4. [5 points] A sphere with center $O$ is inscribed in a trihedral angle with vertex $S$ and touches its faces at points $K, L, M$ (all plane angles of the trihedral angle are different). Find the angle $K S O$ and the area of the section of the trihedral angle by the plane $K L M$, given that the areas of the sections of the trihedral angle by planes tangent to the sphere and perpendicular to the line $S O$ are 1 and 16.
Answer: $\angle K S O=\arcsin \frac{3}{5}, S=\frac{64}{25}$.

Solution. Let the points of intersection of the line $S O$ with the sphere be $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). Let the radius of the sphere be $r$. Triangles $O K S, O L S$, and $O M S$ are right triangles (angles at vertices $K, L, M$ are right angles, as tangents are perpendicular to the radii drawn to the point of tangency). These triangles are equal by the leg and hypotenuse ($O K=O L=O M=R, S O$ is common), therefore, $\angle K S O=\angle L S O=\angle M S O$ (let $\angle K S O=\alpha, S O=x$). The heights dropped from points $K, L, M$ to the hypotenuse $S O$ are equal, and their bases are the same point $H$, lying in the plane $K L M$ (let's call this plane $\tau$). Let $\beta$ and $\gamma$ be the tangent planes to the sphere passing through points $P$ and $Q$, and $E$ and $F$ be the points of intersection of these planes with the line $S K$. By the condition, the areas of the sections of the trihedral angle by these planes are $S_{1}=1$ and $S_{2}=16$, respectively. Consider the section of the trihedral angle and the sphere by the plane $S K O$ (see the figure and notation on it). Since $S H \perp H K$ and $S H \perp H L$, then $\tau \perp S H$. Then the sections of the trihedral angle by the planes $\tau, \beta$, and $\gamma$ are similar triangles, the planes of which are parallel (all of them are perpendicular to $S O$).

If $\Sigma$ is the area of the triangle obtained by the section of the trihedral angle by the plane $K L M$, then from the similarity $\Sigma: S_{1}: S_{2}=K H^{2}: E P^{2}: F Q^{2}$. Therefore, $E P: F Q=\sqrt{S_{1}}: \sqrt{S_{2}}$. Then $\sqrt{S_{1}}: \sqrt{S_{2}}=S P: S Q=(x-r):(x+r)$, from which $r=x \frac{\sqrt{S_{2}}-\sqrt{S_{1}}}{\sqrt{S_{2}}+\sqrt{S_{1}}}$, and $\sin \alpha=\frac{r}{x}=\frac{\sqrt{S_{2}}-\sqrt{S_{1}}}{\sqrt{S_{2}}+\sqrt{S_{1}}}=\frac{3}{5}$. Hence, $\angle K S O=\arcsin \frac{3}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-15.jpg?height=1008&width=765&top_left_y=181&top_left_x=657)

Fig. 7: variant 7, problem 4

Next, $O H=r \sin \alpha, S H=S O-O H=\frac{r}{\sin \alpha}-r \sin \alpha, S P=S O-r=\frac{r}{\sin \alpha}-r$. Therefore, $\Sigma: S_{1}=K H^{2}: E P^{2}=S H^{2}: S P^{2}=\left(\frac{1}{\sin \alpha}-\sin \alpha\right)^{2}:\left(\frac{1}{\sin \alpha}-1\right)^{2}=(1+\sin \alpha)^{2}=\frac{64}{25}$, from which $\Sigma=\frac{64}{25}$.