A Complete List of All Convex Polyhedra Made by Gluing Regular Pentagons
AA complete list of all convex polyhedra made by gluingregular pentagons ∗ Elena Arseneva † Stefan Langerman ‡ Boris Zolotov † July 6, 2020
Abstract
We give a complete description of all convex polyhedra whose surface can be con-structed from several congruent regular pentagons by folding and gluing them edge toedge. Our method of determining the graph structure of the polyhedra from a gluingis of independent interest and can be used in other similar settings.
Given a collection of 2D polygons, a gluing describes a closed surface by specifying howto glue (a part of) each edge of these polygons onto (a part of) another edge. Alexandrov’suniqueness theorem [1] states that any valid gluing that is homeomorphic to a sphereand that does not yield a total facial angle greater than 2 π at any point, corresponds tothe surface of a unique convex 3D polyhedron (doubly covered convex polygons are alsoregarded as polyhedra). Note that the original polygonal pieces might need to be folded toobtain this 3D surface.Unfortunately, the proof of Alexandrov’s theorem is highly non-constructive. Theonly known approximation algorithm to find the vertices of this polyhedron [8] has (pseu-dopolynomial) running time really large in n , where n is the total complexity of the gluing.In particular, its running time depends on n as ˜ O ( n . ), and it also depends on the as-pect ratio of the polyhedral metric, the Gaussian curvature at its vertices, and the desiredprecision of the solution. There is no known exact algorithm for reconstructing the 3Dpolyhedron, and in fact the coordinates of the vertices of the polyhedron might not evenbe expressible as a closed formula [7].Enumerating all possible valid gluings is also not an easy task, as the number of gluingscan be exponential even for a single polygon [4]. However one valid gluing can be found inpolynomial time using dynamic programming [6, 9]. Complete enumerations of gluings andthe resulting polyhedra are only known for very specific cases such as the Latin cross [5]and a single regular convex polygon [6]. ∗ E. A. was supported in part by F.R.S.-FNRS, and by the SNF grant P2TIP2-168563 of the EarlyPostDoc Mobility program. E. A. and B. Z. are partially supported by the Foundation for the Advancementof Theoretical Physics and Mathematics “BASIS” and by “Native towns”, a social investment program ofPJSC “Gazprom Neft”. S. L. is directeur de recherches du F.R.S.-FNRS. † St. Petersburg State University (SPbU). Emails: [email protected], [email protected]. ‡ Universit´e libre de Bruxelles (ULB). Email: [email protected]. a r X i v : . [ c s . C G ] J u l he special case when the polygons to be glued together are all identical regular k -gons, and the gluing is edge-to-edge was recently studied by the first two authors of thispaper [2]. For k >
6, the only two possibilities are two k -gons glued into a doubly-covered k -gon, or one k -gon folded in half (if k is even). When k = 6, the number of hexagons thatcan be glued into a convex polyhedron is unbounded. However, for non-flat polyhedra ofthis type there are at most ten possible graph structures. For six structures out of theseten, the gluings realizing them have been found. For doubly-covered 2D polygons, all thepossible polygons and the gluings forming them have been characterized.In this paper we continue this study by thoroughly considering the case of k = 5,i.e., gluing regular pentagons edge to edge. This setting differs substantially from the caseof hexagons, since it is not possible to produce a flat vertex by gluing regular pentagons.Therefore both the number of possible graph structures and the number of possible gluingsis finite and little enough to study each one of them individually.We start by enumerating all edge-to-edge gluings of regular pentagons satisfying theconditions of the Alexandrov’s Theorem (Section 3). After that we solve the problem of es-tablishing the graph structure of the convex polyhedra corresponding to each such gluing G .Using the existing methods (implementation [10] of the Bobenko-Izmestiev algorithm [3]),we obtain an approximate polyhedron P for gluing G . With the help