Search found 98 matches
- Mon Dec 05, 2016 5:55 pm
- Forum: Junior Level
- Topic: Calculate area of a triangle.
- Replies: 1
- Views: 2701
Re: Calculate area of a triangle.
The answer is $216$.
- Mon Dec 05, 2016 4:02 pm
- Forum: Junior Level
- Topic: Need help to solve number theory problems
- Replies: 5
- Views: 4573
Re: Need help to solve number theory problems
In problem (1) take mod $5$. And remember that $y^2 \equiv 0,1,4$ $(mod $ $ 5)$.
- Sun Oct 30, 2016 7:55 pm
- Forum: Combinatorics
- Topic: n-series
- Replies: 1
- Views: 5544
n-series
Given any integer $n\geq 3$. A finite series is called $n$-series if it satisfies the following two conditions $1)$ It has at least $3$ terms and each term of it belongs to $\{ 1,2,...,n\}$ $2)$ If series has $m$ terms $a_1,a_2,...,a_m$ then $(a_{k+1}-a_k)(a_{k+2}-a_k)<0$ for all $k=1,2,...,m-2$ How...
- Mon Oct 10, 2016 9:06 pm
- Forum: Geometry
- Topic: Iran MO 1997
- Replies: 2
- Views: 3874
Iran MO 1997
In triangle $ABC$, angles $B,C$ are acute. Point $D$ is on the side $BC$ such that $AD\perp{BC}$. Let the interior bisectors of $\angle B,\angle C$ meet $AD$ at $E,F$, respectively. If $BE=CF$, prove that $ABC$ is isosceles.
- Thu Aug 04, 2016 11:47 pm
- Forum: Geometry
- Topic: Right Angles on Incircle Chord
- Replies: 2
- Views: 3126
Right Angles on Incircle Chord
The incircle of $\triangle ABC$ with incentre $I$ touches the sides $BC,CA$ and $AB$ at $D,E$ and $F$ respectively. Now, let $K=BI\cap EF$. Show that $BK\perp CK$.
- Thu Aug 04, 2016 8:08 pm
- Forum: Social Lounge
- Topic: Chat thread
- Replies: 53
- Views: 79584
Re: Chat thread
I was encouraged in a funny way. On 2012/2013, when my brother was going to the Olympiad, I also went there(because I was too small to stay at home!!!), I did not find anything interesting. (and yes,I completely failed). Next year I went there again, and I became 2nd runner-up. Then I found it inter...
- Wed Aug 03, 2016 7:14 pm
- Forum: Social Lounge
- Topic: Chat thread
- Replies: 53
- Views: 79584
Re: Chat thread
I am Zahin, from Dhaka. I wish my little contribution will help you to make this forum active.
- Wed Aug 03, 2016 7:06 pm
- Forum: Geometry
- Topic: Russian Olympiad 1996
- Replies: 2
- Views: 3389
Russian Olympiad 1996
Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $\angle BAE =\angle CDF$ and $\angle EAF =\angle FDE$. Prove that $\angle FAC =\angle EDB$.