NZMO Round Two

Geometry Difficulty 5.7 AIME, harder Prove it New Zealand

Problem:

Let $AB$ be a chord of circle $\Gamma$ . Let $O$ be the centre of a circle which is tangent to $AB$ at $C$ and internally tangent to $\Gamma$ at $P$ . Point $C$ lies between $A$ and $B$ . Let the circumcircle of triangle $POC$ intersect $\Gamma$ at distinct points $P$ and $Q$ . Prove that $\angle AQP = \angle CQB$ .

Solution

Solution:

Construct the tangent line to $\Gamma$ at $P$ . Note that this line is also tangent to the circle through points $C$ and $P$ with centre $O$ . Also construct point $E$ on this tangent line to the right of $P$ . Note that $\angle EPO = 90^\circ$ and $\angle OCA = 90^\circ$ because the radii and tangents are perpendicular.

Let $x = \angle PBQ$

$\angle EPQ = x$ (by alternate segment theorem)

$\angle OPQ = x - 90^\circ$ (because $\angle EPO = 90^\circ$ )

$\angle QCO = 180^\circ - \angle OPQ$ (because opposite angles in a cyclic quad $= 270^\circ - x$ are supplementary)

$\angle ACQ = 360^\circ - \angle QCO - \angle OCA$ (angles around point $C$ are $360^\circ$ )

$= 360^\circ - (270^\circ - x) - 90^\circ$

$= x$ .

We also have $\angle QPB = \angle QAB$ (angles subtended by chord $QB$ ) in cyclic quad $QAPB$ .

Therefore we have similar triangles

$\triangle QBP \sim \triangle QCA \qquad (\angle PBQ = \angle ACQ \mathrm{~and~} \angle QPB = \angle QAC)$

Hence $\angle AQC = \angle PQB$ . Therefore

$\angle AQP = \angle AQC + \angle CQP = \angle PQB + \angle CQP = \angle CQB.$

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