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

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

https://en.wikipedia.org/wiki/Mass_point_geometry

- Mon Feb 15, 2016 12:04 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national higher secondery -prb 07
- Replies:
**2** - Views:
**872**

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!

Best of luck!

- Sun Feb 07, 2016 8:03 pm
- Forum: Introductions
- Topic: My Intooduction
- Replies:
**1** - Views:
**800**

I think it would better be 9 pm. That way none will miss it.

- Sun Aug 23, 2015 11:55 pm
- Forum: National Math Camp
- Topic: ONT Camp Day 1
- Replies:
**12** - Views:
**1401**

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)$a_i\leq (i-2)$

(i)$a_1=a_2=0$

(ii)$a_1\leq a_2\leq ...\leq a_n$

(iii)$a_i\leq (i-2)$

- Sun Dec 28, 2014 8:47 pm
- Forum: Combinatorics
- Topic: Sequence count
- Replies:
**2** - Views:
**393**

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...

- Sun Nov 16, 2014 11:34 am
- Forum: Geometry
- Topic: collinearity from russia
- Replies:
**2** - Views:
**527**

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 13, 2014 2:38 pm
- Forum: Number Theory
- Topic: Perfect Square ratio
- Replies:
**5** - Views:
**815**

Hint:

- Thu Nov 06, 2014 10:14 pm
- Forum: Geometry
- Topic: Show it parallel
- Replies:
**1** - Views:
**535**

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. $IO\cap EF\in \omega$.

- Sun Oct 26, 2014 10:17 pm
- Forum: Secondary Level
- Topic: A circle through incenter
- Replies:
**2** - Views:
**565**

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...

- Wed Oct 15, 2014 12:12 pm
- Forum: Number Theory
- Topic: Arithmetic series in Fibonacci
- Replies:
**5** - Views:
**743**

Hint 1:

Hint 2:

Hint 2:

- Fri Oct 03, 2014 10:57 pm
- Forum: Number Theory
- Topic: A Fact
- Replies:
**1** - Views:
**464**