4. The base of the triangular pyramid $S A B C$ is an equilateral triangle $A B C$ with a side length of 4. It is known that for any point $M$ on the extension of the height of the pyramid $S H$ (point $S$ is between points $M$ and $H$), the angles $M S A, M S B, M S C, A S B, A S C$, and $B S C$ are equal to each other. A sphere with a radius of $1 \mathrm{c}$ is constructed with its center at point $S$. Find the volume of the common part of the pyramid $S A B C$ and the sphere (the volume of a sphere with radius $R$ is calculated using the formula $V=\frac{4}{3} \pi R^{3}$.)
Solution. Let's choose point $M$ (from the problem statement) such that $M A=4$. Then point $S$ will be the center of the circumscribed sphere around the regular tetrahedron $M A B C$. Consequently, each of the four identical pyramids $S A B C, S M A B, S M A C, S A B C$ will cut off an equal part from the sphere of radius 1 centered at point $S$. It is not difficult to calculate that the distance from point $S$ to the plane $A B C$ is $\frac{\sqrt{6}}{3}$. Since $\frac{\sqrt{6}}{3}<1$, to find the volume of the common part of the pyramid $S A B C$ and the sphere of radius $R=1$ centered at point $S$, we need to subtract the volume of the spherical cap of height $h=1-\frac{\sqrt{6}}{3}$ from one quarter of the volume of the sphere $\frac{1}{3} \pi R^{3}$. As is known, the latter is found using the formula $V_{c}=\pi h^{2}\left(R-\frac{h}{3}\right)$. For $R=1, h=1-\frac{\sqrt{6}}{3}$, we get $V_{c}=\pi \cdot \frac{18-7 \sqrt{6}}{27}$. Therefore, the desired volume is $\pi \cdot \frac{7 \sqrt{6}-9}{27}$.

Answer: $\pi \cdot \frac{7 \sqrt{6}-9}{27}$.