Let's determine the radius of the sphere that touches three faces of a regular tetrahedron with unit edge length, as well as the three sides of the fourth face.
Solution. Use the notations in the figure!

![](https://cdn.mathpix.com/cropped/2024_05_02_92d316f7d3ce2dd49257g-1.jpg?height=441&width=412&top_left_y=224&top_left_x=846)

The sphere touches the faces $ABD$, $BCD$, and $ACD$ of the tetrahedron, as well as the sides $AB$, $BC$, and $AC$ of the face $ABC$.

The center $O$ of the sphere, due to the symmetry properties of the tetrahedron, lies on the axis of rotational symmetry $DT$. The sphere touches the sides of the equilateral triangle $ABC$ at their midpoints. Since these midpoints are also points on the lateral faces, the sphere can only touch the lateral faces at these points. (This is possible.) $T$ is the centroid of the base, so $AT: TF = 2: 1$.

The radius $OF$ of the sphere is perpendicular to the tangent $DF$, so the triangle $OTF$ is similar to the triangle $OFD$ (both are right triangles and share one angle), from which $OT: r = TF: FD (= TF: AF) = 1: 3$ follows.

Applying the Pythagorean theorem to the sides of the triangle $OTF$: $r^2 = \left(\frac{r}{3}\right)^2 + TF^2$, where $TF = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}$, so $\frac{8}{9} r^2 = \frac{3}{36}$, $r^2 = \frac{3}{32}$, $r = \frac{1}{4} \cdot \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{8}$. Therefore, the radius of the sphere in question is $\frac{\sqrt{6}}{8}$ units long.