
Ignite168: 11

Hurt #H34N1, heal now a ranger

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 16
13375P34K43V312: 20
mustafa: 20
ckck: 21
now a ranger: 23
Ignite168: 11

Hurt #H34N1, heal ckck

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 15
13375P34K43V312: 20
mustafa: 20
ckck: 22
now a ranger: 23
Ignite168: 11

hurt 13375P34K43V312, heal ckck

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 15
13375P34K43V312: 19
mustafa: 20
ckck: 23
now a ranger: 23
Ignite168: 11

Hurt #H34N1, Heal ckck

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 14
13375P34K43V312: 19
mustafa: 20
ckck: 24
now a ranger: 23
Ignite168: 11

Hurt #H34N1, heal now a ranger

ProtestanT: 20
D3m0n Shad0w: 20
Treething: 20
jli: 19
#H34N1: 13
13375P34K43V312: 19
mustafa: 20
ckck: 24
now a ranger: 24
Ignite168: 11

hurt nar twice heal tree

Treething: 6
nar: 10

hurt treething twice, heal nar

Treething: 4
nar: 11

hurt treething twice heal nar

Treething: 2
nar: 12

hurt treething twice, heal nar

DEAD

nar wins

woot

hurt 2x on treething

treething:0
nar:12

end game

sorry, i was forced to make this message

-jorian

what kind of sick person forced you to make that message?  :mad:

[quote="Klebian"]I'm proud to see the number of alt accounts created for this game.[/quote]
It really is shameful.

People made double accounts and healed themselves/hurt others more with them.

lol, double accounts
well gg

and i can do better than 2nd place in a game of wc3 (tft) if you wanna play that =p

[quote="Walk Around The River"]No, no new accounts were made for this.  We just have a little alliance between us.

As in
ckck-chris kim
me
now a ranger-alex
ragnarok -pratik
asianman-daniel
bfrock2000-jinzi
asianpower-patrick

They all go to fairview except nar, who lives in alabama.

no new accounts were used.  or at least to my knowledge?[/quote]
Uh, aren't alliances cheating as well?

Anybody want to start another game? I  forgot to sign up for the last one.....


Fie K un corp comutativ cu proprietatea ca x^2 +1 este diferit de 0 pentru orice x din K. Sa se arate ca daca x si y sunt elemente din K astfel incat
x^2+y^2=0 atunci x=y=0

[hide="solutie"]deoarece $\mathbb{K}$ este corp comutativ, relatia
$x^{2}+y^{2}=0$
se scrie :
$x^{2}(1+(x^{-1}y)^{2})=0$
cum paranteza este nenula, rezulta $x=0$. apoi $y^{2}=0$, deci $y=0$.[/hide]
mentionez ca eu inca nu m`am apucat de studiat serios partea aceasta a algebrei, deci s`ar putea sa zic o prostie mai mare decat mine  :blush:

[hide="solutie"]Din $x^{2}+y^{2}=0$ rezulta $x^{2}=-y^{2}$. Daca, prin absurd, $y\neq 0$, rezulta ca $y$ e inversabil. Inmultim atunci ambii membri cu $y^{-2}$ si, tinand cont de comutativitate, obtinem:
$(xy^{-1})^{2}=-1$, contradictie cu ipoteza.

Deci $y=0$. Analog $x=0$.[/hide]


limit
integration
Euler
Functional Analysis
advanced fields
advanced fields unsolved

A functional $I \in C^{1}(X,\mathbf R)$ on a Banach space $X$ is said to satisfy Palais-Smale condition at level $c \in \mathbf R$ ($\mathrm{(PS)}_{c}$-condition) iff each sequence $(u_{n}) \subset X$ such that
\[\lim I(u_{n}) = c \text{ and }\lim \|I'(u_{n})\|_{*}= 0\]
contains convergent in $X$ subsequence.

The same functional $I$ is said to satisfy Palais-Smale condition ((PS)-condition) iff it satisfies $\mathrm{(PS)}_{c}$-condition for every $c \in \mathbf R$.

Now consider the functional on $X = H_{0}^{1}(\Omega)$, $\Omega$ being bounded open subset of $\mathbf R^{n}$:
\[I(u) = \frac{1}{2}\int_\Omega |\nabla u|^{2}-\frac 1{p+1}\int_\Omega (u^{+})^{p+1}\]
where $u^{+}: = \sup (u,0)$.

