## Search found 18 matches

Sat Oct 07, 2017 4:32 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2012/04
Replies: 3
Views: 1281

We'll have to show, $BF/CF=AB/AC$ which means $BF/CF=sin\angle{BCF}/sin\angle{CBF}=(BK*AC)/(AB*CK)$ We have to show now, $AB^2/AC^2=BK/CK$ or $AF$ is a symmedian. We get $A,O,P$ collinear Then $AEDM$ is cyclic Again, $\angle{AMC}=180-\angle{AMB}=\angle{AEF}=\angle{ABF}$ Also $\angle{AFB}=\angle{ACM}... Wed Oct 04, 2017 10:49 pm Forum: Geometry Topic: CGMO 2002/4 Replies: 2 Views: 246 ### Re: CGMO 2002/4 I found this in EGMO(Euclidean geometry in mathematical Olympiad). Then it's their mistake. Not mine. Sat Sep 30, 2017 12:02 pm Forum: Geometry Topic: CGMO 2002/4 Replies: 2 Views: 246 ### CGMO 2002/4 Circles$T_1$and$T_2$intersect at two points$B $and$C$, and$BC$is the diameter of$T_1$. Construct a tangent line to circle$T_1$at$C$intersecting$T_2$at another point$A$. Line$AB$meets$T_1$again at$E $and line$CE $meets$T_2$again at$F $. Let$H $be an arbitrary point on seg... Sat Jul 29, 2017 12:38 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2017 P4 Replies: 1 Views: 703 ### Re: IMO 2017 P4 We get$TA$parallel to$KR$because$\angle{ATS}=\angle{SJK}=\angle{SRK}$We extend$KS$to$P$where$KS$intersects$TA$at$P$Now, It's easy to prove that$TPRK$is a rombus So,$\angle{TPK}=\angle{PKR}$Again,$\angle{ARS}=\angle{SKR}$So,$\angle{TPK}=\angle{ARS}$So,$APSR$is cyclic.$\angl...
Sat Jul 29, 2017 12:29 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2017 P4
Replies: 1
Views: 703

### IMO 2017 P4

Let R and S be dierent points on a circle Ω such that RS is not a diameter. Let be the tangent line to Ω at R. Point T is such that S is the midpoint of the line segment RT. Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects at two distinct points. Let...
Sat Jun 03, 2017 3:21 pm
Forum: Geometry
Topic: CGMO 2007/5
Replies: 1
Views: 233

### CGMO 2007/5

Point $D$ lies inside triangle $ABC$ such that $$\angle{DAC}=\angle{DCA}=30^o$$ and $$\angle{DBA}=60^o$$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE$ I $EF$.
Fri Jun 02, 2017 3:04 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2001 Problem1
Replies: 2
Views: 237