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
Search found 550 matches
- Mon Feb 15, 2016 12:04 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO national higher secondery -prb 07
- Replies: 2
- Views: 4686
- Sun Feb 07, 2016 8:03 pm
- Forum: Introductions
- Topic: My Intooduction
- Replies: 1
- Views: 9877
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!
Best of luck!
- Sun Aug 23, 2015 11:55 pm
- Forum: National Math Camp
- Topic: ONT Camp Day 1
- Replies: 12
- Views: 15988
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: 4370
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)$a_i\leq (i-2)$
(i)$a_1=a_2=0$
(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: 4157
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: 5924
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: 3473
Re: Show it parallel
Hint:
- Sun Oct 26, 2014 10:17 pm
- Forum: Secondary Level
- Topic: A circle through incenter
- Replies: 2
- Views: 4643
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: 6076
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: 3270
Re: A Fact
Hint 1:
Hint 2: