A pyramid with a triangular base has side edges of unit length, the angles between the side edges are $60^{\circ}, 90^{\circ}$, and $120^{\circ}$, respectively. What is the volume of the pyramid?
Let the base of the pyramid be the triangle $BCD$, and its fourth vertex be $A$. Let us choose the notation such that $\angle CAD = 60^\circ$, $\angle BAD = 90^\circ$, and $\angle BAC = 120^\circ$. Let the reflection of $B$ over the line $AD$ be $B'$. Since $AD$ and $BA$ are perpendicular, $B$, $A$, and $B'$ lie on the same line, and $B'A = AB = 1$. Since $\angle BAC = 120^\circ$, $\angle CAB' = 60^\circ$, and since $CA = AB' = 1$, the triangle $CAB'$ is equilateral. The triangle $CAD$ is also equilateral because $CA = DA = 1$ and $\angle CAD = 60^\circ$. Therefore, the point $C$ is equidistant from all three vertices of the isosceles right triangle $DAB'$, and thus lies on the perpendicular line to the plane of the triangle through the center of the circle circumscribed around the triangle.

The center of the circle circumscribed around the triangle $DAB'$ is the midpoint $F$ of the segment $DB'$, and $FA = FB' = FD = \frac{DB'}{2} = \frac{\sqrt{2}}{2}$. The length of the segment $CF$ can be calculated using the Pythagorean theorem in the right triangle $CFA$:

$$
CF = \sqrt{AC^2 - AF^2} = \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2}
$$

The height of the pyramid $ABCD$ corresponding to the face $ABD$ is $CF$, so the volume of the pyramid is:

$$
V = \frac{T_{ABD} \cdot CF}{3} = \frac{\left(\frac{1}{2} \cdot 1 \cdot 1\right) \cdot \frac{\sqrt{2}}{2}}{3} = \frac{\sqrt{2}}{12}
$$

Remark. If the reflection of point $A$ over the line $B'D$ is $A'$, it is easy to see that the solid $AD A' B' C$ is a regular pyramid with a square base (Figure 2). It is known that the height from $C$ in this pyramid is $a \cdot \frac{\sqrt{2}}{2}$, where $a$ is the length of the edges.

Based on the work of Gergely Horváth (Fonyód, Mátyás Király Gymnasium, 12th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_d8fc01bf37ab98059c0cg-1.jpg?height=798&width=834&top_left_y=1140&top_left_x=840)