Common Edge-Unzippings for Tetrahedra

Common Edge-Unzippings for T etrahedra Joseph O’Rourk e ∗ August 13, 2021 Abstract It is sho wn that there are examples of distinct p olyhedra, each with a Hamiltonian path of edges, which when cut, unfolds the surfaces to a common net. In particular, it is established for infinite classes of triples of tetrahedra. 1 In tro duction The limited fo cus of this note is to establish that there are an infinite collec- tion of “edge-unfolding zipp er pairs” of conv ex p olyhedra. A net for a conv ex p olyhedron P is an unfolding of its surface to a planar simple (nonov erlapping) p olygon, obtained by cutting a spanning tree of its edges (i.e., of its 1-skeleton); see [DO07, Sec. 22.1]. Shephard explored in the 1970’s the sp ecial case where the spanning tree is a Hamiltonian path of edges on P [She75]. Such Hamiltonian unfoldings were futher studied in [DDLO02] (see [DO07, Fig. 25.59 ]), and most recen tly in [LDD + 10], where the natural term zipp er unfolding was in tro duced. Define tw o p olyhedra to b e an e dge-unfolding zipp er p air if they hav e a zipper unfolding to a common net. Here we emphasize edge-unfolding, as opp osed to an arbitrary zipper path that may cut through the interior of faces, which are easier to identify . Th us we are considering a sp ecial case of more general net p airs : pairs of p olyhedra that may b e cut op en to a common net. In general there is little understanding of whic h polyhedra form net pairs un- der an y definition. See, for example, Op en Problem 25.6 in [DO07], and [O’R10] for an exploration of Platonic solids. Here we establish that there are infinite classes of con vex p olyhedra that form edge-unfolding zipp er pairs. Th us one of these p olyhedra can b e cut op en along an edge zipp er path, and rezipp ed to form a differen t polyhedron with the same property . In particular, we pro v e this theorem: Theorem 1 Every e quilater al c onvex hexagon, with e ach angle in the r ange ( π / 3 , π ) , and e ach p air of angles line arly indep endent over Q , is the c ommon e dge-unzipping of thr e e inc ongruent tetr ahe dr a. ∗ Department of Computer Science, Smith College, Northampton, MA 01063, USA. orourke@cs.smith.edu . 1 The angle restrictions in the statement of the theorem are, in some sense, inciden tal, included to matc h the pro of techniques. The result is quite narrow, and although certainly generalizations hold, there are imp edimen ts to proving them formally . This issue will b e discussed in Section 6. 2 Example An example is shown in Figure 1. The equilateral hexagon is folded via a p erimeter halving folding [DO07, Sec. 25.1.2]: half the p erimeter determined b y opp osite vertices is glued (“zipp ed”) to the other half, matching the other four v ertices in t w o pairs. Because the hexagon is equilateral, the corresp onding edge lengths match. The resulting shap e is a conv ex p olyhedron by Alexandro v’s theorem ([DO07, Sec. 23.3]), and a tetrahedron b ecause it has four p oin ts at which the curv ature is nonzero. Choosing to halv e the p erimeter at eac h of the three pairs of opp osite v ertices leads to the three tetrahedra illustrated. Note that each has three unit- length edges that form a Hamiltonian zipp er path, as claimed in the theorem. The main challenge is to show that the edges of the hexagon glued together in pairs b ecome edges of the p olyhedron, and so the unzipping is an edge- unzipping. 3 Distinct T etrahedra First we sho w that the linear independence condition in Theorem 1 ensures the tetrahedra will b e distinct. W e define tw o angles x and y to b e line arly indep endent over Q if there are no rationals a and b such that y = aπ + bx . W e will abbreviate this condition to “linear indep endence” b elow. W e will show that the curv atures (2 π minus the inciden t face angle) at the v ertices of the tetrahedra differ at one vertex or more, whic h ensures that the tetrahedra are not congruent to one another. Let the vertices of the hexagon b e v 0 , . . . , v 5 , with vertex angles α 0 , . . . , α 5 . W e identify the three tetrahedra by their halving diagonals, and name them