## IMO 2017 P4

- Ananya Promi
**Posts:**18**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:**18**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