4. On the sides $AB$ and $AC$ of triangle $ABC$, points $D$ and $E$ are chosen such that the area of triangle $ADE$ is 0.5. The incircle of quadrilateral $BDEC$ touches side $AB$ at point $K$, and $AK=3$. Find the tangent of angle $BAC$, if a circle can be circumscribed around quadrilateral $BDEC$, and $BC=15$. (20 points)
Solution: Let $O_{1}$ be the center of the incircle of triangle $ABC$, $R$ its radius, $O_{2}$ the center of the incircle of triangle $ADE$, $r$ its radius. Denote $p_{1}$ as the semiperimeter of triangle $ABC$, and $p_{2}$ as the semiperimeter of triangle $ADE$. Then $A K=p_{2}, A K=p_{1}-BC$. Indeed, by the properties of tangents to a circle, we have (see figure)

$$
\begin{gathered}
2 A K=A K+A H=A D+D K+E H+A E=A D+(D M+M E)+A E=2 p_{2} \\
2 A K=(A B-B K)+(A C-C H)=A B+A C+(B F+C F)-2(B F+C F)=2\left(p_{1}-B C\right) .
\end{gathered}
$$

Then $r=S_{\text {ADE }} / p_{2}=S_{\text {ADE }} / A K=1 / 6$.

Since a circle can be circumscribed around quadrilateral $BDEC$, $\angle ABC + \angle DEC = 180^{\circ}$, and $\angle AED = \angle ABC$. Therefore, triangles $ADE$ and $ACB$ are similar, and $\frac{r}{R}=\frac{p_{2}}{p_{1}}, R=\frac{r(A K+B C)}{A K}=\frac{3+15}{6 \cdot 3}=1$.

Let $\angle BAC = \alpha$. Then $\operatorname{tg}(\alpha / 2) = R / A K = 1 / 3$, and $\operatorname{tg} \alpha = 2 \operatorname{tg}(\alpha / 2) / \left(1 - \operatorname{tg}^{2}(\alpha / 2)\right) = 3 / 4$.

Answer: $3 / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_52b2f7cff5dc0b15c76dg-04.jpg?height=653&width=920&top_left_y=1284&top_left_x=274)