Let $PABCD$ be a given quadrilateral pyramid with vertex $P, ABCD$ - a square. Suppose that

$$
\angle PAB = \angle PBC = \angle PCD = \angle PDA = 90^{\circ}
$$

Since the hypotenuse of a right triangle is less than the leg,

$$
PA < PB < PC < PD < PA, \quad
$$

which is impossible. Similarly, we can prove that all cases of the arrangement of right angles, "symmetric" to those considered (i.e., every other one), are impossible. The only possibility is when two right angles are adjacent to one vertex. Let this vertex be $A$. Since $PA \perp AD$ and $PA \perp AB$, the lateral edge $PA$ is perpendicular to the plane of the base, and since $AB \perp BC$ and $AD \perp CD$, by the theorem of three perpendiculars, $PB \perp BC$ and $PD \perp CD$, i.e., $\angle PBC = \angle PDC = 90^{\circ}$. The dihedral angle at the edge $PA$ is right, as its linear angle is the angle $BAD$. The line $BC$ is perpendicular to the plane of the face $PAB$, since $BC \perp AB$ and $BC \perp PB$. Therefore, the plane of the face $PBC$ passes through the perpendicular $BC$ to the plane of the face $PAB$, so the planes of the faces $PAB$ and $PBC$ are perpendicular. Consequently, the dihedral angle at the edge $PB$ is $90^{\circ}$. Similarly, the dihedral angle at the edge $PD$ is also $90^{\circ}$. Thus, the dihedral angle at the edge $PC$ is $120^{\circ}$. Let $K$ be the foot of the perpendicular dropped from vertex $B$ to the edge $PC$. Since $AC \perp BD$ and by the theorem of three perpendiculars, $PC \perp BD$, the line $PC$ is perpendicular to the plane of the triangle $BKD$, so the angle $BKD$ is the linear angle of the dihedral angle at the edge $PC$. By the proven fact, $\angle BKD = 120^{\circ}$. Let $O$ be the center of the square $ABCD$. Then $OK$ is the height, median, and bisector of the triangle $BKD$. Denote by $a$ the side of the square $ABCD$, and by $\alpha$ the angle $ACP$. From the right triangles $KOD$ and $PAC$, we find that

$$
\begin{gathered}
OK = OD \operatorname{tg} \angle KDO = \frac{a \sqrt{2}}{2} \operatorname{tg} 30^{\circ} = \frac{\frac{a \sqrt{2}}{\sqrt{3}}}{\sqrt{2}} = \frac{a \sqrt{6}}{6} \\
\sin \alpha = \sin \angle ACP = \frac{OK}{OC} = \frac{\frac{a \sqrt{6}}{6}}{\frac{a \sqrt{2}}{2}} = \frac{1}{\sqrt{3}} \\
\operatorname{ctg} \alpha = \sqrt{\frac{1}{\sin^2 \alpha} - 1} = \sqrt{3 - 1} = \sqrt{2}, a \sqrt{2} = AC = PA \operatorname{ctg} \alpha = \sqrt{2}
\end{gathered}
$$

from which $a = 1$. Therefore,

$$
V_{PABCD} = \frac{1}{3} AB^2 \cdot PA = \frac{1}{3} \cdot 1 \cdot 1 = \frac{1}{3}
$$

## Answer

$\frac{1}{3}$.
The base of a quadrilateral pyramid is a square, and all lateral faces are right triangles, with the vertices of the right angles lying on the base of the pyramid. Find the volume of the pyramid if its height is 1, and one of the dihedral angles at the vertex is $120^{\circ}$.

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