Search found 3 matches
- Tue Sep 12, 2017 7:41 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2017 P2
- Replies: 2
- Views: 8542
Re: APMO 2017 P2
Let $M$ and $N$ be the midpoints of $AB$ and $AC$.$MN$ meets $DC$ at $X$.Then $\angle AMX=\angle ABC=\angle ADX$.So $A$,$M$,$D$,$X$ are cyclic.Then $\angle NXC=\angle MXD=\angle MAD=\angle DAC=\angle AZN=\angle CZN$ implies $Z$,$N$,$C$,$X$ are cyclic so $\angle CXZ=\angle DXZ=90$,$\angle DAZ=90$. So...
- Wed Mar 29, 2017 4:34 pm
- Forum: National Math Camp
- Topic: The Gonit IshChool Project - Beta
- Replies: 28
- Views: 45583
Re: The Gonit IshChool Project - Beta
Name you'd like to be called: Saad
Course you want to learn: Functional Equations and Number Theory Problem solving.
Preferred methods of communication (Forum, Messenger, Telegram, etc.):Telegram,Messenger.
Do you want to take lessons through PMs or Public?: Public
Course you want to learn: Functional Equations and Number Theory Problem solving.
Preferred methods of communication (Forum, Messenger, Telegram, etc.):Telegram,Messenger.
Do you want to take lessons through PMs or Public?: Public
- Thu Jan 19, 2017 6:46 pm
- Forum: Geometry
- Topic: IGO 2016 Medium/5
- Replies: 1
- Views: 2538
Re: IGO 2016 Medium/5
[AST: Alternate Segment Theorem] We assume that, WLOG radius of $\omega$ is less than radius of $\omega'$(equality case is trivial). Let $P$ be the midpoint of $\widehat{EF}$ not containing $B$ of $\odot BEF$, thus $BP$ is the angle bisector of $\angle EBF$. Let the tangent at $P$ to $BEF$ intersect...