IMO 1960 Problem 5
The problem asks for two loci inside a cube.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m30s
Problem
Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).
a) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$;
b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.
Exploration
The problem asks for two loci inside a cube. Since the objects are affine, coordinates should linearize everything. The midpoint of a segment joining a point on one line segment to a point on another line segment is usually constrained to a plane or smaller affine set. The second part asks for a point dividing the segment in a fixed ratio, which again suggests an affine transformation of the first locus.
Take coordinates adapted to the cube. Since the statement says that face $ABCD$ lies directly above $A'B'C'D'$, it is natural to place the lower face in the plane $z=0$ and the upper face in the plane $z=1$. The segment $AC$ is a diagonal of the top face, and $B'D'$ is a diagonal of the bottom face.
A first coordinate model is
$$A=(0,0,1),\quad B=(1,0,1),\quad C=(1,1,1),\quad D=(0,1,1),$$
$$A'=(0,0,0),\quad B'=(1,0,0),\quad C'=(1,1,0),\quad D'=(0,1,0).$$
Then
$$AC:\ (t,t,1),\qquad 0\le t\le 1,$$
and
$$B'D':\ (1-s,s,0),\qquad 0\le s\le 1.$$
The midpoint becomes
$$M=\left(\frac{t+1-s}{2},\frac{t+s}{2},\frac12\right).$$
The third coordinate is constant, so the locus lies in the midplane of the cube. The first two coordinates satisfy
$$x+y=t+\frac12.$$
Since $0\le t\le 1$, this gives
$$\frac12\le x+y\le \frac32.$$
But there is another relation hidden in eliminating $s$. Compute
$$x-y=\frac{1-2s}{2},$$
hence
$$-\frac12\le x-y\le \frac12.$$
These two inequalities together define a square in the plane $z=\frac12$. Its vertices should come from endpoint combinations:
$$\frac{A+B'}2,\quad \frac{A+D'}2,\quad \frac{C+B'}2,\quad \frac{C+D'}2.$$
Geometrically this is the parallelogram formed by the four midpoint combinations. Because the two diagonals are perpendicular in the chosen coordinates, the figure may actually be a square.
For part b, if $Z$ lies on $XY$ with $ZY=2XZ$, then
$$XZ:ZY=1:2.$$
Hence
$$Z=\frac{2X+Y}{3}.$$
Substitute the same parametrizations:
$$Z=\left(\frac{2t+1-s}{3},\frac{2t+s}{3},\frac23\right).$$
Again the third coordinate is constant. Eliminate parameters:
$$x+y=\frac{4t+1}{3},\qquad x-y=\frac{1-2s}{3}.$$
Thus
$$\frac13\le x+y\le \frac53,\qquad -\frac13\le x-y\le \frac13.$$
This is another square, parallel to the first, lying in the plane $z=\frac23$. Its vertices arise from the four endpoint choices.
The main danger is proving that every point in the claimed square actually occurs from some admissible parameters. The converse direction requires solving explicitly for $s,t$ in terms of $x,y$ and checking that the parameter bounds hold exactly when the point lies in the square.
Problem Understanding
The problem concerns a cube together with two diagonals lying in opposite parallel faces. A point $X$ moves along the diagonal $AC$ of the upper face, and a point $Y$ moves along the diagonal $B'D'$ of the lower face. Part a asks for the set of all midpoints of segments $XY$. Part b asks for the set of all points $Z$ on such segments satisfying the ratio condition
$$ZY=2XZ.$$
This is a Type A problem, because one must determine a complete geometric locus and prove both directions: every constructed point belongs to the locus, and every point of the locus arises from a valid construction.
The objects involved are affine combinations of points on line segments. Since affine combinations preserve linearity, coordinates reduce the problem to solving linear equations and inequalities. The difficulty is not obtaining candidate loci, but proving rigorously that every point in the proposed regions is actually attained by suitable choices of $X$ and $Y$.
The expected answer is that each locus is a square contained in a plane parallel to the faces of the cube. The midpoint condition in part a fixes the height halfway between the two faces, while the ratio condition in part b fixes another constant height. The remaining coordinates vary independently within bounds determined by the endpoints of the two diagonals.
Proof Architecture
The proof uses a coordinate model of the cube and proceeds through four claims.
Claim 1 states that every midpoint of a segment joining a point of $AC$ to a point of $B'D'$ lies in a specific square contained in the plane midway between the two faces. This follows by parametrizing the two diagonals and eliminating the parameters.
Claim 2 states the converse, namely that every point of that square arises as such a midpoint. This requires solving explicitly for the parameters and verifying that they lie in the interval $[0,1]$.
Claim 3 states that every point $Z$ satisfying $ZY=2XZ$ for some admissible segment $XY$ lies in another specific square contained in the plane $z=\frac23$. This again follows from affine parametrization and elimination.
Claim 4 states the converse for part b, namely that every point of the second square arises from suitable points $X$ and $Y$. As in Claim 2, the essential step is reconstructing the parameters and checking their admissibility.
