Asian Pacific Mathematics Olympiad

Let $I$ be the incenter of $\triangle ABC$, and let the feet of the perpendiculars from $I$ to $AB$ and to $AC$ be $D$ and $E$, respectively. (Without loss of generality, we may assume that $AC$ is the longest side. Then $X$ lies on the line segment $AD$. Although $P$ may or may not lie inside $\triangle ABC$, the proof below works for both cases. Note that $P$ is on the line perpendicular to $AB$ passing through $X$.) Let $O$ be the midpoint of $IP$, and let the feet of the perpendiculars from $O$ to $AB$ and to $AC$ be $M$ and $N$, respectively. Then $M$ and $N$ are the midpoints of $DX$ and $EY$, respectively.

The conditions on the points $X$ and $Y$ yield the equations
$$AX = \frac{AB + BC - CA}{2} \quad \text{and} \quad AY = \frac{BC + CA - AB}{2}.$$
From $AD = AE = \frac{CA + AB - BC}{2}$, we obtain
$$BD = AB - AD = AB - \frac{CA + AB - BC}{2} = \frac{AB + BC - CA}{2} = AX.$$
Since $M$ is the midpoint of $DX$, it follows that $M$ is the midpoint of $AB$. Similarly, $N$ is the midpoint of $AC$. Therefore, the perpendicular bisectors of $AB$ and $AC$ meet at $O$, that is, $O$ is the circumcenter of $\triangle ABC$. Since $\angle BAC < 60^{\circ}$, $O$ lies on the same side of $BC$ as the point $A$ and
$$\angle BOC = 2 \angle BAC$$
We can compute $\angle BIC$ as follows:
$$\begin{aligned} \angle BIC &= 180^{\circ} - \angle IBC - \angle ICB = 180^{\circ} - \frac{1}{2} \angle ABC - \frac{1}{2} \angle ACB \\ &= 180^{\circ} - \frac{1}{2}(\angle ABC + \angle ACB) = 180^{\circ} - \frac{1}{2}(180^{\circ} - \angle BAC) = 90^{\circ} + \frac{1}{2} \angle BAC \end{aligned}$$
It follows from $\angle BAC < 60^{\circ}$ that
$$2 \angle BAC < 90^{\circ} + \frac{1}{2} \angle BAC, \quad \text{ i.e., } \quad \angle BOC < \angle BIC.$$
From this it follows that $I$ lies inside the circumcircle of the isosceles triangle $BOC$ because $O$ and $I$ lie on the same side of $BC$. However, as $O$ is the midpoint of $IP$, $P$ must lie outside the circumcircle of triangle $BOC$ and on the same side of $BC$ as $O$. Therefore
$$\angle BPC < \angle BOC = 2 \angle BAC < 120^{\circ}.$$