HMMT Spring, 2021

Solution:

As $\angle PBD = \angle PCD = \angle PAB$, $DB$ is tangent to $(ABP)$. As $DA = DB$, $DA$ is also tangent to $(ABP)$. Let $CB$ intersect $(ABP)$ again at $X \neq B$; it follows that $XD$ is the $X$-symmedian of $\triangle AXB$. As $\angle AXC = \angle DAB = \angle ADC$, $X$ also lies on $(ACD)$. Therefore $\angle PXB = \angle PAB = \angle PCD = \angle AXD$, so $XP$ is the median of $\triangle AXB$, i.e. $XP$ passes through $M$.

Now we have $\triangle MPA \sim \triangle MBX$ and $\triangle ABX \sim \triangle BCD$. Therefore
$$\frac{MP}{AP} = \frac{MB}{XB} = \frac{AB}{2XB} = \frac{BC}{2DC} \Longrightarrow MP = \frac{BC \cdot AP}{2DC} = \frac{27 \cdot 10}{2 \cdot 25} = \frac{27}{5}$$

Solution:

Let $E$ be the point such that $\square BDCE$ is a parallelogram. Then $\square ADCE$ is an isosceles trapezoid, therefore $\angle PAB = \angle CAB = \angle BED = \angle CDE$; this angle is also equal to $\angle PBD = \angle PCD$. Now, $\angle ABP = \angle ABD - \angle PBD = \angle BEC - \angle BED = \angle DEC$, therefore $\triangle PAB \sim \triangle CDE$. Let $F$ be the midpoint of $DE$, which lies on $BC$ because $\square BDCE$ is a parallelogram. It follows that $(\triangle PAB, M) \sim (\triangle CDE, F)$, therefore
$$\frac{MP}{AP} = \frac{FC}{DC} = \frac{BC}{2DC}$$
and again this gives $MP = \frac{27}{5}$.