, 2009

Because $\angle BAP = \angle BQC = 45^\circ$ and $\angle PBA = \angle CQB = 75^\circ$, triangles $BQC$ and $BAP$ are similar to each other, from which it follows that triangles $BCP$ and $BQA$ are similar to each other. Hence, the law of sines gives
$$\frac{CP}{AQ} = \frac{BP}{BA} = \frac{\sin \angle BAP}{\sin \angle APB} = \frac{\sin 45^\circ}{\sin 60^\circ} = \sqrt{\frac{2}{3}}$$
Extend segment $CM$ through $M$ to $S$ with $CM = MS$. Then $CS = 2CM = 2RQ/\sqrt{6}$, from which it follows that
$$\frac{CP}{AQ} = \frac{CS}{QR}.$$
Because triangles $BPC$ and $BAQ$ are similar to each other, we may set $x = \angle CPB = \angle QAB$. Because segments $SC$ and $AP$ bisect each other, $ACPS$ is a parallelogram, implying that $\angle SPA = \angle CAP = \angle CAB + \angle BAP = \angle CAB + 45^\circ$. It follows that
$$\angle SPC = \angle SPA + \angle APB - \angle CPB = \angle CAB + 45^\circ + 60^\circ - x = \angle CAB + 105^\circ - x.$$
On the other hand, $\angle RAQ = \angle RAC + \angle CAB - \angle QAB = 105^\circ + \angle CAB - x$. Hence we have $\angle SPC = \angle RAQ$. Therefore, we have that
$$\angle SPC = \angle RAQ \quad \text{and} \quad \frac{CP}{AQ} = \frac{CS}{QR}$$
and that $\angle SPC = 180^\circ - \angle ACP > 180^\circ - \angle ACB > 90^\circ$ is obtuse because triangle $ABC$ is acute. Therefore, we may conclude that triangles $RAQ$ and $SPC$ are similar. This means that
$$\frac{SP}{AR} = \frac{SC}{QR} = \frac{2CM}{QR} = \frac{2}{\sqrt{6}}.$$
In view of parallelogram $ACPS$, we have that $SP = AC$, so we find that
$$\frac{AC^2}{AR^2} = \frac{SP^2}{AR^2} = \frac{2}{3}.$$