On the Cover of the Rolling Stone∗

Adrian Dumitrescu† Csaba D. T´oth‡

Abstract both inventors and the general public for hundreds of We construct a convex polytope of unit diameter that years. The enormous appeal of perpetual motion resides when placed on a horizontal surface on one of its faces, in the promise of a virtually free and limitless source of it repeatedly rolls over from one face to another until power. The fact that perpetual-motion machines cannot it comes to rest on some face, far away from its start work because they violate the laws of thermodynamics position: that is, the horizontal distance between the has not discouraged inventors and hucksters from at- footprints of the start and final faces can be larger tempting to break, circumvent, or ignore those laws.” than any given threshold. According to the laws of Indeed, this fact did not discourage us either, and the physics, the vertical distance between the center of mass current paper can be viewed as yet another attempt. of the polytope and the horizontal surface continuously It offers a good approximation of such a device in the decreases throughout the entire motion. The speed of sense that the range of the motion can be arbitrarily the motion is irrelevant. Specifically, if the polytope large and prescribed beforehand. is manually stopped after each tumble, the motion Imagine a convex polyhedron standing on a hori- resumes when released (unless it stands on the final zontal surface in Euclidean 3-space. The body is stable stable face). on the facet that it stands on if and only if its center Moreover, such a polytope can be realized so that of mass projects vertically in the interior of that facet. (i) it has a unique stable face, and (ii) it is an arbitrary For example, a tall prism that leans like the Tower of close approximation of a unit ball. As such, this Pisa is unstable (a suggestive example offered by Daw- construction gives a positive answer to a question raised son et al. [10]). Obviously, Platonic solids of uniform by Conway (1969). density are, by symmetry, stable on all their facets. The arbitrarily large rolling distance property inves- Note however that the uniform density is a key assump- tigated here for the first time raises intriguing questions tion: If we place a regular octahedron on a facet (base) and opens new avenues for future research. and concentrate its mass in one of the vertices whose Keywords: perpetuum mobile, rolling distance, projections lie outside of the base facet, the octahedron unistable polytope, center of mass, laws of physics. rolls onto an adjacent facet. A convex polytope will roll from one facet to 1 Introduction another only if it thereby monotonically lowers its center of mass; it therefore follows that no polytope can ever Oh ye seekers after perpetual motion, roll back to a facet from which it has rolled away, and how many vain chimeras have you pursued? so every polytope has at least one facet upon which it Go and take your place with the alchemists. is stable. A polytope that is stable upon only one of — Leonardo da Vinci (1494) its facets is called unistable (or sometimes monostatic, unistatic, or monostable); see, e.g., [10, 11]. The history of perpetual motion machines dates Clearly, if the body is allowed to have nonuniform back to the Middle Ages. The Encyclopaedia Britannica (but positive) density, the center of mass may be entry dedicated to Perpetual Motion reads: “Perpetual anywhere in the interior of the body. As reported in [9], motion, although impossible to produce, has fascinated Conway constructed a tetrahedron in R3 which, with ∗Title inspired by the song The Cover of “Rolling Stone” by a suitably positioned center of mass, is stable only on Dr. Hook & The Medicine Show, 1972. one facet. Moreover, this tetrahedron, when placed †Department of Computer Science, University of Wisconsin– on one of its facets, can roll onto each of its 4 facets Milwaukee, WI, USA. until it reaches a stable position. Conway [16] also ‡Department of Mathematics, California State University Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php constructed a unistable convex polytope (a truncated Northridge, Los Angeles, CA; and Department of Computer Sci- ence, Tufts University, Medford, MA, USA. Research by this cylinder of uniform density) with 19 faces; the cross- author was supported in part by the NSF awards CCF-1422311, section of the cylinder is a 2D convex polygon that CCF-1423615, and DMS-1800734. is unistable with nonuniform density. The cylinder

Copyright © 2020 2575 Copyright for this paper is retained by authors is truncated by slanted planes to move the center of trajectory of the projection of some reference point of mass to the desired position. Bezdek [3] reduced the the polytope (for instance, the center of mass) onto number of faces to 18, while the current record, 14, is the horizontal plane, or simply as (ii) the distance obtained by computer search [24]. See [12, Sec. 4] for between the footprints of the start and final faces in combinatorial properties of small unistable polytopes in the horizontal plane. For convenience, we stick to the R3. Related work regarding unistability of polytopes latter measure that we find more intuitive as well as include [8, 17]; see also [7, Sec. B12]. serving our intent. This definition of stability may be extended, some- Notation and terminology. For brevity, a con- what less physically, to polytopes in other dimensions, vex polyhedron of uniform density is called uniform. For including the plane. Conway showed [6] that in the a convex polytope P in Rd, let ξ(P ) denote the center of plane, no convex polygon of uniform density is unistable, mass. The perimeter of P , denoted per(P ), is the sum and that in 3-space no tetrahedron of uniform density is of the edge lengths of P . The boundary of P is denoted unistable. Heppes [18] constructed a tetrahedron which, by ∂P . placed upon the appropriate face, rolls twice before For a line segment s, let `(s) denote the line reaching one of its two stable faces. It thus tours all of containing s. For two points p, q ∈ Rd, let d(p, q), or its unstable faces before reaching equilibrium. Heppes sometimes |pq|, denote the Euclidean distance between asked in how few dimensions, if any, a full-dimensional them; while the length of a rectifiable curve C is denoted simplex could be unistable. Dawson et al. [10] con- by len(C). For two sets of points A, B ⊂ d, their 10 R structed a unistable simplex in R and showed that no distance is d(A, B) = inf{d(a, b) | a ∈ A, b ∈ B}.A 7 such simplex exists in R ; computational evidence sug- hyperplane H is given by the linear equation H = {x ∈ 8 9 gests that no such simplex exists in R and R . Daw- Rd : a · x = b}, where a ∈ Rd and b ∈ R. The closed son et al. [10] also exhibit a unistable simplex in 11 d R ball of radius r in R centered at point z = (z1, . . . , zd) that sequentially rolls through all 12 facets. d is Bd(z, r) = {x ∈ R : d(z, x) ≤ r}.A unit ball is a Static equilibria of convex bodies (allowing smooth ball of unit radius in Rd. surfaces) have also been studied; stable and unstable equilibria are related to sinks and sources in the gra- dient vector field characterizing the surface. The total ξ0 number of equilibria is governed by the Poincaré-Hopf ξ(P ) p Theorem [2]. For example, a polytope standing on a D vertex may be at equilibrium, but this equilibrium is P f = P ∩ π unstable. Várkonyi and Domokos [26], settling a conjec- 0 ture due to Arnold (1995), constructed a smooth con- A vex body (called gömböc) with a unique stable and a π0 ξ0 unique unstable equilibrium. Their discovery has lead to a systematic study of the topological properties of P stable and unstable points [13]. However, it is not ob- Figure 1: Left: A convex polygon (or polytope) , ξ P ξ0 vious how to approximate a smooth convex body by a its center of mass ( ), and its projection to the π d polyhedron with the same stability: Domokos et al. [14] horizontal hyperplane 0. Right: A polytope in R f P ∩π show that uniformly random discretization may intro- stands on a face = 0, its center of mass vertically ξ0 f P duce additional stable equilibria. projects to in the exterior of , and rolls about the axis A. Although our focus in this paper falls along these lines, it is headed in a different direction. Rules of motion. Consider a convex body P in d, standing on a horizontal hyperplane π . By the laws Question 1. How far can a polytope of unit diameter R 0 of physics, the potential energy of P equals to mgh, roll when placed on a horizontal surface? where m is the mass of P and h denotes the vertical Question 2. For every sufficiently large n, does there distance from ξ(P ) to π0. The body P rolls without exist a polytope with n facets, and an ordering of the slipping only if the height of ξ(P ) strictly decreases. By facets, f1, f2, . . . , fn, so that if P is placed upon fi, it Gauss’ principle of least constraint [15], the direction of will roll through facets fi+1, . . . , fn until resting on its the motion maximizes the rate of descent, i.e., minimizes Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php unique stable facet fn? the slope of the direction vector of the trajectory of ξ(P ). In particular, a polytope P standing on a face The distance traveled by a rolling polytope could f rolls (to another face) if and only if the vertical be measured in ways such as: (i) the length of the projection onto π0 of the center of mass of P , ξ(P ),

