On Polyhedral Realization with Isosceles Triangles
OOn Polyhedral Realization with Isosceles Triangles
David EppsteinComputer Science Department, University of California, Irvine
Abstract
Answering a question posed by Joseph Malkevitch, we prove that there exists a polyhedralgraph, with triangular faces, such that every realization of it as the graph of a convex polyhedronincludes at least one face that is a scalene triangle. Our construction is based on Kleetopes, andshows that there exists an integer i such that all convex i -iterated Kleetopes have a scalene face.However, we also show that all Kleetopes of triangulated polyhedral graphs have non-convexnon-self-crossing realizations in which all faces are isosceles. We answer another question ofMalkevitch by observing that a spherical tiling of Dawson (2005) leads to a fourth infinite familyof convex polyhedra in which all faces are congruent isosceles triangles, adding one to the threefamilies previously known to Malkevitch. We prove that the graphs of convex polyhedra withcongruent isosceles faces have bounded diameter and have dominating sets of bounded size. By Steinitz’s theorem, the graphs of three-dimensional convex polyhedra are exactly the 3-vertex-connected planar graphs [13]. The faces of the polyhedron are uniquely determined as the peripheralcycles of the graph: simple cycles such that every two edges that are not in the cycle can be connectedby a path whose interior vertices are disjoint from the cycle [16]. We call such a graph a polyhedralgraph , and if in addition all faces are triangles we call it a triangulation . If G is the graph of apolyhedron P , we call P a realization of G ; here, P is required only to be non-self-intersecting, notnecessarily convex, but if it is convex we call P a convex realization . We may ask: what constraintsare possible on the shapes of the faces of P [2]?We call a realization or polyhedron isosceles if all faces are isosceles triangles, and monohedral if all faces are congruent. Monohedral isosceles convex polyhedra were investigated by Goldbergin 1936, who observed the existence of an infinite family of these polyhedra (beyond the obviousexample of the bipyramids), obtained by gluing pyramids to the non-triangular faces of antiprisms [7](Figure 1, left). In 2001, Joseph Malkevitch described a triangulation that has no monohedralisosceles convex realization [10]; in response, Branko Gr¨unbaum described a triangulation thathas no monohedral realization, regardless of convexity of the realization and regardless of faceshape [9]. However, Gr¨unbaum’s triangulation can be realized with all faces isosceles or equilatera( Figure 1, right). More recently, Malkevitch has asked again which triangulations have isoscelesconvex realizations, not required to be congruent, and whether all triangulations have such arealization [11]. Isosceles polyhedra have also been studied in connection with unfolding polyhedrainto nets: there exist non-convex isosceles polyhedra with all faces acute that have no such unfolding,while it remains open whether all convex polyhedra have an unfolding [3, 5].In this work, we provide the following results: This terminology should be distinguished from isohedral polyhedra, in which every two faces are symmetric toeach other. a r X i v : . [ c s . C G ] A ug igure 1: Left: A polyhedron of Goldberg [7], with 100 congruent isosceles triangle faces. Right: Atriangulation of Gr¨unbaum that cannot be realized with all faces congruent [9], realized as a convexpolyhedron with edges of two lengths, and with isosceles and equilateral triangle faces. • We prove that some triangulations have no convex isosceles realization. We use the idea of a
Kleetope , a construction named after Victor Klee in which a polyhedron or polyhedral graphis modified by gluing a pyramid onto each face [8]. We show that repeating the Kleetopeconstruction a bounded number of times, starting from any polyhedral graph, will produce atriangulation that has no convex isosceles realization. • We prove that the triangulations produced by this construction, and more generally all theKleetopes of triangulations, have non-convex isosceles realizations. • We answer an open question posed by Malkevitch [11] by finding a fourth infinite family ofmonohedral isosceles convex polyhedra, different from three previously known infinite families. • We investigate the structure of monohedral isosceles convex polyhedra with many faces,proving that they must have very sharp apex angles, bounded graph diameter, and boundeddominating set size. If G is a polyhedral graph, then its Kleetope is another polyhedral graph, obtained by addingto G a vertex for each face of P , adjacent to every vertex of the face. We denote the Kleetope by KG , and we denote the result of repeating the Kleetope operation i times by K i G . For instance,Figure 2 (left) shows a convex realization of K O , where O = K , , is the graph of an octahedron.Geometrically, a realization of a Kleetope can be obtained from a realization of G by attaching apyramid to each face of G . When G is a triangulation, every convex realization of KG leads to aconvex realization of G , obtained by removing the added vertices of KG . (In the non-convex case,this removal may lead to self-intersections. The property that G is a triangulation and not just anarbitrary polyhedral graph is needed to ensure that each pyramid has a flat base.)One of the earliest applications of Kleetopes was in the construction of triangulations in whichall paths or cycles are short. If G is a triangulation with n vertices, then it has 2 n − K O of an octahedron O . Right: Its non-convex isoscelesrealization.and KG has 3 n − KG , no two vertices of KG \ G can beadjacent, and replacing each such vertex by the edge between its two neighbors in G produces asimple cycle of G . Therefore, if the longest simple cycle in G has length c , then the longest simplecycle in KG has length 2 c . Thus, the cycle length only increases by a factor of two, while thenumber of vertices increases by a factor converging (for large n ) to three. Iterating this processproduces n -vertex triangulations in which the longest path or cycle has length O ( n log ) [12], anditerating it a bounded number of times produces the following result in the form that we need it: Lemma 1.
