[ $\left.\frac{\text { angles between bisectors }}{\text { Dihedral angle }}\right]$

The base of the pyramid $S A B C$ is a right triangle $A B C$ ( $C$ - the vertex of the right angle). All lateral faces of the pyramid are inclined to its base at the same angle, equal to $\arcsin \frac{5}{13}$. Find the area of the lateral surface of the pyramid, if $S O-$ is the height of the pyramid, $A O=1, B O=3 \sqrt{2}$.
Since the lateral faces of the pyramid are inclined to its base at the same angle, the height of the pyramid passes through the center of the circle inscribed in the base, i.e., point $O$ is the center of the inscribed circle of triangle $ABC$ (Fig.1). Let this circle touch the legs $AC$ and $BC$ at points $K$ and $L$ respectively, and the hypotenuse $AB$ at point $M$. Since $OK \perp AC$, it follows from the theorem of three perpendiculars that $SK \perp AC$. Therefore, $SKO$ is the linear angle of the dihedral angle between the lateral face $ASC$ and the base of the pyramid. Let $\angle SKO = \beta$. Then $\angle SLO = \angle SMO = \beta$. According to the problem, $\sin \beta = \frac{5}{13}$. Then $\cos \beta = \frac{12}{13}$. Since the center of the circle inscribed in a triangle is the point of intersection of its angle bisectors,

$$
\angle AOB = 90^\circ + \frac{1}{2} \angle ACB = 90^\circ + 45^\circ = 135^\circ
$$

From triangle $AOB$ (Fig.2) by the cosine theorem, we find that

$$
AB = \sqrt{AO^2 + BO^2 - 2 \cdot AO \cdot BO \cos 135^\circ} = \sqrt{1 + 18 + 2 \cdot 1 \cdot 3\sqrt{2} \cdot \frac{1}{\sqrt{2}}} = 5
$$

Let $r$ be the radius of the inscribed circle of triangle $ABC$. Then

$$
S_{\triangle AOB} = \frac{1}{2} AB \cdot r = \frac{5}{2} r.
$$

On the other hand,

$$
S_{\triangle AOB} = \frac{1}{2} OA \cdot OB \sin \angle AOB = \frac{1}{2} \cdot 1 \cdot 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{3}{2}
$$

From the equation $\frac{5}{2} r = \frac{3}{2}$, we find that $r = \frac{3}{5}$. Then

$$
\begin{gathered}
AM = AK = \sqrt{OA^2 - OK^2} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \\
BL = BM = AB - AM = 5 - \frac{4}{5} = \frac{21}{5}, AC = AK + KC = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}, \\
BC = BL + LC = \frac{21}{5} + \frac{3}{5} = \frac{24}{5}
\end{gathered}
$$

Let $p$ be the semiperimeter of triangle $ABC$. Then

$$
p = \frac{1}{2}(AB + BC + AC) = \frac{1}{2}\left(5 + \frac{24}{5} + \frac{7}{5}\right) = \frac{28}{5}.
$$

Thus,

$$
S_{\triangle ABC} = pr = \frac{28}{5} \cdot \frac{3}{5} = \frac{84}{25}
$$

Since the orthogonal projections of the lateral faces of the pyramid onto the plane of the base are triangles $AOB$, $AOC$, and $BOC$, and the planes of the lateral faces form equal angles with the base, the lateral surface area is equal to the area of the base divided by the cosine of the angle between the lateral face and the base, i.e.,

$$
S_{\text{lat.}} = \frac{S_{\triangle ABC}}{\cos \beta} = \frac{\frac{84}{25}}{\frac{12}{13}} = \frac{84}{25} \cdot \frac{13}{12} = \frac{91}{25}
$$

Answer

## $\frac{91}{25}$