Korean Mathematical Olympiad Final Round

Let $M$ be the midpoint of the side $BC$. First, we want to show that $M$ is on the circumcircle of the triangle $DPQ$. For the intersection point $X$ of $PQ$ and $BC$, Menelaus theorem guarantees $$\frac{BX}{XC}=\frac{CE}{EA}=\frac{AF}{FB}=1.$$ Since $AF=AE$, $BD=BF$, $CD=CE$ and $BX\cdot CD=XC\cdot BD$, from which we get $2XB\cdot XC=(XB+XC)\cdot XD$ by replacing $BD$ and $CD$ with $XD-XB$ and $XC-XD$, respectively. It follows that $XB\cdot XC=XM\cdot XD$, as $\frac{XB+XC}{2}=XM$. Hence the point $M$ lies on the circumcircle of the triangle $DPQ$.

Let $O^{\prime }$ be the midpoint of the line segment $O_1{O}_2$. It suffices to show that $O^{\prime }$ is the circumcenter of the triangle $DPQ$. first, we will consider the case where $AB\ne AC$. In this case, we will prove it by showing that $O^{\prime }$ is on both the perpendicular bisector of $PQ$ and the perpendicular bisector of $DM$. In fact, since $AI$ is perpendicular to both $PQ$ and $O_1{O}_2$, $PQ$ and $O_1{O}_2$ are parallel to each other. It follows that $O^{\prime }$ is on the perpendicular bisector of $PQ$. Let $I_A,I_B$ and $I_C$ be the centers of excircles each of which is tangent to $BC$, $AC$ and $AB$, respectively. It is easy to see that the circumcircle of the triangle $ABC$ is the same as the nine point circle of the triangle $I_A{I}_B{I}_C$. Thus, if we let $O_3$ be the circumcenter of the triangle $IBC$ and let $S$ be the intersection point of $O{O}_3$ and $I_B{I}_C$, the point $S$ is the midpoint of the line segment $I_B{I}_C$. Note that the dilatation with center $I$ and ratio 2 sends the triangle $O_1{O}_2{O}_3$ to the triangle $I_C{I}_B{I}_A$, which implies that $O^{\prime }$ is the midpoint of the line segment $IS$. Therefore we can conclude that $O^{\prime }$ is on the perpendicular bisector of $DM$, as both $ID$ and $SM$ are perpendicular to $BC$.

Next, we will consider the case where $AB=AC$. Note that $O^{\prime }$ is the midpoint of $AI$ in this case. Since $\angle AEI=90^{\circ }=\angle AFI$, the points $A,F,I,E$ are on the circle with the center $O^{\prime }$. Let $R$ be the radius of this circle. Then $O^{\prime }{D}^2-R^2=DI\cdot DA=D{A}^2-AI\cdot DA$ holds. The points $B,D,I,F$ being concyclic, we can see $AI\cdot DA=AF\cdot AB$. Since $D{A}^2=A{B}^2-B{D}^2$, $$O^{\prime }{D}^2-R^2=(A{B}^2-B{D}^2)-AF\cdot AB=A{B}^2-B{F}^2-AF\cdot AB=BF\cdot AF.$$ Similarly, we can get $O^{\prime }{P}^2-R^2=PE\cdot PF$. But $BF\cdot AF=QF\cdot PF=PE\cdot PF$, so $$O^{\prime }{D}^2-R^2=O^{\prime }{P}^2-R^2.$$ It easily follows that $O^{\prime }D=O^{\prime }P$. In a similar manner, we can show $O^{\prime }D=O^{\prime }Q$. Since $O^{\prime }P=O^{\prime }D=O^{\prime }Q$, $O^{\prime }$ is the circumcenter of the triangle $DPQ$.