Prove that $I$ satisfies (PS)-condition for $1 < p < \frac{n+2}{n-2}$ if $n\ge3$ and $1<p<+\infty$ if $n=1,2$.

ideas for a proof:
Suppose that $I(u_{n}) \to c$ and $DI(u_{n}) \to 0$ in $H^{-1}$. One wants to extract a convergent subsequence:

1/ The functional $I$ is well defined (Sobolev injection) and it is $C^{1}$, with differential:
\[DI(u).v=\int \nabla{u}\nabla{v}-\int (u^{+})^{p}(v^{+})\]

2/ $DI(u_{n}).u_{n}= (p+1)I(u_{n})-(\frac{p-1}{2}) \|u_{n}\|_{H^{1}_{0}}^{2}$ and $|DI(u_{n}).u_{n}| \leq \|u_{n}\|_{H^{-1}}.\|u_{n}\|_{H^{1}_{0}}$. From here one can conclude that $u_{n}$ is bounded in $H^{1}_{0}$. Because $p+1 < 2^{*}$ (Sobolev exposant), one knows that the injection of $H^{1}_{0}$ into $L^{p+1}$ is compact. This shows(?) that $(u^{+})^{p}$ converges towards $u$ in $L^{\frac{p+1}{p}}$. Duality shows that the injection of $L^{\frac{p+1}{p}}$ into $H^{-1}$ is compact. Hence(?) $v \to \int (u_{n}^{+})^{p}(v^{+})$ converges towards $v \to \int (u^{+})^{p}(v^{+})$ in $H^{-1}$. This is why $v \to \int (\nabla u_{n}) (\nabla v) = \int (\Delta u_{n}) (v)$ tends towards $v \to \int (u^{+})^{p}(v^{+})$. This is why $\Delta u_{n}$ converges in $H^{-1}$, which shows that $u_{n}$ converges in $H^{1}_{0}$ because $\Delta: H^{1}_{0}\to H^{-1}$ is an isomorphism.

Rq: to conclude to something interesting (what is exactly the Euler-Lagrange equation associated to this functional?), it remains to show the existence of a sequence $u_{n}$ such that $I(u_{n}) \to c \neq 0$ and $DI(u_{n}) \to 0$. If the functional were bounded from below, one could use this lemma http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ekeland&t=127075 , but it is not the case  :(  How can one show that such a sequence exists ?

I found your proof a little hard to follow. How do you define $\|u\|_{H^{-1}(\Omega)}$ and why $|\left\langle u_{n},I'(u_{n}) \right\rangle| \le \|u_{n}\|_{H^{-1}}\|u_{n}\|_{H_{0}^{1}}$?

Critical points of $I$ are exactly the solutions of the equation
\[\left\{ \begin{aligned}-\Delta u & = u^{p}& \text{ in }\Omega, \\ u & = 0 & \text{ on }\partial\Omega, \\ u & > 0 & \text{ in }\Omega \end{aligned}\right.\]
or, equivalently, of the equation
\[\left\{ \begin{aligned}-\Delta u & = (u^{+})^{p}& \text{ in }\Omega, \\ u & = 0 & \text{ on }\partial\Omega, \end{aligned}\right.\]
bacause of the strong maximum principle.

Here one should make use of the Mountain-Pass Lemma:

[b]Mountain-Pass Lemma.[/b] Let $X$ be a Banach space, $I \in C^{1}(X,\mathbf R)$. Suppose there exist $x_{1}\in X \setminus \{0\}$ and $0 < R < \|x_{1}\|$ such that
\[\max (I(0),I(x_{1})) < \inf_{\|x\|=R}I(x).\]
Now denote
\[c : = \inf_{\gamma\in\Gamma}\max_{0 \le t \le 1}I(\gamma(t)),\]
where $\Gamma : = \{\gamma \in C([0,1],X) \mid \gamma(0)=0 \text{ and }\gamma(1)=x_{1}\}$.

Then there exists a sequence $(x_{n})$ ("Palais-Smale sequence") in $X$ such that
\[\lim I(x_{n}) = c \text{ and }\lim \|I'(x_{n})\|_{*}= 0.\]

Thus if $I$ satisfies the conditions of Mountain-Pass Lemma and satisfies (PS)-condition, then $I$ has always a critical point.

In our case, $I$ satisfies (PS)-condition for $p$ defined above and satisfies all the conditions of Mountain-Pass Lemma. Hence the system of PDE defined above has a least one solution.

my fault, it was of course a mistake: $|DI(u_{n}).u_{n}| \leq \|DI(u_{n})\|_{H^{-1}}.\|u_{n}\|_{H^{1}_{0}}$. And one can conclude in my 'proof' the boundedness of $u_{n}$ because $\|DI(u_{n})\|_{H^{-1}}$ is bounded (tends towards $0$)

PS: anyone with a proof of this Mountain pass lemma ? very intuitive in finite dimension, but ?!? I find it beautiful to conclude the problem  :)


Could someone help me with this problem?
How many lattice points lie on the hyperbola x^2-y^2=2000^2?