For every ε there is an i such that, for all polyhedral graphs G , the longest simple cyclein K i G includes a fraction of the vertices of K i G that is less than ε .Proof. If G is a polyhedral graph, KG is simplicial and has at least 8 vertices. Each subsequentiteration of the Kleetope operation multiplies the number of vertices by at least 5/2 and multiplesthe longest cycle length by at most 2, so we may take i to be 1 + log / /ε . The discrete Gauss map of a polyhedron (in a fixed position in R ) maps each feature of thepolyhedron to a set of three-dimensional unit vectors, the outward-pointing perpendicular vectorsof supporting planes of the polyhedron (planes that intersect its boundary but not its interior) thatare tangent to the polyhedron at that feature. The set of all unit vectors forms a unit sphere, the Gauss sphere , on which each face of the polyhedron is mapped to a single point, each edge of thepolyhedron is mapped to an arc of a great circle, and each vertex of the polyhedron is mapped to aconvex region bounded by arcs of great circles. That is, the Gauss map is a dimension-reversingmap from the features of the given polyhedron to features of a polygonal subdivision of the sphere,sometimes called the spherical dual of the polyhedron [14].We refer to distances along the surface of the Gauss sphere as geodesic distance , and the maximumdistance between any two points in a given set of unit vectors on the sphere as geodesic diameter . Ifthe Gauss map takes two adjacent faces of the polyhedron to points whose geodesic distance is d ,then the dihedral angle of the faces is exactly π − d .3igure 3: Left: The case of Lemma 2 with two bases on F , viewed in projection onto the planecontaining F . The three edges of T disjoint from f (red) have equal lengths, and meet on a lineperpendicular to the plane of F through the circumcenter of F (seen in projection as the smallyellow circle). Right: The convex hull of two congruent isosceles triangles rotated around a sharededge, used in the case of the lemma with two shared bases and the case with a long side. At anendpoint of the shared edge, either the shared dihedral or the other two equal dihedrals are greaterthan π/ In this section we prove that iterated Kleetopes (with a sufficiently large number of iterations) donot have a convex realization with all faces isosceles. Our proof proceeds in stages, where eachstage adds more iterations to the Kleetope and proves the existence of a face whose shape andneighborhood is more strongly constrained, until the constraints are sufficient to rule out isoscelestriangles as faces. We will eventually show that in the final stage of the Kleetope constructionprocess, it will be necessary to glue on a triangular pyramid with an obtuse base triangle, isoscelesfaces other than the base, and small (very sharp) dihedral angles at the base edges of the pyramid.However, pyramids with these properties do not exist, as the next subsection shows.
Every vertex of every convex polyhedron has an incident edge with dihedral angle > π/
3, but insome polyhedra (for instance flattened pyramids) these large dihedral angles avoid one of the faces.The main result of this section is that in tetrahedra with one face obtuse and the rest isosceles, theobtuse face cannot be avoided by the large dihedrals:
Lemma 2.
Let T be a tetrahedron in which one face F is obtuse, and the other three faces areisosceles. Then at least one of the dihedral angles on an edge of F is greater than π/ .Proof. We divide into cases according to how the base and apex of each isosceles triangle arearranged relative to F . Two bases on F . If at least two of the isosceles triangles have their base on an edge of F , thenthe three edges of T that are disjoint from F have equal length, and the vertex of T wherethese edges meet lies on a line perpendicular to F through the circumcenter of F . Because F lane of projection F pv se q a bcf de
Figure 4: Left: Notation for the case of Lemma 2 of two shared apexes. Right: Notation for thefinal “none of the above” case, shown schematically with equal-length edges in equal colors.is obtuse, its long side separates the circumcenter from F , so the dihedral of T on the longside of F is at least π/
2. (See Figure 3, left. This case requires f to be obtuse.) Two shared bases.
If two of the three isosceles triangles share a base edge with each other, thenin each of them one of the two equal sides is an edge of F . The third isosceles triangle hasedges of these lengths, equal to the lengths of two edges of F , so it is congruent with F , androtated from F around the edge e that it shares with F . Figure 3 (right) depicts the casewhere the shared edge is the base of the isosceles triangle, but it could also be one of the sides.Because the two triangles on e are congruent, the vertex figure of T at either endpoint of e (the intersection of T with a sufficiently small sphere centered on that point) is a sphericalisosceles triangle. Like any spherical triangle its angles sum to greater than π so one angle isat least π/
3. This large angle is either the dihedral at e (the apex of the spherical isoscelestriangle) or is the equal dihedral at the other two edges meeting at the same vertex of T , oneof which is an edge of F . (This case does not require f to be obtuse, and can produce dihedralangles arbitrarily close to π/ Long side.
If the long edge e of F is one of the two equal sides of one of the isosceles triangles,consider the tetrahedron T (cid:48) formed as the convex hull of F and a rotated copy of F by anangle of π/ e (on the same side of the plane through F as T ). As in the case of twoshared bases, the vertex figures at the endpoints of e are spherical isosceles triangles, andthe choice of π/ T (cid:48) , all the dihedrals at the endpoints of e (which include all three dihedrals on the edges of f ) are at least π/ e are the unique farthest pair of points in T (cid:48) , so the edge of T of equallength to e must end at a point outside T (cid:48) . This point is separated from T (cid:48) by one of the faceplanes of T (cid:48) , which meets F at one of its edges. The dihedral of T on that edge is greaterthan the dihedral of T (cid:48) , which is in turn at least π/ Two shared apexes.
