IMO 1962 Problem 3
The motion takes place on two adjacent faces of the cube.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 2m18s
Problem
A cube $ABCDA'B'C'D'$ is given. Point $X$ moves at constant speed along the square $ABCD$ in the direction $A \to B$; point $Y$ moves at the same speed along the square $BCC'B'$ in the direction $B' \to C'$. Initially $X = A$ and $Y = B'$. Find the locus of all midpoints of $XY$.
Exploration
The motion takes place on two adjacent faces of the cube. The first task is to parameterize the positions of the moving points in a coherent coordinate system. Since both points move with the same speed and both paths are piecewise linear with equal edge lengths, the natural parameter is arc length travelled from the initial positions.
Take the cube to have side length $1$. A coordinate model should simplify the midpoint computation. Put
$$A=(0,0,0), \quad B=(1,0,0), \quad C=(1,1,0), \quad D=(0,1,0),$$
and
$$A'=(0,0,1), \quad B'=(1,0,1), \quad C'=(1,1,1), \quad D'=(0,1,1).$$
Then $X$ moves on the perimeter of the square in the plane $z=0$, while $Y$ moves on the perimeter of the square in the plane $x=1$.
The first conjecture is that the locus is polygonal, because each moving point traverses edges linearly in time, hence the midpoint should also move linearly during each interval where both points stay on fixed edges. The difficulty is determining which segments appear and how they connect.
The two motions each have period $4$, measured in side lengths. It is useful to examine the first few intervals carefully. For $0 \le t \le 1$, the point $X$ runs along $AB$ while $Y$ runs along $B'C'$. The midpoint becomes
$$M(t)=\left(1,\frac t2,\frac12\right),$$
which is a vertical segment parallel to the $y$-axis on the face $BCC'B'$.
For $1 \le t \le 2$, the point $X$ runs along $BC$ and $Y$ runs along $C'C$. Then
$$M(t)=\left(1,\frac12+\frac t2,\frac{3-t}{2}\right).$$
This is another straight segment.
A possible mistake is assuming the locus closes after two intervals. Direct computation for $2 \le t \le 3$ and $3 \le t \le 4$ shows two more distinct segments appear. The endpoints should then be checked carefully to ensure continuity.
Another possible conjecture is that the locus is planar. Testing the coordinates disproves this immediately:
$$(1,0,\tfrac12), \quad (1,1,\tfrac12), \quad \left(\tfrac12,1,0\right)$$
do not lie in a common plane parallel to a coordinate plane. However, they may lie in some oblique plane. Computing the affine relation
$$x+z= \frac32$$
for the first two points but not the third shows even that fails. Hence the locus is genuinely spatial.
The structure suggests four line segments forming a skew quadrilateral. The main issue is proving rigorously that these four segments are exactly the locus and determining their endpoints precisely.
Problem Understanding
The problem asks for the set of all midpoints of the segment joining two moving points on a cube. One point traverses the boundary of the bottom face $ABCD$ in cyclic order beginning at $A$, while the other traverses the boundary of the side face $BCC'B'$ beginning at $B'$ and moving in cyclic order toward $C'$. Both move with equal constant speed.
This is a Type D problem. We must construct the locus explicitly and verify that every midpoint produced by the motion lies on the proposed set, and conversely that every point of the proposed set is attained during the motion.
The mathematical objects involved are piecewise linear motions in $\mathbb{R}^3$. Because the motions occur along polygonal paths at equal speeds, the midpoint motion should also be piecewise linear. The challenge is that the moving points change edges at different times, so the midpoint trajectory changes direction several times. A naive geometric visualization easily misses one or more segments of the locus.
The expected answer is a closed broken line consisting of four line segments. Intuitively, during each unit time interval both points move linearly along fixed edges, so the midpoint moves uniformly along a line segment. Since the full motion period is $4$, four such segments should appear.
Proof Architecture
The proof will use four claims.
Claim 1 states that the motions of $X$ and $Y$ admit explicit coordinate parametrizations on each interval
$$[0,1], \quad [1,2], \quad [2,3], \quad [3,4].$$
The reason is that each point moves at unit speed along edges of length $1$.
Claim 2 states that on each of these four intervals, the midpoint $M$ of $XY$ moves along a line segment, and computes the corresponding endpoint coordinates explicitly. This follows because midpoint coordinates are affine averages of the coordinates of $X$ and $Y$.
