1973 IMO Problems/Problem 2

Problem

Determine whether or not there exists a finite set $M$ of points in space not lying in the same plane such that, for any two points $A$ and $B$ of $M$ ; one can select two other points $C$ and $D$ of $M$ so that lines $AB$ and $CD$ are parallel and not coincident.

Solution

In order to solve this problem we can start by finding at least one finite set $M$ that satisfies the condition.

We start by defining our first set $M_{8}$ with the vertices of a cube of side $k$ as follows: $M_{8} = \{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \}$

Since all the faces of this cube have a parallel face, then any two points on one face will have corresponding 2 points on the opposite face that is parallel. However we have four diagonals on this cube that do not have two points that are parallel to any of these diagonals.

By doing a reflection of the points on the $z=k$ plane along the $xy$ -plane these four diagonals will have their respective parallel diagonals on the $z \le 0$ space.

But now we have four more diagonals on the set of two cubes that do not have a parallel line. That is, diagonal $(0,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.

By doing a reflection of the points on the $x=k$ plane along the $yz$ -plane these new four diagonals will have their respective parallel diagonals on the $x \le 0$ space.

But now we have four more diagonals on the set of 4 cubes that do not have a parallel line. That is, diagonal $(-k,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.

By doing a reflection of the points on the $y=k$ plane along the $xz$ -plane these new four diagonals will have their respective parallel diagonals on the $y \le 0$ space.

The new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes.

So, we found a set at least one finite set $M$ that we can define as $M=\{(x,y,z)\}$ where $x,y,z \in \{-k,0,k\}$ giving a total of 27 points. Therefore such a set exists.

Another way to define this set of points is let $M$ be:

Let $V$ be a solid cube or right angled parallelepiped

Let $M_{v}$ be the set of all 8 vertices of $V$

Let $M_{me}$ be the set of all 12 midpoints of the edges of $V$

Let $M_{mf}$ be the set of all 6 midpoints of the faces of $V$

Let $M_{c}$ be the center of $V$

$M$ = $M_{v} \cup M_{me} \cup M_{mf} \cup M_{c}$

It is possible that one can construct many other sets of $M$ using regular tetrahedrons with some reflections and with less points than 27, or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined here. But that is not necessary for this problem because it asks to prove whether there exist a set with the described conditions. By showing that at least one set exists with those conditions, the problem is proved.

~Tomas Diaz. orders@tomasdiaz.com

Remarks (added by pf02, June 2025)

1. Very closely related to the solution above, it is easy to verify that if we delete the center (the point $(0, 0, 0)$ ) from the set $M$ , we obtain a set with $26$ points which still satisfies the requirements of the problem. In the second way of defining this set, we don't need $M_c$ . In other words, we just take $M = M_v \cup M_{me} \cup M_{mf}$ .

2. The last paragraph of the solution above ought to be deleted. The solution would gain from deleting it. Stating that maybe one can find a set by "using regular tetrahedrons with some reflections ... or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined" is vague and useless.

3. Below I will give another solution, inspired by rhombic dodecahedrons. Note that there are several non-equivalent rhombic dodecahedrons. See for example https://en.wikipedia.org/wiki/Rhombic_dodecahedron and https://en.wikipedia.org/wiki/Bilinski_dodecahedron (By non-equivalent, I mean that one can not be obtained from the other by an affine transformation.) Not all of them suggest solutions to this problem.

Solution 2

Let $x, y, z$ be the coordinate in space, and let us consider the set of $14$ points $M = \{(\pm 1, \pm 1, \pm 1), (\pm 2, 0, 0), (0, \pm 2, 0), (0, 0, \pm 2)\}$ .

