BMO 2019 Shortlist

Since $CH=CZ$ we have $OC\perp HZ$. So from the cyclic quadrilateral $SODI$ we get $$CS\cdot CO=CI\cdot CD.(1)$$ Figure 9: G9 We draw the perpendicular line $(v)$ to $HC$ at the point $H$. Let $J$ be the intersection point of lines $(v)$ and $CO$. Then $CJ$ is diameter of the circle $(O,OA)$ and $$CJ=2CO.(2)$$ From the right triangle $JHC$ we have $$H{C}^2=CS\cdot CJ.(3)$$

Therefore, from (1), (2) and (3) we get $$CS\cdot \frac{1}{2}CJ=CI\cdot CD\text{or}H{C}^2=2CI\cdot CD.(3)$$ However $HC=CD$ and thus $CD=2CI$. Thus, $I$ is the midpoint of the segment $CD$. Nevertheless, $LM\parallel OK$, so the points $L,M$ are the midpoints of the sides $CO$ and $CK$ respectively. Therefore, the circumcircle ($k$) of the triangle $LMD$ is the Euler circle of the $COK$ and thus it passes through the point $S$. We have $QS=QU$ and from the right triangles $OSK,OUK$ we get $PS=PU=\frac{OK}{2}$. Therefore, the points $P,Q$ are located on the perpendicular bisector of the segment $SU$. Now, we conclude that $SU\parallel TX$, because $QP\perp (e_3)$. Similarly, we prove that $DU\parallel RT$ and $SD\parallel RX$. Since the triangles $SUD$ and $XTR$ are homothetic we get that the lines $RD,TU,XS$ are concurrent at the center $M$ of homothety. The points $I$ and $Q$ are the incenters of homothetic triangles $SUD$ and $XTR$, respectively. Thus, the line $IQ$ passes through the point $M$. □