## Search found 138 matches

Sun Jan 29, 2017 1:41 pm
Forum: Geometry
Topic: USA TST 2017
Replies: 2
Views: 1703

### Re: USA TST 2017

$\text{Solution of (1):}$ Obviously, $T$ is the point where $A$ tangent of $\bigodot ABC$ meets $BC$. Now, $\bigodot ABC$ and $\bigodot AT$ are orthogonal. Let $OT$ meet $\bigodot AT$ for the second time for $T_1$. So, $T$ and $T_1$ are inverses respect to $\bigodot ABC$.So, applying inverion respec...
Wed Jan 25, 2017 6:20 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63463

### Re: Geometry Marathon : Season 3

$\text{Problem 16:}$ $H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ conc...
Mon Jan 09, 2017 5:49 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63463

If $BKCL$ is cyclic, then $\bigtriangleup ABC$ is oppositely similar to $\bigtriangleup AKL$. Then, generalizing it becomes that, $\bigtriangleup ABC$ is a scalene triangle where the $A$-symmedian is perpendicular to $BC$. And we have to prove that $\angle BAC=90^\circ$. Now,let us assume $\angle BA... Sun Jan 08, 2017 4:16 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 146 Views: 63463 ### Re: Geometry Marathon : Season 3 Problem 9 Let$\{P, P'\}$and$\{Q,Q'\}$be two pairs of isogonal conjugates of$\triangle ABC$. Let$\triangle P_AP_BP_C$be the cevain triangle of$P$wrt$\triangle ABC$. Define$\triangle Q_AQ_BQ_C$similarly. Prove that,$P_B, P_C$and$Q'$are collinear if and only if$Q_B,Q_C$and$P'$are c... Fri Jan 06, 2017 7:14 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 146 Views: 63463 ### Re: Geometry Marathon : Season 3$\text{Problem 3:}$In Acute angled triangle$ABC$, let$D$be the point where$A$angle bisector meets$BC$. The perpendicular from$B$to$AD$meets the circumcircle of$ABD$at$E$. If$O$is the circumcentre of triangle$ABC$then prove that$A,E$and$O$are collinear. Fri Jan 06, 2017 3:54 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 146 Views: 63463 ### Re: Geometry Marathon : Season 3$\text{Solution of Problem 2:}$Let$FD$and$AB$meet at$Y'$. Now,$\angle DBF=\angle DBC=\angle DEC=\angle EDC=\angle FAC=\angle FAD \Rightarrow BFDA$is cyclic$\Rightarrow FDA=Y'DA=90^\circ$. Now,$\angle BY'D=\angle AY'D=90^\circ-\angle BAD=\angle BCD \Rightarrow Y'$lies on circle$BCD$. Now... Fri Jan 06, 2017 12:19 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 146 Views: 63463 ### Re: Geometry Marathon : Season 3$\text {Solution of Problem 1:}$Let$AB$meet$CE$at$X$. SInce$\angle A=90^o$and$\angle EAC= \angle CAF$,$\Rightarrow(X,C;E,F)=-1$. And$B(X,C;E,F) \Rightarrow ADCE$is a harmonic quadrilateral. So, if the perpendicular bisector of$DE$intersects$AC$at$Y$, then$Y$is also the intersect... Thu Jan 05, 2017 7:00 pm Forum: Secondary Level Topic: TJMO 1996/2 Replies: 4 Views: 2026 ### Re: TJMO 1996/2 Trinary should help. Tue Jan 03, 2017 11:43 pm Forum: Geometry Topic: AN IMO PROBLEM Replies: 1 Views: 1765 ### Re: AN IMO PROBLEM Firstly, in the case$AB=BC$, the point$D$is just the antipode of$B$.So let$AB\neq BC$1) We draw point$E$extending$AB$past$B$such that$BE=BC$. 2) Let$d=|AB-BC|$.We draw a circle with radius d and centre$C$. 3) We draw the circle$ACE$. 4) Let the two circles intersect at$X,Y\$ such tha...
Fri Dec 23, 2016 2:16 pm
Forum: Secondary Level
Topic: Two Geometry PROBLEM /hard\
Replies: 1
Views: 1739

### Re: Two Geometry PROBLEM /hard\

For the first one, use ptolemy's theorem.