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Turkey — Team Selection Test

Geometry Difficulty 8.3 Shortlist Prove it Turkey

Let the line segment $[AB]$ be a chord of the circle $\Gamma$ not passing through the center of it and $M$ be the midpoint of $[AB]$ . Let $C$ be a variable point on $\Gamma$ different from $A$ and $B$ , and let $P$ be the point where the tangent line to the circumcircle of the triangle $CAM$ at the point $A$ meets the tangent line to the circumcircle of the triangle $CBM$ at the point $B$ . Show that the lines $CP$ pass through a common fixed point as $C$ varies.

Solution

Let $Q$ be the point where the tangent lines to $\Gamma$ at $A$ and $B$ meet. We will show that $Q$ is the point.
Since $\angle MCA = \angle MAP$ and $\angle MCB = \angle MBP$ , we have $\angle ACB + \angle APB = 180^\circ$ and $P$ lies on $\Gamma$ . Therefore, if $P'$ is the point where $QC$ intersect the circle, it suffices to show that $\angle MCA = \angle BCP'$ as then it will follow that $AP'$ is tangent to the circumcircle of the triangle $CAM$ and $P' = P$ .
The triangles $QAP'$ and $QCA$ , and the triangles $QBP'$ and $QCB$ are similar. Therefore
$\frac{AP'}{CA} = \frac{QA}{QC} = \frac{QB}{QC} = \frac{BP'}{CB}.$
Then by the Ptolemy's Theorem,
$CP' = \frac{CA \cdot BP' + CB \cdot AP'}{BA} = \frac{CA \cdot BP'}{MA} \quad \text{and} \quad \frac{CP'}{BP'} = \frac{CA}{MA}.$
We conclude that the triangles $CP'B$ and $CAM$ are similar, and hence $\angle MCA = \angle BCP'$ .

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