An $r$-gentiling is a dissection of a shape into $r \geq 2$ parts which are all similar to the original shape. An $r$-reptiling is an $r$-gentiling of which all parts are mutually congruent. This article shows that no acute tetrahedron is an $r$-gentile or $r$-reptile for any $r &lt;9$, by showing that no acute spherical diangle can be dissected into less than nine acute spherical triangles

Haverkort, HJ Herman

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No acute tetrahedron is an 8-reptile

An $r$-gentiling is a dissection of a shape into $r \geq 2$ parts which are all similar to the original shape. An $r$-reptiling is an $r$-gentiling of which all parts are mutually congruent. This article shows that no acute tetrahedron is an $r$-gentile or $r$-reptile for any $r <9$, by showing that no acute spherical diangle can be dissected into less than nine acute spherical triangles

Haverkort, H.J.

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 No acute tetrahedron is an 8-reptileCitation for published version (APA):Haverkort, H. J. (2015). No acute tetrahedron is an 8-reptile. (arXiv; Vol. 1508.03773 [cs.CG]). s.n.Document status and date:Published: 01/01/2015Document Version:Publisher’s PDF, also known as Version of Record (includes final page, issue and volume numbers)Please check the document version of this publication:• A submitted manuscript is the version of the article upon submission and before peer-review. There can beimportant differences between the submitted version and the official published version of record. Peopleinterested in the research are advised to contact the author for the final version of the publication, or visit theDOI to the publisher's website.• The final author version and the galley proof are versions of the publication after peer review.• The final published version features the final layout of the paper including the volume, issue and pagenumbers.Link to publicationGeneral rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain            • You may freely distribute the URL identifying the publication in the public portal.If the publication is distributed under the terms of Article 25fa of the Dutch Copyright Act, indicated by the “Taverne” license above, pleasefollow below link for the End User Agreement:www.tue.nl/taverneTake down policyIf you believe that this document breaches copyright please contact us at:openaccess@tue.nlproviding details and we will investigate your claim.Download date: 16. Nov. 2023No acute tetrahedron is an 8-reptileHerman Haverkort∗August 18, 2015AbstractAn r-gentiling is a dissection of a shape into r ≥ 2 parts which are all similar to theoriginal shape. An r-reptiling is an r-gentiling of which all parts are mutually congruent.This article shows that no acute tetrahedron is an r-gentile or r-reptile for any r < 9, byshowing that no acute spherical diangle can be dissected into less than nine acute sphericaltriangles.IntroductionWe call a geometric figure, that is, a set of points in Euclidean space, T an r-gentile if T admitsan r-gentiling, that is, a subdivision of T into r ≥ 2 figures (tiles) T1, ..., Tr, such that each ofthe figures T1, ..., Tr is similar to T . In other words, T is an r-gentile if we can tile it with rsmaller copies of itself. This generalizes the concept of reptiles, coined by Golomb [4]: a figure Tis an r-reptile if T admits an r-reptiling, that is, a subdivision of T into r ≥ 2 figures T1, ..., Tr,such that each of the figures T1, ..., Tr is similar to T and all figures T1, ..., Tr are mutuallycongruent. In other words, T is a r-reptile if we can tile it with r equally large smaller copies ofitself. Interest in reptile tetrahedra (or triangles, for that matter) exists, among other reasons,because of their application in meshes for scientific computing [1, 10]. In this realm certainreptile-based techniques are well-developed in two dimensions [1], but three-dimensional spaceposes great challenges [6].It is known what triangles are r-reptiles [12] and r-gentiles [3, 9] for what r. However,for tetrahedra the situation is much less clear; in fact the identification of reptile and gentiletetrahedra and tetrahedra that tile space has been a long-standing open problem. The shape ofa tetrahedron has five degrees of freedom and these have not been