A sphere passes through the $A$ and $C$ vertices of the unit cube's base $ABCD$ and touches the side line passing through $B$ and the plane of the top face of the cube. Calculate the radius of the sphere.
The center of the sought sphere $G$ lies in the perpendicular bisector plane $S$ of the base diagonal $AC$, which is also the vertical plane through the diagonal $BD$. This plane is perpendicular to the top face plane $V$ of the cube, so the point of tangency $T$ between $V$ and the sphere lies on the intersection line of the two planes, which is the diagonal line $FH$ on the top face, parallel and coincident with $BD$. Therefore, we can say that the sphere is tangent to the line $FH$. $G$ is below $V$ (i.e., $FH$), since the point $A$ on the sphere's surface is below $V$ (see the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_cfb5f549c17aeb95a68dg-1.jpg?height=629&width=919&top_left_y=385&top_left_x=598)

According to our notation, the sphere is tangent to the edge line $FB$ at a point $U$. Since $FB$ is also in $S$, the circle in $S$ that is tangent to the line $FH$ at $T$ and to the half-line $FB$ at $U$ is the orthogonal projection of the sphere onto $S$. Therefore, it contains the projections of $A$ and $C$ onto $S$, which coincide with each other and with the midpoint $K$ of the diagonal $BD$. Thus, $G$ must be in the right-angle domain $BFH$ of $S$ that contains $D$, on its angle bisector $f$.

We determine the radius $r$ of the sphere from the conditions $GA = GT = GU$. Let the projection of $K$ onto $GU$ be $L$, so

$$
\begin{gathered}
GA^2 = GL^2 + LK^2 + KA^2, \quad \text{where} \\
GL = |GU - LU| = |r - KB| = \left|r - \frac{1}{\sqrt{2}}\right| \\
LK = UB = |FB - FU| = |1 - r|, \quad KA = \frac{1}{\sqrt{2}}
\end{gathered}
$$

With these,

$$
\begin{gathered}
r^2 = \left(r - \frac{1}{\sqrt{2}}\right)^2 + (1 - r)^2 + \frac{1}{2} \\
r^2 - (2 + \sqrt{2}) r + 2 = 0
\end{gathered}
$$

Both roots are positive and valid: $r_1 = 0.751$ (as shown in the diagram), $r_2 = 2.663$.

Imre Böösi (Győr, Czuczor G. Bencés Gimn., IV. o. t.)

Remark. Below we outline an approach to solving the problem using the concepts, relationships, and methods of spatial coordinate geometry. We note that such solutions, in the spirit of our competition, can only be considered as solution outlines. We will see that if we add a little explanation to the formally derived steps from the plane, the solution becomes more extensive than the one above, and here the coordinate geometry does not simplify the problem.

Place the cube in a spatial Cartesian coordinate system such that the coordinates of the vertices of the base are

$$
A(0,0,0), \quad B(1,0,0), \quad C(1,1,0), \quad D(0,1,0)
$$

Let the center of the sphere be the point $G$ with coordinates $(a, b, c)$, and its radius be $r$. Thus, the equation of the sphere is

$$
(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2
$$

i.e., the coordinates $(x, y, z)$ of every point on the sphere's surface satisfy this equation.

Since the sphere passes through the points $A$ and $C$, we have

$$
\begin{gathered}
a^2 + b^2 + c^2 = r^2 \\
(1-a)^2 + (1-b)^2 + c^2 = r^2
\end{gathered}
$$

The edge of the cube passing through $B$ is parallel to the $z$-axis, so the first two coordinates of every point on this edge are the same as the first two coordinates of $B$, and the third coordinate is arbitrary: $(1, 0, z)$. For the point of tangency $U$ between the edge and the sphere, the radius $GU$ is perpendicular to the edge, so it is parallel to the plane formed by the $x$ and $y$ axes, and thus the third coordinate of $U$ is $c$. The point $U(1, 0, c)$ lies on the sphere, so

$$
(1-a)^2 + b^2 = r^2
$$

The equation of the top face of the cube is $z=1$. The center $G$ is below this plane because the points $A$ and $C$ on the sphere are below the top face. Let the point of tangency between the top face and the sphere be $T$, with coordinates $(a, b, 1)$, since the radius $GT$ is perpendicular to the top face and parallel to the $z$-axis. Thus, correctly,

$$
\overline{GT} = 1 - c = r
$$

Now, from the system of equations (1) to (4), we can find the desired $r$ by eliminating the coordinates of $G$. From (4),

$$
c = 1 - r
$$

Subtracting (2) from (1) and rearranging,

$$
a + b = 1
$$

(this expresses that $G$ lies in the plane through $B$ and $D$ and parallel to the $z$-axis).

Subtracting (3) from (1) and substituting (5), and rearranging,

$$
a = r - \frac{r^2}{2}
$$

and thus from (6),

$$
b = 1 - r + \frac{r^2}{2}
$$

Finally, substituting (7), (8), and (5) into (1) and rearranging,

$$
r^4 - 4r^3 + 6r^2 - 8r + 4 = 0
$$

The left side can be factored into a difference of squares, and thus into a product:

$$
\left(r^2 - 2r + 2\right)^2 - (\sqrt{2} r)^2 = \left[r^2 - (2 + \sqrt{2}) r + 2\right] \cdot \left[r^2 - (2 - \sqrt{2}) r + 2\right] = 0
$$

Setting the first factor to zero gives the equation and roots of the first solution, while the equation derived from the second factor has no real roots, since its discriminant is $(2 - \sqrt{2})^2 - 8 < 0$. - This completes the answer.

With the radius known, the coordinates of $G$ can also be determined from (7), (8), and (5); for $r_1$: $a_1 = 0.469, b_1 = 0.531, c_1 = 0.249$ and for $r_2$: $a_2 = -0.886, b_2 = 1.886, c_2 = -1.663$ (essentially, these coordinates help visualize the position of the sought sphere).

To determine the coordinates of $U$, we could also say: the edge $B$ is the intersection line of the side face with the equation $x=1$ (in the back in the diagram) and the side face with the equation $y=0$ (on the right).

Based on the work of Erzsébet Kosztán (Kaposvár, Táncsics M. Gimn., IV. o. t.), with significant additions.