Cut a regular tetrahedron from a regular pentagonal pyramid. Construct the base of this tetrahedron so that one of its base edges is parallel to one of the base edges of the original pyramid, and calculate the dimensions of the new tetrahedron.
On the side $AB$ of the regular pentagonal pyramid $ABCDE$, we draw the square $ABFG$, and connect points $F$ and $G$ with $H$, the intersection of $AE$ and $BC$. The extensions of $HF$ and $HG$ intersect $DC$ and $DE$ at $R_3$ and $R_4$, respectively. If the lines parallel to $AG$ through $R_3$ and $R_4$ intersect $BC$ and $AE$ at $R_2$ and $R_1$, respectively, then $R_1 R_2 R_3 R_4$ is the desired square. Indeed, $R_2 R_3 \parallel R_4 R_1$ and $R_1 R_2 \parallel R_3 R_4$, because

$$
H R_2 : H B = H R_3 : H F = H R_4 : H G = H R_1 : H A
$$

Furthermore,

$$
R_2 R_3 : B F = R_3 H : E H = R_3 R_4 : F G \text{. }
$$

But $B F = F G$, so $R_2 R_3 = R_3 R_4$, and thus $R_1 R_2 R_3 R_4$ is indeed the square that meets the problem's requirements.

If $R_1 R_2 = x$ and $AB = a$, then from the triangle $D R_3 R_4$,

$$
D R_3 = x \frac{\sin 36^\circ}{\sin 108^\circ} = 2 x \sin 18^\circ
$$

and from the triangle $C R_2 R_3$,

$$
C R_3 = x \frac{\sin 18^\circ}{\sin 108^\circ} = x \tan 18^\circ
$$

so

$$
D R_3 + C R_3 = D C = a = x (2 \sin 18^\circ + \tan 18^\circ)
$$

or

$$
x = \frac{a}{2 \sin 18^\circ + \tan 18^\circ}
$$

Let $e$ be the edge of the regular pentagonal pyramid, and $S$ be its apex. Then $S R_1 = S R_2$ and $S R_3 = S R_4$. From the triangle $R_1 S E$,

$$
S R_1^2 = E R_1^2 + e^2 - 2 e \cdot E R_1 \cos S E R_1
$$

But

$$
E R_1 = x \frac{\sin 54^\circ}{\sin 198^\circ} = x \frac{\cos 36^\circ}{\cos 18^\circ} = \frac{a \cos 36^\circ}{\sin 36^\circ + \sin 18^\circ} \text{ and } \cos S E R_1 = \frac{a}{2 e},
$$

so

$$
S R_1^2 = \left( \frac{a \cos 36^\circ}{\sin 36^\circ + \sin 18^\circ} \right)^2 + e^2 - \frac{a^2 \cos 36^\circ}{\sin 36^\circ + \sin 18^\circ}
$$

From the triangle $D R_3 R_4$,

$$
S R_3^2 = D R_3^2 + e^2 - 2 e \cdot D R_3 \cos S D R_3
$$

But

$$
D R_3 = x \frac{\sin 36^\circ}{\sin 108^\circ} = x \frac{\sin 36^\circ}{\cos 18^\circ} = \frac{a \sin 36^\circ}{\sin 36^\circ + \sin 18^\circ} \text{ and } \cos S D R_3 = \frac{a}{2 e},
$$

so

$$
S R_3^2 = \left( \frac{a \sin 36^\circ}{\sin 36^\circ + \sin 18^\circ} \right)^2 + e^2 - \frac{a^2 \sin 36^\circ}{\sin 36^\circ + \sin 18^\circ}
$$

(Imre Dömény, Budapest.)

The problem was also solved by: Földes R., Haar A., Kürti I., Messer P., Rosenberg J., Sonnenfeld J.