IMO 2016 Shortlisted Problems

a. Let $A^{\prime }$ be the reflection of $A$ in $BC$ and let $M$ be the second intersection of line $AI$ and the circumcircle $\Gamma$ of triangle $ABC$. As triangles $AB{A}^{\prime }$ and $AOC$ are isosceles with $\angle AB{A}^{\prime }=2\angle ABC=\angle AOC$, they are similar to each other. Also, triangles $AB{I}_A$ and $AIC$ are similar. Therefore we have $$\frac{A{A}^{\prime }}{A{I}_A}=\frac{A{A}^{\prime }}{AB}\cdot \frac{AB}{A{I}_A}=\frac{AC}{AO}\cdot \frac{AI}{AC}=\frac{AI}{AO}.$$ Together with $\angle A^{\prime }A{I}_A=\angle IAO$, we find that triangles $A{A}^{\prime }{I}_A$ and $AIO$ are similar. Denote by $P^{\prime }$ the intersection of line $AP$ and line $OI$. Using directed angles, we have $$\begin{array}{rl}\measuredangle MA{P}^{\prime }& =\measuredangle I_A^{\prime }A{I}_A=\measuredangle I_A^{\prime }A{A}^{\prime }-\measuredangle I_{A}A{A}^{\prime }=\measuredangle A{A}^{\prime }{I}_A-\measuredangle (AM,OM)\\ & =\measuredangle AIO-\measuredangle AMO=\measuredangle MO{P}^{\prime }\end{array}$$ This shows $M,O,A,P^{\prime }$ are concyclic. Denote by $R$ and $r$ the circumradius and inradius of triangle $ABC$. Then $$I{P}^{\prime }=\frac{IA\cdot IM}{IO}=\frac{I{O}^2-R^2}{IO}$$ is independent of $A$. Hence, $BP$ also meets line $OI$ at the same point $P^{\prime }$ so that $P^{\prime }=P$, 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 $$\begin{array}{rl}OD& =IT\cdot \frac{OP}{IP}=r\left(1+\frac{OI}{IP}\right)=r\left(1+\frac{O{I}^2}{OI\cdot IP}\right)=r\left(1+\frac{R^2-2Rr}{R^2-I{O}^2}\right)\\ & =r\left(1+\frac{R^2-2Rr}{2Rr}\right)=\frac{R}{2}=\frac{OX}{2}\end{array}$$ This shows $\angle XZY=60^{\circ }$ and hence $\angle XIY=120^{\circ }$.

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Alternative solution.

a. Note that triangles $A{I}_{B}C$ and $I_{A}BC$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A{I}_B^{\prime }C$, $A{I}_{B}C$, $I_{A}BC$ and $I_A^{\prime }BC$ are all similar. From $△A{I}_B^{\prime }C\sim △I_A^{\prime }BC$, we get $△A{I}_A^{\prime }C\sim △I_B^{\prime }BC$. From $\measuredangle ABP=\measuredangle I_B^{\prime }BC=\measuredangle A{I}_A^{\prime }C$ and $\measuredangle BAP=\measuredangle I_A^{\prime }AC$, the triangles $ABP$ and $A{I}_A^{\prime }C$ 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 $AP\cdot A{I}_A^{\prime }=AB\cdot AC$ so that $P$ is mapped to $I_A^{\prime }$ 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 $A^{\prime }$. Note that $A{A}^{\prime }{I}_A{I}_A^{\prime }$ 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 $△APO\sim △A{A}^{\prime }{I}_A^{\prime }$ and $△A{A}^{\prime }{I}_A\sim △AIO$. Therefore, we find that $$PO=A^{\prime }{I}_A^{\prime }\cdot \frac{AO}{A{I}_A^{\prime }}=A{I}_A\cdot \frac{AO}{A^{\prime }{I}_A}=\frac{A{I}_A}{A^{\prime }{I}_A}\cdot AO=\frac{AO}{IO}\cdot AO$$ 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 $$OD=IT\cdot \frac{PO}{PI}=r\cdot \frac{PO}{PO-IO}=r\cdot \frac{R^2}{R^2-I{O}^2}=r\cdot \frac{R^2}{2Rr}=\frac{R}{2}.$$ This shows $\angle DOX=60^{\circ }$ and hence $\angle XIY=\angle XOY=120^{\circ }$.