Copyright © 2020 2576 Copyright for this paper is retained by authors lies in the exterior of f; see Fig. 1 (left). (i) For every ε > 0, we construct a Hamiltonian Denote this projection point by ξ0. Let D be the unistable convex polygon P in R2 that can be placed largest ball in the horizontal hyperplane π0 centered at on a horizontal line upon one of its sides so that it ξ0 whose interior is disjoint from f. Then ∂D intersects repeatedly rolls over covering a horizontal distance of f in a single point p. The steepest descent of ξ(P ) at least (1 − ε) per(P ) (Theorem 2.1 in Section 2). occurs when P rotates around the (d − 2)-dimensional (ii) For every L ≥ 1, we construct a Hamiltonian A ⊂ π D p 3 axis 0 tangent to at point ; see Fig. 1 (right). unistable convex polytope of unit diameter in R that During the rotation, the face A ∩ P remains fixed, can be placed upon one of its facets so that it repeatedly and the rotation stops when P ∩ π0 becomes a higher- rolls over until it comes to rest on its unique stable 3 dimensional face. In particular in R , if A ∩ P is an face at a distance of at least L from its start position edge e, then P rotates about e to an adjacent face; and if (Theorem 3.1 in Section 3). A∩P is a vertex, then P rotates about A until P “lands” (iii) The polytope in Theorem 3.1 can be an ar- on a facet or an edge. The polytopes we construct in bitrary close approximation of a ball (Theorem 4.1 in Sections 3–4 always roll around edges if initially placed Section 4). This gives a partial answer to a 50-year old on an unstable face. question raised by Conway [16]; see also [7, Sec. B12]. Note that the rolling motion is governed by discrete geometry alone. In particular, it does not take the (iv) For every L ≥ 1, there exists a uniform moment of inertia into account, e.g., as if the convex convex polytope of unit diameter (albeit not necessarily 3 polytope were moving in a fluid of high viscosity. The unistable or Hamiltonian) in R that can be placed rolling motion of a smooth convex body with moment upon one of its facets so that it repeatedly rolls over of inertia is a problem in classical dynamics that can covering a horizontal distance of at least L (Theorem 4.2 be characterized by differential equations. For example, in Section 4). Chaplygin [5] described the motion of a nonuniform ball on a horizontal plane in 1903. We do not pursue that 1.1 Preliminaries The proofs of the following lem- direction here. mas are folklore; refer to Fig. 1. Discussion. In the plane, the distance referred to d Lemma 1.1. Let P be a convex polytope in R and p in Question 1 is clearly bounded: as noted above, a any point contained in P . Then there exists a face convex polygon P stands on each of its faces at most f of P so that the orthogonal projection of p onto its once during a rolling motion, consequently the distance supporting hyperplane is contained in f. traveled cannot exceed the perimeter of P , which is at most π by the isodiametric inequality. Somewhat Proof. Let p ∈ P , let f1 be a facet of the polytope surprisingly, the answer to Question 1 in 3-space is “As whose supporting hyperplane H1 is closest to p, and let far as we like.” Therefore, such a polytope is in some q1 be the orthogonal projection of p onto H1. Note that sense a good approximation of a perpetuum mobile. The dist(p, H1) = |pq1|. If q1 ∈/ f, then the line segment pq1 precise statement is in our Theorem 3.1 (in Section 3). intersects the boundary of P at some point q2 contained A variant of our construction (Theorem 4.1 in in some facet f2, which spans a hyperplane H2. Then Section 4) is related to the following question raised we have dist(p, H2) ≤ |pq2| < |pq1| = dist(p, H1), a by Conway [16, p. 81], [7, Sec. B12], to which it gives contradiction. a partial answer: “What is the set of convex bodies uniformly approximable by unistable polyhedra, and An alternative argument for the proof (in Physics does this contain the sphere?” terms) is as follows. Make p the center of mass of P , To formulate our results it is first convenient to that is, ξ(P ) = p. Let f be a face of the polytope whose strengthen the concept of unistable polytope as follows. supporting hyperplane is closest to ξ(P ). Then f must We say that the pair (P, ξ(P )) of a polytope and its be stable, since rolling to any other face cannot lower center of mass is Hamiltonian unistable if P is unistable the position of ξ(P ). and in addition, there exists an ordering of the facets, Lemma 1.2. Let P be a convex polytope in Rd (uniform f1, f2, . . . , fn, so that if P is placed upon fi, it will roll or not). Then P cannot roll forever. to face fi+1, for i = 1, . . . , n − 1. Perpetuum mobile approximations. We show Proof. Let ξ = ξ(P ) denote the center of mass of P . Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php several constructions of polytopes as described below. Assume that P can roll forever on a horizontal surface The polytopes in constructions (i)–(iii) need not be π0, while the center of mass monotonically gets closer to uniform, and only the one in (iv) yields a uniform π0. Since P has a finite number of facets, some facet f polytope. will repeat; that is, there are two time instances t1 < t2,

