Search found 176 matches
- Mon Nov 12, 2012 2:20 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114162
Re: IMO Marathon
Tahmid, I proved your hint but still can't use it. Give a link please.
- Sun Nov 11, 2012 12:15 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114162
Re: IMO Marathon
At first I'm very sorry for not posting till now. I've been checking and trying to solve the problems but as I'm using a CLASSIC Nokia 6101, (if you know what I mean) I can't post very comfortably. I have a solution for problem 3, but as I dnt have latex, it'd be hard to read for you, if anyone want...
- Fri Nov 09, 2012 1:54 pm
- Forum: News / Announcements
- Topic: Active users for marathon
- Replies: 23
- Views: 17467
Re: Active users for marathon
Though I can't burn my "kheta" of Test exam, but I can do it I guess.
- Thu Aug 09, 2012 10:47 pm
- Forum: Geometry
- Topic: USA TST 2012 Day 3 Problem 1
- Replies: 2
- Views: 2513
USA TST 2012 Day 3 Problem 1
Let $O$ be the circumcircle of $ABC$. Let the angle bisector of angle $A$ intersect $BC$ & $O$ at $D$ & $L$ respectively. $M$ is the midpoint of BC. The circumcircle of $ADM$ intersect $AB$ and $AC$ at $Q$ and $P$ respectively. The midpoint of $PQ$ is $N$. $LH$ is perpendicular on $ND$ and $H$ lies ...
Re: F.E. (2)
My solution outlines: Prove f is bijective. Then prove, f(0)=0 and f(f(x))=x. Then, after some substitutions, we can show, f(x)^2=x^2. Now, we can check and show that either f(x)=x for all x or f(x)=-x for all x. For that, just suppose f(x)=x for some x and f(y)=-y for some y. Then do your calculati...
- Wed Jun 06, 2012 9:39 pm
- Forum: Geometry
- Topic: Balkan-2012-1
- Replies: 1
- Views: 1689
Re: Balkan-2012-1
Hint:
- Tue Jun 05, 2012 7:24 pm
- Forum: Algebra
- Topic: Functional equation
- Replies: 11
- Views: 6408
Re: Functional equation
Umm, I found out f(0)=0.
- Tue Jun 05, 2012 6:44 pm
- Forum: Algebra
- Topic: Functional equation
- Replies: 11
- Views: 6408
Re: Functional equation
My solution is quite as same as Adib. But I proved f(x)=cx in an uglier way than him. I put y=-f(x)/x to get it.
- Thu May 31, 2012 8:55 am
- Forum: Geometry
- Topic: USA TST 2011
- Replies: 9
- Views: 5537
Re: USA TST 2011
Obviously you can't give links if there isn't any proof.
- Wed May 30, 2012 6:22 pm
- Forum: Number Theory
- Topic: Serbia TST 2012 Problem 2
- Replies: 3
- Views: 2324
Serbia TST 2012 Problem 2
Let $g(x)$ denote the sum of divisors of natural number $x$, including $1$ and $x$. For every $n$ belongs to $N$ define $f(n)$ as number of natural numbers $m$, such that $m$ is less than or equal to $n$, for which $g(m)$ is odd number. Prove that there are infinitely many natural numbers $n$, such ...