Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Ireland

Two circles intersect at $A$ and $B$ . A common tangent to the circles touches the circles at $P$ and $Q$ . A circle is drawn through $P$ , $Q$ and $A$ and the line $AB$ meets this circle again at $C$ . Join $CP$ and $CQ$ and extend both to meet the given circles at $F$ and $E$ respectively. Prove $P$ , $Q$ , $F$ and $E$ lie on the circumference of a circle.

Solutions — 2

Solution 1

Because $P$ , $A$ , $Q$ , $C$ are on a circle, $\angle PCA = \angle PQA$ and $\angle APC + \angle CQA = 180^\circ$ . Hence $\angle EPA + \angle AQF = 180^\circ$ . Because $A$ , $B$ , $F$ , $Q$ are on a circle, $\angle AQF + \angle FBA = 180^\circ$ . Finally, as $F$ , $E$ , $B$ , $A$ are concyclic, $\angle EPA + \angle ABE = 180^\circ$ .

The last three equations imply $\angle ABE = \angle AQF$ , $\angle EPA = \angle FBA$ and $\angle ABE + \angle FBA = 180^\circ$ which means that $E$ , $B$ and $F$ are collinear. Hence
$\angle FEP + \angle PQF = \angle FEP + \angle PQA + \angle AQF \\ = \angle FEP + \angle PCA + \angle ABE = 180^\circ$
because the angle sum in triangle $EBC$ is $180^\circ$ . Therefore, the quadrilateral $EFQP$ is cyclic.

Solution 2

The point $C$ is on the radical axis of the two original circles, hence the power of the point $C$ is the same for both circles:
$|CQ| \cdot |CF| = |CA| \cdot |CB| = |CP| \cdot |CE|$
hence $E$ , $F$ , $Q$ and $P$ are on a circle.

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