Copyright © 2020 2577 Copyright for this paper is retained by authors so that the vertical distance from ξ to π0 (and f) at time t2 is smaller than the vertical distance from ξ to π0 (and f) at time t1. This completes the proof. q 2 The Planar Case p o o Let B be a convex body in the plane. As noted above, the distance covered (i.e., the distance between the β initial and ﬁnal footprints) cannot exceed the perimeter of B. Therefore, it is convenient to measure the displacement of a convex body relative to its perimeter.

Theorem 2.1. For every ε > 0, we can construct s1 a convex polygon P with O(ε−2) vertices that (i) is Hamiltonian unistable and (ii) can be placed on a Figure 2: Construction of a unistable polygon P that horizontal line upon one of its sides so that it repeatedly covers a large distance when it rolls. rolls over such that the distance between the initial and the final contact sides on the horizontal line is at least (1 − ε) per(P ). (The polytope P need not be uniform.) We next construct a convex polygon P that mimics the behavior of B(γ). We subdivide the boundary of Proof. Consider two concentric circles, C1 and C2, of B(γ) by choosing a sequence of points that forms the radii 1 − a and 1 respectively, centered at the origin vertex set of P . The first vertex of P is (0, −1). If P o. Let D1 and D2 denote the corresponding disks. See rests upon the current side pq, P will roll to the next Fig. 2. Let γ be an open continuous curve in between the side of P adjacent to q, provided that the angle ∠oqp of two circles, in the form of a spiral and whose parametric the respective triangle is obtuse. Taking this fact into equation is: consideration, if vertex p is fixed on γ, the next vertex a q = q(p) ∈ γ is chosen as follows. By construction, (2.1) 1 − t (sin t, − cos t), 0 ≤ t ≤ 2π. 2π the distance |oq| is strictly decreasing as q moves away from p, and there is a unique point q0 ∈ γ such that Let s be the tangent to γ from the point (0, −1); ∠oq0p = π/2. Indeed, the tangent to γ at point p is let t0 < 2π be the corresponding value of t at the not orthogonal to the segment op, and so γ enters the tangency point. Let B = B(γ) be the planar convex interior of the circle with diameter op at p, and it exits γ body bounded by an initial segment of the spiral and at a unique point q0. As such, the next vertex can be s by the tangent . chosen anywhere on the arc pq0, with q0 specified as We will subsequently construct the polygon P ⊂ B. above. The center of mass of B is the origin and we do not All vertices of P are selected in this way until we assume uniform density. The same assertions will hold reach the point of tangency between s and γ. The last true for P . Observe that D1 ⊂ B ⊂ D2, and so side of P is s; observe that s is the unique stable side 2π(1 − a) ≤ per(B) ≤ 2π. Consider also the tangent a of P . Let x = ∠poq0, and put z = 1 − 2π t; note that s1 to C1 from the point (0, −1), subtending a center- 1 − a ≤ z ≤ 1. We have angle β; its length is √ 1 − a (t + x) z − a x ax p 2 p 2 2π 2π |s1| = 1 − (1 − a) = 2a − a ≤ 2a. (2.2) cos x = a = = 1 − . 1 − 2π t z 2πz Since the tangent s is sandwiched between√D1 and We may assume without loss of generality that x ≤ D2, it follows that |s| ≤ 2|s1| and thus |s| ≤ 8a. Set a = ε2. We have 1/10. The well-known Taylor expansion of cos x around √ √ zero yields |s| 8a 2 2ε ≤ = ≤ ε. (2.3) per(B) 2π(1 − a) 2π(1 − a) x2 x2 x4 47x2 1 1 − ≤ cos x ≤ 1 − + ≤ 1 − , for x ≤ . Consequently, the horizontal distance covered by B 2 2 24 96 10

Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php when rolling clockwise after being placed upon its ﬁrst a Applying this estimate in (2.2) further yields that π ≤ side is a 2 x ≤ 3 . Recall that we have set a = 2ε , thus the length |s| of each side of P can be Θ(ε2); thus P has O(ε−2) sides, per(B) − |s| = 1 − per(B) ≥ (1 − ε) per(B). per(B) as required.

