A sphere of given radius touches all three planes forming a corner of a certain cuboid. What is the geometric locus of the center of the sphere if the sphere occupies all the positions described above?
We denote the circular rim by $k$, its center by $K$, its radius by $r$, the vertex of the angle by $O$, the bounding planes by $S_{1}$, $S_{2}$, and $S_{3}$, the plane of $k$ by $S$, the point of tangency of $k$ on $S_{i}$ by $E_{i}$, the projection of $K$ on $S_{i}$ by $K_{i}$, the line through $O$ perpendicular to $S_{i}$ by $e_{i}$ $(i=1,2,3)$, and the line through $O$ perpendicular to $S$ by $e$. Measure $r$ from $O$ along $e$ in one direction, and denote the resulting endpoint by $R$. Draw a plane through $R$ parallel to $S_{i}$, intersecting $e_{i}$ at $R_{i}$ $(i=1,2,3)$ (see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_275589f95dffc0535644g-1.jpg?height=474&width=550&top_left_y=392&top_left_x=771)

The following observations hold for each $i=1,2,3$. Since $k$ is tangent to the plane $S_{i}$, and $S$ and $S_{i}$ are different, $K_{i}$ is different from $K$ and $R_{i}$ is different from $R$. If $S \perp S_{i}$, then $K_{i}$ coincides with $E_{i}$, and $R_{i}$ coincides with $O$. Otherwise, the points $K, E_{i}, K_{i}$ and the points $R, O, R_{i}$ form real triangles. In these triangles, there is a right angle at $K_{i}$ and at $R_{i}$, and the angle at $E_{i}$ in the first triangle is equal to the angle at $O$ in the second triangle, since the sides of these angles are perpendicular, and both angles are acute. Therefore,

$$
O R_{i}=E_{i} K_{i} \quad \text { and } \quad O R_{i}^{2}+R_{i} R^{2}=E_{i} K_{i}^{2}+K_{i} K^{2}=r^{2}
$$

It is easily seen that (1) is also valid when $S \perp S_{i}$, since in this case $O R_{i}=E_{i} K_{i}=0$.

Denote the projection of $R$ on $S_{1}$ by $R_{1}^{*}$, then

$$
O R^{2}=O R_{1}^{* 2}+R_{1}^{*} R^{2}=O R_{2}^{2}+O R_{3}^{2}+O R_{1}^{2}
$$

Similarly, we obtain

$$
O K^{2}=K K_{1}^{2}+K K_{2}^{2}+K K_{3}^{2}
$$

From (1) and (2),

$$
\begin{gathered}
O K^{2}=\left(r^{2}-E_{1} K_{1}^{2}\right)+\left(r^{2}-E_{2} K_{2}^{2}\right)+\left(r^{2}-E_{3} K_{3}^{2}\right)= \\
=3 r^{2}-\left(O R_{1}^{2}+O R_{2}^{2}+O R_{3}^{2}\right)=3 r^{2}-O R^{2}=2 r^{2}
\end{gathered}
$$

If, therefore, $K$ belongs to the geometric locus, then

$$
O K^{2}=2 r^{2}, \quad 00$, and $O R_{i}<r$, so $R_{i}$ is different from $R$, hence $S$ cannot be parallel to $S_{i}$. Denote the intersection line of $S$ and $S_{i}$ by $m_{i}$, and the perpendicular projection of $K$ on $m_{i}$ by $E_{i}$. Since $O R \perp K E_{i}$, and $R R_{i}=K K_{i}$, the triangles $O R R_{i}$ and $E_{i} K K_{i}$ are congruent (possibly with $O$ coinciding with $R_{i}$, and $E_{i}$ coinciding with $K_{i}$), thus

$$
K E_{i}=O R=r
$$

Therefore, the circle $k$ of radius $r$ centered at $K$ in the plane $S$ is tangent to the line $m$. Hence, $K$ indeed belongs to the geometric locus.

Remark. In the second part of our solution, it was sufficient to find one suitable position for $k$. The fact that we only provided one such position does not mean that there is only one. Generally, there are four suitable circles, except when $K$ lies on the surface of the cube, in which case there are only two.