The most delicate direction is the converse in each part. A careless argument might derive the equations of the containing plane but fail to prove that the entire square, and only that square, is attained.
Solution
Choose Cartesian coordinates so that the cube has side length $1$ and
$$A=(0,0,1),\quad B=(1,0,1),\quad C=(1,1,1),\quad D=(0,1,1),$$
$$A'=(0,0,0),\quad B'=(1,0,0),\quad C'=(1,1,0),\quad D'=(0,1,0).$$
Then the diagonal $AC$ of the upper face has parametrization
$$X=(t,t,1),\qquad 0\le t\le 1,$$
and the diagonal $B'D'$ of the lower face has parametrization
$$Y=(1-s,s,0),\qquad 0\le s\le 1.$$
Claim 1
The midpoint locus in part a is contained in the set
$$\mathcal S_1= \left{ (x,y,\tfrac12): \frac12\le x+y\le \frac32,, -\frac12\le x-y\le \frac12 \right}.$$
Proof
Let $M$ be the midpoint of $XY$. Using the coordinate formulas above,
$$M=\frac{X+Y}{2} = \left( \frac{t+1-s}{2}, \frac{t+s}{2}, \frac12 \right).$$
Hence every such midpoint satisfies
$$z=\frac12.$$
Moreover,
$$x+y = \frac{t+1-s+t+s}{2} = t+\frac12.$$
Since $0\le t\le 1$,
$$\frac12\le x+y\le \frac32.$$
Similarly,
$$x-y = \frac{t+1-s-t-s}{2} = \frac{1-2s}{2}.$$
Since $0\le s\le 1$,
$$-\frac12\le x-y\le \frac12.$$
Thus every midpoint belongs to $\mathcal S_1$. ∎
This establishes all necessary conditions on the midpoint coordinates; stopping here would be insufficient because containment alone does not prove that every point of $\mathcal S_1$ is attained.
Claim 2
Every point of $\mathcal S_1$ is the midpoint of a segment $XY$ with $X\in AC$ and $Y\in B'D'$.
Proof
Take any point
$$M=(x,y,\tfrac12)\in \mathcal S_1.$$
Define
$$t=x+y-\frac12, \qquad s=\frac12-(x-y).$$
From
$$\frac12\le x+y\le \frac32$$
it follows that
$$0\le t\le 1.$$
From
$$-\frac12\le x-y\le \frac12$$
it follows that
$$0\le s\le 1.$$
Hence the points
$$X=(t,t,1)\in AC, \qquad Y=(1-s,s,0)\in B'D'$$
are well defined.
Compute the midpoint:
$$\frac{X+Y}{2} = \left( \frac{t+1-s}{2}, \frac{t+s}{2}, \frac12 \right).$$
Substituting the definitions of $t$ and $s$,
$$\frac{t+1-s}{2} = \frac{ x+y-\frac12+1-\left(\frac12-(x-y)\right) }{2} = x,$$
and
$$\frac{t+s}{2} = \frac{ x+y-\frac12+\frac12-(x-y) }{2} = y.$$
Thus the midpoint equals $M$. ∎
This proves the converse direction rigorously; the key point is that the inequalities defining $\mathcal S_1$ are exactly equivalent to the parameter restrictions $0\le t,s\le 1$.
The set $\mathcal S_1$ is a square in the plane $z=\frac12$. Its vertices are obtained from the endpoint pairs:
$$\left(\frac12,0,\frac12\right), \quad (0,\tfrac12,\tfrac12), \quad (1,\tfrac12,\tfrac12), \quad (\tfrac12,1,\tfrac12).$$
These are respectively the midpoints of
$$AB',\ AD',\ CB',\ CD'.$$
Hence the locus in part a is precisely this square.
Claim 3
The locus in part b is contained in the set
$$\mathcal S_2= \left{ (x,y,\tfrac23): \frac13\le x+y\le \frac53,, -\frac13\le x-y\le \frac13 \right}.$$
Proof
Suppose $Z$ lies on $XY$ and satisfies
$$ZY=2XZ.$$
Then
$$XZ:ZY=1:2.$$
A point dividing a segment internally in the ratio $1:2$ from the first endpoint has vector form
$$Z=\frac{2X+Y}{3}.$$
Using the parametrizations of $X$ and $Y$,
$$Z= \left( \frac{2t+1-s}{3}, \frac{2t+s}{3}, \frac23 \right).$$
Hence every such point satisfies
$$z=\frac23.$$
Also,
$$x+y = \frac{2t+1-s+2t+s}{3} = \frac{4t+1}{3}.$$
Since $0\le t\le 1$,
$$\frac13\le x+y\le \frac53.$$
Further,
$$x-y = \frac{2t+1-s-2t-s}{3} = \frac{1-2s}{3}.$$
Since $0\le s\le 1$,
$$-\frac13\le x-y\le \frac13.$$
Thus every such point belongs to $\mathcal S_2$. ∎
This determines all necessary coordinate conditions for the second locus; the ratio condition fixes the height coordinate uniquely.