fully explored yet. Matoušekand Safernová argued that r-reptilings with tetrahedra exist if and only if r is a cube number [11].In particular, it is known that all so-called Hill tetrahedra (attributed to Hill [8] by Hertel [7] andMatoušek and Safernová [11]) are 8-reptiles. It has been conjectured that the Hill tetrahedra arethe only reptile tetrahedra [7], but this conjecture is false: two non-Hill tetrahedra are knownthat have been recognized as 8-reptiles by Liu and Joe [10]. To the best of our knowledge,the Hill tetrahedra and the two non-Hill tetrahedra from Liu and Joe are the only tetrahedraknown to be reptiles, but there might be others. This paper provides a small contribution tothe answer of the question: exactly what tetrahedra are reptiles?In mesh construction applications one typically needs to enforce certain quality constraintson the mesh elements. This has motivated studies into acute triangles [5] and acute tetrahe-dra [2]:∗Dept. of Mathematics and Computer Science, Eindhoven University of Technology, the Netherlands,cs.herman@haverkort.net1arXiv:1508.03773v1  [cs.CG]  15 Aug 2015Definition 1. A tetrahedron is acute if each pair of its facets has a dihedral angle strictly lessthan π/2.All facets of an acute tetrahedron are acute triangles themselves (Eppstein et al. [2], Lemma2). The Hill tetrahedra, as well as the two non-Hill tetrahedra from Liu and Joe, all have rightdihedral angles. Thus, no acute reptile tetrahedra are known.ResultsIn this article we will prove the following statement, which may serve as evidence that acutereptile tetrahedra are probably hard to find, if they exist at all:Theorem 1. Let T be an acute tetrahedron subdivided into r ≥ 2 acute tetrahedra T1, ..., Tr. Ifthe diameter (longest edge) of each tetrahedron Ti is smaller than the diameter (longest edge)of T , then r ≥ 9.In particular we get:Corollary 1. No acute tetrahedron is an r-gentile for any r < 9.With the result from Matoušek and Safernová that r-reptile tetrahedra can only exist whenr is a cube number [11], we get:Corollary 2. No acute tetrahedron is an r-reptile for any r < 27.The proofNote that if a tetrahedron T is subdivided into tetrahedra T1, ..., Tr with smaller diameter thanT , then at least one tetrahedron Ti, for some i ∈ {1, ..., r}, must have a vertex v on the longestedge of T . For the proof of Theorem 1 we analyse Sv, the subdivision of an infinitesimal spherearound v that is induced by the facets of T and T1, ..., Tr. In such a subdivision, we find:• faces: each face is either a spherical triangle, corresponding to a tetrahedron Ti of whichv is a vertex, or a spherical diangle (also called lune), corresponding to a tetrahedron thathas v on the interior of an edge;• edges: the edges of Sv are segments of great circles and correspond to facets of T1, ..., Trthat contain v; the angle between two adjacent edges on a face of Sv corresponds to thedihedral angle of the corresponding facets of a tetrahedron Ti.• vertices: each vertex of Sv corresponds to an edge of a tetrahedron Ti that contains v.Thus, Sv consists of a spherical diangle D corresponding to T , subdivided into a number ofspherical triangles, and possibly some spherical diangles, that correspond to the tetrahedrafrom T1, ..., Tr that touch v. Below we will see that Sv must contain at least nine faces (notcounting the outer face, that is, the complement of D), which proves Theorem 1.In what follows, when we talk about diangles and triangles, we will mean acute, sphericaldiangles and acute, spherical triangles on a sphere with radius 1. Note that the faces are dianglesor triangles in the geometric sense, but they may have more than two or three vertices on theirboundary. More precisely, a diangle or triangle has, respectively, exactly two or three vertices,called corners, where its boundary has an acute angle, and possibly a number of other verticeswhere its boundary has a straight angle. A chain of edges of a diangle or triangle from one2corner to the next is called a side. Note that Sv contains at least one triangle, since v is avertex of at least one tetrahedron Ti. Therefore, in what follows we consider a subdivision S ofa diangle D into a number of diangles and triangles, among which at least one triangle. We callsuch subdivisions valid. Henceforth, we will