IMO 1960 - Problem 5

Problem: A cube $ABCDA'B'C'D'$ is given.
(a) Find the locus of all midpoints of segments $XY$, where $X$ is any point on segment $AC$ and $Y$ any point on segment $B'D'$.
(b) Find the locus of all point $Z$ on segments $XY$ such that $\vec{ZY}=2\vec{XZ}$.
Solution:
(a) The locus of the points is the square $EFGH$ where these four points are the centers of the faces $ABB'A',BCC'B',CDD'C'$ and $DAA'D'$.
(b) The locus of the points is the rectangle $IJKL$ where these points are on $AB',CB',CD'$ and $AD'$ at a distance of $\frac{AA'}{3}$ with respect to the plane $ABCD$.