4. Determine the volume of a triangular pyramid whose faces have areas $S_{0}, S_{1}, S_{2}, S_{3}$, and the dihedral angles at the face with area $S_{0}$ are equal.
Solution. Let the base $ABC$ of the pyramid $ABCD$ have area $S_{0}$. Let $O$ be the foot of the height of the pyramid, and let $A_{1}, B_{1}, C_{1}$ be the feet of the heights from vertex $D$ to the sides $BC, CA, AB$ of the triangular faces $BCD, CAD, ABD$, respectively. By assumption, the angles $\angle D A_{1} O, \angle D B_{1} O, \angle D C_{1} O$ are equal, so the triangles $D A_{1} O, D B_{1} O, D C_{1} O$ are congruent, see the diagram. Therefore, point $O$ is the center of the inscribed circle in triangle $ABC$, and let $r$ be the radius of this circle. Further, let $DO=H, DA_{1}=DB_{1}=DC_{1}=h, BC=a$, $CA=b, AB=c$, and $S_{1}+S_{2}+S_{3}=S$.

From the triangles $BCD, CAD, ABD$, we obtain $a=\frac{2 S_{1}}{h}, b=\frac{2 S_{2}}{h}, c=\frac{2 S_{3}}{h}$, and for the semiperimeter of triangle $ABC$, we have $s=\frac{a+b+c}{2}=\frac{S}{h}$. From this, we get

$$
S_{0}^{2}=s(s-a)(s-b)(s-c)=\frac{S\left(S-2 S_{1}\right)\left(S-2 S_{2}\right)\left(S-2 S_{3}\right)}{h^{4}}
$$

so

$$
h=\sqrt[4]{\frac{S\left(S-2 S_{1}\right)\left(S-2 S_{2}\right)\left(S-2 S_{3}\right)}{S_{0}^{2}}}
$$

Furthermore,

$$
r=\frac{S_{0}}{s}=\frac{S_{0} h}{S} \text { and } H=\sqrt{h^{2}-r^{2}}=\frac{h}{S} \sqrt{S^{2}-S_{0}^{2}}
$$

so the desired volume is

$$
V=\frac{1}{3} S_{0} H=\frac{1}{3} \sqrt{S_{0}\left(S^{2}-S_{0}^{2}\right)} \cdot \sqrt{\frac{\left(S-2 S_{1}\right)\left(S-2 S_{2}\right)\left(S-2 S_{3}\right)}{S^{3}}}
$$

## 4th Year