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 $\Gamma (\Gamma = \Gamma').$

Let $O$ be the center of $\Gamma.$

Inversion with respect $\omega$ maps $K$ into $K'$ . $K$ belong $KT \implies$ circle $K'SR = \omega_1$ is the image of $KT$ . Let $Q$ be the center of $\omega_1.$

$K'T$ is the image of $\Omega$ at this inversion, $l = AR$ is tangent line to $\Omega$ at $R,$ so $K'T||AR.$

$K'$ is image K at this inversion $\implies K \in RK' \implies RK'||AT \implies ARK'T$ is parallelogramm.

$S$ is the midpoint of $RT \implies S$ is the center of symmetry of $ATK'R \implies$ $\triangle RSK'$ is symmetrical to $\triangle TSA$ with respect to $S \implies$ $\omega_1$ is symmetrical to $\Gamma$ with respect to $S \implies$ $O$ is symmetrycal $Q$ with respect to $S.$

$S$ lies on $\Gamma$ and on $\omega_1 \implies \Gamma$ is tangent $\omega_1 \implies S\Gamma$ is tangent $TK.$

Page

Toolbox

Search

2017 IMO Problems/Problem 4

Solution