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Mon Feb 15, 2016 12:04 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO national higher secondery -prb 07
Replies: 2
Views: 2500

Re: BDMO national higher secondery -prb 07

You can solve this using Mass point geometry as well. Ideas about mass point geometry required for this problem can be found here:
Sun Feb 07, 2016 8:03 pm
Forum: Introductions
Topic: My Intooduction
Replies: 1
Views: 3993

Re: My Intooduction

Welcome to the forum! You can post problems and start discussions on topics of your interest, and take part in previous conversations!

Best of luck!
Sun Aug 23, 2015 11:55 pm
Forum: National Math Camp
Topic: ONT Camp Day 1
Replies: 12
Views: 8997

Re: ONT Camp Day 1

I think it would better be 9 pm. That way none will miss it.
Sun Dec 28, 2014 8:47 pm
Forum: Combinatorics
Topic: Sequence count
Replies: 2
Views: 2117

Sequence count

Find the number of possible sequences $(a_1,a_2,...,a_n)$ in terms of $n$, with the following conditions:
(ii)$a_1\leq a_2\leq ...\leq a_n$
(iii)$a_i\leq (i-2)$
Sun Nov 16, 2014 11:34 am
Forum: Geometry
Topic: collinearity from russia
Replies: 2
Views: 2034

Re: collinearity from russia

In my solution, I have replaced $W$ with $Y$, and $V$ with $Z$. So we need to prove that $Y,X,Z$ are collinear. Let $BX\cap ZY=P$. Now it is clear that $\dfrac{BZ}{ZC}=\dfrac{BA\cdot \sin\angle BAZ}{CA\cdot \sin\angle CAZ}=\dfrac{BA\cdot \sin\angle C}{CA\cdot \sin\angle B}$. Thus by using sine law i...
Thu Nov 13, 2014 2:38 pm
Forum: Number Theory
Topic: Perfect Square ratio
Replies: 5
Views: 3027

Re: Perfect Square ratio

To avoid the confusion, we may write that $x^{2}+2y^{2}\leq |y^{2}-x^{2}|$. Square it, simplify it, then we can get $3x^{2}(x^{2}+2y^{2})\leq 0$.
Thu Nov 06, 2014 10:14 pm
Forum: Geometry
Topic: Show it parallel
Replies: 1
Views: 1665

Re: Show it parallel

In $\triangle ABC$, $D\in AB, E\in AC, DE\parallel BC\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{EB}$.
Sun Oct 26, 2014 10:17 pm
Forum: Secondary Level
Topic: A circle through incenter
Replies: 2
Views: 2161

Re: A circle through incenter

Clearly $EF\perp AI\Rightarrow EF\perp IM$. Let $EF\cap \omega=G$. Then $\angle IMG=90^{\circ}$. So, $IG$ is a diameter of $\omega$ i.e. $IO\cap EF\in \omega$.
Wed Oct 15, 2014 12:12 pm
Forum: Number Theory
Topic: Arithmetic series in Fibonacci
Replies: 5
Views: 3128

Re: Arithmetic series in Fibonacci

But subsequence does not necessarily need to be of consecutive terms of the original sequence, does it? :? Clearly $F_{m}-F_{n}>F_{n+2}-F_{n}>F_{n}-1>F_{n}-F_{k}$ for all $m\geq (n+2), k\leq (n-1)$. This means $F_{n+1}$ is the only possible term which can be in an arithmetic sequence with $F_{n}$ a...
Fri Oct 03, 2014 10:57 pm
Forum: Number Theory
Topic: A Fact
Replies: 1
Views: 1573

Re: A Fact

Hint 1:
$a\equiv b(mod p^{k})\Rightarrow a^{p^{s}}\equiv b^{p^{s}}(mod p^{k+s})$. This should be proved first.
Hint 2:
$\phi(p^{n})=p^{n-1}\cdot (p-1)$