5. On the edge $S B$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, a point $L$ is marked such that $B L: L S=2: 5$. 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 $S A B C D$.

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

b) Suppose it is additionally known that the height of the pyramid $S A B C D$ is 7. Find the volume of the cone.
Answer. a) $\sqrt{\frac{5}{3}}$, b) $V=\frac{125 \pi}{\sqrt{21}}$.

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 $A H L$ and $C H L$ are right, angle $B H L$ is acute, and angle $D H L$ is obtuse, it follows that $D L>A L=C L>B L$.

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

a) Consider triangle $A B S$. Let $B L=2 x, S L=5 x$. Then $A S=7 x, A L=5 x$. From the isosceles triangle $A L S$, we find that $\cos \angle L S A=\frac{A S}{2 S L}=\frac{7}{10}$. Using the cosine rule for triangle $A B S$, we get that $A B=\frac{7 x \sqrt{3}}{\sqrt{5}}$. Therefore, $\frac{A S}{C D}=\frac{A S}{A B}=\sqrt{\frac{5}{3}}$.

b) The height of the cone is the distance from point $L$ to the plane $A S C$. Since $L S: B S=5: 7$, this distance is $\frac{5}{7}$ of the distance from point $B$ to the plane $A S C$, i.e., $\frac{5}{7} B H=\frac{5}{7 \sqrt{2}} A B=x \sqrt{\frac{15}{2}}$. The slant height of the cone is the segment $L S=5 x$. Therefore, the radius of the base of the cone is $\sqrt{25 x^{2}-\frac{15}{2} x^{2}}=x \sqrt{\frac{35}{2}}$. Then its volume $V=\frac{\pi}{3} \cdot \frac{35}{2} x^{2} \cdot x \sqrt{\frac{15}{2}}=\frac{35 \pi \sqrt{15} x^{3}}{6 \sqrt{2}}$.

From the right triangle $B S H$ by the Pythagorean theorem, we get $49 x^{2}=\frac{147 x^{2}}{10}+49$, from which $x=\sqrt{\frac{10}{7}}$. Therefore, $V=\frac{125 \pi}{\sqrt{21}}$.