G12 (4-3, Czechoslovakia) Given a cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Point $X$ moves uniformly along the square $A B C D$ in the direction $A B C D A$, and point $Y$ moves uniformly along the square $B^{\prime} C^{\prime} C B B^{\prime}$ in the same direction and at the same speed. Points $X$ and $Y$ start from $A$ and $B^{\prime}$ simultaneously. Find the trajectory of the midpoint of segment $X Y$.
Let $A$ be the origin, $AB, AD, AA'$ be located on the positive directions of the $x$-axis, $y$-axis, $z$-axis respectively, with the edge length as the unit of measurement, and the time it takes for $X$ to travel along one edge as the unit of time.

When $0 \leqslant t \leqslant 1$, the coordinates of $X, Y$ are $(t, 0,0),(1, t, 1)$, and the midpoint $Z = \frac{X+Y}{2} = \left(\frac{1+t}{2}, \frac{t}{2}, \frac{1}{2}\right)$;

When $1 \leqslant t \leqslant 2$, the coordinates of $X, Y$ are $(1, t-1,0),(1,1,2-t)$, and $Z = \frac{X+Y}{2} = \left(1, \frac{t}{2}, \frac{2-t}{2}\right)$;

When $2 \leqslant t \leqslant 3$, the coordinates of $X, Y$ are $(3-t, 1,0),(1,3-t, 0)$, and $Z = \frac{X+Y}{2} = \left(\frac{4-t}{2}, \frac{4-t}{2}, 0\right)$;

When $3 \leqslant t \leqslant 4$, the coordinates of $X, Y$ are $(0,4-t, 0),(1,0, t-3)$, and $Z = \frac{X+Y}{2} = \left(\frac{1}{2}, \frac{4-t}{2}, \frac{t-3}{2}\right)$.

Next, we prove: If $X=(x, y, z)$ satisfies that $x, y, z$ are linear functions of the parameter $t$, i.e., $x = x(t) = a_1 t + b_1, y = y(t) = a_2 t + b_2, z = z(t) = a_3 t + b_3, t \in [t_0, t_1]$, then the trajectory of $X$ is a line segment with endpoints $X_0(x(t_0), y(t_0), z(t_0))$ and $X_1(x(t_1), y(t_1), z(t_1))$.

This is clearly only necessary to prove that $X_0 X + X_1 X = X_0 X_1$. By the continuity of linear functions, there are no exceptional points on the line segment.
This can be seen from the following equations:
$$
\begin{array}{l}
X_0 X = \sqrt{a_1^2 + a_2^2 + a_3^2} \quad (t - t_0), \\
X_1 X = \sqrt{a_1^2 + a_2^2 + a_3^2} \quad (t_1 - t), \\
X_0 X_1 = \sqrt{a_1^2 + a_2^2 + a_3^2} \quad (t_1 - t_0).
\end{array}
$$

Returning to the previous discussion, when $t = 0, 1, 2, 3, 4$, $Z$ is located at $Z_1\left(\frac{1}{2}, 0, \frac{1}{2}\right)$, $Z_2\left(1, \frac{1}{2}, \frac{1}{2}\right)$, $Z_3(1,1,0)$, $Z_4\left(\frac{1}{2}, \frac{1}{2}, 0\right)$, $Z_1\left(\frac{1}{2}, 0, \frac{1}{2}\right)$. The coordinates $(x, y, z)$ of $Z$ are linear functions of $t$ in each time interval, so in each time interval, $Z$ forms the line segments $Z_1 Z_2, Z_2 Z_3, Z_3 Z_4, Z_4 Z_1$.
$Z_1$ is actually the midpoint of $AB'$, $Z_2$ is the midpoint of $BC'$, $Z_3$ is $C$, and $Z_4$ is the midpoint of $AC$. Since $Z_1 Z_2 \parallel Z_3 Z_4$ and $Z_1 Z_4 = \frac{1}{2} B' C = \frac{1}{2} AC = Z_3 Z_4$, the trajectory of $Z$ is the rhombus $Z_1 Z_2 Z_3 Z_1$.