Search found 461 matches

by sourav das
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...
by sourav das
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
by sourav das
Sun Sep 23, 2012 9:09 am
Forum: Algebra
Topic: Find functions R to R
Replies: 5
Views: 2274

Re: Find functions R to R

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(...
by sourav das
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:
Let $P(x,y)$ be the statement. Now set, i)$P(x,xy)$; then swap $x$ and $y$ and find that $\frac{f(x)}{x^n-\frac{1}{x^n}}$ constant; and $f(0)=0$
by sourav das
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
by sourav das
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...
by sourav das
Fri Sep 07, 2012 12:03 am
Forum: International Olympiad in Informatics (IOI)
Topic: IOI-2012 Bangladesh team
Replies: 7
Views: 4080

Re: IOI-2012 Bangladesh team

Congrats :D . Best of luck.
by sourav das
Wed Sep 05, 2012 7:58 pm
Forum: Geometry
Topic: Indian tst Day 3 Problem 1
Replies: 6
Views: 2793

Re: Indian tst Day 3 Problem 1

Yeah,sorry I should have noticed that. :oops: :oops: Here is what I have done yet,and I haven't completed. We have that $AD\times AC=AE\times AG$ and $\angle A$ common. So, $\triangle ACE\sim \triangle AGD$. (This could also be done by angle chasing.) Now by symmetry $\triangle ABE\cong \triangle A...
by sourav das
Wed Sep 05, 2012 7:45 pm
Forum: Geometry
Topic: Indian tst Day 3 Problem 1
Replies: 6
Views: 2793

Re: Indian tst Day 3 Problem 1

Let $M$ be the midpoint of $BC$. Fact:1) $AD=DF=\frac{1}{2}AC$ Proof: Using cyclic property $\angle DFA=\angle MEB =\frac{1}{2}\angle BEC=\frac{1}{2}\angle BDC$ ; It implies, $\angle DFA=\angle DAF$;So, $AD=DF$ Fact:2) $BI$ bisects $\angle ABK$ Proof: By symmetry $\angle EBA=\angle ECA = \angle EBD$...
by sourav das
Wed Sep 05, 2012 2:27 pm
Forum: Junior Level
Topic: Fourth Power
Replies: 3
Views: 2153

Re: Fourth Power

Quite interesting one. Solution: $ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=14$ $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ Implies, $abc=\frac{41}{3}$ Now comes the bingo: $(a^4+b^4+c^4)-(a+b+c)(a^3+b^3+c^3)+(ab+bc+ca)(a^2+b^2+c^2)$ $-abc(a+b+c)=0$ And which implies that: $a^4+b^4+c^4=0$ Plea...