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 $BAC$.
To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see the figure). We get that $S_{A B M} = S_{B C K}$. Therefore,

${ }^{A M} /_{A C} \cdot S_{A B C}={ }^{B K} /{ }_{A B} \cdot S_{A B C}$, that is, $B K: A B = A M: A C$. 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, $B K: K A = A M: M C$. (*)

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-33.jpg?height=503&width=606&top_left_y=557&top_left_x=723)

Notice now that $\angle B P K = \angle K P A = \angle A P M = \angle M P C = 60^{\circ}$. 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, $B K: K A = B P: P A$ and $A M: M C = A P: P C$. Considering the equality (*), we get $B P: P A = A P: P C$.

Thus, triangles BPA and APC are similar (by two sides and the angle between them). Therefore, $\angle P A C = \angle P B A$. Hence,

$\angle B A C = \angle B A P + \angle P A C = \angle B A P + \angle P B A = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

## Answer

$60^{\circ}$.