It doesn't seem to be a hard problem but I just can't get the right answer.

[hide]Try factoring the left side and then finding factors of the right side.[/hide]

[hide]I got that (x+y)(x-y)=2^8*5^6 but i cant seem to do anything else. This means that 2000^2 has 63 factors but I am still missing what to do next.[/hide]

[hide]It means that one of the factors has to be x+y and one has to be x-y.[/hide]

[hide]

(x+y)(x-y)=2^8*5^6

(x+y) and (x-y) could be both positive or both negative.

Both positive -

(x+y), (x-y) must be of same parity or else x and y would not be integers, so they're both even.

Let (x+y)=2(a), (x-y)=2(b).

2(a)*2(b)=2^8*5^6. So ab=2^6*5^6.

So (x+y) could be (6+1)(6+1)=49 numbers.

Each value of (x+y) corresponds to a distinct lattice point, so 49 lattice points.

Both negative-

Same as above.

So there's 49*2=98 lattice points.

[/hide]

Ahh, I see.  Thanks for the explanation.  I understand it now.


How many ordered triples $ (x,y,z)$ of integers satisfy the system of equations below?

\[ x^2 \minus{} 3xy \plus{} 2yz \minus{} z^2 \equal{} 31\]
\[ \minus{}x^2\plus{}6yz\plus{}2z^2 \equal{} 44\]
\[ x^2 \plus{} xy \plus{} 8z \equal{} 100\]

$ \textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } \text{a finite number greater than } 2 \qquad \textbf{(E) } \text{infinitely many}$

Already shown here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=401136#401136[/url]

Sorry, searched through resources for this problem, but couldn't find it.  This is solely for that.


i have found this inequality: consider a,b: integer and 3<=a<=b then we have  : a^b>=b^a. I know 2 solutions which don't use logarith.How many solutions you have? :D   :P

I think I've seen this topic somewhere before  :?

We can do it by induction.
Suppose $a^b\geq b^a$ and $b\geq a$. We need to show that $a^{b+1}\geq (b+1)^a$, but $a^{b+1}=a\cdot a^b\geq ab^{a}$. Thus we obtained new inequlity to prove: $ab^a\geq (b+1)^a$, but
\[(b+1)^a=\sum_{k=0}^{a}C_a^kb^{a-k}<1+\sum_{k=0}^{a-1}a^kb^{a-k}\leq 1+\sum_{k=0}^{a-1} b^a=1+ab^a,\]
hence result :D
I suspect that there is a nice short proof.

we can prove series inequalities :
   3^(1/3)>=4^(1/4)>=...>=n^(1/n) by this simple inequality:
   for all a>=3 then a^(a+1)>=(a+1)^a by induction.
           from this series inequalities, we get the solution

I'm bolzano-ykl, I have forgotten my password. :D  :lol: is it a nice short solution? Can you give another one?


calculus
real analysis
real analysis unsolved

Hello,
    I read somewhere that a bounded sequence is convergent iff it is monotonic.Is it correct?But I have a counter-example for this statement.Consider the sequence (-1/2)^n i.e., -1/2, 1/4, -1/8, 1/16....Clearly this is bounded and convergent but not monotonic.

No, the statement is not correct, as you have yourself demonstrated.

Maybe you are thinking of one way to formulate the completeness axiom for real numbers: A sequence of real numbers is convergent if it is monotone and bounded.

I like to formulate that particular equivalent to the completeness of the real numbers as the "monotone sequence alternative": a nondecreasing sequence either converges or tends to $+\infty.$ (Of course, you can also say that a nonincreasing sequence either converges or tends to $-\infty.$)

The partial sums of a series with positive terms falls under this alternative - hence the power of comparison tests.

But as thealchemist has realized, this is not an "iff." With series, it is certainly possible to have convergent series whose terms vary in sign.

Thanks for clarifying.And Kalle,is that the completeness 'axiom' or can it be proved?If I am not wrong,axioms cannot be proved.

It is common in calculus or analysis courses to state (as axioms) the properties of a complete ordered field, and then to call those the axioms for the real numbers. There's a set of algebraic axioms which are the axioms for a field (and there are many, many other fields, explored in detail in an algebra course).

The use of axioms is a way to generality: any theorem you can prove using only the field axioms would then be true for any field whatsoever. And that's the way the word "axiom" is most often used in mathematics - not to mean "cannot be proved" but to give a starting place for a class of mathematical objects.

Then there's a set of order axioms; put that together with the field axioms and you have an ordered field. This is more restrictive, but there are a number of ordered fields, including $\mathbb{Q}$ (every ordered field must contain a copy of $\mathbb{Q}$) and some more exotic examples, including non-archimedean ordered fields.

