$\left.\begin{array}{l}\text { [Maximum/Minimum Distance] } \\ {\left[\begin{array}{l}\text { Cube } \\ {[\text { Pythagorean Theorem in Space }}\end{array}\right]}\end{array}\right]$

On the diagonal $A C$ of the lower face of the unit cube $A B C D A_{1} B_{1} C_{1} D_{1}$, a segment $A E$ of length $l$ is laid out. On the diagonal $B_{1} D_{1}$ of its upper face, a segment $B_{1} F$ of length $m l$ is laid out. For what $l$ (and a fixed $m>0$) will the length of the segment $E F$ be the smallest?
Segment $F E$ is determined from triangle $F K E$, where $F K$ is perpendicular to the plane $A B C D, F K=1, F E=$ $\sqrt{F K^{2}+E K^{2}}$ (fig.). From triangle $K E O$, where angle $K O E$ is $90^{\circ}$ as the angle between the diagonals of a square, we find $E K^{2}=K O^{2}+O E^{2}, O K=F L=B_{1} F-L B_{1}=m l-\sqrt{2} / 2 ; O E=A O-A E=\sqrt{2} / 2-l$. Therefore, $E K^{2}=(\mathrm{ml}-\sqrt{2} / 2)^{2}+(\sqrt{2} / 2-$ $l)^{2}=l^{2}\left(m^{2}+1\right)-l \sqrt{2}(m+1)+1$. Substituting the value of $E K^{2}$ and $F K^{2}$ into the formula for $F E:$

$$
F E=\sqrt{2+l^{2}\left(m^{2}+1\right)-l \sqrt{2}(m+1)}
$$

Segment $F E$ will reach its minimum at the same values of $l$ as the expression under the square root. Let's complete the square in the last expression:

$$
\begin{gathered}
2+l^{2}\left(m^{2}+1\right)-l \sqrt{2}(m+1)= \\
=\left(m^{2}+1\right)\left(l^{2}-2 \sqrt{2} / 2 \cdot(m+1) /\left(m^{2}+1\right) l+\left((m+1)^{2}\right) /\left(2\left(m^{2}+1\right)^{2}\right)-\left((m+1)^{2}\right) /\left(2\left(m^{2}+1\right)^{2}\right)+2 /\left(m^{2}+1\right)\right)= \\
\left.=\left(m^{2}+1\right)\left((l-\sqrt{2} / 2 \cdot(m+1))\left(m^{2}+1\right)\right)^{2}+\left(4\left(m^{2}+1\right)-(m+1)^{2}\right) /\left(2\left(m^{2}+1\right)^{2}\right)\right)
\end{gathered}
$$

From this formula, it is clear that $F E$ reaches its minimum value at $l=\sqrt{2} / 2 \cdot(m+1) /\left(m^{2}+1\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6a190adc6812227a77adg-04.jpg?height=551&width=675&top_left_y=-1&top_left_x=700)

## Answer

at $l=\sqrt{2} / 2 \cdot(m+1) /\left(m^{2}+1\right)$.