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BdMO Online Forum • Search

## Search found 18 matches

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 $\angle{AFB}=\angle{ACM}... Sat Oct 07, 2017 4:32 pm Forum: Asian Pacific Math Olympiad (APMO) Topic: APMO 2012/04 Replies: 3 Views: 1148 ### Re: CGMO 2002/4 I found this in EGMO(Euclidean geometry in mathematical Olympiad). Then it's their mistake. Not mine. Wed Oct 04, 2017 10:49 pm Forum: Geometry Topic: CGMO 2002/4 Replies: 2 Views: 147 ### CGMO 2002/4 Circles$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 $be an arbitrary point on seg... Sat Sep 30, 2017 12:02 pm Forum: Geometry Topic: CGMO 2002/4 Replies: 2 Views: 147 ### Re: IMO 2017 P4 We get$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 12:38 pm

Topic: IMO 2017 P4
Replies: 1
Views: 599

### 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 Jul 29, 2017 12:29 pm

Topic: IMO 2017 P4
Replies: 1
Views: 599

### 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 $EF$.
Sat Jun 03, 2017 3:21 pm

Forum: Geometry
Topic: CGMO 2007/5
Replies: 1
Views: 165

$$\angle{A}+\angle{COP} <90° \Rightarrow \angle{COP} <90°-\angle{COD}$$ $$\Rightarrow \angle{COP} <\angle{OCP} \Rightarrow CP<PO$$ is enough to prove. $\angle C -\angle B \geq 30°$ as both $(\angle{C}-\angle{B})$ and 30° are less than $90°$, \begin{align*}&\sin(\angle{C}-\angle{B})\geq \sin30°... Fri Jun 02, 2017 3:04 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2001 Problem1 Replies: 2 Views: 176 ### IMO 2001 Problem1 Consider an acute angled triangle ABC. Let P be the foot of the altitude of triangle ABC issuing from the vertex A, and let O be the circumvented of triangle ABC. Assume that\angle{C}\geq\angle{B}+30°$$. Prove that$$\angle{A}+\angle{COP}<90°$$Fri Jun 02, 2017 3:01 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2001 Problem1 Replies: 2 Views: 176 ### Re: IMO 2007 Problem 4 Let's draw RX perpendicular on PK which intersects thevcircumcirle of triangle ABC at M RX is parallel to BC So, BRMC is a trapizoid. So, BR=MC Again, BR=AR is known as CR in the angle bisector of angle ACB So, AR=MC and it makes ARCM cyclic trapizoid Then MR=b X... Fri Jun 02, 2017 1:59 pm Forum: Geometry Topic: IMO 2007 Problem 4 Replies: 4 Views: 802 ### Re: BDMO 2017 National round Secondary 2 Here, AD is perpendicular on BC Hence, BE=CE We can show that$$\triangle$$BDE and$$\triangle$CFE$ are congruent. So, $FE=ED=OF=\frac{1}{3}OD=\frac{1}{3}OB$ In the right angle triangle $OBE$, $OE^2+BE^2=OB^2$ or,$4OF^2+5=9OF^2$ or,$OF=1=DE$ in rigth angle triangle $CED$, $CE^2+DE^2=CD^2$ ...
Thu Apr 27, 2017 4:31 pm