Copyright © 2020 2578 Copyright for this paper is retained by authors 3 The Three-Dimensional Case and let H+(t) be the halfspace bounded by H(t) that In this section we prove the following result. contains the origin o. We define the convex body   Theorem 3.1. For every L ≥ 1, one can construct 0 \ + a Hamiltonian unistable polytope P = P (L) of unit B = B ∩  H (t) . diameter and with O(L6) faces in 3-space and specify t∈[0,T ] an ordering f , f , . . . , f , of these faces, so that if P is 1 2 n We next specify the parameter a > 0. For t = 0, placed on a horizontal plane π in contact with f , the 0 1 the line `(0) is tangent to B. For t ∈ (0,T ], the line following hold: `(t) intersects the sphere ∂B at two points at distance p − − at/T 2 (i) P successively rolls around the edge fi ∩ fi+1, for 2 1 (1 ) apart. The two intersection points each i = 1, 2, . . . , n − 1. trace out two arcs on ∂B, that we denote by γleft and γright. The plane H(T ) intersects ∂B in a circle (ii) P comes to rest on the last facet fn, which is the C := H(T ) ∩ ∂B. Let a > 0 be sufficiently small only stable facet of P . so that both γleft and γright are simple smooth curves (iii) The horizontal distance between the initial and final disjoint from each other that intersect the circle C only contact faces is at least L. at the endpoints γleft(T ) and γright(T ). By construction, β ⊂ H+(t) for all t ∈ [0,T ], thus β is a smooth arc on (The polytope P need not be uniform.) the surface of B0. As in the proof of Theorem 2.1, the center of mass Proof. We first describe a general construction of a of B0 is the origin o and uniform density is not assumed. Hamiltonian unistable polytope with center of mass at The same assertions will hold for the polytope P that the origin, and then adjust the parameters to ensure will be constructed from B0. If B0 is placed on a that it has unit diameter and travels a distance at horizontal surface making contact at γ(0) = β(0), it least L away from its initial position. We will assume will start rolling continuously while its center of mass (for the proof) that L is sufficiently large, i.e., L ≥ L0 continuously gets closer to the plane π0, with a contact for a suitable L0 ∈ N; then the statements of the segment γleft(t)γright(t) at time t ∈ [0,T ). The body theorem automatically hold for any smaller value of L. B0 stops rolling at time t = T , when the contact face General construction. We start with a unit ball is bounded by γleft(T )γright(T ) and a circular segment B centered at the origin o = 0. Let T ≥ 1 and of C. γ : [0,T ] → ∂B be a simple smooth arc (open curve) Discretization. We construct a polytope P from drawn on the surface of B, parameterized by its arc B0 as follows. Let δ > 0 be a sufficiently small 0 length, that is, its length is T and |γ (t)| = 1 for all parameter (specified below). Consider a subdivision t ∈ [0,T ]. Intuitively, we “scrap off” some neighborhood 0 = t1 < t2 < . . . < tn = T of the interval [0,T ] such of γ, while gradually going deeper to the interior of B. that ti+1 − ti < δ for i = 1, . . . , n − 1. For i = 1, . . . , n, We define an arc β : [0,T ] → B that follows γ but goes we use the shorthand notation Hi = H(ti). We can now gradually deeper into the interior of B, and then use define P as an intersection of halfspaces as follows. hyperplanes tangent to β(t), for all t ∈ [0,T ] to scrap off a neighborhood of γ. n We continue with the technical details. Let 0 < \ (3.5) P = H+. a ≤ 1/10 be a sufficiently small constant (to be specified i i=1 later), and let β : [0,T ] → B be a curve defined by At this point P is a (possibly unbounded) convex at (3.4) β(t) = 1 − γ(t) for t ∈ [0,T ]. polyhedron; we later choose γ to ensure that P is a T (bounded) convex polytope. Note that neither B nor P contains the other body. Since the plane Hi contains Note that β(0) = γ(0) and β(T ) = (1 − a)γ(T ). Also the point β(tj) if and only if i = j, the polytope P has observe that |γ(t)| = |γ0(t)| = 1 for all t ∈ [0,T ]; and precisely n faces, one in each plane Hi (i = 1, . . . , n); 0.9 ≤ |β(t)| ≤ 1 for all t ∈ [0,T ]. denote by fi the face of polytope P contained in Hi. For every t ∈ [0,T ], let β0(t) be the direction vector Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php of β at t; let `(t) be the line incident to β(t) and Lemma 3.1. If δ > 0 is sufficiently small, as specified 0 orthogonal to both the direction vector β (t) and the in (3.12), then for all i = 1, . . . , n − 1, position vector β(t). We also define the plane H(t) 0 incident to β(t) and parallel to both β (t) and `(t); (i) face fi is unstable, and

γ(t) `(T ) β(t) ∂B H(T )

0 ∂B γ (t)

0 β (t) o γleft o C β(t) γ(t) γright β(T ) γ(T )

Figure 3: Left: The vectors γ(t), γ0(t), β(t), and β0(t) in the special case where γ is a circular arc along the equator. Right: The arcs γleft and γright traced out by the intersection points `(t) ∩ ∂B; and the circle C = H(T ) ∩ ∂B. The shaded surface has been scrapped oﬀ from the ball B.

(ii) if P stands on face fi, then it rolls to face fi+1. Recall that T ≥ 1, and consequently we have (a/T )|γ(t)| a/T a tan (β0(t), γ0(t)) = = ≥ We ﬁrst summarize our strategy for the proof of ∠ (1 − at/T )|γ0(t)| 1 − at/T T Lemma 3.1. Let hi ∈ Hi be the orthogonal projection of the origin o to the plane Hi (that is, ohi ⊥ Hi). for all t ∈ [0,T ]. 0 We show that hi ∈/ fi for i = 1, . . . , n − 1, which Recall that `(t) is orthogonal to both β(t) and β (t). establishes part (i). Speciﬁcally, we give a lower bound Since `(t) is parallel to H(t), the orthogonal projection 0 for the distance |β(ti)hi|, and then show that the line of o to H(t) lies in the plane hβ(t), β (t)i. In particular, Hi ∩Hi+1 separates points β(ti) and hi in the plane Hi. for i = 1, . . . , n − 1, point hi ∈ Hi lies in the plane 0 Since fi ⊂ Hi, this implies part (i). For the proof of hβ(ti), β (ti)i. Consider the right triangle ∆(β(ti)ohi). 0 part (ii), we further show that fi and fi+1 are adjacent Since β(ti) and γ (ti) are orthogonal, we have (their common edge is contained in H ∩ H ). We i i+1 β(t )oh = (β0(t ), γ0(t )) ≥ arctan(a/T ). also show—by eliminating all other options—that if P ∠ i i ∠ i i stands on face fi, it rolls around this common edge. Consequently, Importantly, we give a quantitative upper bound 0 0 for δ in terms of the second derivative of β; see (3.12). |hi − β(ti)| = |β(ti)| · sin ∠(β (ti), γ (ti))) We use this bound later for estimating the number of ≥ |β(ti)| · sin (arctan(a/T )) vertices of P in terms of L. We continue with the details. 9 a/T a Recall that γ is a simple smooth arc on the surface (3.8) > · > . 10 2 3T of B. Consequently, its direction vector γ0(t) at t ∈ [0,T ] is tangent to ∂B, hence orthogonal to the position For every s, t ∈ [0,T ], the position β(t) can be vector γ(t); see Fig. 3 for an illustration. The direction approximated using Taylor’s estimate with Lagrange vector of β(t) can be computed as follows: remainder 00 0 β (ξ) 2 0 (3.9) β(t) = β(s) + (t − s)β (s) + (t − s) , β0(t) = [(1 − at/T )γ(t)] 2 0 (3.6) = (1 − at/T )γ (t) − (a/T )γ(t), where ξ ∈ [min{s, t}, max{s, t}]. Let

