Partial Results on Convex Polyhedron Unfoldings

Transcription

1 Partial Results on Convex Polyhedron Unfoldings Brendan Lucier 004/1/11 Abstract This paper is submitted for credit in the Algorithms for Polyhedra course at the University of Waterloo. We discuss a long-standing open problem in computational geometry and prove some partial results. 1 Introduction In this paper we consider the following open problem, first posed formally by Shephard in [3]: Problem 1.1. Can every convex polyhedron be cut along its edges and unfolded into the plane such that the resulting surface does not self-intersect? It has long been believed that the answer is yes, but there have been distressingly few proven results on this topic. In this paper we describe some methods of approach towards a solution and prove some partial results. The arguments, proofs, and conjectures in this paper are thought to be novel unless otherwise stated. Definitions and Known Results This section is a quick overview of assumed background in the field of convex polyhedron unfoldings. I do not claim that any of the results or arguments presented in this section are new. Given a convex polyhedron P, we wish to define an edge unfolding of P. Let F(P), E(P), and V (P) be the sets of faces, edges, and vertices of P, respectively. A cut tree (denoted C) is a connected acyclic subset of E(P) that spans V (P). The cut tree has a natural dual: the Adjacency Tree denoted A. The adjacency tree contains precisely the edges in the dual of P whose adjacent faces do not share a common edge in C. Note that this is the same structure used in von Staudt s 1847 proof of Euler s Formula [4]. Then A will be a spanning tree of the set of faces of P, as demonstrated in that proof. Consider now the set F(P) as polygons. Given any two faces that share a common edge e in P, we attach those faces at edge e iff e C. That is, we topologically identify pairs of corresponding edges that are not in the cut tree. Call the resulting complex of polygons P. But now A defines the structure of P : two faces are connected in P iff their duals are connected by an edge in A. Since A is a spanning tree, we conclude that P is connected and has no cycles of connected polygons. P is therefore homeomorphic to a disc and hence sits in the plane. We call P an edge unfolding of P. The polygonal complex P can also be viewed as a net. A net is a set of convex polygons with edges identified topologically, such that 1. identified edges have the same length,. the set of polygons is topologically connected, and 3. every edge of a polygon is identified with at most one edge of another polygon. 1

2 (a) (b) (c) Figure 1: [a] A net. [b] A closed net. [c] Representation of the net in (b). A closed net is a net in which every edge is identified with exactly one other edge. See Figure 1. P is not a closed net, since images of cut edges are not identified. A representation of net N is an arrangement of the polygons of N in the plane such that the following properties hold: 1. The polygons have consistent orientation; e.g. the side of a face that points towards the exterior of P will point up in the plane.. For each polygon there is at least one other polygon such that they intersect only at a common identified edge. 3. The union of all polygons is a connected region. Every net has a representation, since the set of polygons in a net is connected. P has exactly one representation, since the identified edges form a tree. This representation is what we visualize as an unfolding into the plane. A celebrated theorem of Aleksandrov states that every net that is homeomorphic to a sphere and whose total angle sum at every vertex is π corresponds to a closed convex polyhedron [1]. In a recent CS860 lecture, Craig Kaplan noted that a polyhedron whose curvature at every vertex is 0 must have genus zero. But recall that curvature of a vertex is precisely π minus the total angle sum at that vertex. So any closed polyhedron whose angle sum at every vertex is π must have genus zero, and hence be homeomorphic to a sphere. Kaplan s observation allows us to restate Alexsandrov s Theorem as Every closed net whose angle sum at every vertex is π corresponds to a closed convex polyhedron. This in turn allows us to restate the unfolding problem as follows: Problem.1. Does every closed net whose angle sum at every vertex is π have a representation that does not self-overlap? 3 Local Overlaps In polyhedron unfoldings there is a common type of overlap, called a local overlap []. I make some use of local overlaps in my proofs, but I have been unable to find any formal discussion of them in the literature. I shall therefore formalize the idea here. Consider an edge e = (v,w) in the cut tree C of convex polyhedron P. Suppose that w has degree 1 in C. Then e is the only cut adjacent to w, so we do not cut through w. Therefore w will have only one image, say w, in the unfolding P. Then w is a vertex on the boundary of P, so must be adjacent to two edges on the boundary of P. Say these edges are e 1 = (w,v 1 ) and e = (w,v ). Let e be the other edge adjacent to v 1 on the exterior of P. See Figure for an illustration. Claim 3.1. Let φ be the exterior angle at v 1, and let θ be the curvature at w. Then P will overlap if 1. θ + φ π. e e sin θ sin(π θ φ).

