5. On the edge $B B_{1}$ of a regular triangular prism $A B C A_{1} B_{1} C_{1}$, a point $T$ is taken such that $B T: B_{1} T=2: 3$. The point $T$ is the vertex of a right circular cone such that three vertices of the prism lie on the circumference of its base.

a) Find the ratio of the height of the prism to the edge of its base.

b) Suppose it is additionally known that $C C_{1}=5$. Find the volume of the cone.
Answer. a) $\sqrt{5}$; b) $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.

Solution. If three vertices of the prism lie on the circumference of the base of the cone, this means that the three vertices of the prism are equidistant from point $T$, i.e., three of the segments $T A, T B, T C, T A_{1}, T B_{1}, T C_{1}$ are equal to each other. Note that $T A_{1}=T C_{1}>T A=T C$; moreover, $T A>T B, T A_{1}>T B_{1}$. From these inequalities, it follows that the segments $T A_{1}$ and $T C_{1}$ are the longest, and the segment $T B$ is the shortest. Therefore, the segments $T A, T C$, and $T B_{1}$ are equal to each other.

a) Let $T B_{1}=3 x, T B=2 x$. Then $T A=3 x$. By the Pythagorean theorem for triangle $A B T$, we find that $A B=x \sqrt{5}$. Thus, the desired ratio is $\frac{5 x}{x \sqrt{5}}=\sqrt{5}$.

b) From triangle $B B_{1} A$ by the Pythagorean theorem, we get that $A B_{1}=x \sqrt{30}$. The radius of the base of the cone is the radius of the circle circumscribed around triangle $A B_{1} C$. Its sides are: $A B_{1}=C B_{1}=x \sqrt{30}, A C=x \sqrt{5}$. Then we find its height: $B_{1} H=\frac{x \sqrt{115}}{2}$, area: $S=\frac{1}{2} \cdot x \sqrt{5} \cdot \frac{x \sqrt{115}}{2}=\frac{5 x^{2} \sqrt{23}}{4}$, radius of the circumscribed circle: $R=\frac{x \sqrt{30} \cdot x \sqrt{30} \cdot x \sqrt{5}}{4 S}=\frac{6 x \sqrt{5}}{\sqrt{23}}$.

The slant height of the cone is the segment $A T$, i.e., it is equal to $3 x$. Then the height of the cone $h=\sqrt{9 x^{2}-\frac{180 x^{2}}{23}}=\frac{3 x \sqrt{3}}{\sqrt{23}}$.

We find the volume of the cone: $V=\frac{\pi}{3} \cdot \frac{180 x^{2}}{23} \cdot \frac{3 x \sqrt{3}}{\sqrt{23}}=\frac{180 \pi x^{3} \sqrt{3}}{23 \sqrt{23}}$. Since $x=1$, we finally get $V=\frac{180 \pi \sqrt{3}}{23 \sqrt{23}}$.