12.4.3 ** Let $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a cube, $X$ be a moving point on line segment $A C$, and $Y$ be a moving point on line segment $B^{\prime} D^{\prime}$.
(1) Find the locus of the midpoint of line segment $X Y$;
(2) If $Z$ is a point on line segment $X Y$ such that $Z Y=2 X Z$, find the locus of point $Z$.
We notice a fact: If plane $\pi \parallel \pi^{\prime}$, and $X$, $Y$ are moving points in $\pi$ and $\pi^{\prime}$ respectively, $Z$ is on the line segment $XY$, and $XZ:ZY = c:1$, then the trajectory of point $Z$ is a plane $\pi^{\prime \prime}$ parallel to both $\pi$ and $\pi^{\prime}$, and the distance ratio from $\pi^{\prime \prime}$ to $\pi$ and $\pi^{\prime}$ is $c:1$, as shown in Figure a. Therefore, we know that the trajectories in (1) or (2) are both in a plane parallel to plane $ABCD$. Let the edge length of $ABCD-A^{\prime}B^{\prime}C^{\prime}D^{\prime}$ be $a$.
Due to a
Group b
(1) Let $Z$ be the midpoint of $XY$. As shown in Figure b, first fix point $X$, and let $Y$ move on the line segment $B^{\prime}D^{\prime}$. It is easy to see that the trajectory of point $Z$ is the midline of $\triangle XB^{\prime}D^{\prime}$ parallel to side $B^{\prime}D^{\prime}$, with a length of $\frac{\sqrt{2}}{2}a$. Then let $X$ move on $AC$, and we can see that the trajectory of point $Z$ is a square with side length $\frac{\sqrt{2}}{2}a$ and equal distances to the two bases.
(2) Similarly to (1), the trajectory of point $Z$ is a rectangle with side lengths $\frac{\sqrt{2}}{3}a$ and $\frac{2\sqrt{2}}{3}a$, and the distance ratio from the upper base $A^{\prime}B^{\prime}C^{\prime}D^{\prime}$ and the lower base $ABCD$ is $2:1$.