2. The base of the triangular prism $A B C A^{\prime} B^{\prime} C^{\prime}$ is a right isosceles triangle $A B C$ with legs $A B, A C$ of length $a$. The lateral edges $A A^{\prime}, B B^{\prime}, C C^{\prime}$ form an angle of $60^{\circ}$ with the planes of the bases. The diagonal $B C^{\prime}$ of the lateral face $B C C^{\prime} B^{\prime}$ has a length of $a \sqrt{6}$ and is perpendicular to the edge $A C$. Determine the volume of the prism.
Solution. The volume of our prism is $V=a^{2} \cdot v / 2$, where $v$ is the unknown distance between its bases. Both lines $B C^{\prime}$ and $A B$, and thus the plane $A B C^{\prime}$, are perpendicular to the line
$A C$. If $P$ is the orthogonal projection of point $C^{\prime}$ onto the line $A B$, then both lines $A B$ and $A C$, and thus the plane $A B C$, are perpendicular to the line $C^{\prime} P$. The sought distance $v$ is therefore equal to $\left|C^{\prime} P\right|$ and both angles $C P C^{\prime}, A P C^{\prime}$ are right. Moreover, $\left|\varangle P C C^{\prime}\right|=60^{\circ}$, so $v=|C P| \sqrt{3}$. Let $x$ be the coordinate of point $P$ on the line $A B$ in a system where $A=[0]$ and $B=[-a]$(i.e., $x= \pm|A P|$, where the sign - or + is taken depending on whether $P$ falls on the ray $A B$ or on the opposite ray.) Then from $\triangle A C P$ it follows that $|C P|^{2}=a^{2}+x^{2}$, so $v^{2}=3\left(a^{2}+x^{2}\right)$. Since $\left|B C^{\prime}\right|=a \sqrt{6}$, the Pythagorean theorem for $\triangle B P C^{\prime}$ has the form $6 a^{2}=(a+x)^{2}+3\left(a^{2}+x^{2}\right)$. This equation has two roots $x_{1}=a / 2$, $x_{2}=-a$. The conditions of the problem are therefore satisfied by two prisms (fig. 1,2) with heights

$$
v_{1}=\sqrt{3\left(a^{2}+\frac{a^{2}}{4}\right)}=\frac{a \sqrt{15}}{2} \quad \text { and } \quad v_{2}=\sqrt{3\left(a^{2}+a^{2}\right)}=a \sqrt{6},
$$

verify. Their volumes are $V_{1}=a^{3} \sqrt{15} / 4$ and $V_{2}=a^{3} \sqrt{6} / 2$.