IMO 2016 Shortlisted Problems

Let the perpendicular from $S$ to $XY$ meet line $QM$ at $S^{\prime }$. Let $E$ be the foot of altitude from $B$ to side $AC$. Since $Q$ and $S$ lie on the perpendicular bisector of $AD$, using directed angles, we have $$\begin{array}{rl}\measuredangle SDQ& =\measuredangle QAS=\measuredangle XAS-\measuredangle XAQ=\left(\frac{\pi }{2}-\measuredangle AYX\right)-\measuredangle BAP=\measuredangle CBA-\measuredangle AYX\\ & =(\measuredangle CBA-\measuredangle ACB)-\measuredangle BCA-\measuredangle AYX=\measuredangle PEM-(\measuredangle BCA+\measuredangle AYX)\\ & =\measuredangle PQM-\measuredangle (BC,XY)=\frac{\pi }{2}-\measuredangle \left(S^{\prime }Q,BC\right)-\measuredangle (BC,XY)=\measuredangle S{S}^{\prime }Q\end{array}$$ This shows $D,S^{\prime },S,Q$ are concyclic.



Let the perpendicular from $N$ to $BC$ intersect line $S{S}^{\prime }$ at $O_1$. (Note that the two lines coincide when $S$ is the midpoint of $AO$, in which case the result is true since the circumcentre of triangle $XSY$ must lie on this line.) It suffices to show that $O_1$ is the circumcentre of triangle $XSY$ since $N$ lies on the perpendicular bisector of $PM$. From $$\measuredangle D{S}^{\prime }{O}_1=\measuredangle DQS=\measuredangle SQA=\angle (SQ,QA)=\angle \left(OD,O_{1}N\right)=\measuredangle DN{O}_1$$ since $SQ\parallel OD$ and $QA\parallel O_{1}N$, we know that $D,O_1,S^{\prime },N$ are concyclic. Therefore, we get $$\measuredangle SD{S}^{\prime }=\measuredangle SQ{S}^{\prime }=\angle \left(SQ,Q{S}^{\prime }\right)=\angle \left(ND,N{S}^{\prime }\right)=\measuredangle DN{S}^{\prime }$$ so that $SD$ is a tangent to the circle through $D,O_1,S^{\prime },N$. Then we have $$\begin{array}{c}S{S}^{\prime }\cdot S{O}_1=S{D}^2=S{X}^2\end{array}\text{\hspace{1em}(1)}$$ Next, we show that $S$ and $S^{\prime }$ are symmetric with respect to $XY$. By the Sine Law, we have $$\frac{S{S}^{\prime }}{\sin \angle SQ{S}^{\prime }}=\frac{SQ}{\sin \angle S{S}^{\prime }Q}=\frac{SQ}{\sin \angle SDQ}=\frac{SQ}{\sin \angle SAQ}=\frac{SA}{\sin \angle SQA}.$$ It follows that $$S{S}^{\prime }=SA\cdot \frac{\sin \angle SQ{S}^{\prime }}{\sin \angle SQA}=SA\cdot \frac{\sin \angle HOA}{\sin \angle OHA}=SA\cdot \frac{AH}{AO}=SA\cdot 2\cos A,$$ which is twice the distance from $S$ to $XY$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $PM$ if and only if $\angle SAC<\angle OAC$ if and only if $\angle YXA>\angle CBA$. This shows $S$ and $O_1$ lie on different sides of $XY$. As $S^{\prime }$ lies on ray $S{O}_1$, it follows that $S$ and $S^{\prime }$ cannot lie on the same side of $XY$. Therefore, $S$ and $S^{\prime }$ are symmetric with respect to $XY$. Let $d$ be the diameter of the circumcircle of triangle $XSY$. As $S{S}^{\prime }$ is twice the distance from $S$ to $XY$ and $SX=SY$, we have $S{S}^{\prime }=2\frac{S{X}^2}{d}$. It follows from (1) that $d=2S{O}_1$. As $S{O}_1$ is the perpendicular bisector of $XY$, point $O_1$ is the circumcentre of triangle $XSY$.

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Alternative solution.

Denote the orthocentre and circumcentre of triangle $ABC$ by $H$ and $O$ respectively. Let $O_1$ be the circumcentre of triangle $XSY$. Consider two other possible positions of $S$. We name them $S^{\prime }$ and $S^{\prime \prime }$ and define the analogous points $X^{\prime },Y^{\prime },O_1^{\prime },X^{\prime \prime },Y^{\prime \prime }{O}_1^{\prime \prime }$ accordingly. Note that $S,S^{\prime },S^{\prime \prime }$ lie on the perpendicular bisector of $AD$. As $X{X}^{\prime }$ and $Y{Y}^{\prime }$ meet at $A$ and the circumcircles of triangles $AXY$ and $A{X}^{\prime }{Y}^{\prime }$ meet at $D$, there is a spiral similarity with centre $D$ mapping $XY$ to $X^{\prime }{Y}^{\prime }$. We find that $$\measuredangle SXY=\frac{\pi }{2}-\measuredangle YAX=\frac{\pi }{2}-\measuredangle Y^{\prime }A{X}^{\prime }=\measuredangle S^{\prime }{X}^{\prime }{Y}^{\prime }$$ and similarly $\measuredangle SYX=\measuredangle S^{\prime }{Y}^{\prime }{X}^{\prime }$. This shows triangles $SXY$ and $S^{\prime }{X}^{\prime }{Y}^{\prime }$ are directly similar. Then the spiral similarity with centre $D$ takes points $S,X,Y,O_1$ to $S^{\prime },X^{\prime },Y^{\prime },O_1^{\prime }$. Similarly, there is a spiral similarity with centre $D$ mapping $S,X,Y,O_1$ to $S^{\prime \prime },X^{\prime \prime },Y^{\prime \prime },O_1^{\prime \prime }$. From these, we see that there is a spiral similarity taking the corresponding points $S,S^{\prime },S^{\prime \prime }$ to points $O_1,O_1^{\prime },O_1^{\prime \prime }$. In particular, $O_1,O_1^{\prime },O_1^{\prime \prime }$ are collinear.



It now suffices to show that $O_1$ lies on the perpendicular bisector of $PM$ for two special cases. Firstly, we take $S$ to be the midpoint of $AH$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $XSY$ is the nine-point circle of triangle $ABC$. Then $O_1$ is the nine-point centre and $O_{1}P=O_{1}M$. Indeed, $P$ and $M$ also lies on the nine-point circle. Secondly, we take $S^{\prime }$ to be the midpoint of $AO$. Then $X^{\prime }$ and $Y^{\prime }$ are the midpoints of $AB$ and $AC$ respectively. Then $X^{\prime }{Y}^{\prime }\parallel BC$. Clearly, $S^{\prime }$ lies on the perpendicular bisector of $PM$. This shows the perpendicular bisectors of $X^{\prime }{Y}^{\prime }$ and $PM$ coincide. Hence, we must have $O_1^{\prime }P=O_1^{\prime }M$.