3 Figure : A section of a polyhedron unfolding for Claim 3.1. Shaded area represents interiors of faces. Proof. The proof of this claim has been moved to Appendix A, due to space constraints. Note that, by symmetry, Claim 3.1 holds if we apply the conditions to v instead of v 1. An overlap of the form described in Claim 3.1 is a local overlap. Informally, the features of the construction that cause the overlap to occur are the large ( 3π ) interior angle at v 1, the sufficiently small curvature at w, and the sufficiently long edge e. 4 Directional Algorithms Given a convex polyhedron P, an unfolding algorithm generates an unfolding P, by creating either a cut tree or an adjacency tree. Our goal is to find an unfolding algorithm that produces non-overlapping unfoldings. A directional algorithm is an unfolding algorithm whose first step is to choose a direction vector. Heuristically, a directional algorithm fixes an orientation of the given polyhedron and generates an unfolding based upon that orientation. 4.1 Steepest Edge In [], Schlickenreider performed empirical tests to compare different types of unfolding algorithms. He put forth a conjecture regarding a particular directional algorithm: the Steepest Edge algorithm. In this section we shall describe this algorithm, present Schlickenreider s conjecture, and prove that it is false Algorithm Description The Steepest Edge algorithm proceeds as follows: Pick a unit direction vector c. Without loss of generality we assume that c = (0,0,1) (by reorienting space). We shall informally refer to vector (0,0,1) as up. Let v + be the vertex of the polyhedron with maximal z-coordinate (we choose our orientation so that this v + is unique). Now for each vertex v in V (P) {v + }, and for each edge (v,w), consider the unit vector d(v,w) := w v v,w. We shall say that the steepest edge at v is the edge (v,w ) for which the z-coordinate of d(v,w ) is maximal. The Steepest Cut algorithm chooses C to be the set of steepest edges for each vertex in V (P) {v + }. Heuristically, we are cutting the most upward that we can at each vertex. Schlickenreider shows that C is indeed a cut tree []. Conjecture 4.1. (Schlickenreider) For every convex polyhedron P there exists some unit vector c such that the Steepest Edge algorithm with direction c generates a non-overlapping unfolding Counterexample We shall disprove Conjecture 4.1 by presenting a counterexample. 3

4 (a) (b) Figure 3: [a] Map M 1 used in the counterexample to the Steepest Edge Algorithm. [b] Overlapping unfolding of M 1 when curvature is small. [c] Embedding of M 1 into a triangle. Consider the planar graph illustrated in Figure 3(a). We can convert this graph into a convex terrain M 1 by raising the interior vertices (marked as bold circles). Note that we can raise the interior vertices as little as we desire, so indeed the curvature at these two vertices (and differences in orientation of the faces) can be made arbitrarily small. Note also that once we choose the amount to raise one vertex, the height for the other vertex is determined, since the quadrilateral in M 1 must remain planar. Suppose that M 1 appears as part of a convex polyhedron P, and we wish to use the Steepest Edge algorithm on P. Suppose further that the directional vector c points towards the top of the page, as illustrated in Figure 3(a). Then the bold edges in the diagram must be cut (as they are the steepest for the interior vertices), and the hashed edges cannot be cut (since there are steeper edges for both adjacent vertices). If we choose the curvatures of the interior vertices small enough, a local overlap occurs when we unfold. See Figure 3(b) and Section 3. Note that this intersection was caused only by the curvature of the points and the choice of which edges to cut. Any c that causes the pattern of cuts and non-cuts in Figure 3(a) will generate an intersection. There are no parts of M 1 where a small enough perturbation of c will change the edge choices, so there is some open range D(c) of choices for direction vectors on the unit sphere that will cause an intersection. Choose some set of direction vectors c i such that the open sets D(c i ) cover the unit sphere. We now wish to construct a polyhedron P that contains a copy of M 1 for each range of direction vectors D(c i ). These copies of M 1 will be positioned and oriented in such a way that if the Steepest Edge algorithm is applied with c D(c i ), then the corresponding copy of M 1 will generate an overlap as discussed above. Since the D(c i ) cover the unit sphere, we can then conclude that the Steepest Edge algorithm will always cause an overlap when applied to P. P is therefore the counterexample that we require. The actual construction of the polyhedron P is tedious and unenlightening, so we have moved that portion of this proof to Appendix A. 4. Normal Order Unfoldings Though Schlickenreider s conjecture is false, he has empirical evidence that suggests that the Steepest Edge algorithm works for almost all convex polyhedra. I therefore believed that Schlickenreider s conjecture could be made true if it were weakened slightly. However, it turns out that even a greatly weakened version of Conjecture 4.1 is still false, as we shall show. We define a class of unfoldings: the Normal Order unfoldings. Take a convex polyhedron P and choose a direction vector c. Reorient space so that c = (0,0,1). Given two faces F 1 and F in P, we shall say that F 1 is higher (resp. lower) than F if the z-coordinate of the exterior-facing unit normal of F 1 is larger (smaller) than that of F. That is, higher faces are those whose normals point more towards c. Now consider some unfolding of P given by some adjacency tree A. We shall say that this is a normal order unfolding if there is some choice of root for A and some direction c so that every face in A is higher than its parent. That is, a normal order unfolding is built by choosing the lowest face, then adding faces to the unfolding in order from lowest to highest, always attaching a face to a lower face. Conjecture 4.. Every convex polyhedron admits a normal order unfolding that does not overlap. 4

