International Mathematical Olympiad

In the circles $\Omega$ and $\Gamma$ we have $\angle JRS=\angle JIS=\angle A{R}^{\prime }S$. On the other hand, since $RA$ is tangent to $\Omega$, we get $\angle SJR=\angle SRA$. So the triangles $AR{R}^{\prime }$ and $SJR$ are similar, and $$\frac{R^{\prime }R}{RJ}=\frac{A{R}^{\prime }}{SR}=\frac{A{R}^{\prime }}{S{R}^{\prime }}.$$ The last relation, together with $\angle A{R}^{\prime }S=\angle JR{R}^{\prime }$, yields $△AS{R}^{\prime }\sim △R^{\prime }JR$, hence $\angle SA{R}^{\prime }=\angle R{R}^{\prime }J$. It follows that $J{R}^{\prime }$ is tangent to $\Gamma$ at $R^{\prime }$.

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Alternative solution.

As in Solution 1, we notice that $\angle JRS=\angle JIS=\angle A{R}^{\prime }S$, so we have $RJ\parallel A{R}^{\prime }$. Let $A^{\prime }$ be the reflection of $A$ about $S$; then $AR{A}^{\prime }{R}^{\prime }$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R{A}^{\prime }$. From $\angle S{R}^{\prime }{A}^{\prime }=\angle SRA=\angle SJR$ we get that the points $S,J,A^{\prime },R^{\prime }$ are concyclic. This proves that $\angle S{R}^{\prime }J=\angle S{A}^{\prime }J=\angle S{A}^{\prime }R=\angle SA{R}^{\prime }$, so $J{R}^{\prime }$ is tangent to $\Gamma$ at $R^{\prime }$.