assume that S has the smallest number of facesout of all possible valid subdivisions of all possible diangles D. Our goal is now to prove thatS contains at least 9 faces.Lemma 1. Each face of S is a triangle.Proof. If S would have any diangular face F , it must have the same corners as D, because thecorners of any diangle must be an antipodal pair and there is only one antipodal pair withinD. The removal of F would separate S into at most two diangular components with the samecorners as D. By construction, at least one of these components contains a triangle. Thatcomponent would then constitute a valid subdivision that has fewer faces than S, contradictingour choice of S.In S, we distinguish boundary vertices (vertices on the boundary of D) and interior vertices(vertices in the interior of D). Among the boundary vertices, we distinguish poles (the cornersof D) and side vertices (the remaining boundary vertices). Among the interior vertices wedistinguish full vertices and hanging vertices: a vertex v is a full vertex if it is a corner of eachface incident on v; a vertex v is a hanging vertex if it is a non-corner vertex of one of the facesincident on v.We will now derive a few properties of S from the acuteness of its angles.Lemma 2. Each side of each face of S has length strictly less than π/2.Proof. Consider any face F of S. Let a be the length of a particular side of F , let α be theangle in the opposite corner of F , and let β and γ be the angles in the other two corners ofF . Since F is acute, the sines and cosines of α, β and γ are all positive. By the supplementarycosine rule ([13], Art. 47) we have cosα = − cosβ cos γ + sinβ sin γ cos a, so cos a = (cosα +cosβ cos γ)/(sinβ sin γ) > 0. It follows that a < π/2.Lemma 3. There are at least four side vertices: two on each side of D.Proof. For the sake of contradiction, suppose one side of D contains only one side vertex. Thenthis side would consist of two edges, at least one of which has length at least π/2, contradictingLemma 2.Lemma 4. Each pole is incident on at least two edges.Each hanging vertex and each side vertex is incident on at least four edges.Each full vertex is incident on at least five edges.Proof. The poles are incident on at least two edges by definition. If a side vertex or a hangingvertex would be incident on only three edges, then two of these edges make a straight angleon one side, while the third edge divides the straight angle on the other side. Thus, at leastone of the angles that results from this division would be non-acute. If a full vertex wouldbe incident on at most four edges, then at least two of those edges must make an angle of atleast 2π/4 = π/2 on their common face, again contradicting the assumption that all faces areacute.Now we can combine these properties with Euler’s formula and find:Lemma 5. The number of faces equals 2f + h+ s, where f ≥ 2 is the number of full vertices,h is the number of hanging vertices, and s ≥ 4 is the number of side vertices.3Figure 1: In grey: the boundary of D sketched as a smooth loop, without indicating the locationof the poles. In black: the only network of edges that subdivides D into at most eight faces andcomplies with Lemma 5. However, this contradicts Lemma 4.Proof. Let v = f + h+ s+ 2 be the number of vertices, let e be the number of edges and r bethe number of triangles of S. By Lemma 1 all faces are triangles, so by Euler’s formula we havev + r = f + h + s + 2 + r = e + 1, hence 2e = 2f + 2h + 2s + 2 + 2r. We say that a hangingvertex is owned by the triangle of which it is a non-corner vertex; each hanging vertex is ownedby exactly one triangle. If we add up the edges of all triangles, we count, for each triangle,three edges plus the number of hanging vertices it owns, making 3r+h in total. This counts alledges double, except the s+ 2 edges on the boundary, which are counted only once. Thereforewe have 2e− s− 2 = 3r+ h. Hence we have 2e = 3r+ h+ s+ 2 = 2f + 2h+ 2s+ 2 + 2r, whichsolves to r = 2f + h+ s.Thus we have 2e = 3r + h + s + 2 = 6f + 4h + 4s + 2. By Lemma 4 we also have2e ≥ 5f + 4h+ 4s+ 4, so 6f + 4h+ 4s+ 2 ≥ 5f + 4h+ 4s+ 4, which solves to f ≥ 2.The condition s ≥ 4 is given by Lemma 3.Lemma 6. The number of faces of S is at least nine.Proof. Suppose, for the sake of contradiction, that S has at most eight