00 0 M = max |β (ξ)|. where γ(t) and γ (t) are orthogonal unit vectors. In ξ∈[0,T ] Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php particular, Then (3.9) yields at 0 at a a M (3.7) 1−a ≤ 1− ≤ |β (t)| ≤ 1 − + ≤ 1+ . (3.10) |β(t) − (β(s) + (t − s)β0(s))| ≤ (t − s)2. T T T T 2

β(t ) Di+1 i+1 Di+1 Di+1 γright(ti+1)

γleft(ti+1) β(ti+1) `(ti+1) Hi ∩ Hi+1

0 β (ti) Di+1 qi+1 `(ti) `(ti) γleft(ti) β(ti) γright(ti) β(ti)

Figure 4: Points β(ti) and hi, and segment γleft(ti)γright(ti) in the plane Hi. The points marked with empty dots are orthogonal projections to Hi. An overview (left), and a detailed view around Di+1 (right).

We can use Taylor’s estimate for the direction vectors which is orthogonal to `(ti) (by deﬁnition). Therefore 0 β (t), as well. It yields the segment β(ti)hi is orthogonal to `(ti), and parallel 0 to the vector β (ti). 0 0 (3.11) |β (t) − β (s)| ≤ M |t − s|. Next we approximate the orthogonal projection of γ (t )γ (t ) and its midpoint β(t ) to H . We are now ready to specify δ > 0. Let left i+1 right i+1 i+1 i From (3.13), β(ti+1) lies in a ball of radius a 1 (3.12) δ = min 1, . a(ti+1 − ti) 10T M (3.15) ri+1 := . 20T In particular, note that centered at

1 a a 0 δ ≤ , δ ≤ , and δ ≤ . (3.16) ci+1 := β(ti) + (ti+1 − ti)β (ti). 10 10T 10TM Denote this ball by Di+1. Its center ci+1 is on the ray The Taylor estimate (3.10) with s = ti and t = ti+1 −−−−→ gives β(ti)hi, and by (3.13) and (3.7) we have 0 0 |β(ti)ci+1| = |(ti+1 − ti)β (ti)| |β(ti+1) − (β(ti) + (ti+1 − ti)β (ti))| ≤ ≥ (ti+1 − ti)(1 − a) M 2 M ≤ (ti+1 − ti) ≤ δ(ti+1 − ti) 2 2 (3.17) ≥ 0.9(ti+1 − ti). M a a |β(t )c | = |(t − t )β0(t )| (3.13) ≤ (t − t ) ≤ (t − t ), i i+1 i+1 i i MT i+1 i T i+1 i 2 10 20 ≤ (ti+1 − ti) · (1 + a/T ) and (3.11) similarly gives ≤ δ(1 + a/T ) a 0 0 ≤ (1 + a/T ) · |β (ti+1) − β (ti)| ≤ M(ti+1 − ti) ≤ Mδ 10T a a a a (3.14) ≤ M = . (3.18) ≤ (1 + 0.1) · < . 10TM 10T 10T 9T

Proof Lemma 3.1. (i) We show that β(ti) and hi By (3.8) and (3.18), we have H ∩ H lie on opposite sides of the line i i+1. Instead a a a of computing H ∩ H explicitly, we approximate |β(ti)hi| ≥ > 1 + ≥ |β(ti)ci+1|. i i+1 3T T 10T its location using γleft(ti)γright(ti) and the orthogonal projection of γleft(ti+1)γright(ti+1) to the plane Hi. It follows that ci+1 (the center of Di+1) is on the line Assume that P stands on facet fi for some i ∈ segment β(ti)hi. Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php {1, . . . , n − 1}. The plane Hi contains facet fi, line The orthogonal projection of β(ti+1) to Hi lies `(ti), segment γleft(ti)γright(ti), and its midpoint β(ti); in the disk Hi ∩ Di+1. The direction vector of `(ti) 0 refer to Figs. 4 and 5. The vertical projection of o to is given by the cross product β(ti) × β (ti); and the 0 0 Hi is hi. Recall that hi lies in the plane hβ(ti), β (ti)i, direction vector of `(ti+1) is given by β(ti+1) × β (ti+1).

Hi = π0 H ∩ H β(ti) i i+1 hi Di+1 Di+1

Figure 5: Points β(ti) and hi in a plane perpendicular to Hi ∩ Hi+1. The points marked with empty dots are orthogonal projections to this plane.

