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China Hong Kong Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Hong Kong

Two circles intersect at points $A$ and $B$ . Through the point $B$ a straight line is drawn, intersecting the first circle at $K$ and the second circle at $M$ . A line parallel to $AM$ is tangent to the first circle at $Q$ . The line $AQ$ intersects the second circle again at $R$ .

a. Prove that the tangent to the second circle at $R$ is parallel to $AK$ .

b. Prove that these two tangents are concurrent with $KM$ .

Solution

a. We only provide a proof for the configuration as shown. The other cases are similar.
Let $X$ and $Y$ be any points on the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ respectively as shown. We have
$\angle MBR = \angle MAR = \angle YQA = \angle QBA,$
and so
$\begin{aligned} \angle XRA &= \angle RBA = \angle RBQ + \angle QBA = \angle RBQ + \angle MBR \\ &= \angle MBQ = \angle KAQ = \angle KAR. \end{aligned}$
This implies $RX \parallel KA$ .

b. Let the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ meet at $S$ . Since
$\angle BRS = \angle BAR = \angle BAQ = \angle BQS,$
the points $Q$ , $B$ , $S$ , $R$ are concyclic. This implies
$\angle SBR = \angle SQR = \angle MBR,$
and hence $S$ , $M$ , $B$ are collinear. Thus, the two tangents and $KM$ are concurrent.

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