7. Given a truncated pyramid $A B C A_{1} B_{1} C_{1}$ with lateral edges $A A_{1}, B B_{1}, C C_{1}\left(A B C \| A_{1} B_{1} C_{1}\right)$, such that the triangle $A B A_{1}$ is equilateral. On the edge $C C_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $M$ such that $C M: M C_{1}=1: 2$. The sphere $\Omega$ with radius $\sqrt{5}$ passes through the vertices of the triangle $A B A_{1}$ and touches the segment $C C_{1}$ at point $M$.

a) Find the length of the edge $A B$.

b) Suppose additionally that $\angle B A C=\arcsin \frac{1}{\sqrt{3}}$. Find the angle between the line $C C_{1}$ and the plane $A B A_{1}$, as well as the length of the edge $A_{1} C_{1}$.
Answer: a) $AB=\sqrt{15}$; b) $\angle\left(CC_{1}, ABA_{1}\right)=\arcsin \frac{1}{\sqrt{6}}, A_{1}C_{1}=\sqrt{\frac{5}{2}}$.

Solution. a) Let $O$ be the center of the sphere $\Omega$, $Q$ be the center of the triangle $ABA_{1}$, and $A_{1}X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $(ABA_{1})$ passing through $Q$. On the other hand, $O$ lies in the plane $\alpha$ passing through $M$ and perpendicular to $CC_{1}$, which is parallel to $(ABC)$. Note that $Q$ lies in $\alpha$ since $XQ:QA_{1}=1:2=CM:MC_{1}$. But $Q$ also lies on $\ell$. If the plane $(AB B_{1}A_{1})$ were perpendicular to the plane $(ABC)$, then all three lateral faces would be perpendicular to the base, which is impossible in a truncated pyramid. Therefore, $\alpha$ and $\ell$ intersect at exactly one point, which is $Q$. Thus, $Q=O$.

This means that the circle circumscribed around the triangle $ABA_{1}$ is a great circle on $\Omega$, and its radius is $\sqrt{5}$. Hence, $AB=R\sqrt{3}=\sqrt{15}$.

b) Let $Y$ and $Z$ be the projections of points $A_{1}$ and $O$ onto the plane $(ABC)$; then $Z$ divides the segment $XY$ in the ratio $XZ:ZY=1:2$. Since $CC_{1} \perp (ABC)$, the point $Y$ lies on the line $AC$.

The right triangles $A_{1}YA$ and $A_{1}YB$ are equal by the leg and hypotenuse, so $YA=YB$. Therefore, the height $YX$ of the triangle $ABY$ is $XA \operatorname{tg} \angle BAC=\frac{\sqrt{15}}{2} \cdot \sqrt{\frac{1}{2}}=\sqrt{\frac{15}{8}}$. From the right triangle $A_{1}YX$ we find

$$
\angle\left(CC_{1}, ABA_{1}\right)=\angle YA_{1}X=\arcsin \frac{YX}{A_{1}X}=\arcsin \left(\sqrt{\frac{15}{8}}:\left(\frac{\sqrt{3}}{2} \cdot \sqrt{15}\right)\right)=\arcsin \frac{1}{\sqrt{6}}
$$

Moreover, $YZ=\frac{2}{3} YX=\sqrt{\frac{5}{6}}$.

Let $T$ be the projection of $Z$ onto $AC$. Then $ZT=YZ \cos \angle YZT=YZ \cos \angle BAC=\frac{\sqrt{5}}{3}$ and $YT=$ $YZ \sin \angle YZT=\frac{\sqrt{5}}{3 \sqrt{2}}$. On the other hand, since $CZ=MO=\sqrt{5}$, we have $CT=\sqrt{CZ^{2}-ZT^{2}}=$ $\sqrt{5-\frac{5}{9}}=\frac{2 \sqrt{10}}{3}$. Therefore, $A_{1}C_{1}=YC=CT-YT=\frac{2 \sqrt{10}}{3}-\frac{5}{3 \sqrt{2}}=\sqrt{\frac{5}{2}}$.

## Ticket 14