7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 12. A sphere $\Omega$ with radius $r=\frac{3 \sqrt{5}}{2}$ touches all the lateral faces of the prism. Points $T$ and $F$ are chosen on segments $C C_{1}$ and $B B_{1}$, respectively, such that $F T \| B C$, and the planes $A B T$ and $F A_{1} C_{1}$ touch the sphere $\Omega$. Find the volume of the prism and the length of segment $C T$.
Answer: $V=405 \sqrt{3}, C T=9$ or $C T=3$.

“Fiztekh-2016”, mathematics, solutions for ticket 35

Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=3 \sqrt{15}$, the area of the base $S=\frac{\sqrt{3}}{4}(3 \sqrt{15})^{2}=\frac{135 \sqrt{3}}{4}$, and the volume of the prism is $S h=\frac{135 \sqrt{3}}{4} \cdot 12=405 \sqrt{3}$.

From the fact that the prism is regular and the sphere touches all its faces, it follows that the plane $F A C$ also touches the sphere; in this case, the points of tangency of the sphere with the planes $F A C$ and $F A_{1} C_{1}$ lie on the segments $F Q$ and $F Q_{1}$, where the points $Q$ and $Q_{1}$ are the midpoints of the edges $A C$ and $A_{1} C_{1}$.

Consider the plane of the rectangle $B B_{1} Q_{1} Q$. Let the center of the circle $\omega$, obtained by the section of the sphere by this plane, be denoted by $O$, and the points of tangency of the segments $F Q_{1}, Q Q_{1}$, and $F Q$ with the circle be denoted by $G, J$, and $P$ respectively. The height $F H$ of the triangle $F Q Q_{1}$ is equal to the height of the triangle lying in the base of the prism, that is, $F H=3 r$.

Let's write the area $S_{0}$ of the triangle $F Q Q_{1}$ in two ways: $S_{0}=p r$ and $S_{0}=\frac{1}{2} F H \cdot Q Q_{1}=\frac{1}{2} \cdot 3 r \cdot 12=18 r$, where $p$ is the semiperimeter of the triangle $F Q Q_{1}$. Therefore, $p=18$. Let $Q J=x$; then $Q_{1} G=Q_{1} J=12-x$, $P Q=x, F G=F P=\frac{1}{2}(36-2 x-2(12-x))=6$.

Thus, $Q Q_{1}=12, Q F=6+x, Q_{1} F=18-x$. Then Heron's formula gives $S_{0}=\sqrt{18 \cdot 6 \cdot x \cdot(12-x)}$, from which $\sqrt{18 \cdot 6 \cdot x \cdot(12-x)}=18 r, x(12-x)=3 r^{2}$. Substituting the value of the radius from the condition, we get the equation $x^{2}-12 x+\frac{135}{4}=0$, hence $x=\frac{9}{2}$ or $x=\frac{15}{2}$.

Then $Q F=\frac{21}{2}$ or $Q F=\frac{27}{2}$, and since $B F=\sqrt{Q F^{2}-B Q^{2}}=\sqrt{Q F^{2}-9 r^{2}}$, we have $B F=3$ or $B F=9$. It remains to note that $C T=B F=Q H$.