Solution.

$$
\begin{aligned}
& b=\sqrt{a(a+c)} \Leftrightarrow b^{2}=a^{2}+a c \Leftrightarrow b^{2}-a^{2}=a c \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} \beta-\sin ^{2} \alpha=\sin \alpha \cdot \sin \gamma
\end{aligned}
$$

The last transition is valid by the Law of Sines: $a=2 R \sin \alpha, b=2 R \sin \beta, c=2 R \sin \gamma$.

$(1) \Leftrightarrow(\sin \beta+\sin \alpha)(\sin \beta-\sin \alpha)=\sin \alpha \cdot \sin [\pi-(\alpha+\beta)] \Leftrightarrow$

$\Leftrightarrow 2 \sin \frac{\beta+\alpha}{2} \cos \frac{\beta-\alpha}{2} 2 \sin \frac{\beta-\alpha}{2} \cos \frac{\beta+\alpha}{2}=\sin \alpha \cdot \sin (\alpha+\beta) \Leftrightarrow$

$\Leftrightarrow \sin (\alpha+\beta) \sin (\beta-\alpha)=\sin \alpha \cdot \sin (\alpha+\beta) \Leftrightarrow$

$\Leftrightarrow \sin (\alpha+\beta)[\sin (\beta-\alpha)-\sin \alpha]=0 \Leftrightarrow \sin (\alpha+\beta) \sin \frac{\beta-2 \alpha}{2} \cos \frac{\beta}{2}=0 \Leftrightarrow$

$\Leftrightarrow\left[\begin{array}{c}\sin (\alpha+\beta)=0 \\ \sin \frac{\beta-2 \alpha}{2}=0 \\ \cos \frac{\beta}{2}=0\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left[\begin{array}{c}\alpha+\beta=\pi n, n \in \mathbb{Z} \\ \beta-2 \alpha=2 \pi k, k \in \mathbb{Z} \\ \beta=\pi+2 \pi m, m \in \mathbb{Z}\end{array}\right.$

From the obtained options, only $\beta=2 \alpha$ is possible in a triangle, when $k=0$.

From the condition $\alpha=15^{\circ}$ we get $\beta=30^{\circ}, \gamma=135^{\circ}$.

Then in triangle $\mathrm{ABC}$ we get

$a=2 R \sin 15^{\circ}=\frac{1}{2} R(\sqrt{6}-\sqrt{2}) ; b=2 R \sin 30^{\circ}=R$;

$c=2 R \sin \left(90^{\circ}+45^{\circ}\right)=2 R \cos 45^{\circ}=R \sqrt{2}$;

Moreover, $\sin \left(15^{\circ}\right)=\sin \left(45^{\circ}-30^{\circ}\right)$ is calculated honestly or through the half-angle.

$p=R \frac{2+\sqrt{6}+\sqrt{2}}{4}$,

$S_{A B C}=\frac{1}{2} a \cdot c \cdot \sin \beta=p \cdot r \Leftrightarrow \frac{1}{4} R^{2}(2 \sqrt{3}-2)=\frac{1}{4} R(2+\sqrt{6}+\sqrt{2}) \Leftrightarrow$

$\Leftrightarrow R=\frac{2+\sqrt{6}+\sqrt{2}}{2 \sqrt{3}-2}$

Then $S_{A B C}=\frac{1}{2} a \cdot c \cdot \sin \beta=\frac{1}{4} R^{2}(2 \sqrt{3}-2)=\frac{1}{4} \cdot \frac{(2+\sqrt{6}+\sqrt{2})^{2}}{2 \sqrt{3}-2}=\frac{3}{2}+\sqrt{2}+\sqrt{3}+\frac{\sqrt{6}}{2}$

Answer: $S_{A B C}=\frac{3}{2}+\sqrt{2}+\sqrt{3}+\frac{\sqrt{6}}{2}$
1. (10 points) In triangle $\mathrm{ABC}$ with angle $\alpha=15^{\circ}$ and unknown side lengths $a, b, c$, the relation $b=\sqrt{a(a+c)}$ holds. Find the area of triangle $\mathrm{ABC}$ if the radius of the inscribed circle $r=1$. Present the answer as a sum of numbers.