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BdMO Online Forum • Search

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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:

https://en.wikipedia.org/wiki/Mass_point_geometry
by SANZEED
Mon Feb 15, 2016 12:04 pm
 
Forum: National Math Olympiad (BdMO)
Topic: BDMO national higher secondery -prb 07
Replies: 2
Views: 963

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!
by SANZEED
Sun Feb 07, 2016 8:03 pm
 
Forum: Introductions
Topic: My Intooduction
Replies: 1
Views: 889

Re: ONT Camp Day 1

I think it would better be 9 pm. That way none will miss it.
by SANZEED
Sun Aug 23, 2015 11:55 pm
 
Forum: National Math Camp
Topic: ONT Camp Day 1
Replies: 12
Views: 1697

Sequence count

Find the number of possible sequences $(a_1,a_2,...,a_n)$ in terms of $n$, with the following conditions:
(i)$a_1=a_2=0$
(ii)$a_1\leq a_2\leq ...\leq a_n$
(iii)$a_i\leq (i-2)$
by SANZEED
Sun Dec 28, 2014 8:47 pm
 
Forum: Combinatorics
Topic: Sequence count
Replies: 2
Views: 453

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...
by SANZEED
Sun Nov 16, 2014 11:34 am
 
Forum: Geometry
Topic: collinearity from russia
Replies: 2
Views: 590

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 $3x^{2}(x^{2}+2y^{2})\leq 0$.
by SANZEED
Thu Nov 13, 2014 2:38 pm
 
Forum: Number Theory
Topic: Perfect Square ratio
Replies: 5
Views: 897

Re: Show it parallel

Hint:
In $\triangle ABC$, $D\in AB, E\in AC, DE\parallel BC\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{EB}$.
by SANZEED
Thu Nov 06, 2014 10:14 pm
 
Forum: Geometry
Topic: Show it parallel
Replies: 1
Views: 570

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. $IO\cap EF\in \omega$.
by SANZEED
Sun Oct 26, 2014 10:17 pm
 
Forum: Secondary Level
Topic: A circle through incenter
Replies: 2
Views: 610

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...
by SANZEED
Wed Oct 15, 2014 12:12 pm
 
Forum: Number Theory
Topic: Arithmetic series in Fibonacci
Replies: 5
Views: 845

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:
$\phi(p^{n})=p^{n-1}\cdot (p-1)$
by SANZEED
Fri Oct 03, 2014 10:57 pm
 
Forum: Number Theory
Topic: A Fact
Replies: 1
Views: 506
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