T 03 , T 14 , T 25 . W e name the curv atures at the four vertices of the tetrahedra ω 1 , ω 2 , ω 3 , ω 4 . These curv atures for the three tetrahedra (see Figure 1) are as follo ws: T etrahedron V ertex Curv atures halving diagonal ω 1 ω 2 ω 3 ω 4 T 03 : v 0 v 3 1 2 (2 π − α 0 ) 1 2 (2 π − α 3 ) 2 π − ( α 1 + α 5 ) 2 π − ( α 2 + α 4 ) T 14 : v 1 v 4 1 2 (2 π − α 1 ) 1 2 (2 π − α 4 ) 2 π − ( α 0 + α 2 ) 2 π − ( α 3 + α 5 ) T 25 : v 2 v 5 1 2 (2 π − α 2 ) 1 2 (2 π − α 5 ) 2 π − ( α 0 + α 1 ) 2 π − ( α 3 + α 4 ) No w we explore under what conditions could the four curv atures of T 03 b e iden tical to the four curv atures of T 14 . (There is no need to explore other p os- 2 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 Figure 1: Three tetrahedra that edge-unzip (via the dashed Hamiltonian path) to a common equilateral hexagon. 3 sibilies, as they are all equiv alen t to this situation by relab eling the hexagons.) Let us lab el the curv atures of T 03 without primes, and those of T 14 with primes. First note that we cannot hav e ω 1 = ω 0 1 or ω 1 = ω 0 2 , for then tw o angles m ust b e equal: α 0 = a 1 or α 0 = α 4 resp ectiv ely . And equal angles are not linearly indep enden t. So we are left with these p ossibilities: ω 1 = ω 0 3 , or ω 1 = ω 0 4 . The first leads to the relationship α 2 = π − 1 2 α 0 , a violation of linear independence. The second p ossibilit y , ω 1 = ω 0 4 , requires further analysis. It is easy to eliminate all but these tw o possibilities, which map indices (1 , 2 , 3 , 4) to either (4 0 , 3 0 , 1 0 , 2 0 ) or (4 0 , 3 0 , 2 0 , 1 0 ): ω 1 = ω 0 4 , ω 2 = ω 0 3 , ω 3 = ω 0 1 , ω 4 = ω 0 2 and ω 1 = ω 0 4 , ω 2 = ω 0 3 , ω 3 = ω 0 2 , ω 4 = ω 0 1 Explicit calculation shows that the first set implies that α 4 = 2 π − 2 α 2 , and the second implies that α 1 = 3 4 π − α 0 . Th us we hav e reached the desired conclusion: Lemma 1 Line ar indep endenc e over Q of p airs of angles of the hexagon (as state d in The or em 1) implies that the thr e e tetr ahe dr a have distinct vertex cur- vatur es, and ther efor e ar e inc ongruent p olyhe dr a. W e hav e phrased the condition as linear indep endence of pairs of angles for simplicit y , but in fact a collection of linear relationships must hold for the four curv atures to be equal. So the restriction could b e phrased more narro wly . Note also we hav e not used in the pro of of Lemma 1 the restriction that the angles all b e “fat”, α i > π / 3. This will b e used only in Lemma 3 b elow. It would b e equally p ossible to rely on edge lengths rather than angles to force distinctness of the tetrahedra. F or example, we could demand that no tw o diagonals of the hexagon hav e the same length (but that w ould lea v e further w ork). Another alternative is to av oid angle restrictions, p ermitting all equi- lateral hexagons, but only conclude that “generally” the three tetrahedra are distinct. Obviously a regular hexagon leads to three identical tetrahedra. The form of Theorem 1 as stated has the adv antage of easily implying that an uncoun table num b er of hexagons satisfy its conditions (see Corollary 1). 4 Shortest P aths are Edges No w we know that the hexagons fold to three distinct tetrahedra. So there is a zipp er path on each tetrahedron that unfolds it to that common hexagon. All the remaining w ork is to show that the zipp er path is an edge path—comp osed of p olyhedron edges. This seems to b e less straightforw ard than one might exp ect, largely b ecause there are not many to ols av ailable b eyond Alexandrov’s existence theorem. It is easy to see that ev ery edge of a polyhedron P is a shortest path b et w een its endpoints. The rev erse is far from true in general, but it holds for tetrahedra: 4 Lemma 2 Each shortest p ath b etwe en vertic es of a tetr ahe dr on is r e alize d by an e dge of the tetr ahe dr on. Pro of: Note that there are  4 2  = 6 shortest paths b etw een the four vertices, and six edges in a tetrahedron, so the combinatorics are correct. Supp ose a path ρ = xy is a shortest path b etw een vertices x and y of a tetrahedron T , but not an edge of T . Because eac h pair of vertices of a tetrahedron