If two of the isosceles faces of T both have their apex at the same vertex v of F , then the three edges of T meeting at that vertex all have the same length (cid:96) . As twoof them are edges of F , F must be isosceles, with these edges as its sides. We can view this5ase in projection onto a plane perpendicular to F through its center line. Of the three equalshared edges, the edge e that is disjoint from F lies in this plane, while the other two lie insymmetric positions on either side of it, and project to a point p at the midpoint of the baseof F . Triangle F itself projects to a line segment s connecting v to p . Let q be the non-sharedendpoint of edge e (Figure 4, left).Then, in triangle vpq , edge vq is longer than edge vp , because vp is the projection of athree-dimensional line segment of equal length to vq . Let θ = max( ∠ qvp, ∠ qpv ), the largerof the two angles on side s of this triangle. Then θ ≥ θ ∗ where θ ∗ is the base angle of theisosceles triangle obtained by moving q to lie on the perpendicular bisector of s (keeping thelength of edge vp the same). Because vq is longer than vp , θ ≥ θ ∗ > π/
3. If θ = ∠ qpv , it isalso the dihedral angle on the base of f , which is therefore greater than π/
3. If θ = ∠ qvp , it isless than the dihedral angles on the two sides of f (because these sides are not perpendicularto the plane of projection), which are therefore both greater than π/
3. (This case does notrequire F to be obtuse.) None of the above.
In the remaining case, let a , b , and c be the sides of F , with c being thelongest side, and let d , e , and f be the other three sides of T , with d opposite a , e opposite b ,and f opposite c . In order to avoid the case of a long side, isosceles triangle cde must have c as its base, so d and e must have equal length. In order to avoid the cases of a shared baseor a shared basis, exactly one of the other two triangles must have d or e as its base; bysymmetry, we may assume without loss of generality that isosceles triangle aef has e as itsbase (so a and f are equal) and that isosceles triangle bdf has f as its base (so b , d , and e areequal). Figure 4 shows this case schematically (not to scale), with edges that in T have equallengths depicted as having equal colors and textures. We divide into sub-cases according tothe comparison between these lengths: • If a and f have the same length as b , d , and e , then we would be in the case of twoshared bases (shared at f ). • If a and f are shorter than b , d , and e , then consider the tetrahedron T (cid:48) formed from T by lengthening a and f to have the same length as b , d , and e , but keeping all other edgelengths the same. T (cid:48) has two equilateral triangle faces aef and bdf , and two isoscelesfaces with apex angle greater than equilateral, abc and cde (although these faces neednot be obtuse). The same analysis as the case of two shared bases shows that, in T (cid:48) ,the dihedrals on a , b , d , and e are all greater than π/
3. Lengthening a and f causes thedihedral on b to decrease (among other changes), so in T as well, b is greater than π/ • If a and f are longer than b , d , and e , then consider the tetrahedron T (cid:48) formed from T by lengthening b to equal a and f , keeping all other edge lengths the same. Then T (cid:48) fallsinto the case of two shared apexes, although in T (cid:48) triangle abc may not be obtuse. Fromthe analysis of that case, T (cid:48) has a dihedral greater than π/ a and c .Lengthening b causes these two dihedrals to decrease, so in T as well, at least one of a and c has a dihedral greater than π/ We define the sharpness of a face F of a polyhedron to be π − θ where θ is the smallest dihedralangle of an edge of F . Equivalently, on the Gaussian sphere, the sharpness of F is the length ofthe longest edge connecting the Gaussian image of F to one of its neighbors. In a face with smallsharpness, all dihedrals are close to π . In terms of this quantity, Lemma 2 from the previous section6an be interpreted as stating that one cannot glue an isosceles pyramid onto an obtuse triangle ofsharpness less than π/ Lemma 3.
For every ε there exists i such that, for all polyhedral graphs G and all convex realizationsof K i G , the realization includes at least one face whose sharpness is ≤ ε .Proof. Construct an arrangement of c = O (1 /ε ) great circles on the Gauss sphere that partitionthe sphere into cells with geodesic diameter at most (cid:15) , for instance by choosing two points on thesphere at geodesic distance π/ i be such that the longest cycle in K i G includes a fraction less than 1 / c of the edges of K i G . We claim that i has the stated property.To prove this, consider any given choice of G and convex realization of K i G , and rotate thearrangement of great circles (if necessary) so that none of them passes through a vertex of thespherical dual of K i G . Then each of the great circles passes through a sequence of edges and facesof the spherical dual corresponding to a simple cycle in K i G , the cycle formed as the silhouetteof K i G in a perpendicular projection onto the plane of the great circle. By Lemma 1 and the choiceof i , these silhouette cycles together use a fraction less than 1 / K i G . It followsthat at least one face F of K i G is bounded by edges that are not in any silhouette cycle, for if allfaces had at least one edge in a cycle the total fraction of edges in cycles would be at least 1 / F and its three neighboring triangles in K i G are all in the same cell ofthe arrangement of great circles, of geodesic diameter at most ε . It follows that the edge from theGaussian image of F to the farthest of its neighbors has length at most ε .If we glue a pyramid onto a face F of a convex polyhedron to produce another convex polyhedron(as happens for realizations of Kleetopes), the faces of the pyramid that replace F can have greatersharpness than F had. However, the sharpness cannot increase too much: Lemma 4.
Let P be a triangulation, let Q be a convex realization of K i P for some i , and let F be a face of P that (in the realization of P derived from Q by removing the vertices of Q \ P ) hassharpness ε . Then every face of Q formed by subdividing F has sharpness at most ε .Proof. On the Gaussian sphere, the image of F is a point, incident to three great circle arcs of lengthat most ε corresponding to the dihedrals along the edges of F . The spherical triangle T formingthe convex hull of these three arcs has geodesic diameter at most 2 ε , by the triangle inequality.Subdividing F to form Q replaces the Gaussian image of F (a single vertex in the spherical dualof P ) by a subgraph, attached to the rest of the graph at three vertices on the same three arcs.The sharpness of any face of the subdivision equals the length of the longest edge incident to itsGaussian image, which may either be an edge of this subgraph or one of the three edges connectingthis subgraph to the rest of the graph. All of these edges lie within spherical triangle T , so theirlengths are at most the diameter of T , which is at most 2 ε . The tetrahedron formed as the convex hull of the origin and the three perpendicular unit basisvectors of three-dimensional space has the same shape as the truncated corner of a cube, withthree isosceles right triangle faces and one equilateral face. When the right triangle faces areprojected perpendicularly onto the equilateral face, their projections are isosceles triangles with7 oints that projectfrom an acute vertexpoints that projectto a vertex of angle > 2π/3 ec Figure 5: A semi-ellipse (projected from a semicircle) and a circular arc through the same threepoints (endpoints of e and the center c of an equilateral triangle on edge e ) separate points thatproject from an acute angle and points that project to an angle > π/ π/ π/
6. The dihedral angles of this tetrahedron are π/ cos − ( − ) ≈ . π between the right triangles and the equilateraltriangles; we use ϕ to denote this latter angle. The familiar geometry of this example forms theextreme case for the following lemma: Lemma 5.
