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 $\sqrt{7}$ passes through the vertices of the triangle $A A_{1} B$ and touches the segment $C C_{1}$ at point $M$.

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

b) Suppose additionally that $\angle B A C=\arcsin \sqrt{\frac{2}{3}}$. Find the angle between the line $C_{1} C$ and the plane $A A_{1} B$, as well as the length of the edge $A_{1} C_{1}$.
Answer: a) $AB=\sqrt{21}$; b) $\angle\left(C_{1} C, A A_{1} B\right)=\arcsin \sqrt{\frac{2}{3}}, A_{1} C_{1}=\frac{7-2 \sqrt{7}}{3}$.

Solution. a) Let $O$ be the center of the sphere $\Omega, Q$ be the center of the triangle $ABA_{1}, A_{1}X$ 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 $(ABB_{1}A_{1})$ 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 - the point $Q$. Thus, $Q=O$.

This means that the circle circumscribed around the triangle $ABA_{1}$ is a great circle on $\Omega$, and its radius is $\sqrt{7}$. Hence, $AB=R\sqrt{3}=\sqrt{21}$.

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 equal to $XA \operatorname{tg} \angle BAC=\frac{\sqrt{21}}{2} \cdot \sqrt{2}=\sqrt{\frac{21}{2}}$. From the right triangle $A_{1}YX$ we find

$$
\angle\left(CC_{1}, ABA_{1}\right)=\angle YA_{1}X=\arcsin \frac{YX}{A_{1}X}=\arcsin \left(\sqrt{\frac{21}{2}}:\left(\frac{\sqrt{3}}{2} \cdot \sqrt{21}\right)\right)=\arcsin \sqrt{\frac{2}{3}}
$$

In addition, $YZ=\frac{2}{3} YX=\sqrt{\frac{14}{3}}$.

Let $T$ be the projection of $Z$ onto $AC$. Then $ZT=YZ \cos \angle YZT=YZ \cos \angle BAC=\frac{\sqrt{14}}{3}$ and $YT=$ $YZ \sin \angle YZT=\frac{2 \sqrt{7}}{3}$. On the other hand, since $CZ=MO=\sqrt{7}$, we have $CT=\sqrt{CZ^{2}-ZT^{2}}=$ $\sqrt{7-\frac{14}{9}}=\frac{7}{3}$. Therefore, $A_{1}C_{1}=YC=CT-YT=\frac{7}{3}-\frac{2 \sqrt{7}}{3}$.

## Ticket 16