IMO 1961 Problem 6
The problem asks for the locus of the centroid of a triangle whose vertices are the midpoints of three segments joining fixed points above a plane to arbitrary points on the plane.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 3m04s
Problem
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 $\epsilon$?
Exploration
The problem asks for the locus of the centroid of a triangle whose vertices are the midpoints of three segments joining fixed points above a plane to arbitrary points on the plane. Since the data are affine in nature, vector methods appear natural. The centroid is itself an affine combination, and midpoints are affine combinations as well, so the entire construction should collapse to a simple linear relation.
A first experiment is to place the plane $\epsilon$ as $z=0$ in a Cartesian coordinate system. Let
$$A=(a_1,a_2,a_3), \quad B=(b_1,b_2,b_3), \quad C=(c_1,c_2,c_3),$$
with $a_3,b_3,c_3>0$, and let
$$A'=(x_1,x_2,0), \quad B'=(y_1,y_2,0), \quad C'=(z_1,z_2,0).$$
Then
$$L=\frac{A+A'}2,\quad M=\frac{B+B'}2,\quad N=\frac{C+C'}2.$$
The centroid is
$$G=\frac{L+M+N}{3} =\frac{A+B+C+A'+B'+C'}6.$$
Its third coordinate equals
$$\frac{a_3+b_3+c_3}{6},$$
which is independent of the choices of $A',B',C'$. This strongly suggests that the locus is a plane parallel to $\epsilon$.
The next question is whether every point of that parallel plane can actually occur. The horizontal coordinates of $G$ are
$$\frac{a_1+b_1+c_1+x_1+y_1+z_1}{6}, \qquad \frac{a_2+b_2+c_2+x_2+y_2+z_2}{6}.$$
Since the six variables $x_i,y_i,z_i$ are arbitrary apart from the condition that $L,M,N$ form a triangle, it appears that any horizontal position can be achieved. The degeneracy condition on $L,M,N$ should not obstruct the locus because one may choose $A',B',C'$ generically.
A second approach is purely geometric. Let $P$ be the centroid of triangle $ABC$ and let $Q$ be the centroid of triangle $A'B'C'$. Since midpoint and centroid constructions commute with affine averaging,
$$G$$
should be the midpoint of segment $PQ$. The point $P$ is fixed, while $Q$ ranges over $\epsilon$, because every point of a plane is the centroid of infinitely many triangles contained in that plane. Then the locus of the midpoint of $PQ$ as $Q$ varies over $\epsilon$ is precisely the image of $\epsilon$ under the homothety centered at $P$ with ratio $\tfrac12$. Hence the locus is a plane parallel to $\epsilon$.
The geometric approach is cleaner and reveals the structure immediately. The delicate point is proving rigorously that every point of $\epsilon$ can occur as the centroid of some nondegenerate triangle $A'B'C'$. A careless argument might ignore the nondegeneracy condition.
Problem Understanding
We are given a fixed plane $\epsilon$ and three fixed non-collinear points $A,B,C$ lying on the same side of $\epsilon$. The plane through $A,B,C$ is not parallel to $\epsilon$. For arbitrary points $A',B',C'$ in $\epsilon$, define $L,M,N$ as the midpoints of $AA',BB',CC'$ respectively. Let $G$ be the centroid of triangle $LMN$. The task is to determine the locus of all possible positions of $G$.
This is a Type A problem, because the problem asks for a complete characterization of all possible points $G$.
The objects involved are affine constructions: midpoints, centroids, and parallel planes. The main difficulty is not computing one position of $G$, but proving both directions of the classification. One must prove first that every admissible point $G$ lies on a certain geometric set, and second that every point of that set is actually attained by some nondegenerate choice of $A',B',C'$.
The expected answer is a plane parallel to $\epsilon$. Intuitively, the centroid operation averages the three midpoint constructions, so the dependence on the arbitrary points $A',B',C'$ reduces to the centroid of a triangle contained in $\epsilon$. Since centroids of triangles in $\epsilon$ fill the whole plane $\epsilon$, the resulting points $G$ should fill a translated copy of $\epsilon$ halfway toward the fixed centroid of $ABC$.
Proof Architecture
The proof will proceed through three lemmas.
Lemma 1 states that if $P$ and $Q$ are respectively the centroids of triangles $ABC$ and $A'B'C'$, then $G$ is the midpoint of segment $PQ$. This follows from direct vector averaging.