5 (a) (b) (c) Figure 4: [a] A widget used in the counterexample for Normal Order Unfoldings. [b] A possible unfolding when curvatures are small. [c] Another possible unfolding Relationship to Steepest Edge Conjecture 4. is weaker than Conjecture 4.1, as demonstrated by the following proposition. Proposition 4.3. Every steepest edge unfolding is a normal order unfolding. Proof. We have moved the proof of this proposition to Appendix A due to space constraints. 4.. Counterexample We shall disprove conjecture 4. by constructing a counterexample. This example is very similar to the one constructed in section Consider the planar graph illustrated in Figure 4(a). The labels a f refer to the faces in which they appear. We can again convert this graph into a convex terrain M by raising the interior vertices. As before, the curvature at these vertices can be made arbitrarily small. If we take the direction vector to be pointing to the top of the page, the hashed edges cannot be cut. This is because face b will be the only lower face adjacent to faces a, c, and f (Justification: the hashed edges are the only edges marked down for faces a, c, and f, as defined in the proof of Proposition 4.3). The trick here is that face d can be placed in two different locations; it can be made adjacent to b or to c. If it is made adjacent to c (i.e. the edge between d and b is cut) then face d will intersect face f. See Figure 4(b). The overlap occurs because θ < π, so a local overlap occurs if curvatures are sufficiently small. If, on the other hand, face d is made adjacent to b, then the edge between d and c is cut. Now if the edge between d and e is also cut then we are in the same situation discussed in Section 4.1., so we will have an overlap. See Figure 3(b). Therefore face e must be made adjacent to d. But then, taking curvatures sufficiently small, there will be an overlap between face e and face a. See Figure 4(c). Again, this intersection occurs because φ < π. So there is no way to unfold terrain M while respecting the normal order induced by c. We do not have any boundary conditions for the face ordering in this example, so there is an (admittedly quite small) open range of direction vectors c for which M cannot be unfolded. Using the same trick as in the previous section, we embed copies of this widget onto a spherical polyhedron until no possible direction vectors can be used. Such a polyhedron will not have a normal order unfolding. 5 Partial Nets Another avenue of research is the analysis of subparts of unfoldings. A partial unfolding is a complex N of convex polygons that is contained in a representation of a net N, where N corresponds to a closed convex polyhedron. A natural question that arises is how to categorize partial nets. The reason we wish to study partial nets is to make progress in developing a repair algorithm. In a repair algorithm, we take a (possibly overlapping) unfolding of a convex polyhedron. We then modify this unfolding by altering the adjacency tree. In this way, we come up with new representations for the net defined by the unfolding. The goal of the repair algorithm is to modify the representation until it no longer overlaps. 5