We will prove that this set satisfies the conditions of the problem. One possible proof (the one I will give below) is purely algebraic. For any two points $A, B$ , consider the vectors $v_A = \overrightarrow{OA}, v_B = \overrightarrow{OB}$ . The segment $AB$ is parallel to the vector $v_B - v_A$ . We have to show that we can find two distinct points $C, D$ , non-colinear with $A, B$ such that $v_B - v_A = \alpha (v_D - v_C)$ for some $\alpha$ , where $v_C, v_D$ are the vectors given by $C, D$ .

Before proceeding to do this, let us get a geometric feel for what our set $M$ looks like.

The $x, y$ plane is the plane of the screen of your computer, and the $z$ axis is perpendicular to it, with the positive direction towards you. The red points are the $6$ points $\{(\pm 2, 0, 0), (0, \pm 2, 0), (0, 0, \pm 2)\}$ and the blue points are the $8$ points $\{(\pm 1, \pm 1, \pm 1)\}$ . All the faces are congruent rhombi.

It is clear that each edge is parallel to several other edges. It is intuitively clear that each diagonal of a face is parallel to a diagonal of another face. And it is intuitively clear that each interior diagonal is parallel to another interior diagonal, or the diagonal of a face, or an edge. But this is not a proof, so we will proceed to give a formal proof.

We will simply compute all the difference vectors, and show that each difference vector is a scalar multiple of at least one other difference vector. We will do this in a table. We will label all difference vectors with (a), (b), (c), ..., and use the same label when vectors are scalar multiples of each other.

Below is the table. On the top row (the column headers) and in the left column (the row headers) we list all the points in $M$ . In the cells of the table we write the vector $\overrightarrow{OB} - \overrightarrow{OA}$ , where $B$ is in the top row, and $A$ is in the left-most column. We don't compute the vector for $A = B$ (which would be $(0, 0, 0)$ ), or, if the index of $B$ is $\le$ the index of $A$ in the list of points, we don't compute the vector $\overrightarrow{OB} - \overrightarrow{OA}$ .

	(1,1,1)	(1,1,-1)	(1,-1,1)	(-1,1,1)	(1,-1,-1)	(-1,1,-1)	(-1,-1,1)	(-1,-1,-1)	(0,0,2)	(0,0,-2)	(0,2,0)	(0,-2,0)	(2,0,0)	(-2,0,0)
(1,1,1)		(0,0,-2)(a)	(0,-2,0)(b)	(-2,0,0)(c)	(0,-2,-2)(d)	(-2,0,-2)(e)	(-2,-2,0)(f)	(-2,-2,-2)(j)	(-1,-1,1)(k)	(-1,-1,-3)(n)	(-1,1,-1)(l)	(-1,-3,-1)(o)	(1,-1,-1)(m)	(-3,-1,-1)(p)
(1,1,-1)			(0,-2,2)(g)	(-2,0,2)(h)	(0,-2,0)(b)	(-2,0,0)(c)	(-2,-2,2)(k)	(-2,-2,0)(f)	(-1,-1,3)(q)	(-1,-1,-1)(j)	(-1,1,1)(m)	(-1,-3,1)(u)	(1,-1,1)(l)	(-3,-1,1)(v)
(1,-1,1)				(-2,2,0)(i)	(0,0,-2)(a)	(-2,2,-2)(l)	(-2,0,0)(c)	(-2,0,-2)(e)	(-1,1,1)(m)	(-1,1,-3)(t)	(-1,3,-1)(r)	(-1,-1,-1)(j)	(1,1,-1)(k)	(-3,1,-1)(y)
(-1,1,1)					(2,-2,-2)(m)	(0,0,-2)(a)	(0,-2,0)(b)	(0,-2,-2)(d)	(1,-1,1)(l)	(1,-1,-3)(w)	(1,1,-1)(k)	(1,-3,-1)(x)	(3,-1,-1)(s)	(-1,-1,-1)(j)
{1,-1,-1)						(-2,2,0)(i)	(-2,0,2)(h)	(-2,0,0)(c)	(-1,1,3)(w)	(-1,1,-1)(l)	(-1,3,1)(x)	(-1,-1,1)(k)	(1,1,1)(j)	(-3,1,1)(s)
(-1,1,-1)							(0,-2,2)(g)	(0,-2,0))(b)	(1,-1,3)(t)	(1,-1,-1)(m)	(1,1,1)(j)	(1,-3,1)(r)	(3,-1,1)(y)	(-1,-1,1)(k)
(-1,-1,1)								(0,0,-2)(a)	(1,1,1)(j)	(1,1,-3)(q)	(1,3,-1)(u)	(1,-1,-1)(m)	(3,1,-1)(v)	(-1,1,-1)(l)
(-1,-1,-1)									(1,1,3)(n)	(1,1,-1)(k)	(1,3,1)(o)	(1,-1,1)(l)	(3,1,1)(p)	(-1,1,1)(m)
(0,0,2)										(0,0,-4)(a)	(0,2,-2)(g)	(0,-2,-2)(d)	(2,0,-2)(h)	(-2,0,-2)(e)
(0,0,-2)											(0,2,2)(d)	(0,-2,2)(g)	(2,0,2)(e)	(-2,0,2)(h)
(0,2,0)												(0,-4,0)(b)	(2,-2,0)(i)	(-2,-2,0)(f)
(0,-2,0)													(2,2,0)(f)	(-2,2,0)(i)
(2,0,0)														(-4,0,0)(c)
(-2,0,0)