of a computer pro-gram, we generate a certificate that the edges of these approximate polyhedra are presentin the sought polyhedra. In particular, we upper bound the discrepancy in vertex coor-dinates between the unique convex polyhedron corresponding to G a given approximatepolyhedron (Theorem 4), which implies a sufficient condition for the polyhedron to havea certain edge (Theorem 5). Our computer program checks this condition automatically.For non-simplicial approximate polyhedra P , to prove that there are no additional edgespresent in the sought polyhedra, we resort to ad-hoc geometric methods, using symmetryarguments and reconstructing the process of gluing the polyhedron (Section 6).While the main outcome of this work is the full list of the convex polyhedra that areobtained by gluing regular pentagons edge to edge (Section 4), the methods for obtainingit are of independent interest and may be applied to other problems of the same flavour. In this section we review definitions and previous results that are necessary for therest of this paper. We start with some basic notions.By a polyhedron we mean a three-dimensional polytope, and, unless stated otherwise,all the polyhedra we are considering are convex. Doubly-covered convex polygon is alsoregarded as a convex polyhedron. A polyhedron is called simplicial if all its faces aretriangles.Consider an edge e of a polyhedron; and let f and f be the two faces of the polyhedronthat are incident to e . We call a vertex in f or f opposite to e if it is not incident to e .If f and f are triangles, then there are exactly two vertices opposite to e , see Figure 1. Definition 1.
Let P be a convex polyhedron. The Gaussian curvature at a vertex v of P equals (cid:16) π − (cid:80) tj =1 α vj (cid:17) , where t is the number of faces of P incident to v , and α vj is theangle at v of the j -th face incident to v . 2igure 1: Vertices u and u are oppositeto edge e of polyhedron P . (cid:0) − − (cid:1) π = π π π π (cid:0) − (cid:1) π = π Figure 2: Gaussian curvature ofthe vertices of a convex pentahedron.Since P is convex, the Gaussian curvature at each vertex of P is non-negative. Theorem 1 (Gauss, Bonnet 1848) . The total sum of the Gaussian curvature of all verticesof a 3D polyhedron P equals π . For an example, see Figure 2 that shows a convex pentahedron and the values ofGaussian curvature at each of its vertices.
Definition 2.
A gluing G is a collection of polygons T . . . T n equipped with an equivalencerelation ∼ on their border describing how the polygons should be glued to one another. Definition 3.
The polyhedral metric M of a gluing G is the intrinsic metric of the simplicialcomplex corresponding to G : the distance between two points of the gluing is the infimumof the lengths of the polygonal lines joining the points such that each vertex of it is withinone of the polygons T . . . T n .We denote the distance between points p , q of G by | pq | . Definition 4.
Gluing G (and the polyhedral metric corresponding to it) is said to satisfy Alexandrov’s conditions if:a) the topological space produced by G is homeomorphic to a sphere, andb) the total sum of angles at each of the vertices of G is at most 2 π . Theorem 2 (Alexandrov, 1950, [1]) . If a gluing G satisfies Alexandrov’s conditions thenthis gluing corresponds to a unique convex polyhedron P ( G ) : that is, the polyhedral metricof G and the shortest-path metric of the surface of P ( G ) are equivalent. Correspondence to a polyhedron discribed in this theorem intuitively means that P ( G )can be glued from polygons of G in accordance with relation ∼ . Note that polygons of G need not correspond to faces of P ( G ).Recall that a chord of a polygon Q is any segment connecting two points on the borderof Q that lies completely inside Q . Definition 5.
For a polyhedron P , a net of P is a gluing G = ( T . . . T n , ∼ ) of P togetherwith the set of chords of the polygons T i that do not intersect each other except possibly atendpoints. Those chords represent creases, i.e. lines along which P should be folded fromthis polygon. 3 Gluing regular pentagons together
In this section, we describe how to enumerate all the edge-to-edge gluings of regularpentagons.