Lemma 2 states that as $A',B',C'$ vary among non-collinear points of $\epsilon$, their centroid $Q$ ranges over the whole plane $\epsilon$. This is proved by explicit construction: for any prescribed $Q\in\epsilon$, choose two arbitrary distinct points in $\epsilon$ and define the third so that the centroid equals $Q$.
Lemma 3 states that the set of midpoints of segments joining a fixed point $P$ to arbitrary points of a plane $\epsilon$ is a plane parallel to $\epsilon$, namely the image of $\epsilon$ under the homothety centered at $P$ with ratio $\tfrac12$. This follows from elementary affine geometry.
After proving the lemmas, the classification follows immediately. Lemma 1 converts the problem into the study of midpoints of $PQ$. Lemma 2 identifies the possible positions of $Q$, and Lemma 3 identifies the resulting locus.
The most delicate part is Lemma 2. A careless proof may accidentally choose collinear or coincident points, violating the condition that $L,M,N$ form a triangle. The construction must guarantee non-collinearity explicitly.
Solution
Let $P$ denote the centroid of triangle $ABC$, and let $Q$ denote the centroid of triangle $A'B'C'$. Thus
$$P=\frac{A+B+C}{3}, \qquad Q=\frac{A'+B'+C'}{3}.$$
We first relate $G$ to $P$ and $Q$.
Lemma 1
The point $G$ is the midpoint of segment $PQ$.
Proof
Since $L,M,N$ are the midpoints of $AA',BB',CC'$ respectively,
$$L=\frac{A+A'}2,\qquad M=\frac{B+B'}2,\qquad N=\frac{C+C'}2.$$
Because $G$ is the centroid of triangle $LMN$,
$$G=\frac{L+M+N}{3}.$$
Substituting the expressions for $L,M,N$ gives
$$G = \frac1{3}\left( \frac{A+A'}2 + \frac{B+B'}2 + \frac{C+C'}2 \right).$$
Combining terms,
$$G = \frac{A+B+C+A'+B'+C'}6.$$
Using the definitions of $P$ and $Q$,
$$A+B+C=3P, \qquad A'+B'+C'=3Q.$$
Hence
$$G = \frac{3P+3Q}{6} = \frac{P+Q}{2}.$$
Thus $G$ is the midpoint of segment $PQ$. ∎
This lemma establishes the affine structure of the problem; merely computing coordinates without identifying this midpoint relation would obscure the geometric locus.
Lemma 2
As $A',B',C'$ vary among non-collinear points of $\epsilon$, their centroid $Q$ ranges over the entire plane $\epsilon$.
Proof
Fix an arbitrary point $Q\in\epsilon$. We construct non-collinear points $A',B',C'\in\epsilon$ having centroid $Q$.
Choose two distinct points $U,V\in\epsilon$ such that the line $UV$ does not pass through $Q$. This is possible because infinitely many lines in $\epsilon$ avoid $Q$.
Define
$$A'=U,\qquad B'=V,\qquad C'=3Q-U-V.$$
Since $\epsilon$ is a plane and affine combinations of points of $\epsilon$ remain in $\epsilon$, the point $C'$ also belongs to $\epsilon$.
Now
$$\frac{A'+B'+C'}3 = \frac{U+V+(3Q-U-V)}3 = Q.$$
Hence the centroid of triangle $A'B'C'$ is $Q$.
It remains to verify that $A',B',C'$ are non-collinear. Suppose they were collinear. Since $A'=U$ and $B'=V$, all three points would lie on line $UV$. The centroid of three collinear points lies on the same line, so $Q$ would lie on line $UV$. This contradicts the choice of $U$ and $V$.
Therefore $A',B',C'$ are non-collinear, and every point $Q\in\epsilon$ occurs as the centroid of such a triangle. ∎
This lemma establishes surjectivity onto $\epsilon$; omitting the non-collinearity verification would leave a genuine gap.
Lemma 3
Let $P$ be a fixed point not lying in plane $\epsilon$. The set of midpoints of segments joining $P$ to arbitrary points of $\epsilon$ is a plane parallel to $\epsilon$.
Proof
Define
$$\Sigma=\left{\frac{P+Q}{2};:;Q\in\epsilon\right}.$$
Consider the homothety with center $P$ and ratio $\tfrac12$. For every point $Q$, its image is the midpoint of $PQ$. A homothety maps planes not passing through the center to parallel planes. Since $P\notin\epsilon$, the image of $\epsilon$ under this homothety is a plane parallel to $\epsilon$. By definition of $\Sigma$, this image is precisely $\Sigma$.
Hence $\Sigma$ is a plane parallel to $\epsilon$. ∎
This lemma identifies the geometric nature of the locus; treating the midpoint condition only algebraically would conceal the affine transformation governing the construction.
