Let's calculate the radius of the sphere inscribed in a regular tetrahedron and the radius of the sphere circumscribed around the tetrahedron.
# First solution.

The intersection point of the heights is at an equal distance from the vertices of the tetrahedron, as well as from the faces, which is why this point is the center of the spheres that can be inscribed in and circumscribed around the tetrahedron. If we connect this point to the vertices of the tetrahedron, we obtain four pyramids of equal volume, each with one of the tetrahedron's faces as its base and the radius of the inscribed sphere as its height. The volume of the tetrahedron is:

$$
V=\frac{a^{3}}{12} \sqrt{2}
$$

Thus, the volume of one pyramid is:

$$
v=\frac{a^{3}}{48} \sqrt{2}
$$

The area of one face of the tetrahedron is:

$$
t=\frac{a^{2}}{4} \sqrt{3}
$$

and thus the height of one pyramid, which is also the radius of the inscribed sphere, is:

$$
r=\frac{a}{4} \sqrt{\frac{2}{3}}=\frac{a}{12} \sqrt{6}
$$

The radius of the circumscribed sphere around the tetrahedron is obtained by subtracting the radius of the inscribed sphere from the height of the tetrahedron; the height of the tetrahedron is: $a \sqrt{\frac{2}{3}}, \mathrm{~s}$ so

$$
R=a \sqrt{\frac{2}{3}}-\frac{a}{4} \sqrt{\frac{2}{3}}=\frac{3 a}{4} \sqrt{\frac{2}{3}}=\frac{a}{4} \sqrt{6}
$$

(Hofbauer Ervin, secondary grammar school VII. grade, Budapest, autumn half-year.)

Second solution.

The radius of the sphere circumscribed around the tetrahedron is $R=O A_{1}$, the radius of the sphere inscribed in the tetrahedron is $O B=r ; O$ is the intersection point of the heights.

![](https://cdn.mathpix.com/cropped/2024_05_02_75390e162ff03d8f72bcg-1.jpg?height=458&width=502&top_left_y=1457&top_left_x=779)

The height of the tetrahedron is: $a \sqrt{\frac{2}{3}}$; thus

$$
A_{1} B=a \sqrt{\frac{2}{3}}=A_{1} O+O B=R+r
$$

Since the triangles $A_{3} B D$ and $A_{1} C O$ are similar, it follows that

$$
\begin{gathered}
A_{1} O: A_{1} D=O C: B D \\
R: A_{1} D=r: \frac{A_{1} D}{3}
\end{gathered}
$$

and thus

$$
R=3 r
$$

From (1) and (2) we get:

$$
R=\frac{3 a}{4} \sqrt{\frac{2}{3}}, r=\frac{a}{4} \sqrt{\frac{2}{3}}
$$

(Suschnik József, secondary real school VIII. grade, Kecskemét.)

The problem was also solved by: Friedmann Bernát, S.-A.-Ujhely; Grünhut Béla and Visnya Aladár, Pécs.