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### IMO 2017 P4

Posted: 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.

### Re: IMO 2017 P4

Posted: 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

### Re: IMO 2017 P4

Posted: Thu Mar 15, 2018 12:31 pm
very easy problem for IMO

### Re: IMO 2017 P4

Posted: Thu Oct 25, 2018 5:12 pm
TPRK is parallelogram

### Re: IMO 2017 P4

Posted: Thu Oct 25, 2018 5:17 pm
TPRK is parallelogram