faces. Then, by Lemma 5,we have four side vertices, two full vertices, and no hanging vertices. Each of the two interiorvertices is incident on five faces, so in order to have at most eight faces in total, the interiorvertices must share two of their incident faces, see Figure 1. Adding further edges in theinterior of D is not possible, as this would increase the number of faces. Thus there are atleast six boundary vertices, four of which are incident on only one interior edge and thereforeon only three edges in total, and at least two of these must be side vertices. This contradictsLemma 4.This concludes the proof of Theorem 1.Bold conjecturesThe proof of Theorem 1 is based on the following observation: any dissection of an acutespherical diangle into acute spherical triangles requires at least nine triangles. I do not knowwhether this bound is tight. It seems that a dissection into ten triangles is easy to achieve: inFigure 1, add an edge connecting the two side vertices on the left, add an edge connecting thetwo side vertices on the right, place the poles on the left and the right end of the figure, anddistort the construction to make all angles acute. So far, I have not been able to find a solutionwith nine triangles, so I conjecture:Conjecture 1. Any dissection of an acute spherical diangle into acute spherical triangles re-quires at least ten triangles.4Note that proving this would immediately strengthen Corollary 1 to: no acute tetrahedronis an r-gentile for any r < 10.Now let b be this lower bound on the required number of triangles, proven to be at least9 and conjectured to be 10. Suppose we can prove Conjecture 1, then how could we go aboutimproving Corollary 2? A first step could be the following. If T is an acute tetrahedron withdiameter d, and we can identify three segments x, y and z of length d/3 on the edges of T ,then the arguments presented in this paper tell us that any 27-reptiling of T must contain atleast b tiles that intersect x, b tiles that intersect y, and b tiles that intersect z. If additionally,one can ensure that x, y and z lie at distance more than d/3 from each other, these three setsof b tiles must be mutually disjoint, so there must be at least 3b tiles in total. With b = 10,this would contradict the existence of a 27-reptiling and thus improve Corollary 2 to: no acutetetrahedron is an r-reptile for any r < 64.However, given that no acute tetrahedron can be an r-reptile or r-gentile for small valuesof r (and given, in general, that the past hundred years did not turn up any reptile tetrahedrawithout right dihedral angles), we may rather restate the obvious:Conjecture 2. There are no reptile acute tetrahedra.A stronger conjecture would be:Conjecture 3. There are no gentile acute tetrahedra.We conclude with an even bolder conjecture:Conjecture 4. There are no gentile tetrahedra that do not have a dihedral angle of exactly π/2.References[1] M. Bader and C. Zenger, Efficient storage and processing of adaptive triangular grids usingSierpinski curves, 6th International Conference on Computational Science (ICCS), LectureNotes in Computer Science, vol. 3991, 2006, pp. 673–680.[2] D. Eppstein, J. M. Sullivan, and A. Üngör, Tiling space and slabs with acute tetrahedra,Computational Geometry 27 (2004), no. 3, 237–255.[3] R. W. Freese, A. K. Miller, and Z. Usiskin, Can every triangle be divided into n trianglessimilar to it?, Am. Math. Monthly 77 (1990), 867–869.[4] S. W. Golomb, Replicating figures in the plane, The Mathematical Gazette 48 (1964),no. 366, 403–412.[5] M. Gottschau, H. Haverkort, and K. Matzke, Reptilings and space-filling curves for acutetriangles, Manuscript, please send e-mail to cs.herman@haverkort.net for current version,2015.[6] H. Haverkort, Space-filling curve properties for 3D mesh traversals, Presentation at Int.Conf. on Parallel Computing (ParCo) 2013, slides available from http://haverkort.net,2013.[7] E. Hertel, Self-similar simplices, Beiträge zur Algebra und Geometrie 41 (2000), no. 2,589–595.[8] M. J. M. Hill, Determination of the volumes of certain species of tetrahedra, Proc. LondonMathematical Society 27 (1896), 39–53.5[9] H. Kaiser, Selbstähnliche Dreieckszerlegungen, Tech. report, Friedrich-Schiller-UniversitätJena, 1990.[10] A. Liu and B. Joe, On the shape of tetrahedra from bisection, Mathematics of Computation63 (1994), no. 207, 141–154.[11] J. Matoušek and Z. Safernová, On the nonexistence of k-reptile tetrahedra, Discrete &Computational Geometry 46 (2011), no. 3, 599–609.[12] S. L. Snover, C. Waiveris, and J. K. Williams, Rep-tiling for triangles, Discrete Mathematics91 (1991), 193–200.[13] I. Todhunter, Spherical trigonometry, 5th ed., MacMillan and Co., 1886, http://www.gutenberg.org/ebooks/19770.6

