We will round the vertices of a regular octahedron with the sphere that touches all its edges. What is the surface area and volume of the resulting body?

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First, we examine whether the sphere in question (let's call it a semi-inscribed sphere) actually exists and is uniquely defined.

Both the regular octahedron and the sphere are centrally symmetric bodies. The center of the semi-inscribed sphere must coincide with the center of the octahedron. If this were not the case, then by reflecting the semi-inscribed sphere over the center of the octahedron, we would obtain another semi-inscribed sphere, which could also be obtained by translating the original one. However, during the translation, the sphere cannot move parallel to every edge of the octahedron, so the new sphere could not be a semi-inscribed sphere of our octahedron.

This implies that any plane passing through the center of the octahedron intersects the desired sphere in a great circle. Such planes include those that contain four edges of the octahedron. Four edges in a plane form a square. The circle inscribed in this square is a great circle of the sphere. Let \( a \) be the length of the edges of the octahedron. The radius of the semi-inscribed sphere is: \( R = \frac{a}{2} \).

![](https://cdn.mathpix.com/cropped/2024_05_02_9e221ed11d53712f7e67g-1.jpg?height=890&width=1422&top_left_y=612&top_left_x=340)

Figure 1

By rounding the vertices of the octahedron, we obtain the common part of the octahedron and the semi-inscribed sphere. This is also what we get when we remove the spherical caps cut off by the faces of the octahedron from the semi-inscribed sphere (Figure 1). Due to the symmetry of the regular octahedron, these spherical caps are congruent. The bases of the spherical caps are the incircles of the faces of the octahedron, so their radii are: \( \varrho = \frac{a \sqrt{3}}{6} \).

The height of the spherical caps can be determined from a plane section where the cutting plane is a plane of symmetry of the spherical cap.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e221ed11d53712f7e67g-2.jpg?height=1079&width=773&top_left_y=130&top_left_x=665)

Figure 2

In Figure 2, the plane section through four face altitudes of the octahedron is shown. The cutting plane is perpendicular to the faces that cut off the spherical caps and contains the center of the sphere, thus it is a plane of symmetry of the spherical caps. The segment \( x \) on the diagram is the distance of the faces from the center of the octahedron, i.e., the radius of the inscribed sphere of the octahedron. According to the Pythagorean theorem:

\[
x = \sqrt{R^2 - \varrho^2} = \frac{\sqrt{6}}{6} a
\]

The height of the spherical caps is:

\[
m = R - x = \frac{3 - \sqrt{6}}{6} a
\]

The surface area of the body is obtained by subtracting the surface area of the 8 cut-off spherical cap shells (spherical segments) from the surface area of the semi-inscribed sphere, and adding the area of their base circles to the resulting value. Thus,

\[
F = 4 \pi R^2 - 8(2 \pi R m) + 8 \pi \varrho^2 = \pi \frac{4 \sqrt{6} - 7}{3} a^2 \left(= a^2 \cdot 2.93\right)
\]

The volume of the body is obtained by subtracting the volume of the 8 cut-off spherical caps from the volume of the semi-inscribed sphere.

\[
V = \frac{4}{3} \pi R^3 - 8 \cdot \frac{\pi}{3} m^2 (3 R - m) = \pi \frac{14 \sqrt{6} - 27}{54} a^3 \left(= a^3 \cdot 0.4243\right)
\]

(The formulas for the surface area of a spherical segment and the volume of a spherical cap can be found in the school function tables.) (L.L.)

Reviczki Zoltán (Mezőkövesd, I. László Gymnasium, 4th grade)