We now determine the locus of $G$.
By Lemma 1,
$$G=\frac{P+Q}{2},$$
where $P$ is fixed and $Q$ is the centroid of triangle $A'B'C'$.
By Lemma 2, the point $Q$ ranges over the entire plane $\epsilon$.
Therefore $G$ ranges over the set of midpoints of segments joining $P$ to arbitrary points of $\epsilon$.
By Lemma 3, this set is a plane parallel to $\epsilon$.
More precisely, the locus is the image of $\epsilon$ under the homothety centered at $P$ with ratio $\tfrac12$, where $P$ is the centroid of triangle $ABC$.
Thus the locus of $G$ is exactly the plane parallel to $\epsilon$ obtained by taking midpoints of segments joining $P$ to points of $\epsilon$.
$$\boxed{\text{The locus of }G\text{ is the image of }\epsilon\text{ under the homothety centered at the centroid of }ABC\text{ with ratio }\frac12.}$$
Verification of Key Steps
The first delicate step is the identity
$$G=\frac{P+Q}{2}.$$
An independent derivation proceeds directly from coordinates. Choose any affine coordinate system. Then
$$L=\frac{A+A'}2,\quad M=\frac{B+B'}2,\quad N=\frac{C+C'}2.$$
Averaging,
$$G = \frac{A+B+C+A'+B'+C'}6.$$
Since
$$P=\frac{A+B+C}3,\qquad Q=\frac{A'+B'+C'}3,$$
substitution yields
$$G=\frac{P+Q}{2}.$$
A careless argument could incorrectly average the coefficients and obtain $\tfrac13(P+Q)$ or $\tfrac14(P+Q)$.
The second delicate step is proving that every point of $\epsilon$ occurs as a centroid of a nondegenerate triangle. Fix $Q\in\epsilon$. Choose a line $\ell$ in $\epsilon$ not passing through $Q$, and choose distinct points $U,V\in\ell$. Define
$$C'=3Q-U-V.$$
Then
$$Q=\frac{U+V+C'}3.$$
If $C'$ also lay on $\ell$, then the centroid of the three collinear points would belong to $\ell$, forcing $Q\in\ell$, contradiction. Hence the triangle is nondegenerate. A careless proof might choose arbitrary $U,V$ without excluding the case that $Q$ lies on $UV$.
The third delicate step is identifying the locus as a plane parallel to $\epsilon$. In coordinates, let $\epsilon$ be given by
$$z=0,$$
and let
$$P=(p_1,p_2,p_3).$$
Then any point $Q\in\epsilon$ has the form
$$Q=(x,y,0).$$
The midpoint of $PQ$ is
$$\left(\frac{p_1+x}{2},\frac{p_2+y}{2},\frac{p_3}{2}\right).$$
Its third coordinate is constant, equal to $\tfrac{p_3}{2}$. Hence the locus is the plane
$$z=\frac{p_3}{2},$$
parallel to $\epsilon$. A careless argument might assert parallelism without proving that every point of the translated plane is attained.
Alternative Approaches
A purely coordinate solution is possible from the outset. Place $\epsilon$ in the plane $z=0$. Write
$$A=(a_1,a_2,a_3),\quad B=(b_1,b_2,b_3),\quad C=(c_1,c_2,c_3),$$
with positive third coordinates, and let
$$A'=(x_1,x_2,0),\quad B'=(y_1,y_2,0),\quad C'=(z_1,z_2,0).$$
After computing $L,M,N$ and their centroid, one finds
$$G= \left( \frac{a_1+b_1+c_1+x_1+y_1+z_1}{6}, \frac{a_2+b_2+c_2+x_2+y_2+z_2}{6}, \frac{a_3+b_3+c_3}{6} \right).$$
The third coordinate is constant, while the first two coordinates vary freely, so the locus is a plane parallel to $\epsilon$.
Another approach uses vector spaces directly. Regard $\epsilon$ as an affine plane with direction space $V$. If $P$ is the centroid of $ABC$, then every centroid $Q$ of a triangle in $\epsilon$ has the form
$$Q=Q_0+v,\qquad v\in V,$$
for some fixed $Q_0\in\epsilon$. Since
$$G=\frac{P+Q}{2},$$
the set of all such $G$ is
$$\frac{P+Q_0}{2}+\frac12V,$$
which is an affine plane parallel to $\epsilon$.
The geometric midpoint-homothety argument is preferable because it explains immediately why the answer must be a parallel plane and isolates the essential affine structure without lengthy coordinate calculations.