Search found 461 matches

by sourav das
Wed Jan 16, 2013 5:36 pm
Forum: Social Lounge
Topic: Any Way
Replies: 11
Views: 5979

Re: Any Way

nafistiham wrote:A non residential camp for SSC examinees was held in 2011.
I don't know when it will be held again.
That was my first camp. I wonder what could have happened if the camp wasn't held that year. I strongly in favor of arranging extra facilities for $S.S.C.$ examines.
by sourav das
Wed Jan 16, 2013 5:13 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 64073

Re: IMO Marathon

Part (a) $K,L$ are the reflection of ($E$ on $BI$), ($F$ on $CI$) respectively. But $D$ is the reflection of ($E$ on $CI$) and ($F$ on $BI$). So, $\angle EDF = \angle KFD= \angle KLD$ and also $\angle FDE= \angle LED= \angle LKD$ which implies $DL=DK$ and so $DI$ is the perpendicular bisector of $KL...
by sourav das
Tue Jan 15, 2013 7:27 pm
Forum: Algebra
Topic: A Square Expression
Replies: 4
Views: 2041

Re: A Square Expression

I just wanted to solve viewtopic.php?f=27&t=2568&p=12849#p12849
and wanted to represent the smallest number( $c$ ) with something else. First $p+2y$ and then $p+2gy$ with co-prime condition.
by sourav das
Tue Jan 15, 2013 7:21 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 64073

Re: IMO Marathon

Sorry for being late to attend the marathon. For problem 14 Solution: Take prime $p$ such that $p$ is co-prime to both $a,b$. Then take $n=p-1$ and use Fermat's little theorem to prove $p|a-b$. But there are infinite $p$ primes co-prime to $a-b$. implies $a=b$. P.S.:: My solution is identical to off...
by sourav das
Tue Jan 15, 2013 7:20 pm
Forum: Algebra
Topic: A Take on Pythagorean-Triples
Replies: 2
Views: 1502

Re: A Take on Pythagorean-Triples

$2c^2-a^2=b^2$ (assuming $a>b$) and now substitute $a=x+y$ and $c=x-y$ and our problem transform into,
viewtopic.php?f=27&p=12845&sid=f2621629 ... 17d#p12845
by sourav das
Tue Jan 15, 2013 12:20 am
Forum: Social Lounge
Topic: একটি জিজ্ঞাসা এবং একটি অনুরধ
Replies: 4
Views: 3220

একটি জিজ্ঞাসা এবং একটি অনুরধ

আমারা কি এই ফোরাম এ গণিত/ বিজ্ঞান বিষয়ক বিষয় গুলা ছাড়া অন্য কোন বিষয় আলোচনার ব্যাবহার করতে পারব (জ্যেষ্ঠ মডারেটরদের কাছে প্রশ্ন)? আমি চাই এই ফোরাম এ "দর্শন" বিষয় এর জন্য আলাদা একটা সাব-ফোরাম খোলা হোক। "দর্শন" এর অনেক পাঙ্খা পাঙ্খা বিষয় আছে যা অসম্ভব দারুন।
by sourav das
Tue Jan 15, 2013 12:05 am
Forum: Algebra
Topic: A Square Expression
Replies: 4
Views: 2041

Re: A Square Expression

W.L.O.G. let $a>b$. Let $a^2-6ab+b^2=2(a-b)^2-(a+b)^2=c^2$. Now, substitute $a-b=p$...(i), $a+b=p+2gy$....(ii), $c=p-2gx$...(iii) with $x,y$ co-prime. Then, you will find out that $a-b=p=g(x-y)+ \frac{2gxy}{x-y}$. Using (i),(ii) we will get $a=gx+\frac{2gxy}{x-y}$ and $b=gy$ . But since $(xy,x-y)=1$...
by sourav das
Mon Jan 14, 2013 8:33 pm
Forum: Geometry
Topic: China National-2013-1
Replies: 7
Views: 3287

Re: China National-2013-1

Solution: Let $K,L$ be midpoints of $BE$ and $BF$. Let $P'$ be the center of circle $BEF$. Now $CA.CD=CB.CF=CL^2-LF^2=CP'^2-P'F^2$ and $DA.DC=DE.DB=DK^2-BK^2=DP'^2-BP'^2$ using power of point. But as $CA=AD$ and $P'F=P'B$; so $CP'=P'D$ and so $P=P'$. Also, since $2CA^2=CA.CD=CL^2-LF^2=CP'^2-P'F^2=AP...
by sourav das
Mon Jan 14, 2013 8:13 pm
Forum: Geometry
Topic: SD=SM
Replies: 5
Views: 2312

Re: SD=SM

Let $P'$ and $Q'$ be the midpoints of $AP$ and $AQ$. Let $AM$ meet $P'Q'$ at $M'$. $AS$ meets circle $APQ$ at $Y$. $YP=YQ$ ,$BP=CQ$ [use sine law], $\angle P=\angle Q =90$. So, $YB=YC$ and thus $YM||AD$. But a homothety with ratio 2 sends $\triangle SP'Q'$ to $\triangle YPQ$. And so, $SM'||YM||AD$ a...
by sourav das
Tue Jan 01, 2013 11:50 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National 2012: Primary 1
Replies: 8
Views: 4955

Re: BdMO National 2012: Primary 1

Actually there are twenty of them. 111,123, 135, 147, 210, 222, 234, 246, 321, 333, 345, 420, 432, 444, 531, 543, 630, 642, 741, 840. Now prove it without calculating ;)