Now the proof is done. Indeed, all we have to do is examine the table, and ascertain that each label is there at least twice, and it is attached to proportional vectors. The reader can easily do this task.

For the sake of making things explicit, and in order to understand the geometry of this rhombic dodecahedron a little better, and to assist the diligent reader who would want to verify that each label appears twice and is attached to proportional vectors, I will give a table showing information about the labels.

label	difference vectors	no. of proportional vectors	description of segments yielding this difference vector
(a)	(0,0,-2)	5	red point to red point interior diagonal, blue point diagonals of 4 faces
(b)	(0,-2,0)	5	red point to red point interior diagonal, blue point diagonals of 4 faces
(c)	(-2,0,0)	5	red point to red point interior diagonal, blue point diagonals of 4 faces
(d)	(0,-2,-2)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(e)	(-2,0,-2)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(f)	(-2,-2,0)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(g)	(0,-2,2)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(h)	(-2,0,2)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(i)	(-2,2,0)	4	2 blue point to blue point short interior diagonals, red point diagonals of 2 faces
(j)	(-2,-2,-2)	7	6 sides, blue point to blue point long interior diagonal
(k)	(-1,-1,1)	7	6 sides, blue point to blue point long interior diagonal
(l)	(-1,1,-1)	7	6 sides, blue point to blue point long interior diagonal
(m)	(1,-1,-1)	7	6 sides, blue point to blue point long interior diagonal
(n)	(-1,-1,-3)	2	red point to blue point interior diagonals
(o)	(-1,-3,-1)	2	red point to blue point interior diagonals
(p)	(-3,-1,-1)	2	red point to blue point interior diagonals
(q)	(-1,-1,3)	2	red point to blue point interior diagonals
(r)	(-1,3,-1)	2	red point to blue point interior diagonals
(s)	(3,-1,-1)	2	red point to blue point interior diagonals
(t)	(-1,1,-3)	2	red point to blue point interior diagonals
(u)	(-1,-3,1)	2	red point to blue point interior diagonals
(v)	(-3,-1,1)	2	red point to blue point interior diagonals
(w)	(1,-1,-3)	2	red point to blue point interior diagonals
(x)	(1,-3,-1)	2	red point to blue point interior diagonals
(y)	(-3,1,-1)	2	red point to blue point interior diagonals

[Solution by pf02, June 2025]

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1973 IMO (Problems) • Resources
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1973 IMO Problems/Problem 2

Contents

Problem

Solution

Remarks (added by pf02, June 2025)

Solution 2

See Also