Let T be a triangle in R , let e be an edge of T , and P be a plane through e forming anangle less than ϕ with the plane of T . Suppose that the perpendicular projection T (cid:48) of T onto P is atriangle T (cid:48) whose largest angle is at least π/ , at the vertex opposite e . Then T is obtuse.Proof. Increasing the angle between the plane of T and P can only increase the largest angle of T (cid:48) ,so we may assume without loss of generality that the angle between the plane of T and P is exactly ϕ , and prove that in this case T is either right or obtuse. In the plane of T , the points p that forman acute vertex of a triangle with opposite edge e are separated from the points that form an obtusevertex by a semicircle having e as its diameter. The midpoint of this semicircle is a point that,together with e , forms an isosceles right triangle, and its projection onto P is the center point c ofan equilateral triangle that has e as one of its three sides. The projection of the semicircle itselfonto P is a semi-ellipse having e as its diameter and passing through c (the boundary of the outeryellow region in Figure 5).In P , the points q that form a vertex of angle > π/ e are separatedfrom the points that form a vertex of angle < π/ π/ e through c (the boundary of the inner blue region in Figure 5). This circular arclies inside the semi-ellipse, except for the endpoints of e (where the circle and ellipse, if continued,would cross) and the point c where the two curves are tangent (counting as a double contact andusing up all four possible points of intersection of a circle and ellipse). For T (cid:48) to have angle atleast 2 π/ e , its vertex must be on or inside the circular arc, and therefore either insidethe semi-ellipse or on it at c . To project to a point on or inside the semi-ellipse, the vertex of T opposite v must be on or inside the semi-circle, and therefore T must be obtuse or right.Using this lemma, we can prove that sufficiently iterated Kleetopes have an obtuse face, and,moreover, one with low sharpness: 8 emma 6. Let ε be a positive number with ε < ϕ , let i be an integer large enough that, by Lemma 3,every convex realization of K i P has a face of sharpness less than (cid:15) , and let P be any triangulation.Then every convex realization of K i +1 P has an obtuse-triangle face of sharpness less than ε .Proof. Given a realization of K i +1 P , form from it a realization of K i P by removing the extravertices added in going from K i P to K i +1 P , and apply Lemma 3 to this realization to find a face F of K i P whose sharpness is less than ε . Now consider the three faces that subdivide F in K i +1 P , andproject these three faces perpendicularly onto the plane of F . Their projections must be containedwithin F , for otherwise the angle between one of these faces and F would be greater than π/
2, toobig to form a convex dihedral with the neighboring face across the edge of F given the low sharpnessof F itself. Therefore, within the plane of projection, these three projected triangles subdivide F and meet at a shared vertex interior to F . If F (cid:48) is the face of K i +1 whose projection has the largestangle at this shared vertex, then this angle is least 2 π/
3. By the assumption on the sharpness of F ,the plane of F (cid:48) makes an angle less than ϕ with the plane of F , so we can apply Lemma 5 to F (cid:48) and its projection. By this lemma, F (cid:48) is obtuse, and by Lemma 4 its sharpness is less than 2 ε . Combining the results from this section, we have:
Theorem 7.
There exists an integer j such that, for every triangulation P , every convex realizationof K j P has a non-isosceles face.Proof. Let i be an integer large enough that, by Lemma 3, every convex realization of K i P has aface of sharpness less than π/
6, and let j = i + 2. Let P be an arbitrary triangulation. Considerany convex realization of K j P , and the convex realization of K j − P = K i +1 P obtained from it bydeleting the last layer of added vertices. By Lemma 6, the realization of K j − P has an obtuse face F of sharpness less than π/
3. If the realization of K j P had all-isosceles faces, Lemma 2 (applied tothe tetrahedron formed by F and the three isosceles triangles that subdivide it) would show thatone of these isosceles triangles lies in a plane making an angle greater than π/ F . Because this angle is greater than the flatness of F , it is impossible for such a triangle tomake a convex dihedral angle with whatever other triangle neighbors it across the edge it shareswith F , contradicting the assumption that the realization is convex. This contradiction shows thatthe assumption that the realization has only isosceles triangles as its faces cannot be true. In this section we prove that, whenever G is a triangulation, KG has a non-convex realization inwhich all faces are isosceles. In particular, this is true of the polyhedra that we we constructed asexamples that have no convex isosceles realization. Theorem 8.
