HMMT February, 2025

Solution:

Let $X$ and $Y$ be the bottom and top point on $\omega$ (i.e., the tangents of $X$ and $Y$ to $\omega$ are parallel to $BC$, and $Y$ and $A$ lie on the same side of $BC$). Note that $A$, $P$, $D$, and $X$ are collinear by homothety between the incircle and $\omega$. The key claim is the following.

Claim 1. Line $IY$ is tangent to $\omega$.

Proof. Let the line through $I$ parallel to $BC$ meet $AB$ and $AC$ at $B'$, $C'$, respectively. Notice that $B'L$ is the perpendicular bisector of $BI$, so $B'L$ externally bisects $\angle AB'C'$. Similarly, $C'L$ externally bisects $\angle AC'B'$. Hence, $L$ is the excenter of $\triangle AB'C'$, which means that $B'C'$ is tangent to $\omega$. $\square$

Now, we note that $LY \perp BC$, so $L$, $Y$, and $M$ are collinear (on the perpendicular bisector of $BC$). Since $\angle YPX = 90^{\circ}$ and $\angle YMD = 90^{\circ}$, $PDMY$ is cyclic. However, $IYMD$ is a rectangle, so $IPDMY$ is a cyclic pentagon. Hence, $\angle IPM = \angle IDM = 90^{\circ}$.

Solution:

Let the incircle touch $AC$ and $AB$ at $E$ and $F$, respectively. Let $DI$ intersect $EF$ at $X$. Let $D'$ be the other intersection of $AD$ and the incircle.

Claim 2. $PM \parallel D'X$.

Proof. Consider the homothety at $A$ that sends $\omega$ to the incircle. It sends $L$ to $I$ and $P$ to $D'$. Furthermore, it's well-known that $X$ lies on $AM$. Because $IX \parallel LM$, we also have that the homothety sends $M$ to $X$. These facts imply that $D'X \parallel PM$. $\square$

Let $T$ be the antipode of $D$ on the incircle. Let $AT$ intersect the incircle again at $T'$. Since $X$ lies on the polar of $A$ with respect to the incircle, by Brocard's theorem, we have $D'$, $X$, and $T'$ are collinear. It is well-known that $AT \parallel IM$. Therefore, $\angle DPM = \angle DD'X = \angle DD'T' = \angle DTT' = \angle DIM$. Consequently, $IMDP$ is cyclic, and $\angle IPM = \angle IDM = 90^{\circ}$.

Solution:

Let $\omega$ be tangent to $AB$ and $AC$ at $E$ and $F$, respectively. Note that these are the feet of the altitudes from $L$ to $AB$ and $AC$, and $L$ lies on the circumcircle of $\triangle ABC$ by Fact 5. As $M$ is clearly the foot from $L$ to $BC$, it follows that $E$, $F$, and $M$ are collinear on the Simson Line of $L$ with respect to $\triangle ABC$.

Lastly, we want $P$ to be on the circle with diameter $IM$. This circle intersects $EF$ again at the foot from $M$ to $AI$, which is the midpoint of $EF$. Let this point be $M'$. Consider the homothety sending the incircle to $\omega$. This clearly sends $D$ to the second intersection of $AD$ and $\omega$, which is $P'$, and it sends $I$ to $L$. Note that $AP \cdot AP' = AE^2 = AM \cdot AL$, as the circle with diameter $LE$ is tangent to $AE$. Thus, $PP'M'L$ is cyclic. Since $ID \parallel LP'$, $I$ lies on $M'L$, and $D$ lies on $PP'$. By Reim's, we also have $PDM'I$ is cyclic. As $IM$ is a diameter of $(DM'I)$, we have $\angle IPM = 90^{\circ}$.