6 (a) (b) Figure 5: [a] A vertex with total face angle > π. [b] A complex that is not a partial net. A, B, and C mark the polygons, integer i marks point p i. In order to design such an algorithm, we must have some notion of what kinds of overlaps can appear in a general unfolding. This study of partial nets aims to provide some intuition into this problem. Let N be a complex of convex polygons. We shall prove some propositions. Proposition 5.1. If N does not overlap then N is a partial unfolding. Proof. Embed N in a sufficiently large square S. Triangulate int(s) ext(n) so that S is decomposed into convex polygons (including those of N). We can now create an unfolding of a cube, using this decomposed square as one of the pieces. The result is an unfolding that contains N. Proposition 5.. If N contains a vertex with total face angle > π then N is not a partial unfolding. Proof. Such a vertex would violate a condition of Alexsandrov s Theorem. See Figure 5(a) for an example of such an N. Proposition 5.3. There exists an N in which every vertex has total face angle π but N is not a partial unfolding. Proof. We construct a particular N. See Figure 5(b). The vertices are located at points (p i ) 6 i=1 = ((, ),(4, ),(0,0),(4,0),(6,8),(,8)). We wish to prove that N is not a partial unfolding. This proof has been moved to Appendix A, due to space constraints. The proof of the above proposition leads us to believe that it is the structure of overlapping polygons that causes a complex N to not be a partial net. This informal conjecture leads us to the following definition. Recall that a polygonal complex N is formed by an adjacency tree A acting upon a set of polygons. Thus, given any two polygons F 1 and F in N, there is a unique shortest path in A that connects F 1 and F. Let C(F 1,F ) := (C i ) k i=1 be the sequence of all polygons on that path; with C 1 = F 1 and C k = F k. We call C(F 1,F ) the connecting chain for F 1 and F. I made the following conjecture: Conjecture 5.4. A polygonal complex N is a partial net iff C(F 1,F ) is a partial net for each pair of overlapping faces F 1,F. But this conjecture is false, even in the case that N has only a single overlap. Proposition 5.5. There exists a polygonal complex N in which only two faces F 1 and F overlap and C(F 1,F ) is a partial net, but N is not a partial net. 6

7 Figure 6: A connecting chain is drawn in black, and an additional face is drawn in grey. The chain is a partial net, but the entire complex is not. Proof. (Sketch) Let N be the complex depicted in Figure 6. It has only one pair of overlapping polygons, and their connecting chain C is drawn in black. The one remaining face is drawn in grey. Note that C is a local overlap (the labels from our local overlap discussion in Section 3 have been marked in Figure 6) and therefore folds into a convex terrain. Thus C is a partial net, since it can fold onto the surface of a convex polyhedron. N, however, is not a partial net. Informally, the addition of the extra triangle makes it impossible for v 1 and v to fold together, since this would result in a total face angle > π. We claim without proof that there is no other way to fold N onto the surface of a polyhedron, and therefore N is not a partial net. Despite the negative resolution of Conjecture 5.4, I still feel that the notion of a connecting chain is useful. My intuition is that in a folding of a convex polyhedron, every connecting chain of overlapping polygons must curl into itself. A polygonal chain is then a partial net only if these curlings can occur. I have not yet had any luck in formalizing this intuition, but I feel that it is the next step in the line of reasoning that I am pursuing. 6 Future Work We shall continue to work on characterizing partial nets of convex polyhedra, as defined in Section 5. We feel that this will be a fundamental step towards designing a successful overlap repair algorithm, and thereby solving the convex polyhedron unfolding problem. We would also like to see a non-analytical proof of Proposition 5.3. Preferably, one would prove that any extension of N in which every edge is associated with some other edge must contain a vertex whose total face angle is > π. My intuition is that a proof of this nature could be very clean, and could provide some insight into proving other properties of unfoldings and partial nets. References [1] A. D. Aleksandrov. Konvexe Polyeder. Akademie-Verlag, Berlin, Math. Lehrbucher und Monographien. [] W. Schlickenreider. Nets of Polyhedra. Unpublished. Technische Universitat Berlin. June [3] C. Shephard. Convex polytopes with convex nets. Math. Proc. Camb. Phil. Soc., 78: , [4] G. K. C. Von Staudt, Geometrie der Lage, Nurnberg,