Claim 4
Every point of $\mathcal S_2$ arises from suitable points $X\in AC$ and $Y\in B'D'$ satisfying $ZY=2XZ$.
Proof
Take any point
$$Z=(x,y,\tfrac23)\in \mathcal S_2.$$
Define
$$t=\frac{3(x+y)-1}{4}, \qquad s=\frac12-\frac32(x-y).$$
From
$$\frac13\le x+y\le \frac53$$
we obtain
$$0\le t\le 1.$$
From
$$-\frac13\le x-y\le \frac13$$
we obtain
$$0\le s\le 1.$$
Hence
$$X=(t,t,1)\in AC, \qquad Y=(1-s,s,0)\in B'D'.$$
Now compute
$$\frac{2X+Y}{3} = \left( \frac{2t+1-s}{3}, \frac{2t+s}{3}, \frac23 \right).$$
Substituting the definitions of $t$ and $s$ gives
$$\frac{2t+1-s}{3} = x, \qquad \frac{2t+s}{3} = y.$$
Hence
$$Z=\frac{2X+Y}{3}.$$
Therefore $Z$ lies on $XY$ and satisfies
$$XZ:ZY=1:2,$$
equivalently
$$ZY=2XZ.$$
Thus every point of $\mathcal S_2$ belongs to the required locus. ∎
This completes the converse direction for part b; without reconstructing $t$ and $s$, one would only know that the locus is contained in a plane strip rather than exactly equal to the square.
The set $\mathcal S_2$ is a square in the plane $z=\frac23$. Its vertices are
$$\left(\frac13,0,\frac23\right), \quad \left(0,\frac13,\frac23\right), \quad \left(1,\frac23,\frac23\right), \quad \left(\frac23,1,\frac23\right).$$
Hence:
For part a, the locus is the square in the plane midway between the two faces, with vertices equal to the midpoints of
$$AB',\ AD',\ CB',\ CD'.$$
For part b, the locus is the square in the plane parallel to the faces at height $\frac23$ above the lower face, with the vertices listed above.
$$\boxed{ \text{Part a: } \mathcal S_1= \left{ (x,y,\tfrac12): \frac12\le x+y\le \frac32,, -\frac12\le x-y\le \frac12 \right} }$$
$$\boxed{ \text{Part b: } \mathcal S_2= \left{ (x,y,\tfrac23): \frac13\le x+y\le \frac53,, -\frac13\le x-y\le \frac13 \right} }$$
Verification of Key Steps
The first delicate step is the elimination of parameters in part a. Starting from
$$M= \left( \frac{t+1-s}{2}, \frac{t+s}{2}, \frac12 \right),$$
one computes independently
$$t=x+y-\frac12, \qquad s=\frac12-(x-y).$$
The parameter conditions
$$0\le t,s\le 1$$
translate exactly into
$$\frac12\le x+y\le \frac32, \qquad -\frac12\le x-y\le \frac12.$$
A careless argument might derive only the plane equation
$$z=\frac12,$$
which would incorrectly enlarge the locus from a square to the entire plane section.
The second delicate step is the ratio formula in part b. If a point $Z$ divides the segment from $X$ to $Y$ internally in the ratio
$$XZ:ZY=1:2,$$
then
$$Z=X+\frac13(Y-X)=\frac{2X+Y}{3}.$$
A common error is to interchange the coefficients and write
$$\frac{X+2Y}{3},$$
which would place the point one third of the way from $Y$ instead of one third of the way from $X$.
The third delicate step is verifying that the converse construction in part b always yields admissible parameters. From
$$t=\frac{3(x+y)-1}{4},$$
the bounds
$$\frac13\le x+y\le \frac53$$
give
$$0\le t\le 1.$$
If one checked only one endpoint inequality, points outside the intended square could incorrectly be admitted.
Alternative Approaches
A synthetic affine-geometric approach avoids coordinates almost entirely. The midpoint of a segment joining a point on $AC$ to a point on $B'D'$ depends affinely on the two parameters along the diagonals. Since the midpoint operation preserves parallelism and ratios, the image of the rectangle of parameter pairs is a parallelogram. Evaluating the four endpoint combinations shows that the vertices are the midpoints of
$$AB',\ AD',\ CB',\ CD'.$$
The same reasoning applies to part b, where the map
$$(X,Y)\mapsto \frac{2X+Y}{3}$$
is again affine.
Another possible approach uses vector geometry. Let the cube edges from $A'$ define basis vectors $\mathbf e_1,\mathbf e_2,\mathbf e_3$. Then
$$X=t(\mathbf e_1+\mathbf e_2)+\mathbf e_3, \qquad Y=(1-s)\mathbf e_1+s\mathbf e_2.$$
The loci follow by direct vector combination. This method is compact, but the coordinate form used in the main proof makes the converse direction and the boundary inequalities more transparent.