## Search found 461 matches

Mon Nov 05, 2012 1:45 pm
Forum: News / Announcements
Topic: Active users for marathon
Replies: 23
Views: 9989

### Active users for marathon

I'm Interested to start or restart problem solving marathon. But I'm not quite sure how many of you can participate. So those who want to participate and ready to visit those topics at least one time everyday please reply. I'll be grateful to you if you can take the responsibility of the publicity o...
Sat Oct 13, 2012 1:04 am
Forum: Higher Secondary Level
Topic: IMO Mocks
Replies: 4
Views: 2526

### Re: IMO Mocks

Mock-5.Rock-n-roll
Sun Sep 23, 2012 9:09 am
Forum: Algebra
Topic: Find functions R to R
Replies: 5
Views: 2274

All the substitutions are in the main equation. :arrow: Set $x=y=0$. Then $f(f(0))=f(0)$. :arrow: Set $y=0$. Then $f(f(x))=xf(0)+f(x)$. Now if $f(0)\ne 0$, then $f(x)$ is injective, which would imply $f(0)=0$. This contradiction implies that $f(0)=0,f(f(x))=f(x)$. :arrow: Set $x=f(x),y=x$. Then $f(... Wed Sep 19, 2012 2:30 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO Mock-1 Problem 1 Replies: 2 Views: 2071 ### Re: IMO Mock-1 Problem 1 My solution steps: Mon Sep 17, 2012 3:11 pm Forum: Higher Secondary Level Topic: IMO Mocks Replies: 4 Views: 2526 ### IMO Mocks I'll post some mocks here. You can discuss them creating other threads, but not here. First Mock from me: Mock-1.pdf Mock-1 (26 KiB) Downloaded 245 times Sat Sep 08, 2012 10:46 pm Forum: Geometry Topic: Circumcenter Replies: 4 Views: 2095 ### Re: Rectangle I think$ABCD$is parallelogram, not rectangle. Proof:$\angle ABD=\angle BDL=\angle BAL$, and since$\angle ABN = 90; \angle ADB=\angle ANB$so,$\angle BAN = 90-\angle ADB$It implies,$\angle LAN=\angle LAB- \angle NAB =\angle ABD + \angle ADB -90 =90-\angle BAD$Similarly we can show that,$\ang...
Fri Sep 07, 2012 12:03 am
Forum: International Olympiad in Informatics (IOI)
Replies: 7
Views: 4080