Track / Stage 6 / 97 of 400 #1097 of 2000

Problem 1097

National olympiad, first round

Geometry Difficulty 6.1 Prove it

4. Let $R$ and $S$ be two distinct points on circle $\Gamma$ , and $RS$ is not a diameter. Let $l$ be the tangent line to circle $\Gamma$ at point $R$ . A point $T$ in the plane satisfies that $S$ is the midpoint of segment $RT$ . $J$ is a point on the minor arc $\overparen{RS}$ of circle $\Gamma$ such that the circumcircle $\Gamma_{1}$ of $\triangle JST$ intersects $l$ at two distinct points. Denote the intersection point of circle $\Gamma_{1}$ and $l$ closer to $R$ as $A$ , and the intersection of line $AJ$ with circle $\Gamma$ as another point $K$ . Prove: Line $KT$ is tangent to circle $\Gamma_{1}$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

4. As shown in Figure 2.

From the fact that points $R, K, S, J, S, J, A, T$ are concyclic, we know $\angle K R S = \angle K J S = \angle S T A$ .
From the tangency of $A R$ to circle $\Gamma$ , we get $\angle R K S = \angle T R A$ .
Thus, $\triangle R K S \backsim \triangle T R A \Rightarrow \frac{R K}{R S} = \frac{T R}{T A}$ .
Since $S$ is the midpoint of segment $R T$ , we have $R S = S T$ .
Therefore, $\frac{R K}{T S} = \frac{R T}{T A}$ .
Combining this with $\angle K R T = \angle S T A$ , we get
$\triangle K R T \backsim \triangle S T A \Rightarrow \angle S A T = \angle S T K$ .
This indicates that line $K T$ is tangent to circle $\Gamma_{1}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.