0 0 0 From (3.10) and (3.14), we have |β (ti) − β (ti+1)| ≤ Consequently, Di+1 contains neither β(ti) nor hi, that 2 2 0 a/10T and |β(ti) − β(ti+1)| ≤ (M/2)δ ≤ (a/T ) /200. is, Di+1 lies between β(ti) and hi. This implies that the line `(ti+1) is nearly parallel to Since line ì := Hi ∩ Hi+1 separates fi and the `(ti), and so its orthogonal projection to Hi is also (orthogonal) projection of fi+1 to Hi, ì crosses the nearly parallel to it. segment β(ti)hi between β(ti) and the projection of 0 We may assume that the angle between `(ti) and γleft(ti+1)γright(ti+1), which lies in Di+1. Thus ì crosses the projection of `(ti+1) is less than π/4. We would the segment β(ti)hi between β(ti) and hi. As β(ti) ∈ like to locate the intersection point of β(ti)hi and the γleft(ti)γright(ti) ⊂ fi, we conclude that fi and hi lie on orthogonal projection of `(ti+1) to Hi; denote this point opposite sides of ì. by q Since β(t ) ∈ D , the point q lies in the i+1 i+1 i+1 i+1 (ii) Assume that if P stands on face f for some projection of the the disk H ∩ D to line β(t )h , i i i+1 i+1 i i ∈ {1, . . . , n−1}, it rolls to face f . Initially, the center where the direction of the projection makes an angle j of mass, ξ(P ), is at distance |oh | < |β(t )| ≤ 1 from the at most π/4 with ` (i.e., an angle more than π/4 i i i+1 ground, and it can only move closer to the ground, hence with β(t )h ); see Fig 4(right). In particular, either i+1 i |h | < |h |. The line H ∩H must intersect B, otherwise q ∈ H ∩ D or the tangents from q to the disk j i i j i+1 i i+1 i+1 the center of mass would be at distance more than 1 H ∩ D subtend an angle of at least 2(π/4) = π/2. i i+1 from the ground during the motion. Since B0 ⊂ P , this In both cases, q lies in a circumscribed square of i+1 further implies that H ∩ H intersects B \ B0. H ∩ D . Let D0 be the ball concentric with D i j i i+1 i+1 i+1 Recall that ∂B∩∂B0 is a Jordan curve S composed with radius 2r . Then the projection of `(t ) crosses 0 i+1 i+1 of γ , γ , and a halfcircle centered at β(T ); and the β(t )h in H ∩ D0 . left right i i i i+1 boundary ∂(B \ B0) is composed of two surface patches: From (3.15) and (3.17) we obtain 0 a surface patch S1 of ∂B that consists of the surface a(ti+1 − ti) swept by the segment γleft(t)γright(t) for all t ∈ [0,T ] 2ri+1 = ≤ 0.01(ti+1 − ti) 10T and a halfdisk centered at β(T ); and a surface patch S2 of the sphere ∂B bounded by S . Refer to Fig. 3. (3.19) < 0.9(ti+1 − ti) ≤ |β(ti)ci+1|. 0 Consider now P ∩ (B \ B0). Since B0 ⊂ P , the 0 Further, from Equations (3.15), (3.12), and (3.8), boundary of P ∩ (B \ B ) contains S0 and S1, but the we obtain surface patch S2 is replaced by a surface patch S3 of P ∩ B S 3a(ti+1 − ti) 3aδ bounded by the Jordan curve 0. The line 6ri+1 = ≤ 10T 10T segments γleft(tk)γright(tk), for k = 2, . . . , n, each lie 3a a in both S1 and S3. Since these n − 1 line segments (3.20) ≤ < ≤ |β(ti)hi|. 100T 3T are pairwise disjoint chords of the Jordan curve S0, the deletion of these segments partitions P ∩ (B \ B0) into n Combined with (3.18) and (3.19) this yields components, each of which is incident to at most two segments, γleft(tk)γright(tk) and γleft(tk+1)γright(tk+1) |ci+1hi| ≥ |β(ti)hi| − |β(ti)ci+1| k , . . . , n f Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php for some = 1 . In particular, has only two a a 2a 2a i ≥ − = > edges that intersect B, namely fi ∩ fi−1 and fi ∩ fi+1. 3T 9T 9T 10T Since |h | > |h | > |h |, the polytope cannot roll 2a i−1 i i+1 (3.21) > (ti+1 − ti) = 4ri+1. from fi to fi−1, consequently it rolls from fi to fi+1. 2 10T

Copyright © 2020 2582 Copyright for this paper is retained by authors Distance between initial and ﬁnal footprints. path Γ = (p1, p2, . . . , pn) ⊂ π0. Alternatively, Γ is the We choose the arc γ to ensure that P covers a distance development of the geodesic path (β(t1), . . . , β(tn)) on at least L when it successively rolls from face f1 to fn ∂P . It remains to estimate the total length of Γ and the (i.e., the distance between the initial and ﬁnal footprints distance between its two endpoints. is at least L). We prove that |p1pn| ≥ 3.2L, by establishing a lower Let C0 be a great circle of B in the plane x = 0; bound len(Γ) > 3.8 L, and then bounding the deviation and let C1 and C2 be two circles in ∂B that lie in two of Γ from a straight-line path. −1 1 parallel planes x = 5L and x = 5L , respectively. The The portion of the geodesic path (β(t1), . . . , β(tn)) tangent lines to C0 form a cylinder of radius 1; and the on ∂P between β(ti) and β(ti+1), for i = 1, . . . , n − 1, tangent lines to C1 (resp., C2) form a cone of aperture is a polygonal path contained in fi ∪ fi+1, with a bend 1 2 arcsin( 5L ). If B rolls once along C0, its contact point point at fi ∩ fi+1. Recall that with π0 traces out a straight line segment of length

len(C0) = 2π; and similarly, if B rolls once along Ci, n−1 s it traces out a circular arc of length len(Ci) of the circle X 1 √ len(γ) = |γ(ti)γ(ti+1)| ≥ 4L 1 − > 3.9 L. 2 (5L)2 Cbi (i = 1, 2) of radius 25L − 1 centered at the apex i=1 of the cone of tangent lines. We choose γ as a dense and long spiral that lies It follows that strictly between C1 and C2. Let T = 4L, and γ : [0, 4L] → ∂B be deﬁned as follows: n−1 X (3.22) len(Γ) > len(β) > |β(ti)β(ti+1)| t − 2L i=1 x(t) = , n−1 10L2 X > (1 − a) |γ(ti)γ(ti+1)| p 2 p 2 γ(t) = x(t), cos(t) 1 − x (t), sin(t) 1 − x (t) . i=1 = (1 − a) len(γ) > (1 − a) 3.9 L > 3.8 L. A loop of γ is the image of an interval of length 2π in [0, 4L]. Note that (i) −1/(5L) ≤ x(t) ≤ 1/(5L); and (ii) the x-coordinate of γ(t), x(t), increases by Let γb : [0,T ] → π0 be the development of γ as B 2 2 2π/10L > 1/(2L ) after each loop. The arclength of continuously rolls on the plane π0 with a contact point p 2 every loop is at least len(C1) = 2π 1 − 1/(5L) , and so at γ(t); similarly, let βb : [0,T ] → π0 be the development p 0 the arclength of γ is at least 4L 1 − 1/(5L)2 > 3.9 L. of β as the convex body B rolls on the plane π0 with a We re-parameterize γ by arclength, as the machin- contact point at β(t) for t ∈ [0,T ]. 2 ery introduced above for the general construction re- In general, for a continuous curve α : [0,T ] → R , quires such a parameterization. Each loop of the spiral the turning angle ϕ(t) at t ∈ [0,T ] equals the angle α0 α0 t is close to the great circle C0 = {(0, cos t, sin t): t ∈ between the direction vectors (0) and ( ) modulo [0, 2π)}, which is parameterized by arclength and its 2π. If α is parameterized by arclength, and its signed ﬁrst and second derivatives are unit vectors. After re- curvature is κ : [0,T ] → R, then the turning angle at t parameterization, T equals the arclength of T , that is can be computed as follows; see [23, Sec. 2.2]. T > 3.9 L, and M is close to 1, consequently we may assume that M ≤ 2. Z t Set a = 1/(6400 L4); in particular a ≤ 1/10. (3.23) ϕ(t) = κ(s) ds. 1/2 2 s=0 We have |γleft(t)γright(t)| ≤ 4a = 1/(20L ), for all t ∈ [0,T ]. Since the x-coordinates of γ(t) increase by more than 1/(2L2) after each loop (regardless of the For example, the turning angle of a simple closed curve re-parameterization), γleft and γright are disjoint simple (parameterized counterclockwise) is 2π, and a turning arcs, and they both lie between C1 and C2. Recall angle of a circular arc of central angle θ is exactly θ. We a 4 use this machinery to approximate the turning angle of that δ = 10T min{1, 1/M}, thus δ = O(1/L · 1/L) = O(1/L5). Since T = Θ(L), a subdivision of [0,T ] into the polygonal path Γ. intervals of length at most δ requires Θ(L6) subdivision Since γ lies between the circles C1 and C2, the Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php points; this yields n = Θ(L6). curvature of γb is less than that of Cb1 and Cc2, i.e., p 2 As P rolls through the faces f1, f2, . . . , fn, the less than 1/ (5L) − 1 ≈ 1/(5L). The curvature of points β(ti)(i = 1, . . . , n) are successively in contact βb is close to that of γb at every t ∈ [0,T ]: Since 00 with the horizontal plane π0, and trace out a polygonal β(t) = (1 − at/T )γ(t) and |γ (t)| ≤ 2, Equation (3.6)

