, 2017

Solution:

Set up a coordinate system where the coordinates of the stations are $\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)$, $\left(-\frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)$, $\left(\frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}}\right)$, and $\left(-\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}},-\frac{1}{2 \sqrt{2}}\right)$. The sum of squares of the distances is then
$$\sum_{\mathrm{sym}} 2\left(x-\frac{1}{2 \sqrt{2}}\right)^{2}+2\left(x+\frac{1}{2 \sqrt{2}}\right)^{2}=4\left(x^{2}+y^{2}+z^{2}\right)+\frac{3}{2}=4 r^{2}+\frac{3}{2}$$
where $r$ is the distance from the center of the tetrahedron. It follows from $4 r^{2}+\frac{3}{2} \leq 15$ that $r \leq \frac{3 \sqrt{6}}{4}$, so the set is a ball with radius $R=\frac{3 \sqrt{6}}{4}$, and the volume is $\frac{4 \pi}{3} R^{3}=\frac{27 \sqrt{6}}{8} \pi$.

Solution 2:

Let $P$ be the location of the base; $S_{1}, S_{2}, S_{3}, S_{4}$ be the stations; and $G$ be the center of the tetrahedron. We have:
$$\begin{gathered} \sum_{i=1}^{4} P S_{i}^{2}=\sum_{i=1}^{4} \overrightarrow{P S}_{i} \cdot \overrightarrow{P S_{i}} \\ \sum_{i=1}^{4} P S_{i}^{2}=\sum_{i=1}^{4}\left(\overrightarrow{P G}+\overrightarrow{G S_{i}}\right) \cdot\left(\overrightarrow{P G}+\overrightarrow{G S_{i}}\right) \\ \sum_{i=1}^{4} P S_{i}^{2}=\sum_{i=1}^{4}\left(\overrightarrow{P G} \cdot \overrightarrow{P G}+2 \overrightarrow{P G} \cdot \overrightarrow{G S_{i}}+\overrightarrow{G S_{i}} \cdot \overrightarrow{G S}_{i}\right) \end{gathered}$$
Since $G S_{1}=G S_{2}=G S_{3}=G S_{4}$, we have:
$$\begin{aligned} & \sum_{i=1}^{4} P S_{i}^{2}=4 P G^{2}+4 G S_{1}^{2}+2 \sum_{i=1}^{4}\left(\overrightarrow{P G} \cdot \overrightarrow{G S_{i}}\right) \\ & \sum_{i=1}^{4} P S_{i}^{2}=4 P G^{2}+4 G S_{1}^{2}+2 \overrightarrow{P G} \cdot\left(\sum_{i=1}^{4} \overrightarrow{G S_{i}}\right) \end{aligned}$$
Since $G$ is the center of the tetrahedron, $\overrightarrow{G S}_{1}+\overrightarrow{G S}_{2}+\overrightarrow{G S}_{3}+\overrightarrow{G S}_{4}=\overrightarrow{0}$. Thus:
$$\sum_{i=1}^{4} P S_{i}^{2}=4 P G^{2}+4 G S_{1}^{2}$$
Since $G S_{1}^{2}=\frac{3}{8}$, we have $P G^{2} \leq \frac{27}{8}$. Thus, the locus of all good points is a ball centered at $G$ with radius $r=\sqrt{\frac{27}{8}}$. Then, the volume is $V=\frac{4}{3} \pi r^{3}=\frac{27 \sqrt{6} \pi}{8}$.