MAPPINGS WHICH PRESERVE REGULAR DODECAHEDRONS

for all x, y ∈ X . A distance r > 0 is said to be preserved by a mapping f : X → Y if ‖ f (x)− f (y)‖ = r for all x, y ∈ X whenever ‖x− y‖ = r. If f is an isometry, then every distance r > 0 is preserved by f , and conversely. We can now raise a question whether each mapping that preserves certain distances is an isometry. Indeed, Aleksandrov [1] had raised a question whether a mapping f : X → X preserving a distance r > 0 is an isometry, which is now known to us as the Aleksandrov problem. Beckman and Quarles [2] solved the Aleksandrov problem for finite-dimensional real Euclidean spaces X =Rn (see also [3–6, 10, 12–17]). Theorem 1.1 (Beckman and Quarles). If a mapping f :Rn→Rn (2≤ n <∞) preserves a distance r > 0, then f is a linear isometry up to translation.


Introduction
Let us begin with the properties of an isometry. For normed spaces X and Y , a mapping f : X → Y is called an isometry if f satisfies the equality for all x, y ∈ X. A distance r > 0 is said to be preserved by a mapping f : X → Y if f (x) − f (y) = r for all x, y ∈ X whenever x − y = r. If f is an isometry, then every distance r > 0 is preserved by f , and conversely. We can now raise a question whether each mapping that preserves certain distances is an isometry. Indeed, Aleksandrov [1] had raised a question whether a mapping f : X → X preserving a distance r > 0 is an isometry, which is now known to us as the Aleksandrov problem.
It is an interesting question whether the "distance r > 0" in the above theorem can be replaced by some properties characterized by "geometrical figures" without loss of its validity. In [7][8][9], the authors proved that if a one-to-one mapping f : R n → R n maps every unit circle (or unit sphere, tetrahedron) onto a unit circle (or a unit sphere, tetrahedron, resp.), then f is a linear isometry up to translation.
In this connection, we will extend these results to the more general three-dimensional objects, that is, we prove in this note that if a one-to-one mapping f : R 3 → R 3 maps every regular dodecahedron onto a regular dodecahedron, then f is a linear isometry up to translation.

Main theorem
In the following, by a regular dodecahedron we mean a regular dodecahedron with its side length one. We first make our terms precise as follows. In Figure 2.1, we will call the point a a "vertex" and the line ab an "edge" and the plane bounded by the five edges ab, bc, cd, df , f a "face abcdf " or simply a "face." Further by a dodecahedron we will mean the 12 faces only and not the three-dimensional open set bounded by those 12 faces. Let us denote the three-dimensional open set bounded by the dodecahedron A as "Inside of A" or simply as InsA. Now we begin with the following lemma.
Proof. First, we show that if q / ∈ InsA, then f (q) / ∈ Ins f (A). In other words, we show that if f (q) ∈ Ins f (A), then q ∈ InsA. Suppose that q ∈ A. Then f (q) ∈ f (A) and so f (q) / ∈ Ins f (A). Suppose that q / ∈ InsA and q / ∈ A. Then choose another dodecahedron B such that q ∈ B and We show now that if any one-to-one mapping preserves regular dodecahedrons, then it is actually an isometry. More precisely, we have the following.
Proof. We show f preserves the distance 1. Then the theorem of Beckman and Quarles implies that f is an isometry. We will use solid angle arguments. Let A be a regular dodecahedron. For any p ∈ A, let us denote the solid angle that InsA subtends with respect to p ∈ A as Ω(A, p). We first find the solid angles at a vertex and at an edge point. To do so, let us choose suitable coordinate axes so that the vertex a of Figure 2.1 is located at the origin (see Figure 2.2). To get the coordinate of b, first we note that the ∠ f ab is a part of a regular pentagon abcdf with side length one and therefore it is 108 • . Therefore the length of b f is 2sin54 • and the triangle b f g is a regular triangle with side length 2sin54 • . Then the x coordinate of b is sin54 • , the y coordinate is (1/3) × √ 3sin54 • , and we can find the z coordinate If we call C 1 the pentagon abcdf , then the radius r of the circle which circumscribes C 1 satisfies 2r cos 54 • = 1 and therefore r = 1/2cos54 • = 0.851. Now the center o 1 of C 1 is the distance r = 0.851 away from the origin a and located along the direction of ( ab + a f )/| ab + a f | = (0,(1/ The two pentagons C 2 and C 3 meet at the edge ag and the vectors eo 2 and eo 3 are both perpendicular to ag. Therefore, to find the (ordinary) angle between the two planes C 2 and C 3 , we have only to find the angle between the two vectors eo 2 and eo 3 . Since eo 2 = ao 2 − ae = ao 2 − (1/2) ag and eo 3 = ao 3 − ae = ao 3 − (1/2) ag, the angle θ between these two vectors is given by and we get θ = 116.5 • or 2.034 rad. This angle is often called the dihedral angle of the dodecahedron. Now we are ready to find the solid angle Ω(A,e). We use the unit of the solid angle such that the solid angle of the whole sphere with respect to its center is 4π. Then the solid angle that Ins A subtends with respect to e ∈ A is Having found the solid angle Ω e at the edge point of the regular dodecahedron, let us now find the solid angle that InsA subtends at the vertex a. For that we use the result of [11], where it is shown that given any three vectors T 1 , T 2 , T 3 , all starting from the origin, the solid angle Ω that these three vectors subtend with respect to the origin (i.e., the solid angle that the (not necessarily regular) tetrahedron whose vertices are the end tips of these three vectors and the origin subtends with respect to the origin) is . Note that this expression does not depend on the lengths of the three vectors. For example, when T 1 , T 2 , T 3 are mutually orthogonal, we get Ω = π/2. Now we are ready to find the solid angle that InsA subtends at the vertex a, Ω(A,a). To find Ω, let T 1 = ag, T 2 = a f , and T 3 = ab. Then using (2.3) we can compute the solid angle we want. We find [T 1 T 2 T 3 ] = 0.809, T 1 · T 2 = T 2 · T 3 = T 3 · T 1 = −0.309, and T 1 = T 2 = T 3 = 1. Therefore we get Let us summarize our results above. Suppose that p ∈ A where p is a point and A is a regular dodecahedron. If p is a vertex, say p = a, then the solid angle that InsA subtends with respect to p is Ω(A, p) = Ω v = 2.962. If p is a point which belongs to an edge and is not a vertex, then Ω(A, p) = Ω e = 4.068. If p ∈ A is neither a vertex nor an edge point, then Ω(A, p) = 2π.
Suppose now that a is a vertex of a regular dodecahedron A = A 1 . We show then that f Therefore the distance between f (a) and f (b) is again one too. Since f preserves the unit distance, by the theorem of Beckman and Quarles, we conclude that f is a linear isometry up to translation.