Search found 138 matches

by Raiyan Jamil
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...
by Raiyan Jamil
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...
by Raiyan Jamil
Mon Jan 09, 2017 5:49 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 63463

Re: Geometry Marathon : Season 3

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...
by Raiyan Jamil
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...
by Raiyan Jamil
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.
by Raiyan Jamil
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...
by Raiyan Jamil
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...
by Raiyan Jamil
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.
by Raiyan Jamil
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...
by Raiyan Jamil
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.