From the midpoint $K$ of one edge of the cube, we draw a tangent to the inscribed sphere, which intersects an opposite edge. Let the point of tangency be $E$, and the intersection point be $F$. What is the ratio $K E: E F$?
The ratio of distances does not change if we choose the edge of the cube to be 2 units. In this case, the radius of the inscribed sphere is $R=1$ unit.

Let's denote three vertices of the cube as $A, B, C$ (as shown in the figure), and the center of the sphere as $O$. If we cut the cube with a plane parallel to the base and passing through point $K$, the resulting square contains the center of the sphere, $O$, and due to the central symmetry of the sphere and the cube, $K K^{\prime}$ is the diagonal of the square, so its length is $2 \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6008b242ce7bdb907d7dg-1.jpg?height=650&width=788&top_left_y=434&top_left_x=644)

Since the radius drawn at the point of tangency is always perpendicular to the tangent, the triangle $K O E$ is a right triangle, so

$$
K O=\sqrt{2}, \quad O E=R=1, \quad \sin O K E \angle=\sin \alpha=\frac{\sqrt{2}}{2},
$$

thus

$$
\alpha=45^{\circ}, \quad \text { so } \quad K E=1
$$

Next, let's calculate the distance $K F$; let $B F=x$.

From the right triangle $A B F$,

$$
A F^{2}=x^{2}+2^{2}
$$

and from the right triangle $K A F$,

$$
K F^{2}=A F^{2}+A K^{2}=x^{2}+2^{2}+1^{2}=x^{2}+5
$$

From the right triangle $F C K^{\prime}$,

$$
F K^{\prime 2}=(2-x)^{2}+1^{2}=5-4 x+x^{2}
$$

Finally, in the triangle $F K^{\prime} K$, by the cosine rule,

$$
K^{\prime} F^{2}=F K^{2}+K K^{\prime 2}-2 F K \cdot K K^{\prime} \cos 45^{\circ}
$$

Substituting the values obtained earlier,

$$
5-4 x+x^{2}=x^{2}+5+(2 \sqrt{2})^{2}-2 \cdot 2 \sqrt{2} \cdot \sqrt{x^{2}+5} \cdot \frac{\sqrt{2}}{2}
$$

Performing the operations and simplifications,

$$
x+2=\sqrt{5+x^{2}}
$$

squaring both sides,

$$
x=B F=\frac{1}{4}
$$

From this, $K F=\sqrt{x^{2}+5}=\frac{9}{4}, E F=\frac{5}{4}$, so the desired ratio is

$$
K E: E F=1: \frac{5}{4}=4: 5
$$