IMO Team Selection Test 3

Let $\ell$ be the second tangent through $P$ to the circle aside from $BC$. Note that $$\begin{array}{rl}\angle ABP& =2\angle IBP=2(90^{\circ }-\angle BPI)=180^{\circ }-2\angle BPI\\ & =180^{\circ }-\angle (BP,\ell )=\angle (CP,\ell ),\end{array}$$ where we use the fact that $BI$ is the angular bisector of $\angle ABP$, and that $PI$ is the angular bisector of the angle between $PB$ and $\ell$. It follows that $AB\parallel \ell$. The distance between these two parallel lines is $2r$ with $r$ the radius of the incircle. Denote the distance from $C$ to $AB$ by $h$. Then we see that the area of $△ABC$ equals $rs$ on the one hand, and $\frac{1}{2}ch$ on the other hand. It now follows from $rs=\frac{1}{2}ch$ that $$\frac{|BC|}{|BP|}=\frac{d(AB,C)}{d(AB,\ell )}=\frac{h}{2r}=\frac{s}{c}.$$ We conclude that $$\frac{|PC|}{|BP|}=\frac{|BC|-|BP|}{|BP|}=\frac{|BC|}{|BP|}-1=\frac{s}{c}-1=\frac{s-c}{c}.\square$$