8 (a) (b) (c) Figure 7: [a] An unfolding around w. [b] Angles at w and v 1. [c] A local overlap, illustrating the edge length condition. Appendix A: Omitted Proofs Proof of Claim 3.1: Proof. Edges e 1 and e are images of cuts adjacent to w, so e 1 and e are both images of e. Therefore e 1 = e = e. Suppose that the conditions of the claim are satisfied. Note that since w is the only image of w, the interior angle at w is precisely the sum of face angles at w. Therefore the exterior angle at w is the curvature at w, which is θ. See Figure 7(b). Note that, by the first condition in the claim, θ π and φ π. Now consider the isoceles triangle formed by v 1, v, and w. It will have angle θ at w, and angle ψ := π θ at v 1 and v. But we know that θ + φ π, so φ π θ = ψ. This means that v will be on the interior side of the line containing edge e. Thus, if edge e is sufficiently long, it will intersect e. We now determine what sufficiently long means. Extend edge e from v 1 until it intersects e. Call that point of intersection q. The exterior angle at q will be π θ φ. See Figure 7(c). Then, by the sine rule, we have that q v 1 sin θ = e sin(π θ φ). We conclude that e will contain point q, and hence intersect e, if as required. e q v 1 = e Construction of Polyhedron P for Section 4.1.: sin θ sin(π θ φ) Embed M 1 into an equilateral triangle, as shown in Figure 8(a). We can lower the vertices of the triangle so that we still have a convex terrain, say M. Since we raised the vertices of M 1 by an arbitrarily small amount, the amount we lower the vertices of the triangle can be small as well, so the differences in face orientations for M will still be arbitrarily small. We now construct a spherical polyhedron P 1. Let S be the surface of the unit sphere. Choose some set of direction vectors c i such that the open sets D(c i ) cover S. For each vector c i, we shall consider three points {p 1 i,p i,p3 i } on S that form the corners of an equilateral triangle. We wish to place these points so that c i is precisely the direction vector illustrated in Figure 8(a) with respect to p 1 i, p i, and p3 i. To achieve this, we must consider the equator E i of S lying on a plane perpendicular to c i. Choose any point q i on that equator and place the {p j i } around q i so that the resulting triangle is perpendicular to E i. See Figure 8(b). For each i, let B i S be the smallest spherical disc centred at q i that contains p 1 i, p i, and p3 i. See Figure 8(b). We want to achieve the property that all B i are disjoint. To guarantee this, we place restrictions on the size of each B i. For each i, place the points p j i close enough to q i so that the diameter of B i is no more 8

9 (a) (b) Figure 8: Illustrations for the counterexample of Section [a] Embedding of M 1 into a triangle. [b] Placing a triangle on the unit sphere. than 1. We can do this since our triangles can be arbitrarily small. Now suppose for contradiction that i the kth set of points cannot be placed so that B k is disjoint from k 1 i=1 B i. Then it must be the case that k 1 i=1 B i covers all possible choice of q k on E k. But E k is an equator of S, and therefore has length π. The sum of the diameters of the B i is k 1 i=1 1 i < 1 and therefore the B i regions cannot cover all of E k, a contradiction. We conclude that a choice of {p j i } exists that satisfies our required properties. Now take P 1 to be the convex hull of all the p j i. This convex polyhedron will contain all the triangles associated with triples of points (since the triples occupy distinct regions of S). We remove each triangle from P 1 and replace with a copy of M, oriented to match the corresponding direction vector c i. We shall choose the curvatures of M 1 (and hence M ) to be small enough that the dihedral angles within M are much closer to π than the largest dihedral angle in P 1. Call the resulting convex polyhedron P. But now, for any choice of direction vector c, there is some c i for which c D(c i ), so there is some copy of M in P that will generate an overlap when Steepest Edge algorithm is applied to P with direction c. The Steepest Edge algorithm will therefore always cause an overlap when applied to P. Proof of Proposition 4.3: Proof. Suppose for contradiction that there exists a polyhedron P under a certain orientation (i.e. c = (0,0,1)) such that its steepest-edge unfolding is not a normal order unfolding. That is, there is a face F of P that is not the lowest face, but all edges between F and adjacent lower faces are cut. Now orient P so that the normal of F has x-coordinate 0 (we can do this while maintaining c = (0,0,1)). Under this orientation, project F onto the xz-plane. If the result is a line segment, then F must have normal (0,0,1) or (0,0, 1). If the latter, F must be the lowest face, a contradiction. If the former, then every face adjacent to F must be lower. But not all edges adjacent to F can be cut, since the cuts form a tree. We therefore assume that our projection is a polygon, say F. See Figures 9(a) and 9(b). Mark each edge of F (and the corresponding edges in P) as up if it above the interior of the polygon, or down if it is below the interior of the polygon. See figure 9(b). Any edges parallel to the z-axis are marked up. We make the following informal observations: 1. The set of up edges is connected, as is the set of down edges; 9

