Continuous flattening of the 2-dimensional skeleton of a regular 24-cell

Bellow theorem says that any polyhedron with rigid faces cannot change its volume even if it is flexible. The problem on continuous flattenig of polyhedra with non-rigid faces proposed by Demaine et al. was solved for all convex polyhedra by using the notion of moving creases to change some of the faces. This problem was extended to a problem on continuous flattening of the 2-dimensional skeleton of higher dimensional polytopes. This problem was solved for all regular polytopes except three types, the 24-cell, the 120-cell, and the 600-cell. This article addresses the 24-cell and gives a continuous flattening motion for its 2-skeleton, which is related to the Jitterbug by Buckminster Fuller.


Introduction
We use the term polyhedron for a closed polyhedral surface that is permitted to touch but not cross itself. We assume a polyhedron can always be folded by creases like a piece of paper ignoring its thickness, but not allowing stretching. A flat folding of a polyhedron is a folding by creases into a multi-layered planar shape so that the number of creases is finite. In [2] it was proved that any polyhedron of genus zero has a flat folded state. The topic presented here is related to the problem by Demaine et al. [4] (Open Problems 18.1 and 18.3 in [5]): Can every flat folded state of a polyhedron be reached by a continuous folding process?
An important limitation to continuous flattening is Bellow theorem [3,13] that the volume of any polyhedron with rigid faces is invariant, even if the dihedral The first author is supported by JSPS KAKENHI 17K05222. The second author is supported by JSPS KAKENHI 20K03726. Figure 1 From left to right: a rectangular box, the moving creases shown by long-dotted line segments, and its flat folded state angles can be changed by hinges so that the polyhedron is flexible. Hence, to obtain a flat folded state of a polyhedron continuously, the polyhedral surface should change by moving creases (edges).
For example, a rectangular box can be flat folded by creases as shown in Fig. 1. The problem asks to find a continuous motion to reach the flat folded state from the original box. The long-dotted line segments are moving creases used in [8].
We say a polyhedron Q is isometric to a polyhedron P with respect to the intrinsic metric if there is a polyhedron P obtained from P by subdividing some faces of P (i.e., some faces of P may be included in the same face of P , but P is congruent to P ) such that Q is combinatorially equivalent to P and the corresponding faces of P and Q are congruent. In [8] the authors defined a continuous flat-folding motion of a polyhedron P by a family of polyhedra {P t : 0 ≤ t ≤ 1} such that each P t is isometric to P with respect to the intrinsic metric, and P t converges to P 1 continuously, where P 0 = P and P 1 is a flat folded state of P .
The first paper [8] on continuous flattening of polyhedra gives results for the Platonic polyhedra, and we explain more in detail the continuous motion of the regular octahedron in Sect. 2, which will be used for the proof of Theorem 4.1. The case of any convex polyhedron was solved by two methods [1,11]. However, the set of moving creases in those two methods covers almost the entire surface of a convex polyhedron, but the method used in [8], namely the kite method, only requires small portions of the convex polyhedron's surface, and maximizes the number of rigid edges and faces in some sense. So, we use the kite method in this article because of these advantages.
This problem was extended to higher dimensional polytopes in [9,10].
Problem. Find a continuous flat-folding motion of the 2-dimensional skeleton (2-skeleton for short), denoted by C(P ), of any given polytope P .
We concentrate on regular polytopes. There are three types of n-dimensional regular polytopes for n ≥ 5, that is, the hypercube, the simplex, and the cross-polytope, and three more types for n = 4, that is, the 24-cell, the 120cell, and the 600-cell. We showed in [9] that for any given hypercube there is a continuous flat-folding motion of the 2-skeleton onto any of its square face F such that all square faces parallel to F are rigid, that is, have no crease during the motion, and in [10] that for any given regular simplex or cross-polytope there is a continuous flat-folding motion of the 2-skeleton such that at least two thirds of the edges and two ninths of the triangular faces are rigid.
Here we address the regular 24-cell. Let P be a regular 24-cell. Then P has 24 regular octahedral facets. First, we show that the 2-skeleton of P can be continously folded onto any of its octahedral facets (Theorem 4.1). We show that the 2-skeleton of P contains the eight triangular faces of a cuboctahedron, denoted by Q, and Q can be continuously transformed into an octahedron so that all edges are rigid. Such motion of Q is same as the motion of the Jitterbug by Buckminster Fuller [6,7]. We give a precise calculation of this motion in Sect. 4, which is used to prove Theorem 4.1. Next, we show the existence of a continuous flattening motion of the 2-skeleton of P onto any of its triangular face F such that 54 out of the 96 edges and all triangular faces parallel to F are rigid during the motion (Theorem 5.1).

