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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 6.4 National Olympiad Prove it Singapore

Let $CD$ be a chord of a circle $\Gamma_1$ and $AB$ a diameter of $\Gamma_1$ perpendicular to $CD$ at $N$ with $AN > NB$ . A circle $\Gamma_2$ centred at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$ . The line $PQ$ intersects $CD$ at $M$ and $AC$ at $K$ ; and the extension of $NK$ meets $\Gamma_2$ at $L$ . Prove that $PQ$ is perpendicular to $AL$ .

Solution

1. Extend $DC$ meeting $\Gamma_2$ at $H$ . Let the radius of $\Gamma_2$ be $r$ . Note that $DN = NC = CH = r$ . Since $M$ is of equal power with respect to $\Gamma_1$ and $\Gamma_2$ . Thus $MN \cdot NH = MC \cdot MD$ . That is $MN(MC + r) = MC(MN + r)$ giving $MN = MC$ . Thus $M$ is the midpoint of $NC$ .

As $K$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$ , the points $C$ , $N$ , $A$ , $L$ are concyclic. Thus $\angle ALC = \angle ANC = 90^\circ$ so that $AL$ is tangent to $\Gamma_2$ . It follows that $AC$ is perpendicular to $NL$ at $K$ , and hence $MN = MC = MK$ .

Now let $PQ$ intersect $AL$ at $T$ . We have $\angle TAK = \angle KNM = \angle NKM = \angle LKT$ and similarly $\angle TLK = \angle AKT$ . Consequently, $2\angle KTL = 2(\angle TAK + \angle AKT) = \angle TAK + \angle AKT + \angle LKT + \angle TLK = 180^\circ$ , which means $\angle KTL = 90^\circ$ .

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