Claim 3 states that consecutive segments join continuously at their endpoints and together form a closed broken line. This must be checked directly at the transition times
$$t=1,2,3,4.$$
Claim 4 states that every point on each segment is attained by the midpoint for a unique parameter value within the corresponding interval. This proves that the constructed broken line is exactly the locus.
The most delicate part is keeping the parametrizations consistent when the moving points switch edges. A sign error in one coordinate would produce an incorrect segment and destroy continuity.
Solution
Choose Cartesian coordinates so that the cube has side length $1$ and
$$A=(0,0,0), \quad B=(1,0,0), \quad C=(1,1,0), \quad D=(0,1,0),$$
$$A'=(0,0,1), \quad B'=(1,0,1), \quad C'=(1,1,1), \quad D'=(0,1,1).$$
Let $t$ denote the elapsed time. Since both points move with constant speed $1$, each traverses one edge during each unit interval of time. The full period is $4$.
Claim 1
For $0 \le t \le 4$, the coordinates of $X$ and $Y$ are given as follows.
For $0 \le t \le 1$,
$$X(t)=(t,0,0), \qquad Y(t)=(1,t,1).$$
For $1 \le t \le 2$,
$$X(t)=(1,t-1,0), \qquad Y(t)=(1,1,2-t).$$
For $2 \le t \le 3$,
$$X(t)=(3-t,1,0), \qquad Y(t)=(1,3-t,0).$$
For $3 \le t \le 4$,
$$X(t)=(0,4-t,0), \qquad Y(t)=(1,0,t-3).$$
Proof
During the interval $0 \le t \le 1$, the point $X$ moves from $A$ to $B$ along the edge $AB$ at speed $1$. Since $AB$ is the segment from $(0,0,0)$ to $(1,0,0)$, the coordinate description is
$$X(t)=(t,0,0).$$
Similarly, $Y$ moves from $B'$ to $C'$ along the edge $B'C'$, giving
$$Y(t)=(1,t,1).$$
During the interval $1 \le t \le 2$, the point $X$ moves from $B$ to $C$, so
$$X(t)=(1,t-1,0).$$
At the same time, $Y$ moves from $C'$ to $C$, hence
$$Y(t)=(1,1,2-t).$$
During the interval $2 \le t \le 3$, the point $X$ moves from $C$ to $D$, giving
$$X(t)=(3-t,1,0),$$
while $Y$ moves from $C$ to $B$, giving
$$Y(t)=(1,3-t,0).$$
During the interval $3 \le t \le 4$, the point $X$ moves from $D$ to $A$, so
$$X(t)=(0,4-t,0),$$
while $Y$ moves from $B$ to $B'$, so
$$Y(t)=(1,0,t-3).$$
This establishes the coordinate parametrizations for the entire motion. The tempting shortcut would be to describe the paths qualitatively, but the midpoint computation requires exact formulas. ∎
Claim 2
The midpoint $M(t)$ of $X(t)Y(t)$ moves successively along the following four line segments:
For $0 \le t \le 1$,
$$M(t)=\left(\frac{t+1}{2},\frac t2,\frac12\right).$$
For $1 \le t \le 2$,
$$M(t)=\left(1,\frac t2,\frac{2-t}{2}\right).$$
For $2 \le t \le 3$,
$$M(t)=\left(\frac{4-t}{2},\frac{4-t}{2},0\right).$$
For $3 \le t \le 4$,
$$M(t)=\left(\frac12,\frac{4-t}{2},\frac{t-3}{2}\right).$$
Proof
The midpoint formula in $\mathbb{R}^3$ gives
$$M(t)=\frac{X(t)+Y(t)}2.$$
For $0 \le t \le 1$,
$$M(t)=\frac{(t,0,0)+(1,t,1)}2 =\left(\frac{t+1}{2},\frac t2,\frac12\right).$$
For $1 \le t \le 2$,
$$M(t)=\frac{(1,t-1,0)+(1,1,2-t)}2 =\left(1,\frac t2,\frac{2-t}{2}\right).$$
For $2 \le t \le 3$,
$$M(t)=\frac{(3-t,1,0)+(1,3-t,0)}2 =\left(\frac{4-t}{2},\frac{4-t}{2},0\right).$$
For $3 \le t \le 4$,
$$M(t)=\frac{(0,4-t,0)+(1,0,t-3)}2 =\left(\frac12,\frac{4-t}{2},\frac{t-3}{2}\right).$$
Each coordinate depends linearly on $t$, hence each portion of the trajectory is a line segment. This establishes the precise equations of the four segments; omitting the explicit midpoint calculation would leave the spatial geometry ambiguous. ∎
Claim 3
These four segments join consecutively at the points
$$P_1=\left(\frac12,0,\frac12\right), \quad P_2=\left(1,\frac12,\frac12\right), \quad P_3=(1,1,0),$$
$$P_4=\left(\frac12,\frac12,0\right), \quad P_5=\left(\frac12,0,\frac12\right),$$
and therefore form a closed broken line.
