IMO 1961 - Problem 6

Problem: A plane $\epsilon$ is given and on one side of the plane three nocollinear points $A,B$ and $C$ such that the plane determined by them is not parallel to $\epsilon$. Three arbitrary points $A',B'$ and $C'$ in $\epsilon$ are selected. Let $L,M$ and $N$ be the midpoints of $AA',BB'$ and $CC'$ and $G$ the centroid of $\triangle{LMN}$. Find the locus of all points obtained for $G$ as $A',B'$ and $C'$ are varied (independently of each other) across $\epsilon$.
Solution: Let $h(X)$ denote the distance of a point $X$ from $\epsilon$, $X$ restricted to being on the same side of $\epsilon$ as $A,B$ and $C$. Let $G_1$ be the (fixed) centroid of $\triangle{ABC}$ and $G_1'$ the centroid of $\triangle{A'B'C'}$. It is trivial to prove that $G$ is the midpoint of $G_1G_1'$. Hence varying $G_1'$ across $\epsilon$, we get that the locus of $G$ is the plane $\alpha$ parallel to $\epsilon $ such that:
$X\in \alpha\iff h(X)=\frac{h(G_1)}{2}=\frac{h(A)+h(B)+h(C)}{6}$.