XLIII OM - II - Problem 5

Determine the supremum of the volumes of spheres contained in tetrahedra with all altitudes no longer than $1$.
Suppose a sphere with center $ Q $ and radius $ r $ is contained within a tetrahedron with all altitudes no longer than $ 1 $. Let the volume of the tetrahedron be denoted by $ V $, and the areas of its faces by $ S_1 $, $ S_2 $, $ S_3 $, $ S_4 $; we arrange them such that $ S_1 \leq S_2 \leq S_3 \leq S_4 $.
The height of the tetrahedron perpendicular to the $ i $-th face has a length of $ 3V/S_i $ (by the formula: the volume of a pyramid is $ 1/3 $ of the product of the base area and the height). According to the given condition, these lengths do not exceed $ 1 $. In particular, $ 3V/S_1 \leq 1 $.
Notice that $ V $ is the sum of the volumes of four tetrahedra, whose bases are the faces of the given tetrahedron, and the common vertex is the point $ Q $ (i.e., the center of the considered sphere). The height of each of these four tetrahedra, drawn from the point $ Q $, has a length of at least $ r $.
From this, we conclude that

Therefore,

When the tetrahedron is regular and the sphere is inscribed in it, all the inequalities written above become equalities. Hence, the upper bound of possible values of $ r $ is $ 1/4 $. A sphere with radius $ 1/4 $ has a volume of $ \pi/48 $, and this number is the upper bound of the volumes of all considered spheres.