|β(t) − γ(t)| = (at/T )|γ(t)| = at/T ≤ a, |β0(t) − γ0(t)| ≤ (at/T )|γ0(t)| + (a/T )|γ(t)| ≤ a + a/T ≤ 2a, |β00(t) − γ00(t)| ≤ (at/T )|γ00(t)| + 2(a/T )|γ0(t)| Figure 6: A bird’s eye view of the polygonal path ≤ 2a + 2a/T ≤ 4a. Γ ∈ π0, the trail of a unistable polytope P whose flips cover a large distance. While it would be tedious to compute the curvatures of γ and βb exactly, the curvature of a curve can be b Based on the choice of γ from (3.22) and moreover, expressed as a rational function of the coordinates of the due to the two spirals attached to its endpoints, P is curve itself and its first and second derivatives (in terms bounded and its diameter is close to 1. Indeed, the of cross products and determinants); see [25, Sec. 1.4]. normal vectors of the planes H and H at the two It follows that the curvatures of γ and β differ by at 1 n b b endpoints β(t ) and β(t ) are nearly parallel to the x- O a O L−4 L > 1 n √ most ( ) = ( ). For sufficiently large 0, the P < −4 axis, consequently diam( ) / 2 2 3. A suitable curvature of βb is less than 1/(5L) + O(L ) < 1/(4L). scaling by a ratio Since len(βb) = len(β) < len(γ) < 4L, Eq. (3.23) implies 1 1 that the turning angle of βb does not exceed % = ∈ , 1 diam(P ) 3 Z 4L 1 π ds = 1 ≤ . produces a polytope of unit diameter. The scaling pre- s=0 4L 3 serves unistability and Hamiltonicity, while the distance Moreover, the turning angle of any subarc of βb is also between the initial and final footprints only decreases by 0 1 bounded by π/3, and so all direction vectors βb (t), a factor of %; this distance is at least % 3.2 L > 3 3.2 L > t ∈ [0,T ], are in a cone of aperture at most π/3. L, as required. In the polygonal path Γ = (p1, p2, . . . , pn), −−−−→ −−−−→ 4 Other Variants of Rolling Polytopes the turning angles ∠(pi−1pi, pipi+1) are close to 0 ∠(βb (ti), βb(ti+1)) for i = 2, . . . , n − 1. Therefore, we In this section, we prove that a Hamiltonian unistable −−−−→ may assume that the vectors pi−1pi, i = 1, . . . , n − 1, polytope can be an arbitrarily close approximation of a are in a cone of aperture at most π/3. If ` is the angle ball. For ε > 0, a polytope P is an ε-approximation of bisector of this cone, ` makes an angle of at most π/6 the unit ball B if (1 − ε)B ⊂ P ⊂ (1 + ε)B. with every edge of Γ. If xi > 0 is the length of the pro- Theorem 4.1. For every ε > 0, there is a Hamiltonian jection of pipi+1 onto `, we have |pipi+1| ≤ xi/ cos(π/6), Pn−1 unistable (nonuniform) polytope P that ε-approximates for i = 1, . . . , n − 1. Since xi = |p1pn|, this yields i=1 the unit ball B while retaining the arbitrarily large n−1 n−1 rolling distance property. (The polytope P need not be X X xi len(Γ) = |pipi+1| ≤ uniform.) cos(π/6) i=1 i=1 n−1 Proof. Let ε ∈ (0, π/2) be given. We use the construc- X 2 2 ≤ √ xi = √ |p1pn| < 1.16 |p1pn|. tion from the proof of Theorem 3.1 with a suitable arc i=1 3 3 γ : [0,T ] → ∂B, and parameter a > 0. First note that Q ⊆ ∂B if is a finite set such that the spherical caps Consequently, |p1pn| ≥ len(Γ) 1.16 > 3.2 L, as claimed. of radius ε centered at points in Q cover ∂B, then the Unit diameter. If γ is chosen as in (3.22), then planes tangent to B at the points in Q define a polytope the normal vectors of the planes Hi, i = 1, . . . , n, are P that satisfies the inclusions B ⊂ P ⊂ (1 + ε)B. nearly orthogonal to the x-axis, and the diameter of Lay down a raster of circles in ∂B that lie in planes P might be quite large. To ensure that diam(P ) = 1, orthogonal to the x-axis such that the spherical distance we modify γ, by attaching two additional x-monotone between two consecutive circles is less than ε/2. Since spirals to each of its endpoints that lead to the points the spherical distance between two antipodal points, Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php (−1, 0, 0) and (1, 0, 0) of ∂B, respectively. The length (−1, 0, 0) and (1, 0, 0), is π, we use 2dπ/εe − 1 circles. of these spirals is Θ(1), and the spirals could decrease Let γ : [0,T ] → ∂B be a spiral from (−1, 0, 0) to the distance between the initial and final footprints by (1, 0, 0) that makes one full rotation about the x-axis at most O(1). See Fig. 6. between consecutive raster circles. Then every point in

