Hong Kong Preliminary Selection Contest

As shown in the figure, let $D$ be the point for which $ABPD$ is an isosceles trapezium with $AB\parallel DP$. Let also $E$ be the point for which $APED$ is a parallelogram, and $F$ be the point such that $A$ and $F$ lie on different sides of $BC$ and for which $CBF$ is an equilateral triangle.

Note that by our construction and the given condition $AP=BC$, we have

$DE=BD=BF$ as each of these three segments has the same length as $AP$.

We have $\angle CBD=\angle CBA-\angle DBA=(180^{\circ }-54^{\circ }-24^{\circ })-54^{\circ }=48^{\circ }$. Therefore, we get $\angle BDE=\angle BDP+\angle PDE=\angle BDP+\angle DPA=54^{\circ }+54^{\circ }=108^{\circ }$ and $\angle FBD=\angle FBC+\angle CBD=60^{\circ }+48^{\circ }=108^{\circ }$.

These show that $E,D,B,F$ are consecutive vertices of a regular pentagon. It follows that $C$ and $E$ both lie on the perpendicular bisector of $BF$.



Thus we have $\angle DEC=\frac{108^{\circ }}{2}=54^{\circ }=\angle PDE$. Together with $PC\parallel DE$, we see that $PCED$ is an isosceles trapezium. In particular, we have $CD=PE=AD$. Hence we have $$\angle PAD=\angle DCA=\angle DCB-\angle ACB=\frac{180^{\circ }-48^{\circ }}{2}-24^{\circ }=42^{\circ }$$ and so $\angle CBP=\angle CBA-\angle PBD-\angle DBA=102^{\circ }-\angle PAD-54^{\circ }=6^{\circ }$.