Search found 244 matches

by Nadim Ul Abrar
Thu Jan 31, 2013 12:45 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57095

Re: Secondary and Higher Secondary Marathon

$n^2-n=n(n-1) \equiv 0 (mod10^5)$
by Nadim Ul Abrar
Wed Jan 30, 2013 11:35 pm
Forum: Social Lounge
Topic: আকাশ দেখবে ? (must read)
Replies: 0
Views: 1433

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...
by Nadim Ul Abrar
Wed Jan 30, 2013 10:13 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

Re: IMO Marathon

You may see this .@ Mahi
http://www.math.ust.hk/excalibur/v17_n3.pdf
(page 4)
by Nadim Ul Abrar
Wed Jan 30, 2013 6:48 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

Re: IMO Marathon

*Mahi* wrote: I found a proof of problem $26$ with inversion and some ugly stuff
It will be great if you share your solution .
by Nadim Ul Abrar
Wed Jan 30, 2013 12:54 pm
Forum: Secondary Level
Topic: Little Trigonometry
Replies: 2
Views: 1406

Re: Little Trigonometry

$\displaystyle \cos 3.\cos 6.\cos12 ... \cos 384$ $\displaystyle =\frac{2^7.\sin3.\cos 3.\cos 6.\cos12 ... \cos 384}{2^7.\sin3}=\frac{\sin 768}{2^7.\sin3}=\frac{\sin 48}{2^7.\sin3}$ $\displaystyle =\frac{\sin3+\cos3}{2^7 \sqrt {2}.\sin 3}=\frac{1}{2^7 \sqrt {2}}+\frac{\cot 3}{2^7 \sqrt {2}}$
by Nadim Ul Abrar
Tue Jan 29, 2013 6:59 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

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...
by Nadim Ul Abrar
Mon Jan 28, 2013 11:42 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

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
by Nadim Ul Abrar
Mon Jan 28, 2013 10:47 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

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...
by Nadim Ul Abrar
Thu Jan 24, 2013 12:15 am
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

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
by Nadim Ul Abrar
Tue Jan 22, 2013 8:50 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 62188

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 ...