Copyright © 2020 2584 Copyright for this paper is retained by authors ∂B lies between two raster circles, and is within distance above the points q1, q2, q3, q4 by raising the four points ε/2 from some point on γ. We choose a subdivision and updating the convex hull. Specifically, we construct 0 0 0 0 0 0 0 = s1 < s2 < . . . < sm = T such that the point set P = conv(P, q1, q2, q3, q4), where qi is a mountain top, 0 Q = {γ(si): i = 1, . . . , n} defines a covering of ∂B by and oqi ⊆ oqi, for i = 1,..., 4. The heights of the 0 spherical caps of radius ε. mountains (i.e., hi = |oqi| for i = 1,..., 4) are variables Next, choose a sufficiently small a ∈ (0, ε/2) such that we can set so as to annihilate the shift of the center Tm + that the polytope P0 = i=1 H(si) lies in (1 + ε)B, of mass in the opposite direction. The volume of each where the halfspaces Hi, i = 1, . . . , n are defined for mountain monotonically increases with its height, and the curve β(t) = (1 − at/T )γ(t) as in the proof of is concentrated to a neighborhood of radius at most p 2 2 p Theorem 3.1. Finally, refine the subdivision 0 = s1 < hi − (1 − ε) . If hi ≤ 1.105 < 10/9, then this s2 < . . . < sm = T into a subdivision 0 = t1 < t2 < radius is less than 1/3, and the mountain at qi is disjoint . . . < tn = T , if necessary, to meet all constraints of that from all faces f1, . . . , fn. Each coordinate of the center construction. None of the additional tangent planes of mass ξ(P 0) is a piecewise rational function of the H(ti), i = 1, . . . , n, intersects (1 − ε)B. Consequently, heights hi (i = 1,..., 4). If we set h1 = 1.05 and Tn + 0 0 the polytope P1 = i=1 H(ti) contains (1 − ε)B, and qi = qi for i = 2, 3, 4, then ξ(P ) shifts roughly towards is contained in (1 + ε)B, as required. qi by a small constant (independent of ε). Provided that ε > 0 is sufficiently small, by the (multidimensional) Uniform density. A variant of the polytope in intermediate value theorem, there exit suitable heights 0 Theorem 3.1 can be constructed from material of uni- hi ∈ [1 − ε, 1.1], i = 1,..., 4, for which ξ(P ) is the form density. origin.

Theorem 4.2. For every L ≥ 1, there exits a uniform 5 Concluding Remarks polytope P of unit diameter and a facet f1 such that if We suspect that the properties in Theorems 4.1 and 4.2 P stands on face f1 it rolls to distance at least L. (The polytope P need not be unistable or Hamiltonian.) can be combined; we leave this as an open problem: Is there a Hamiltonian unistable uniform polytope of unit Proof. We start with a uniform unit ball B centered diameter that can roll to an arbitrary given distance at the origin 0 and let ε ∈ (0, 0.01) be sufficiently away? small as specified below. By Theorem 4.1, there is a nonuniform unistable Hamiltonian polytope P = The perpetuum mobile desideratum. We P (L, ε) that ε-approximates B, and P has a sequence know (by Lemma 1.2) that no polytope exists that can of facets f1, . . . , fn that all lie in the slab between the roll on forever; indeed, this is so because every poly- parallel planes x = −1/(5L) and x = 1/(5L) such that tope has a finite number of faces. This leaves open the if P stands on facet fi (i = 1, . . . , n − 1), it rolls to face existence of a (smooth) convex body that can roll on fi+1, and the distance between the footprints at f1 and forever. While we cannot rule out this possibility, we fn is at least L. can immediately observe that our construction method However, unlike in construction in the proofs of (i.e., scraping off a neighborhood of a spherical arc from Theorem 3.1 and Theorem 4.1, the faces of P outside of a ball) is due to fail: Indeed, the width of the neighbor- the parallel slab between x = −1/(5L) and x = 1/(5L) hood of the arc γ is positive and strictly increasing, and lie in the exterior of B; i.e., we do not use the any spirals by construction, this thickened arc cannot cross itself. from (−1, 0, 0) to (1, 0, 0). In particular, P never rolls As such, if γ were an infinite arc, then this neighborhood to any of these faces from the faces f1, . . . , fn, and so P would also have infinite area, contradicting the fact that is neither unistable nor Hamiltonian. the surface area of the ball is bounded. Note that neither B nor P contains the other body. The volume of the symmetric difference of B and P Diameter versus girth. Conway [16] (see also [7, 4π 3 3 is bounded above by 3 [(1 + ε) − (1 − ε) ] = O(ε). Sec. B12]) constructed a convex unistable polytope Consequently, the center of mass ξ(P ) shifts by O(ε) as whose diameter to girth ratio is at most 3/π + ε, for well, i.e., ξ(P ) lies in a ball of radius O(ε) centered at 0. any ε > 0. He also asked: What is the smallest To move the center of mass of P back to the origin possible diameter to girth ratio for a convex unistable we first determine four points q1, q2, q3, q4 ∈ ∂P so polyhedron? (The girth is the minimum length of Downloaded 07/10/20 to 209.6.90.196. Redistribution subject SIAM license or copyright; see http://www.siam.org/journals/ojsa.php that the tetrahedron ∆ = conv(q1, q2, q3, q4) closely the perimeter of a projection onto a plane.) Our approximates a regular tetrahedron inscribed in B, and construction (Theorem 4.1) shows that a ratio of 1/π+ε, they are each at distance at least 1/3 from the plank for any ε > 0, can be attained with nonuniform between x = ±1/(5L). We now stack four “mountains” polytopes.

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