## Search found 461 matches

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.
Wed Jan 16, 2013 5:13 pm
Topic: IMO Marathon
Replies: 184
Views: 64073

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 $KL... Tue Jan 15, 2013 7:27 pm Forum: Algebra Topic: A Square Expression Replies: 4 Views: 2041 ### Re: A Square Expression I just wanted to solve viewtopic.php?f=27&t=2568&p=12849#p12849 and wanted to represent the smallest number($c$) with something else. First$p+2y$and then$p+2gy$with co-prime condition. Tue Jan 15, 2013 7:21 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO Marathon Replies: 184 Views: 64073 ### Re: IMO Marathon Sorry for being late to attend the marathon. For problem 14 Solution: Take prime$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... 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 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। 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$(xy,x-y)=1$... 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...
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...
Tue Jan 01, 2013 11:50 pm
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 