is connected by an edge, there is an edge e = xy of T . Because ρ is not an edge of T , it cannot b e realized as a straight segment in R 3 , b ecause all of those v ertex-v ertex segments are edges of T . How ever, e is a straight segmen t in R 3 , and so | e | < | ρ | , contradicting the assumption that ρ is a shortest path. The reason this pro of works for tetrahedra but can fail for a p olyhedron of n > 4 vertices is that the condition that ev ery pair of vertices are connected by an edge fails in general. The p olyhedra for whic h that holds are the “neigh b orly p olyhedra” (more precisely , the 2-neighborly 3-p olytop es). 5 Hexagon Edges are T etrahedron Edges With Lemma 2 in hand, it only remains to show that the pairs of matched unit- length hexagon edges are shortest paths on the manifold M obtained b y zipping the hexagon. W e use the lab eling and folding shown in Figure 2(a), where v 1 and v 0 1 are identified, as are v 2 and v 0 2 . W e need to show that b oth v 0 v 1 and v 1 v 2 are shortest paths ( v 2 v 3 is symmetric to v 0 v 1 and follo ws by relab eling). Let D i b e the geo desic disk of unit radius centered on v i , on the zipp ed manifold M . If the disks D 0 and D 1 , illustrated in Figure 2(a), are empt y of other v ertices, then the desired shortest paths are established. It is here that w e will employ the assumption that the hexagon angles are greater than π / 3. First, for any unit-equilateral con v ex hexagon, without constraints on the angles, diagonals connecting opp osite vertices are at least length 1: | v 0 v 3 | ≥ 1, | v 1 v 0 2 | ≥ 1, and | v 2 v 0 1 | ≥ 1. W e no w argue for this elemen tary fact (whic h is lik ely known in some guise in the literature). Concen trating on D 0 and v 0 v 3 (all others are equiv alent by relab eling), con- sider the quadrilateral ( v 0 , v 1 , v 2 , v 3 ). Because the hexagon is unit-equilateral, | v 0 v 1 | = 1, | v 1 v 2 | = 1, and | v 2 v 3 | = 1. Assume for a con tradiction that the diagonal is short: | v 0 v 3 | < 1. Then it is not difficult to prov e that the sum of the quadrilateral angles at the endp oints of this shorter side exceed π : ∠ v 3 v 0 v 1 + ∠ v 2 v 3 v 0 > π . Applying the same logic to the other half of the hexagon sharing diagonal v 0 v 3 , the quadrilateral ( v 3 , v 0 2 , v 0 1 , v 0 ), we reach the conclusion that the sum of the hexagon angles at v 0 and v 3 exceeds 2 π . Thus one of those angles exceeds π , contradicting the fact that the hexagon is con v ex. 1 W e may conclude from this analysis is that v 3 cannot lie inside D 0 . Supp ose now that v 0 2 is inside D 0 , as in Figure 2(b). Then, the angle α 0 1 at v 0 1 m ust b e smaller than π / 3, contradicting the angle minim um assumed in 1 I thank Mirela Damian for this argument, whic h is simpler than my original pro of. 5 v 1 v' 1 v 0 v 3 v 2 v' 2 v 1 v 0 v 3 v 2 v' 2 v' 1 (a) (c) (b) v 1 v' 1 v 0 v 3 v 2 v' 2 Figure 2: (a) Lab els for the zipping of a hexagon. The dashed edge shows the p erimeter halving line, but there is no assumption that diagonal is a crease that b ecomes an edge of the tetrahedron. (b) When D 0 is not empty , some angle (here, α 0 1 ) is smaller than π / 3. (c) A “degenerate” hexagon, with tw o angles π , whic h folds to a doubly cov ered square. 6 the theorem. The same clearly holds for v 2 inside D 0 , as well as the other com binations. So the assumption that the hexagon is “fat” in the sense that none of its angles are small, guarantees that the disks D 0 and D 1 (and b y symmetry , D 2 and D 3 ) are empty of vertices of the hexagon. As is evident from Figure 2(b), ho wev er, even when no v ertex is inside D 0 , a p ortion of D 0 can fall outside the hexagon, which, b ecause it is all zipp ed to a closed manifold M , means it re-enters the hexagon at the other copy of that edge. How ev er, three facts are easily established. First, the “ov erhang” has width at most h = 1 − √ 3 / 2; see Figure 3. Second, no vertex of the “next” hexagon cop y can lie inside that ov erhang (without violating conv exity of the hexagon). Third, to even ha v e a hexagon edge b e partially interior (and so o verhanging the disk in to a third hexagon copy) requires some angle ( β in the figure) to b e very small, violating the π / 3 minimum angle. Th us the ov erhang of a D i disk b ey ond the original hexagon cannot encompass a vertex. 