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Iranian Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Iran

Let $C$ be a circle and $P$ a point outside of it. $PA$ and $PB$ are the two tangent lines to this circle and point $K$ is chosen arbitrarily on the segment $AB$ . The circumcircle of triangle $PBK$ intersects circle $C$ for the second time at $T$ . Let $P'$ be the reflection of $P$ with respect to $A$ . Show that $\angle PBT = \angle P'KA$ .

Solution

In this solution, all of the arcs considered are from circle $C$ .

Since quadrilateral $KTPB$ is cyclic, $\angle AKT = \angle BPT$ .
$\left. \begin{array}{l} \angle TAK = \frac{\widehat{TB}}{2} = \angle TBP \\ \angle AKT = \angle BPT \end{array} \right\} \Rightarrow T'AK \sim \widehat{\Delta} TBP.$
Therefore,
$\frac{TA}{TB} = \frac{AK}{BP} = \frac{AK}{AP'} \Rightarrow \frac{AP'}{TB} = \frac{AK}{TA} \quad (1).$
On the other hand,
$\text{. P'AK = AYB 2 = BTA (1) P'AK BTA.}$
So, $\angle P'KA = \angle BAT = \angle PBT$ , and the assertion is proved.

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