XXI OM - I - Problem 12

Find the smallest positive number $ r $ with the property that in a regular tetrahedron with edge length 1, there exist four points such that the distance from any point of the tetrahedron to one of them is $ \leq r $.
Let $ O $ be the center of the sphere circumscribed around the regular tetrahedron $ ABCD $ with edge $ 1 $ (Fig. 9). The radius of the sphere circumscribed around $ ABCD $ is $ R = \frac{1}{4} \sqrt{6} $. The tetrahedron $ ABCD $ is divided into tetrahedra with vertices: $ O $, one of the vertices of the tetrahedron, the midpoint of one of the edges emanating from that vertex of the tetrahedron, and the foot of the perpendicular from one of the faces containing that edge. Since from each of the $ 4 $ vertices of the tetrahedron, $ 3 $ edges emanate, and each of them belongs to two faces of the tetrahedron, the number of considered tetrahedra is $ 4 \cdot 3 \cdot 2 = 24 $.
We will prove that each of these $ 24 $ tetrahedra is inscribed in one of the four spheres with radius $ \frac{R}{2} $ and centers $ P_A $, $ P_B $, $ P_C $, $ P_D $, which are the midpoints of the segments $ \overline{OA} $, $ \overline{OB} $, $ \overline{OC} $, $ \overline{OD} $, respectively. Let $ E $ be the midpoint of the edge $ \overline{AB} $, and $ F $ the foot of the perpendicular dropped from point $ D $ to the plane of triangle $ ABC $. Consider the tetrahedron $ AEFO $. Since angle $ AFO $ is a right angle, $ P_AF = AP_A = OP_A = \frac{R}{2} $.
Similarly, to prove that $ P_AE = \frac{1}{2} R $, it suffices to state that angle $ AEO $ is a right angle. This follows from the fact that the plane defined by points $ D $, $ E $, $ F $ is perpendicular to the edge $ \overline{AB} $, because $ DE \bot AB $, $ CE \bot AB $, and $ S \in CE $.
In this way, we prove that each of the $ 6 $ tetrahedra, one of whose vertices is $ A $, is contained in a sphere with center $ P_A $ and radius $ \frac{R}{2} $. By symmetry, it follows that the remaining tetrahedra are also contained in the corresponding spheres with radius $ \frac{R}{2} $.
Thus, every point of the tetrahedron $ ABCD $ is at a distance from one of the points $ P_A $, $ P_B $, $ P_C $, $ P_D $ not greater than $ \frac{R}{2} $.
Suppose that certain points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ and a number $ S $ have the property that every point of the tetrahedron $ ABCD $ is at a distance from one of the points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ not greater than $ S $. In particular, each of the five points $ O $, $ A $, $ B $, $ C $, $ D $ is at a distance from one of the points $ Q_1 $, $ Q_2 $, $ Q_3 $, $ Q_4 $ not greater than $ S $. Therefore, two of the points $ O $, $ A $, $ B $, $ C $, $ D $ are at a distance from some point $ Q_i $ not greater than $ S $. If these points are vertices of the tetrahedron, for example, $ A $ and $ B $, then we obtain $ 1 = AB \leq AQ_i + BQ_i \leq 2S $, i.e., $ S \geq \frac{1}{2} $. If, however, these points are $ O $ and one of the vertices of the tetrahedron, for example, $ A $, then we obtain $ R = OA \leq OQ_i + AQ_i \leq 2S $, i.e., $ S \geq \frac{R}{2} $.
Since $ \frac{R}{2} < \frac{1}{2} $, in each case we obtain that $ S \geq \frac{R}{2} $. Therefore, the smallest positive number $ r $ with the property given in the problem is $ r = \frac{R}{2} = \frac{1}{8} \sqrt{6} $.
Note. It can be proved that the points $ P_A $, $ P_B $, $ P_C $, $ P_D $ are the only points such that every point of the tetrahedron is at a distance from one of them not greater than $ \frac{\sqrt{6}}{8} $.