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Taiwan IMO Selection Camp

Geometry Difficulty 4.6 AIME Prove it Taiwan

Let $AB$ be a chord on a circle $O$ , $M$ be the midpoint of the smaller arc $AB$ . From a point $C$ outside the circle $O$ draw two tangents to the circle $O$ at the points $S$ and $T$ . Suppose $MS$ intersects with $AB$ at the point $E$ , $MT$ intersects with $AB$ at the point $F$ . From $E$ , $F$ draw a line perpendicular to $AB$ that intersects with $OS$ , $OT$ at the points $X$ , $Y$ , respectively. Draw another line from $C$ which intersects with the circle $O$ at the points $P$ and $Q$ . Let $R$ be the intersection point of $MP$ and $AB$ . Finally, let $Z$ be the circumcenter of $\triangle PQR$ .
Prove that $X$ , $Y$ , and $Z$ are collinear.

Solution

作 $AB$ 的中垂線 $OM$ 。故 $\triangle XES \sim \triangle OMS$ ，於是 $SX = XE$ 。

Draw the perpendicular bisector $OM$ of $AB$ . Thus $\triangle XES \sim \triangle OMS$ , so $SX = XE$ .

畫以 $XE$ 為半徑的圓 $X$ 。圓 $X$ 與弦 $AB$ 及直線 $CS$ 均相切。又作 $\triangle PQR$ 的外接圓，以及直線 $MA$ 與 $MC$ ，如圖所示。

Draw circle $X$ with radius $XE$ . Circle $X$ is tangent to both the chord $AB$ and the line $CS$ . Also draw the circumcircle of $\triangle PQR$ , together with the lines $MA$ and $MC$ , as shown in the figure.

因為 $\triangle AMR \sim \triangle PMA$ ，所以有
$MR \cdot MP = MA^2 = ME \cdot MS.$
又由圓幂定理知 $CQ \cdot CP = CS^2$ 。故 $M$ , $C$ 兩點皆位於圓 $Z$ 與圓 $X$ 的根軸上，得 $ZX \perp MC$ 。同理可知 $ZY \perp MC$ 。所以 $X$ , $Y$ , $Z$ 三點共線，得證。

Since $\triangle AMR \sim \triangle PMA$ , we have
$MR \cdot MP = MA^2 = ME \cdot MS.$
Also, by the power of a point theorem, $CQ \cdot CP = CS^2$ . Hence both $M$ and $C$ lie on the radical axis of circle $Z$ and circle $X$ , giving $ZX \perp MC$ . By the same reasoning, $ZY \perp MC$ . Therefore $X$ , $Y$ , $Z$ are collinear, as required.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.