Proof
Substituting the interval endpoints into the formulas from Claim 2 gives:
At $t=0$,
$$M(0)=\left(\frac12,0,\frac12\right).$$
At $t=1$,
$$M(1)=\left(1,\frac12,\frac12\right).$$
At $t=2$,
$$M(2)=(1,1,0).$$
At $t=3$,
$$M(3)=\left(\frac12,\frac12,0\right).$$
At $t=4$,
$$M(4)=\left(\frac12,0,\frac12\right)=M(0).$$
Hence the endpoint of each segment coincides with the starting point of the next segment, and the final point coincides with the initial point. Thus the locus is a closed broken line with vertices
$$\left(\frac12,0,\frac12\right), \quad \left(1,\frac12,\frac12\right), \quad (1,1,0), \quad \left(\frac12,\frac12,0\right).$$
This establishes the global shape of the locus; failing to check continuity at the switching times could incorrectly produce disconnected pieces. ∎
Claim 4
Every point of the broken line described in Claim 3 occurs as the midpoint of $XY$ for some time $t \in [0,4]$.
Proof
Fix one of the four segments.
For the first segment,
$$M(t)=\left(\frac{t+1}{2},\frac t2,\frac12\right), \qquad 0 \le t \le 1.$$
As $t$ varies continuously from $0$ to $1$, the midpoint traverses the entire segment joining
$$\left(\frac12,0,\frac12\right) \quad \text{and} \quad \left(1,\frac12,\frac12\right).$$
The same argument applies on each remaining interval. Since the coordinate functions are affine in $t$ and the parameter interval is closed, the midpoint traverses every point of each corresponding segment exactly once.
Consequently, the entire broken line is attained by the midpoint motion, and no additional points occur because Claim 2 exhausts all time intervals in one full period.
We have therefore constructed the complete locus. The midpoint of $XY$ traces the closed broken line with vertices
$$\left(\frac12,0,\frac12\right), \quad \left(1,\frac12,\frac12\right), \quad (1,1,0), \quad \left(\frac12,\frac12,0\right).$$
Verification of Key Steps
The first delicate step is the parametrization during the interval $2 \le t \le 3$. Re-derive it independently. At time $t=2$, the point $X$ is at $C=(1,1,0)$ and moves toward $D=(0,1,0)$ with unit speed. After elapsed time $t-2$, its $x$-coordinate decreases by exactly $t-2$, giving
$$x=1-(t-2)=3-t.$$
The $y$-coordinate remains $1$. Thus
$$X(t)=(3-t,1,0).$$
A careless argument might write $X(t)=(t-2,1,0)$, which would move the point outside the cube.
The second delicate step is the midpoint formula on the same interval. Using
$$X(t)=(3-t,1,0), \qquad Y(t)=(1,3-t,0),$$
the midpoint is
$$M(t)=\left(\frac{4-t}{2},\frac{4-t}{2},0\right).$$
Both coordinates are equal, so the segment lies on the diagonal line
$$x=y$$
in the plane $z=0$. A sign error in either coordinate would destroy this symmetry and produce the wrong endpoint at $t=3$.
The third delicate step is verifying closure. Evaluating at $t=4$ gives
$$M(4)=\left(\frac12,0,\frac12\right),$$
which matches the initial point at $t=0$. Without this explicit check, one could incorrectly conclude that the locus is merely an open polygonal chain.
Alternative Approaches
A more geometric approach unfolds the two faces traversed by $X$ and $Y$ into a planar net. In the unfolded picture, both motions become motions along the boundary of a planar rectangle. The midpoint of two points moving linearly on edges of this rectangle again traces line segments. This method reduces the three-dimensional geometry to planar affine geometry, but keeping track of the folding identifications requires additional care.
Another approach uses vector notation. Let
$$M=\frac{X+Y}{2}.$$
Since both $X$ and $Y$ move piecewise linearly with constant velocity vectors on each interval, the midpoint velocity is the average of two constant vectors. Hence the midpoint trajectory is piecewise linear automatically. One then computes only the four vertices and deduces the entire locus from affine interpolation. This is shorter conceptually, but the coordinate computation in the main solution is preferable because it leaves no ambiguity about the exact segments traced.