Let P be a convex polyhedron obtained by gluing several regular pentagons edge toedge. Vertices of P are clearly vertices of the pentagons. The sum of facial angles arounda vertex v of P equals 3 π/ v . Since the Gaussian curvature at v is in (0 , π ), the numberof pentagons glued at v can be either one, two, or three. This yields the Gaussian curvatureat v to be respectively 7 π/
5, 4 π/
5, or π/ P ) by gluing severalpentagons. Therefore all the vertices of the pentagons must correspond to vertices of P . Proposition 3.
Suppose P is a convex polyhedron obtained by gluing edge-to-edge N regularpentagons. Then: (a) P has . N vertices in total. In particular, N must be even. (b) N is at most .Proof. From the above discussion, the vertices of P can be subdivided into three typesaccording to their Gaussian curvature: (1) the ones of curvature 7 π/
5, (2) 4 π/
5, and (3) π/
5. Let us denote the number of vertices type 1, 2 and 3, respectively, as x, y, z . Then wehave the following system of two equations: x + 4 y + z = 20 x + 2 y + 3 z = 5 N The first equation is implied by the Gauss-Bonnet theorem; the second one is obtainedby counting the vertices of pentagons, since each polyhedron vertex of type 1, 2 and 3corresponds to respectively one, two or three pentagon vertices.(a) By summing up the equations after multiplying the first one by 0 . .
3, we obtain that x + y + z = 2 + 1 . N .(b) Since x, y, z are non-negative integers, from the first equation we derive that themaximum number of vertices is obtained when x = 0 , y = 0 , z = 20. This assignmentcorresponds to N = 12 by the second equation. We used a computer program to list all the non-isomorphic gluings of this type. Ourprogram is a simple modification of the one that enumerates the gluings of hexagons [2].The gluings are depicted in Figures 3c, 3d, 4d, 4e, 4f, 5d, 5e, 5f.
Below is the list of all polyhedra that can be obtained by gluing regular pentagons.For those polyhedra that are simplicial, their graph structure is confirmed by applying4 a) P , (b) P , (c) Net of P , (d) Net of P , Figure 3: Polyhedra glued from two regular pentagons and their nets. Hereand further black lines are creases along which the polyhedron is folded.Dark red lines always denote borders between the polygons of the gluing.method of Section 5, for the others the proof is geometric and is done in Section 6. • – doubly-covered regular pentagon, see Figures 3a, 3c. – simplicial hexahedron with 5 vertices (3 vertices of degree 4, and 2 vertices ofdegree 3), see Figures 3b, 3d. • – simplicial dodecahedron with 8 vertices (2 vertices of degree 5 and 6 vertices ofdegree 4), see Figures 4a, 4d. – octohedron with 8 vertices (4 vertices of degree 4 and 4 vertices of degree 3) and 4quadrilateral and 4 triangular faces. It is a truncated biprism, see Figures 4b, 4e. – hexahedron with 8 vertices each of degree 3 and 6 quadrilateral faces. This is aparallelepiped, see Figures 4c, 4f.Note that P , , P , , P , can be glued from a single common polygon by altering therelation ∼ . • • a) P , (b) P , (c) P , (d) Net of P , (e) Net of P , (f) Net of P , Figure 4: Polyhedra glued from four regular pentagons and their nets. •
12 pentagons: regular dodecahedron with 20 vertices of degree 3 and 12 pentagonalfaces, see Figure 5c, 5f.We now proceed with a description of how to determine the graph structures of thepolyhedra in this list. We separately confirm the presence of the edges (Section 5) and provethat no additional edges are present in the quadrilateral faces of P , and P , (Section 6). Consider a polyhedral metric M that satisfies the Alexandrov’s conditions and thuscorresponds to a unique polyhedron P . Suppose we have a polyhedron P that approximates P . That is, vertices of P are in one-to-one correspondence with the cone points of M (andthus with the vertices of P ). In this section we show how to check whether the graphstructure of P contains all the edges of P .We will be using the following notation: v , v , v , . . . for the vertices of P ; u , u , u , . . . for the corresponding vertices of P ; V , E , F for the number of vertices, edges andfaces of P respectively; D for the maximum degree of a vertex of P ; L for the length of thelongest edge of P ; B r ( u ) for the ball in R of radius r centered at the point u .6 a) P (b) P (c) P (d) Net of P (e) Net of P (f) Net of P Figure 5: Polyhedra glued from six or more regular pentagons and theirnets. 7e also know the lengths of edges and distances between vertices of P since those arelengths of shortest paths between cone points of metric M . Let the discrepancy of an edge u i u j of P be the absolute value of the difference between the length of that edge and thedistance between the corresponding vertices v i and v j of P . Let maximum edge discrepancy µ of P be the maximum discrepancy for all edges of P .Similarly, for any facial angle u j u i u k of P , let discrepancy of this angle be the absolutevalue of the difference between the values of u j u i u k and of the angle between the corre-sponding shortest paths in P ; let the maximum angle discrepancy γ of P be the maximumdiscrepancy for all the facial angles of P .We base our check on the following theorem. Theorem 4.
