## Search found 36 matches

Mon Apr 09, 2018 4:54 pm
Forum: Geometry
Topic: Secondary Special Camp 2011: Geometry P 4
Replies: 6
Views: 11493

We will have to prove that $PD.DC=RD.DQ$ $\angle{AFE}=\angle{BRQ}=\angle{BCQ}$ So, $BRCQ$ is cyclic. Then, $RD.DQ=BD.DC$ We wanna prove, $PD.DM=BD.DC$ Or, $(PB+a/2-DM)DM=(a/2-DM)(a/2+DM)$ Or, $BP.DM=a^2/4-a/2.DM$ We get $(P,D;B,C)$ is a harmonic bundle. So, $BP.CD=BD.CP$ Or, $BP(a/2+DM)=(a/2-DM)(BP+... Tue Apr 03, 2018 2:11 pm Forum: Geometry Topic: What is the distance? Replies: 3 Views: 8018 ### Re: What is the distance? I'm avoiding the calculations how I got that$PX=2R=65/2$(By Heron's formula we got the area of the triangle$ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.) Using the fact that,$CS/PS=XS/CS$we get$CS=10$I think you've made a typo here. ... Mon Apr 02, 2018 12:16 pm Forum: Geometry Topic: incircles and excircles Replies: 1 Views: 7649 ### Re: incircles and excircles Let$R_1$and$P_1$be the incenter and the C-excenter of triangle$AMC$.$R_2$and$P_2$be the incenter and the C-excenter of triangle$BMC$. So, we get$AP_1MR_1$and$BP_2MR_2$cyclic. $$\angle{AMR_1}=\angle{R_1MC}=90^o-\angle{R_2MC} \angle{R_2MC}=\angle{R_2MB} \angle{AMP_1}=90^o-\angle{R_1MC}$$... Mon Apr 02, 2018 12:34 am Forum: Geometry Topic: A cool Geo! Replies: 1 Views: 7492 ### Re: A cool Geo! $$\angle{BAP}=\angle{PAC}=\angle{PCX}$$ Again, $$\angle{PAC}=\angle{YAC}=\angle{YCA}$$ So, $$\angle{ACY}=\angle{PCX}$$ In triangle$ABC$,$CY$is the isogonal of$CX$So, $$\frac{AC^2}{CP^2}=\frac{AY*AX}{YP*XP}$$ Again, $$\frac{AC}{CP}=\frac{AB}{BP}$$ So, $$\frac{AB^2}{BP^2}=\frac{AY*AX}{YP*XP}$$ It... Fri Mar 30, 2018 10:53 pm Forum: Geometry Topic: What is the distance? Replies: 3 Views: 8018 ### Re: What is the distance? We use Pascal's theorem on$YXZCPB$and get$M, N, I$collinear where$I$is the intersection point of$ZC$and$BY$. In fact,$I$is the incenter of triangle$ABC$. So,$XC=XI$Again, $$\angle{CXY}=\angle{AXY}=\angle{CXN}=\angle{IXN}$$ And $$\angle{PCX}=\angle{NCX}=90^o$$ Ultimately we get$CNIX$c... Fri Mar 30, 2018 10:07 pm Forum: Geometry Topic: What is the distance? Replies: 3 Views: 8018 ### What is the distance? Let$ABC$be a triangle with$AB=26, AC=28, BC=30$. Let$X, Y, Z$be the midpoints of arcs$BC, CA, AB$(not containg the opposite vertics) respectively on the circumcircle of triangle$ABC$. Let$P$be the midpoint of arc$BC$containing point A. Suppose, lines$BP$and$XZ$meet at$M$, while line... Fri Mar 30, 2018 10:03 pm Forum: Geometry Topic: What is the distance? Replies: 1 Views: 7136 ### What is the distance? Let$ABC$be a triangle with$AB=26, AC=28, BC=30$. Let$X, Y, Z$be the midpoints of arcs$BC, CA, AB$(not containg the opposite vertics) respectively on the circumcircle of triangle$ABC$. Let$P$be the midpoint of arc$BC$containing point A. Suppose, lines$BP$and$XZ$meet at$M$, while line... Thu Mar 29, 2018 1:01 pm Forum: Geometry Topic: Hard geo? Replies: 0 Views: 7652 ### Hard geo?$ABCDEF$is a convex hexagon inscribed in a circle so that the diagonals$AD, BE$and$CF$are concurrent at$J$and $$\angle{AJB}=\angle{BJC}=60^o$$. Prove that,$AJ+CJ+EJ=BJ+DJ+FJ$. Sat Oct 07, 2017 4:32 pm Forum: Asian Pacific Math Olympiad (APMO) Topic: APMO 2012/04 Replies: 3 Views: 7021 ### Re: APMO 2012/04 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: 8223

### Re: CGMO 2002/4

I found this in EGMO(Euclidean geometry in mathematical Olympiad). Then it's their mistake. Not mine.