5. (7 points) On side $AB$ of triangle $ABC$, a point $K$ is marked, and on side $AC$ - a point $M$. 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 the angle $BAC$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_75d07bfe930f434a7986g-2.jpg?height=420&width=508&top_left_y=1732&top_left_x=840)

Add the area of triangle $B P K$ to both sides of the equality $S_{\text {AKPM }}=S_{\text {BPC }}$:

$$
\begin{aligned}
& S_{A K P M}+S_{B P K}=S_{B P C}+S_{B P K}, \\
& S_{A B M}=S_{B C K} .
\end{aligned}
$$

Therefore, $\frac{1}{2} A B \cdot A M \cdot \sin \angle A=\frac{1}{2} B K \cdot B C \cdot \sin \angle B$.

Then $\frac{B K}{A M}=\frac{A B \cdot \sin \angle A}{B C \cdot \sin \angle B}$.

By the Law of Sines: $\quad \frac{B C}{\sin \angle A}=\frac{A C}{\sin \angle B}$. Thus, $\quad \frac{\sin \angle A}{\sin \angle B}=\frac{B C}{A C}$, $\frac{B K}{A M}=\frac{A B \cdot B C}{B C \cdot A C} \Leftrightarrow \frac{B K}{A B}=\frac{A M}{A C}$.

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, $\frac{B K}{K A}=\frac{A M}{M C}$.

$\angle B P K=\angle K P A=\angle A P M=\angle M P C=60^{\circ}$ (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 Angle Bisector Theorem, we get: $\frac{B K}{K A}=\frac{B P}{P A}$ and $\frac{A M}{M C}=\frac{A P}{P C}$. Considering equality (1), 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, $\angle P A C=\angle P B A$.

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

Answer. $60^{0}$.

Maximum points - 35.