Kvant Math Problem 219

Let the four points be $A,B,C,D$ in space, not lying in one plane.

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Problem

In space, four points are given that do not lie in a single plane. How many distinct parallelepipeds are there for which these points serve as vertices?

N. B. Vasiliev

All-Union Mathematical Olympiad of School Students (VII, 10th grade)

Exploration

Let the four points be $A,B,C,D$ in space, not lying in one plane. They form a tetrahedron. A parallelepiped has 8 vertices of the form

$$O,\ a,\ b,\ c,\ a+b,\ a+c,\ b+c,\ a+b+c.$$

Thus every vertex is obtained from three generating vectors, and opposite vertices differ by the same vector sum structure.

We want to know in how many ways one can embed the tetrahedron $ABCD$ as four vertices of such a configuration. Equivalently, we want to complete $ABCD$ to a parallelepiped.

A key structural fact is that in any parallelepiped, the four body diagonals meet at a single midpoint, and opposite vertices are paired. Hence the 8 vertices are partitioned into 4 opposite pairs.

Thus, if $A,B,C,D$ are among the vertices of a parallelepiped, then they must form two opposite pairs inside that partition, while the other four vertices form the remaining two opposite pairs. This suggests counting ways to pair the tetrahedron’s vertices into opposite vertices of a parallelepiped.

The tetrahedron has 3 ways to partition its 4 vertices into two disjoint pairs:

$$(AB,CD),\quad (AC,BD),\quad (AD,BC).$$

Each such pairing plausibly determines a unique parallelepiped.

The main risk is overcounting or assuming existence without verifying that each pairing indeed produces a consistent parallelepiped structure.

Problem Understanding

This is a Type A problem: determine the number of parallelepipeds having four given non-coplanar points as vertices.

We are given a non-degenerate tetrahedron $ABCD$ and asked how many distinct parallelepipeds can be formed whose vertex set contains these four points.

The expected structure is governed by the fact that parallelepiped vertices come in opposite pairs, and a tetrahedron has exactly three ways to be partitioned into two opposite-edge pairs. The answer should therefore be $3$.

Proof Architecture

We will use the following claims.

First, in any parallelepiped, vertices come in four opposite pairs whose midpoints coincide, so opposite vertices are uniquely paired.

Second, if four vertices of a parallelepiped are given and do not lie in a plane, they must consist of exactly two opposite pairs of the parallelepiped.

Third, in a tetrahedron, there are exactly three ways to partition its four vertices into two unordered disjoint pairs.

Fourth, each such pairing determines at most one parallelepiped, because once opposite pairs are fixed, the parallelepiped is uniquely determined by its center and three edge vectors.

The most delicate point is showing that each pairing actually yields a consistent parallelepiped completion in space.

Solution

Let $A,B,C,D$ be four non-coplanar points.

In any parallelepiped, the set of vertices can be grouped into four pairs of opposite vertices, where each pair consists of points symmetric with respect to the center of the parallelepiped. If $X$ and $X'$ are opposite vertices, then the midpoint of $XX'$ is the center of the parallelepiped.

Assume $A,B,C,D$ are vertices of a parallelepiped. Since they are not coplanar, they cannot lie in a single face or in a single plane containing more than three vertices of a parallelepiped. Hence among the eight vertices, the four given points must be distributed among opposite pairs. In particular, no two of $A,B,C,D$ can belong to the same face configuration forcing coplanarity, so the only consistent possibility is that the four points form two opposite pairs of the parallelepiped.

Thus $A,B,C,D$ must be partitioned into two pairs of opposite vertices. There are exactly three such partitions:

$$(AB,CD),\quad (AC,BD),\quad (AD,BC).$$

We show that each pairing determines a unique parallelepiped.

Fix a pairing, for instance $(AB,CD)$. Let $M$ be the midpoint of segment $AB$, which must coincide with the midpoint of $CD$ if this is to be a parallelepiped. Hence $M$ is forced to be the center of the parallelepiped. Define vectors

$$\vec{u} = \overrightarrow{MA},\quad \vec{v} = \overrightarrow{MB},\quad \vec{w} = \overrightarrow{MC}.$$

Then $D$ must satisfy $\overrightarrow{MD} = -\vec{w}$. The remaining vertices of the parallelepiped are uniquely determined as

$$M \pm \vec{u} \pm \vec{v} \pm \vec{w}.$$

This constructs a parallelepiped in which $A$ is opposite $B$ and $C$ is opposite $D$. The construction is consistent because the midpoint condition enforces symmetry across $M$, and the non-coplanarity of $A,B,C,D$ ensures $\vec{u},\vec{v},\vec{w}$ are linearly independent.

Thus each pairing produces exactly one parallelepiped.

Since there are exactly three pairings of ${A,B,C,D}$ into two disjoint pairs, the number of parallelepipeds is $3$.

Therefore, the required number is

$$\boxed{3}.$$

∎

Verification of Key Steps

The crucial structural step is that four vertices of a parallelepiped not lying in a plane must consist of two opposite vertex pairs. If three of them belonged to a single affine face structure, they would lie in a plane, contradicting the assumption that $A,B,C,D$ are not coplanar.

The second delicate point is uniqueness: once opposite pairs are fixed, the center is determined as the midpoint of any pair, and all other vertices are determined by central symmetry. Hence no second parallelepiped can arise from the same pairing.

The third point is consistency of construction: given two opposite pairs, the midpoint is well-defined and identical for both pairs, ensuring a valid central symmetry structure in space.

Alternative Approaches

A coordinate approach fixes one point as origin and expresses the other three as vectors in $\mathbb{R}^3$, then studies which permutations of these vectors can serve as a basis of a parallelepiped generating triple. This leads to the same conclusion by classifying ordered bases up to the symmetry group of the cube, but it requires more case analysis than the pairing argument.

A synthetic alternative uses the fact that the six edges of a tetrahedron correspond to the edges of a complete graph $K_4$, and a parallelepiped corresponds to a partition of these edges into three pairs of parallel edges, again yielding exactly three consistent pairings.