Let P be a convex polyhedron whose graph G is a triangulation. Choose distance R ,for each face f of P place a new vertex at distance R from each of the three vertices of f , separatedfrom P by the plane of f , and replace f by three triangles through the new vertex and two of thethree vertices of f . Then for sufficiently large values of R , the result is a non-self-crossing realizationof KG , in which all faces are isosceles triangles.Proof. Given any triangle f in three-dimensional space, two points at distance R from the threevertices of f will exist whenever R is greater than the circumradius r f of f . These two points aresymmetric from each other across the plane of f , and the constraint that this plane separates each9igure 6: Left: Bipyramid. Right: Biarc hull.new point from P uniquely determines the placement used in the theorem. Therefore, whenever R isgreater than max f r f , the construction given in the statement of the theorem is well-defined. Eachnew face is a triangle with two sides of length R , so the isosceles shape of the faces is immediate. Thenontrivial part of the theorem is the claim that the resulting system of triangles is non-self-crossing.The new point p for f lies on a line through the circumcenter of f , perpendicular to f . Thecircumcenter has distance ≤ r f from each edge of f , and the new point is at distance at least R − r f away from the plane of f , by the triangle inequality. It follows that, although the dihedral anglebetween f and each triangle incident to p may be obtuse (as happens when f itself is obtuse), itmay not be very obtuse: it is π/ O ( r f /R ). By choosing R sufficiently large, we may make thisangle as close to π/ P (the partition of the space outside P intocells within which one face is nearer than all others) has cells bounded by the planes bisecting thedihedral angles of P . Let θ < π be the largest dihedral angle of P . Then, as long as we choose R large enough that each dihedral angle between a new triangle and the face f that it replaces is lessthan π − θ/
2, each new point and its incident triangles will lie within the Voronoi cell of the facethat it replaces. Since these Voronoi cells are disjoint, the new triangles cannot cross each other.Figure 2 (right) depicts a non-convex isosceles realization of the iterated Kleetope of theoctahedron, obtained by this construction.
There are only eight convex polyhedra with congruent equilateral triangle faces, called deltahedra ,but infinitely many non-convex deltahedra [6, 15]. Generalizing this concept, Malkevitch [11] asksabout convex polyhedra with congruent isosceles triangle faces. He states that there are three knowninfinite families of these monohedral isosceles convex polyhedra (leaving as a puzzle for the readerfinding these families) and poses as an open question whether there are any more. We answer thisquestion by identifying four families: • The bipyramids are convex hulls of equally spaced points on a circle, together with twoadditional points on the axis through the center of the circle, equally far from it on each side(Figure 6, left). • Let X and Y be two arcs of circles, each equidistant from the endpoints of the other arc, suchthat both arcs lie on the boundary of the convex hull of their union. Then the convex hull10igure 7: An order-25 twisted gyroelongated bipyramid.of any finite set of points on both arcs that includes the endpoints of each arc consists ofequilateral triangles whose base connects two consecutive points on one arc and whose apexis on the other arc. For any two integers x ≥ y ≥ x + 1 equally-spaced points on arc X and y + 1 equally-spaced points on arc Y all have the same distance between consecutive points; for instance, as in Figure 6 (right),this can be done by selecting X and Y to be concentric semicircles on perpendicular planeswith a carefully-chosen ratio of radii. The convex hull of the resulting point set has 2( x + y )congruent isosceles triangles as faces; we call this shape a biarc hull . • Goldberg [7] constructed an infinite family of convex polyhedra with congruent isosceles facesby attaching pyramids to opposite faces of an antiprism (Figure 1, left). For an antiprismover a k -gon, the result has 4 k faces. The cases k = 4 and k = 5 can also be realized byequilateral triangle faces, producing a gyroelongated square bipyramid and regular icosahedronrespectively. As the name “Goldberg polyhedra” already refers to an unrelated class ofpolyhedra, we call the general case an order- k gyroelongated bipyramid . • In a gyroelongated bipyramids of odd order k , the faces can be partitioned into two subsets,separating each pair of opposite faces. For certain realizations of these polyhedra, the edgesseparating these two subsets form a non-planar hexagon with order-6 symmetry. Twisting oneof the two subsets by an angle of 2 π/ k twisted gyroelongated bipyramid (Figure 7).We detail the construction below. To understand the twisted gyroelongated bipyramids, it is helpful to into a little more detail aboutGoldberg’s method for realizing order- k gyroelongated bipyramids in such a way that all faces arecongruent and isosceles. In Goldberg’s construction, all of the points lie on a unit sphere, with thetwo pyramid apexes at its poles and the other points spaced equally around two circles of equal and11pposite latitude. Goldberg chooses these latitudes so that the spherical angles of the great-circlearcs on the sphere, connecting the points in the same pattern as the polyhedron, are 2 π/k at theapex of each triangle and (1 / − / k ) π at each base angle. In this way, k of these spherical trianglesmeet with total angle 2 π at each pole, and the five triangles meeting at each remaining point havetotal angle (2 /k + 4(1 / − / k )) π = 2 π . The angular excess of each triangle, the amount by whichits total angle exceeds π , is π/k ; for spherical triangles, the angular excess equals the area, so thearea of each triangle is exactly 1 / k times the surface area of the entire sphere. The subdivision ofthe