## Search found 176 matches

Tue Apr 18, 2017 12:22 am
Forum: Social Lounge
Topic: BDMO Forum Mafia #2
Replies: 30
Views: 24629

### Re: BDMO Forum Mafia #2

2. Raiyan Jamil
3. dshasan
4. Epshita32
5. Ananya Promi
6. TashkiManda
7. nahinmunkar
8. Thamim Zahin
9. Thanic Samin

This setup is something I can't but join.
Mon Apr 17, 2017 8:33 pm
Forum: Geometry
Topic: A Lemma?
Replies: 8
Views: 2479

### Re: A Lemma?

Mallika Prova wrote: surely it wants to be $MN||AO$ ??
Thanks I edited that.
Mon Apr 17, 2017 12:40 pm
Forum: Geometry
Topic: A Lemma?
Replies: 8
Views: 2479

### Re: A Lemma?

Also, an elegant proof for part 2 is presented below. Take the homothety with center $A$ and ratio $\dfrac{1}{2}$. This sends $(HBC)$ to the nine point circle. Let the center of the nine point circle be $N$, and let the midpoint of $AH$ be $M$. Now, since $N$ is the midpoint of $OH$, $MN||AO$[typo e...
Mon Apr 17, 2017 12:27 pm
Forum: Geometry
Topic: A Lemma?
Replies: 8
Views: 2479

### Re: A Lemma?

This note might be relevant.
Mon Apr 17, 2017 1:56 am
Topic: IMO 2016 Problem 1
Replies: 3
Views: 5996

### Re: IMO 2016 Problem 1

Let $\angle FAB=x$. Since $\triangle AFB \sim \triangle ADC$, $\triangle AFD \sim \triangle ABC$ [sprial similarity] Thus $\angle AFD = 90^{\circ}+x$, and so $FD\perp AB$. Since $FB=FA$, $FD$ is the perpendicular biscector of $AB$. Now, $\angle DBF = x$. Since $\angle AED=180^{\circ}-2x$, $ABDE$ is ...
Sat Apr 08, 2017 3:30 pm
Forum: Algebra
Topic: Find largest pos int n with special condition
Replies: 1
Views: 3487

### Re: Find largest pos int n with special condition

Let $a_i=\tan ^{-1} x_i$. Now, $(3x_i-x_j)(x_i-3x_j)\ge (1-x_ix_j)^2\Rightarrow \dfrac{x_i-x_j}{1+x_ix_j}\ge \dfrac{1}{\sqrt{3}}$ when $x_i> x_j$. But this implies $\tan(a_i-a_j)\ge \dfrac{1}{\sqrt{3}}$ which means $a_i-a_j\ge 30^{\circ}$, whereas $0^{\circ}< a_k< 90^{\circ}$. If we take $n\ge 4$, t...
Sat Apr 08, 2017 3:07 pm
Forum: Junior Level
Topic: geomerty
Replies: 10
Views: 3417

Here is an easy solution: Take $K$ on $BC$ so that $CD=CK$. Now, since $\triangle CDK$ is isosceles and $\angle DCK=20^{\circ}$, $\angle DKC=80^{\circ}$ implying $ADKC$ is cyclic. Now, $\angle BDK=\angle BCA=\angle ABC$, so $BK=DK$. Again, $CD$ is the angle biscector, so $AD=DK$. So $AD+CD=BK+KC=BC=... Mon Apr 03, 2017 8:51 pm Forum: Algebra Topic: FE: Brahmagupta-Fibonacci identity!!! Replies: 1 Views: 3745 ### FE: Brahmagupta-Fibonacci identity!!! Find all functions from reals to reals so that $$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$ holds. Sun Apr 02, 2017 9:50 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2015 - Problem 5 Replies: 1 Views: 5179 ### Re: IMO 2015 - Problem 5 The solutions are$2-x$and$x$. They satisfy the FE. Let$P(x,y)$denote the FE. $$P(0,0)\Rightarrow f(f(0))=0$$ $$P(0,f(0))\Rightarrow f(0)=0,2$$. Case$1$:$f(0)=2$$$P(0,x)\Rightarrow f(f(x))-f(x)=2(x-1)$$ Which implies$f(x)=x$is only possible when$x=1$.$\$P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+...
Fri Mar 31, 2017 9:58 pm
Forum: Social Lounge
Topic: Math
Replies: 7
Views: 4769

### Re: Math

ahmedittihad wrote: Oh and take the bottle to the exams, I believe that it's a good luck charm.
Its not bringing the bottle the reason you flunked APMO?