Search found 53 matches

by Prosenjit Basak
Fri Sep 04, 2015 11:06 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2000/1
Replies: 6
Views: 2478

Re: IMO 2000/1

Yeah, I meant $EM$. And I meant $\triangle ABM$ too. It was just a typo.I don't know what happened to my typing. :( I edited it though. Thanks for pointing out the mistakes. :)
by Prosenjit Basak
Fri Sep 04, 2015 12:00 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2000/1
Replies: 6
Views: 2478

Re: IMO 2000/1

As $AB || CD$, we get $\angle EBA = \angle EDC$. Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$. So, $\angle EBA = \angle ABM$. Similarly we get $\angle EAB = \angle BAM$. So,we get $\triangle EBA \cong \triangle ABM$ by the ASA theorem. Then, we get $EM \perp AB.$ Now, let the segmen...
by Prosenjit Basak
Tue Mar 31, 2015 9:31 pm
Forum: Combinatorics
Topic: An interesting combinatorial identity
Replies: 3
Views: 1621

Re: An interesting combinatorial identity

Here's a Combinatorial argument: Consider a $m \times n$ chessboard. Now we want to place $2$ rooks such that no two rooks attack each other. In how many ways can we do that? We count it in two different ways. The first way is to choose the rows and columns. First we choose $2$ rows from the $m$ row...
by Prosenjit Basak
Sun Dec 15, 2013 10:15 am
Forum: News / Announcements
Topic: Combinatorics Workshop: Day 9 (15.12.13)
Replies: 2
Views: 2197

Re: Combinatorics Workshop: Day 9 (15.12.13)

যে এক্সামটা নেয়ার কথা ছিল সেইটা আর নিবা না? :?:
by Prosenjit Basak
Wed Dec 04, 2013 8:40 pm
Forum: News / Announcements
Topic: Combinatorics Workshop: Day 1 (04.12.13)
Replies: 7
Views: 6194

Re: Combinatorics Workshop: Day 1 (04.12.13)

আচ্ছা ভাইয়া, প্রোডাক্ট রুলের দ্বিতীয় উদাহরনের ক্ষেত্রে প্রথম ডিজিট বেছে নেওয়ার অপশন ১০ টি কেন হবে ? '0' নিলে তো তা দুই ডিজিটের হয়ে যায়। হ্যা, ভাইয়া এই প্রশ্ন টা আমারও । বলা হয়েছে তিন ডিজিটের সংখ্যা তো আমি যদি প্রথম স্থানে $0$ বসাই তো এইটা দুই ডিজিটের একটা সংখ্যা হয়ে যাবে। যেমনঃ একটা $012$ এ...
by Prosenjit Basak
Fri May 03, 2013 4:48 pm
Forum: Junior Level
Topic: Prove irrational
Replies: 1
Views: 1355

Re: Prove irrational

Let us assume that the given expression is rational. Then $\sqrt{\sqrt{37}+6}+\sqrt{\sqrt{37}-6} = \frac{a}{b}$ where $a,b$ are integer and $b$ is not equal to zero. We can substitute $\sqrt{\sqrt{37}+6} = m$ and $\sqrt{\sqrt{37}-6} = n$ Now $\sqrt{m} + \sqrt{n} = \frac{a}{b}$ $\Rightarrow (\sqrt{m}...
by Prosenjit Basak
Wed Mar 27, 2013 9:02 pm
Forum: Junior Level
Topic: Solve this PLEASE!
Replies: 5
Views: 2488

Re: Solve this PLEASE!

যেইদিন পোষ্ট করলাম সেইদিনই রাতে একখান বইয়ে দেখি আরে এইতো Cauchy - schwarz inequality. যদিও ঐখানে এত সহজ কইরা দেয় নাই
একটু কটমটে ইকুয়েশন দিছিল। যাই হোক পরে বুঝছি।
by Prosenjit Basak
Mon Mar 25, 2013 10:26 pm
Forum: Junior Level
Topic: Solve this PLEASE!
Replies: 5
Views: 2488

Re: Solve this PLEASE!

HM মানে হারমনিক মিন না? তাইলে ঠিক আছে । আর সানজিদ ভাই এর সলিউশনটা সহজ ছিল। আসলে আমি Cauchy-Schwarz's inequality-টা কি তা জানি না তাই প্রবলেম হইছে।
by Prosenjit Basak
Fri Mar 22, 2013 10:37 pm
Forum: Junior Level
Topic: Solve this PLEASE!
Replies: 5
Views: 2488

Solve this PLEASE!

If $a , b , c > 0$, prove that
$\displaystyle \frac{9}{a + b + c} \leq 2(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
by Prosenjit Basak
Tue Feb 26, 2013 10:09 pm
Forum: Algebra
Topic: INEQUALITY PROBLEM
Replies: 2
Views: 1247

Re: INEQUALITY PROBLEM

Hmm... I like the solution.It's quite easy.But can we solve it using AM-GM inequality? Actually I want that very solution.I gave this problem in a particular chapter regarding AM-GM. So I want that kind of solution. :)