sphere into spherical triangles of this shape can be transformed into a polyhedron by takingthe convex hull of its vertices, and (as is true whenever a spherical triangulation consists only ofacute triangles) the combinatorial structure of the resulting convex hull is the same as the sphericaltriangulation. In this way, one obtains a gyroelongated bipyramid with congruent isosceles facesthat is, moreover, both inscribed in the unit sphere and circumscribes another smaller sphere. Thepolyhedron illustrated in Figure 1 (left) was constructed in this way.Now, divide the resulting polyhedron into two polyhedral surfaces by a non-planar hexagonthrough both poles, separating each face from its opposite face. The same subdivision into twosurfaces can also be made on the spherical triangulation from which the polyhedron was derived.On the surface of the sphere, at each pole, the hexagon subdivides ( k − / k + 1) / − /k ) π with respect to each other at the poles. At each non-pole vertex, the hexagon subdividestwo spherical triangles (meeting at their base) on one side from three spherical triangles (at twobase angles and an apex) on the other, so its angle is 2(1 / − / k ) π = (1 − /k ) π . That is, thishexagon lies on a sphere, has equal edge lengths, has equal angles, and splits the sphere into twoequal areas. It follows that it has order-6 symmetry: it can be twisted by an angle of 2 π/ π/ π/ k >
5. For k = 3 it produces a triangulation ofa cube by some of the diagonals of its squares (with all faces right and isosceles, but not forminga strictly convex polyhedron) and for k = 5 the twisting process leaves the regular icosahedronunchanged. In relation to the work of Dawson, we remark that (for any n ≥
2) it is also possible totile the sphere by 4 n isosceles spherical triangles with base angle π/n and apex angle π (1 − /n ),meeting edge-to-edge in the pattern of a bipyramid. (This is the spherical analog of the non-convexpolyhedron depicted in Figure 8.) When n is odd, the resulting spherical triangulation again has askew hexagon with order-6 symmetry, and twisting one side of this skew hexagon by an angle of2 π/ π . It follows from Alexandrov’s theorem that12he result is the surface of a convex polyhedron, but the theorem does not tell us whether theedges of the polyhedron follow the edges of the triangles from which the surface was derived, orare rearranged into a different combinatorial structure. If the edges of the resulting polyhedroncoincide with the edges of the triangles, the result would be a generalized asymmetric form of thetwisted gyroelongated bipyramid, with different numbers of congruent isosceles triangles on each ofits two sides. We leave as open for future research whether this generalized twisted gyroelongatedbipyramid actually exists. The known infinite families of monohedral isosceles convex polyhedra all have very sharp apexangles when they have large numbers of faces. This is not a coincidence: as we show, in such apolyhedron with n faces, the apex angle must be O (1 /n ).Our proof depends on two ideas: first, that in any convex polyhedron, the total angular deficit(the amount by which the sum of angles at each vertex fall short of 2 π , summed over all vertices) isexactly 4 π ; this is Descartes’ theorem on total angular defect, a discrete form of the Gauss–Bonnettheorem. Second, in a monohedral isosceles convex polyhedron, there are (usually) only a smallnumber of distinct types of vertex, as follows: • If a vertex has only the apexes of isosceles triangles incident to it, we call it a pyramidal vertex . • If a vertex has two bases of isosceles triangles and one or more apexes incident to it, we call ita semipyramidal vertex . • If a vertex has only four bases of isosceles triangles incident to it, we call it a basic vertex . • If a vertex has four bases and one apex incident to it, we call it a semibasic vertex .As a shorthand, we call a convex polyhedron with congruent isosceles faces whose vertices areonly of these four types well-behaved . Observation 9.
At any vertex of a monohedral isosceles convex polyhedron, the number of incidentbase angles of faces is even, and these angles can be grouped together in pairs, adjacent across ashared base edge of two isosceles triangles.
Observation 10.
In a monohedral isosceles convex polyhedron, a vertex incident to four or morebase angles can be incident to at most one apex angle, so the well-behaved vertex types exhaust thepossible vertices that can be incident to zero, two, or four base angles.Proof.
A vertex of a monohedral isosceles convex polyhedron with four or more base angles can haveonly one incident apex angle, because four base angles and two apex angles together add to 2 π , toomuch for the vertex of a convex polyhedron. The well-behaved vertex types allow arbitrarily manyapex angles together with zero or two base angles, and zero or one apex angles with four base angles,so they include all the vertices that have at most four base angles and obey this constraint. Lemma 11.
In a semipyramidal vertex incident to exactly two apex angles of isosceles triangles,with all incident triangles congruent and having apex angle greater than π/ , the dihedral anglebetween the two incident apex angles is less than π/ .Proof. For two isosceles triangles with apex angle θ > π/ − (cos ( π − θ )) >π − θ . However, two base angles of the same triangle can together span an angle of at most π − θ ,13ot enough to connect two apex angles with this dihedral. Therefore, in order to form a connectedsurface at the vertex, the dihedral must be smaller.Monohedral isosceles convex polyhedra with more than four base angles at some vertices doexist. For instance, the Kleetopes of the regular tetrahedron, octahedron, or icosahedron have six,eight, or ten bases meeting at some vertices respectively. However, there are only finitely many suchcases: Theorem 12.
There are only a finite number of triangulations that can be realized as non-well-behaved monohedral isosceles convex polyhedra.Proof.