1 1 1 β h Figure 3: A disk D i “o verhangs” the original hexagon and enters another copy . Here β = 2 sin − 1 ( h/ 2) < 8 ◦ . W e may conclude: Lemma 3 The thr e e unit-length e dges of the hexagon, { v 0 v 1 , v 1 v 2 , v 2 v 3 } ar e e ach shortest p aths on the folde d manifold M b etwe en their endp oint vertic es. I ha ve firm empirical evidence (via explorations in Cinderella) that Lemma 3 holds just as stated without an y assumption that the hexagon is fat. But pro ving this formally seems difficult. Figure 4 illustrates a path ρ that spirals around the manifold from v 0 to v 1 . Each such geo desic path candidate must b e established to b e at least length 1. 6 Discussion W e hav e now prov ed Theorem 1: Lemma 3 shows that the three unit-length edges forming a Hamiltonian path are shortest paths, Lemma 2 then implies that they are edges of the tetrahedron, and Lemma 1 establishes that the three tetrahedra are distinct. Corollary 1 Ther e ar e an unc ountable numb er of hexagons that satisfy the c onditions of The or em 1. 7 v 1 v' 1 x v 2 v' 2 v 3 v 0 v 3 v 1 v 0 v 1 v' 1 v 2 v' 2 v 0 v 3 x x (a) (b) ρ Figure 4: (a) A “spiral” path from v 0 to v 1 . (b) The path flattened, crossing sev eral copies of the hexagon. Pro of: The constraints that P α i = 4 π and π / 3 < α i < π clearly leav e an uncoun table num ber of solutions, in fact a 5-dimensional op en set in R 5 (the sixth angle is determined b y the other fiv e). (F or example, a small 5-ball around ( α 0 , α 1 , α 2 , α 3 , α 4 ) = ( π − 1 2 , π − 1 2 , π − 1 2 , π 3 + 1 4 , π 3 + 1 4 ) with α 5 = π 3 + 1, is inside this set.) The constrain t requiring indep endence o ver Q only excludes a countable num b er of 4-dimensional h yp erplanes (e.g., α 2 = π − 1 2 α 0 ). These hyperplanes hav e zero measure in R 5 , and a coun table union of sets of zero measure has zero measure, 2 lea ving an uncoun table num b er of solutions after excluding the hyperplanes. The construction considered here generalizes to arbitary even n , although I do not see how to prov e that the zipp er path follows edges of the polyhedron: Prop osition 1 F or any even n ≥ 4 , an e quilater al c onvex n -gon is the c ommon zipp er-unfolding of n/ 2 gener al ly distinct p olyhe dr a of ( n − 2) vertic es e ach. The regular-p olygon version of this construction folds to what were called “pita p olyhedra” in [DO07, Sec. 25.7.2]. I conjecture that all the zipp er-paths in Prop osition 1, for strictly conv ex equilateral conv ex n -gons, in fact follo w p olyhedron edges. Resolving this con- jecture w ould require to ols to determine when particular geo desic paths on an Alexandro v-glued manifold are edges of the resulting conv ex p olyhedron. References [DDLO00] Erik D. Demaine, Martin L. Demaine, Anna Lubiw, and Joseph O’Rourk e. Examples, counterexamples, and en umeration re- sults for foldings and unfoldings b et w een p olygons and p olytop es. 2 I thank Qiao c hu Y uan and Theo Buehler for guidance here, http://math.stackexchange. com/questions/41494/ . 8 T echnical Rep ort 069, Smith College, Northampton, July 2000. arXiv:cs.CG/0007019. [DDLO02] Erik D. Demaine, Martin L. Demaine, Anna Lubiw, and Joseph O’Rourk e. Enumerating foldings and unfoldings b etw een p olygons and polytop es. Gr aphs and Combin. , 18(1):93–104, 2002. See also [DDLO00]. [DO07] Erik D. Demaine and Joseph O’Rourke. Ge ometric F olding Algo- rithms: Linkages, Origami, Polyhe dr a . Cambridge Universit y Press, July 2007. http://www.gfalop.org . [LDD + 10] Anna Lubiw, Erik Demaine, Martin Demaine, Arlo Shallit, and Jonah Shallit. Zipp er unfoldings of p olyhedral complexes. In Pr o c. 22nd Canad. Conf. Comput. Ge om. , pages 219–222, August 2010. [O’R10] Joseph O’Rourk e. Flat zipper-unfolding pairs for platonic solids. http://arxiv.org/abs/1010.2450 , Octob er 2010. [She75] Geoffrey C. Shephard. Conv ex p olytop es with conv ex nets. Math. Pr o c. Camb. Phil. So c. , 78:389–403, 1975. 9