5. Two equal cones are arranged so that the axis of each is the generatrix of the other. The angles at the vertices in the axial sections of these cones are each $90^{\circ}$. Find the angle between the two generatrices along which these cones intersect.
Problem 5.

Answer: $2 \arccos \sqrt{\frac{2}{2+\sqrt{2}}}$.

Solution. Let $S$ be the common vertex of the considered cones, $S A_{1}$ and $S A_{2}$ be their axes. Denote by $S B$ and $S C$ their common generators and by $\alpha$ the desired angle $\angle B S C$. The configuration described in the problem has two planes of symmetry: one $-S A_{1} A_{2}-$ contains the axes of the cones, the other - $S B C$ - contains their common generators. Then these planes are perpendicular, let $S O$ be the line of their intersection.

Let $\varphi$ be the angle in the axial section of each cone. Since $S A_{2}$ is a generator for the cone with axis $S A_{1}$ and vice versa, then $\angle A_{2} S O=\angle O S A_{1}=\varphi / 4$. In addition, $\angle A_{2} S B=\angle A_{2} S C=\angle A_{1} S B=\angle A_{1} S C=\varphi / 2, \angle O S B=\angle O S C=\alpha / 2$.

We will assume that the points $A_{1}, A_{2}, B, C, O$ lie in some plane perpendicular to the line $S O$ and located at a distance $h$ from the vertex $S$. Then from the pyramid $S O A_{1} C$, in which all planar angles at the vertex $O$ are right, we have

$$
\left.\begin{array}{c}
S A_{1}=\frac{h}{\cos (\varphi / 4)}, \quad O A_{1}=h \operatorname{tg}(\varphi / 4), \quad S C=\frac{h}{\cos (\alpha / 2)}, \quad O C=h \operatorname{tg}(\alpha / 2) \\
\triangle O A_{1} C: \quad A_{1} C^{2}=h^{2} \operatorname{tg}^{2}(\varphi / 4)+h^{2} \operatorname{tg}^{2}(\alpha / 2) \\
\triangle S A_{1} C: \quad A_{1} C^{2}=\frac{h^{2}}{\cos ^{2}(\varphi / 4)}+\frac{h^{2}}{\cos ^{2}(\alpha / 2)}-2 \frac{h^{2}}{\cos (\varphi / 4) \cos (\alpha / 2)} \cos (\varphi / 2)
\end{array}\right\} \Rightarrow
$$

Since $\varphi=90^{\circ}$, then $\cos (\alpha / 2)=\sqrt{\frac{2}{2+\sqrt{2}}}$.