English

 No acute tetrahedron is an 8-reptileCitation for published version (APA):Haverkort, H. J. (2015). No acute tetrahedron is an 8-reptile. (arXiv; Vol. 1508.03773 [cs.CG]). s.n.Document status and date:Published: 01/01/2015Document Version:Publisher’s PDF, also known as Version of Record (includes final page, issue and volume numbers)Please check the document version of this publication:• A submitted manuscript is the version of the article upon submission and before peer-review. There can beimportant differences between the submitted version and the official published version of record. Peopleinterested in the research are advised to contact the author for the final version of the publication, or visit theDOI to the publisher's website.• The final author version and the galley proof are versions of the publication after peer review.• The final published version features the final layout of the paper including the volume, issue and pagenumbers.Link to publicationGeneral rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain            • You may freely distribute the URL identifying the publication in the public portal.If the publication is distributed under the terms of Article 25fa of the Dutch Copyright Act, indicated by the “Taverne” license above, pleasefollow below link for the End User Agreement:www.tue.nl/taverneTake down policyIf you believe that this document breaches copyright please contact us at:openaccess@tue.nlproviding details and we will investigate your claim.Download date: 17. Nov. 2023No acute tetrahedron is an 8-reptileHerman Haverkort∗August 18, 2015AbstractAn r-gentiling is a dissection of a shape into r ≥ 2 parts which are all similar to theoriginal shape. An r-reptiling is an r-gentiling of which all parts are mutually congruent.This article shows that no acute tetrahedron is an r-gentile or r-reptile for any r < 9, byshowing that no acute spherical diangle can be dissected into less than nine acute sphericaltriangles.IntroductionWe call a geometric figure, that is, a set of points in Euclidean space, T an r-gentile if T admitsan r-gentiling, that is, a subdivision of T into r ≥ 2 figures (tiles) T1, ..., Tr, such that each ofthe figures T1, ..., Tr is similar to T . In other words, T is an r-gentile if we can tile it with rsmaller copies of itself. This generalizes the concept of reptiles, coined by Golomb [4]: a figure Tis an r-reptile if T admits an r-reptiling, that is, a subdivision of T into r ≥ 2 figures T1, ..., Tr,such that each of the figures T1, ..., Tr is similar to T and all figures T1, ..., Tr are mutuallycongruent. In other words, T is a r-reptile if we can tile it with r equally large smaller copies ofitself. Interest in reptile tetrahedra (or triangles, for that matter) exists, among other reasons,because of their application in meshes for scientific computing [1, 10]. In this realm certainreptile-based techniques are well-developed in two dimensions [1], but three-dimensional spaceposes great challenges [6].It is known what triangles are r-reptiles [12] and r-gentiles [3, 9] for what r. However,for tetrahedra the situation is much less clear; in fact the identification of reptile and gentiletetrahedra and tetrahedra that tile space has been a long-standing open problem. The shape ofa tetrahedron has five degrees of freedom and these have not been fully explored yet. Matoušekand Safernová argued that r-reptilings with tetrahedra exist if and only if r is a cube number [11].In particular, it is known that all so-called Hill tetrahedra (attributed to Hill [8] by Hertel [7] andMatoušek and Safernová [11]) are 8-reptiles. It has been conjectured that the Hill tetrahedra arethe only reptile tetrahedra [7], but this conjecture is false: two non-Hill tetrahedra are knownthat have been recognized as 8-reptiles by Liu and Joe [10]. To the best of our knowledge,the Hill