A cube with edge length $a$ has two parallel square faces $A B C D$ and $E F G H$. The center of the latter face is $M$. How far is the line $M A$ from the line $B C$?
The distance between two skew lines is understood as the smallest of the distances determined by a point on each line. The minimum distance is when it forms a right angle with both lines (because in any other case, the distance can be shortened by replacing the vertex of an acute angle with the vertex of a right angle), that is, if it lies on the so-called normal-transversal: $t_{n}$.

Let the two skew lines be $a$ and $b$. A simple construction method for $t_{n}$ is: 1) Through one of the lines, for example, $a$, we draw a plane $\mathbf{S}$ which is parallel to $b$. (If we only want to determine the distance between the two skew lines, it is sufficient to determine the distance of any point on $b$ from $\mathbf{S}$.) - 2) The plane $\mathbf{N}$, which is perpendicular to $\mathbf{S}$ and passes through $b$, intersects $a$ at a point $P$ on $t_{n}$. - 3) The perpendicular line to $\mathbf{S}$ through $P$ is the desired $t_{n}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b8c29bf2eeb16024e3f7g-1.jpg?height=708&width=639&top_left_y=546&top_left_x=732)

In this case, the labeling is shown in the diagram. The line $B_{1} C_{1}$ through the point $M$ on $a$ defines the plane $A D C_{1} B_{1} = \mathbf{S}$, which is parallel to $b$. From any point on $b$, for example, $B$, we drop a perpendicular to $\mathbf{S}$: $B B_{0}$. (Since both $b$ and $\mathbf{S}$ are perpendicular to the base of the cube $A B F E$, the intersection line $A B_{1}$ of $\mathbf{S}$ with this base forms a right angle with $B B_{0}$.) The plane $\mathbf{N}$ defined by $b$ and $B B_{0}$ intersects $\mathbf{S}$ in the line $B_{0} C_{0}$. The intersection of this line with $a$ is the point $P$ of $t_{n}$. The line through $P$ parallel to $B_{0} B$ intersects $b$ at $Q$.

The sought distance is $P Q = B_{0} B$.

$$
\triangle A B B_{0} \sim \triangle B_{1} A E
$$

because both are right-angled, and $\angle A = \angle B$, as angles with perpendicular sides.

Thus,

$$
\frac{B B_{0}}{A B_{0}} = \frac{A E}{B_{1} E} = 2
$$

and therefore,

$$
a^{2} = x^{2} + (2x)^{2} = 5x^{2}, \quad \text{so} \quad x = \frac{a}{\sqrt{5}}
$$

hence,

$$
B B_{0} = 2x = \frac{2a}{\sqrt{5}} = \frac{2\sqrt{5}}{5}a
$$

Maria Detre (Esztergom, 9th grade, III. class)