IMO 2017 P4

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IMO 2017 P4

Post Number:#1  Unread postby Ananya Promi » Sat Jul 29, 2017 12:29 pm

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.
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Ananya Promi
 
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Re: IMO 2017 P4

Post Number:#2  Unread postby Ananya Promi » Sat Jul 29, 2017 12:38 pm

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
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