10 (a) (b) (c) Figure 9: [a] A convex polyhedron with a face aligned to the xz-plane. [b] Projection of face F onto the xz-plane, with up and down edges labelled. [c] An up edge of F connecting to lower face H, and a down edge connecting to higher face G. Either case violates the convexity of P.. For every up edge, the corresponding edge in P cannot connect F to a lower face. 3. For every down edge, the corresponding edge in P cannot connect F to a higher face. See figure 9(c) for an informal justification of points and 3. This figure is a projection of F onto the yz-plane; it demonstrates that an up (down) edge connecting to a lower (higher) face will violate convexity. We now conclude that all down edges of F are cut by the steepest-edge algorithm. Note that F must have at least one down edge. If there are k down edges, they are bounded by k + 1 vertices adjacent to F. But two of these vertices, say v 1 and v, are also adjacent to up edges. See Figure 9(b). Since, in the projection of F, every down edge is below the interior of F and every up edge is above the interior of F, it must be the case that the up edges point more upwards than the down edges, with respect to v 1 and v. Therefore neither v 1 nor v will choose down edges by the steepest-cut algorithm. But the k down edges cannot all be chosen by the k 1 remaining vertices, so some down edge will not be cut. This is a contradiction. Proof of Proposition 5.3: Proof. We wish to show that the polygonal complex N shown in Figure 5(b) is not a partial net. The vertices are located at points (p i ) 6 i=1 = ((, ),(4, ),(0,0),(4,0),(6,8),(,8)). Note that any polygonal complex is a partial unfolding iff it can be folded along its interior edges such that the resulting three-dimensional complex can be part of the surface of a convex polyhedron (proof: if it s contained in a net corresponding to a convex polyhedron, then a folded version will be contained in the folded net that forms the surface of a polyhedron). We shall use some messy analytical geometry to prove that N cannot satisfy this condition. Suppose side A is folded up by an angle of θ, and B is folded up by an angle of φ. We keep face C in the plane. We shall call the three-dimensional images of A and B A and B, respectively. Let p a denote the image of p 5 by this folding, and let p b denote the image of p 6. We now wish to describe A and p a. Note that 0 θ < π. If θ = 0 then p a = (8,6,0) and if θ = π then p a = (8,6,0). The possible images of p a will form a half-circle rising above the plane, with (6,8,0) and (8,6,0) being endpoints of a diameter. We conclude that p a will be some point on the circle with centre (7,7,0), containing points (6,8,0) and (8,6,0). Specifically, p a will map to (7 cos θ,7+cos θ, sinθ). By a similar sort of reasoning, the normal to plane A will be ( sin θ,sin θ, cos θ). Since A contains point (0,0,0), the equation of that plane will be (sin θ)x + (sin θ)y (cos θ)z = 0. I am omitting some details of these arguments for brevity. Similarly, we have that p b is mapped to ( 3 + cos φ,7 cos φ, sin φ). Also, the normal of B will be (sin φ,sin φ, cos φ), so equation of the resulting plane will be (sinφ)x + (sin φ)y (cos φ)z 4sin φ = 0 (since point (4,0,0) lies on B ). Now if A and B are faces of a convex polyhedron, it must be the case that no vertex of A is exterior to the halfspace of the plane for B, and vice-versa. In particular, we require that p a not be on the exterior 10

11 of the halfspace for B, and p b not be on the exterior of the halfspace for A. That is, we require that and But () implies that and hence We therefore conclude that sin θ( 3 + cos φ) + sinθ(7 cos φ) cos θ( sin φ) 0 (1) sin φ(7 cos θ) + sinφ(7 + cos θ) cos φ( sin θ) 4sin φ 0. () 14sin φ 4sin φ cos φsinθ 0 sin φ 10 cos φsin θ. (3) sin θ( 3 + cos φ) + sinθ(7 cos φ) cos θ( sin φ) =10sin θ sin θ cos φ cos θ sin φ 10sin θ sin θ cos φ cos θ( cos φsinθ) 10 = sinθ[10 cos φ 1 cos θ cos φ] 5 7sin θ (4) since sin θ 0. Then (1) implies that 7sin θ 0, so in fact sinθ = 0, hence θ = 0 or θ = π. We certainly can t have θ = π, since then face A would intersect face C. We conclude θ = 0. But now by (3), sinφ cos φsin θ = cos φ(sin π) = and we know that sinφ 0, so we must have sin φ = 0, and hence φ = 0 as above. But θ = φ = 0 gives us precisely the net drawn in Figure 5(b), which self-intersects! This cannot be part of the surface of a polyhedron. We conclude that no folding of N can be part of the surface of a convex polyhedron, so N is not a partial net. 11