Taiwan IMO Selection Camp

Let $A'$ be the reflection of $A$ with respect to $M$, and let $I'$ be the reflection of $I$ with respect to $P$. Then $A'$ lies on $BC$, and $R$ is the circumcenter of $\triangle AII'$, so $\angle AMR$ is a right angle if and only if $A, I, A', I'$ are concyclic.

Below we prove that $\triangle AIP \sim \triangle IA'P$ (where $\sim$ denotes direct similarity): Take $J$ such that $AIA'J$ is a parallelogram, take $P'$ such that $\triangle AIP' \sim \triangle JIA'$, then
$$\triangle IA'P' \sim \triangle IJA' \sim \triangle JIA' \sim \triangle AIP',$$
and from
$$\angle IAP' = \angle JIA = \angle QIA = \angle IAQ,$$
we know that $A, Q, P'$ are collinear. Take $Y$ such that $\triangle AIP' \sim \triangle YBP'$; from $AYIP' \sim IBA'P'$ we obtain
$$\angle AYI = \angle IBA' = \angle ABI,$$
that is, $A, Y, B, I$ are concyclic. Therefore
$$\begin{aligned} \angle CBP' &= \angle IYP' = \angle BYP' - \angle BYI = \angle IAP' - \angle BAI \\ &= \angle IAP' - \angle IAC = \angle CAP', \end{aligned}$$
hence $A, B, C, P'$ are concyclic, so $P'$ is the other intersection point of $AQ$ with $\odot(ABC)$, that is, $P = P'$.

From $\triangle AIP \sim \triangle IA'P$ we easily obtain $\triangle AI'P \sim \triangle I'A'P$, so
$$\angle IAI' = \angle IAP + \angle PAI' = \angle A'IP + \angle PI'A' = \angle IA'I'$$
that is, $A, I, A', I'$ are concyclic, and thus the original proposition is proved.