Team selection tests for IMO 2018

Denote $(I_a),(I_b),(I_c)$ as the ex-circles with respect to angles $A,B,C$ of triangle $ABC$. It is easy to check that ${\mathcal{P}}_{M/(I)}={\mathcal{P}}_{M/(I_a)}$ and ${\mathcal{P}}_{M/(I_b)}={\mathcal{P}}_{M/(I_c)}$. Let $O,K$ be the circumcenters of triangles $ABC$ and $DEF$.

Hence, $d_1$ is the radical axis of $(I),(I_a)$ since $d_1\perp I{I}_a$. Similarly for $d_2,d_3$, which implies that $D=d_2\cap d_3$ is the radical center of the three circles $(I),(I_b),(I_c)$. Thus, $MD$ is the radical axis of $(I_b),(I_c)$.

So we have $MD\perp I_b{I}_c$, but $I_b{I}_c\perp AI$, $EF\perp AI$, then $MD\perp EF$ leads to $MD$ being the altitude in triangle $DEF$.

Similarly for $NE,PF$, so the three lines $MD,NE,PF$ concur at point $T$, which is the orthocenter of $DEF$. The two triangles $ABC$ and $DEF$ share the same nine-point circle $(MNP)$, so $OH$ and $KT$ share the midpoint, which is the circumcenter of triangle $(MNP)$. This implies that $\overrightarrow{HK}=\overrightarrow{TO}$.

On the other hand, it is easy to check that $T$ is also the incenter of triangle $MNP$. Then consider the homothety with center $G$ as the centroid of triangle $ABC$, ratio $-2$, then $\mathcal{H}:M\to A,N\to B,P\to C$ implies that $\mathcal{H}:T\to I,O\to H$. Hence, $\overrightarrow{HI}=2\overrightarrow{TO}$.

Combine with the previous equality, we have $\overrightarrow{HI}=2\overrightarrow{HK}$, so $K$ is the midpoint of the segment $HI$. $\square$