4. [5 points] a) A sphere with center $O$ touches the lateral edges $S A, S B, S C$ of the pyramid $S A B C$ at points $K, L, M$ respectively, and also touches its base $A B C$. A plane is drawn through the point on the sphere closest to point $S$, tangent to the sphere. The area of the section of the pyramid $S A B C$ by this plane is $5, \angle K S O=\arccos \frac{\sqrt{21}}{5}$. Find the area of triangle $K L M$.

b) Suppose it is additionally known that $S O=36$, and the planes $K L M$ and $A B C$ are parallel. Find the volume of the pyramid $2 A B C$.
Answer: a) $S_{K L M}=9.8 ;$ b) $V_{S A B C}=\frac{1372}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-27.jpg?height=963&width=740&top_left_y=1186&top_left_x=675)

Fig. 1: Variant 13, Problem 4

Solution. a) Let the radius of the sphere be $R$. Denote the points of intersection of the line $S O$ with the sphere as $P$ and $Q$ (point $P$ lies on the segment $S O$, and $Q$ lies outside it). 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 ($O K=O L=O M=R, S O$ - common), therefore, $\angle K S O=\angle L S O=\angle M S O$ (denote these angles as $\alpha ; \sin \alpha=\frac{2}{5})$; 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 $K L M$ (call this plane $\tau$). Let $\sigma$ be the tangent plane to the sphere passing through point $P$. Denote the point of intersection of $\sigma$ and $S A$ as $E$. Consider the section of the pyramid and the sphere by the plane $A S O$ (see the figure; only part of this section is shown).

From the right triangle $K S O$ we get $S O=\frac{R}{\sin \alpha}$. Then $S P=S O-O P=R\left(\frac{1}{\sin \alpha}-1\right)$. Let the area of the section of the pyramid by the plane $\sigma$ be $S_{0}=5$, and by the plane $\tau-S_{K L M}$. From similarity, it follows that $S_{K L M}: S_{0}=(K H: E P)^{2}=(S H: S P)^{2}=(S O-O H)^{2}: S P^{2}=\left(\frac{R}{\sin \alpha}-R \sin \alpha\right)^{2}:$ $R^{2}\left(\frac{1}{\sin \alpha}-1\right)^{2}=(1+\sin \alpha)^{2}$. Therefore, $S_{K L M}=S_{0}(1+\sin \alpha)^{2}=9.8$.

b) If the planes $\tau$ and $A B C$ are parallel, then point $A$ coincides with point $A_{1}$ such that $A_{1} Q \perp$ $S O$ (see the figure). Then, denoting the area of triangle $A B C$ as $S_{A B C}$, we get $V_{S A B C}=$ $\frac{1}{3} \cdot S Q \cdot S_{A B C}=\frac{1}{3} \cdot(S O+R) \cdot S_{0} \cdot\left(\frac{S Q}{S P}\right)^{2}=\frac{1}{3} \cdot(S O+S O \sin \alpha) \cdot S_{0} \cdot(S O+S O \sin \alpha)^{2}:(S O-S O \sin \alpha)^{2}=$ $\frac{1}{3} \cdot S O \cdot S_{0} \cdot \frac{(1+\sin \alpha)^{3}}{(1-\sin \alpha)^{2}}=\frac{1372}{3}$.