7.1. In triangle $A B C$, the median $A M$ is drawn, point $O$ is the center of the circumscribed circle around it, and point $Q$ is the center of the inscribed circle in it. Segments $A M$ and $O Q$ intersect at point $S$, and $2 \frac{O S}{M S}=3 \sqrt{3} \frac{Q S}{A S}$. Find the sum of the sines of the measures of angles $A B C$ and $A C B$, given that $\angle B A C=\frac{\pi}{3}$. Round your answer to the nearest hundredth if necessary.
Answer: 1.13 (exact value: 9/8).

Solution. Point $Q$ is the intersection point of the angle bisectors of triangle $ABC$. Draw the angle bisector of $\angle BAC$, and denote the intersection point of this bisector with the circumcircle of triangle $ABC$ by $L$. Connect points $C$ and $Q$.

Since $\angle BAL = \angle CAL = \pi / 6$, arcs $BL$ and $CL$ are equal and have measures $\pi / 3$. Therefore, $BL = CL$, making triangle $BLC$ isosceles. Thus, its median $LM$ is also its altitude, meaning that line $LM$ is the perpendicular bisector of side $BC$. Therefore, point $O$ also lies on line $LM$, and since arc $BLC$ is less than $\pi$, points $O$ and $L$ lie in different half-planes relative to line $BC$.

Applying Menelaus' theorem twice (in triangle $AML$ with transversal $QS$ and in triangle $LOQ$ with transversal $MS$), we have

$$
\frac{MS}{AS} \cdot \frac{AQ}{QL} \cdot \frac{LO}{OM} = 1; \quad \frac{QS}{OS} \cdot \frac{OM}{LM} \cdot \frac{AL}{AQ} = 1
$$

Multiplying these two relations, we get

$$
\frac{MS}{AS} \cdot \frac{QS}{OS} \cdot \frac{AL}{QL} \cdot \frac{LO}{LM} = 1
$$

$(\star)$

From the given conditions, it follows that $\frac{MS}{AS} \cdot \frac{QS}{OS} = \frac{2}{3 \sqrt{3}}$. Additionally, note that triangle $LOC$ is isosceles (since segments $OL$ and $OC$ are radii of the circumcircle of triangle $ABC$). However, $\angle LOC = \pi / 3$, so it is equilateral; $CM$ is its height and thus its median, so $LO: LM = 2$. With this, equation $(\star)$ becomes $\frac{4}{3 \sqrt{3}} \cdot \frac{AL}{QL} = 1$, from which $\frac{AL}{QL} = \frac{3 \sqrt{3}}{4}$.

Let the measure of angle $ACB$ be $2\gamma$, then $\angle ABC = 2\pi / 3 - 2\gamma$. By the properties of inscribed angles, $\angle ALC = \angle ABC = 2\pi / 3 - 2\gamma$, $\angle BCL = \angle BAL = \pi / 6$, and $\angle ACL = \angle ACB + \angle BCL = \pi / 6 + 2\gamma$. After this, we find

$$
\begin{gathered}
\angle QCL = \angle QCB + \angle BCL = \frac{\pi}{6} + \gamma \\
\angle LQC = \pi - \angle QCL - \angle ALC = \frac{\pi}{6} + \gamma
\end{gathered}
$$

Thus, triangle $LQC$ is isosceles, $QL = CL$. Using the Law of Sines for triangle $ALC$, we have

$$
\frac{AL}{QL} = \frac{AL}{CL} = \frac{\sin \angle ACL}{\sin \angle CAL} = \frac{\sin (2\gamma + \pi / 6)}{1 / 2} = 2 \sin \left(2\gamma + \frac{\pi}{6}\right)
$$

from which we find $\sin (2\gamma + \pi / 6) = \frac{1}{2} \cdot \frac{AL}{QL} = \frac{3 \sqrt{3}}{8}$. Therefore,

$$
\sin \angle ABC + \sin \angle ACB = \sin 2\gamma + \sin \left(\frac{2\pi}{3} - 2\gamma\right) = 2 \sin \frac{\pi}{3} \cos \left(\frac{\pi}{3} - 2\gamma\right) = \sqrt{3} \sin \left(2\gamma + \frac{\pi}{6}\right) = \frac{9}{8}
$$