IMO 1996 SL G2

Origin: CAN | Category: Geometry

Problem

Let $P$ be a point inside $\triangle ABC$ such that

$\angle APB - \angle C = \angle APC - \angle B.$

Let $D, E$ be the incenters of $\triangle APB$ and $\triangle APC$, respectively. Show that $AP$, $BD$, and $CE$ meet in a point.

Solution

Let $X, Y, Z$ respectively be the feet of the perpendiculars from $P$ to $BC$, $CA$, and $AB$. Examining the cyclic quadrilaterals $AZPY$, $BXPZ$, $CYPX$, one can easily see that

$\angle XZY = \angle APB - \angle C \quad \text{and} \quad XY = PC \sin \angle C.$

The first relation gives that $\triangle XYZ$ is isosceles with $XY = XZ$, so from the second relation

$PB \sin \angle B = PC \sin \angle C.$

Hence

$\frac{AB}{PB} = \frac{AC}{PC}.$

This implies that the bisectors $BD$ and $CE$ of $\angle ABP$ and $\angle ACP$ divide the segment $AP$ in equal ratios; i.e., they concur with $AP$.

Second solution.

Take $X, Y, Z$ as the points of intersection of $AP, BP, CP$ with the circumscribed circle of $\triangle ABC$ instead. We similarly obtain $XY = XZ$. If we write

$AP \cdot PX = BP \cdot PY = CP \cdot PZ = k,$

from the similarity of $\triangle APC$ and $\triangle ZPX$ we get

$\frac{AC}{XZ} = \frac{AP}{PZ} = \frac{AP \cdot CP}{k},$

i.e.,

$XZ = \frac{k \cdot AC \cdot BP}{AP \cdot BP \cdot CP}.$

It follows again that

$\frac{AC}{AB} = \frac{PC}{PB}.$

Third solution.

Apply an inversion with center at $A$ and radius $r$, and denote by $Q'$ the image of any point $Q$. Then the given condition becomes

$\angle BCP' = \angle CBP', \quad \text{i.e., } BP = PC.$

But

$PB = \frac{r^2}{AP \cdot AB} PB,$

$\frac{AC}{AB} = \frac{PC}{PB}.$

Remark. Moreover, it follows that the locus of $P$ is an arc of the circle of Apollonius through $C$.