Suppose µ is the maximum edge discrepancy between P and P , γ is themaximum angle discrepancy between P and P , D is the maximum degree of a vertex of P . If D γ < π/ , then each vertex of P lies within an r –ball centered at the correspondingvertex of P , where r = E · L · D γ/
2) +
Eµ. (1)We defer its proof to the Section 5.1, and for now we focus on describing our check,using the theorem as a black box.Let u i u j be an edge of P and let u a , u b be the two vertices of P opposite to the edge u i u j (see Figure 1). We want to check that there does not exist a plane intersecting all four r –balls centered at u i , u j , u a , u b respectively.Assume without loss of generality that the plane passing through u a , u i , u j is notvertical and that P lies below that plane (otherwise apply a rigid transformation to P sothat it becomes true). Note that we always can do this since P is convex.Consider three planes Π , Π , Π tangent to B r ( u i ), B r ( u j ), B r ( u a ) such that: • Π is below B r ( u i ), B r ( u j ) and above B r ( u a ), • Π is below B r ( u i ) and above B r ( u j ), B r ( u a ), • Π is below B r ( u j ) and above B r ( u i ), B r ( u a ). Theorem 5. If u b lies below Π , Π and Π and the distance from u b to each of the planes Π , Π and Π is greater than r , then there must be the edge v i v j in P . An example can be seen on Figure 6: plane Π is tangent to B r ( u i ), B r ( u j ), B r ( u a ).Point u b, is below Π , and point u b, is above Π , the distance from each of the points toΠ is greater than r .To prove this theorem, we need the following lemma. Lemma 6.
Given two disks B r ( u left ) , B r ( u right ) in R ; points u left , u right lie on x axis.Given a point u , x u > x u right , y u < . If u lies below the common tangent of the disksthat is above B r ( u left ) and below B r ( u right ) , than there is no line passing through B r ( u left ) , B r ( u right ) , and u . The example for this lemma can be seen in Figure 7. Point u is above the tangent,so there may be a line passing through it and the two disks. Point u is below the tangent,so no lines through B r ( u left ), B r ( u right ), u are possible.8 u a u i u j u b, u b, t Figure 6: Plane Π tangent to B r ( u i ), B r ( u j ), B r ( u a ). xu u u left u right Figure 7: An example for Lemma 6.
Proof.
Consider the set of points in R covered by all lines passing through B r ( u left ), B r ( u right ). We are looking for the lower border of it which corresponds to the lowest linepassing through these disks.Consider a line passing through the disks. If it is not tangent to B r ( u left ) from above,it can be made lower by raising its intersection with B r ( u left ), see Figure 8a. If it is nottangent to B r ( u right ) from below, it also can be made lower by lowering its intersectionwith B r ( u right ), see Figure 8b.Therefore, any line passing through B r ( u left ), B r ( u right ) is higher than the commontangent of these disks when x > x u right . (a) (b) Figure 8: Common tangent of disks is lower than any line passing throughthem. 9 roof of Theorem 5.
