32. [23] Let $A B C$ be a triangle with $\angle B A C>90^{\circ}$. Let $D$ be the foot of the perpendicular from $A$ to side $B C$. Let $M$ and $N$ be the midpoints of segments $B C$ and $B D$, respectively. Suppose that $A C=2$, $\angle B A N=\angle M A C$, and $A B \cdot B C=A M$. Compute the distance from $B$ to line $A M$.
Answer:
$\frac{\sqrt{285}}{38}$
Solution:
Extend $A M$ to meet the circumcircle of $\triangle A B C$ at $X$. Then, we have $\triangle A B M \sim \triangle C X M$, which implies that $\frac{C X}{C M}=\frac{A B}{A M}$. Using the condition $A B \cdot B C=A M$, we get that $C X=\frac{1}{2}$.
Now, the key observation is that $\triangle A N B \sim \triangle A C X$. Thus, if we let $Y$ be the reflection of $X$ across point $C$, we get that $\angle A Y C=90^{\circ}$. Thus, Pythagorean's theorem gives $A Y=\sqrt{A C^{2}-C Y^{2}}=\sqrt{15 / 4}$ and $A X^{2}=\sqrt{A Y^{2}+X Y^{2}}=\sqrt{19 / 4}$.
Finally, note that the distance from $B$ and $C$ to line $A M$ are equal. Thus, let $H$ be the foot from $C$ to $A M$. Then, from $\triangle X C H \sim \triangle X A Y$, we get that
$$
C H=A Y \cdot \frac{C X}{A X}=\frac{\sqrt{15}}{2} \cdot \frac{1 / 2}{\sqrt{19} / 2}=\frac{\sqrt{285}}{38} .
$$