IMO Shortlist, 2016

a. Let $A'$ be the reflection of $A$ in $BC$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\Gamma$ of triangle $ABC$. As triangles $ABA'$ and $AOC$ are isosceles with $\angle ABA' = 2\angle ABC = \angle AOC$, they are similar to each other. Also, triangles $ABI_{A}$ and $AIC$ are similar. Therefore we have
$$\frac{AA'}{AI_{A}} = \frac{AA'}{AB} \cdot \frac{AB}{AI_{A}} = \frac{AC}{AO} \cdot \frac{AI}{AC} = \frac{AI}{AO}.$$
Together with $\angle A'AI_{A} = \angle IAO$, we find that triangles $AA'I_{A}$ and $AIO$ are similar.

Denote by $P'$ the intersection of line $AP$ and line $OI$. Using directed angles, we have
$$\begin{aligned} \measuredangle MAP' &= \measuredangle I_{A}'AI_{A} = \measuredangle I_{A}'AA' - \measuredangle I_{A}AA' = \measuredangle AA'I_{A} - \measuredangle (AM, OM) \\ &= \measuredangle AIO - \measuredangle AMO = \measuredangle MOP' \end{aligned}$$
This shows $M, O, A, P'$ are concyclic.
Denote by $R$ and $r$ the circumradius and inradius of triangle $ABC$. Then
$$IP' = \frac{IA \cdot IM}{IO} = \frac{IO^{2} - R^{2}}{IO}$$
is independent of $A$. Hence, $BP$ also meets line $OI$ at the same point $P'$ so that $P' = P$, and $P$ lies on $OI$.

b. By Poncelet's Porism, the other tangents to the incircle of triangle $ABC$ from $X$ and $Y$ meet at a point $Z$ on $\Gamma$. Let $T$ be the touching point of the incircle to $XY$, and let $D$ be the midpoint of $XY$. We have
$$\begin{aligned} OD &= IT \cdot \frac{OP}{IP} = r\left(1 + \frac{OI}{IP}\right) = r\left(1 + \frac{OI^{2}}{OI \cdot IP}\right) = r\left(1 + \frac{R^{2} - 2Rr}{R^{2} - IO^{2}}\right) \\ &= r\left(1 + \frac{R^{2} - 2Rr}{2Rr}\right) = \frac{R}{2} = \frac{OX}{2} \end{aligned}$$
This shows $\angle XZY = 60^{\circ}$ and hence $\angle XIY = 120^{\circ}$.

a. Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}' C$, $A I_{B} C$, $I_{A} B C$ and $I_{A}' B C$ are all similar. From $\triangle A I_{B}' C \sim \triangle I_{A}' B C$, we get $\triangle A I_{A}' C \sim \triangle I_{B}' B C$. From $\measuredangle ABP = \measuredangle I_{B}' B C = \measuredangle A I_{A}' C$ and $\measuredangle BAP = \measuredangle I_{A}' AC$, the triangles $ABP$ and $A I_{A}' C$ are directly similar.

Consider the inversion with centre $A$ and radius $\sqrt{AB \cdot AC}$ followed by the reflection in $AI$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.
From the similar triangles obtained, we have $AP \cdot AI_{A}' = AB \cdot AC$ so that $P$ is mapped to $I_{A}'$ under the transformation. In addition, line $AO$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $BC$, which we call point $A'$. Note that $AA'I_{A}I_{A}'$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear.

b. Denote by $R$ and $r$ the circumradius and inradius of triangle $ABC$. Note that by the above transformation, we have $\triangle APO \sim \triangle AA'I_{A}'$ and $\triangle AA'I_{A} \sim \triangle AIO$. Therefore, we find that
$$PO = A'I_{A}' \cdot \frac{AO}{AI_{A}'} = AI_{A} \cdot \frac{AO}{A'I_{A}} = \frac{AI_{A}}{A'I_{A}} \cdot AO = \frac{AO}{IO} \cdot AO$$
This shows $PO \cdot IO = R^{2}$, and it follows that $P$ and $I$ are mapped to each other under the inversion with respect to the circumcircle $\Gamma$ of triangle $ABC$. Then $PX \cdot PY$, which is the power of $P$ with respect to $\Gamma$, equals $PI \cdot PO$. This yields $X, I, O, Y$ are concyclic.
Let $T$ be the touching point of the incircle to $XY$, and let $D$ be the midpoint of $XY$. Then
$$OD = IT \cdot \frac{PO}{PI} = r \cdot \frac{PO}{PO - IO} = r \cdot \frac{R^{2}}{R^{2} - IO^{2}} = r \cdot \frac{R^{2}}{2Rr} = \frac{R}{2}.$$
This shows $\angle DOX = 60^{\circ}$ and hence $\angle XIY = \angle XOY = 120^{\circ}$.