Let P be such a polyhedron, with n faces. By Observation 10, P must have a vertex v atwhich six or more base angles meet. Therefore, the base angle must be less than π/
3, from which itfollows that the apex angle must be greater than π/ P , as in any icosahedral polyhedron) the number of base angles of faces is exactly twice thenumber of apex angles of faces, and because the large apex angle allows at most five apex angles tomeet at a vertex, there must be Ω( n ) faces with at least half as many apex angles as base angles.According to Observation 10, such a vertex can only be pyramidal or semi-pyramidal, with at mostfive incident faces. In order to have total angular defect 4 π , these faces need to have angular defect O (1 /n ). We distinguish the following cases: • A semipyramidal vertex with one incident apex angle has angular defect exactly π , which canonly be O (1 /n ) if n is itself bounded. • If there are Ω( n ) semipyramidal vertices with two incident apex angles, then to achieve angulardefect O (1 /n ) for each of these vertices, the apex angle must be π − δ and the base anglemust be δ for some δ = O (1 /n ).Because these vertices use equal numbers of apex and base angles, but the faces of thepolyhedron have twice as many base angles as apex angles, there must also be a linear numberof base angles that participate in other vertices with zero or one incident apex angles. If thereare k of these other vertices, then each apex angle contributes less than π to the total angle ofeach of these other vertices, and all of the base angles together contribute O (1) to the totalangle, so the total angle is ≤ kπ + O (1), but in order to achieve total angular deficit at most4 π the total angle must be ≥ (2 k − π . Combining these two inequalities shows that k = O (1)and that, therefore, at least one of these other vertices has Ω( n ) incident base angles.At a vertex with Ω( n ) incident base angles, each pair of consecutive base angles that meetalong a side of their faces (rather than sharing a base edge) must have a semipyramidal vertexwith two incident apex angles at the other end of that side. When n is large enough thatthe apex angle is greater than π/
2, by Lemma 11, the dihedral angle between these twoconsecutive base angles must be less than π/
2. But a vertex of a convex polyhedron can haveat most three dihedral angles less than π/
2, so n must be bounded. • If there are Ω( n ) pyramidal vertices with three incident apex angles, then to achieve angulardefect O (1 /n ) for each of these vertices, the apex angle must be 2 π/ − δ and the base anglemust be π/ δ for some δ = O (1 /n ). However, the base angles of the faces must also appearin a linear number of non-pyramidal vertices, and for these apex and base angles, all othervertex types have angular deficit at least π/ − O ( δ ), bounded away from zero. Therefore, tokeep the total angular defect below 4 π , n must be bounded.14igure 8: A polyhedron with 16 faces, part of an infinite family of non-well-behaved non-convexmonohedral isosceles bipyramids in which every vertex has positive angular deficit. • If there are Ω( n ) pyramidal vertices with four incident apex angles, or semipyramidal verticeswith three incident apex angles, then to achieve angular defect O (1 /n ) for each of thesevertices, the apex angle must be π/ − δ and the base angle must be π/ δ for some δ = O (1 /n ). Because vertices of these types use fewer than twice as many base angles as apexangles, the base angles of the faces must also appear in a linear number of vertices of othertypes, and for these apex and base angles, all other vertex types have angular deficit at least π/ − O ( δ ), bounded away from zero. Therefore, to keep the total angular defect below 4 π , n must be bounded. • If there are Ω( n ) pyramidal vertices with five incident apex angles, then to achieve angulardefect O (1 /n ) for each of these vertices, the apex angle must be 2 π/ − δ and the base anglemust be 3 π/
10 + δ for some δ = O (1 /n ). Because vertices of these types use fewer than twiceas many base angles as apex angles, the base angles of the faces must also appear in a linearnumber of vertices of other types, and for these apex and base angles, all other vertex typeshave angular deficit at least π/ − O ( δ ), bounded away from zero. Therefore, to keep the totalangular defect below 4 π , n must be bounded. • There can be no pyramidal vertices with six or more incident apex angles, or semipyramidalvertices with four or more incident apex angles, because the total angle of such a vertex wouldbe greater than 2 π .No case allows n to be arbitrarily large, so it must be bounded, and the number of triangulationswith n faces must also be bounded.We remark that this result applies only to convex realizations. There exist non-convex non-well-behaved monohedral isosceles polyhedra, combinatorially equivalent to bipyramids, with arbitrarilymany faces. It is even possible to realize these polyhedra in such a way that all vertices have positiveangular defect (Figure 8). Theorem 13.
A monohedral isosceles convex polyhedron with n faces must have apex angle O (1 /n ) .Proof. By Theorem 12 we may assume without loss of generality that the polyhedron is well-behaved,because the finitely many non-well-behaved possibilities do not affect the asymptotic statement ofthe theorem.Because the polyhedron has twice as many base angles as apex angles, and (by the assumptionthat it is well-behaved) each vertex has O (1) base angles, it follows that there are Ω( n ) vertices thathave at least twice as many incident base angles as apex angles. To keep the total angular deficit15qual to 4 π , Ω( n ) of these vertices must have deficit O (1 /n ). This rules out the semipyramidalvertices with one incident apex angle, for sufficiently large n , because they have deficit exactly π .The only remaining possibilities are the basic and semibasic vertices. If the base angle is θ and theapex angle is π − θ , then a basic vertex has angular deficit 2 π − θ and a semibasic vertex hasangular deficit π − θ . This can only be O (1 /n ) if θ = π/ − O (1 /n ), for which the apex angle is O (1 /n ). The infinite families of monohedral isosceles convex polyhedra listed at the start of this section allhave many basic or semi-basic vertices, but few pyramidal or semipyramidal vertices. This can beformalized and proven more generally:
Lemma 14.
In a well-behaved monohedral isosceles convex polyhedron, there is a nonzero butbounded number of pyramidal or semipyramidal vertices.Proof.