tetrahedra and the two non-Hill tetrahedra from Liu and Joe are the only tetrahedraknown to be reptiles, but there might be others. This paper provides a small contribution tothe answer of the question: exactly what tetrahedra are reptiles?In mesh construction applications one typically needs to enforce certain quality constraintson the mesh elements. This has motivated studies into acute triangles [5] and acute tetrahe-dra [2]:∗Dept. of Mathematics and Computer Science, Eindhoven University of Technology, the Netherlands,cs.herman@haverkort.net1arXiv:1508.03773v1  [cs.CG]  15 Aug 2015Definition 1. A tetrahedron is acute if each pair of its facets has a dihedral angle strictly lessthan π/2.All facets of an acute tetrahedron are acute triangles themselves (Eppstein et al. [2], Lemma2). The Hill tetrahedra, as well as the two non-Hill tetrahedra from Liu and Joe, all have rightdihedral angles. Thus, no acute reptile tetrahedra are known.ResultsIn this article we will prove the following statement, which may serve as evidence that acutereptile tetrahedra are probably hard to find, if they exist at all:Theorem 1. Let T be an acute tetrahedron subdivided into r ≥ 2 acute tetrahedra T1, ..., Tr. Ifthe diameter (longest edge) of each tetrahedron Ti is smaller than the diameter (longest edge)of T , then r ≥ 9.In particular we get:Corollary 1. No acute tetrahedron is an r-gentile for any r < 9.With the result from Matoušek and Safernová that r-reptile tetrahedra can only exist whenr is a cube number [11], we get:Corollary 2. No acute tetrahedron is an r-reptile for any r < 27.The proofNote that if a tetrahedron T is subdivided into tetrahedra T1, ..., Tr with smaller diameter thanT , then at least one tetrahedron Ti, for some i ∈ {1, ..., r}, must have a vertex v on the longestedge of T . For the proof of Theorem 1 we analyse Sv, the subdivision of an infinitesimal spherearound v that is induced by the facets of T and T1, ..., Tr. In such a subdivision, we find:• faces: each face is either a spherical triangle, corresponding to a tetrahedron Ti of whichv is a vertex, or a spherical diangle (also called lune), corresponding to a tetrahedron thathas v on the interior of an edge;• edges: the edges of Sv are segments of great circles and correspond to facets of T1, ..., Trthat contain v; the angle between two adjacent edges on a face of Sv corresponds to thedihedral angle of the corresponding facets of a tetrahedron Ti.• vertices: each vertex of Sv corresponds to an edge of a tetrahedron Ti that contains v.Thus, Sv consists of a spherical diangle D corresponding to T , subdivided into a number ofspherical triangles, and possibly some spherical diangles, that correspond to the tetrahedrafrom T1, ..., Tr that touch v. Below we will see that Sv must contain at least nine faces (notcounting the outer face, that is, the complement of D), which proves Theorem 1.In what follows, when we talk about diangles and triangles, we will mean acute, sphericaldiangles and acute, spherical triangles on a sphere with radius 1. Note that the faces are dianglesor triangles in the geometric sense, but they may have more than two or three vertices on theirboundary. More precisely, a diangle or triangle has, respectively, exactly two or three vertices,called corners, where its boundary has an acute angle, and possibly a number of other verticeswhere its boundary has a straight angle. A chain of edges of a diangle or triangle from one2corner to the next is called a side. Note that Sv contains at least one triangle, since v is avertex of at least one tetrahedron Ti. Therefore, in what follows we consider a subdivision S ofa diangle D into a number of diangles and triangles, among which at least one triangle. We callsuch subdivisions valid. Henceforth, we will assume that S has the smallest number of facesout of all possible valid subdivisions of all possible diangles D. Our goal is now to prove thatS contains at least 9 faces.Lemma 1. Each face of S is a triangle.Proof. If S would have any diangular face F , it must have the same corners as D, because thecorners of any diangle must be an antipodal pair and there is only one antipodal pair withinD. The removal of F would separate S into at most two diangular components with the samecorners as D. By construction, at least one of