5. On the edge $S A$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $K$ is marked such that $A K: K S=1: 4$. The point $K$ is the vertex of a right circular cone, on the circumference of the base of which lie three vertices of the pyramid $2 A B C D$.

a) Find the ratio $D S: B C$.

b) Suppose it is additionally known that the height of the pyramid $S A B C D$ is 5. Find the volume of the cone.
Answer. a) $\frac{2}{\sqrt{3}}$, b) $V=\frac{64 \pi}{\sqrt{15}}$.

Solution. Let point $H$ be the center of the base of the pyramid. Consider triangles $A H K, B H K, C H K, D H K$. They have the side $H K$ in common, and the sides $A H, B H, C H, D H$ are equal to each other. Since angles $B H K$ and $D H K$ are right angles, angle $A H K$ is acute, and angle $C H K$ is obtuse, it follows that $C K>B K=D K>A K$.

Since point $K$ is equidistant from three vertices of the pyramid, we get that $B K=D K=S K$. Therefore, points $B, D$, and $S$ lie on the circumference of the base of the cone.

a) Consider triangle $A B S$. Let $A K=x, S K=4 x$. Then $B S=5 x, B K=4 x$. From the isosceles triangle $B K S$, we find that $\cos \angle A S B=\frac{B S}{2 B K}=\frac{5}{8}$. Using the cosine theorem for triangle $A B S$, we get that $A B=\frac{5 x \sqrt{3}}{2}$. Therefore, $\frac{D S}{B C}=\frac{A S}{A B}=\frac{2}{\sqrt{3}}$.

b) The height of the cone is the distance from point $K$ to the plane $B S D$. Since $K S: A S=4: 5$, this distance is $\frac{4}{5}$ of the distance from point $A$ to the plane $B S D$, i.e., $\frac{4}{5} A H=\frac{2 \sqrt{2}}{5} A B=x \sqrt{6}$. The slant height of the cone is the segment $K S=4 x$. Therefore, the radius of the base of the cone is $\sqrt{16 x^{2}-6 x^{2}}=x \sqrt{10}$. Then its volume $V=\frac{\pi}{3} \cdot 10 x^{2} \cdot x \sqrt{6}=\frac{10 \pi \sqrt{2} x^{3}}{\sqrt{3}}$.

From the right triangle $A S H$ using the Pythagorean theorem, we get $25 x^{2}=\frac{75 x^{2}}{8}+25$, from which $x=\frac{2 \sqrt{2}}{\sqrt{5}}$. Therefore, $V=\frac{64 \pi}{\sqrt{15}}$.