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 $B B_{1} C$ is equilateral. On the edge $A A_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $N$ such that $A N: N A_{1}=1: 2$. The sphere $\Omega$ with radius $\sqrt{5}$ passes through the vertices of the triangle $B B_{1} C$ and touches the segment $A A_{1}$ at point $N$.

a) Find the length of the edge $B B_{1}$.

b) Suppose additionally that $\angle A B C=\arccos \sqrt{\frac{2}{5}}$. Find the angle between the line $A A_{1}$ and the plane $B B_{1} C$, as well as the length of the edge $A_{1} B_{1}$.
Answer: a) $B B_{1}=\sqrt{15}$; b) $\angle\left(A A_{1}, B B_{1} C\right)=\pi / 4, A_{1} B_{1}=2-\sqrt{\frac{3}{2}}$.

Solution. a) Let $O$ be the center of the sphere $\Omega, Q$ be the center of the triangle $B C B_{1}, B_{1} X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $\left(B C B_{1}\right)$ passing through $Q$. On the other hand, $O$ lies in the plane $\alpha$ passing through $N$ perpendicular to $A A_{1}$ - that is, parallel to ( $A B C$ ). Note that $Q$ lies in $\alpha$, since $X Q: Q B_{1}=1: 2=A N: N A_{1}$. But $Q$ also lies on $\ell$. If the plane $\left(B C C_{1} B_{1}\right)$ were perpendicular to the plane $(A B C)$, 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 $B C B_{1}$ is a great circle on $\Omega$, and its radius is $\sqrt{5}$. Hence, $B C=R \sqrt{3}=\sqrt{15}$.

b) Let $Y$ and $Z$ be the projections of points $B_{1}$ and $O$ onto the plane $(A B C)$; then $Z$ divides the segment $X Y$ in the ratio $X Z: Z Y=1: 2$. Since $A A_{1} \perp(A B C)$, the point $Y$ lies on the line $A B$.

Right triangles $B_{1} Y B$ and $B_{1} Y C$ are equal by leg and hypotenuse, so $Y C=Y B$. Therefore, the height $Y X$ of the triangle $C B Y$ is equal to $X B \operatorname{tg} \angle A B C=\frac{\sqrt{15}}{2} \cdot \sqrt{\frac{3}{2}}=\sqrt{\frac{45}{8}}$. From the right triangle $A_{1} Y X$ we find

$$
\angle\left(A A_{1}, B C B_{1}\right)=\angle Y B_{1} X=\arcsin \frac{Y X}{B_{1} X}=\arcsin \left(\sqrt{\frac{45}{8}}:\left(\frac{\sqrt{3}}{2} \cdot \sqrt{15}\right)\right)=\arcsin \frac{1}{\sqrt{2}}
$$

from which $\angle\left(A A_{1}, B C B_{1}\right)=\frac{\pi}{4}$. In addition, $Y Z=\frac{2}{3} Y X=\sqrt{\frac{5}{2}}$.

Let $T$ be the projection of $Z$ onto $A B$. Then $Z T=Y Z \cos \angle Y Z T=Y Z \cos \angle A B C=1$ and $Y T=$ $Y Z \sin \angle Y Z T=\sqrt{\frac{3}{2}}$. On the other hand, since $A Z=N O=\sqrt{5}$, we have $A T=\sqrt{A Z^{2}-Z T^{2}}=$ $\sqrt{5-1}=2$. Hence $A_{1} B_{1}=A Y=A T-Y T=2-\sqrt{\frac{3}{2}}$.

## Ticket 15