Let's consider two faces of a cube with edge $a$: the squares $A B C D$ and $A B E F$. (See L. diagram.) Measure the distances $A M = F N = x$ from points $A$ and $F$ to the diagonals $A C$ and $F B$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_e4af9be20e107f0ad2fdg-1.jpg?height=388&width=353&top_left_y=234&top_left_x=867)

1. Show that as $x$ varies, $M N$ remains parallel to a fixed plane and determine the geometric locus of the midpoint of $M N$.
2. Express $M N^{2}-\mathrm{t}$ as a function of $a$ and $x$. Plot this function and analyze the variation of $M N$.
3. Let $\alpha$ and $\beta$ be the acute angles that the segment $M N$ makes with the diagonals $A C$ and $B F$, respectively. Calculate $\alpha$ and $\beta$ when $M N$ is minimal.
4. Express $\cos \alpha$ and $\cos \beta$ generally as functions of $a$ and $x$, and show that $M N$ cannot be perpendicular to both diagonals simultaneously.
5. Determine $\alpha$ when $x = \frac{a}{2}$.
1. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_4e54d539d22836a973a6g-1.jpg?height=529&width=585&top_left_y=207&top_left_x=759)

Let $M G \perp A D$ and $N H \perp A F$, then $A G M$ and $F H N$ are isosceles right triangles, whose hypotenuse $(x)$ is equal. Therefore,

$$
A G M_{\Delta} \simeq F H N_{\Delta}
$$

and thus, independently of $x$,

$$
M G=N H
$$

which implies that $M N \| G H$. However, since $G H$ is always in the plane $A D F$, $M N$ is always parallel to the plane $A D F$. (A line is parallel to a plane if it is parallel to a line in the plane.)

Since $A B \perp$ the plane $A D F$ (because it is perpendicular to two distinct lines in this plane, $A D$ and $A F$), the line $M N$, which is parallel to the plane $A D F$, is always perpendicular to $A B$. Let the plane perpendicular to $A B$ through $M N$ intersect $A B$ at point $T$. Thus, $M T \perp A B$ and $N T \perp A B$.

The perpendicular projection of the midpoint $O$ of $M N$ onto the plane $A B C D$ is necessarily the midpoint $O^{\prime}$ of $M T$, and the projection of $O$ onto the square face $A B E F$ is $O^{\prime \prime}$, which bisects the distance $N T$. ($O O^{\prime}$ and $O O^{\prime \prime}$ are the medians of the right triangle $M T N$.) Therefore, as $x$ varies from 0 to $a \sqrt{2}$, the midpoint $O$ traces a geometric locus in space whose perpendicular projection onto the square $A B C D$ is the median $A A_{1}$ of the triangle $A B C$, while the projection onto the plane $A B E F$ traces the median $B B_{1}$ of the triangle $B A F$. This implies that the geometric locus of the points $O$ is the distance $B_{1} A_{1}$, where $B_{1}$ is the midpoint of the side $A F$ and $A_{1}$ is the midpoint of the side $B C$.

2. Using the Pythagorean theorem,

$$
\begin{aligned}
M N^{2}=M T^{2} & +N T^{2}=\left(\frac{A M}{\sqrt{2}}\right)^{2}+\left(\frac{B N}{\sqrt{2}}\right)^{2}=\left(\frac{x}{\sqrt{2}}\right)^{2}+\left(\frac{a \sqrt{2}-x}{\sqrt{2}}\right)^{2}= \\
& =\frac{x^{2}}{2}+\frac{2 a^{2}-2 a x \sqrt{2}+x^{2}}{2}=x^{2}-a \sqrt{2} x+a^{2}
\end{aligned}
$$

The graph of this function is a parabola in the interval $0 \leq x \leq a \sqrt{2}$. The endpoints of this parabola are the points $\left(0, a^{2}\right)$ and $\left(\alpha \sqrt{2}, a^{2}\right)$.

$M N^{2}$ can also be written as:

$$
M N^{2}=\left(x-\frac{a \sqrt{2}}{2}\right)^{2}+\frac{\alpha^{2}}{2}
$$

from which it is clear that $M N^{2}$, and thus $M N$, is minimized when the non-negative first term is 0, i.e., when

$$
x=\frac{\alpha \sqrt{2}}{2}, \text { in which case } M N^{2}=\frac{\alpha^{2}}{2}, \text { and } M N=\frac{\alpha \sqrt{2}}{2}
$$

Thus, the focus of the parabola is the line $x=\frac{\alpha \sqrt{2}}{2}$, and its vertex is the point $\left(\frac{\alpha \sqrt{2}}{2}, \frac{\alpha^{2}}{2}\right)$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_4e54d539d22836a973a6g-1.jpg?height=387&width=507&top_left_y=2322&top_left_x=798)

## Figure 2

3. As we have seen, $M N$ is minimized when $x=\frac{a \sqrt{2}}{2}$, i.e., when $M$ and $N$ are the midpoints of the squares, and the minimum value of $M N$ is $\frac{a \sqrt{2}}{2}$. Therefore, in this case, the triangles $A M N$ and $B N M$ are equilateral, and we can state that if $M N$ is minimized, then $\angle A M N = \alpha = 60^{\circ}$, and $\angle B N M = \beta = 60^{\circ}$.
4. Since $T A = T M$ and $T B = T N$, the triangles

$$
A T N_{\Delta} \simeq M T N_{\Delta} \simeq M T B_{\Delta}
$$

are congruent because all three are right triangles with equal legs. This implies the equality of the hypotenuses, i.e., the triangles $M N A_{\Delta}$ and $N M B_{\Delta}$ are isosceles with bases $A M$ and $B N$. Therefore,

$$
\begin{aligned}
\cos \alpha & =\frac{A M}{2 M N}=\frac{x}{2 \sqrt{x^{2}-a x \sqrt{2}+a^{2}}} \\
\text { and } \cos \beta & =\frac{B N}{2 M N}=\frac{a \sqrt{2}-x}{2 \sqrt{x^{2}-a x \sqrt{2}+a^{2}}}
\end{aligned}
$$

From this, it is clear that the assumption $\alpha = \beta = 90^{\circ}$ would imply that $x$ and $a \sqrt{2} - x$ are both zero simultaneously, which is impossible. Therefore, $M N$ cannot be perpendicular to both diagonals at the same time.
5. When $x = \frac{a}{2}$,

$$
\cos \alpha=\frac{a}{4 \sqrt{\frac{a^{2}}{4}-\frac{a^{2} \sqrt{2}}{2}+a^{2}}}=\frac{1}{\sqrt{4-8 \sqrt{2}+16}}=\frac{1}{2 \sqrt{5-2 \sqrt{2}}}=\frac{1}{2 \sqrt{3.172}}
$$

from which

$$
\lg \cos \alpha=9.5305-10
$$

and thus

$$
\alpha=70^{\circ} 10^{\prime}
$$