4 $[\quad$ Equifacetal Tetrahedron $\quad]$

In the triangular pyramid $A B C D$, the sums of the three dihedral angles at each of the vertices $B$ and $C$ are $180^{\circ}$ and $A D=B C$. Find the volume of the pyramid if the area of the face $B C D$ is 100, and the distance from the center of the circumscribed sphere to the plane of the base $A B C$ is 3.
Consider the unfolding $D 1 A D 2 B D 3 C$ of the tetrahedron $A B C D$ onto the plane of triangle $A B C$, where points $D 1, D 2$, and $D 3$ are the vertices of triangles with bases $A C, A B$, and $B C$ respectively (Fig.2). Since the sums of the three plane angles at each of the vertices $B$ and $C$ of the tetrahedron $A B C D$ are $180^{\circ}$, point $C$ lies on the segment $D 1 D 3$, and point $B$ lies on the segment $D 2 D 3$, with $C$ and $D$ being the midpoints of these segments. Therefore, $B C$ is the midline of triangle $D 1 D 2 D 3$. This means $D 1 D 2=2 B C$, and since $A D=B C$, we have $A D 1=A D 2=B C$, so $A D 1+A D 2=D 1 D 2$. This implies that point $A$ lies on the segment $D 1 D 2$, and $A$ is the midpoint of this segment. Thus, $A C, A B$, and $B C$ are the midlines of triangle $D 1 D 2 D 3$. Consequently,

$$
B D=B D 2=A C, C D=C D 1=A B
$$

i.e., the opposite edges of the tetrahedron $A B C D$ are pairwise equal. Therefore, all faces of the tetrahedron are equal triangles (by three sides) (Fig.3). We will prove that the distances from the center of the circumscribed sphere of this tetrahedron to all its faces are equal. Let $O$ be the center of the sphere of radius $R$ circumscribed around the given tetrahedron.

Perpendiculars dropped from point $O$ to the faces of the tetrahedron pass through the centers of the circumscribed circles of these faces (Fig.1). Since all faces are equal triangles, the radii of their circumscribed circles are equal. Denote them by $R 1$. Then the distances from point $O$ to the planes of the faces are $\sqrt{R^{2}-R_{1}^{2}}=r$. This means that point $O$ is equidistant from all faces of the tetrahedron at a distance $r$. By connecting point $O$ to all vertices of the tetrahedron, we divide it into four tetrahedra, the bases of which are the faces of the original tetrahedron, and the heights dropped to the bases are equal to $r$. The volume of the original tetrahedron is equal to the sum of the volumes of the four tetrahedra of our partition. Since the total surface area $S$ of the tetrahedron $A B C D$ is $4 \cdot 100=400$, and the distances from point $O$ to the faces of the tetrahedron are equal to $r$, we have

$$
V_{A B C D}=\frac{1}{3} S \cdot r=\frac{1}{3} \cdot 400 \cdot 3=400
$$

## Answer

400.00