1. ahmedittihad
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.
Search found 176 matches
- Tue Apr 18, 2017 12:22 am
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies: 30
- Views: 47202
Re: A Lemma?
Thanks I edited that.Mallika Prova wrote: surely it wants to be $MN||AO$ ??
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...
Re: A Lemma?
This note might be relevant.
- Mon Apr 17, 2017 1:56 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2016 Problem 1
- Replies: 3
- Views: 8564
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: 4954
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: 8262
Re: geomerty
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: 5493
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.
$$[f(x)+f(y)][f(z)+f(t)]=f(xz+yt)+f(xt-yz)$$
holds.
- Sun Apr 02, 2017 9:50 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2015 - Problem 5
- Replies: 1
- Views: 6751
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: 9309
Re: Math
Its not bringing the bottle the reason you flunked APMO?ahmedittihad wrote: Oh and take the bottle to the exams, I believe that it's a good luck charm.