A unit cube has two adjacent vertices through which a sphere passes, which touches the planes of the faces not containing these vertices. What is the radius of this sphere?
Let the center of the sphere be $O$, two adjacent vertices on the sphere be $A$ and $B$, the points of tangency on the mentioned plane be $K$ and $L$, and the desired radius be $r$. Let $N$ be the midpoint of the segment $AB$, and $M$ be the intersection point of the line $OK$ with the opposite face of the cube. Use the additional notations from the diagram. Since the sphere passes through the points $A$ and $B$, the points $O, K$, and $L$ lie in the perpendicular bisector plane of the segment $AB$, and clearly $OA = OB = OK = OL = r$. From the isosceles right triangle $OMN$,

$$
ON = OM \sqrt{2} = (1 - r) \sqrt{2}
$$

Considering that $ON$ lies in the perpendicular bisector plane and $\angle ANO = 90^\circ$, we have $[(1 - r) \sqrt{2}]^2 + \left(\frac{1}{2}\right)^2 = r^2$, which simplifies to

$$
r^2 - 4r^2 + \frac{9}{4} = 0
$$

The solutions to this equation are: $r_1 = 2 - \frac{\sqrt{7}}{2}, r_2 = 2 + \frac{\sqrt{7}}{2}$.

The first solution is the radius of the sphere centered at $O$, which touches the two faces of the cube at one internal point each - the points $K$ and $L$. The second solution is the radius of a sphere centered at $O'$, for example, touching the face $DCGH$ at $L'$. In this case, $O'N = (r - 1) \sqrt{2}$, which also leads to the same equation (2). Therefore, the problem has two solutions.

Based on the work of Tamás Terpai (Fazekas M. Fóv. Gyak. Gimn., I. o.t.)

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