4. Circle $\omega$ with radius 6 and center $O$ is inscribed in an acute-angled triangle $C F M$ and touches its sides $C M$ and $F M$ at points $P$ and $K$ respectively. Circle $\Omega$ with radius $\frac{5 \sqrt{13}}{2}$ and center $T$ is circumscribed around triangle $P K M$.

a) Find $O M$.

b) Suppose it is additionally known that the ratio of the area of triangle $C F T$ to the area of triangle $C F M$ is $\frac{5}{8}$. Find the length of the angle bisector $M A$ of triangle $C F M$, as well as its area.
Answer: a) $O M=5 \sqrt{13}$, b) $M A=\frac{20 \sqrt{13}}{3}, S=204$.

Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O K M$ and $O P M$ are right angles, i.e., from points $K$ and $P$, the segment $O M$ is seen at a right angle. Therefore, the circle constructed on the segment $O M$ as a diameter also passes through points $K$ and $P$, meaning this is circle $\Omega$. The segment $O M$ is equal to twice the radius of this circle, i.e., $O M=5 \sqrt{13}$.

b) Let the point of tangency of the circle with side $C F$ be $B$, and the height of the triangle from vertex $M$ be $M H$. Let $\angle A M H=\gamma$. The center of the inscribed circle is the point of intersection of the angle bisectors of the triangle, so $O \in M A$. From triangle $P O M$, we find that $P M=\sqrt{O M^{2}-O P^{2}}=17$.

Since triangles $C F T$ and $C F M$ share the base $C F$, their areas are in the ratio of the heights drawn to this base, and the ratio of these heights is $T A: M A$. From this, we get $\frac{5}{8}=\frac{T A}{M A}=\frac{M A-M T}{M A}$, $5 M A=8 M A-8 M T, M A=\frac{8}{3} M T=\frac{8}{3} \cdot \frac{5 \sqrt{13}}{2}=\frac{20 \sqrt{13}}{3}, A O=A M-O M=\frac{5 \sqrt{13}}{3}$.

Since $O B \| A H, \angle A O B=\angle A M H=\gamma$; from triangle $A O B$, we get $\cos \gamma=\frac{B O}{A O}=\frac{18}{5 \sqrt{13}}$. From triangle $A M H$, we find that $M H=M A \cos \gamma=24$.

We express the area of triangle $C M F$ in two ways. On one hand, $S_{C M F}=\frac{1}{2} \cdot C F \cdot M H=12 \cdot C F$. On the other hand, $S_{C M F}=p r$, where $r=6$ is the radius of the inscribed circle, and $p$ is the semiperimeter of the triangle. Since $F B=F K, \quad M K=M P, \quad C B=C P, \quad$ we get $\quad p=F B+C B+M P=C F+M P=C F+17$, $S_{C M F}=6(C F+17)$. We obtain the equation $12 \cdot C F=6(C F+17)$, from which $C F=17, S_{C M F}=12 \cdot C F=204$.