Search found 244 matches
- Thu Jan 31, 2013 12:45 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 299139
Re: Secondary and Higher Secondary Marathon
$n^2-n=n(n-1) \equiv 0 (mod10^5)$
- Wed Jan 30, 2013 11:35 pm
- Forum: Social Lounge
- Topic: আকাশ দেখবে ? (must read)
- Replies: 0
- Views: 2307
আকাশ দেখবে ? (must read)
ছোটবেলা থেকেই টেলিস্কোপ দিয়ে একবার আকাশের দিকে তাকানোর খুব শখ ছিল। কয়েক বছর আগে গণিত অলিম্পিয়াডে মাত্র কয়েক সেকেন্ডের জন্য শুক্র গ্রহ দেখতে পেরেছিলাম। তাই আমার ইচ্ছাটা অপূর্ণই থেকে গিয়েছিল। যারা আমার মত এরকম আকাশ দেখার ব্যাপারে আগ্রহী কিন্তু সুযোগ পাচ্ছেন না তাদের জন্য দারুন একটা খবর আছে। আগামি...
- Wed Jan 30, 2013 10:13 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
- Wed Jan 30, 2013 6:48 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
It will be great if you share your solution .*Mahi* wrote: I found a proof of problem $26$ with inversion and some ugly stuff
- Wed Jan 30, 2013 12:54 pm
- Forum: Secondary Level
- Topic: Little Trigonometry
- Replies: 2
- Views: 2492
Re: Little Trigonometry
$\displaystyle \cos 3.\cos 6.\cos12 ... \cos 384$ $\displaystyle =\frac{2^7.\sin3.\cos 3.\cos 6.\cos12 ... \cos 384}{2^7.\sin3}=\frac{\sin 768}{2^7.\sin3}=\frac{\sin 48}{2^7.\sin3}$ $\displaystyle =\frac{\sin3+\cos3}{2^7 \sqrt {2}.\sin 3}=\frac{1}{2^7 \sqrt {2}}+\frac{\cot 3}{2^7 \sqrt {2}}$
- Tue Jan 29, 2013 6:59 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
My বিশ্রী solution: I don't see why this is ugly . Is it for dirichlet ? :P problem $\boxed {26}$ Let $I$ be the incenter of acute $\triangle ABC$. Let $\omega$ be a circle with center $I$ that lies inside $\triangle ABC$ . $D, E, F$ are the intersection points of circle $\omega$ with the perpendic...
- Mon Jan 28, 2013 11:42 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
Problem $\boxed {25}$
Let $m$ be a positive odd integer. Prove that there exist infinitely many positive integer $n$ such that $\displaystyle \frac{2^n−1}{mn+1}$ is an integer.
source : Mongolia TST 2011
Let $m$ be a positive odd integer. Prove that there exist infinitely many positive integer $n$ such that $\displaystyle \frac{2^n−1}{mn+1}$ is an integer.
source : Mongolia TST 2011
- Mon Jan 28, 2013 10:47 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
Iran 10-5.PNG Solution $\boxed {24}$ Let $(ABC)\cap W_1=D, DC \cap W_1=E$ Observe some facts (Can be proved easily ) 1.$XY||BE$ 2.$XE||AC$ $\angle KEX=\angle KPC=\angle KYC$ imply $Y,K,E$ are colinear . Let $QY \cap BE =G$ . Now $\angle QGB =\angle QYX=180- \angle QAB$ imply $G$ lie on $(ABC)$ . So...
- Thu Jan 24, 2013 12:15 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
Another proof for $\boxed {22}$
Edit [ solution removed due to some bug . ( I dunno why i did consider CL ,CK as bisector ) ]
I'll fix it ,when my laptop be fixed
Edit [ solution removed due to some bug . ( I dunno why i did consider CL ,CK as bisector ) ]
I'll fix it ,when my laptop be fixed
- Tue Jan 22, 2013 8:50 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 113832
Re: IMO Marathon
$\boxed {19}$ My Solution . (I did post this because it seemed interesting to me) Let $l',l_p,l_q$ be the line that is tangent to $\omega$ at $B,P,Q$ respectively. Now $l' \cap PQ=I , l_p \cap BQ=G, l_q \cap BP=H, l_p \cap l_q=J$ , Its well known that $G,H,I$ are colinear. Let $BJ \cap IG=K,BJ \cap ...