67th NMO Selection Tests for JBMO

a) Let $O$ be the circumcenter of $ABCD$. It is well known that the midpoints of the sides of a quadrilateral $ABCD$ are the vertices of a parallelogram, hence the line segments joining the midpoints of two opposite sides have the same midpoint, $G$. The perpendicular lines from $O$ to $AB$ and $CD$ pass through the midpoints of these sides, therefore the perpendiculars dropped from $O$ and from the midpoints of two opposite sides onto their opposite side form a parallelogram whose center is $G$. It follows that the two perpendicular lines dropped from the midpoints of two opposite sides onto their opposite side intersect at the reflection of $O$ in $G$. The other two perpendiculars intersect at the same point.

b) We assume the angle $AXB$ to be acute, the other case being similar. The quadrilaterals $AB{A}^{\prime }{B}^{\prime }$, $CD{C}^{\prime }{D}^{\prime }$ and $ABCD$ being cyclic, we have $\angle XD{C}^{\prime }\equiv \angle XDC\equiv \angle XAB\equiv \angle X{A}^{\prime }{B}^{\prime }$, hence $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is cyclic.

c) Let $H_1$ and $H_2$ be the orthocenters of triangles $XAB$, and $XCD$, respectively. If $O^{\prime \prime }$ is the midpoint of $[H_1{H}_2]$, as $O^{\prime \prime }$ belongs to the midsegment of the trapezoid $H_1{B}^{\prime }{H}_2{D}^{\prime }$, $O^{\prime \prime }$ belongs to the perpendicular bisector of the line segment $[B^{\prime }{D}^{\prime }]$. Similarly, $O^{\prime \prime }$ belongs to the midsegment of the trapezoid $A^{\prime }{H}_1{C}^{\prime }{H}_2$, hence to the perpendicular bisector of $[A^{\prime }{C}^{\prime }]$. As $A^{\prime }{C}^{\prime }$ and $B^{\prime }{D}^{\prime }$ are not parallel, it follows that $O^{\prime \prime }$ is precisely the circumcenter of $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$, i.e. $O^{\prime \prime }$ coincides with $O^{\prime }$.

d) We have $\angle A^{\prime }{B}^{\prime }X\equiv \angle ABX\equiv \angle DCX$, hence $A^{\prime }{B}^{\prime }\parallel CD$. If $N$ is the midpoint of $[AB]$, then $N{A}^{\prime }=N{B}^{\prime }$, hence $N$ belongs to the perpendicular bisector of $[A^{\prime }{B}^{\prime }]$. But so does $O^{\prime }$, therefore it follows that $N{O}^{\prime }\perp CD$. Similarly, $O^{\prime }$ belongs to the perpendicular dropped from the midpoint of $[CD]$ on $AB$, hence $O^{\prime }$ is the Mathot point of the quadrilateral.