We can assume that points u i , u j lie on y axis, see Figure 6. For eachpair ( x, y ) we want to find minimum z such that there is a plane passing through B r ( u i ), B r ( u j ), B r ( u a ), and ( x, y, z ). Let us consider three cases: (1) y u i ≤ y ≤ y u j , (2) y ≤ y u i ,(3) y u j ≤ y .Consider case 1. Project everything on plane y = 0. The projections of B r ( u i ) and B r ( u j ) coincide, and a plane λ passing through these disks can be lowered by matching theprojections of its intersections with the disks, thus making projection of λ a line. Now wecan apply Lemma 6 to the projection to get plane Π from the statement of the Theorem.Consider case 2. Project everything on a plane orthogonal to the segment u j u a .Using similar argument, applying Lemma 6 we get plane Π from the statement. Case 3 issymmetric to case 2 and gives us plane Π .Therefore, all points of B r ( u b ) should lie below the planes Π , Π , Π , which yieldsthe condition of distance between u b and the planes being greater than r .The check suggested in Theorem 5 requires O (1) time, and has to be performed oncefor every edge u i u j of P . This implies the following. Theorem 7.
Given a polyhedral metric M satisfying Alexandrov’s conditions and an ap-proximation P for the polyhedron P that corresponds to M , there is a procedure to verifyfor each edge of P if it is present in P . The procedure answers “yes” only for those edgesthat are present in P , and it answers “inconclusive” if the approximation P is not preciseenough. The procedure requires time O ( E ) . Inconclusive answers occur if a plane exists that intersects all four r –balls even thoughthere is an edge connecting two of the vertices. In such case, precision has to be increasedby replacing P with a polyhedron that has smaller discrepancy in edge lengths and valuesof angles and repeating the procedure.Theorem 7 yields that if P is simplicial we can in time O ( E ) verify whole its graphstructure without any additional effort. However, if there are faces in P with four or morevertices, the absence of the edges that are diagonals of these faces has to be proved, whichrequires some creativity. For non-simplicial shapes glued from pentagons such proofs aregiven in Section 6.To obtain polyhedron P one can use the algorithm developed by Kane et al. [8] orthe one by Bobenko, Izmestiev [3]. Each of them outputs a polyhedron P which is anapproximation of P . In this work we used the implementation of the latter presented bySechelmann [10]. It gave us approximation with µ ∼ − , γ ∼ − , L ∼ .
5. These pa-rameters allowed for r ∼ − , which was enough to verify the presence of all the suggestededges.To do so, we developed a program that checks the condition of Theorem 5. Its sourcecode can be found in our bitbucket repository . We now proceed with the proof of Theorem 4. To prove it, we need the followinglemma. bitbucket.org/boris-a-zolotov/diplomnaia-rabota-19/src/master/praxis/haskell emma 8. Let pq , pq (cid:48) be line segments in R , | pq | = (cid:96) . If there are two real numbers ε , θ with ε > and < θ < π such that (cid:96) − ε ≤ | pq (cid:48) | ≤ (cid:96) + ε and (cid:93) qpq (cid:48) ≤ θ, then | qq (cid:48) | ≤ (cid:96) sin θ + ε. (2) Proof. pq (cid:48) can be obtained from pq , as shown in Figure 9, by a composition ρ ◦ τ of(1) rotation ρ around p by an angle at most θ ,(2) homothety τ with center p and ratio λ , where λ is some real number with (cid:96) − ε(cid:96) ≤ λ ≤ (cid:96) + ε(cid:96) . p qρ ( q ) (cid:96)(cid:96) ≤ θ q (cid:48) ε ε ρ Figure 9: After a segment is rotated by at most θ and its length changedby at most ε , its endpoint q moves by at most (cid:96) · θ + ε .First, it is clear that | ρ ( q ) , τ ( ρ ( q )) | ≤ ε , since τ is defined so as to add not more than ε to a segment of length (cid:96) . Now we estimate dist( q, ρ ( q )). It is at most (cid:96) · θ/ (cid:96) and angle at the apex θ . Combining the above estimations with the triangle inequality concludes the proof. Proof of Theorem 4.
