4. Circle $\omega$ with radius 4 and center $O$ is inscribed in an acute-angled triangle $E F Q$ and touches its sides $F Q$ and $E Q$ at points $M$ and $P$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{65}}{2}$ and center $T$ is circumscribed around triangle $P Q M$.

a) Find $O Q$.

b) Suppose it is additionally known that the ratio of the area of triangle $F T E$ to the area of triangle $E F Q$ is $\frac{2}{3}$. Find the length of the angle bisector $Q A$ of triangle $E F Q$, as well as its area.
Answer: a) $O Q=\sqrt{65}$, b) $Q A=\frac{3 \sqrt{65}}{2}, S=84$.

Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $O M Q$ and $O P Q$ are right angles, i.e., from points $M$ and $P$, the segment $O Q$ is seen at a right angle. Therefore, the circle constructed on the segment $O Q$ as a diameter also passes through points $M$ and $P$, meaning this is the circle $\Omega$. The segment $O Q$ is equal to twice the radius of this circle, i.e., $O Q=\sqrt{65}$.

b) Let the point of tangency of the circle with side $E F$ be $B$, and the height of the triangle from vertex $Q$ be $Q H$. Let $\angle A Q H=\gamma$. The center of the inscribed circle is the point of intersection of the angle bisectors of the triangle, so $O \in Q A$. From triangle $P O Q$, we find that $P Q=\sqrt{O Q^{2}-O P^{2}}=7$.

Since triangles $E F T$ and $E F Q$ have the same base $E F$, their areas are proportional to the heights drawn to this base, and the ratio of these heights is $T A: Q A$. From this, we get $\frac{2}{3}=\frac{T A}{Q A}=\frac{Q A-Q T}{Q A}$, $2 Q A=3 Q A-3 Q T, Q A=3 Q T=3 \cdot \frac{\sqrt{65}}{2}=\frac{3 \sqrt{65}}{2}, A O=A Q-O Q=\frac{\sqrt{65}}{2}$.

Since $O B \| Q H, \angle A O B=\angle A Q H=\gamma$; from triangle $A O B$, we get $\cos \gamma=\frac{B O}{A O}=\frac{8}{\sqrt{65}}$. From triangle $A Q H$, we find that $Q H=Q A \cos \gamma=12$.

We can express the area of triangle $E F Q$ in two ways. On one hand, $S_{E F Q}=\frac{1}{2} \cdot E F \cdot Q H=6 \cdot E F$. On the other hand, $S_{E F Q}=p r$, where $r=4$ is the radius of the inscribed circle, and $p$ is the semiperimeter of the triangle. Since $F B=F M, M Q=Q P, B E=E P$, we get $p=F B+B E+P Q=E F+P Q=E F+7, S_{E F Q}=4(E F+7)$. We obtain the equation $6 \cdot E F=4(E F+7)$, from which $E F=14, S_{C M F}=6 \cdot E F=84$.