IMO 2017 P4
 Ananya Promi
 Posts: 36
 Joined: Sun Jan 10, 2016 4:07 pm
 Location: Naogaon, Bangladesh
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 A be the common point of Γ and that is closer to R. Line AJ meets Ω again at K. Prove that the line KT is tangent to Γ.
 Ananya Promi
 Posts: 36
 Joined: Sun Jan 10, 2016 4:07 pm
 Location: Naogaon, Bangladesh
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.
$\angle{PRT}=\angle{RTK}$
$\angle{STK}=\angle{SAT}$
So, $KT$ is tangent to the circle.
We are done
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.
$\angle{PRT}=\angle{RTK}$
$\angle{STK}=\angle{SAT}$
So, $KT$ is tangent to the circle.
We are done

 Posts: 3
 Joined: Mon Mar 26, 2018 2:20 pm
Re: IMO 2017 P4
TPRK is parallelogram

 Posts: 3
 Joined: Mon Mar 26, 2018 2:20 pm
Re: IMO 2017 P4
TPRK is parallelogram