5. On the edge $S C$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $C L: L S=1: 5$. The point $L$ 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 $A S: A B$.

b) Suppose it is additionally known that the height of the pyramid $S A B C D$ is 6. Find the volume of the cone.
Answer. a) $\frac{\sqrt{5}}{2}$, b) $V=\frac{125 \pi \sqrt{2}}{3 \sqrt{3}}$.

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

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

a) Consider triangle $C B S$. Let $C L=x, S L=5 x$. Then $B L=5 x, B S=6 x$. From the isosceles triangle $B L S$, we find that $\cos \angle L S B=\frac{B S}{2 S L}=\frac{3}{5}$. Using the cosine theorem for triangle $C B S$, we get that $B C=\frac{12 x}{\sqrt{5}}$. Therefore, $\frac{A S}{A B}=\frac{B S}{B C}=\frac{\sqrt{5}}{2}$.

b) The height of the cone is the distance from point $L$ to the plane $B S D$. Since $L S: C S=5: 6$, this distance is $\frac{5}{6}$ of the distance from point $C$ to the plane $B S D$, i.e., $\frac{5}{6} C H=\frac{5}{6 \sqrt{2}} B C=x \sqrt{10}$. The slant height of the cone is the segment $L S=5 x$. Therefore, the radius of the base of the cone is $\sqrt{25 x^{2}-10 x^{2}}=x \sqrt{15}$. Then its volume $V=\frac{\pi}{3} \cdot 15 x^{2} \cdot x \sqrt{10}=5 \pi \sqrt{10} x^{3}$.

From the right triangle CSH, by the Pythagorean theorem, we get $36 x^{2}=\frac{72 x^{2}}{5}+36$, from which $x=\sqrt{\frac{5}{3}}$. Therefore, $V=\frac{125 \pi \sqrt{2}}{3 \sqrt{3}}$.