these components contains a triangle. Thatcomponent would then constitute a valid subdivision that has fewer faces than S, contradictingour choice of S.In S, we distinguish boundary vertices (vertices on the boundary of D) and interior vertices(vertices in the interior of D). Among the boundary vertices, we distinguish poles (the cornersof D) and side vertices (the remaining boundary vertices). Among the interior vertices wedistinguish full vertices and hanging vertices: a vertex v is a full vertex if it is a corner of eachface incident on v; a vertex v is a hanging vertex if it is a non-corner vertex of one of the facesincident on v.We will now derive a few properties of S from the acuteness of its angles.Lemma 2. Each side of each face of S has length strictly less than π/2.Proof. Consider any face F of S. Let a be the length of a particular side of F , let α be theangle in the opposite corner of F , and let β and γ be the angles in the other two corners ofF . Since F is acute, the sines and cosines of α, β and γ are all positive. By the supplementarycosine rule ([13], Art. 47) we have cosα = − cosβ cos γ + sinβ sin γ cos a, so cos a = (cosα +cosβ cos γ)/(sinβ sin γ) > 0. It follows that a < π/2.Lemma 3. There are at least four side vertices: two on each side of D.Proof. For the sake of contradiction, suppose one side of D contains only one side vertex. Thenthis side would consist of two edges, at least one of which has length at least π/2, contradictingLemma 2.Lemma 4. Each pole is incident on at least two edges.Each hanging vertex and each side vertex is incident on at least four edges.Each full vertex is incident on at least five edges.Proof. The poles are incident on at least two edges by definition. If a side vertex or a hangingvertex would be incident on only three edges, then two of these edges make a straight angleon one side, while the third edge divides the straight angle on the other side. Thus, at leastone of the angles that results from this division would be non-acute. If a full vertex wouldbe incident on at most four edges, then at least two of those edges must make an angle of atleast 2π/4 = π/2 on their common face, again contradicting the assumption that all faces areacute.Now we can combine these properties with Euler’s formula and find:Lemma 5. The number of faces equals 2f + h+ s, where f ≥ 2 is the number of full vertices,h is the number of hanging vertices, and s ≥ 4 is the number of side vertices.3Figure 1: In grey: the boundary of D sketched as a smooth loop, without indicating the locationof the poles. In black: the only network of edges that subdivides D into at most eight faces andcomplies with Lemma 5. However, this contradicts Lemma 4.Proof. Let v = f + h+ s+ 2 be the number of vertices, let e be the number of edges and r bethe number of triangles of S. By Lemma 1 all faces are triangles, so by Euler’s formula we havev + r = f + h + s + 2 + r = e + 1, hence 2e = 2f + 2h + 2s + 2 + 2r. We say that a hangingvertex is owned by the triangle of which it is a non-corner vertex; each hanging vertex is ownedby exactly one triangle. If we add up the edges of all triangles, we count, for each triangle,three edges plus the number of hanging vertices it owns, making 3r+h in total. This counts alledges double, except the s+ 2 edges on the boundary, which are counted only once. Thereforewe have 2e− s− 2 = 3r+ h. Hence we have 2e = 3r+ h+ s+ 2 = 2f + 2h+ 2s+ 2 + 2r, whichsolves to r = 2f + h+ s.Thus we have 2e = 3r + h + s + 2 = 6f + 4h + 4s + 2. By Lemma 4 we also have2e ≥ 5f + 4h+ 4s+ 4, so 6f + 4h+ 4s+ 2 ≥ 5f + 4h+ 4s+ 4, which solves to f ≥ 2.The condition s ≥ 4 is given by Lemma 3.Lemma 6. The number of faces of S is at least nine.Proof. Suppose, for the sake of contradiction, that S has at most eight faces. Then, by Lemma 5,we have four side vertices, two full vertices, and no hanging vertices. Each of the two interiorvertices is incident on five faces, so in order to have at most eight faces in total, the interiorvertices must share two of their incident faces, see Figure 1. Adding further edges in theinterior of D is not possible, as this would increase the number of faces. Thus there are atleast six boundary vertices, four of which are incident on only one interior edge and thereforeon only three edges in total, and