Convex Polyhedra Realizing Given Face Areas

When does A represent the face areas of a convex polyhedron in R 3 ? For example, suppose A = (100, 1, 1, 1). It is clear there is no tetrahedron realizing these areas, because the face of area 100 is too large to be “covered” by the three faces of area 1. So A 1 ≤ i>1 A i is an obvious necessary condition. The main result of this note is that this is also a sufficient condition. Enroute to establishing this we connect the question to robot arm linkages and to 3D polygons.

The main tool we use is Minkowski’s 1911 theorem. Here is a version from Alexandrov, who devotes an entire chapter to the theorem and variations in his book [Ale05,Chap. 7,p. 311ff].

Theorem 1 (Minkowski (a)) Let A i be positive faces areas and n i distinct, noncoplanar unit face normals, i = 1, . . . , n. Then if i A i n i = 0, there is a closed polyhedron whose faces areas uniquely realize those areas and normals.

Here, uniqueness is up to translation.

In our situation, we are given the areas A i , and the task is to determine if there exist normals n i that satisfy Minkowski’s theorem. Although this superficially may seem like a complex problem, we will see it has a simple solution. Although I have not been able to find this result in the literature, it seems likely that it is known, because the proof is not difficult.

It will be more convenient for our purposes to follow Grünbaum’s (equivalent) formulation of Minkowski’s theorem [Grü03,p. 332] phrased in terms of “fully equilibrated” vectors. Vectors are equilibrated if they sum to zero and no two are positively proportional. They are fully equilibrated in R k if they in addition span R k .

Theorem 1 (Minkowski (b)) Let v i = A i n i be vectors whose lengths are A i , |v i | = A i , and whose directions are unit normal vectors n i , i = 1, . . . , n. Then if the vectors are fully equilibrated in R 3 , there is a unique closed polyhedron P with faces areas A i and normal vectors n i .

Note that for n vectors to be fully equilibrated in R 3 , n must be at least 4: It requires 3 vectors to span R 3 , but any three vectors that sum to zero form a triangle and so lie in a plane. Thus 3 vectors cannot both be equilibrated and span R 3 . Four clearly suffices: P is a tetrahedron.

Theorem 2 If A 1 , A 2 , . . . , A n are positive real numbers with n ≥ 4 and A i ≥ A i+1 , then there is a closed convex polyhedron P with these A i as its face areas if and only if A 1 ≤ i>1 A i . When equality holds, we permit the polyhedron to be flat, i.e., its faces tessellate and doubly cover a planar convex polygon.

Let F i be the face whose area is A i . As previously mentioned, the condition is necessary, because if A 1 > i>1 A i , then the F 1 face cannot be covered by all the others, so it is not possible to form a closed surface. In the case of equality, A 1 = i>1 A i , the areas are realized by a flat polyhedron that can be constructed as follows. For F 1 , select a square of side length √ A 1 . (Any other convex polygon would serve as well.) This becomes one side of P . For the other side, partition F 1 into strips of width A i / √ A 1 , so that each strip has area A i , and serves as F i .

Henceforth assume A 1 is strictly smaller than the sum of the other areas. In this circumstance we can always obtain a non-flat polyhedron P , with none of the faces F i coplanar.

Let C be a polygonal chain (sometimes called a “robot arm” or just “arm”) whose n links have lengths A 1 , A 2 , …, A n . Then it is a corollary to a theorem of Hopcroft, Joseph, and Whitesides [HJW84] that the chain can close (i.e., the “hand” can touch the “shoulder”) iff the longest link is not longer than all the other links together. See, e.g., [O’R98, Thm. 8.6.3, p. 326] or [DO07, Thm. 5.1.2, p. 61]. The similarity to the statement of Theorem 2 should be evident. Our plan is to form a chain C from fully equilibrated vectors, and apply Minkowski’s theorem, Theorem 1(b). This can be accomplished in three stages: (1) arrange that the vectors sum to zero, (2) ensure that none is a positive multiple of another, and (3) ensure that they span R 3 . We know from the robot-arm theorem that (1) is achievable, but that theorem is an existence theorem.

Satisfying (1) can be accomplished with the “Two Kinks” theorem [O’R98, Thm. 8.6.5, p. 329] [DO07, Thm. 5.1.4, p. 62], which would result in the links arranged to form a triangle (and so summing to 0), with (in general) many vectors aligned. Although it is then not so difficult to break all the alignments and achieve (2), instead we opt for a method that achieves (1) and (2) simultaneously.

Lay out all the links in a straight line of length i A i . View the links as inscribed in a circle of infinite radius. Now imagine shrinking the radius from R = ∞ down toward R = 0, maintaining the chain inscribed at all times. Knowing from the Hopcr