HMMT February, 2023

Solution:

Extend $AM$ to meet the circumcircle of $\triangle ABC$ at $X$. Then, we have $\triangle ABM \sim \triangle CXM$, which implies that $\frac{CX}{CM} = \frac{AB}{AM}$. Using the condition $AB \cdot BC = AM$, we get that $CX = \frac{1}{2}$.

Now, the key observation is that $\triangle ANB \sim \triangle ACX$. Thus, if we let $Y$ be the reflection of $X$ across point $C$, we get that $\angle AYC = 90^{\circ}$. Thus, Pythagorean's theorem gives $AY = \sqrt{AC^2 - CY^2} = \sqrt{15/4}$ and $AX^2 = \sqrt{AY^2 + XY^2} = \sqrt{19/4}$.

Finally, note that the distance from $B$ and $C$ to line $AM$ are equal. Thus, let $H$ be the foot from $C$ to $AM$. Then, from $\triangle XCH \sim \triangle XAY$, we get that
$$CH = AY \cdot \frac{CX}{AX} = \frac{\sqrt{15}}{2} \cdot \frac{1/2}{\sqrt{19}/2} = \frac{\sqrt{285}}{38}$$