Continuous flattening of a regular octahedron
We denote by |uv| length of the line segment uv. The following property, called the kite property, plays a key role for continuous flattening of a regular octahedron.
Definition 2.1. Let K = abcd be a kite with |ab| = |ad|, |cb| = |cd|, and h the midpoint of bd. Let q and q be points on bh and dh, respectively, so that |qh| = |q h|. Fold K by valley creases along dh, aq, bq, and cq, and mountain creases along ah, ch, and hq so that the point q attaches to the point q (Fig. 2). We call the resulting figure a folded kite with wing-shape at q. This figure is flexible. If the distances of two pairs of diagonal vertices of K in the resulting figure are given, such as dist(a, c) = l and dist(b, d) = m, the resulting figure is uniquely determined up to congruence. We denote it by K(l, m).
Proposition 2.2 [8,12]. Let K = abcd be a kite with |ab| = |ad|, |cb| = |cd|, and h the midpoint of bd. For any pair {l, m} with ||ah| − |ch|| ≤ l ≤ |ac| and 0 ≤ m ≤ |bd|, there is a unique point q in the line segment bh so that one of its folded kites with wing-shape at q is congruent to K(l, m). For any two folded kites K(l 1 , m 1 ) and We denote by S(p, r) the sphere with center p of radius r in R 3 with the xyzaxes. (We will use the same symbols to spheres in higher dimensional Euclidean spaces.) We denote by v the point symmetric to a point v about the origin in R 3 . For any real number t (0 ≤ t ≤ 1), we denote by p(t) the position of a point p in the space considered.

Proposition 2.3
There are six side faces and they compose three folded rhombuses Applying the kite property to those folded rhombuses, we can continuously fold the octahedron onto v 1 v 2 v 3 with moving creases shown in Fig. 3c.
Remark. In the above proof since v 2 moves to v 1 , the edge v 2 v 1 is folded in half at its midpoint h.
We use the motion of the regular octahedron mentioned in the proof of Proposition 2.3 later, so we describe it as a function of the time t (0 ≤ t ≤ 1). Let

Regular 24-cell
Let P be a regular n-polytope in R n and C = C(P ) the 2-skeleton of P . For C, we call a 2-skeleton C a folded C if C is intrinsically isometric to C, is composed of flat polygons, allowed to touch each other without self-crossing.
Definition 3.1. We say the 2-skeleton C of a regular n-polytope P is folded continuously to a folded C, which we call C 1 , if there is a continuous family of {C t : 0 ≤ t ≤ 1} satisfying the following conditions: (1) for each t, C t is a folded C, (2) C 0 = C and the mapping from t ∈ {0 ≤ t ≤ 1} to C t is continuous.
Moreover, if C 1 is a flat folded state of C, we say C is flat folded continuously.
We also call the family {C t : 0 ≤ t ≤ 1} a continuous folding (or flat-folding) process from C to C 1 .
In the definition above, the stacking order of the flat folded state C 1 is not taken into account, but in this paper we take care of it for the case of a regular 24-cell P by first folding the 2-skeleton C onto an octahedral facet of P in R 4 .
A regular 24-cell has 24 regular octahedral facets. We can realize it in R 4 with vertices (x, y, z, w) such that two of the coordinates are 1 or −1 and the others 0, we denote it by P through this section. Then there are 24 vertices in P . The edges of P are the line segments joining two vertices of distance √ 2, and there are 96 edges. The 2-skeleton is the set of the triangular faces of P , and there are 96 triangular faces.
Note that the vertices of W 0 are the centroids of six square faces of the cuboctahedron, and W + and W − are images of W 0 by translation parallel to the w-axis with +1 and −1, respectively (Fig. 5b).
Hence the 24 vertices of C(P ) are a i , a i (1 ≤ i ≤ 6) and For each triangular face F of Q, there is a unique triangular face G of W + (W − ) parallel to F with distance 2/ √ 3, so that F and G compose an octahedral

Theorem 4.1. Let P be a regular 24-cell. There is a continuous folding process from the 2-dimensional skeleton of P onto any of the octahedral facets of P so that 72 edges are rigid and each of the other 24 edges is folded at its midpoint, and furthermore, the 24 faces are rigid during the motion.
Without loss of generality, we can assume that a regular 24-cell is the 24-cell P described in the previous section and W + is the chosen facet. We fix W 0 and translate W + to W 0 in the following motion. We divide the proof into three steps. First, we define the motion of each vertex in P and then, we determine the motion of each triangular face in P so as to synchronize the motion of the vertices. Finally, combining those two motions we prove that the 2-skeleton of P can be continuously folded onto W 0 so that all requirements in the theorem are satisfied.
We first define the motion of vertices of Q. Figure 6 shows the edge graph of the cuboctahedron together with vertices of W 0 . The (curved) dotted line segment with arrow shows where each vertex in Q moves, precisely, a 1 and a 4 to v 1 , a 2 and a 6 to v 2 , a 3 and a 5 to v 3 , a 1 and a 4 to v 1 , a 2 and a 6 to v 2 , and a 3 and a 5 to v 3 . Fig. 5a onto the octahedron W 0 satisfying the following conditions for each face F of Q, e.g. F = a 1 a 2 a 3 .

