## 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**

### 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(...

- 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:

- 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)
- Topic: IOI-2012 Bangladesh team
- Replies:
**7** - Views:
**4080**

### Re: IOI-2012 Bangladesh team

Congrats . Best of luck.

- 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...

- 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 $\angle EBA=\angle ECA = \angle EBD$...

- 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...