Place P and P in such a way that1) a pair of their corresponding vertices, u in P and v in P , coincide,2) a pair of corresponding edges, e (cid:48) incident to u in P and e incident to v in P , lie onthe same ray, and3) a pair of corresponding faces, f (cid:48) in P incident to u and e (cid:48) and f in P incident to v and e , lie on the same half-plane.Figure 10: The angle between the edge of P and the edge of P is less than D γ . 11 a) (b) Figure 11: Illustration for the proof of Theorem 4: (a)
Rotation by theangle less than D γ is applied to the path w (cid:48) i . . . w (cid:48) k . (b) The edge w (cid:48) i w (cid:48) i +1 isbeing lengthened or shortened by not more than µ .Consider a pair of corresponding vertices, u in P and v in P . In order to estimate | uv | consider a shortest path π = u w (cid:48) w (cid:48) . . . w (cid:48) k u in the graph structure of polyhedron P . It iscomprised of edges of P and is not the geodesic shortest path from u to u . Vertices of π correspond to the vertices of another path π = v w w . . . w k v in P . Since π is a simplepath, it contains at most E edges and therefore its total length is at most EL .We now focus on the paths themselves, not on the polyhedra. Path π can be obtainedfrom π by a sequence of changes of edge directions (see Figures 10, 11a) and edge lengths(see Figure 11b). Let us estimate by how much endpoint u of path π can move when thissequence of changes is applied.Denote w (cid:48) := u , w := v and assume that for each j = 1 , . . . , i edge w (cid:48) j − w (cid:48) j is parallelto w j − w j . Then, by the triangle inequality, the angle α between w (cid:48) i w (cid:48) i +1 and w i w i +1 is atmost D γ , see Figure 10. Rotate the path w (cid:48) i . . . w (cid:48) k u around w (cid:48) i by angle α so w (cid:48) i w (cid:48) i +1 and w i w i +1 become parallel.Distance | w (cid:48) i u | is at most EL , so, by Lemma 8, every time we apply such rotation,the endpoint u of path π moves by at most EL · D γ/ E vertices in the path and E rotations are applied, the endpoint u moves by at most E · L · (cid:18) D γ (cid:19) . (3)Now that the directions of all the edges in path π coincide with the directions of theedges in path π , we can make the lengths of corresponding edges match. If the length ofa single edge of a path in P is changed by at most µ , and other edges are not changed (asshown in Figure 11b), then the end of the path also moves by not more than µ . Thereforeafter we adjust the length of all the edges, the endpoint u of path π moves by at most E · µ. (4)Combining (3) and (4) implies that in total point u moved by at most E · L · D γ/
2) +
Eµ. (5)This completes the proof. 12
Geometric methods to determine graph structure
In this section we give the last part of the proof that the polyhedra correspondingto the gluings listed in Section 4 have the same graph structure as the polyhedra listedin the same section. That is, we prove that quadrilateral faces of P , , P , correspond toquadrlateral faces of P , , P , , i. e., that certain edges are not present in P , , P , . P , Recall that P , is the polyhedron that corresponds to the gluing G , (see Figure 4e).Let A, B, . . . , H denote the vertices of G , , see Figure 13. We have already established bythe methods of Section 5 that P , has edges that are shown in the net on Figure 4e (blacklines). We now prove the following. Theorem 9.
