3. In a sphere with center $O$, a regular tetrahedron $S A B C$ with edge length $a$ is inscribed. Through the vertex $A$, a plane is drawn passing through $O$ and intersecting the face $S B C$ along a line parallel to the edge $B C$. Find the area of the part of the plane that is inside the sphere but outside the tetrahedron.
Solution. The center $O$ of the sphere is the center of the circumscribed circle around the triangle $A B_{1} C_{1}$. The required area $P$ will be

$$
P=r^{2} \pi - P_{A B_{1} C_{1}}
$$

where $r$ is the radius of the sphere. Triangle $A B_{1} C_{1}$ is isosceles with base $B_{1} C_{1}$ and height $\overline{A S_{1}}=H$, the height of the tetrahedron. From triangle ASS' (see diagram) we have

![](https://cdn.mathpix.com/cropped/2024_06_05_d61831f2f852c6e23b39g-5.jpg?height=642&width=642&top_left_y=1611&top_left_x=842)

$$
H^{2}=a^{2}-\left(\frac{2}{3} \frac{a \sqrt{3}}{2}\right)^{2}=\frac{2}{3} a^{2}
$$

i.e., $H=\frac{a \sqrt{6}}{3}$.

From triangle $A O S'$ we have:

$$
r^{2}=(H-r)^{2}+\left(\frac{2}{3} \frac{a \sqrt{3}}{2}\right)^{2}
$$

i.e., $r=\frac{a \sqrt{6}}{4}$. Substituting into (1) we get

$$
P=r^{2} \pi - \frac{1}{2} \overline{B_{1} C_{1}} \cdot \overline{A S_{1}}=\frac{3 a^{2}}{8} \pi - \frac{1}{2} \frac{2}{3} a \frac{a \sqrt{6}}{3}=\frac{3 a^{2}}{8} \pi - \frac{a^{2} \sqrt{6}}{9}=\frac{a^{2}}{72}(27 \pi - 8 \sqrt{6})
$$