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Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Geometry Difficulty 8.1 Shortlist Prove it Romania

Let $\gamma$ be a circle, and let $P$ be a point in its plane, not situated on $\gamma$ . Two variable lines $\ell$ and $\ell'$ through $P$ meet $\gamma$ at $X$ and $Y$ , and $X'$ and $Y'$ , respectively. Show that the line through the centres of the circles $PXY'$ and $PX'Y$ passes through a fixed point.

Solution

Let the circles $PXY'$ and $PX'Y$ meet again at $Q$ . A suitable inversion of pole $P$ sends the circles $PXY'$ and $PX'Y$ onto the lines $XY'$ and $X'Y$ , respectively, while leaving $\gamma$ invariant. The image of $Q$ under this inversion is the point $R$ where the lines $XY'$ and $X'Y$ meet. Upon inversion, the locus of $R$ — the polar of $P$ relative to $\gamma$ — transforms into the circle on diameter $OP$ , where $O$ is the centre of $\gamma$ . Consequently, the lines $OQ$ and $PQ$ are perpendicular, so the line through the centres of the circles $PXY'$ and $PX'Y$ , which is the perpendicular bisector of the segment $PQ$ , passes through the midpoint of the segment $OP$ . The conclusion follows.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.