Team Selection Test for IMO 2023

First observe that $\angle MHI=\angle MIO$ and $\angle MIH=\angle MOI$, hence the similarity $MIH\sim MOI$; thus $MI/MO=IH/IO$. Now let $N$ be the midpoint of the segment $[OH]$ (so $N$ is the center of the 9-point circle), and let $S$ be the reflection of $I$ over $N$. Thus $SOIH$ is a parallelogram; therefore $\angle IMO=\angle HIO=\angle IHS$ and $IM/MO=IH/IO=IH/HS$, which implies the similarity $IMO\sim IHS$. Consequently $$MO=\frac{HS\cdot IO}{IS}=\frac{I{O}^2}{2\cdot IN}=\frac{R(R-2r)}{2(R/2-r)}=R.$$ Note: The equality $IN=R/2-r$ is Feuerbach's theorem.