For the polyhedron P , = P ( G , ) , each of the 4-tuples of vertices ( G, H, C, D ) , ( A, B, H, G ) , ( E, F, C, B ) , ( A, D, F, E ) forms a quadrilateral face of P , . ζA BD CE FG HM M M Figure 12: P , is symmetric with respect to 2 vertical planes, which yieldsfour quadrilateral faces.Figure 13: The net of P , . Proof.
Observe first that there are two vertical planes such that P , is symmetric withrespect to both of them: (1) the plane ζ that passes through edge GH (the common sideof two pentagons), and the midpoints M , M , M of edges AD , EF , BC respectively (see13igure 12); and (2) the plane ζ (cid:48) that passes through edge EF and the midpoints of edges AB , GH and DC . Indeed, polyhedron P , is symmetric with respect to plane ζ , since thesegment HM cuts in half the pentagon EF CHB (colored orange in Figures 12 and 13),and so does the segment GM does with pentagon F EAGD (colored yellow in Figures 12and 13). The argument for the plane ζ (cid:48) is analogous.Suppose for the sake of contradiction that BF is an edge of P , . Then segment EC must also be an edge due to the symmetry with respect to plane ζ . However, segments BF and EC cross inside the pentagon EF CHB and thus cannot be both the chords of the netof P , . We arrive to a contradiction. By the same argument EC cannot be an edge of P , . Therefore EF CB is a quadrilateral face of P , .The existence of quadrilateral faces GHCD, ABHG, ADF E is implied by a sym-metric argument. This completes the proof. P , Polyhedron P , is the polyhedron that corresponds to the gluing G , (see Figure 4f).Again let A, B, . . . H denote the vertices of G , , see Figure 15. The chords shown in thenet on Figure 4f (black lines) are already proven to be corresponding to the edges of P , .We now prove the following. Theorem 10.
For the polyhedron P , = P ( G , ) , each of the 4-tuples of vertices ( E, A, B, F ) , ( E, A, D, H ) , ( C, G, F, B ) , ( C, G, H, D ) , ( A, B, C, D ) , ( E, F, G, H ) forms a quadrilateralface of P , . In particular, each of these faces is a parallelogram. A B CDE F GHM M Figure 14: P , is symmetric with respect to the plane EACG . There aresix faces which are all parallelograms.
Proof.
We show that there is a convex polyhedron with the net as in Figure 15 that satisfiesthe claim. By Alexandrov’s theorem such polyhedron is unique and is exactly P , .The pentagon EAF HA (colored green in Figure 15) is folded along its diagonals EF and EH and glued along its edge EA . We use one degree of freedom to place it so thatit is symmetric with respect to the plane through EAM , where M is the midpoint of HF . Let us now take another pentagon AF BDH and glue one of its vertices to A . Placethis pentagon in a way that the plane ADB is parallel to the plane
EHF (see the orangepentagon in Figure 15). Now we glue these two pentagons along the edges AF and AH without changing the position of the triangle ADB . Since (cid:93)
F EA + (cid:93) EAF + (cid:93) F AB = π ,the points E, A, B, F are coplanar and form a parallelogram. By analogous arguments,
EADH , CGF B , and
CGHD are parallelograms as well.14igure 15: The net of P , .It is easy to see that the shape we just obtained by gluing the pentagons EAF HA and
AHDBF is still symmetric with respect to the plane
EAM , and the planes EHF and
ADB are parallel.Now let us show that points
H, D, B, F are coplanar and form a square
HDBF .Indeed, all of its sides are have equal length as sides of a regular pentagon, and it has anaxis of symmetry passing through the midpoints M and M of its opposite sides. Now ifwe glue the two halves of the polyhedron along this common square, the triangles CDB and
ADB will be coplanar, since (cid:93) CM M = (cid:93) EM M and (cid:93) EM M + (cid:93) M M A = π .Since | AD | = | DC | = | CB | = | BA | as diagonals of a regular pentagon, ADCB isa rhombus. By a similar argument,
EHGF is a rhombus as well. This completes theproof.
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