10.5. On side $AB$ of triangle $ABC$, point $K$ is marked, and on side $AC$, point $M$ is marked. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find angle $BAC$.
Answer: $60^{\circ}$.

Solution. To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see Fig. 10.5). We get that $S_{A B M}=S_{B C K}$. Therefore, $\frac{1}{2} B C \cdot B K \sin \angle B = \frac{1}{2} A B \cdot A M \sin \angle A$. Then $\frac{B K}{A M} = \frac{A B \sin \angle A}{B C \sin \angle B}$. From the Law of Sines: $\frac{\sin \angle A}{\sin \angle B} = \frac{B C}{A C}$, so $\frac{B K}{A M} = \frac{A B}{A C} \Leftrightarrow \frac{B K}{A B} = \frac{A M}{A C}$. Thus, points $K$ and $M$ divide segments $B A$ and $A C$ in the same ratio, counting from vertices $B$ and $A$ respectively, that is, $\frac{B K}{K A} = \frac{A M}{M C}$ (*).

![](https://cdn.mathpix.com/cropped/2024_05_06_c991e5a02c6db2b67b15g-2.jpg?height=363&width=440&top_left_y=1166&top_left_x=1597)

Fig. 10.5

Notice now that $\angle B P K = \angle K P A = \angle A P M = \angle M P C = 60^{\circ}$ (angles adjacent to the given angles of $120^{\circ}$). Therefore, $P K$ and $P M$ are the angle bisectors of triangles $A P B$ and $A P C$ respectively. By the property of the angle bisector of a triangle, we get: $\frac{B K}{K A} = \frac{B P}{P A}$ and $\frac{A M}{M C} = \frac{A P}{P C}$. Then, taking into account the equality (*), we get that $\frac{B P}{P A} = \frac{A P}{P C}$.

Moreover, $\angle B P A = \angle A P C = 120^{\circ}$. Thus, triangles $B P A$ and $A P C$ are similar (by two sides and the angle between them). Therefore, $\angle P A C = \angle P B A$. Hence, $\angle B A C = \angle B A P + \angle P A C = \angle B A P + \angle P B A = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

## Grading Criteria:

## + a complete and justified solution  干 only that points K and M divide the sides in the same ratio  - only the answer is provided