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### Re: Geometry Marathon : Season 3

$\text{Problem 50:}$ Let $\vartriangle ABC$ be a triangle, $O$ be the circumcenter of $\vartriangle ABC$, $N$ be the center of nine point circle of $\vartriangle ABC$ and $X$ be the midpoint of the line segment $ON$. Let $A',B',C'$ be the midpoints of the line segments $BC,CA,AB$, respectively. Let ...
Sun Nov 26, 2017 11:16 pm

Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 109
Views: 4822

$\text{Solution to Problem 49:}$ Let $BH \cap AC=E, CH \cap AB=F, AP \cap BC=X$. Now, by miquel's theorem on circle $BCEF, M,H,P$ are collinear. Also by radical axis theorem on circle $ABC, APFHE, BCEF$, we get $X,F,E$ are collinear. Now, we get that $H$ is the orthocentre of triangle $AXM$ so $XH \... Sun Nov 26, 2017 10:48 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: Geometry Marathon : Season 3$\text{Problem 48:}$Prove that for a scalene triangle$ABC$, one can never find two pairs of isogonal conjugates${P,P'}$and${Q,Q'}$such that$P,P',Q,Q'$are all collinear and given that they're not self isogonal conjugates. Sat Nov 25, 2017 11:59 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: Geometry Marathon : Season 3$\text{Solution to Problem 47:}$Let$AC \cap BD=X$,$AC \cap EF=Y$and$BD \cap EF=Z$. Now$P,M,N$are collinear by newton line. Then,$XM.XZ=XB.XD=XA.XC=XN.XY \Rightarrow MNYZ$is cyclic and$PE^2=PY.PZ=PN.PM$. Sat Nov 25, 2017 11:56 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: Geometry Marathon : Season 3$\text{Problem 45}$Let$ABC$be a triangle with orthocentre$H$and circumcircle$\omega$centered at$O$. Let$M_a,M_b,M_c$be the midpoints of$BC,CA,AB$. Lines$AM_a,BM_b,CM_c$meet$\omega$again at$P_a,P_b,P_c$. Rays$M_aH,M_bH,M_cH$intersect$\omega$at$Q_a,Q_b,Q_c$. Prove that$P_aQ_a,P_b...
Sat Oct 21, 2017 4:20 pm

Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 109
Views: 4822

Let $BH,CH$ meet $(O)$ at $X,Y$. We get $CXNH,BYMH$ are similar rhombuses. Let the perpendicular bisectors of $HM,HN$ meet at $K$ and perpendicular bisectors of $BY,CX$ meet at $O$. Let perpendicular bisectors of $BY,CX$ meet $HM,HN$ at $P,Q$ repectively. Its enough to prove that $\frac{HP}{HM/2}= \... Sat Oct 21, 2017 3:58 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: CGMO 2002/4 Correction:$G$lies on$T_1$, not$T_2$. Solution:$I$be a point on line$AF$such that it is the reflection of$D$under$AB$. We get$HEBI$is cyclic. So,$AH \times AI=AE \times AB=AF^2$from which we can get the result. Sun Oct 01, 2017 2:59 pm Forum: Geometry Topic: CGMO 2002/4 Replies: 2 Views: 147 ### Re: Geometry Marathon : Season 3 Problem 43: An equilateral pentagon$AMNPQ$is inscribed in triangle$ABC$such that$M\in\overline{AB}$,$Q\in\overline{AC}$, and$N,P\in\overline{BC}$. Let$S$be the intersection of$ \overleftrightarrow{MN}$and$ \overleftrightarrow{PQ}$. Denote by$\ell$the angle bisector of$\angle MSQ$. Pr... Tue Aug 08, 2017 12:58 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: Geometry Marathon : Season 3 Solution to Problem 42:$a)P$is just the reflection of$F$under$AC$and$Q$is the reflection of$E$under$AB$.$AFDCP$and$AEDBQ$cyclic. The rest is trivial by taking homothety with centre$A$and ratio 2.$b)$Let$H_1$be the reflection of$H$under$A$. We can find that the perpendicul... Tue Aug 08, 2017 12:37 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 109 Views: 4822 ### Re: Geometry Marathon : Season 3$\text{Problem 40:}$Let the Nagel point of triangle$ABC$be$N$. We draw lines from$B$and$C$to$N$so that these lines intersect sides$AC$and$AB$in$D$and$E$respectively.$M$and$T$are midpoints of segments$BE$and$CD$respectively.$P\$ is the second intersection point of circumcirc...
Sat Apr 29, 2017 11:01 pm

Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 109
Views: 4822
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