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, 2016

Geometry Difficulty 6.0 AIME, harder Prove it Hong Kong

Let $\Gamma$ be a circle and $AB$ be a diameter. Let $\ell$ be a line outside the circle, and is perpendicular to $AB$ . Let $X, Y$ be two points on $\ell$ . If $X'$ and $Y'$ are two points on $\ell$ such that $AX$ and $BX'$ intersect on $\Gamma$ and such that $AY$ and $BY'$ intersect on $\Gamma$ , prove that the circumcircles of the triangles $AXY$ and $AX'Y'$ intersect at a point on $\Gamma$ other than $A$ , or the three circles are tangent at $A$ .

Solution

Let $AX$ meet $\Gamma$ again at $P$ , and let $AY$ meet $\Gamma$ again at $Q$ . If $PQ \parallel \ell$ , the figure is symmetric with respect to $AB$ , and so $(AXY)$ and $(AX'Y')$ are tangent at $A$ . In the following, we only consider the configuration as shown.

Firstly, since
$\angle AQP = \angle ABP = 90^\circ - \angle PAB = \angle YXP,$
the points $Q, P, X, Y$ are concyclic. Similarly, $P, Q, X', Y'$ are concyclic.

Now, let $PQ$ meet $\ell$ at $C$ . Then we have
$CX \times CY = CP \times CQ = CX' \times CY'.$
Hence, $C$ has the same power with respect to $(AXY)$ , $(AX'Y')$ and $\Gamma$ . As all three circles pass through $A$ , the line $AC$ is the common radical axis of these circles. Thus, the circles are coaxial as desired.

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