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Singapore International Mathematical Olympiad Committee National Team Selection Test

Geometry Difficulty 8.1 Shortlist Prove it Singapore

Let $OT$ be a diameter of a circle. Let $A$ and $B$ be two distinct points on the circle both on the same side of $OT$ , and let $C$ be the intersection of the tangents to the circle at $A$ and $B$ . The tangent to the circle at $T$ meet the lines $OA$ , $OB$ and $OC$ at $A'$ , $B'$ and $C'$ respectively. Prove that $C'$ is the midpoint of $A'B'$ .

Solution

Consider the inversion in the circle $\omega$ centred at $O$ with radius $OT$ . The circle $\alpha$ with diameter $OT$ is inverted into the tangent line $\alpha'$ to $\alpha$ at $T$ . Thus $A'$ and $B'$ are the inverses of $A$ and $B$ respectively. The circle $\beta$ centred at $C$ with radius $CA$ or $CB$ is orthogonal to $\alpha$ so that its inverse $\beta'$ remains orthogonal to $\alpha'$ . This implies that the segment $A'B'$ is the diameter of $\beta'$ and its midpoint is the centre of $\beta'$ . As the points $O$ and the centres of $\beta$ and $\beta'$ are collinear under the inversion, the point $C'$ is the midpoint of $A'B$ '.

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