152. In the tetrahedron $ABCD$, it is given that $|BC|=|CD|=$ $=|DA|, |BD|=|AC|, |BD|>|BC|$, and the dihedral angle at edge $AB$ is $\pi / 3$. Find the sum of the other dihedral angles.
152. First, let's prove that the dihedral angles at the edges $DB$ and $AC$ are equal to $\pi / 2$. Let $|AD| = |CD| = |BC| = a$, $|BD| = |AC| = b$, $|AB| = c$, and $b > a$. Drop perpendiculars $DK$ and $CL$ from $D$ and $C$ to the edge $AB$ (Fig. 31, a). Denote

$|AK| = |BL| = |x|, \quad |KL| = |c - 2x|, \quad |DK| = |CL| = h$.

Since the dihedral angle at the edge $AB$ is $\pi / 3$, we have $|DC|^2 = |DK|^2 + |CL|^2 - |DK| \cdot |CL| + |KL|^2$, i.e., $a^2 = h^2 + (c - 2x)^2$; substituting $h^2 = a^2 - x^2$, we get $3x^2 - 4cx + c^2 = 0$, from which $x_i = c / 3, x_2 = c$. From the condition $b > a$, it follows that $x < c / 2$, hence $x = c / 3$. Thus, the quantities $a, b$, and $c$ are related by the equation $c^2 = 3(b^2 - a^2)$.

Now, let's find the areas of triangles $ABD$ and $ACD$:

$$
\begin{aligned}
& S_{ABD} = S_{ABC} = \frac{1}{2} c \sqrt{a^2 - \frac{c^2}{9}} = \frac{1}{2} c \sqrt{\frac{4a^2 - b^2}{3}} \\
& S_{ACD} = S_{BDC} = \frac{1}{4} b \sqrt{4a^2 - b^2}.
\end{aligned}
$$

Expressing the volume of the tetrahedron $ABCD$ using the formula from problem 11 through the dihedral angle at the edge $AB$ and the areas of the faces $ABD$ and $ABC$,
![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-091.jpg?height=326&width=770&top_left_y=730&top_left_x=242)

Fig. 31,

and then through $\varphi$ - the dihedral angle at the edge $AC$ (which is also equal to the angle at the edge $BD$) and the areas of the faces $ABC$ and $ACD$, we get

$$
V_{ABCD} = \frac{1}{3} \frac{S_{ABD} S_{ABC}}{|AB|} \cdot \frac{\sqrt{3}}{2} = \frac{1}{3} \frac{S_{ACD} S_{ABC}}{|AC|} \sin \varphi
$$

from which

$\sin \varphi = \frac{S_{ABD}}{S_{ACD}} \frac{|AC|}{|AB|} \cdot \frac{\sqrt{3}}{2} = $

$$
= \frac{2c \sqrt{\frac{4a^2 - b^2}{3}}}{b \sqrt{4a^2 - b^2}} \cdot \frac{b}{c} \cdot \frac{\sqrt{3}}{2} = 1.
$$

Thus, $\varphi = \frac{\pi}{2}$.

To determine the sum of the three remaining dihedral angles, consider the prism $BCDMNA$ (Fig. 31, b). The tetrahedron $ABCN$ is equal to the tetrahedron $ABCD$ because the plane $ABC$ is perpendicular to the plane $ADCN$, and $ADCN$ is a rhombus, hence the tetrahedra $ABCD$ and $ABCN$ are symmetric with respect to the plane $BCA$. Similarly, the tetrahedron $ABMN$ is symmetric to the tetrahedron $ABCN$ with respect to the plane $ABN$ (the angle at the edge $BN$ in the tetrahedron $ABCN$ is equal to the angle at the edge $BD$ in the tetrahedron $ABCD$, i.e., equal to $\pi / 2$), hence the tetrahedron $ABMN$ is equal to the tetrahedron $ABCN$ and is equal to the original tetrahedron $ABCD$.

The dihedral angles of the prism at the edges $CN$ and $BM$ are equal to the dihedral angles at the edges $DC$ and $BC$ of the tetrahedron $ABCD$. Since the sum of the dihedral angles at the lateral edges of a triangular prism is $\pi$, the sum of the dihedral angles at the edges $AD$, $DC$, and $CB$ of the tetrahedron $ABCD$ is also $\pi$, and the sum of all dihedral angles of the tetrahedron, excluding the given angle at the edge $AB$, is $2\pi$.