What is the geometric locus of points $P$ in space from which three tangent planes can be drawn to a given sphere of radius $r$, such that these planes are pairwise perpendicular to each other?

The three points of tangency determine a circle $k$ on the sphere. Determine:

a) the surface area of the smaller spherical cap associated with $k$,
b) the surface area of the conical frustum determined by $k$ and $P$.

How do these two surface areas compare to each other?
Let the center of the given sphere be denoted by $O$. Consider a suitable point $P$, and three mutually perpendicular tangent planes passing through it. By drawing planes tangent to the sphere and parallel to each of these tangent planes, we arrive at a cube circumscribed around the sphere, with edge length $2r$. Each vertex of this cube satisfies the conditions of the problem. These vertices, and thus all points that satisfy the conditions, lie on the surface of a sphere $\mathbf{G}$ concentric with the given sphere, whose radius is $OP$, half the space diagonal of the cube, i.e., $OP = r \sqrt{3}$. By rotating such a cube around $O$, one of its vertices can be rotated to any point on the surface of $\mathbf{G}$ (with the other vertices also moving on the surface of $\mathbf{G}$), meaning that every point on $\mathbf{G}$ is a suitable point. Therefore, the geometric locus of points $P$ is the surface of the sphere $\mathbf{G}$.

a) The points of tangency on the given sphere of radius $r$ form an equilateral triangle, whose side is the distance between the centers of two adjacent cube faces: $r \sqrt{2}$. Since the radius of the circumscribed circle of an equilateral triangle with side length $a$ is $\frac{2}{3} \cdot \frac{a}{2} \sqrt{3} = \frac{a}{3} \sqrt{3}$, the radius of the circle $k$ is

$$
\varrho = \frac{1}{3} r \sqrt{2} \cdot \sqrt{3} = \frac{r \sqrt{6}}{3}
$$

The surface area of the spherical cap is

$$
F_{1} = 2 r \pi m = 2 r \pi \left( r - \sqrt{r^{2} - \varrho^{2}} \right) = \frac{2(3 - \sqrt{3})}{3} r^{2} \pi
$$

b) The length of the generator of the cone is the distance from the points of tangency to $P$, which is nothing other than half the diagonal of the cube face: $r \sqrt{2}$. (Thus, the three points of tangency and point $P$ form the four vertices of a regular tetrahedron.)

The lateral surface area of the cone is

$$
F_{2} = \varrho \pi r \sqrt{2} = \frac{r \sqrt{6}}{3} r \sqrt{2} \pi = \frac{2 \sqrt{3}}{3} r^{2} \pi
$$

The ratio of the two surface areas is

$$
\frac{F_{1}}{F_{2}} = \frac{3 - \sqrt{3}}{\sqrt{3}} = \sqrt{3} - 1
$$

Károly Danassy (Mosonmagyaróvár, Kossuth L. g. II. o. t.)