2017 IMO Problems/Problem 4

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ 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 $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$ . Line $AJ$ meets $\Omega$ again at $K$ . Prove that the line $KT$ is tangent to $\Gamma$ .

Solution

We construct inversion which maps $KT$ into the circle $\omega_1$ and $\Gamma$ into $\Gamma.$ Than we prove that $\omega_1$ is tangent to $\Gamma.$

Quadrungle $RJSK$ is cyclic $\implies \angle RSJ = \angle RKJ.$ Quadrungle $AJST$ is cyclic $\implies \angle RSJ = \angle TAJ \implies AT||RK.$ We construct circle $\omega$ centered at $R$ which maps $\Gamma$ into $\Gamma.$ Let $C = \omega \cap RT \implies RC^2 = RS \cdot RT.$ Inversion with respect $\omega$ swap $T$ and $S \implies \Gamma$ maps into $\Gamma.$ Inversion with respect $\omega$ maps $K$ into $K'$ .

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2017 IMO Problems/Problem 4

Solution