Consider a plane $\epsilon$ and three non-collinear points $A,B,C$ on the same side of $\epsilon$; suppose the plane determined by these three points is not parallel to $\epsilon$. In plane $\epsilon$ take three arbitrary points $A',B',C'$. Let $L,M,N$ be the midpoints of segments $AA', BB', CC'$; Let $G$ be the centroid of the triangle $LMN$. (We will not consider positions of the points $A', B', C'$ such that the points $L,M,N$ do not form a triangle.) What is the locus of point $G$ as $A', B', C'$ range independently over the plane $\epsilon$?
Problem
Source: IMO 1961, Day 2, Problem 6
Tags: geometry, 3D geometry, Locus, Locus problems, IMO, IMO 1961
yetti
30.10.2005 03:12
Let $J \in\epsilon',\ J' \in \epsilon$ be the centroids of the triangles $\triangle ABC \in \epsilon'$, $\triangle A'B'C' \in \epsilon$. If we put equal masses m to the points $A, B, C, A', B', C'$, we can pile the 3 equal masses m at A, B, C into the centroid J of the triangle $\triangle ABC$ and and the 3 equal masses m at A', B', C' into the centroid J' of the triangle $\triangle A'B'C'$ without changing the center of mass of the mass points $A, B, C, A', B', C'$. Then we can pile the 2 equal masses 3m at the centroids $J, J'$ into the midpoint of the segment $JJ'$ without changing the system center of mass. Alternately, we can pile the 2 equal masses m at $A, A'$ into the midpoint K of AA', the 2 equal masses m at $B, B'$ into the midpoint L of BB' and the 2 equal masses m at $C, C'$ into the midpoint M of CC' without changing the center of mass of the system. Then we can pile the 3 equal masses 2m at the midpoints K, L, M into the centroid G of the triangle $\triangle KLM$ without changing the center of mass of the system. It follows that the centroid G of the triangle $\triangle KLM$ is the midpoint of the segment $JJ'$. The centroid $J \in \epsilon'$ is fixed, because the points $A, B, C \in \epsilon'$ are fixed. The centroid $J' \in \epsilon$ moves in this entire plane, because the points $A', B', C' \in \epsilon$ move in this entire plane. Thus the problem is reduced to finding the locus of the midpoint G of the segment $JJ'$, where the point $J \in \epsilon'$ is fixed and the point $J' \in \epsilon$ moves arbitrarily. Let F be the foot of a normal from the point J to the plane $\epsilon$. The locus of the midpoints G of the segments $JJ'$ obviously is a plane $\gamma$ through the midpoint N of the segment JF parallel to the plane $\epsilon$, because G is one endpoint of the midline $NG \parallel FJ' \in \epsilon$ of the right angle triangle $\triangle JFJ'$.