Lemma 4.2. There is a continuous motion of Q (the cuboctahedron without square faces) shown in
(1) F is rotated and moved toward v 1 v 2 v 3 along the line l joining the centroids of F and v 1 v 2 v 3 . (2) F always touches the cylinder T (F ), that is, F is always orthogonal to l.
Proof. Fix the vertices v i and v i (1 ≤ i ≤ 3). Let F i (1 ≤ i ≤ 4) be four triangular faces F 1 = a 1 a 2 a 3 , F 2 = a 3 a 6 a 4 , F 3 = a 4 a 2 a 5 , and F 4 = a 1 a 5 a 6 (Fig. 6). Then the central axes of those four cylinders T (F i ) (1 ≤ i ≤ 4) meet at the origin O, so that any pair of them makes the angle arccos(1/3) at O, same as the supplement of the angle made by the two vectors from the center of a regular tetrahedron to any two vertices. Note that the intersection of any pair of those cylinders are two ellipsoids (Fig. 7a).
Choose the ellipsoid in T (F 1 )∩T (F 4 ) including the vertices a 1 and v 1 . Move the vertex a 1 along the shorter curve joining a 1 and v 1 of the ellipsoid divided by a 1  and v 1 . Move a 2 and a 3 by a congruent motion of a 1 , and move a i (1 ≤ i ≤ 3) by a symmetric motion of a i about O. Since F 1 and F 4 move onto v 1 v 2 v 3 and v 1 v 2 v 3 , respectively, there are no self-crossing for F 1 and F 4 during the motion. For any other pair of triangular faces having a common vertex in Q, there are no self-crossing by a similar reason. Therefore, there are no self-crossings during the motion of Q onto W 0 .
Note that the motion of Q mentioned above is same as the one used in the model, so-called, Octabug or Jitterbug [6,7]. Let θ(t) = (π/3)t for 1 ≤ t ≤ 1 be the rotation angle of the face F = F 1 = a 1 a 2 a 3 in the motion defined in the proof of Lemma 4.2 (Fig. 7b). Denote by m(t) the distance between a 1 (t)a 2 (t)a 3 (t) and v 1 v 2 v 3 .

Lemma 4.3. The following holds
Proof. Let E be the ellipsoid in T (F 1 )∩T (F 4 ) defined in the proof of Lemma 4.2.
Since the central axes of T (F 1 ) and T (F 4 ) makes the angle α = arccos 1/3, the angle is the angle of the plane including E and the plane, setting H, including O and parallel to F 1 , and let a * 1 (t) be the orthogonal projection of a 1 (t) to H (Fig. 7b).
By cos α = 1/3, we have Substituting this in the above, we get the equality required.
Lemma 4.4. The distance between the centroid of a 1 (t)a 2 (t)a 3 (t) and Proof. Let c 1 , c 2 = c 2 (t) and c 3 = c 3 (t) be the centroids of v 1 v 2 v 3 , Since the midpoint h of a 1 v 1 + moves to the midpoint of v 1 v 3 , H 1 and H 2 are folded in half and the other one H 3 is folded with moving creases as shown in Fig. 3. More precisely, H 3 can be continuously flat-folded as a part of the rhombus, obtained together with either H 1 or H 2 , by the kite property.
Move each of the other triangular faces in {0 < w < 1} and {−1 < w < 0} as was done with faces with the edge a 1 v + 1 , as defined above. There is no self-crossing during the motion; because for any pair E 1 , E 2 of rigid edges in {0 < w < 1} there is a hyperplane Γ such that E 1 and E 2 are included in the different half-spaces divided by Γ, and each edge moves in the one of the half-spaces. For example, when E 1 = v + 1 a 2 and E 2 = v + 1 a 6 , Γ is the hyperplane determined by a 1 v 1 v 3 and v + 1 (see Figs. 5, 6).
Proof of Theorem 4.1. By defining the motion of P onto W 0 as mentioned in Steps 1 through Step 3, there are no self-crossings. For any given octahedral facet of P we can assume, without loss of generality that it is W + . In this motion, 24 edges in and E obtained from E by replacing a i and v i to a i and v i , respectively for 1 ≤ i ≤ 6, are folded at their midpoints and the other 72 edges are rigid (see Figs. 5 and 6). So the proof of Theorem 4.1 has completed.
Remark. In the proof of Theorem 4.1, since the 2-skeleton C, of the regular 24cell P , are folded so that two octahedral facets W + and W − move onto different sides of W 0 , the stacking order of the resulting figure can be determined in such process. The remaining triangular faces of P in the half-space {w > 0} (respectively,{w < 0}) are folded between W 0 and W + (respectively, W − ), according to the motion defined in the proof of Proposition 2.3.

Flattening the 24-cell
Theorem 5.1. Let P be a regular 24-cell. There is a continuous folding process from the 2-dimensional skeleton of P onto any of its faces F , so that 54 edges and all six faces parallel to F are rigid during the motion.
Proof. Combining Theorem 4.1 and Proposition 2.3 we can prove the result, because each edge of the resulting octahedron is a sextuple and hence 18 edges are folded at their midpoints to flatten the octahedron. Therefore 54 edges are rigid and six parallel faces are rigid during the motion.
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