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International Mathematical Olympiad

Geometry Difficulty 8.4 Shortlist Prove it IMO

Let $R$ and $S$ be distinct points on circle $\Omega$ , and let $t$ denote the tangent line to $\Omega$ at $R$ . Point $R'$ is the reflection of $R$ with respect to $S$ . A point $I$ is chosen on the smaller arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $ISR'$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$ . Line $AI$ meets $\Omega$ again at $J$ . Show that $JR'$ is tangent to $\Gamma$ .

Solutions — 2

Solution 1

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

Solution 2

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

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