at least two of these must be side vertices. This contradictsLemma 4.This concludes the proof of Theorem 1.Bold conjecturesThe proof of Theorem 1 is based on the following observation: any dissection of an acutespherical diangle into acute spherical triangles requires at least nine triangles. I do not knowwhether this bound is tight. It seems that a dissection into ten triangles is easy to achieve: inFigure 1, add an edge connecting the two side vertices on the left, add an edge connecting thetwo side vertices on the right, place the poles on the left and the right end of the figure, anddistort the construction to make all angles acute. So far, I have not been able to find a solutionwith nine triangles, so I conjecture:Conjecture 1. Any dissection of an acute spherical diangle into acute spherical triangles re-quires at least ten triangles.4Note that proving this would immediately strengthen Corollary 1 to: no acute tetrahedronis an r-gentile for any r < 10.Now let b be this lower bound on the required number of triangles, proven to be at least9 and conjectured to be 10. Suppose we can prove Conjecture 1, then how could we go aboutimproving Corollary 2? A first step could be the following. If T is an acute tetrahedron withdiameter d, and we can identify three segments x, y and z of length d/3 on the edges of T ,then the arguments presented in this paper tell us that any 27-reptiling of T must contain atleast b tiles that intersect x, b tiles that intersect y, and b tiles that intersect z. If additionally,one can ensure that x, y and z lie at distance more than d/3 from each other, these three setsof b tiles must be mutually disjoint, so there must be at least 3b tiles in total. With b = 10,this would contradict the existence of a 27-reptiling and thus improve Corollary 2 to: no acutetetrahedron is an r-reptile for any r < 64.However, given that no acute tetrahedron can be an r-reptile or r-gentile for small valuesof r (and given, in general, that the past hundred years did not turn up any reptile tetrahedrawithout right dihedral angles), we may rather restate the obvious:Conjecture 2. There are no reptile acute tetrahedra.A stronger conjecture would be:Conjecture 3. There are no gentile acute tetrahedra.We conclude with an even bolder conjecture:Conjecture 4. There are no gentile tetrahedra that do not have a dihedral angle of exactly π/2.References[1] M. Bader and C. Zenger, Efficient storage and processing of adaptive triangular grids usingSierpinski curves, 6th International Conference on Computational Science (ICCS), LectureNotes in Computer Science, vol. 3991, 2006, pp. 673–680.[2] D. Eppstein, J. M. Sullivan, and A. Üngör, Tiling space and slabs with acute tetrahedra,Computational Geometry 27 (2004), no. 3, 237–255.[3] R. W. Freese, A. K. Miller, and Z. Usiskin, Can every triangle be divided into n trianglessimilar to it?, Am. Math. Monthly 77 (1990), 867–869.[4] S. W. Golomb, Replicating figures in the plane, The Mathematical Gazette 48 (1964),no. 366, 403–412.[5] M. Gottschau, H. Haverkort, and K. Matzke, Reptilings and space-filling curves for acutetriangles, Manuscript, please send e-mail to cs.herman@haverkort.net for current version,2015.[6] H. Haverkort, Space-filling curve properties for 3D mesh traversals, Presentation at Int.Conf. on Parallel Computing (ParCo) 2013, slides available from http://haverkort.net,2013.[7] E. Hertel, Self-similar simplices, Beiträge zur Algebra und Geometrie 41 (2000), no. 2,589–595.[8] M. J. M. Hill, Determination of the volumes of certain species of tetrahedra, Proc. LondonMathematical Society 27 (1896), 39–53.5[9] H. Kaiser, Selbstähnliche Dreieckszerlegungen, Tech. report, Friedrich-Schiller-UniversitätJena, 1990.[10] A. Liu and B. Joe, On the shape of tetrahedra from bisection, Mathematics of Computation63 (1994), no. 207, 141–154.[11] J. Matoušek and Z. Safernová, On the nonexistence of k-reptile tetrahedra, Discrete &Computational Geometry 46 (2011), no. 3, 599–609.[12] S. L. Snover, C. Waiveris, and J. K. Williams, Rep-tiling for triangles, Discrete Mathematics91 (1991), 193–200.[13] I. Todhunter, Spherical trigonometry, 5th ed., MacMillan and Co., 1886, http://www.gutenberg.org/ebooks/19770.6

No acute tetrahedron is an 8-reptile

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