4. Circle $\omega$ with radius 6 and center $O$ is inscribed in an acute triangle $ABC$ and touches its sides $AB$ and $BC$ at points $F$ and $T$ respectively. Circle $\Omega$ with radius $\frac{\sqrt{85}}{2}$ and center $M$ is circumscribed around triangle $BFT$.

a) Find $BO$.

b) Suppose it is additionally known that the ratio of the area of triangle $ACM$ to the area of triangle $ABC$ is $\frac{7}{10}$. Find the length of the angle bisector $BQ$ of triangle $ABC$, as well as its area.
Answer: a) $BO=\sqrt{85}$, b) $BQ=\frac{5 \sqrt{85}}{3}, S=210$.

Solution. a) The radius drawn to the point of tangency is perpendicular to the tangent, so angles $OFB$ and $OTB$ are right angles, i.e., from points $F$ and $T$, segment $OB$ is seen at a right angle. Therefore, the circle constructed on segment $OB$ as a diameter also passes through points $F$ and $T$, meaning this is circle $\Omega$. Segment $OB$ is equal to twice the radius of this circle, i.e., $OB=\sqrt{85}$.

b) Let the point of tangency of the circle with side $AC$ be $L$, and the height of the triangle from vertex $B$ be $BH$. Let $\angle QBH=\gamma$. The center of the inscribed circle is the point of intersection of the angle bisectors of the triangle, so $O \in BQ$. From triangle $BOT$, we find that $BT=\sqrt{OB^2-OT^2}=7$.

Since triangles $AMC$ and $ABC$ have the same base $AC$, their areas are proportional to the heights drawn to this base, and the ratio of these heights is equal to $MQ: BQ$. From this, we get $\frac{7}{10}=\frac{MQ}{BQ}=\frac{BQ-BM}{BQ}$, $7BQ=10BQ-10BM, BQ=\frac{10}{3}BM=\frac{10}{3} \cdot \frac{\sqrt{85}}{2}=\frac{5 \sqrt{85}}{3}, OQ=BQ-OB=\frac{2 \sqrt{85}}{3}$.

Since $OL \parallel BH, \angle LOQ=\angle HBQ=\gamma$; from triangle $LOQ$ we get that $\cos \gamma=\frac{LO}{QO}=\frac{9}{\sqrt{85}}$. From triangle $BQH$ we find that $BH=BQ \cos \gamma=15$.

We can express the area of triangle $ABC$ in two ways. On one hand, $S_{ABC}=\frac{1}{2} \cdot AC \cdot BH=\frac{15}{2} \cdot AC$. On the other hand, $S_{ABC}=pr$, where $r=6$ is the radius of the inscribed circle, and $p$ is the semiperimeter of the triangle. Since $AF=AL, BF=BT, CL=CT$, we get that $p=AL+LC+BT=AC+BT=AC+7, S_{ABC}=6(AC+7)$. We obtain the equation $7.5 \cdot AC=6(AC+7)$, from which $AC=28, S_{ABC}=7.5 \cdot AC=210$.