Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it China

Let $AB$ be a chord of circle $O$ , $M$ the midpoint of arc $AB$ , and $C$ a point outside of the circle $O$ . From $C$ draw two tangents to the circle at points $S$ , $T$ . $MS \cap AB = E$ , $MT \cap AB = F$ . From $E$ , $F$ draw a line perpendicular to $AB$ , and intersecting $OS$ , $OT$ at $X$ , $Y$ respectively. Now draw a line from $C$ which intersects the circle $O$ at $P$ and $Q$ . Let $Z$ be the circumcenter of $\triangle PQR$ . Prove that $X$ , $Y$ , $Z$ are collinear.

Solution

Proof Refer to the figure, join points $O$ and $M$ . Then $OM$ is the perpendicular bisector of $AB$ . So $\triangle XES \sim \triangle OMS$ , and thus $SX = XE$ .
Now draw a circle with center $X$ whose radius is $XE$ . Then the circle $X$ is tangent to chord $AB$ and line $CS$ . Draw the circumcircle of $\triangle PQR$ , line $MA$ and line $MC$ .

It is easy to see ( $\triangle AMR \sim \triangle PMA$ etc)
$MR \cdot MP = MA^2 = ME \cdot MS. \qquad \textcircled{1}$
By the Power of a Point theorem,
$CQ \cdot CP = CS^2. \qquad \textcircled{2}$
So $M$ , $C$ are on the radical axis of circle $Z$ and circle $X$ . Thus
$ZX \perp MC.$
Similarly, we have $ZY \perp MC$ .
So $X$ , $Y$ , $Z$ are collinear.

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