Baltic Way 1996 · Problem 2

Let $r$ and $s$ be the radii of the half-circles with diameters $A P$ and $B P$. Then we have

$$39 \pi=\frac{\pi}{2}\left((r+s)^{2}-r^{2}-s^{2}\right)-9 \pi$$

hence $r s=48$. Let $M$ be the midpoint of the diameter $A B, N$ be the midpoint of $P B, O$ be the centre of the circle $C$, and let $F$ be the orthogonal projection of $O$ on $A B$. Since the radius of $C$ is 3 , we have $|M O|=r+s-3,|M F|=r-s+3,|O N|=s+3$, and $|F N|=s-3$.

Applying the Pythagorean theorem to the triangles $M F O$ and $N F O$ yields

$$(r+s-3)^{2}-(r-s+3)^{2}=|O F|^{2}=(s+3)^{2}-(s-3)^{2},$$

which implies $r(s-3)=3 s$, so that $3(r+s)=r s=48$. Hence $|A B|=2(r+s)=32$.