The number of apex angles is half the number of base angles, but the basic and semi-basicvertices use more than twice as many base angles as apex angles, so the leftover apex angles mustbe incident to pyramidal or semipyramidal vertices. If there are k pyramidal or semipyramidalvertices, then the base angles incident to these vertices contribute total angle less than πk (less than π for each pair of base angles incident to a semipyramidal vertex) and the apex angles contribute O (1) (because there are ≤ n of them, each with angle O (1 /n ). Therefore, the total angle is at most πk + O (1). However, the total angle must be at least 2 πk − π in order to achieve total angulardeficit 4 π for the whole polyhedron. Combining these inequalities, k must be O (1).We can use this fact to show that the graphs of monohedral isosceles convex polyhedra havesmall dominating sets. Recall that a dominating set, in a graph, is a subset of the vertices such thatevery vertex in the graph either belongs to this set or has a neighbor in the set. Theorem 15. If G is a triangulation that can be realized as a monohedral isosceles convex polyhedron,then G has a dominating set of size O (1) .Proof. If the realization is not well-behaved, then G has O (1) vertices, and we may take the entirevertex set as the dominating set. Otherwise, we claim that the pyramidal and semipyramidal verticesform a dominating set, which by Lemma 14 has size O (1). By the assumption that the realizationis well-behaved, every vertex is pyramidal or semipyramidal (and belongs to this set), or is basicor semibasic. In a basic or semibasic vertex, there are four incident base angles, and at most oneapex angle to separate them, so some two base angles must be adjacent across a shared side oftheir two faces rather than across a shared base edge. The vertex adjacent to the basic or semibasicvertex at the other end of this shared side has two incident apex angles, so it must be pyramidal orsemipyramidal. Theorem 16. If G is a triangulation that can be realized as a monohedral isosceles convex polyhedron,then G has unweighted graph diameter O (1) .Proof. Choose a starting vertex v within the small dominating set for G , and partition the verticesinto subsets S i of the vertices at distance i from v (the layers of G in a breadth-first search from v ).Then at most two consecutive layers can be disjoint from the dominating set, because three or moreconsecutive layers would leave the vertices in the middle layer without a neighbor in the dominatingset. Therefore, the number of layers is at most three times the size of the dominating set.16 Conclusions and open problems
We have shown that not every triangulated polyhedral graph has a convex isosceles realization, butthat a wide family of polyhedral graphs (including all the ones that we use to prove have no convexisosceles realizations) do have non-convex isosceles realizations. It would be of interest to findtriangulated polyhedral graphs for which even a non-convex but non-self-crossing isosceles realizationis impossible. More ambitiously, we would like to understand the computational complexity oftesting whether a (convex or non-convex) isosceles realization exists, and of finding one when itexists, but such a result seems far beyond the methods we have used in this paper.We have also added to the zoo of known infinite families of polyhedra with congruent isoscelesfaces. As well as these infinite families, there also some number of sporadic examples of suchpolyhedra, presumably many more than the eight convex deltahedra (convex polyhedra withcongruent equilateral triangle faces). Is it possible to classify all such polyhedra rather than merelyfinding more examples? Can our graph-theoretic analysis assist in this classification?The example of Gr¨unbaum’s polytope (Figure 1, right) shows that there are additional familiesof convex polyhedra with edges of two lengths (necessarily, with all faces isosceles or equilateral)than the families of monohedral isosceles polyhedra already listed. Can these families be classified?To what extent can our structural analysis of monohedral isosceles polyhedra be extended to them?Our triangulations that cannot be realized as isosceles convex polyhedra show that, for convexrealizations, at least three edge lengths may sometimes be needed. Is the minimum number ofedge lengths of convex realizations of polyhedral graphs bounded or unbounded? What about fornon-convex realizations?
Acknowledgements
We thank Joseph Malkevitch for directing our attention to his
Consortium paper, and Philip Rideoutfor the svg3d Python package with which many of the figures were made.
References [1] A. D. Alexandrov.
Convex Polyhedra . Springer Monographs in Mathematics. Springer-Verlag,Berlin, 2005.[2] David Barnette and Branko Gr¨unbaum. Preassigning the shape of a face.
Pacific J. Math. ,32:299–306, 1970. URL: https://projecteuclid.org/euclid.pjm/1102977361 .[3] Marshall Bern, Erik D. Demaine, David Eppstein, Eric Kuo, Andrea Mantler, and Jack Snoeyink.Ununfoldable polyhedra with convex faces.
Computational Geometry: Theory & Applications ,24(2):51–62, 2003. doi:10.1016/S0925-7721(02)00091-3 .[4] Robert J. MacG. Dawson. Some new tilings of the sphere with congruent triangles. In RezaSarhangi and Robert V. Moody, editors,
Renaissance Banff: Mathematics, Music, Art, Culture ,pages 489–496, Southwestern College, Winfield, Kansas, 2005. Bridges Conference. URL: https://archive.bridgesmathart.org/2005/bridges2005-489.html .[5] Erik D. Demaine, Martin L. Demaine, and David Eppstein. Acutely triangulated, stacked, andvery ununfoldable polyhedra. In
Proceedings of the 32nd Canadian Conference on ComputationalGeometry (CCCG 2020) , 2020. 176] Hans Freudenthal and B. L. van der Waerden. Over een bewering van Euclides.
Simon Stevin ,25:115–128, 1947.[7] Michael Goldberg. Questions, discussions, and notes: New equilateral polyhedra.
AmericanMathematical Monthly , 43(3):172–174, 1936. doi:10.2307/2300360 .[8] Branko Gr¨unbaum. Unambiguous polyhedral graphs.
Israel Journal of Mathematics , 1(4):235–238, 1963. doi:10.1007/BF02759726 .[9] Branko Gr¨unbaum. A convex polyhedron which is not equifacettable.
Geombinatorics , 10:165–171, 2001.[10] Joseph Malkevitch. Convex isosceles triangle polyhedra.
Geombinatorics , 10:122–132, 2001.[11] Joseph Malkevitch. Convex isosceles triangle polyhedra.
Consortium , 116:28–30, Spring–Summer2019.[12] J. W. Moon and L. Moser. Simple paths on polyhedra.
Pacific J. Math. , 13:629–631, 1963.URL: https://projecteuclid.org/euclid.pjm/1103035751 .[13] Ernst Steinitz. Polyeder und Raumeinteilungen. In
Encyclop¨adie der mathematischenWissenschaften, Band 3 (Geometries) , volume IIIAB12, pages 1–139. 1922. URL: https://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN360609767{&}DMDID=dmdlog203 .[14] John M. Sullivan. Minkowski sums and spherical duals. In Reza Sarhangi and John Sharp,editors,
Bridges London: Mathematics, Music, Art, Architecture, Culture , pages 117–122,London, 2006. Tarquin Publications. URL: https://archive.bridgesmathart.org/2006/bridges2006-117.html .[15] Charles W. Trigg. An infinite class of deltahedra.
Mathematics Magazine , 51(1):55–57, 1978. doi:10.2307/2689647 .[16] W. T. Tutte. How to draw a graph.
Proceedings of the London Mathematical Society (ThirdSeries) , 13:743–767, 1963. doi:10.1112/plms/s3-13.1.743doi:10.1112/plms/s3-13.1.743