Finally, there's the axiom of completeness. Most of the textbooks I know state it as "every nonempty set that is bounded above has a least upper bound." Start there, and you can prove as a theorem that "monotone bounded sequences convege." But it's also possible to start with "monotone bounded sequences converge" and get back to the least upper bound property as a theorem based on that. I can think of several other places to start.

Finally: can any of this be proved? That question turns out to be equivalent to asking if the real numbers can be constructed. They can. There are several possible constructions, the most popular being "Dedekind cuts." One fallout from the construction is that one can also prove this: given any two complete ordered fields, then there is an order-preserving isomorphism from one field onto the other. In effect, there is (up to isomorphism) only one complete orderd field. That is, there is only one object that satisfies that list of properties you called the axioms of $\mathbb{R}.$


geometry
trigonometry
complex numbers
geometry proposed

The well-known Newton's theorem. Posted many times before  :wink:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=19549[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=147923[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=208680[/url]

Let $ S \equiv{} EG \cap HF.$ $ X \equiv{} BC \cap AD$ is the pole of line $ HF$ WRT incircle $(I)$ and $ Y \equiv{} AB \cap DC$ is the pole of  $ EG$ WRT $(I)$ $\Longrightarrow$ $ XY$ is the polar of $S$ WRT  $(I).$ Thus, points $ W \equiv EF \cap HG$ and  $V \equiv{} EH \cap FG$ lie on $XY.$ $ W$ and $V$ are the poles of $DB$ and $AC$ WRT $(I).$ If $ X,Y,V,W$ are collinear, then $ EG,HF,AC,DB$ concur.

Actually the proof is quite simple: let's call P the intersection of AC and EG. See that AE = AH ( 1 ), CG = CF ( 2 ),
angle AEP + angle CGP = 180 degs ( 3 ). Applying the sine theorem to the triangles AEP and PCG, then to the triangles AHP and CFP we get that both GE and FH intersect AC at the same point. Similarly we can prove they are concurrent on BD.

Best regards,
sunken rock

[quote="Johan Gunardi"]Let $ ABCD$ be a quadrilateral which has an incircle. The incircle touches sides $ AB, BC, CD, DA$ at $ E,F,G,H$ respectively. Prove that $ AC, BD, EG, FH$ concur.[/quote]
Let $ a,b,c,d,e,f,g,h$ be the corresponding complex numbers to $ A,B,C,D,E,F,G,H$, and consider the complex plane with the same center as the incircle. (Assume wlog $ |e|\equal{}|f|\equal{}|g|\equal{}|h|\equal{}1$)
Then we easily get:
$ a \equal{} \frac{2he}{h\plus{}e}, b\equal{} \frac{2ef}{e\plus{}f}, c \equal{} \frac{2fg}{f\plus{}g}, d \equal{} \frac{2gh}{g\plus{}h}$.
$ AC,EG,FH$ are concurrent iff:
$ ((a\minus{}c) \minus{} (e\minus{}g))(\overline{(e\minus{}g)\minus{}(f\minus{}h)}) \equal{} (\overline{(a\minus{}c) \minus{} (e\minus{}g)})((e\minus{}g)\minus{}(f\minus{}h))$, which should be true :)
(I don't feel like doing it by hand right now, but notice that $ \overline a \equal{} \frac{2}{h\plus{}e}, \overline c \equal{} \frac{2}{f\plus{}g}$... It should be pretty straightforward.. )


[hide]we need to get y by itself and have x with the same coefficients so that they are parallel. x here is 2/4=5/c, 2c=20, so c=10[/hide]

[quote="13375P34K43V312"]The system of equations

$2x+4y=7$
$5x+cy=\pi^{\pi}\cdot67^{34}\sqrt{3}$

has no solutions. Find the value of $c$.[/quote]
[hide="a comment"]If you wanted this to be trickier you shouldn't have made it obvious that there is a smooth solution. YOu should have made the rhs of the second equation something like $c^{3}$ so that the person might hope that they can brute force it...[/hide]
Just a comment...
The same thing goes for the bonus in you last round of SD III, if you had said an average american male more people would have instantly turned to wikipedia and bam gotten the wrong answer. Just a comment...


Prove that there exist infinitely many $n\in N$ such that the largest prime divisor of $n^4+1$ is greater than $2n$.


$ a,b,c\geq 0$
$ \sum_{cyc} a\sqrt{a^2\plus{}3bc}\geq 2(ab\plus{}bc\plus{}ca)$

[quote="Bloody-Joker"]$ a,b,c\geq 0$
$ \sum_{cyc} a\sqrt {a^2 \plus{} 3bc}\geq 2(ab \plus{} bc \plus{} ca)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33515